QUOTE (ubavontuba+Feb 3 2008, 07:57 PM)
It's "hypocrite," not "hypocrit."
Do you have anything relevant to add to the discussion?
Do you?
You ceartainly haven't yet, and so far today all you've done is post irrelevant spam.
Do you have anything relevant to add to the discussion?
Do you?
You ceartainly haven't yet, and so far today all you've done is post irrelevant spam.
QUOTE (Cecil P Abstract+Feb 3 2008, 06:57 AM)
Au contraire .....you frigging mo' 
No. You're thinking of au as in au jus (meaning with juice). I meant it as an exclamation/interjection.
No. You're thinking of au as in au jus (meaning with juice). I meant it as an exclamation/interjection.
QUOTE (ubavontuba+Feb 3 2008, 07:08 AM)
No. You're thinking of au as in au jus (meaning with juice). I meant it as an exclamation/interjection.
Bollocks you mo', no such definition in any dictionary on Earth ...... quit the drugs and promise to die!
Bollocks you mo', no such definition in any dictionary on Earth ...... quit the drugs and promise to die!
how predictable.
QUOTE (Trippy+Feb 3 2008, 07:14 AM)
Waddya reckon, Cranial Fecaloma? Or Megamandibular Acephaly?
Both, but with additional brain pulping from anyone who's had the severest misfortune to have actually met him.
Both, but with additional brain pulping from anyone who's had the severest misfortune to have actually met him.
QUOTE (Trippy+Feb 3 2008, 06:59 AM)
Go for it, report it..
Of course, bare in mind, by any definition, your posts are blatantly inflammatory and baiting, so maybe that will be recognized.
You forget that you started this with this post:
Of course, bare in mind, by any definition, your posts are blatantly inflammatory and baiting, so maybe that will be recognized.
You forget that you started this with this post:
QUOTE (Trippy+Jan 31 2008, 04:49 PM)
Actually, I have no idea how you manage to turn you computer on each morning.
But hang on, according to Peanuts post...
You really are a friggan mo, you know that?
The only thing around here that's useless is you.
Besides, you're the one that thinks the mass of the blackhole after the collision is the same as the mass of the particles before the collision (a stance that's blatantly wrong).
And now you're griping about me assuming you were correct and doing calculations?
Sheesh, yet again you contradict yourself.
So who's posts are "blatantly inflammatory and baiting?" Why, it's yours!
Proper spelling is a requirement to proper grammar.
Proper spelling is a requirement to proper grammar.
So... I guess that makes you the idiot, thinking that a dropped letter in a word is a grammatical error.
And there's a simple explanation as to why it was missed.
Like 90% of the human race, I read what I thought I had typed, rather then what I had actually typed (in other words, what I read out had the missing R in place).
Unless you're going to start claiming that you're not human as well.
Granted that one was a typo, what about the others?
What about "bare in mind" (as in naked in mind)? You use it regularly. It should be, "bear in mind," (as in carry it in mind).
But hang on, according to Peanuts post...
You really are a friggan mo, you know that?
The only thing around here that's useless is you.
Besides, you're the one that thinks the mass of the blackhole after the collision is the same as the mass of the particles before the collision (a stance that's blatantly wrong).
And now you're griping about me assuming you were correct and doing calculations?
Sheesh, yet again you contradict yourself.
So who's posts are "blatantly inflammatory and baiting?" Why, it's yours!
QUOTE
And for the record, it was a spelling mistake, not a grammar mistake ya friggin mo.
Proper spelling is a requirement to proper grammar.
QUOTE (->
| QUOTE |
| And for the record, it was a spelling mistake, not a grammar mistake ya friggin mo. |
Proper spelling is a requirement to proper grammar.
So... I guess that makes you the idiot, thinking that a dropped letter in a word is a grammatical error.
And there's a simple explanation as to why it was missed.
Like 90% of the human race, I read what I thought I had typed, rather then what I had actually typed (in other words, what I read out had the missing R in place).
Unless you're going to start claiming that you're not human as well.
Granted that one was a typo, what about the others?
What about "bare in mind" (as in naked in mind)? You use it regularly. It should be, "bear in mind," (as in carry it in mind).
QUOTE (Cecil P Abstract+Feb 3 2008, 07:00 AM)
Only, that you're a frigging mo'
So then you don't have anything relevant to add to the discussion?
I thought not.
So then you don't have anything relevant to add to the discussion?
I thought not.
QUOTE (Trippy+Feb 3 2008, 07:02 AM)
Do you?
You ceartainly haven't yet, and so far today all you've done is post irrelevant spam.
I've only directly replied to your posts. If they're "irrelevant spam," then the problem is the source material, isn't it?
You ceartainly haven't yet, and so far today all you've done is post irrelevant spam.
I've only directly replied to your posts. If they're "irrelevant spam," then the problem is the source material, isn't it?
QUOTE (ubavontuba+Feb 3 2008, 08:30 PM)
Granted that one was a typo, what about the others?
what part of "i have a three week old infant in one arm, and am typing one handed" did you not understand?
i don't know about you, but when i type one handed, capital letters are the first to go (and experience tells me this is true for most people.
edit:
so you don't consider picking away at the same irrelevant point, inspite of it having been explained to you, to be baiting, or inflammatory. shoulda figured as much.
what part of "i have a three week old infant in one arm, and am typing one handed" did you not understand?
i don't know about you, but when i type one handed, capital letters are the first to go (and experience tells me this is true for most people.
edit:
so you don't consider picking away at the same irrelevant point, inspite of it having been explained to you, to be baiting, or inflammatory. shoulda figured as much.
QUOTE (Cecil P Abstract+Feb 3 2008, 07:17 AM)
Bollocks you mo', no such definition in any dictionary on Earth ...... quit the drugs and promise to die!
Actually, "au contraire" isn't in my dictionary, and MS Word didn't correct it either. I guess French phrases aren't my forte'.
Actually, "au contraire" isn't in my dictionary, and MS Word didn't correct it either. I guess French phrases aren't my forte'.
QUOTE (Trippy+Feb 3 2008, 07:42 AM)
what part of "i have a three week old infant in one arm, and am typing one handed" did you not understand?
What part of, "I have a BA, and a first class MA sitting in the room with me at the moment, both of whom majored in English (one of whom was born in London), both of whom say that specific sentence is grammatically correct, so if you know otherwise, I challenge you to be specific." am I supposed to ignore?
Do you always have a three week old infant in your arm when you type? It seems it must be so.
Capitalization isn't the primary concern. I never even mentioned it.
Capitalization isn't the primary concern. I never even mentioned it.
edit:
so you don't consider picking away at the same irrelevant point, inspite of it having been explained to you, to be baiting, or inflammatory. shoulda figured as much.
Again, I'm only responding to your posts. If you can't take it, don't dish it out.
What part of, "I have a BA, and a first class MA sitting in the room with me at the moment, both of whom majored in English (one of whom was born in London), both of whom say that specific sentence is grammatically correct, so if you know otherwise, I challenge you to be specific." am I supposed to ignore?
Do you always have a three week old infant in your arm when you type? It seems it must be so.
QUOTE
i don't know about you, but when i type one handed, capital letters are the first to go (and experience tells me this is true for most people.
Capitalization isn't the primary concern. I never even mentioned it.
QUOTE (->
| QUOTE |
| i don't know about you, but when i type one handed, capital letters are the first to go (and experience tells me this is true for most people. |
Capitalization isn't the primary concern. I never even mentioned it.
edit:
so you don't consider picking away at the same irrelevant point, inspite of it having been explained to you, to be baiting, or inflammatory. shoulda figured as much.
Again, I'm only responding to your posts. If you can't take it, don't dish it out.
QUOTE (ubavontuba+Feb 3 2008, 09:05 PM)
What part of, "I have a BA, and a first class MA sitting in the room with me at the moment, both of whom majored in English (one of whom was born in London), both of whom say that specific sentence is grammatically correct, so if you know otherwise, I challenge you to be specific." am I supposed to ignore?
Do you always have a three week old infant in your arm when you type? It seems it must be so.
Capitalization isn't the primary concern. I never even mentioned it.
Again, I'm only responding to your posts. If you can't take it, don't dish it out.
in context? i expected you to elaborate on the specific error in the specific post you were referring to. get that? i was challenging you to be specific with the error in the post in question.
the fact that you then went and did it to my entire next post is what is blatantly baiting and inflammatory, ESPECIALLY after having been provided with an explanation for the occasional dropped letter.
and yes, you did, one of the errors you highlighted was a dropped capital letter.
and no, you're not simply responding to my posts, you're continually stepping it up levels, and provoking things further, as you have right from your first post to me.
Do you always have a three week old infant in your arm when you type? It seems it must be so.
Capitalization isn't the primary concern. I never even mentioned it.
Again, I'm only responding to your posts. If you can't take it, don't dish it out.
in context? i expected you to elaborate on the specific error in the specific post you were referring to. get that? i was challenging you to be specific with the error in the post in question.
the fact that you then went and did it to my entire next post is what is blatantly baiting and inflammatory, ESPECIALLY after having been provided with an explanation for the occasional dropped letter.
and yes, you did, one of the errors you highlighted was a dropped capital letter.
and no, you're not simply responding to my posts, you're continually stepping it up levels, and provoking things further, as you have right from your first post to me.
QUOTE (Trippy+Feb 3 2008, 09:12 AM)
in context? i expected you to elaborate on the specific error in the specific post you were referring to. get that? i was challenging you to be specific with the error in the post in question.
I was.
Don't make bold challenges then. Instead, simply ask what the problem is... or better yet, don't make such a big fuss to begin with.
Don't make bold challenges then. Instead, simply ask what the problem is... or better yet, don't make such a big fuss to begin with.
and yes, you did, one of the errors you highlighted was a dropped capital letter.
I merely highlighted it in the context of multiple errors. It wasn't being singled out.
Every response I've given has been in direct reply to your comments. You started the flame war, and you continue it. Why don't you just drop it and get back to the science?
Ok
Ok
The Peanuts post only considers the velocity of the protons towards each other,
Why would I consider anything else, nothing else is relevant to the point I was making.
Who cares about the x-axis, in the scenario I presented the micro black hole will already has more than enough velocity along the y-axis to zip right through earth.
A micro black hole could be traveling at exactly the same velocity as earth in 2 axis but if it is moving at half the speed of light relative to earth in the other axis it will not be gravitationally captured by earth.
Who cares about the x-axis, in the scenario I presented the micro black hole will already has more than enough velocity along the y-axis to zip right through earth.
A micro black hole could be traveling at exactly the same velocity as earth in 2 axis but if it is moving at half the speed of light relative to earth in the other axis it will not be gravitationally captured by earth.
and although the components in the y axis can be considered to cancel, the components in the x-axis, in most cases will not,
How does this invalidate my argument? The components in the y-axis don't cancel in earths reference frame, they do in the center of mass frame but so what? Are you confusing the two? In earths reference frame one proton has zero velocity along the y-axis and the other is traveling at the speed of light, there is no way they can cancel to produce a MBH traveling at a velocity below earths escape velocity.
this means that there will be a narrow set of angles that the two protons can approach each other on that will result in a residual velocity, relative to the earth, with an absolute magnitude that is equal to, or less than the earths escape velocity.
It appears you are confusing the COM frame with earths reference frame in the scenario I presented in my other post. There are no angles that protons A and B in that scenario can collide that produces a micro black hole with a velocity lower than earths escape velocity.
I was.
QUOTE
the fact that you then went and did it to my entire next post is what is blatantly baiting and inflammatory, ESPECIALLY after having been provided with an explanation for the occasional dropped letter.
Don't make bold challenges then. Instead, simply ask what the problem is... or better yet, don't make such a big fuss to begin with.
QUOTE (->
| QUOTE |
| the fact that you then went and did it to my entire next post is what is blatantly baiting and inflammatory, ESPECIALLY after having been provided with an explanation for the occasional dropped letter. |
Don't make bold challenges then. Instead, simply ask what the problem is... or better yet, don't make such a big fuss to begin with.
and yes, you did, one of the errors you highlighted was a dropped capital letter.
I merely highlighted it in the context of multiple errors. It wasn't being singled out.
QUOTE
and no, you're not simply responding to my posts, you're continually stepping it up levels, and provoking things further, as you have right from your first post to me.
Every response I've given has been in direct reply to your comments. You started the flame war, and you continue it. Why don't you just drop it and get back to the science?
QUOTE (ubavontuba+Feb 3 2008, 06:30 AM)
...
...
RC,
I'm sorry. It seems we are again at a communications impasse. I don't really know what you're expecting from me.
All I can tell you is that the earth's gravity can't forcibly dissolve a micro black hole. It might cause objects with very weak gravity gradients to lose cohesion (as with Saturn and its rings), but it can't "erase" the gravity of the mass within its system.
Hi uba!
Sorry for the delay, but my attention has been elsewhere of late.
Of course it cannot 'erase the gravity of the mass within its system'. It DOES however DISTORT and REDIRECT any 'surface mass component' gravity PROFILE so the overall 'curvature strength' is towards the Earth mass overall. Which is why when you throw your 'two stones' and they fall (one stone above the other), THEIR OVERALL gravity profile as a 'two-stone system' is DISTORTED and overwhelmed in STRENGTH and DIRECTION towards the Earth's overwhelming strength/profile gradient.
The Earth gravity DOES NOT have to 'directly' "forcibly dissolve a micro black hole" as you infer, but only has to 'disrupt' the symmetry and strength OVERALL on the side of the (still putative) micro-hole that faces the earth as it falls.
Since your (putative) micro-hole is supposedly held together by its 'puny' gravity against its ULTRA-HIGH ENERGY DENSITY which supposedly forms its 'gradient' in the first place, then any 'disruption' to the 'sharp and ultra-thin cross-section 'horizon', no matter how slight, will be OVERWHELMINGLY disastrous for the micro-holes ability to contain its energy density...which will 'break out' from the weakened/distorted 'underside gradient' point where the interaction between the massive gravity well profile of the Earth distorts the miniscule gravity profile of the (putative) micro-hole.
Like I pointed out before, uba, the free-collision TWO BODY 'hole formation' EVENT is a comparatively much more MESSY 'event' than the MACRO-hole formation event (in which the central core 'black' density is at all times during formation symmetrically contained and UNdistorted by its self gravity as a ONE BODY 'hole formation' EVENT.
Which is why the (putative) Micro-hole formation event involves ASYMMETRIC DYNAMIC FORCES that overwhelm any miniscule, plank-scale 'mass/gravity' tendency to constrain the free-collision products to 'black' densities in the first place....and THEN on top of THAT, in the LHC at the surface of the Earth, there is the ever-present EXTERNAL GRAVITY (ie, 'external' to the 'event components') which have pre/during/post DISTORTION effects on any gravity/mass PROFILE/DISTRIBUTION of the 'free-collision' event components at the EDGE OF THE OVERWHELMING EARTH MASS distribution and attendent GRAVITY profile.
Think about the interation of EDGE TO EDGE gravity profiles between TWO BODIES OF EXTREMELY DISPARATE mass and their respective gravity wells/gradients.
Remember: Consider only the gravity well to gravity well INDIRECT EFFECTS ....and NOT any direct Earth-gravity to Micro-mass effects per se.....since any distortion/weakening of the micro-hole gravity 'containment' of its supposed extrem energy DENSITY would result in the micro-hole energy EXPLODING ITSELF due to its OWN UNSYMMETRIC/WEAKENED gravity at the edge of the earth's own gravity.
See? Your initial 'not soluable' comment is irrelevant....since only a 'destabilising effect' rather 'soluability' is involved in the gravity/gravity scenario I describe. (And all THAT 'gravity/gravity' disruption is, of course, IN ADDITION to all the 'quantum scale' disruption effects I previously mentioned!)
Back in a few days, uba. Cheers!
RC.
.
...
RC,
I'm sorry. It seems we are again at a communications impasse. I don't really know what you're expecting from me.
All I can tell you is that the earth's gravity can't forcibly dissolve a micro black hole. It might cause objects with very weak gravity gradients to lose cohesion (as with Saturn and its rings), but it can't "erase" the gravity of the mass within its system.
Hi uba!
Sorry for the delay, but my attention has been elsewhere of late.
Of course it cannot 'erase the gravity of the mass within its system'. It DOES however DISTORT and REDIRECT any 'surface mass component' gravity PROFILE so the overall 'curvature strength' is towards the Earth mass overall. Which is why when you throw your 'two stones' and they fall (one stone above the other), THEIR OVERALL gravity profile as a 'two-stone system' is DISTORTED and overwhelmed in STRENGTH and DIRECTION towards the Earth's overwhelming strength/profile gradient.
The Earth gravity DOES NOT have to 'directly' "forcibly dissolve a micro black hole" as you infer, but only has to 'disrupt' the symmetry and strength OVERALL on the side of the (still putative) micro-hole that faces the earth as it falls.
Since your (putative) micro-hole is supposedly held together by its 'puny' gravity against its ULTRA-HIGH ENERGY DENSITY which supposedly forms its 'gradient' in the first place, then any 'disruption' to the 'sharp and ultra-thin cross-section 'horizon', no matter how slight, will be OVERWHELMINGLY disastrous for the micro-holes ability to contain its energy density...which will 'break out' from the weakened/distorted 'underside gradient' point where the interaction between the massive gravity well profile of the Earth distorts the miniscule gravity profile of the (putative) micro-hole.
Like I pointed out before, uba, the free-collision TWO BODY 'hole formation' EVENT is a comparatively much more MESSY 'event' than the MACRO-hole formation event (in which the central core 'black' density is at all times during formation symmetrically contained and UNdistorted by its self gravity as a ONE BODY 'hole formation' EVENT.
Which is why the (putative) Micro-hole formation event involves ASYMMETRIC DYNAMIC FORCES that overwhelm any miniscule, plank-scale 'mass/gravity' tendency to constrain the free-collision products to 'black' densities in the first place....and THEN on top of THAT, in the LHC at the surface of the Earth, there is the ever-present EXTERNAL GRAVITY (ie, 'external' to the 'event components') which have pre/during/post DISTORTION effects on any gravity/mass PROFILE/DISTRIBUTION of the 'free-collision' event components at the EDGE OF THE OVERWHELMING EARTH MASS distribution and attendent GRAVITY profile.
Think about the interation of EDGE TO EDGE gravity profiles between TWO BODIES OF EXTREMELY DISPARATE mass and their respective gravity wells/gradients.
Remember: Consider only the gravity well to gravity well INDIRECT EFFECTS ....and NOT any direct Earth-gravity to Micro-mass effects per se.....since any distortion/weakening of the micro-hole gravity 'containment' of its supposed extrem energy DENSITY would result in the micro-hole energy EXPLODING ITSELF due to its OWN UNSYMMETRIC/WEAKENED gravity at the edge of the earth's own gravity.
See? Your initial 'not soluable' comment is irrelevant....since only a 'destabilising effect' rather 'soluability' is involved in the gravity/gravity scenario I describe. (And all THAT 'gravity/gravity' disruption is, of course, IN ADDITION to all the 'quantum scale' disruption effects I previously mentioned!)
Back in a few days, uba. Cheers!
RC.
.
QUOTE (ubavontuba+Feb 3 2008, 08:40 PM)
I've only directly replied to your posts. If they're "irrelevant spam," then the problem is the source material, isn't it?
you know, you really are a friggin mo, in this post, you make this claim, but, if you look back, we find in the post that you claim started the flamewar, me referencing how your claims in relation to The Peanuts post contradict claims you've made elsewhere, and I've tried to point out to you before that elements of the peanuts post contradict other claims you have made.
Also, in amongst all the garbage you've spouted, we have you claiming that I contradict myself, when in fact I have not, neither of the mechanisms I have examinedare contradictory, in fact, I must say that I am somewhat surprised that you would be naive enough to believe that there would only be a single mechanism involved, but then again, the depths of ignorance and stupidity some people on this forum insist on plumb never ceases to amaze me.
As far as the Peanuts post goes, there is a glaring inconsistency that I have endeavoured to point out to you before, but you have thus far refused to address.
Consider, in the peanuts post, the plane which encompases the three points represented by the center of each proton, and the center of mass of the system. The Peanuts post only considers the velocity of the protons towards each other, but, because one of the protons is at rest wrt the earth (or nearly so) it shares the earths motion. In other words, if we consider the motion of the protons towards each other to be representative of the y-axis on our plane, then there must also be some component along the x-axis, and although the components in the y axis can be considered to cancel, the components in the x-axis, in most cases will not, this means that there will be a narrow set of angles that the two protons can approach each other on that will result in a residual velocity, relative to the earth, with an absolute magnitude that is equal to, or less than the earths escape velocity.
Thus contradicting everything you claim about thepeanuts post.
you know, you really are a friggin mo, in this post, you make this claim, but, if you look back, we find in the post that you claim started the flamewar, me referencing how your claims in relation to The Peanuts post contradict claims you've made elsewhere, and I've tried to point out to you before that elements of the peanuts post contradict other claims you have made.
Also, in amongst all the garbage you've spouted, we have you claiming that I contradict myself, when in fact I have not, neither of the mechanisms I have examinedare contradictory, in fact, I must say that I am somewhat surprised that you would be naive enough to believe that there would only be a single mechanism involved, but then again, the depths of ignorance and stupidity some people on this forum insist on plumb never ceases to amaze me.
As far as the Peanuts post goes, there is a glaring inconsistency that I have endeavoured to point out to you before, but you have thus far refused to address.
Consider, in the peanuts post, the plane which encompases the three points represented by the center of each proton, and the center of mass of the system. The Peanuts post only considers the velocity of the protons towards each other, but, because one of the protons is at rest wrt the earth (or nearly so) it shares the earths motion. In other words, if we consider the motion of the protons towards each other to be representative of the y-axis on our plane, then there must also be some component along the x-axis, and although the components in the y axis can be considered to cancel, the components in the x-axis, in most cases will not, this means that there will be a narrow set of angles that the two protons can approach each other on that will result in a residual velocity, relative to the earth, with an absolute magnitude that is equal to, or less than the earths escape velocity.
Thus contradicting everything you claim about thepeanuts post.
QUOTE
Consider, in the peanuts post, the plane which encompases the three points represented by the center of each proton, and the center of mass of the system.
Ok
QUOTE (->
| QUOTE |
| Consider, in the peanuts post, the plane which encompases the three points represented by the center of each proton, and the center of mass of the system. |
Ok
The Peanuts post only considers the velocity of the protons towards each other,
Why would I consider anything else, nothing else is relevant to the point I was making.
QUOTE
because one of the protons is at rest wrt the earth (or nearly so) it shares the earths motion. In other words, if we consider the motion of the protons towards each other to be representative of the y-axis on our plane, then there must also be some component along the x-axis
Who cares about the x-axis, in the scenario I presented the micro black hole will already has more than enough velocity along the y-axis to zip right through earth.
A micro black hole could be traveling at exactly the same velocity as earth in 2 axis but if it is moving at half the speed of light relative to earth in the other axis it will not be gravitationally captured by earth.
QUOTE (->
| QUOTE |
| because one of the protons is at rest wrt the earth (or nearly so) it shares the earths motion. In other words, if we consider the motion of the protons towards each other to be representative of the y-axis on our plane, then there must also be some component along the x-axis |
Who cares about the x-axis, in the scenario I presented the micro black hole will already has more than enough velocity along the y-axis to zip right through earth.
A micro black hole could be traveling at exactly the same velocity as earth in 2 axis but if it is moving at half the speed of light relative to earth in the other axis it will not be gravitationally captured by earth.
and although the components in the y axis can be considered to cancel, the components in the x-axis, in most cases will not,
How does this invalidate my argument? The components in the y-axis don't cancel in earths reference frame, they do in the center of mass frame but so what? Are you confusing the two? In earths reference frame one proton has zero velocity along the y-axis and the other is traveling at the speed of light, there is no way they can cancel to produce a MBH traveling at a velocity below earths escape velocity.
QUOTE
this means that there will be a narrow set of angles that the two protons can approach each other on that will result in a residual velocity, relative to the earth, with an absolute magnitude that is equal to, or less than the earths escape velocity.
It appears you are confusing the COM frame with earths reference frame in the scenario I presented in my other post. There are no angles that protons A and B in that scenario can collide that produces a micro black hole with a velocity lower than earths escape velocity.
QUOTE (RealityCheck+Feb 7 2008, 12:14 AM)
Hi uba!
Sorry for the delay, but my attention has been elsewhere of late.
I'm feeling your pain.
Sure, their mass contributes to the system's overall gravity.
Sure, their mass contributes to the system's overall gravity.
The Earth gravity DOES NOT have to 'directly' "forcibly dissolve a micro black hole" as you infer, but only has to 'disrupt' the symmetry and strength OVERALL on the side of the (still putative) micro-hole that faces the earth as it falls.
How's it going to do that? Gravity doesn't "flow" from a greater potential to a lesser potential.
Consider: The earth's gravity gradient is quite flat, as opposed to the black hole's gravity gradient.
Pish. It doesn't work like that. This would only matter if the black hole could be held away from the center of mass by resting on the earth's surface (even then, it's so slight a tug as to be meaningless). The black hole moves freely through mass. It's always in freefall. Therefore your point is altogether irrelevant. Gravity isn't distorting it. Gravity's only affecting its trajectory as it moves through the curved space-time.
Pish. It doesn't work like that. This would only matter if the black hole could be held away from the center of mass by resting on the earth's surface (even then, it's so slight a tug as to be meaningless). The black hole moves freely through mass. It's always in freefall. Therefore your point is altogether irrelevant. Gravity isn't distorting it. Gravity's only affecting its trajectory as it moves through the curved space-time.
Like I pointed out before, uba, the free-collision TWO BODY 'hole formation' EVENT is a comparatively much more MESSY 'event' than the MACRO-hole formation event (in which the central core 'black' density is at all times during formation symmetrically contained and UNdistorted by its self gravity as a ONE BODY 'hole formation' EVENT.
I never argued otherwise. I've said many times that collisions are messy.
Some energy escapes the system, but that doesn't mean all of it does.
Some energy escapes the system, but that doesn't mean all of it does.
....and THEN on top of THAT, in the LHC at the surface of the Earth, there is the ever-present EXTERNAL GRAVITY (ie, 'external' to the 'event components') which have pre/during/post DISTORTION effects on any gravity/mass PROFILE/DISTRIBUTION of the 'free-collision' event components at the EDGE OF THE OVERWHELMING EARTH MASS distribution and attendent GRAVITY profile.
Irrelevant (see above).
I have.
I have.
Remember: Consider only the gravity well to gravity well INDIRECT EFFECTS ....and NOT any direct Earth-gravity to Micro-mass effects per se.....since any distortion/weakening of the micro-hole gravity 'containment' of its supposed extrem energy DENSITY would result in the micro-hole energy EXPLODING ITSELF due to its OWN UNSYMMETRIC/WEAKENED gravity at the edge of the earth's own gravity.
Co-joined gravity wells add to the total space-time curvature, not subtract from it.
I give up. It seems apparent that I can't make you understand. You seem to have decided how it is, without regard to the facts. Believe what you will.
Sorry for the delay, but my attention has been elsewhere of late.
I'm feeling your pain.
QUOTE
Of course it cannot 'erase the gravity of the mass within its system'. It DOES however DISTORT and REDIRECT any 'surface mass component' gravity PROFILE so the overall 'curvature strength' is towards the Earth mass overall. Which is why when you throw your 'two stones' and they fall (one stone above the other), THEIR OVERALL gravity profile as a 'two-stone system' is DISTORTED and overwhelmed in STRENGTH and DIRECTION towards the Earth's overwhelming strength/profile gradient.
Sure, their mass contributes to the system's overall gravity.
QUOTE (->
| QUOTE |
| Of course it cannot 'erase the gravity of the mass within its system'. It DOES however DISTORT and REDIRECT any 'surface mass component' gravity PROFILE so the overall 'curvature strength' is towards the Earth mass overall. Which is why when you throw your 'two stones' and they fall (one stone above the other), THEIR OVERALL gravity profile as a 'two-stone system' is DISTORTED and overwhelmed in STRENGTH and DIRECTION towards the Earth's overwhelming strength/profile gradient. |
Sure, their mass contributes to the system's overall gravity.
The Earth gravity DOES NOT have to 'directly' "forcibly dissolve a micro black hole" as you infer, but only has to 'disrupt' the symmetry and strength OVERALL on the side of the (still putative) micro-hole that faces the earth as it falls.
How's it going to do that? Gravity doesn't "flow" from a greater potential to a lesser potential.
Consider: The earth's gravity gradient is quite flat, as opposed to the black hole's gravity gradient.
QUOTE
Since your (putative) micro-hole is supposedly held together by its 'puny' gravity against its ULTRA-HIGH ENERGY DENSITY which supposedly forms its 'gradient' in the first place, then any 'disruption' to the 'sharp and ultra-thin cross-section 'horizon', no matter how slight, will be OVERWHELMINGLY disastrous for the micro-holes ability to contain its energy density...which will 'break out' from the weakened/distorted 'underside gradient' point where the interaction between the massive gravity well profile of the Earth distorts the miniscule gravity profile of the (putative) micro-hole.
Pish. It doesn't work like that. This would only matter if the black hole could be held away from the center of mass by resting on the earth's surface (even then, it's so slight a tug as to be meaningless). The black hole moves freely through mass. It's always in freefall. Therefore your point is altogether irrelevant. Gravity isn't distorting it. Gravity's only affecting its trajectory as it moves through the curved space-time.
QUOTE (->
| QUOTE |
| Since your (putative) micro-hole is supposedly held together by its 'puny' gravity against its ULTRA-HIGH ENERGY DENSITY which supposedly forms its 'gradient' in the first place, then any 'disruption' to the 'sharp and ultra-thin cross-section 'horizon', no matter how slight, will be OVERWHELMINGLY disastrous for the micro-holes ability to contain its energy density...which will 'break out' from the weakened/distorted 'underside gradient' point where the interaction between the massive gravity well profile of the Earth distorts the miniscule gravity profile of the (putative) micro-hole. |
Pish. It doesn't work like that. This would only matter if the black hole could be held away from the center of mass by resting on the earth's surface (even then, it's so slight a tug as to be meaningless). The black hole moves freely through mass. It's always in freefall. Therefore your point is altogether irrelevant. Gravity isn't distorting it. Gravity's only affecting its trajectory as it moves through the curved space-time.
Like I pointed out before, uba, the free-collision TWO BODY 'hole formation' EVENT is a comparatively much more MESSY 'event' than the MACRO-hole formation event (in which the central core 'black' density is at all times during formation symmetrically contained and UNdistorted by its self gravity as a ONE BODY 'hole formation' EVENT.
I never argued otherwise. I've said many times that collisions are messy.
QUOTE
Which is why the (putative) Micro-hole formation event involves ASYMMETRIC DYNAMIC FORCES that overwhelm any miniscule, plank-scale 'mass/gravity' tendency to constrain the free-collision products to 'black' densities in the first place
Some energy escapes the system, but that doesn't mean all of it does.
QUOTE (->
| QUOTE |
| Which is why the (putative) Micro-hole formation event involves ASYMMETRIC DYNAMIC FORCES that overwhelm any miniscule, plank-scale 'mass/gravity' tendency to constrain the free-collision products to 'black' densities in the first place |
Some energy escapes the system, but that doesn't mean all of it does.
....and THEN on top of THAT, in the LHC at the surface of the Earth, there is the ever-present EXTERNAL GRAVITY (ie, 'external' to the 'event components') which have pre/during/post DISTORTION effects on any gravity/mass PROFILE/DISTRIBUTION of the 'free-collision' event components at the EDGE OF THE OVERWHELMING EARTH MASS distribution and attendent GRAVITY profile.
Irrelevant (see above).
QUOTE
Think about the interation of EDGE TO EDGE gravity profiles between TWO BODIES OF EXTREMELY DISPARATE mass and their respective gravity wells/gradients.
I have.
QUOTE (->
| QUOTE |
| Think about the interation of EDGE TO EDGE gravity profiles between TWO BODIES OF EXTREMELY DISPARATE mass and their respective gravity wells/gradients. |
I have.
Remember: Consider only the gravity well to gravity well INDIRECT EFFECTS ....and NOT any direct Earth-gravity to Micro-mass effects per se.....since any distortion/weakening of the micro-hole gravity 'containment' of its supposed extrem energy DENSITY would result in the micro-hole energy EXPLODING ITSELF due to its OWN UNSYMMETRIC/WEAKENED gravity at the edge of the earth's own gravity.
Co-joined gravity wells add to the total space-time curvature, not subtract from it.
QUOTE
See? Your initial 'not soluable' comment is irrelevant....since only a 'destabilising effect' rather 'soluability' is involved in the gravity/gravity scenario I describe. (And all THAT 'gravity/gravity' disruption is, of course, IN ADDITION to all the 'quantum scale' disruption effects I previously mentioned!)
I give up. It seems apparent that I can't make you understand. You seem to have decided how it is, without regard to the facts. Believe what you will.
QUOTE (Trippy+Feb 7 2008, 04:22 AM)
you know, you really are a friggin mo, in this post, you make this claim, but, if you look back, we find in the post that you claim started the flamewar, me referencing how your claims in relation to The Peanuts post contradict claims you've made elsewhere, and I've tried to point out to you before that elements of the peanuts post contradict other claims you have made.
What's that got to do with the attitude of your post?
Fine. Explain how they're different.
Fine. Explain how they're different.
As far as the Peanuts post goes, there is a glaring inconsistency that I have endeavoured to point out to you before, but you have thus far refused to address.
Because there isn't any inconsistency. It's only your misinterpretation that makes you think so.
It happens to be co-moving with the earth, but that's inconsequential to the argument. The collision energy (resulting in a micro black hole) removes it from the earth's rest frame.
It happens to be co-moving with the earth, but that's inconsequential to the argument. The collision energy (resulting in a micro black hole) removes it from the earth's rest frame.
In other words, if we consider the motion of the protons towards each other to be representative of the y-axis on our plane, then there must also be some component along the x-axis, and although the components in the y axis can be considered to cancel, the components in the x-axis, in most cases will not,
What's the x-axis got to do with anything?
I think you've now confused ThePeanut's post about cosmic ray/earth collisions with the two cosmic rays colliding above the earth argument.
I think you've now confused ThePeanut's post about cosmic ray/earth collisions with the two cosmic rays colliding above the earth argument.
Thus contradicting everything you claim about thepeanuts post.
Thus demonstrating your inability to maintain a coherent argument.
EDIT: Oops! I see ThePeanut just posted a reply too! What a coincidence! He said it better than I. Sorry if my post seems redundant.
Try again, both the centre of mass and earth are on the line drawn between proton A and proton B.
Try again, both the centre of mass and earth are on the line drawn between proton A and proton B. [/QUOTE]
So now you're talking about a collision between two cosmic ray protons?
The same collision that Ubavontuba has stated repeatedly is represented by "INfinities within infinities" and therefore could not have happened in the earths life time?
No. We're talking about a collision between two particles. One happens to initially be co-moving with the earth (which is actually an irrelevant point).
What's that got to do with the attitude of your post?
QUOTE
Also, in amongst all the garbage you've spouted, we have you claiming that I contradict myself, when in fact I have not, neither of the mechanisms I have examinedare contradictory, in fact, I must say that I am somewhat surprised that you would be naive enough to believe that there would only be a single mechanism involved, but then again, the depths of ignorance and stupidity some people on this forum insist on plumb never ceases to amaze me.
Fine. Explain how they're different.
QUOTE (->
| QUOTE |
| Also, in amongst all the garbage you've spouted, we have you claiming that I contradict myself, when in fact I have not, neither of the mechanisms I have examinedare contradictory, in fact, I must say that I am somewhat surprised that you would be naive enough to believe that there would only be a single mechanism involved, but then again, the depths of ignorance and stupidity some people on this forum insist on plumb never ceases to amaze me. |
Fine. Explain how they're different.
As far as the Peanuts post goes, there is a glaring inconsistency that I have endeavoured to point out to you before, but you have thus far refused to address.
Because there isn't any inconsistency. It's only your misinterpretation that makes you think so.
QUOTE
Consider, in the peanuts post, the plane which encompases the three points represented by the center of each proton, and the center of mass of the system. The Peanuts post only considers the velocity of the protons towards each other, but, because one of the protons is at rest wrt the earth (or nearly so) it shares the earths motion.
It happens to be co-moving with the earth, but that's inconsequential to the argument. The collision energy (resulting in a micro black hole) removes it from the earth's rest frame.
QUOTE (->
| QUOTE |
| Consider, in the peanuts post, the plane which encompases the three points represented by the center of each proton, and the center of mass of the system. The Peanuts post only considers the velocity of the protons towards each other, but, because one of the protons is at rest wrt the earth (or nearly so) it shares the earths motion. |
It happens to be co-moving with the earth, but that's inconsequential to the argument. The collision energy (resulting in a micro black hole) removes it from the earth's rest frame.
In other words, if we consider the motion of the protons towards each other to be representative of the y-axis on our plane, then there must also be some component along the x-axis, and although the components in the y axis can be considered to cancel, the components in the x-axis, in most cases will not,
What's the x-axis got to do with anything?
QUOTE
this means that there will be a narrow set of angles that the two protons can approach each other on that will result in a residual velocity, relative to the earth, with an absolute magnitude that is equal to, or less than the earths escape velocity.
I think you've now confused ThePeanut's post about cosmic ray/earth collisions with the two cosmic rays colliding above the earth argument.
QUOTE (->
| QUOTE |
| this means that there will be a narrow set of angles that the two protons can approach each other on that will result in a residual velocity, relative to the earth, with an absolute magnitude that is equal to, or less than the earths escape velocity. |
I think you've now confused ThePeanut's post about cosmic ray/earth collisions with the two cosmic rays colliding above the earth argument.
Thus contradicting everything you claim about thepeanuts post.
Thus demonstrating your inability to maintain a coherent argument.
EDIT: Oops! I see ThePeanut just posted a reply too! What a coincidence! He said it better than I. Sorry if my post seems redundant.
ThePeanut/Ubavontuba.
I can't argue with people willing to ignore part of the story.
You're treating the velocity as if it's a 1-d vector. It's not, it's a 3-d vector, as such there are three directional components to consider.
I tried to elaborate on this, but apparently you can't understand it.
Until you're willing this basic fact, there can be no further discussion.
I can't argue with people willing to ignore part of the story.
You're treating the velocity as if it's a 1-d vector. It's not, it's a 3-d vector, as such there are three directional components to consider.
I tried to elaborate on this, but apparently you can't understand it.
Until you're willing this basic fact, there can be no further discussion.
QUOTE (Trippy+Feb 11 2008, 05:02 AM)
ThePeanut/Ubavontuba.
I can't argue with people willing to ignore part of the story.
You're treating the velocity as if it's a 1-d vector. It's not, it's a 3-d vector, as such there are three directional components to consider.
I tried to elaborate on this, but apparently you can't understand it.
Until you're willing this basic fact, there can be no further discussion.
How is a two-particle collision a 3-D vector?
Let's see... There's point A and there's point B. We can draw a straight line from point A to point B. We can draw another straight line from point B to point A. Oh look! They're the same line!
You need a third reference point to have 2-D, and four to have 3-D.
I can't argue with people willing to ignore part of the story.
You're treating the velocity as if it's a 1-d vector. It's not, it's a 3-d vector, as such there are three directional components to consider.
I tried to elaborate on this, but apparently you can't understand it.
Until you're willing this basic fact, there can be no further discussion.
How is a two-particle collision a 3-D vector?
Let's see... There's point A and there's point B. We can draw a straight line from point A to point B. We can draw another straight line from point B to point A. Oh look! They're the same line!
You need a third reference point to have 2-D, and four to have 3-D.
QUOTE (ubavontuba+Feb 11 2008, 06:12 PM)
How is a two particle collision a 3-D vector?
Because we're considering it relative to the earth (and center of mass of the system) (or the fixed and distant stars, take your pick).
Because we're considering it relative to the earth (and center of mass of the system) (or the fixed and distant stars, take your pick).
QUOTE (Trippy+Feb 11 2008, 05:17 AM)
Because we're considering it relative to the earth (and center of mass of the system) (or the fixed and distant stars, take your pick).
How is the earth relevant? How is the center of mass relevant?
The center of momentum can be relative to the earth as a reference point, but it's not "stuck" to the earth.
How is the earth relevant? How is the center of mass relevant?
The center of momentum can be relative to the earth as a reference point, but it's not "stuck" to the earth.
QUOTE (ubavontuba+Feb 11 2008, 06:27 PM)
How is the earth relevant? How is the center of mass relevant?
The center of momentum can be relative to the earth as a reference point, but it's not "stuck" to the earth.
Think about that question, for a moment.
We're considering whether or not the velocity of the final product relative to the earth is below or above a certain value, and you're asking me how the earth is relevant.
We're talking about a collision, and you're asking how the center of mass is relevant?
The center of momentum can be relative to the earth as a reference point, but it's not "stuck" to the earth.
Think about that question, for a moment.
We're considering whether or not the velocity of the final product relative to the earth is below or above a certain value, and you're asking me how the earth is relevant.
We're talking about a collision, and you're asking how the center of mass is relevant?
QUOTE
QUOTE (->
| QUOTE |
| How is a two particle collision a 3-D vector? Because we're considering it relative to the earth (and center of mass of the system) |
Try again, both the centre of mass and earth are on the line drawn between proton A and proton B.
QUOTE (ThePeanut+Feb 11 2008, 07:04 PM)
Because we're considering it relative to the earth (and center of mass of the system)[/QUOTE]
Try again, both the centre of mass and earth are on the line drawn between proton A and proton B.
So now you're talking about a collision between two cosmic ray protons?
The same collision that Ubavontuba has stated repeatedly is represented by "INfinities within infinities" and therefore could not have happened in the earths life time?
Try again, both the centre of mass and earth are on the line drawn between proton A and proton B.
So now you're talking about a collision between two cosmic ray protons?
The same collision that Ubavontuba has stated repeatedly is represented by "INfinities within infinities" and therefore could not have happened in the earths life time?
QUOTE (Trippy+Feb 11 2008, 06:12 AM)
Try again, both the centre of mass and earth are on the line drawn between proton A and proton B. [/QUOTE]
So now you're talking about a collision between two cosmic ray protons?
The same collision that Ubavontuba has stated repeatedly is represented by "INfinities within infinities" and therefore could not have happened in the earths life time?
No. We're talking about a collision between two particles. One happens to initially be co-moving with the earth (which is actually an irrelevant point).
No I am, and always have been talking about two protons A and B, A is stationary relative to earth, B is a cosmic ray traveling towards earth at virtually the speed of light.
Are you disputing that in this setup both earth and the COM are on the line AB?
Are you disputing that in this setup both earth and the COM are on the line AB?
QUOTE (Trippy+Feb 11 2008, 05:49 AM)
Think about that question, for a moment.
We're considering whether or not the velocity of the final product relative to the earth is below or above a certain value, and you're asking me how the earth is relevant.
We're talking about a collision, and you're asking how the center of mass is relevant?
I'm not saying they aren't relevant. I'm asking you to explain your interpretation of their relevance.
We're considering whether or not the velocity of the final product relative to the earth is below or above a certain value, and you're asking me how the earth is relevant.
We're talking about a collision, and you're asking how the center of mass is relevant?
I'm not saying they aren't relevant. I'm asking you to explain your interpretation of their relevance.
QUOTE (ubavontuba+Feb 11 2008, 03:49 AM)
I'm feeling your pain.
Sure, their mass contributes to the system's overall gravity.
How's it going to do that? Gravity doesn't "flow" from a greater potential to a lesser potential.
Consider: The earth's gravity gradient is quite flat, as opposed to the black hole's gravity gradient.
Pish. It doesn't work like that. This would only matter if the black hole could be held away from the center of mass by resting on the earth's surface (even then, it's so slight a tug as to be meaningless). The black hole moves freely through mass. It's always in freefall. Therefore your point is altogether irrelevant. Gravity isn't distorting it. Gravity's only affecting its trajectory as it moves through the curved space-time.
I never argued otherwise. I've said many times that collisions are messy.
Some energy escapes the system, but that doesn't mean all of it does.
Irrelevant (see above).
I have.
Co-joined gravity wells add to the total space-time curvature, not subtract from it.
I give up. It seems apparent that I can't make you understand. You seem to have decided how it is, without regard to the facts. Believe what you will.
The gravity gradient of Earth may be 'flat', but it is CUMULATIVELY much stronger overall that the 'puny' mass gravity strength.....so the strength of Earth's gravity affects the UNDERSIDE 'gradient' of the purtative micro-hole feature.
And it falls in 'free fall' precisely BECAUSE of the Earth's overwhelming gravity 'curvature' strength ob=verall.
Moreover, when you say...
Yes
Yes
The velocity of the cosmic ray photon is (nominally) radial (WRT the earth)
You're saying that the velocity of the proton is along the (arbitrary) z-axis, and the velocities in the x and y axis are zero correct?
That's a possible scenario sure
Yes
Yes
But, what changes relative to the earth in the other two directions?
Nothing changes - the resulting black hole is at rest in the COM frame.
How do you make the leap from the MBH being at rest in the COM frame to being at rest relative to earth. Earths frame and the center of mass frame are not the same - I fear you're confusing the two.
In earths reference frame proton A is stationary , proton B travels along the z axis at the speed of light - the velocities do not cancel, the resulting MBH still has significant velocity along the z-axis.
In the COM frame both protons are moving in opposite directions along the z-axis at the same speed so that the resulting MBH is stationary.
How do you make the leap from the MBH being at rest in the COM frame to being at rest relative to earth. Earths frame and the center of mass frame are not the same - I fear you're confusing the two.
In earths reference frame proton A is stationary , proton B travels along the z axis at the speed of light - the velocities do not cancel, the resulting MBH still has significant velocity along the z-axis.
In the COM frame both protons are moving in opposite directions along the z-axis at the same speed so that the resulting MBH is stationary.
Aside from which, your scenario is unrealistic - the average velocity of atoms in the earths atmosphere is 450 m/s.
I was trying to keep the argument simple, obviously molecules in earths atmosphere are not exactly stationary relative to earth. In any event, giving the "stationary" particle some small amount of velocity does not change the argument, the resulting MBH would still have more than enough velocity to zip through earth without becoming gravitationally bound. A lone proton in the atmosphere is quite random as well, it would probably be an oxygen or nitrogen molecule that the cosmic ray collides with but I wanted to keep the example simple.
The original point was brought up to counter RealityCheck's claim of some "splatter" effect bleeding away the MBH's velocity so that it would get captured by earth. I was trying to demonstrate that this violates the conservation of momentum.
No, uba has already explained this to you several times, do you think the MBH is slowed down in earths reference frame due to mass gain?
So that in our setup with A being stationary in earths frame and B the cosmic ray, you think the resulting MBH will slow due to mass gain right? Do you understand that this implies an acceleration in the COM frame in the direction that A was initially traveling? What causes that acceleration?
In the COM frame the MBH product is now traveling along the z-axis in the initial direction proton A was traveling. How can this be so when the two protons were colliding head on along the z-axis at exactly the same speed? Where has the momentum come from when there is no net momentum initially in the COM frame?
Shouldn't the resulting MBH be stationary in the COM frame? If we reverse the setup so earth is in B's reference frame so proton B is now stationary, does the MBH also slow? This implies back in the COM frame that the MBH is now traveling in the initial direction of B. How can we have two results simply by changing our frame of reference? Again where has the momentum come from?
Mind you, looking back at your previous posts it doesn't surprise me - you also think that parton radiation would cause some sort of rocket effect on the MBH causing it to speed up in earths reference frame.
Care to explain how that one works?
Sure, their mass contributes to the system's overall gravity.
How's it going to do that? Gravity doesn't "flow" from a greater potential to a lesser potential.
Consider: The earth's gravity gradient is quite flat, as opposed to the black hole's gravity gradient.
Pish. It doesn't work like that. This would only matter if the black hole could be held away from the center of mass by resting on the earth's surface (even then, it's so slight a tug as to be meaningless). The black hole moves freely through mass. It's always in freefall. Therefore your point is altogether irrelevant. Gravity isn't distorting it. Gravity's only affecting its trajectory as it moves through the curved space-time.
I never argued otherwise. I've said many times that collisions are messy.
Some energy escapes the system, but that doesn't mean all of it does.
Irrelevant (see above).
I have.
Co-joined gravity wells add to the total space-time curvature, not subtract from it.
I give up. It seems apparent that I can't make you understand. You seem to have decided how it is, without regard to the facts. Believe what you will.
The gravity gradient of Earth may be 'flat', but it is CUMULATIVELY much stronger overall that the 'puny' mass gravity strength.....so the strength of Earth's gravity affects the UNDERSIDE 'gradient' of the purtative micro-hole feature.
And it falls in 'free fall' precisely BECAUSE of the Earth's overwhelming gravity 'curvature' strength ob=verall.
Moreover, when you say...
QUOTE (uba+)
Co-joined gravity wells add to the total space-time curvature, not subtract from it.
That is EXACTLY what I am saying....only you seem to think that 'addition' is 'symmetrical'....when it actually is A-symmetrical...with the UPPER side of the putative micro-hole gravity profile adding constructively (stronger) DOWNWARDS INTO the micro-hole TOP SIDE) while the UNDERSIDE adding constructively ALSO DOWNWARDS but AWAY FROM the micro-hole and towards the Earth.
That is why the hole 'falls' at all. And since the putative micro-hole feature does not fall INSTANTANEOUSLY into the Erath, then the 'edge-to-edge' gravity CURVATURE distortions/redirections between the micro-feature gravity and Earth gravity occur at THE SPEED OF LIGHT...plenty of time to cause havoc for the micro-hole gravitic/symmetric stability.
And I again repeat, the totality of disruptor effects from UNCONTAINED DYNAMIC COLLISIONS and QUANTUM PROCESSES AT THAT SCALE and EARTH GRAVITY-DISTORTION mitigate against the formation of such micro-holes in the first place in such a manner.
Anyhow, that's all I have time for this for a little while, mate.
Thanks for the stimulating discussions so far, uba! Catch ya later.
Cheers!
RC.
.
That is EXACTLY what I am saying....only you seem to think that 'addition' is 'symmetrical'....when it actually is A-symmetrical...with the UPPER side of the putative micro-hole gravity profile adding constructively (stronger) DOWNWARDS INTO the micro-hole TOP SIDE) while the UNDERSIDE adding constructively ALSO DOWNWARDS but AWAY FROM the micro-hole and towards the Earth.
That is why the hole 'falls' at all. And since the putative micro-hole feature does not fall INSTANTANEOUSLY into the Erath, then the 'edge-to-edge' gravity CURVATURE distortions/redirections between the micro-feature gravity and Earth gravity occur at THE SPEED OF LIGHT...plenty of time to cause havoc for the micro-hole gravitic/symmetric stability.
And I again repeat, the totality of disruptor effects from UNCONTAINED DYNAMIC COLLISIONS and QUANTUM PROCESSES AT THAT SCALE and EARTH GRAVITY-DISTORTION mitigate against the formation of such micro-holes in the first place in such a manner.
Anyhow, that's all I have time for this for a little while, mate.
Thanks for the stimulating discussions so far, uba! Catch ya later.
Cheers!
RC.
.
QUOTE (ThePeanut+Feb 11 2008, 07:19 PM)
No I am, and always have been talking about two protons A and B, A is stationary relative to earth, B is a cosmic ray traveling towards earth at virtually the speed of light.
Are you disputing that in this setup both earth and the COM are on the line AB?
Right...
So, you're considering a Proton approaching another proton with a velocity of 0 m/s relative to the earth.
The velocity of the cosmic ray photon is (nominally) radial (WRT the earth)
You're saying that the velocity of the proton is along the (arbitrary) z-axis, and the velocities in the x and y axis are zero correct?
And you claim that in the reference frame of the center of mass, the velocities cancel.
But, what changes relative to the earth in the other two directions? The product is, by your own logic, still at rest relative to the earth.
Aside from which, your scenario is unrealistic - the average velocitiy of atoms in the earths atmosphere is 450 m/s.
Are you disputing that in this setup both earth and the COM are on the line AB?
Right...
So, you're considering a Proton approaching another proton with a velocity of 0 m/s relative to the earth.
The velocity of the cosmic ray photon is (nominally) radial (WRT the earth)
You're saying that the velocity of the proton is along the (arbitrary) z-axis, and the velocities in the x and y axis are zero correct?
And you claim that in the reference frame of the center of mass, the velocities cancel.
But, what changes relative to the earth in the other two directions? The product is, by your own logic, still at rest relative to the earth.
Aside from which, your scenario is unrealistic - the average velocitiy of atoms in the earths atmosphere is 450 m/s.
QUOTE
So, you're considering a Proton approaching another proton with a velocity of 0 m/s relative to the earth.
Yes
QUOTE (->
| QUOTE |
| So, you're considering a Proton approaching another proton with a velocity of 0 m/s relative to the earth. |
Yes
The velocity of the cosmic ray photon is (nominally) radial (WRT the earth)
You're saying that the velocity of the proton is along the (arbitrary) z-axis, and the velocities in the x and y axis are zero correct?
That's a possible scenario sure
QUOTE
And you claim that in the reference frame of the center of mass, the velocities cancel.
Yes
QUOTE (->
| QUOTE |
| And you claim that in the reference frame of the center of mass, the velocities cancel. |
Yes
But, what changes relative to the earth in the other two directions?
Nothing changes - the resulting black hole is at rest in the COM frame.
QUOTE
The product is, by your own logic, still at rest relative to the earth.
How do you make the leap from the MBH being at rest in the COM frame to being at rest relative to earth. Earths frame and the center of mass frame are not the same - I fear you're confusing the two.
In earths reference frame proton A is stationary , proton B travels along the z axis at the speed of light - the velocities do not cancel, the resulting MBH still has significant velocity along the z-axis.
In the COM frame both protons are moving in opposite directions along the z-axis at the same speed so that the resulting MBH is stationary.
QUOTE (->
| QUOTE |
| The product is, by your own logic, still at rest relative to the earth. |
How do you make the leap from the MBH being at rest in the COM frame to being at rest relative to earth. Earths frame and the center of mass frame are not the same - I fear you're confusing the two.
In earths reference frame proton A is stationary , proton B travels along the z axis at the speed of light - the velocities do not cancel, the resulting MBH still has significant velocity along the z-axis.
In the COM frame both protons are moving in opposite directions along the z-axis at the same speed so that the resulting MBH is stationary.
Aside from which, your scenario is unrealistic - the average velocity of atoms in the earths atmosphere is 450 m/s.
I was trying to keep the argument simple, obviously molecules in earths atmosphere are not exactly stationary relative to earth. In any event, giving the "stationary" particle some small amount of velocity does not change the argument, the resulting MBH would still have more than enough velocity to zip through earth without becoming gravitationally bound. A lone proton in the atmosphere is quite random as well, it would probably be an oxygen or nitrogen molecule that the cosmic ray collides with but I wanted to keep the example simple.
The original point was brought up to counter RealityCheck's claim of some "splatter" effect bleeding away the MBH's velocity so that it would get captured by earth. I was trying to demonstrate that this violates the conservation of momentum.
QUOTE (ThePeanut+Feb 11 2008, 10:54 PM)
Yes
That's a possible scenario sure
Yes
Nothing changes - the resulting black hole is at rest in the COM frame.
How do you make the leap from the MBH being at rest in the COM frame to being at rest relative to earth. Earths frame and the center of mass frame are not the same - I fear you're confusing the two.
In earths reference frame proton A is stationary , proton B travels along the z axis at the speed of light - the velocities do not cancel, the resulting MBH still has significant velocity along the z-axis.
In the COM frame both protons are moving in opposite directions along the z-axis at the same speed so that the resulting MBH is stationary.
I was trying to keep the argument simple, obviously molecules in earths atmosphere are not exactly stationary relative to earth. In any event, giving the "stationary" particle some small amount of velocity does not change the argument, the resulting MBH would still have more than enough velocity to zip through earth without becoming gravitationally bound. A lone proton in the atmosphere is quite random as well, it would probably be an oxygen or nitrogen molecule that the cosmic ray collides with but I wanted to keep the example simple.
The original point was brought up to counter RealityCheck's claim of some "splatter" effect bleeding away the MBH's velocity so that it would get captured by earth. I was trying to demonstrate that this violates the conservation of momentum.
You're also ignoring the change in mass due to the collision.
That's a possible scenario sure
Yes
Nothing changes - the resulting black hole is at rest in the COM frame.
How do you make the leap from the MBH being at rest in the COM frame to being at rest relative to earth. Earths frame and the center of mass frame are not the same - I fear you're confusing the two.
In earths reference frame proton A is stationary , proton B travels along the z axis at the speed of light - the velocities do not cancel, the resulting MBH still has significant velocity along the z-axis.
In the COM frame both protons are moving in opposite directions along the z-axis at the same speed so that the resulting MBH is stationary.
I was trying to keep the argument simple, obviously molecules in earths atmosphere are not exactly stationary relative to earth. In any event, giving the "stationary" particle some small amount of velocity does not change the argument, the resulting MBH would still have more than enough velocity to zip through earth without becoming gravitationally bound. A lone proton in the atmosphere is quite random as well, it would probably be an oxygen or nitrogen molecule that the cosmic ray collides with but I wanted to keep the example simple.
The original point was brought up to counter RealityCheck's claim of some "splatter" effect bleeding away the MBH's velocity so that it would get captured by earth. I was trying to demonstrate that this violates the conservation of momentum.
You're also ignoring the change in mass due to the collision.
QUOTE
You're also ignoring the change in mass due to the collision.
No, uba has already explained this to you several times, do you think the MBH is slowed down in earths reference frame due to mass gain?
So that in our setup with A being stationary in earths frame and B the cosmic ray, you think the resulting MBH will slow due to mass gain right? Do you understand that this implies an acceleration in the COM frame in the direction that A was initially traveling? What causes that acceleration?
In the COM frame the MBH product is now traveling along the z-axis in the initial direction proton A was traveling. How can this be so when the two protons were colliding head on along the z-axis at exactly the same speed? Where has the momentum come from when there is no net momentum initially in the COM frame?
Shouldn't the resulting MBH be stationary in the COM frame? If we reverse the setup so earth is in B's reference frame so proton B is now stationary, does the MBH also slow? This implies back in the COM frame that the MBH is now traveling in the initial direction of B. How can we have two results simply by changing our frame of reference? Again where has the momentum come from?
Mind you, looking back at your previous posts it doesn't surprise me - you also think that parton radiation would cause some sort of rocket effect on the MBH causing it to speed up in earths reference frame.
Care to explain how that one works?
Aren't you guys discussing a completely inelastic collision at relativistic speeds? It shouldn't be as hard as you are making it out to be.
Now the website LHCconcerns.com is trying to raise awareness out there, honestly there are quite a few people who are generally concerned and realize that not everything adds up to a safe expirement.
There is even an article on Digg right now that is getting some attention.
http://digg.com/general_sciences/Dangers_o...C_this_May_2008
There is even an article on Digg right now that is getting some attention.
http://digg.com/general_sciences/Dangers_o...C_this_May_2008
Making micro black holes from Gold would increase the cost of gold on the world markets.
QUOTE (barakn+Feb 12 2008, 11:31 AM)
Aren't you guys discussing a completely inelastic collision at relativistic speeds? It shouldn't be as hard as you are making it out to be.
Well, I thought we were, but apparenly p≠mγv anymore, so go figure.
Well, I thought we were, but apparenly p≠mγv anymore, so go figure.
QUOTE (ThePeanut+Feb 12 2008, 11:08 AM)
No, uba has already explained this to you several times, do you think the MBH is slowed down in earths reference frame due to mass gain?
So that in our setup with A being stationary in earths frame and B the cosmic ray, you think the resulting MBH will slow due to mass gain right? Do you understand that this implies an acceleration in the COM frame in the direction that A was initially traveling? What causes that acceleration?
In the COM frame the MBH product is now traveling along the z-axis in the initial direction proton A was traveling. How can this be so when the two protons were colliding head on along the z-axis at exactly the same speed? Where has the momentum come from when there is no net momentum initially in the COM frame?
Shouldn't the resulting MBH be stationary in the COM frame? If we reverse the setup so earth is in B's reference frame so proton B is now stationary, does the MBH also slow? This implies back in the COM frame that the MBH is now traveling in the initial direction of B. How can we have two results simply by changing our frame of reference? Again where has the momentum come from?
Mind you, looking back at your previous posts it doesn't surprise me - you also think that parton radiation would cause some sort of rocket effect on the MBH causing it to speed up in earths reference frame.
Care to explain how that one works?
Because the Center of Mass is not stationary in the Earths Reference Frame.
The momentum of the center of mass is constant, but the total mass before and after the collision is not, therefore the velocity of the center of mass relative to the earth is not constant.
In your scenario, the center of mass has a velocity relative to the earth along the line connecting A, the COM, B, and the center of the earth, before, and after the collision, all I have been saying is that that velocity is reduced, and in some cases may be below the escape velocity of the earth.
It really isn't hard to understand.
I don't know where you've plucked this mythical acceleration from.
And, if you had actually read my posts, you would have seen that what I originally said was that IF Partons removed total energy from the forming black hole, without affecting its momentum, then the total mass of the black hole will be lower, therefore, according to the conservation of momentum, its velocity relative to the earth must increase (as I explained at the time).
It really isn't that hard to understand p=mγv.
If Momentum is constant, then as mass changes, velocity must also change.
So that in our setup with A being stationary in earths frame and B the cosmic ray, you think the resulting MBH will slow due to mass gain right? Do you understand that this implies an acceleration in the COM frame in the direction that A was initially traveling? What causes that acceleration?
In the COM frame the MBH product is now traveling along the z-axis in the initial direction proton A was traveling. How can this be so when the two protons were colliding head on along the z-axis at exactly the same speed? Where has the momentum come from when there is no net momentum initially in the COM frame?
Shouldn't the resulting MBH be stationary in the COM frame? If we reverse the setup so earth is in B's reference frame so proton B is now stationary, does the MBH also slow? This implies back in the COM frame that the MBH is now traveling in the initial direction of B. How can we have two results simply by changing our frame of reference? Again where has the momentum come from?
Mind you, looking back at your previous posts it doesn't surprise me - you also think that parton radiation would cause some sort of rocket effect on the MBH causing it to speed up in earths reference frame.
Care to explain how that one works?
Because the Center of Mass is not stationary in the Earths Reference Frame.
The momentum of the center of mass is constant, but the total mass before and after the collision is not, therefore the velocity of the center of mass relative to the earth is not constant.
In your scenario, the center of mass has a velocity relative to the earth along the line connecting A, the COM, B, and the center of the earth, before, and after the collision, all I have been saying is that that velocity is reduced, and in some cases may be below the escape velocity of the earth.
It really isn't hard to understand.
I don't know where you've plucked this mythical acceleration from.
And, if you had actually read my posts, you would have seen that what I originally said was that IF Partons removed total energy from the forming black hole, without affecting its momentum, then the total mass of the black hole will be lower, therefore, according to the conservation of momentum, its velocity relative to the earth must increase (as I explained at the time).
It really isn't that hard to understand p=mγv.
If Momentum is constant, then as mass changes, velocity must also change.
QUOTE (RealityCheck+Feb 11 2008, 06:48 AM)
The gravity gradient of Earth may be 'flat', but it is CUMULATIVELY much stronger overall that the 'puny' mass gravity strength.....so the strength of Earth's gravity affects the UNDERSIDE 'gradient' of the purtative micro-hole feature.
So? Gravity is curved space-time. It's passive. It's not like it's attacking the MBH.
No, it's in freefall because of the black hole's properties, not the earth's gravity.
No, it's in freefall because of the black hole's properties, not the earth's gravity.
Moreover, when you say...
So? Gravity is curved space-time. It's passive. It's not like it's attacking the MBH.
QUOTE
And it falls in 'free fall' precisely BECAUSE of the Earth's overwhelming gravity 'curvature' strength ob=verall.
No, it's in freefall because of the black hole's properties, not the earth's gravity.
QUOTE (->
| QUOTE |
| And it falls in 'free fall' precisely BECAUSE of the Earth's overwhelming gravity 'curvature' strength ob=verall. |
No, it's in freefall because of the black hole's properties, not the earth's gravity.
Moreover, when you say...
QUOTE (uba+)
Co-joined gravity wells add to the total space-time curvature, not subtract from it.
That is EXACTLY what I am saying....only you seem to think that 'addition' is 'symmetrical'....when it actually is A-symmetrical...with the UPPER side of the putative micro-hole gravity profile adding constructively (stronger) DOWNWARDS INTO the micro-hole TOP SIDE) while the UNDERSIDE adding constructively ALSO DOWNWARDS but AWAY FROM the micro-hole and towards the Earth.
"Up" and "down" are meaningless in freefall.
Gravity isn't dynamic. It's passive. The MBH simply glides through space that happens to be curved by gravity. It's no different to the MBH than gliding through open space, except the trajectory is curved.
Gravity isn't dynamic. It's passive. The MBH simply glides through space that happens to be curved by gravity. It's no different to the MBH than gliding through open space, except the trajectory is curved.
And I again repeat, the totality of disruptor effects from UNCONTAINED DYNAMIC COLLISIONS and QUANTUM PROCESSES AT THAT SCALE and EARTH GRAVITY-DISTORTION mitigate against the formation of such micro-holes in the first place in such a manner.
"Disruptor?" Isn't that the preferred Klingon weapon? ...I digress...
You seem to think gravity is a dynamic, active force. It isn't. It's merely bent space-time.
That is EXACTLY what I am saying....only you seem to think that 'addition' is 'symmetrical'....when it actually is A-symmetrical...with the UPPER side of the putative micro-hole gravity profile adding constructively (stronger) DOWNWARDS INTO the micro-hole TOP SIDE) while the UNDERSIDE adding constructively ALSO DOWNWARDS but AWAY FROM the micro-hole and towards the Earth.
"Up" and "down" are meaningless in freefall.
QUOTE
That is why the hole 'falls' at all. And since the putative micro-hole feature does not fall INSTANTANEOUSLY into the Erath, then the 'edge-to-edge' gravity CURVATURE distortions/redirections between the micro-feature gravity and Earth gravity occur at THE SPEED OF LIGHT...plenty of time to cause havoc for the micro-hole gravitic/symmetric stability.
Gravity isn't dynamic. It's passive. The MBH simply glides through space that happens to be curved by gravity. It's no different to the MBH than gliding through open space, except the trajectory is curved.
QUOTE (->
| QUOTE |
| That is why the hole 'falls' at all. And since the putative micro-hole feature does not fall INSTANTANEOUSLY into the Erath, then the 'edge-to-edge' gravity CURVATURE distortions/redirections between the micro-feature gravity and Earth gravity occur at THE SPEED OF LIGHT...plenty of time to cause havoc for the micro-hole gravitic/symmetric stability. |
Gravity isn't dynamic. It's passive. The MBH simply glides through space that happens to be curved by gravity. It's no different to the MBH than gliding through open space, except the trajectory is curved.
And I again repeat, the totality of disruptor effects from UNCONTAINED DYNAMIC COLLISIONS and QUANTUM PROCESSES AT THAT SCALE and EARTH GRAVITY-DISTORTION mitigate against the formation of such micro-holes in the first place in such a manner.
"Disruptor?" Isn't that the preferred Klingon weapon? ...I digress...
You seem to think gravity is a dynamic, active force. It isn't. It's merely bent space-time.
You know what Ubavontubs?
You really are an Arrogant ####wit you know that?
I'm sorry that the conservation of Momentum is so hard for you to understand.
For a Proton travelling at 0.9999c towards the earth and another proton.
It sees the earth and the other proton travelling towards it at 0.9999c
In the reference frame of the center of mass, the two protons (and the earth) are travelling towards each other at about 0.985c.
Therefore, if there is no change in total mass of the system, then the center of mass of the system must carry on it's merry way at 0.985c through the center of the earth, and out the other side.
If there is a change of mass in the system, then this value must decrease.
At the somewhat sluggish speed of 0.985c, the lorentz factor is a measily 6.
In the earths frame, the lorentz factor on the black hole is 6, and the lorentz factor on the initial proton is 70.
The total energy of the collision provides enough energy (in the center of mass frame) to create a blackhole with a mass of 11.59 protons.
The momentum of the initial proton is 3.55x10^-17 kgm/s
The momentum of the blackhole is 3.32x10^-17 kgm/s
Conservation of momentum is violated in the earths reference frame in your scenario.
Looks like you're wrong...
Where as my velocity figures were derived from the conservation of momentum.
You really are an Arrogant ####wit you know that?
I'm sorry that the conservation of Momentum is so hard for you to understand.
For a Proton travelling at 0.9999c towards the earth and another proton.
It sees the earth and the other proton travelling towards it at 0.9999c
In the reference frame of the center of mass, the two protons (and the earth) are travelling towards each other at about 0.985c.
Therefore, if there is no change in total mass of the system, then the center of mass of the system must carry on it's merry way at 0.985c through the center of the earth, and out the other side.
If there is a change of mass in the system, then this value must decrease.
At the somewhat sluggish speed of 0.985c, the lorentz factor is a measily 6.
In the earths frame, the lorentz factor on the black hole is 6, and the lorentz factor on the initial proton is 70.
The total energy of the collision provides enough energy (in the center of mass frame) to create a blackhole with a mass of 11.59 protons.
The momentum of the initial proton is 3.55x10^-17 kgm/s
The momentum of the blackhole is 3.32x10^-17 kgm/s
Conservation of momentum is violated in the earths reference frame in your scenario.
Looks like you're wrong...
Where as my velocity figures were derived from the conservation of momentum.
QUOTE (Trippy+Feb 12 2008, 10:46 PM)
You know what Ubavontubs?
You really are an Arrogant ####wit you know that?
...said the primitive, cursing that which he could not comprehend.
Ha!
Ha!
For a Proton travelling at 0.9999c towards the earth and another proton.
It sees the earth and the other proton travelling towards it at 0.9999c
In the reference frame of the center of mass, the two protons (and the earth) are travelling towards each other at about 0.985c.
Therefore, if there is no change in total mass of the system, then the center of mass of the system must carry on it's merry way at 0.985c through the center of the earth, and out the other side.
O-o-kay...
No. The momentum of the system would decrease, but the velocity wouldn't.
No. The momentum of the system would decrease, but the velocity wouldn't.
At the somewhat sluggish speed of 0.985c, the lorentz factor is a measily 6.
In the earths frame, the lorentz factor on the black hole is 6, and the lorentz factor on the initial proton is 70.
The total energy of the collision provides enough energy (in the center of mass frame) to create a blackhole with a mass of 11.59 protons.
The momentum of the initial proton is 3.55x10^-17 kgm/s
The momentum of the blackhole is 3.32x10^-17 kgm/s
Conservation of momentum is violated in the earths reference frame in your scenario.
Looks like you're wrong...
Where as my velocity figures were derived from the conservation of momentum.
You obviously don't know what that means.
Again, if the earth was co-moving with the other particle, what would an earthbound observer see? Would the collision result still stop relative to the earth? How could it when you've just stated it would stop relative to the other particle's motion? Which is it?
You really are an Arrogant ####wit you know that?
...said the primitive, cursing that which he could not comprehend.
QUOTE
I'm sorry that the conservation of Momentum is so hard for you to understand.
Ha!
QUOTE (->
| QUOTE |
| I'm sorry that the conservation of Momentum is so hard for you to understand. |
Ha!
For a Proton travelling at 0.9999c towards the earth and another proton.
It sees the earth and the other proton travelling towards it at 0.9999c
In the reference frame of the center of mass, the two protons (and the earth) are travelling towards each other at about 0.985c.
Therefore, if there is no change in total mass of the system, then the center of mass of the system must carry on it's merry way at 0.985c through the center of the earth, and out the other side.
O-o-kay...
QUOTE
If there is a change of mass in the system, then this value must decrease.
No. The momentum of the system would decrease, but the velocity wouldn't.
QUOTE (->
| QUOTE |
| If there is a change of mass in the system, then this value must decrease. |
No. The momentum of the system would decrease, but the velocity wouldn't.
At the somewhat sluggish speed of 0.985c, the lorentz factor is a measily 6.
In the earths frame, the lorentz factor on the black hole is 6, and the lorentz factor on the initial proton is 70.
The total energy of the collision provides enough energy (in the center of mass frame) to create a blackhole with a mass of 11.59 protons.
The momentum of the initial proton is 3.55x10^-17 kgm/s
The momentum of the blackhole is 3.32x10^-17 kgm/s
Conservation of momentum is violated in the earths reference frame in your scenario.
Looks like you're wrong...
Where as my velocity figures were derived from the conservation of momentum.
You obviously don't know what that means.
Again, if the earth was co-moving with the other particle, what would an earthbound observer see? Would the collision result still stop relative to the earth? How could it when you've just stated it would stop relative to the other particle's motion? Which is it?
QUOTE (ubavontuba+Feb 12 2008, 05:14 AM)
So? Gravity is curved space-time. It's passive. It's not like it's attacking the MBH.
No, it's in freefall because of the black hole's properties, not the earth's gravity.
No, it's in freefall because of the black hole's properties, not the earth's gravity.
QUOTE (RC+)
That is EXACTLY what I am saying....only you seem to think that 'addition' is 'symmetrical'....when it actually is A-symmetrical...with the UPPER side of the putative micro-hole gravity profile adding constructively (stronger) DOWNWARDS INTO the micro-hole TOP SIDE) while the UNDERSIDE adding constructively ALSO DOWNWARDS but AWAY FROM the micro-hole and towards the Earth.
"Up" and "down" are meaningless in freefall.
Gravity isn't dynamic. It's passive. The MBH simply glides through space that happens to be curved by gravity. It's no different to the MBH than gliding through open space, except the trajectory is curved.
"Disruptor?" Isn't that the preferred Klingon weapon? ...I digress...
You seem to think gravity is a dynamic, active force. It isn't. It's merely bent space-time.
Hi uba!
Heheh....of course it's not 'attacking' the micro-hole per se. Obviously you STIIL missed where I EXPLICITLY asked you to consider ONLY the earth GRAVITY profile to Micro-hole GRAVITY profile effects...and NOT to the micro-hole per se.
And please don't go cute/obtuse on me now, mate.....re "up"/"down". hehehe
You know perfectly well by the context that I was using "DOWN" in the sense/context of 'direction' TOWARDS EARTH...hence the 'underside' of the micro-hole feature. OK?
And the spacetime 'curvature' AROUND the micro-hole is what is being AFFECTED by the Earth's overwhelming CUMULATIVE STRENGTH 'curvature' along which the micro-hole is made to 'follow' even as it's OWN 'curvature' profile is 'weakened/distorted' by the overall 'curvature' of the Earth towards which the micro-hole is STRONGLY directed BECAUSE ITS own gravity 'curvature' on the micro-hole 'underside' (side facing the Earth) is overwhelmed in STRENGTH/DIRECTION...and so is 'blended in' with the Earths' CURVATURE PROFILE/STRENGTH/DIRECTION STRONGLY AWAY from that 'underside' of the micro-hole and towards the Earth.
Like I told you before, the Earth's gravity does NOT have to 'cancel' any of the micro-hole gravity....merely DISTORT/DESTABILISE it MOMENTARILY such that the EXTREME ENERGY DENSITY putatively compressed intom the micro-feature BY its OWN gravity will BREAK OUT ON ITS OWN.
See?
The normal forces which hold any classical particles together on the surface of the Earth are NOT 'gravity', but 'charge' etc effects.
That is why the usual particles are NOT 'disrupted' by Earth's gravity profile effects on the particle's own, since the gravity profile is NOT what keeps their 'integrity'...simply because Earth's gravity distortion effects is relatively many orders of magnitude WEAKER than the forces holding 'normal' PARTICLES TOGETHER'.....
.....UNLIKE the micro-feature 'particle, which SUPPOSEDLY IS a 'gravity contained' extreme energy density feature prevented from exploding IMMEDIATELY ONLY by its purported gravity...BUT which gravity IS vulnerable to 'gravitic profile' DESTABILISATION and hence micro-hole SELF-DISINTEGRATION FROM that 'gravity-only' CONTAINMENET STATE which supposedly formed the micro-hole in the first place (which I have already explained cannot happen anyway in such a scenario).
EXAMPLE: Think what would happen to the particles/energy on the surface of the MOON facing the Earth IF it was trying to EXP[LODE DUE TO ITS ENERGY DENSITY (like a micro-hole)....and there was ONLY its gravity preventing such explosion? Ask yourself, would the gravuity effects from EARTH that creates the TIDAL EFFECT on the moon's "underside' (facing earth) DISRUPT the moon's gravity-only 'containment' symmetry? Of course it would. And what would happen? Well, at present, the moon is NOT of an energy density/feature which is 'tending to explode itself', so the tides merely distort the crust as the 'bulge' of material raised by the Earth gravity effects (AGAINST THE MOON'S OWN GRAVITY 'strength/profile!)...BUT IF IT WERE like a micro-hole of TINY MASS, and was 'tending to explode' ITSELF excerpt for its gravity, then the Earth's tidal/strong gravity effects would DISTORT it disastrously...and since it was only held together by its gravity, then the 'side' facing the earth would 'bulge' into the Earth's gravity well DIRECTION. and so NOT be 'contained' on that underside as effectively against its tendency to explode ITSELF. See? What I speak of is happening NOW to every particle at the 'edge' of the Earth...but CLASSICAL/NORMAL particles ARE NOT EXTREME DENSITY FEATURES which are prevented from 'exploding' by their 'self-gravity. See? ONLY when the feature WANTS to explode against its own 'gravity compression', will the effects of my observations apply.
Anyhow, please stop thinking about the micro-hole being 'attacked'.
It is the gravity that contains it that is being 'affected/destabilised' as I explained. The micro-hole then DESTROYS ITSELF. See?
So have another go at it, mate! But this time, please DO take some time to think about what I actually SAID and NOT what you THINK I said! hehehe.
Cheers!.....and see ya in a couple of days, uba!
RC.
.
So you think the velocity of the COM is not stationary in Earths reference frame?
Lets go back to our setup with proton A and B with A being stationary relative to earth.
Picture an observer, lets call him "Andy" in the initial COM frame of the two protons and an observer on earth, lets call him "Bob" in proton A's reference frame.
Lets go through this step by step.
1) Andy sees the two protons traveling at the same initial speed collide head on with each other right?
2) Surely a resulting MBH will be stationary in Andy's frame? (Obviously I'm simplifying things and assuming an improbably "clean" collision, even if we assume "messy" collisions which produce MBH's of varying velocities we would still expect the distribution to be centered around zero velocity though right?)
3) Has the COM before the collision changed after the collision for our observer Andy? No, there was no net momentum in this frame before the collision, there is none after the collision.
4) Has the velocity of Andy changed relative to Bob? No, the two are uninvolved in the collision - the collision has not had an effect on their velocites.
So if the COM hasn't changed in Andy's frame and if Andy's velocity hasn't changed in Bob's frame. How exactly does the COM of this collision change relative to Bob (earth)?
So you think the velocity of the COM is not stationary in Earths reference frame?
Lets go back to our setup with proton A and B with A being stationary relative to earth.
Picture an observer, lets call him "Andy" in the initial COM frame of the two protons and an observer on earth, lets call him "Bob" in proton A's reference frame.
Lets go through this step by step.
1) Andy sees the two protons traveling at the same initial speed collide head on with each other right?
2) Surely a resulting MBH will be stationary in Andy's frame? (Obviously I'm simplifying things and assuming an improbably "clean" collision, even if we assume "messy" collisions which produce MBH's of varying velocities we would still expect the distribution to be centered around zero velocity though right?)
3) Has the COM before the collision changed after the collision for our observer Andy? No, there was no net momentum in this frame before the collision, there is none after the collision.
4) Has the velocity of Andy changed relative to Bob? No, the two are uninvolved in the collision - the collision has not had an effect on their velocites.
So if the COM hasn't changed in Andy's frame and if Andy's velocity hasn't changed in Bob's frame. How exactly does the COM of this collision change relative to Bob (earth)?
In your scenario, the center of mass has a velocity relative to the earth along the line connecting A, the COM, B, and the center of the earth, before, and after the collision, all I have been saying is that that velocity is reduced, and in some cases may be below the escape velocity of the earth.
So you admit now it is a line rather than a plane as you previously stated connecting A the COM and B? That's a start I guess..
Apparently it is..
Apparently it is..
I don't know where you've plucked this mythical acceleration from.
An object slowing down is experiencing negative acceleration - an acceleration vector is effectively being applied to it in the opposite direction to its velocity. Do you understand if the MBH slows down in earths reference frame there are other reference frames where the MBH will appear to accelerate. For example in the scenario above, you think the MBH will slow down in Bob's (earths) reference frame. In Andy's reference frame the MBH would then no longer stationary but will start accelerating in some direction as if by magic. How can this be?
What you originally said was that in order for the MBH's speed relative to earth to increase 50-fold (to beyond earth's escape velocity) it would need to radiate away 98% of it's energy.
What you originally said was that in order for the MBH's speed relative to earth to increase 50-fold (to beyond earth's escape velocity) it would need to radiate away 98% of it's energy.
The black hole would need to radiate away (about) 98% of it's energy as Partons and secondary particles in order to achieve escape velocity (making it's rest mas one fiftieth of what I've calculated it).
Do you still think this statement is correct? I see you have added that the velocity of the MBH will only increase in earths reference frame IF the parton radiation doesn't affect total momentum.
Here's a question for you - Do you think that in earths reference frame - the parton radiation from the MBH will have
1) No net momentum
2) Some net momentum
So you think the velocity of the COM is not stationary in Earths reference frame?
Too tired to read your post in it's entirety (although I will anyway, but) (And after reading it, it all comes back to this point anyway)...
If the COM is stationary in the earths reference frame (as you just implied it should be).
And the residual blackhole is stationary in the COM reference frame.
Then how is the residual black hole not stationary in the Earths reference frame?
The velocity decreases relative to what? Earth? In your scenario an observer in the cosmic rays initial reference frame will see the velocity of the MBH increase relative to them. What makes earths reference frame so special?
You are creating an imaginary mass that for some reason always has zero velocity relative to earth.
It's just like your parton distribution example where you think the MBH losing 98% of its mass through secondary particle/parton emission will cause its velocity to increase 50-fold relative to earth. You just think mass decreases, velocity increases (relative to earth) without considering the system as a whole. Why should the radiating partons/secondary particles emitted from the system have zero velocity relative to earth?
2) Surely a resulting MBH will be stationary in Andy's frame?
Not according to the untested 2 TeV Black hole = radius 0.0001 fm hypothesis being pushed by ubavontuba. Not according to the tested 2 TeV Black hole = radius 5.3 × 10^-36 fm theory called GR.
A proton has a RMS charge radius of 0.875 fm.
Since the collision in both predictions involves only 2 quarks, and the quarks are relativistic, the fact that the total momentum is zero tells you nothing about momentum of the black hole.
You stated that you think the MBH radiating away 98% of its mass will cause its velocity to increase 50-fold relative to earth.
In order for this to be true, the parton radiation cannot have any net effect on the momentum of the MBH (relative to earth). i.e. The net momentum (and therefore velocity) of the emitted partons has to be zero relative to earth in your scenario.
You stated that you think the MBH radiating away 98% of its mass will cause its velocity to increase 50-fold relative to earth.
In order for this to be true, the parton radiation cannot have any net effect on the momentum of the MBH (relative to earth). i.e. The net momentum (and therefore velocity) of the emitted partons has to be zero relative to earth in your scenario.
Did you just ignore the part where I specified that for this to be true, the partons had to be emitted in OPPOSING PAIRS - in other words, with equal and opposite momentums (assuming equal mass of course) - thus reducing the total mass of the MBH without affecting its momentum.
This would be the case in the reference frame where the forming MBH is stationary (different to the reference frame where earth is stationary). In this frame the MBH has zero momentum so reducing the mass of the MBH (without affecting the momentum) does not increase velocity (velocity, like momentum remains zero).
"Up" and "down" are meaningless in freefall.
Gravity isn't dynamic. It's passive. The MBH simply glides through space that happens to be curved by gravity. It's no different to the MBH than gliding through open space, except the trajectory is curved.
"Disruptor?" Isn't that the preferred Klingon weapon? ...I digress...
You seem to think gravity is a dynamic, active force. It isn't. It's merely bent space-time.
Hi uba!
Heheh....of course it's not 'attacking' the micro-hole per se. Obviously you STIIL missed where I EXPLICITLY asked you to consider ONLY the earth GRAVITY profile to Micro-hole GRAVITY profile effects...and NOT to the micro-hole per se.
And please don't go cute/obtuse on me now, mate.....re "up"/"down". hehehe
You know perfectly well by the context that I was using "DOWN" in the sense/context of 'direction' TOWARDS EARTH...hence the 'underside' of the micro-hole feature. OK?
And the spacetime 'curvature' AROUND the micro-hole is what is being AFFECTED by the Earth's overwhelming CUMULATIVE STRENGTH 'curvature' along which the micro-hole is made to 'follow' even as it's OWN 'curvature' profile is 'weakened/distorted' by the overall 'curvature' of the Earth towards which the micro-hole is STRONGLY directed BECAUSE ITS own gravity 'curvature' on the micro-hole 'underside' (side facing the Earth) is overwhelmed in STRENGTH/DIRECTION...and so is 'blended in' with the Earths' CURVATURE PROFILE/STRENGTH/DIRECTION STRONGLY AWAY from that 'underside' of the micro-hole and towards the Earth.
Like I told you before, the Earth's gravity does NOT have to 'cancel' any of the micro-hole gravity....merely DISTORT/DESTABILISE it MOMENTARILY such that the EXTREME ENERGY DENSITY putatively compressed intom the micro-feature BY its OWN gravity will BREAK OUT ON ITS OWN.
See?
The normal forces which hold any classical particles together on the surface of the Earth are NOT 'gravity', but 'charge' etc effects.
That is why the usual particles are NOT 'disrupted' by Earth's gravity profile effects on the particle's own, since the gravity profile is NOT what keeps their 'integrity'...simply because Earth's gravity distortion effects is relatively many orders of magnitude WEAKER than the forces holding 'normal' PARTICLES TOGETHER'.....
.....UNLIKE the micro-feature 'particle, which SUPPOSEDLY IS a 'gravity contained' extreme energy density feature prevented from exploding IMMEDIATELY ONLY by its purported gravity...BUT which gravity IS vulnerable to 'gravitic profile' DESTABILISATION and hence micro-hole SELF-DISINTEGRATION FROM that 'gravity-only' CONTAINMENET STATE which supposedly formed the micro-hole in the first place (which I have already explained cannot happen anyway in such a scenario).
EXAMPLE: Think what would happen to the particles/energy on the surface of the MOON facing the Earth IF it was trying to EXP[LODE DUE TO ITS ENERGY DENSITY (like a micro-hole)....and there was ONLY its gravity preventing such explosion? Ask yourself, would the gravuity effects from EARTH that creates the TIDAL EFFECT on the moon's "underside' (facing earth) DISRUPT the moon's gravity-only 'containment' symmetry? Of course it would. And what would happen? Well, at present, the moon is NOT of an energy density/feature which is 'tending to explode itself', so the tides merely distort the crust as the 'bulge' of material raised by the Earth gravity effects (AGAINST THE MOON'S OWN GRAVITY 'strength/profile!)...BUT IF IT WERE like a micro-hole of TINY MASS, and was 'tending to explode' ITSELF excerpt for its gravity, then the Earth's tidal/strong gravity effects would DISTORT it disastrously...and since it was only held together by its gravity, then the 'side' facing the earth would 'bulge' into the Earth's gravity well DIRECTION. and so NOT be 'contained' on that underside as effectively against its tendency to explode ITSELF. See? What I speak of is happening NOW to every particle at the 'edge' of the Earth...but CLASSICAL/NORMAL particles ARE NOT EXTREME DENSITY FEATURES which are prevented from 'exploding' by their 'self-gravity. See? ONLY when the feature WANTS to explode against its own 'gravity compression', will the effects of my observations apply.
Anyhow, please stop thinking about the micro-hole being 'attacked'.
It is the gravity that contains it that is being 'affected/destabilised' as I explained. The micro-hole then DESTROYS ITSELF. See?
So have another go at it, mate! But this time, please DO take some time to think about what I actually SAID and NOT what you THINK I said! hehehe.
Cheers!.....and see ya in a couple of days, uba!
RC.
.
QUOTE (ubavontuba+Feb 13 2008, 01:03 PM)
...said the primitive, cursing that which he could not comprehend.
Ha!
O-o-kay...
No. The momentum of the system would decrease, but the velocity wouldn't.
You obviously don't know what that means.
Again, if the earth was co-moving with the other particle, what would an earthbound observer see? Would the collision result still stop relative to the earth? How could it when you've just stated it would stop relative to the other particle's motion? Which is it?
Idiot.
If you had half a brain, you'd realize that ALL I have done is provide the calculations for thepeanuts post.
You think they're wrong?
Prove it, do them yourself, and show the world where my error is.
If I'm as wrong as you claim I am, and you're as right as you seem to think you are it shouldn't be that hard.
Ha!
O-o-kay...
No. The momentum of the system would decrease, but the velocity wouldn't.
You obviously don't know what that means.
Again, if the earth was co-moving with the other particle, what would an earthbound observer see? Would the collision result still stop relative to the earth? How could it when you've just stated it would stop relative to the other particle's motion? Which is it?
Idiot.
If you had half a brain, you'd realize that ALL I have done is provide the calculations for thepeanuts post.
You think they're wrong?
Prove it, do them yourself, and show the world where my error is.
If I'm as wrong as you claim I am, and you're as right as you seem to think you are it shouldn't be that hard.
QUOTE
Because the Center of Mass is not stationary in the Earths Reference Frame.
The momentum of the center of mass is constant, but the total mass before and after the collision is not, therefore the velocity of the center of mass relative to the earth is not constant.
The momentum of the center of mass is constant, but the total mass before and after the collision is not, therefore the velocity of the center of mass relative to the earth is not constant.
So you think the velocity of the COM is not stationary in Earths reference frame?
Lets go back to our setup with proton A and B with A being stationary relative to earth.
Picture an observer, lets call him "Andy" in the initial COM frame of the two protons and an observer on earth, lets call him "Bob" in proton A's reference frame.
Lets go through this step by step.
1) Andy sees the two protons traveling at the same initial speed collide head on with each other right?
2) Surely a resulting MBH will be stationary in Andy's frame? (Obviously I'm simplifying things and assuming an improbably "clean" collision, even if we assume "messy" collisions which produce MBH's of varying velocities we would still expect the distribution to be centered around zero velocity though right?)
3) Has the COM before the collision changed after the collision for our observer Andy? No, there was no net momentum in this frame before the collision, there is none after the collision.
4) Has the velocity of Andy changed relative to Bob? No, the two are uninvolved in the collision - the collision has not had an effect on their velocites.
So if the COM hasn't changed in Andy's frame and if Andy's velocity hasn't changed in Bob's frame. How exactly does the COM of this collision change relative to Bob (earth)?
QUOTE (->
| QUOTE |
| Because the Center of Mass is not stationary in the Earths Reference Frame. The momentum of the center of mass is constant, but the total mass before and after the collision is not, therefore the velocity of the center of mass relative to the earth is not constant. |
So you think the velocity of the COM is not stationary in Earths reference frame?
Lets go back to our setup with proton A and B with A being stationary relative to earth.
Picture an observer, lets call him "Andy" in the initial COM frame of the two protons and an observer on earth, lets call him "Bob" in proton A's reference frame.
Lets go through this step by step.
1) Andy sees the two protons traveling at the same initial speed collide head on with each other right?
2) Surely a resulting MBH will be stationary in Andy's frame? (Obviously I'm simplifying things and assuming an improbably "clean" collision, even if we assume "messy" collisions which produce MBH's of varying velocities we would still expect the distribution to be centered around zero velocity though right?)
3) Has the COM before the collision changed after the collision for our observer Andy? No, there was no net momentum in this frame before the collision, there is none after the collision.
4) Has the velocity of Andy changed relative to Bob? No, the two are uninvolved in the collision - the collision has not had an effect on their velocites.
So if the COM hasn't changed in Andy's frame and if Andy's velocity hasn't changed in Bob's frame. How exactly does the COM of this collision change relative to Bob (earth)?
In your scenario, the center of mass has a velocity relative to the earth along the line connecting A, the COM, B, and the center of the earth, before, and after the collision, all I have been saying is that that velocity is reduced, and in some cases may be below the escape velocity of the earth.
So you admit now it is a line rather than a plane as you previously stated connecting A the COM and B? That's a start I guess..
QUOTE
It really isn't hard to understand.
Apparently it is..
QUOTE (->
| QUOTE |
| It really isn't hard to understand. |
Apparently it is..
I don't know where you've plucked this mythical acceleration from.
An object slowing down is experiencing negative acceleration - an acceleration vector is effectively being applied to it in the opposite direction to its velocity. Do you understand if the MBH slows down in earths reference frame there are other reference frames where the MBH will appear to accelerate. For example in the scenario above, you think the MBH will slow down in Bob's (earths) reference frame. In Andy's reference frame the MBH would then no longer stationary but will start accelerating in some direction as if by magic. How can this be?
QUOTE
And, if you had actually read my posts, you would have seen that what I originally said was that IF Partons removed total energy from the forming black hole, without affecting its momentum, then the total mass of the black hole will be lower, therefore, according to the conservation of momentum, its velocity relative to the earth must increase (as I explained at the time).
What you originally said was that in order for the MBH's speed relative to earth to increase 50-fold (to beyond earth's escape velocity) it would need to radiate away 98% of it's energy.
QUOTE (->
| QUOTE |
| And, if you had actually read my posts, you would have seen that what I originally said was that IF Partons removed total energy from the forming black hole, without affecting its momentum, then the total mass of the black hole will be lower, therefore, according to the conservation of momentum, its velocity relative to the earth must increase (as I explained at the time). |
What you originally said was that in order for the MBH's speed relative to earth to increase 50-fold (to beyond earth's escape velocity) it would need to radiate away 98% of it's energy.
The black hole would need to radiate away (about) 98% of it's energy as Partons and secondary particles in order to achieve escape velocity (making it's rest mas one fiftieth of what I've calculated it).
Do you still think this statement is correct? I see you have added that the velocity of the MBH will only increase in earths reference frame IF the parton radiation doesn't affect total momentum.
Here's a question for you - Do you think that in earths reference frame - the parton radiation from the MBH will have
1) No net momentum
2) Some net momentum
QUOTE (ThePeanut+Feb 15 2008, 01:28 PM)
So you think the velocity of the COM is not stationary in Earths reference frame?
Too tired to read your post in it's entirety (although I will anyway, but) (And after reading it, it all comes back to this point anyway)...
If the COM is stationary in the earths reference frame (as you just implied it should be).
And the residual blackhole is stationary in the COM reference frame.
Then how is the residual black hole not stationary in the Earths reference frame?
Replace stationary with constant in the first line - my mistake.
I also notice that no-ones critiscized the calculations that say your scenario is in error.
QUOTE (Trippy+Feb 15 2008, 05:12 AM)
I also notice that no-ones critiscized the calculations that say your scenario is in error.
That's because your whole premise is wrong. We're trying to teach you something more fundamental.
In your "calculations" you state: "If there is a change of mass in the system, then this (velocity) value must decrease."
This si so-o-o-o wrong on a basic level. This is why ThePeanut asked you if you think it's some sort of rocket effect. It doesn't work like that. You keep confusing velocity and momentum. They aren't the same thing.
So by starting with an erroneous presumption, you end up with a meaningless calculation. No matter how neat and tidy the numbers are, they're wrong!
That's because your whole premise is wrong. We're trying to teach you something more fundamental.
In your "calculations" you state: "If there is a change of mass in the system, then this (velocity) value must decrease."
This si so-o-o-o wrong on a basic level. This is why ThePeanut asked you if you think it's some sort of rocket effect. It doesn't work like that. You keep confusing velocity and momentum. They aren't the same thing.
So by starting with an erroneous presumption, you end up with a meaningless calculation. No matter how neat and tidy the numbers are, they're wrong!
QUOTE (ubavontuba+Feb 15 2008, 07:33 PM)
That's because your whole premise is wrong. We're trying to teach you something more fundamental.
In your "calculations" you state: "If there is a change of mass in the system, then this (velocity) value must decrease."
This si so-o-o-o wrong on a basic level. This is why ThePeanut asked you if you think it's some sort of rocket effect. It doesn't work like that. You keep confusing velocity and momentum. They aren't the same thing.
So by starting with an erroneous presumption, you end up with a meaningless calculation. No matter how neat and tidy the numbers are, they're wrong!
No. The laws of physics say that if the mass increases then the velocity must decrease.
And if you actually bothered to read through the calculations instead of closing your eyes (or whatever it is that you do) you would see that, as I have already stated, those calculations were SOLELY for the peanuts scenario.
Not mine.
So the only error you're able to pick is actually your own.
You really are a friggin idiot.
In your "calculations" you state: "If there is a change of mass in the system, then this (velocity) value must decrease."
This si so-o-o-o wrong on a basic level. This is why ThePeanut asked you if you think it's some sort of rocket effect. It doesn't work like that. You keep confusing velocity and momentum. They aren't the same thing.
So by starting with an erroneous presumption, you end up with a meaningless calculation. No matter how neat and tidy the numbers are, they're wrong!
No. The laws of physics say that if the mass increases then the velocity must decrease.
And if you actually bothered to read through the calculations instead of closing your eyes (or whatever it is that you do) you would see that, as I have already stated, those calculations were SOLELY for the peanuts scenario.
Not mine.
So the only error you're able to pick is actually your own.
You really are a friggin idiot.
QUOTE (Trippy+Feb 15 2008, 07:31 AM)
No. The laws of physics say that if the mass increases then the velocity must decrease.
And if you actually bothered to read through the calculations instead of closing your eyes (or whatever it is that you do) you would see that, as I have already stated, those calculations were SOLELY for the peanuts scenario.
Not mine.
So the only error you're able to pick is actually your own.
You really are a friggin idiot.
It's p=mv, not v=mv. Decreasing the mass changes the momentum, not necessarily the velocity. The mass and the velocity aren't dependent on each other.
And if you actually bothered to read through the calculations instead of closing your eyes (or whatever it is that you do) you would see that, as I have already stated, those calculations were SOLELY for the peanuts scenario.
Not mine.
So the only error you're able to pick is actually your own.
You really are a friggin idiot.
It's p=mv, not v=mv. Decreasing the mass changes the momentum, not necessarily the velocity. The mass and the velocity aren't dependent on each other.
QUOTE (ubavontuba+Feb 17 2008, 07:50 AM)
It's p=mv, not v=mv. Decreasing the mass changes the momentum, not necessarily the velocity. The mass and the velocity aren't dependent on each other.
If p=mv
THEN
v=p/m
if p is constant and m increases, what do you think v is going to do?
Only YOU would try and twist what I said into that.
If p=mv
THEN
v=p/m
if p is constant and m increases, what do you think v is going to do?
Only YOU would try and twist what I said into that.
QUOTE (Trippy+Feb 16 2008, 10:00 PM)
If p=mv
THEN
v=p/m
if p is constant and m increases, what do you think v is going to do?
Only YOU would try and twist what I said into that.
What makes you think p is constant? It's only conserved in isolated systems. Throwing off mass that's no longer included in the system, breaks isolation. Adding mass from an external source also breaks isolation.
Besides, I've already explained the mass increase you keep referring to is nonexistent. Where do you think the extra mass comes from? Does it appear out of nowhere, as if by magic?
THEN
v=p/m
if p is constant and m increases, what do you think v is going to do?
Only YOU would try and twist what I said into that.
What makes you think p is constant? It's only conserved in isolated systems. Throwing off mass that's no longer included in the system, breaks isolation. Adding mass from an external source also breaks isolation.
Besides, I've already explained the mass increase you keep referring to is nonexistent. Where do you think the extra mass comes from? Does it appear out of nowhere, as if by magic?
QUOTE
No. The laws of physics say that if the mass increases then the velocity must decrease.
The velocity decreases relative to what? Earth? In your scenario an observer in the cosmic rays initial reference frame will see the velocity of the MBH increase relative to them. What makes earths reference frame so special?
You are creating an imaginary mass that for some reason always has zero velocity relative to earth.
It's just like your parton distribution example where you think the MBH losing 98% of its mass through secondary particle/parton emission will cause its velocity to increase 50-fold relative to earth. You just think mass decreases, velocity increases (relative to earth) without considering the system as a whole. Why should the radiating partons/secondary particles emitted from the system have zero velocity relative to earth?
QUOTE (ubavontuba+Feb 17 2008, 11:30 PM)
Where do you think the extra mass comes from? Does it appear out of nowhere, as if by magic?
Forgotten mass-energy equivalence, haven't you. Did you learn nothing from Einstein?
Forgotten mass-energy equivalence, haven't you. Did you learn nothing from Einstein?
QUOTE (ThePeanut+Feb 15 2008, 12:28 AM)
2) Surely a resulting MBH will be stationary in Andy's frame?
Not according to the untested 2 TeV Black hole = radius 0.0001 fm hypothesis being pushed by ubavontuba. Not according to the tested 2 TeV Black hole = radius 5.3 × 10^-36 fm theory called GR.
A proton has a RMS charge radius of 0.875 fm.
Since the collision in both predictions involves only 2 quarks, and the quarks are relativistic, the fact that the total momentum is zero tells you nothing about momentum of the black hole.
QUOTE (ThePeanut+Feb 18 2008, 03:04 PM)
It's just like your parton distribution example where you think the MBH losing 98% of its mass through secondary particle/parton emission will cause its velocity to increase 50-fold relative to earth. You just think mass decreases, velocity increases (relative to earth) without considering the system as a whole. Why should the radiating partons/secondary particles emitted from the system have zero velocity relative to earth?
Did you just ignore the part where I specified that for this to be true, the partons had to be emitted in OPPOSING PAIRS - in other words, with equal and opposite momentums (assuming equal mass of course) - thus reducing the total mass of the MBH without affecting its momentum.
NOWHERE did I state that the partons had zero velocity relative to the earth.
Please try and get your facts a little bit straight please.
Did you just ignore the part where I specified that for this to be true, the partons had to be emitted in OPPOSING PAIRS - in other words, with equal and opposite momentums (assuming equal mass of course) - thus reducing the total mass of the MBH without affecting its momentum.
NOWHERE did I state that the partons had zero velocity relative to the earth.
Please try and get your facts a little bit straight please.
QUOTE (ubavontuba+Feb 18 2008, 12:30 PM)
What makes you think p is constant? It's only conserved in isolated systems. Throwing off mass that's no longer included in the system, breaks isolation. Adding mass from an external source also breaks isolation.
Besides, I've already explained the mass increase you keep referring to is nonexistent. Where do you think the extra mass comes from? Does it appear out of nowhere, as if by magic?
1) Your entire argument against the cosmic ray defense is based around the conservation of momentum.
2) In a multi-bodi collision where there are a different number of bodies before and after the collision, momentum can be shown to be conserved by summing the momentum vectors before and after the collision (mass shed and mass gained are included as part of the system).
3) I (and others) have already explained to you that the mass of the black-hole after the collision is dependent on the total energy of the bodies involved in the collision, before the collision (this is the whole point of the scenario - simply put, kinetic energy is converted into mass).
Besides, I've already explained the mass increase you keep referring to is nonexistent. Where do you think the extra mass comes from? Does it appear out of nowhere, as if by magic?
1) Your entire argument against the cosmic ray defense is based around the conservation of momentum.
2) In a multi-bodi collision where there are a different number of bodies before and after the collision, momentum can be shown to be conserved by summing the momentum vectors before and after the collision (mass shed and mass gained are included as part of the system).
3) I (and others) have already explained to you that the mass of the black-hole after the collision is dependent on the total energy of the bodies involved in the collision, before the collision (this is the whole point of the scenario - simply put, kinetic energy is converted into mass).
QUOTE (rpenner+Feb 18 2008, 03:48 PM)
Since the collision in both predictions involves only 2 quarks, and the quarks are relativistic, the fact that the total momentum is zero tells you nothing about momentum of the black hole.
I had often wondered about this - whether or not, seeing as how the alleged MBH is on the scale of quarks, if the motion of the quarks could impart/remove momentum from the black-hole.
If so (and if I'm interpreting things correctly) then a series of identical collisions, with identical cross sections and identical energies could produce a series of black holes with a thermal distribution of velocities relative to your object of choice correct?
I had often wondered about this - whether or not, seeing as how the alleged MBH is on the scale of quarks, if the motion of the quarks could impart/remove momentum from the black-hole.
If so (and if I'm interpreting things correctly) then a series of identical collisions, with identical cross sections and identical energies could produce a series of black holes with a thermal distribution of velocities relative to your object of choice correct?
Actually, I expect the distribution to be significantly different than thermal (since thermal implies an equilibrium of many particles over time), but it's model dependent and you need to make semi-classical assumptions based on the parton momentum distribution functions to get answers.
But all of the models (other than classical GR) require the black hole to evaporate to end-point. ubavontuba needs a self-consistent model which allows for the existence of quarks, protons and black holes before any calculations can be made and this requires quantum physics and we think this requires black holes to evaporate. If ubavontuba were to come up with a model about how it could be dangerous, then I think people, even me, would listen to him. But all we have are disconnected assumptions and no tight chain of logic connecting them. ubavontuba is reduced to nitpicking at the posts of others rather than putting forward his ideas in a scientific paper.
But all of the models (other than classical GR) require the black hole to evaporate to end-point. ubavontuba needs a self-consistent model which allows for the existence of quarks, protons and black holes before any calculations can be made and this requires quantum physics and we think this requires black holes to evaporate. If ubavontuba were to come up with a model about how it could be dangerous, then I think people, even me, would listen to him. But all we have are disconnected assumptions and no tight chain of logic connecting them. ubavontuba is reduced to nitpicking at the posts of others rather than putting forward his ideas in a scientific paper.
QUOTE (barakn+Feb 18 2008, 02:32 AM)
Forgotten mass-energy equivalence, haven't you. Did you learn nothing from Einstein?
Are you a chatbot? Do you know what a rhetorical question is?
Are you a chatbot? Do you know what a rhetorical question is?
QUOTE (rpenner+Feb 18 2008, 07:26 PM)
Actually, I expect the distribution to be significantly different than thermal (since thermal implies an equilibrium of many particles over time), but it's model dependent and you need to make semi-classical assumptions based on the parton momentum distribution functions to get answers.
But all of the models (other than classical GR) require the black hole to evaporate to end-point. ubavontuba needs a self-consistent model which allows for the existence of quarks, protons and black holes before any calculations can be made and this requires quantum physics and we think this requires black holes to evaporate. If ubavontuba were to come up with a model about how it could be dangerous, then I think people, even me, would listen to him. But all we have are disconnected assumptions and no tight chain of logic connecting them. ubavontuba is reduced to nitpicking at the posts of others rather than putting forward his ideas in a scientific paper.
What do you feel is missing?
What if Hawking forgot to account for a significant source of energy in his calculations regarding black hole evaporation? Would that pique your interest?
But all of the models (other than classical GR) require the black hole to evaporate to end-point. ubavontuba needs a self-consistent model which allows for the existence of quarks, protons and black holes before any calculations can be made and this requires quantum physics and we think this requires black holes to evaporate. If ubavontuba were to come up with a model about how it could be dangerous, then I think people, even me, would listen to him. But all we have are disconnected assumptions and no tight chain of logic connecting them. ubavontuba is reduced to nitpicking at the posts of others rather than putting forward his ideas in a scientific paper.
What do you feel is missing?
What if Hawking forgot to account for a significant source of energy in his calculations regarding black hole evaporation? Would that pique your interest?
QUOTE
NOWHERE did I state that the partons had zero velocity relative to the earth.
You stated that you think the MBH radiating away 98% of its mass will cause its velocity to increase 50-fold relative to earth.
In order for this to be true, the parton radiation cannot have any net effect on the momentum of the MBH (relative to earth). i.e. The net momentum (and therefore velocity) of the emitted partons has to be zero relative to earth in your scenario.
QUOTE (->
| QUOTE |
| NOWHERE did I state that the partons had zero velocity relative to the earth. |
You stated that you think the MBH radiating away 98% of its mass will cause its velocity to increase 50-fold relative to earth.
In order for this to be true, the parton radiation cannot have any net effect on the momentum of the MBH (relative to earth). i.e. The net momentum (and therefore velocity) of the emitted partons has to be zero relative to earth in your scenario.
Did you just ignore the part where I specified that for this to be true, the partons had to be emitted in OPPOSING PAIRS - in other words, with equal and opposite momentums (assuming equal mass of course) - thus reducing the total mass of the MBH without affecting its momentum.
This would be the case in the reference frame where the forming MBH is stationary (different to the reference frame where earth is stationary). In this frame the MBH has zero momentum so reducing the mass of the MBH (without affecting the momentum) does not increase velocity (velocity, like momentum remains zero).
QUOTE (rpenner+Feb 18 2008, 02:48 AM)
Not according to the untested 2 TeV Black hole = radius 0.0001 fm hypothesis being pushed by ubavontuba. Not according to the tested 2 TeV Black hole = radius 5.3 × 10^-36 fm theory called GR.
First, you took ThePeanut's comment out of context.
Second, I agree with him that in Andy's frame (as described by ThePeanut) "we would still expect the (relative velocity) distribution to be centered around zero."
It tells you the momentums of the thousands of black holes expected to be created will be centererd around zero (relative to the earth).
First, you took ThePeanut's comment out of context.
Second, I agree with him that in Andy's frame (as described by ThePeanut) "we would still expect the (relative velocity) distribution to be centered around zero."
QUOTE
A proton has a RMS charge radius of 0.875 fm.
Since the collision in both predictions involves only 2 quarks, and the quarks are relativistic, the fact that the total momentum is zero tells you nothing about momentum of the black hole.
Since the collision in both predictions involves only 2 quarks, and the quarks are relativistic, the fact that the total momentum is zero tells you nothing about momentum of the black hole.
It tells you the momentums of the thousands of black holes expected to be created will be centererd around zero (relative to the earth).
QUOTE (Trippy+Feb 18 2008, 04:07 AM)
Did you just ignore the part where I specified that for this to be true, the partons had to be emitted in OPPOSING PAIRS - in other words, with equal and opposite momentums (assuming equal mass of course) - thus reducing the total mass of the MBH without affecting its momentum.
NOWHERE did I state that the partons had zero velocity relative to the earth.
Please try and get your facts a little bit straight please.
Wow! This is so-o-o-o wrong! Just what do you think happens to the momentum of the ejected mass/energy?
Ejecting mass/energy in opposing proportions reduces the momenum, not the velocity. The ejected mass/energy takes its share of the momentum away with it.
NOWHERE did I state that the partons had zero velocity relative to the earth.
Please try and get your facts a little bit straight please.
Wow! This is so-o-o-o wrong! Just what do you think happens to the momentum of the ejected mass/energy?
Ejecting mass/energy in opposing proportions reduces the momenum, not the velocity. The ejected mass/energy takes its share of the momentum away with it.
QUOTE (Trippy+Feb 18 2008, 05:34 PM)
1) Your entire argument against the cosmic ray defense is based around the conservation of momentum.
That's correct, but you have to understand the conservation of momentum to understand the argument.
True also (are you getting this from Wikipedia?)
True also (are you getting this from Wikipedia?)
3) I (and others) have already explained to you that the mass of the black-hole after the collision is dependent on the total energy of the bodies involved in the collision, before the collision (this is the whole point of the scenario - simply put, kinetic energy is converted into mass).
That's not exactly true. Mass and energy are equivalent. The mass (in the form of kinetic energy) is always there. It doesn't just appear. It's also there in the form of relativistic mass.
That's correct, but you have to understand the conservation of momentum to understand the argument.
QUOTE
2) In a multi-bodi collision where there are a different number of bodies before and after the collision, momentum can be shown to be conserved by summing the momentum vectors before and after the collision (mass shed and mass gained are included as part of the system).
True also (are you getting this from Wikipedia?)
QUOTE (->
| QUOTE |
| 2) In a multi-bodi collision where there are a different number of bodies before and after the collision, momentum can be shown to be conserved by summing the momentum vectors before and after the collision (mass shed and mass gained are included as part of the system). |
True also (are you getting this from Wikipedia?)
3) I (and others) have already explained to you that the mass of the black-hole after the collision is dependent on the total energy of the bodies involved in the collision, before the collision (this is the whole point of the scenario - simply put, kinetic energy is converted into mass).
That's not exactly true. Mass and energy are equivalent. The mass (in the form of kinetic energy) is always there. It doesn't just appear. It's also there in the form of relativistic mass.
QUOTE (ThePeanut+Feb 19 2008, 05:27 PM)
You stated that you think the MBH radiating away 98% of its mass will cause its velocity to increase 50-fold relative to earth.
I also stated that this was ONLY TRUE if the particles were emitted in OPPOSING PAIRS that had ZERO NET effect on the momentum of the alleged MBH. I did NOT however state that the partons had zero velocity relative to the earth.
I also stated that this was ONLY TRUE if the particles were emitted in OPPOSING PAIRS that had ZERO NET effect on the momentum of the alleged MBH. I did NOT however state that the partons had zero velocity relative to the earth.
QUOTE (ThePeanut+Feb 19 2008, 05:27 PM)
In order for this to be true, the parton radiation cannot have any net effect on the momentum of the MBH (relative to earth). i.e. The net momentum (and therefore velocity) of the emitted partons has to be zero relative to earth in your scenario.
No, it doesn't. If I sit in an office Chair (mass 120kgs) and am stationary (net momentum zero) and I manage to emit two 30 kg masses, one traveling in the +x direction at 10 m/s and one travelling in the -x direction at 10 m/s, what do you think the change in my momentum is going to be? Here's a clue. It's Zero. Don't believe me? Do the math for yourself, all the information you need is there. (Assume I'm holding to 30kg sacks of coal while in the office chair).
No, it doesn't. If I sit in an office Chair (mass 120kgs) and am stationary (net momentum zero) and I manage to emit two 30 kg masses, one traveling in the +x direction at 10 m/s and one travelling in the -x direction at 10 m/s, what do you think the change in my momentum is going to be? Here's a clue. It's Zero. Don't believe me? Do the math for yourself, all the information you need is there. (Assume I'm holding to 30kg sacks of coal while in the office chair).
QUOTE (ThePeanut+Feb 19 2008, 05:27 PM)
This would be the case in the reference frame where the forming MBH is stationary (different to the reference frame where earth is stationary). In this frame the MBH has zero momentum so reducing the mass of the MBH (without affecting the momentum) does not increase velocity (velocity, like momentum remains zero).
No, actually, it would be true in a moving frame as well.
No, actually, it would be true in a moving frame as well.
QUOTE (rpenner+Feb 18 2008, 07:26 PM)
Actually, I expect the distribution to be significantly different than thermal (since thermal implies an equilibrium of many particles over time), but it's model dependent and you need to make semi-classical assumptions based on the parton momentum distribution functions to get answers.
But all of the models (other than classical GR) require the black hole to evaporate to end-point. ubavontuba needs a self-consistent model which allows for the existence of quarks, protons and black holes before any calculations can be made and this requires quantum physics and we think this requires black holes to evaporate. If ubavontuba were to come up with a model about how it could be dangerous, then I think people, even me, would listen to him. But all we have are disconnected assumptions and no tight chain of logic connecting them. ubavontuba is reduced to nitpicking at the posts of others rather than putting forward his ideas in a scientific paper.
Hey rpenner,
Since you're getting all chummy with Trippy now, perhaps you might be willing to tell us if his collision model (where the cosmic ray induced MBH would always be at rest with the earth) is correct?
You've ducked this question before... would you please take a stand, now?
But all of the models (other than classical GR) require the black hole to evaporate to end-point. ubavontuba needs a self-consistent model which allows for the existence of quarks, protons and black holes before any calculations can be made and this requires quantum physics and we think this requires black holes to evaporate. If ubavontuba were to come up with a model about how it could be dangerous, then I think people, even me, would listen to him. But all we have are disconnected assumptions and no tight chain of logic connecting them. ubavontuba is reduced to nitpicking at the posts of others rather than putting forward his ideas in a scientific paper.
Hey rpenner,
Since you're getting all chummy with Trippy now, perhaps you might be willing to tell us if his collision model (where the cosmic ray induced MBH would always be at rest with the earth) is correct?
You've ducked this question before... would you please take a stand, now?
QUOTE (ubavontuba+Feb 19 2008, 06:30 PM)
That's correct, but you have to understand the conservation of momentum to understand the argument.
I do. Apparently better then you.
I do. Apparently better then you.
QUOTE (ubavontuba+Feb 19 2008, 06:30 PM)
True also (are you getting this from Wikipedia?)
No. Unlike you, I don't need to rely on wikipedia for my physics knowledge. I've actually studied physics at University level (and passed it with a B odd average). I also still possess my Physics textbooks, one of which I have alluded to repeatedly (not that I actually need to look up this highschool physics you're referring to).
In order for your statement to be true, the emitted partons would need to take away zero net momentum in earths reference frame. Do you know the consequences of this? It means the emitted partons must have zero velocity/momentum in earths frame.
Once again - in order for your 50-fold increase in velocity relative to earth to occur, the emitted partons would need to have zero net velocity relative to earth. The fact you don't understand this says a lot about your (lack of) physics knowledge.
In order for your statement to be true, the emitted partons would need to take away zero net momentum in earths reference frame. Do you know the consequences of this? It means the emitted partons must have zero velocity/momentum in earths frame.
Once again - in order for your 50-fold increase in velocity relative to earth to occur, the emitted partons would need to have zero net velocity relative to earth. The fact you don't understand this says a lot about your (lack of) physics knowledge.
No, it doesn't. If I sit in an office Chair (mass 120kgs) and am stationary (net momentum zero) and I manage to emit two 30 kg masses, one traveling in the +x direction at 10 m/s and one traveling in the -x direction at 10 m/s, what do you think the change in my momentum is going to be? Here's a clue. It's Zero. Don't believe me? Do the math for yourself, all the information you need is there. (Assume I'm holding to 30kg sacks of coal while in the office chair).
Gee you think?
How precisely does your example contradict the statement you quoted?
Your example is analogous to a reference frame where the MBH is stationary.
However in a frame where the MBH is moving the emitted partons have net momentum.
HINT: Consider a second observer moving at a speed of 10m/s in the (+) x direction relative to you in the office chair. It sees one stationary mass emitted and the other emitted mass moving at 20m/s in the -x direction. Does this observer see the masses emitted with a net momentum of zero? Here's a clue. No.
Oh dear
Oh dear, zero. Your observer would see me moving at 10 m/s in the -x direction before and after the collision.
Exactly, you have lost mass without affecting momentum (in your reference frame) however your velocity relative to me has remained unchanged. So why does the MBH's velocity increase relative to earth 50-fold when it has emitted 98% of its mass in parton radiation?
You admitted in your chair example "emitting" the bricks in the way you did leaves your momentum in this frame unchanged (zero before and after the emitting event). I assume you also believe it wouldn't change your velocity in this frame (again zero before and after the event).
Surely you can see that it also leaves your velocity unchanged relative to some observer? Certainly the emitting of the partons won't affect the observes velocity and we have agreed that the event hasn't changed our velocity.
This is analogous to the MBH scenario where you sitting on the chair is the MBH and the "observer" is earth. The MBH simply won't speed up relative to earth as you claimed.
p=mv
In earths frame the momentum and mass of the black hole are decreased proportionally, velocity is unchanged.
In the frame where the MBH is stationary (the chair example), momentum and velocity are both unchanged (both zero before and after "emission") and mass is decreased.
Why not just admit you are wrong with your comment about the parton emission increasing the MBH's velocity relative to earth?
If you have indeed done physics papers as you claim maybe you should brush up on the conservation of momentum.
You admitted in your chair example "emitting" the bricks in the way you did leaves your momentum in this frame unchanged (zero before and after the emitting event). I assume you also believe it wouldn't change your velocity in this frame (again zero before and after the event).
Surely you can see that it also leaves your velocity unchanged relative to some observer? Certainly the emitting of the partons won't affect the observes velocity and we have agreed that the event hasn't changed our velocity.
This is analogous to the MBH scenario where you sitting on the chair is the MBH and the "observer" is earth. The MBH simply won't speed up relative to earth as you claimed.
p=mv
In earths frame the momentum and mass of the black hole are decreased proportionally, velocity is unchanged.
In the frame where the MBH is stationary (the chair example), momentum and velocity are both unchanged (both zero before and after "emission") and mass is decreased.
Why not just admit you are wrong with your comment about the parton emission increasing the MBH's velocity relative to earth?
If you have indeed done physics papers as you claim maybe you should brush up on the conservation of momentum.
Oh stuff off and read what I actually said you friggin idiot.
Did I, or did I not in the VERY POST YOU FRIGGING QUOTED admit that I had made in error.
Moron.
It's not me that needs to brush up on my conservation of momentum.
I at least have attempted the calculations.
Get over yourself.
First of all calm down.
First of all calm down.
Did I, or did I not in the VERY POST YOU FRIGGING QUOTED admit that I had made in error.
Oh you were referring to the whole parton emission argument then? Cool.
I note that you say that "it has absolutely no bearing on anything else I have said." It in fact does have a lot of bearing on other things you have said (HINT: think about where you went wrong in this argument and you should be able to see why an 1150 TeV cosmic ray hitting a relatively stationary molecule in earths atmosphere won't produce MBH's with a residual velocity (or distributed around a mean velocity) of 239 m/s relative to earth - a figure you calculated on this page.
Lets see...
1) You point out an "obvious error" in my initial post because I was only considering the collision in one dimension, asking me to consider the plane between the two protons and the center of mass of the system.
You were incorrect (there is no plane, only a line - the relative velocities of the protons and the COM along this line are the only important factors in my example).
2) You claim that the COM of the two-proton isolated system moves in earths reference frame.
Lets see...
1) You point out an "obvious error" in my initial post because I was only considering the collision in one dimension, asking me to consider the plane between the two protons and the center of mass of the system.
You were incorrect (there is no plane, only a line - the relative velocities of the protons and the COM along this line are the only important factors in my example).
2) You claim that the COM of the two-proton isolated system moves in earths reference frame.
Because the Center of Mass is not stationary in the Earths Reference Frame.
The momentum of the center of mass is constant, but the total mass before and after the collision is not, therefore the velocity of the center of mass relative to the earth is not constant.
This blatantly violates the conservation of momentum law. In an isolated system (no net external force acting on it), the COM moves with uniform velocity (relative to an inertial observer) along a straight line. Would you like me to derive the particular equation for you?
3) You argued that the velocity of the MBH would increase relative to earth due to parton radiation, which you now admit was incorrect.
Thats three statements, just from the last few pages of this thread that suggest your knowledge of some basic physics principles is less than sound.
Would you like me to go back a few more pages to some of your discussions with ubavontuba? There are numerous other examples where your lack of physics knowledge shows.
No. Unlike you, I don't need to rely on wikipedia for my physics knowledge. I've actually studied physics at University level (and passed it with a B odd average). I also still possess my Physics textbooks, one of which I have alluded to repeatedly (not that I actually need to look up this highschool physics you're referring to).
QUOTE (ubavontuba+Feb 19 2008, 06:30 PM)
That's not exactly true. Mass and energy are equivalent. The mass (in the form of kinetic energy) is always there. It doesn't just appear. It's also there in the form of relativistic mass.
Actually, it's exactly true, and you might understand this if you had ever actually bothered to try and do anything other then flap your arms (like you know, calculations).
Actually, it's exactly true, and you might understand this if you had ever actually bothered to try and do anything other then flap your arms (like you know, calculations).
QUOTE (ubavontuba+Feb 19 2008, 06:39 PM)
Hey rpenner,
Since you're getting all chummy with Trippy now, perhaps you might be willing to tell us if his collision model (where the cosmic ray induced MBH would always be at rest with the earth) is correct?
You've ducked this question before... would you please take a stand, now?
Get over yerself bub.
Why should he?
1) I haven't asked him for the affirmation.
2) Rpenner has essentialy said (already) that I am correct in principle.
3) I haven't provided any substantial workings (I would not expect anyone with a reputation in physics to support an answer that contained little/incomplete workings, and no substantive justifications of assumptions).
4) I have stated (in my first post addressing this issue, or it might have been the second) that I was uncertain of the exact value of my result, because I feared I might have botched a step in re-arranging an equation.
I'm sure there are other reasons why your request is demonstrably unreasonable, but...
Since you're getting all chummy with Trippy now, perhaps you might be willing to tell us if his collision model (where the cosmic ray induced MBH would always be at rest with the earth) is correct?
You've ducked this question before... would you please take a stand, now?
Get over yerself bub.
Why should he?
1) I haven't asked him for the affirmation.
2) Rpenner has essentialy said (already) that I am correct in principle.
3) I haven't provided any substantial workings (I would not expect anyone with a reputation in physics to support an answer that contained little/incomplete workings, and no substantive justifications of assumptions).
4) I have stated (in my first post addressing this issue, or it might have been the second) that I was uncertain of the exact value of my result, because I feared I might have botched a step in re-arranging an equation.
I'm sure there are other reasons why your request is demonstrably unreasonable, but...
QUOTE (Trippy+Feb 19 2008, 05:34 AM)
No, it doesn't. If I sit in an office Chair (mass 120kgs) and am stationary (net momentum zero) and I manage to emit two 30 kg masses, one traveling in the +x direction at 10 m/s and one travelling in the -x direction at 10 m/s, what do you think the change in my momentum is going to be? Here's a clue. It's Zero. Don't believe me? Do the math for yourself, all the information you need is there. (Assume I'm holding to 30kg sacks of coal while in the office chair).
But what would happen to your momentum if another fellow in another office chair (also holding 30 sacks of coal) collided with you?
But what would happen to your momentum if another fellow in another office chair (also holding 30 sacks of coal) collided with you?
QUOTE (Trippy+Feb 19 2008, 05:52 AM)
Get over yerself bub.
Why should he?
1) I haven't asked him for the affirmation.
2) Rpenner has essentialy said (already) that I am correct in principle.
3) I haven't provided any substantial workings (I would not expect anyone with a reputation in physics to support an answer that contained little/incomplete workings, and no substantive justifications of assumptions).
4) I have stated (in my first post addressing this issue, or it might have been the second) that I was uncertain of the exact value of my result, because I feared I might have botched a step in re-arranging an equation.
I'm sure there are other reasons why your request is demonstrably unreasonable, but...
Afraid of the truth, are we? Why don't you just let rpenner step up to the plate now?
Why should he?
1) I haven't asked him for the affirmation.
2) Rpenner has essentialy said (already) that I am correct in principle.
3) I haven't provided any substantial workings (I would not expect anyone with a reputation in physics to support an answer that contained little/incomplete workings, and no substantive justifications of assumptions).
4) I have stated (in my first post addressing this issue, or it might have been the second) that I was uncertain of the exact value of my result, because I feared I might have botched a step in re-arranging an equation.
I'm sure there are other reasons why your request is demonstrably unreasonable, but...
Afraid of the truth, are we? Why don't you just let rpenner step up to the plate now?
QUOTE
I also stated that this was ONLY TRUE if the particles were emitted in OPPOSING PAIRS that had ZERO NET effect on the momentum of the alleged MBH. I did NOT however state that the partons had zero velocity relative to the earth.
In order for your statement to be true, the emitted partons would need to take away zero net momentum in earths reference frame. Do you know the consequences of this? It means the emitted partons must have zero velocity/momentum in earths frame.
Once again - in order for your 50-fold increase in velocity relative to earth to occur, the emitted partons would need to have zero net velocity relative to earth. The fact you don't understand this says a lot about your (lack of) physics knowledge.
QUOTE (->
| QUOTE |
| I also stated that this was ONLY TRUE if the particles were emitted in OPPOSING PAIRS that had ZERO NET effect on the momentum of the alleged MBH. I did NOT however state that the partons had zero velocity relative to the earth. |
In order for your statement to be true, the emitted partons would need to take away zero net momentum in earths reference frame. Do you know the consequences of this? It means the emitted partons must have zero velocity/momentum in earths frame.
Once again - in order for your 50-fold increase in velocity relative to earth to occur, the emitted partons would need to have zero net velocity relative to earth. The fact you don't understand this says a lot about your (lack of) physics knowledge.
No, it doesn't. If I sit in an office Chair (mass 120kgs) and am stationary (net momentum zero) and I manage to emit two 30 kg masses, one traveling in the +x direction at 10 m/s and one traveling in the -x direction at 10 m/s, what do you think the change in my momentum is going to be? Here's a clue. It's Zero. Don't believe me? Do the math for yourself, all the information you need is there. (Assume I'm holding to 30kg sacks of coal while in the office chair).
Gee you think?
How precisely does your example contradict the statement you quoted?
Your example is analogous to a reference frame where the MBH is stationary.
However in a frame where the MBH is moving the emitted partons have net momentum.
HINT: Consider a second observer moving at a speed of 10m/s in the (+) x direction relative to you in the office chair. It sees one stationary mass emitted and the other emitted mass moving at 20m/s in the -x direction. Does this observer see the masses emitted with a net momentum of zero? Here's a clue. No.
QUOTE
No, actually, it would be true in a moving frame as well.
Oh dear
QUOTE (ubavontuba+Feb 19 2008, 06:53 PM)
But what would happen to your momentum if another fellow in another office chair (also holding 30 sacks of coal) collided with you?
Irrelevant.
That's not the point being discussed here.
Irrelevant.
That's not the point being discussed here.
QUOTE (ubavontuba+Feb 19 2008, 06:57 PM)
Afraid of the truth, are we? Why don't you just let rpenner step up to the plate now?
Get over yourself.
Get over yourself.
QUOTE (ThePeanut+Feb 19 2008, 07:39 PM)
HINT: Consider a second observer moving at a speed of 10m/s in the (+) x direction relative to you in the office chair. It sees one stationary mass emitted and the other emitted mass moving at 20m/s in the -x direction. Does this observer see the masses emitted with a net momentum of zero? Here's a clue. No.
IN this specific case case, what do you think the net effect on my momentum would be?
Oh dear, zero. Your observer would see me moving at 10 m/s in the -x direction before and after the collision.
Oh dear, you're wrong again.
IN this specific case case, what do you think the net effect on my momentum would be?
Oh dear, zero. Your observer would see me moving at 10 m/s in the -x direction before and after the collision.
Oh dear, you're wrong again.
QUOTE (Trippy+Feb 19 2008, 06:39 AM)
Irrelevant.
That's not the point being discussed here.
It is relevant and it is exactly the point. We're discussing a collision. Your attempts to modify the velocity of the collision by expelling mass in equal opposing proportions is secondary to the point (not to mention wrong).
That's not the point being discussed here.
It is relevant and it is exactly the point. We're discussing a collision. Your attempts to modify the velocity of the collision by expelling mass in equal opposing proportions is secondary to the point (not to mention wrong).
QUOTE
Oh dear, zero. Your observer would see me moving at 10 m/s in the -x direction before and after the collision.
Exactly, you have lost mass without affecting momentum (in your reference frame) however your velocity relative to me has remained unchanged. So why does the MBH's velocity increase relative to earth 50-fold when it has emitted 98% of its mass in parton radiation?
Actually, you know what?
You're right (in this specific instance) but it has absolutely no bearing on anything else I have said.
And you know what? Unless one of you two muppets can tell me exactly what I did wrong, you don't get to brag about it, and you don't get to claim it. Because I can tell you exactly what I did wrong.
You're right (in this specific instance) but it has absolutely no bearing on anything else I have said.
And you know what? Unless one of you two muppets can tell me exactly what I did wrong, you don't get to brag about it, and you don't get to claim it. Because I can tell you exactly what I did wrong.
QUOTE
Actually, you know what?
You're right (in this specific instance) but it has absolutely no bearing on anything else I have said.
And you know what? Unless one of you two muppets can tell me exactly what I did wrong, you don't get to brag about it, and you don't get to claim it. Because I can tell you exactly what I did wrong.
You're right (in this specific instance) but it has absolutely no bearing on anything else I have said.
And you know what? Unless one of you two muppets can tell me exactly what I did wrong, you don't get to brag about it, and you don't get to claim it. Because I can tell you exactly what I did wrong.
You admitted in your chair example "emitting" the bricks in the way you did leaves your momentum in this frame unchanged (zero before and after the emitting event). I assume you also believe it wouldn't change your velocity in this frame (again zero before and after the event).
Surely you can see that it also leaves your velocity unchanged relative to some observer? Certainly the emitting of the partons won't affect the observes velocity and we have agreed that the event hasn't changed our velocity.
This is analogous to the MBH scenario where you sitting on the chair is the MBH and the "observer" is earth. The MBH simply won't speed up relative to earth as you claimed.
p=mv
In earths frame the momentum and mass of the black hole are decreased proportionally, velocity is unchanged.
In the frame where the MBH is stationary (the chair example), momentum and velocity are both unchanged (both zero before and after "emission") and mass is decreased.
Why not just admit you are wrong with your comment about the parton emission increasing the MBH's velocity relative to earth?
If you have indeed done physics papers as you claim maybe you should brush up on the conservation of momentum.
QUOTE (ThePeanut+Feb 19 2008, 10:44 PM)
You admitted in your chair example "emitting" the bricks in the way you did leaves your momentum in this frame unchanged (zero before and after the emitting event). I assume you also believe it wouldn't change your velocity in this frame (again zero before and after the event).
Surely you can see that it also leaves your velocity unchanged relative to some observer? Certainly the emitting of the partons won't affect the observes velocity and we have agreed that the event hasn't changed our velocity.
This is analogous to the MBH scenario where you sitting on the chair is the MBH and the "observer" is earth. The MBH simply won't speed up relative to earth as you claimed.
p=mv
In earths frame the momentum and mass of the black hole are decreased proportionally, velocity is unchanged.
In the frame where the MBH is stationary (the chair example), momentum and velocity are both unchanged (both zero before and after "emission") and mass is decreased.
Why not just admit you are wrong with your comment about the parton emission increasing the MBH's velocity relative to earth?
If you have indeed done physics papers as you claim maybe you should brush up on the conservation of momentum.
Oh stuff off and read what I actually said you friggin idiot.
Did I, or did I not in the VERY POST YOU FRIGGING QUOTED admit that I had made in error.
Moron.
It's not me that needs to brush up on my conservation of momentum.
I at least have attempted the calculations.
Get over yourself.
QUOTE
Oh stuff off and read what I actually said you friggin idiot.
First of all calm down.
QUOTE (->
| QUOTE |
| Oh stuff off and read what I actually said you friggin idiot. |
First of all calm down.
Did I, or did I not in the VERY POST YOU FRIGGING QUOTED admit that I had made in error.
Oh you were referring to the whole parton emission argument then? Cool.
I note that you say that "it has absolutely no bearing on anything else I have said." It in fact does have a lot of bearing on other things you have said (HINT: think about where you went wrong in this argument and you should be able to see why an 1150 TeV cosmic ray hitting a relatively stationary molecule in earths atmosphere won't produce MBH's with a residual velocity (or distributed around a mean velocity) of 239 m/s relative to earth - a figure you calculated on this page.
QUOTE
It's not me that needs to brush up on my conservation of momentum.
Lets see...
1) You point out an "obvious error" in my initial post because I was only considering the collision in one dimension, asking me to consider the plane between the two protons and the center of mass of the system.
You were incorrect (there is no plane, only a line - the relative velocities of the protons and the COM along this line are the only important factors in my example).
2) You claim that the COM of the two-proton isolated system moves in earths reference frame.
QUOTE (->
| QUOTE |
| It's not me that needs to brush up on my conservation of momentum. |
Lets see...
1) You point out an "obvious error" in my initial post because I was only considering the collision in one dimension, asking me to consider the plane between the two protons and the center of mass of the system.
You were incorrect (there is no plane, only a line - the relative velocities of the protons and the COM along this line are the only important factors in my example).
2) You claim that the COM of the two-proton isolated system moves in earths reference frame.
Because the Center of Mass is not stationary in the Earths Reference Frame.
The momentum of the center of mass is constant, but the total mass before and after the collision is not, therefore the velocity of the center of mass relative to the earth is not constant.
This blatantly violates the conservation of momentum law. In an isolated system (no net external force acting on it), the COM moves with uniform velocity (relative to an inertial observer) along a straight line. Would you like me to derive the particular equation for you?
3) You argued that the velocity of the MBH would increase relative to earth due to parton radiation, which you now admit was incorrect.
Thats three statements, just from the last few pages of this thread that suggest your knowledge of some basic physics principles is less than sound.
Would you like me to go back a few more pages to some of your discussions with ubavontuba? There are numerous other examples where your lack of physics knowledge shows.
QUOTE (ThePeanut+Feb 20 2008, 02:07 PM)
I note that you say that "it has absolutely no bearing on anything else I have said." It in fact does have a lot of bearing on other things you have said (HINT: think about where you went wrong in this argument and you should be able to see why an 1150 TeV cosmic ray hitting a relatively stationary molecule in earths atmosphere won't produce MBH's with a residual velocity (or distributed around a mean velocity) of 239 m/s relative to earth - a figure you calculated on this page.
A figure that I still stand beside, as it is based on calculations using p=mγv. My error with the partons and the error that I have freely admitted I could have made are completely unrelated.
You want to prove me wrong? Stop flapping your arms, and do the calculation for yourself.
A figure that I still stand beside, as it is based on calculations using p=mγv. My error with the partons and the error that I have freely admitted I could have made are completely unrelated.
You want to prove me wrong? Stop flapping your arms, and do the calculation for yourself.
QUOTE (ThePeanut+Feb 20 2008, 02:07 PM)
1) You point out an "obvious error" in my initial post because I was only considering the collision in one dimension, asking me to consider the plane between the two protons and the center of mass of the system.
You were incorrect (there is no plane, only a line - the relative velocities of the protons and the COM along this line are the only important factors in my example).
A statement that I still stand beside. Only in one very specific case are your assertions true, when the incoming proton is following a radial vector. That is ALL I have admitted. If the incoming proton is following anything other then a radial path, then my statement, which is a more general solution, remains true.
You were incorrect (there is no plane, only a line - the relative velocities of the protons and the COM along this line are the only important factors in my example).
A statement that I still stand beside. Only in one very specific case are your assertions true, when the incoming proton is following a radial vector. That is ALL I have admitted. If the incoming proton is following anything other then a radial path, then my statement, which is a more general solution, remains true.
QUOTE (ThePeanut+Feb 20 2008, 02:07 PM)
2) You claim that the COM of the two-proton isolated system moves in earths reference frame.
I maintain that this can be true in some circumstances.
I maintain that this can be true in some circumstances.
QUOTE (ThePeanut+Feb 20 2008, 02:07 PM)
This blatantly violates the conservation of momentum law. In an isolated system (no net external force acting on it), the COM moves with uniform velocity (relative to an inertial observer) along a straight line. Would you like me to derive the particular equation for you?
Bull. Again, I refer you to the same page and experiemnt I refered Ubavontuba to (you seem to be enjoying mining my post history for quotes so much, you should be able to find it easy enough).
By your logic, when a Mini hits an 18 wheeler, the center of mass of the mini should keep moving at the same speed in the same direction (which would require the 18-wheeler to do so). Consider what happens when you drop a sack of potatoes on a skateboard as it rolls by you (the skateboard slows down).
Bull. Again, I refer you to the same page and experiemnt I refered Ubavontuba to (you seem to be enjoying mining my post history for quotes so much, you should be able to find it easy enough).
By your logic, when a Mini hits an 18 wheeler, the center of mass of the mini should keep moving at the same speed in the same direction (which would require the 18-wheeler to do so). Consider what happens when you drop a sack of potatoes on a skateboard as it rolls by you (the skateboard slows down).
QUOTE (ThePeanut+Feb 20 2008, 02:07 PM)
3) You argued that the velocity of the MBH would increase relative to earth due to parton radiation, which you now admit was incorrect.
Ironicaly, the mistake I made was considering a 2d vector in only 1d (see why it's so important yet?)
Ironicaly, the mistake I made was considering a 2d vector in only 1d (see why it's so important yet?)
QUOTE (ThePeanut+Feb 20 2008, 02:07 PM)
Thats three statements, just from the last few pages of this thread that suggest your knowledge of some basic physics principles is less than sound.
Bull. Only in your world.
And I have explained to you that your result violates the conservation of momentum. The velocity of the COM in your result is clearly different (relative to any inertial observer), post collision to it was pre-collision despite the lack of an external force acting on your system.
And I have explained to you that your result violates the conservation of momentum. The velocity of the COM in your result is clearly different (relative to any inertial observer), post collision to it was pre-collision despite the lack of an external force acting on your system.
A statement that I still stand beside. Only in one very specific case are your assertions true, when the incoming proton is following a radial vector. That is ALL I have admitted. If the incoming proton is following anything other then a radial path, then my statement, which is a more general solution, remains true.
Go on then, give me a velocity vector for the incoming proton that invalidates my initial argument, it's simply not important.
Bull. Only in your world.
QUOTE (ThePeanut+Feb 20 2008, 02:07 PM)
Would you like me to go back a few more pages to some of your discussions with ubavontuba? There are numerous other examples where your lack of physics knowledge shows.
I know, here's an idea, why don't you get over yourself. Or better yet, why don't you go through your bed-buddy Ubavontubas posts, and see some of the gems he's come up with regarding the conservation of energy.
And no, you nimrod, if you had bothered actually reading through my posts, you's see that I have demonstrated, more then once, that I'm perfectly capable of deriving these results for myself.
Don't mistake missing an error for a lack of knowledge. The fact that I was able to find the specific error says that I know more then you seem to think I do.
Get over yourself.
I know, here's an idea, why don't you get over yourself. Or better yet, why don't you go through your bed-buddy Ubavontubas posts, and see some of the gems he's come up with regarding the conservation of energy.
And no, you nimrod, if you had bothered actually reading through my posts, you's see that I have demonstrated, more then once, that I'm perfectly capable of deriving these results for myself.
Don't mistake missing an error for a lack of knowledge. The fact that I was able to find the specific error says that I know more then you seem to think I do.
Get over yourself.
The silence is astounding.
Here's a hint.
First you need to calculate the velocity of the proton, based on the knowledge of it's rest mass, and it having a total energy of 1150 TeV.
Then you need to calculate the momentum of a proton travelling at that speed.
The Black hole, after the collision has the same momentum as the proton had before the collision. We know it's mass, and we know how much momentum it has, which enables us to calculate its velocity (All relative to the Earth).
Edit: It occurs to me to mention, that when I figured the mass of the (alleged) black hole I assumed that all of the total energy of the proton became the rest energy of the black hole. Another possible interpretation of what I said with the partons is that if only 2% of the total energy of the cosmic ray proton is converted into black hole rest energy, then the result is a black that is light enough, and has sufficient momentum to acheive escape velocity - a point that I have tried to make previously, but managed to mis-state.
Here's a hint.
First you need to calculate the velocity of the proton, based on the knowledge of it's rest mass, and it having a total energy of 1150 TeV.
Then you need to calculate the momentum of a proton travelling at that speed.
The Black hole, after the collision has the same momentum as the proton had before the collision. We know it's mass, and we know how much momentum it has, which enables us to calculate its velocity (All relative to the Earth).
Edit: It occurs to me to mention, that when I figured the mass of the (alleged) black hole I assumed that all of the total energy of the proton became the rest energy of the black hole. Another possible interpretation of what I said with the partons is that if only 2% of the total energy of the cosmic ray proton is converted into black hole rest energy, then the result is a black that is light enough, and has sufficient momentum to acheive escape velocity - a point that I have tried to make previously, but managed to mis-state.
QUOTE
A figure that I still stand beside, as it is based on calculations using p=mγv. My error with the partons and the error that I have freely admitted I could have made are completely unrelated.
You want to prove me wrong? Stop flapping your arms, and do the calculation for yourself.
You want to prove me wrong? Stop flapping your arms, and do the calculation for yourself.
And I have explained to you that your result violates the conservation of momentum. The velocity of the COM in your result is clearly different (relative to any inertial observer), post collision to it was pre-collision despite the lack of an external force acting on your system.
QUOTE (->
| QUOTE |
| A figure that I still stand beside, as it is based on calculations using p=mγv. My error with the partons and the error that I have freely admitted I could have made are completely unrelated. You want to prove me wrong? Stop flapping your arms, and do the calculation for yourself. |
And I have explained to you that your result violates the conservation of momentum. The velocity of the COM in your result is clearly different (relative to any inertial observer), post collision to it was pre-collision despite the lack of an external force acting on your system.
A statement that I still stand beside. Only in one very specific case are your assertions true, when the incoming proton is following a radial vector. That is ALL I have admitted. If the incoming proton is following anything other then a radial path, then my statement, which is a more general solution, remains true.
Go on then, give me a velocity vector for the incoming proton that invalidates my initial argument, it's simply not important.
QUOTE
QUOTE (ThePeanut @ Feb 20 2008, 02:07 PM)
Look, it's the conservation of linear momentum, are you saying the law is wrong? The COM of an isolated system of particles moves at constant speed in a straight line with respect to any system of inertial coordinates. Unless you apply some external force to the isolated system, an observer watching from the COM frame sees the net momenta of the particles sum to zero before and after any collision.
Your first example above, is frankly bizarre, why should the center of mass of the mini keep moving at the same speed in the same direction? Are you confusing the center of mass of the mini with the COM of the system as a whole (which includes the mini and the 18 wheeler). Again, the velocity of the center of mass (momentum) frame of this (isolated) system doesn't change after the collision relative to any inertial observer.
In example two you are correct, the skateboard indeed would slow down. I'm not sure what you are trying to prove though? Do you think it changes the COM of the system (ignore the ground for a moment and just consider an isolated system consisting of the moving skateboard and falling potatoes)? Do you think it's analogous to your MBH slowing down below escape velocity? I'm not sure you've learned your lesson from the parton emission example.
It appears you haven't learned your lesson from the parton emission example.
It appears you haven't learned your lesson from the parton emission example.
I know, here's an idea, why don't you get over yourself. Or better yet, why don't you go through your bed-buddy Ubavontubas posts, and see some of the gems he's come up with regarding the conservation of energy.
Wow your maturity just shines through. I agree with ubavontuba on a couple of points so I'm his "bed-buddy"? I disagree with you and I'm a *****? Grow up.
Oh dear, you clearly haven't learned your lesson from the parton radiation example because this is not the case.
And I have explained to you that your result violates the conservation of momentum. The velocity of the COM in your result is clearly different (relative to any inertial observer), post collision to it was pre-collision despite the lack of an external force acting on your system.
Only if you ignore the CHANGE IN MASS of the system.
Or are you closing your eyes and ignoring how I stated that I did the calculations?
QUOTE (->
| QUOTE |
| QUOTE (ThePeanut @ Feb 20 2008, 02:07 PM) This blatantly violates the conservation of momentum law. In an isolated system (no net external force acting on it), the COM moves with uniform velocity (relative to an inertial observer) along a straight line. Would you like me to derive the particular equation for you? Bull. Again, I refer you to the same page and experiment I referred Ubavontuba to (you seem to be enjoying mining my post history for quotes so much, you should be able to find it easy enough). By your logic, when a Mini hits an 18 wheeler, the center of mass of the mini should keep moving at the same speed in the same direction (which would require the 18-wheeler to do so). Consider what happens when you drop a sack of potatoes on a skateboard as it rolls by you (the skateboard slows down). |
Look, it's the conservation of linear momentum, are you saying the law is wrong? The COM of an isolated system of particles moves at constant speed in a straight line with respect to any system of inertial coordinates. Unless you apply some external force to the isolated system, an observer watching from the COM frame sees the net momenta of the particles sum to zero before and after any collision.
Your first example above, is frankly bizarre, why should the center of mass of the mini keep moving at the same speed in the same direction? Are you confusing the center of mass of the mini with the COM of the system as a whole (which includes the mini and the 18 wheeler). Again, the velocity of the center of mass (momentum) frame of this (isolated) system doesn't change after the collision relative to any inertial observer.
In example two you are correct, the skateboard indeed would slow down. I'm not sure what you are trying to prove though? Do you think it changes the COM of the system (ignore the ground for a moment and just consider an isolated system consisting of the moving skateboard and falling potatoes)? Do you think it's analogous to your MBH slowing down below escape velocity? I'm not sure you've learned your lesson from the parton emission example.
QUOTE
Ironically, the mistake I made was considering a 2d vector in only 1d (see why it's so important yet?)
It appears you haven't learned your lesson from the parton emission example.
QUOTE (->
| QUOTE |
| Ironically, the mistake I made was considering a 2d vector in only 1d (see why it's so important yet?) |
It appears you haven't learned your lesson from the parton emission example.
I know, here's an idea, why don't you get over yourself. Or better yet, why don't you go through your bed-buddy Ubavontubas posts, and see some of the gems he's come up with regarding the conservation of energy.
Wow your maturity just shines through. I agree with ubavontuba on a couple of points so I'm his "bed-buddy"? I disagree with you and I'm a *****? Grow up.
QUOTE
Edit: It occurs to me to mention, that when I figured the mass of the (alleged) black hole I assumed that all of the total energy of the proton became the rest energy of the black hole. Another possible interpretation of what I said with the partons is that if only 2% of the total energy of the cosmic ray proton is converted into black hole rest energy, then the result is a black that is light enough, and has sufficient momentum to acheive escape velocity - a point that I have tried to make previously, but managed to mis-state.
Oh dear, you clearly haven't learned your lesson from the parton radiation example because this is not the case.
QUOTE (ThePeanut+Feb 21 2008, 11:45 AM)
And I have explained to you that your result violates the conservation of momentum. The velocity of the COM in your result is clearly different (relative to any inertial observer), post collision to it was pre-collision despite the lack of an external force acting on your system.
Only if you ignore the CHANGE IN MASS of the system.
Or are you closing your eyes and ignoring how I stated that I did the calculations?
QUOTE (ThePeanut+Feb 21 2008, 11:45 AM)
Go on then, give me a velocity vector for the incoming proton that invalidates my initial argument, it's simply not important.
Almost any tangential vector that doesn't pass through the center of the earth requires consideration of 2 or 3 dimensional vectors.
Almost any tangential vector that doesn't pass through the center of the earth requires consideration of 2 or 3 dimensional vectors.
QUOTE (ThePeanut+Feb 21 2008, 11:45 AM)
Look, it's the conservation of linear momentum, are you saying the law is wrong? The COM of an isolated system of particles moves at constant speed in a straight line with respect to any system of inertial coordinates. Unless you apply some external force to the isolated system, an observer watching from the COM frame sees the net momenta of the particles sum to zero before and after any collision.
Your first example above, is frankly bizarre, why should the center of mass of the mini keep moving at the same speed in the same direction? Are you confusing the center of mass of the mini with the COM of the system as a whole (which includes the mini and the 18 wheeler). Again, the velocity of the center of mass (momentum) frame of this (isolated) system doesn't change after the collision relative to any inertial observer.
No, you're saying that the law is wrong. And the point you raise is precisely the point that I'm tying to make (more or less) that you seem to be making the very mistake that you say that I am. Remember, I stated that that conclusion was with your logic, not based on my understanding of the conservation of momentum, I already know exactly what happens.
Your first example above, is frankly bizarre, why should the center of mass of the mini keep moving at the same speed in the same direction? Are you confusing the center of mass of the mini with the COM of the system as a whole (which includes the mini and the 18 wheeler). Again, the velocity of the center of mass (momentum) frame of this (isolated) system doesn't change after the collision relative to any inertial observer.
No, you're saying that the law is wrong. And the point you raise is precisely the point that I'm tying to make (more or less) that you seem to be making the very mistake that you say that I am. Remember, I stated that that conclusion was with your logic, not based on my understanding of the conservation of momentum, I already know exactly what happens.
QUOTE (ThePeanut+Feb 21 2008, 11:45 AM)
In example two you are correct, the skateboard indeed would slow down. I'm not sure what you are trying to prove though? Do you think it changes the COM of the system (ignore the ground for a moment and just consider an isolated system consisting of the moving skateboard and falling potatoes)? Do you think it's analogous to your MBH slowing down below escape velocity? I'm not sure you've learned your lesson from the parton emission example.
Whatever, it's analagous, but not the same, both deal with a perfectly inelastic system, which is the point that I have been trying to make all along, the collision between the proton and whatever is a perfectly inelastic collision, and in an inelastic collision the velocity changes. In CLASSICAL system:
v=(m_1/M)*v_1
IN a relativistic system, you need to consider γ as well. Even with a γ of (about)5,000 (which is, IIRC what a Proton with a total energy of 1150 TeV experiences) the mass of the proton is still only 5,000 AMU, however, the mass of a black hole with a rest energy of 1150 TeV is 1,226,000 AMU.
Get that? This is the simple fact that Ubavontuba has failed to grasp.
p=mγv before, and after the collision.
p is the same before and after the collision.
m as increased by a factor of about 245.
Therefore, v must decreases by the same amount otherwise MOMENTUM CAN NOT BE CONSERVED IN THE EARTHS REFERENCE FRAME.
It's not me that's trying to violate the laws of physics.
I've already explicitly stated that if the final velocity is in error, then I know exactly where the error is, and it ISN'T in the basic principles I've been using, the most likely source of error was in my workings, trying to rearrange the expression γv to solve for v.
Get that?
Oh please. Consider our two proton example again. In the COM of the two proton system the two protons collide and form a stationary MBH (again simplifying things but at the least MBH production is distributed around zero velocity) right? Has the COM moved in this example after the collision? No. If we define some observer in an inertial reference frame, has the velocity of the COM changed relative to them? No. Does the velocity of the COM in your example change after the collision relative to an intertial observer? Yes. You haven't conserved momentum.
Oh please. Consider our two proton example again. In the COM of the two proton system the two protons collide and form a stationary MBH (again simplifying things but at the least MBH production is distributed around zero velocity) right? Has the COM moved in this example after the collision? No. If we define some observer in an inertial reference frame, has the velocity of the COM changed relative to them? No. Does the velocity of the COM in your example change after the collision relative to an intertial observer? Yes. You haven't conserved momentum.
Almost any tangential vector that doesn't pass through the center of the earth requires consideration of 2 or 3 dimensional vectors.
I fear you're misunderstanding the point I was making in my initial post.
Look it's really simple, the velocity of the center of mass of this system does not change (relative to an inertial observer). This does not mean the mini must keep moving at the same speed in the same direction.
Lets consider a one-dimensional collision - the mini moving at 10 m/s and the 18-wheeler at -10m/s. Lets also assume the 18-wheeler has a mass of 3 tonnes and the mini 1 tonne. What is the velocity of the center of momentum frame of this system relative to me? It is moving at -5m/s relative to me right? Because in this frame the 18-wheeler is moving at -5m/s and the mini at 15m/s and net linear momentum is zero: (-5x3) + (15x1) = 0
If we assume a perfectly inelastic collision, the mini and 18-wheeler are now both stationary in the COM frame (momentum still zero). The COM frame is still moving at -5m/s relative to me as we would expect and therefore I see the two vehicles both moving at -5m/s.
Note that the mini has not kept moving at the same speed relative to me and yet the center of momentum frame HAS kept moving at the same speed.
Again the law of the conservation of linear momentum states that the velocity of the COM frame of an isolated system is constant as seen by observers in any intertial reference frame. Your MBH calculation leads to a COM frame that has a different velocity (relative to an intertial observer) after the collision than it had before the collision, this cannot happen in an isolated system and is a violation of the conservation of momentum.
Look it's really simple, the velocity of the center of mass of this system does not change (relative to an inertial observer). This does not mean the mini must keep moving at the same speed in the same direction.
Lets consider a one-dimensional collision - the mini moving at 10 m/s and the 18-wheeler at -10m/s. Lets also assume the 18-wheeler has a mass of 3 tonnes and the mini 1 tonne. What is the velocity of the center of momentum frame of this system relative to me? It is moving at -5m/s relative to me right? Because in this frame the 18-wheeler is moving at -5m/s and the mini at 15m/s and net linear momentum is zero: (-5x3) + (15x1) = 0
If we assume a perfectly inelastic collision, the mini and 18-wheeler are now both stationary in the COM frame (momentum still zero). The COM frame is still moving at -5m/s relative to me as we would expect and therefore I see the two vehicles both moving at -5m/s.
Note that the mini has not kept moving at the same speed relative to me and yet the center of momentum frame HAS kept moving at the same speed.
Again the law of the conservation of linear momentum states that the velocity of the COM frame of an isolated system is constant as seen by observers in any intertial reference frame. Your MBH calculation leads to a COM frame that has a different velocity (relative to an intertial observer) after the collision than it had before the collision, this cannot happen in an isolated system and is a violation of the conservation of momentum.
Whatever, it's analagous, but not the same, both deal with a perfectly inelastic system, which is the point that I have been trying to make all along, the collision between the proton and whatever is a perfectly inelastic collision, and in an inelastic collision the velocity changes.
The velocity of what changes? The velocity of the COM of the system relative to an inertial observer doesn't. The velocity of the MBH produced will be slower than the velocity of the cosmic ray proton relative to earth obviously, but it will still be much faster than the 239m/s figure you came up with.
Why even say it in the first place? Why resort to name calling at all? Calling me an idiot or a muppet simply for disagreeing with you is pathetic. Name calling is generally what people do when they cannot put together a rational argument in defense of their position.
Why even say it in the first place? Why resort to name calling at all? Calling me an idiot or a muppet simply for disagreeing with you is pathetic. Name calling is generally what people do when they cannot put together a rational argument in defense of their position.
I notice that neither of you two has taken the challenge to do teh calculations for yourselves, even though I've given you all the information you need to do so.
The center of momentum frame of the system consisting of your relatively stationary oxygen atom and your cosmic ray proton is clearly not moving at 239m/s relative to earth, it's moving at a considerable fraction of the speed of light. I don't need to perform any calculations to work that out. I'm happy to work through the calculation for you if you don't believe me though.
The velocity of the COM frame (relative to earth) in your system has changed from a significant fraction of the speed of light pre-collision to 239m/s after the collision despite no external force acting on the system. Collisions between particles inside the system will not change the velocity of the COM frame relative to an inertial observer.
Whatever, it's analagous, but not the same, both deal with a perfectly inelastic system, which is the point that I have been trying to make all along, the collision between the proton and whatever is a perfectly inelastic collision, and in an inelastic collision the velocity changes. In CLASSICAL system:
v=(m_1/M)*v_1
IN a relativistic system, you need to consider γ as well. Even with a γ of (about)5,000 (which is, IIRC what a Proton with a total energy of 1150 TeV experiences) the mass of the proton is still only 5,000 AMU, however, the mass of a black hole with a rest energy of 1150 TeV is 1,226,000 AMU.
Get that? This is the simple fact that Ubavontuba has failed to grasp.
p=mγv before, and after the collision.
p is the same before and after the collision.
m as increased by a factor of about 245.
Therefore, v must decreases by the same amount otherwise MOMENTUM CAN NOT BE CONSERVED IN THE EARTHS REFERENCE FRAME.
It's not me that's trying to violate the laws of physics.
I've already explicitly stated that if the final velocity is in error, then I know exactly where the error is, and it ISN'T in the basic principles I've been using, the most likely source of error was in my workings, trying to rearrange the expression γv to solve for v.
Get that?
QUOTE (ThePeanut+Feb 21 2008, 11:45 AM)
Wow your maturity just shines through. I agree with ubavontuba on a couple of points so I'm his "bed-buddy"? I disagree with you and I'm a *****? Grow up.
If you want to take it that way, go for it, but before you do, I suggest you look up it's other meanings. In otherwords, the fact that you've been getting pal-ly and chummy with Unavontuba makes you bed buddies. It was, in no way, intended as an allusion to your sexuality.
I notice that neither of you two has taken the challenge to do teh calculations for yourselves, even though I've given you all the information you need to do so.
Typical, you'd much rather sit around and whine about something you don't like.
If you want to take it that way, go for it, but before you do, I suggest you look up it's other meanings. In otherwords, the fact that you've been getting pal-ly and chummy with Unavontuba makes you bed buddies. It was, in no way, intended as an allusion to your sexuality.
I notice that neither of you two has taken the challenge to do teh calculations for yourselves, even though I've given you all the information you need to do so.
Typical, you'd much rather sit around and whine about something you don't like.
QUOTE
Only if you ignore the CHANGE IN MASS of the system.
Or are you closing your eyes and ignoring how I stated that I did the calculations?
Or are you closing your eyes and ignoring how I stated that I did the calculations?
Oh please. Consider our two proton example again. In the COM of the two proton system the two protons collide and form a stationary MBH (again simplifying things but at the least MBH production is distributed around zero velocity) right? Has the COM moved in this example after the collision? No. If we define some observer in an inertial reference frame, has the velocity of the COM changed relative to them? No. Does the velocity of the COM in your example change after the collision relative to an intertial observer? Yes. You haven't conserved momentum.
QUOTE (->
| QUOTE |
| Only if you ignore the CHANGE IN MASS of the system. Or are you closing your eyes and ignoring how I stated that I did the calculations? |
Oh please. Consider our two proton example again. In the COM of the two proton system the two protons collide and form a stationary MBH (again simplifying things but at the least MBH production is distributed around zero velocity) right? Has the COM moved in this example after the collision? No. If we define some observer in an inertial reference frame, has the velocity of the COM changed relative to them? No. Does the velocity of the COM in your example change after the collision relative to an intertial observer? Yes. You haven't conserved momentum.
Almost any tangential vector that doesn't pass through the center of the earth requires consideration of 2 or 3 dimensional vectors.
I fear you're misunderstanding the point I was making in my initial post.
QUOTE
No, you're saying that the law is wrong. And the point you raise is precisely the point that I'm tying to make (more or less) that you seem to be making the very mistake that you say that I am. Remember, I stated that that conclusion was with your logic, not based on my understanding of the conservation of momentum, I already know exactly what happens.
Look it's really simple, the velocity of the center of mass of this system does not change (relative to an inertial observer). This does not mean the mini must keep moving at the same speed in the same direction.
Lets consider a one-dimensional collision - the mini moving at 10 m/s and the 18-wheeler at -10m/s. Lets also assume the 18-wheeler has a mass of 3 tonnes and the mini 1 tonne. What is the velocity of the center of momentum frame of this system relative to me? It is moving at -5m/s relative to me right? Because in this frame the 18-wheeler is moving at -5m/s and the mini at 15m/s and net linear momentum is zero: (-5x3) + (15x1) = 0
If we assume a perfectly inelastic collision, the mini and 18-wheeler are now both stationary in the COM frame (momentum still zero). The COM frame is still moving at -5m/s relative to me as we would expect and therefore I see the two vehicles both moving at -5m/s.
Note that the mini has not kept moving at the same speed relative to me and yet the center of momentum frame HAS kept moving at the same speed.
Again the law of the conservation of linear momentum states that the velocity of the COM frame of an isolated system is constant as seen by observers in any intertial reference frame. Your MBH calculation leads to a COM frame that has a different velocity (relative to an intertial observer) after the collision than it had before the collision, this cannot happen in an isolated system and is a violation of the conservation of momentum.
QUOTE (->
| QUOTE |
| No, you're saying that the law is wrong. And the point you raise is precisely the point that I'm tying to make (more or less) that you seem to be making the very mistake that you say that I am. Remember, I stated that that conclusion was with your logic, not based on my understanding of the conservation of momentum, I already know exactly what happens. |
Look it's really simple, the velocity of the center of mass of this system does not change (relative to an inertial observer). This does not mean the mini must keep moving at the same speed in the same direction.
Lets consider a one-dimensional collision - the mini moving at 10 m/s and the 18-wheeler at -10m/s. Lets also assume the 18-wheeler has a mass of 3 tonnes and the mini 1 tonne. What is the velocity of the center of momentum frame of this system relative to me? It is moving at -5m/s relative to me right? Because in this frame the 18-wheeler is moving at -5m/s and the mini at 15m/s and net linear momentum is zero: (-5x3) + (15x1) = 0
If we assume a perfectly inelastic collision, the mini and 18-wheeler are now both stationary in the COM frame (momentum still zero). The COM frame is still moving at -5m/s relative to me as we would expect and therefore I see the two vehicles both moving at -5m/s.
Note that the mini has not kept moving at the same speed relative to me and yet the center of momentum frame HAS kept moving at the same speed.
Again the law of the conservation of linear momentum states that the velocity of the COM frame of an isolated system is constant as seen by observers in any intertial reference frame. Your MBH calculation leads to a COM frame that has a different velocity (relative to an intertial observer) after the collision than it had before the collision, this cannot happen in an isolated system and is a violation of the conservation of momentum.
Whatever, it's analagous, but not the same, both deal with a perfectly inelastic system, which is the point that I have been trying to make all along, the collision between the proton and whatever is a perfectly inelastic collision, and in an inelastic collision the velocity changes.
The velocity of what changes? The velocity of the COM of the system relative to an inertial observer doesn't. The velocity of the MBH produced will be slower than the velocity of the cosmic ray proton relative to earth obviously, but it will still be much faster than the 239m/s figure you came up with.
QUOTE
If you want to take it that way, go for it, but before youdo, I suggest you look up it's other meanings. In other words, the fact that you've been getting pal-ly and chummy with Unavontuba makes you bed buddies. It was, in no way, intended as an allusion to your sexuality.
Why even say it in the first place? Why resort to name calling at all? Calling me an idiot or a muppet simply for disagreeing with you is pathetic. Name calling is generally what people do when they cannot put together a rational argument in defense of their position.
QUOTE (->
| QUOTE |
| If you want to take it that way, go for it, but before youdo, I suggest you look up it's other meanings. In other words, the fact that you've been getting pal-ly and chummy with Unavontuba makes you bed buddies. It was, in no way, intended as an allusion to your sexuality. |
Why even say it in the first place? Why resort to name calling at all? Calling me an idiot or a muppet simply for disagreeing with you is pathetic. Name calling is generally what people do when they cannot put together a rational argument in defense of their position.
I notice that neither of you two has taken the challenge to do teh calculations for yourselves, even though I've given you all the information you need to do so.
The center of momentum frame of the system consisting of your relatively stationary oxygen atom and your cosmic ray proton is clearly not moving at 239m/s relative to earth, it's moving at a considerable fraction of the speed of light. I don't need to perform any calculations to work that out. I'm happy to work through the calculation for you if you don't believe me though.
The velocity of the COM frame (relative to earth) in your system has changed from a significant fraction of the speed of light pre-collision to 239m/s after the collision despite no external force acting on the system. Collisions between particles inside the system will not change the velocity of the COM frame relative to an inertial observer.
QUOTE (ThePeanut+Feb 21 2008, 03:02 PM)
Why even say it in the first place? Why resort to name calling at all? Calling me an idiot or a muppet simply for disagreeing with you is pathetic. Name calling is generally what people do when they cannot put together a rational argument in defense of their position.nertial observer.
Oh for gods sake, get over it.
Would you think it was name calling if I said that corporation A and corporation B were bed buddies? No, you wouldn't. You'd take it in context.
What's the difference.
Oh for gods sake, get over it.
Would you think it was name calling if I said that corporation A and corporation B were bed buddies? No, you wouldn't. You'd take it in context.
What's the difference.
QUOTE (ThePeanut+Feb 21 2008, 03:02 PM)
No. Does the velocity of the COM in your example change after the collision relative to an intertial observer? Yes. You haven't conserved momentum.
Oh please. How do you get "You haven't conserved momentum" when the opening statement is "If we assume the momentum of the black hole after the collision (relative to the earth) is the same as the proton before the collision."
Oh please. How do you get "You haven't conserved momentum" when the opening statement is "If we assume the momentum of the black hole after the collision (relative to the earth) is the same as the proton before the collision."
QUOTE (ThePeanut+Feb 21 2008, 03:02 PM)
I fear you're misunderstanding the point I was making in my initial post.
You know what? I actually no longer care.
You know what? I actually no longer care.
QUOTE (ThePeanut+Feb 21 2008, 03:02 PM)
Look it's really simple, the velocity of the center of mass of this system does not change (relative to an inertial observer). This does not mean the mini must keep moving at the same speed in the same direction.
Lets consider a one-dimensional collision - the mini moving at 10 m/s and the 18-wheeler at -10m/s. Lets also assume the 18-wheeler has a mass of 3 tonnes and the mini 1 tonne. What is the velocity of the center of momentum frame of this system relative to me? It is moving at -5m/s relative to me right? Because in this frame the 18-wheeler is moving at -5m/s and the mini at 15m/s and net linear momentum is zero: (-5x3) + (15x1) = 0
If we assume a perfectly inelastic collision, the mini and 18-wheeler are now both stationary in the COM frame (momentum still zero). The COM frame is still moving at -5m/s relative to me as we would expect and therefore I see the two vehicles both moving at -5m/s.
Really? Oh gosh. And what have I been saying the whole GOD DAM TIME? Gee, let me see, I believe that my initial statement was this:
"Because of the change in mass, the conservation of momentum says the velocity of the black hole after the collision must be less then the velocity of the proton before the collision, so in SOME CASES it MAY be possible for the blackhole to have a residual velocity less then the escape velocity of the earth".
As I said Sherlock, I'm perfectly capable of doing the calculations for myself.
Also notice that I initially considered an entire 'avaerage' oxygen nucleus, rather then just two protons, which changes things significantly.
Lets consider a one-dimensional collision - the mini moving at 10 m/s and the 18-wheeler at -10m/s. Lets also assume the 18-wheeler has a mass of 3 tonnes and the mini 1 tonne. What is the velocity of the center of momentum frame of this system relative to me? It is moving at -5m/s relative to me right? Because in this frame the 18-wheeler is moving at -5m/s and the mini at 15m/s and net linear momentum is zero: (-5x3) + (15x1) = 0
If we assume a perfectly inelastic collision, the mini and 18-wheeler are now both stationary in the COM frame (momentum still zero). The COM frame is still moving at -5m/s relative to me as we would expect and therefore I see the two vehicles both moving at -5m/s.
Really? Oh gosh. And what have I been saying the whole GOD DAM TIME? Gee, let me see, I believe that my initial statement was this:
"Because of the change in mass, the conservation of momentum says the velocity of the black hole after the collision must be less then the velocity of the proton before the collision, so in SOME CASES it MAY be possible for the blackhole to have a residual velocity less then the escape velocity of the earth".
As I said Sherlock, I'm perfectly capable of doing the calculations for myself.
Also notice that I initially considered an entire 'avaerage' oxygen nucleus, rather then just two protons, which changes things significantly.
QUOTE (ThePeanut+Feb 21 2008, 03:02 PM)
Note that the mini has not kept moving at the same speed relative to me and yet the center of momentum frame HAS kept moving at the same speed.
This is precisely my point Sherlock.
This is precisely my point Sherlock.
QUOTE (ThePeanut+Feb 21 2008, 03:02 PM)
Again the law of the conservation of linear momentum states that the velocity of the COM frame of an isolated system is constant as seen by observers in any intertial reference frame. Your MBH calculation leads to a COM frame that has a different velocity (relative to an intertial observer) after the collision than it had before the collision, this cannot happen in an isolated system and is a violation of the conservation of momentum.
ONce again, you're ignoring the fact that the rest mass of the black hole is 254 times more massive then the observed mass of the proton (in a stationary frame).
ONce again, you're ignoring the fact that the rest mass of the black hole is 254 times more massive then the observed mass of the proton (in a stationary frame).
QUOTE (ThePeanut+Feb 21 2008, 03:02 PM)
The velocity of what changes? The velocity of the COM of the system relative to an inertial observer doesn't. The velocity of the MBH produced will be slower than the velocity of the cosmic ray proton relative to earth obviously, but it will still be much faster than the 239m/s figure you came up with.
The center of momentum frame of the system consisting of your relatively stationary oxygen atom and your cosmic ray proton is clearly not moving at 239m/s relative to earth, it's moving at a considerable fraction of the speed of light. I don't need to perform any calculations to work that out. I'm happy to work through the calculation for you if you don't believe me though.
The velocity of the COM frame (relative to earth) in your system has changed from a significant fraction of the speed of light pre-collision to 239m/s after the collision despite no external force acting on the system. Collisions between particles inside the system will not change the velocity of the COM frame relative to an inertial observer.
And once again. We come back to the point that I stated as the first step assume momentum is the same before and after the collision.
It seems to me that you're incapable of understanding what I've been saying. I've been saying "If we assume that momentum is conserved in the earths reference frame, then this must be true."
No, the velocity of the MBH relative to the earth. The ground you're standing on is looking thinner and thinner. And again, if you disagree with the 239 m/s figure, do the freaking calculations yourself, i'm sick of all this armwaving. I've already told you what I did, and where I MOIGHT have screwed it up. Frankly i'm sick to death of all of this bluster. You and Ubavonfatss, you sit there naysaying and disagreeing with everybody else, simply because you don't like it, and yet you refuse to do the calculations yourself. Frankly? I'm bored with it.
No where have I addressed the motion of the center of mass. Quote me. Oh that's right, you can't. Nowhere have I claimed that the motion of the center of mass changes.
Although, having said that, I have said that I am highly dubious of Ubavontubas claim that the Oxygen atom is unaffected by the collision.
Think about it for a minute. The rest mass of the black hole is 1 million times greater then the rest mass of the proton. If we were dealing with a classical system, you'd have no problem with me saying that the final velocity of the black hole had to be 1 millionth of the velocity of the proton would you? If the Proton was moving at 1 km/s, you'd have no qualms with me saying that the black hole produced would be moving at 1mm/s, because the momentum is the same in both cases 1000 amu.m/s
I'm about done with this discussion, in all honesty, I'm sick of dishonest quacks that just sit around pointing fingers, and making accusations, without actually doing the calculations to back them up.
You want to see my level of understanding? http://forum.physorg.com/index.php?showtop...ndpost&p=259924 I derived this from first principles.
I at least can prove that I understand what I'm talking about. I'm not just talking through a bald patch in the top of my head.
The center of momentum frame of the system consisting of your relatively stationary oxygen atom and your cosmic ray proton is clearly not moving at 239m/s relative to earth, it's moving at a considerable fraction of the speed of light. I don't need to perform any calculations to work that out. I'm happy to work through the calculation for you if you don't believe me though.
The velocity of the COM frame (relative to earth) in your system has changed from a significant fraction of the speed of light pre-collision to 239m/s after the collision despite no external force acting on the system. Collisions between particles inside the system will not change the velocity of the COM frame relative to an inertial observer.
And once again. We come back to the point that I stated as the first step assume momentum is the same before and after the collision.
It seems to me that you're incapable of understanding what I've been saying. I've been saying "If we assume that momentum is conserved in the earths reference frame, then this must be true."
No, the velocity of the MBH relative to the earth. The ground you're standing on is looking thinner and thinner. And again, if you disagree with the 239 m/s figure, do the freaking calculations yourself, i'm sick of all this armwaving. I've already told you what I did, and where I MOIGHT have screwed it up. Frankly i'm sick to death of all of this bluster. You and Ubavonfatss, you sit there naysaying and disagreeing with everybody else, simply because you don't like it, and yet you refuse to do the calculations yourself. Frankly? I'm bored with it.
No where have I addressed the motion of the center of mass. Quote me. Oh that's right, you can't. Nowhere have I claimed that the motion of the center of mass changes.
Although, having said that, I have said that I am highly dubious of Ubavontubas claim that the Oxygen atom is unaffected by the collision.
Think about it for a minute. The rest mass of the black hole is 1 million times greater then the rest mass of the proton. If we were dealing with a classical system, you'd have no problem with me saying that the final velocity of the black hole had to be 1 millionth of the velocity of the proton would you? If the Proton was moving at 1 km/s, you'd have no qualms with me saying that the black hole produced would be moving at 1mm/s, because the momentum is the same in both cases 1000 amu.m/s
I'm about done with this discussion, in all honesty, I'm sick of dishonest quacks that just sit around pointing fingers, and making accusations, without actually doing the calculations to back them up.
You want to see my level of understanding? http://forum.physorg.com/index.php?showtop...ndpost&p=259924 I derived this from first principles.
I at least can prove that I understand what I'm talking about. I'm not just talking through a bald patch in the top of my head.
QUOTE (ubavontuba+Feb 19 2008, 05:39 AM)
Hey rpenner,
Since you're getting all chummy with Trippy now, perhaps you might be willing to tell us if his collision model (where the cosmic ray induced MBH would always be at rest with the earth) is correct?
You've ducked this question before... would you please take a stand, now?
This is interesting. All of a sudden rpenner hasn't anything to say!
Rpenner, your silence is at the very least dishonest... if not despicable.
Since you're getting all chummy with Trippy now, perhaps you might be willing to tell us if his collision model (where the cosmic ray induced MBH would always be at rest with the earth) is correct?
You've ducked this question before... would you please take a stand, now?
This is interesting. All of a sudden rpenner hasn't anything to say!
Rpenner, your silence is at the very least dishonest... if not despicable.
Lemme put it to you another way.
Cosmic ray Proton.
Total Energy = 1150 TeV
Rest Mass = 1.671 x10^-27 kg.
1150 TeV = 1.84x10^-4 J
v=0.9999996c
p=5.61x10^-16 kgm/s
p(blackhole) = p(proton)=5.61x10^-16 kgm/s
This collisions (should) produce a blackhole with a mass of 1,220,000 amu.
This will have a velocity of v=274,402 m/s
(The fact that the number is same, but is three orders of magnitude greater suggests a conversion on my part, but my point remains the same. The black holes don't go shooting off at relativistic speeds like some posters seem to think they do)
Don't believe me? Do the math yourself.
Cosmic ray Proton.
Total Energy = 1150 TeV
Rest Mass = 1.671 x10^-27 kg.
1150 TeV = 1.84x10^-4 J
v=0.9999996c
p=5.61x10^-16 kgm/s
p(blackhole) = p(proton)=5.61x10^-16 kgm/s
This collisions (should) produce a blackhole with a mass of 1,220,000 amu.
This will have a velocity of v=274,402 m/s
(The fact that the number is same, but is three orders of magnitude greater suggests a conversion on my part, but my point remains the same. The black holes don't go shooting off at relativistic speeds like some posters seem to think they do)
Don't believe me? Do the math yourself.
QUOTE (ubavontuba+Feb 21 2008, 06:10 PM)
This is interesting. All of a sudden rpenner hasn't anything to say!
Rpenner, your silence is at the very least dishonest... if not despicable.
This is funny, expecially coming from you.
You don't think it could have anything to do with the fact that I said what needs to be said?
Get over yourself.
Rpenner, your silence is at the very least dishonest... if not despicable.
This is funny, expecially coming from you.
You don't think it could have anything to do with the fact that I said what needs to be said?
Get over yourself.
And if we set up a spreadsheet, we can determine that between an energy of 699 PeV, and 699.5 PeV, the residual velocity of the resultant blackhole drops from 11,001 m/s to 10,997 m/s (the escape velocity of the earth is 11,000 m/s).
To put that in perspective, particles with energies as high as 300,000 PeV have been observed, and particles with energies on the order of 700 PeV occur on average twice per km^2/yr
To put that in perspective, particles with energies as high as 300,000 PeV have been observed, and particles with energies on the order of 700 PeV occur on average twice per km^2/yr
Oh, and while you're arguing with me over the behaviour of the center of mass, you might want to consider defining what happens to the atmospheric proton after the collision.
Surely someone who claims to understand as much as you do about physics (either of you actually) should understand the relevance of considering such a thing.
Although, I know that Ubavontuba never understood the relevance of considering the whole 'average' oxygen atom, rather then just one proton.
Surely someone who claims to understand as much as you do about physics (either of you actually) should understand the relevance of considering such a thing.
Although, I know that Ubavontuba never understood the relevance of considering the whole 'average' oxygen atom, rather then just one proton.
QUOTE (Trippy+Feb 21 2008, 06:06 PM)
Oh, and while you're arguing with me over the behaviour of the center of mass, you might want to consider defining what happens to the atmospheric proton after the collision.
Surely someone who claims to understand as much as you do about physics (either of you actually) should understand the relevance of considering such a thing.
Although, I know that Ubavontuba never understood the relevance of considering the whole 'average' oxygen atom, rather then just one proton.
..... seven protons short of an oxygen atom, that's uba' .... not to mention all the neutrons/electrons.
Surely someone who claims to understand as much as you do about physics (either of you actually) should understand the relevance of considering such a thing.
Although, I know that Ubavontuba never understood the relevance of considering the whole 'average' oxygen atom, rather then just one proton.
..... seven protons short of an oxygen atom, that's uba' .... not to mention all the neutrons/electrons.
QUOTE (Trippy+Feb 21 2008, 06:06 PM)
Oh, and while you're arguing with me over the behaviour of the center of mass, you might want to consider defining what happens to the atmospheric proton after the collision.
Surely someone who claims to understand as much as you do about physics (either of you actually) should understand the relevance of considering such a thing.
Although, I know that Ubavontuba never understood the relevance of considering the whole 'average' oxygen atom, rather then just one proton.
Trippy,
An atmospheric proton is stationary relative to the earth. Another proton moving at 10kph bumps into and sticks to it (never mind that they'd actually repel each other).
Does the collision result:
Surely someone who claims to understand as much as you do about physics (either of you actually) should understand the relevance of considering such a thing.
Although, I know that Ubavontuba never understood the relevance of considering the whole 'average' oxygen atom, rather then just one proton.
Trippy,
An atmospheric proton is stationary relative to the earth. Another proton moving at 10kph bumps into and sticks to it (never mind that they'd actually repel each other).
Does the collision result:
- a. accelerate together to 20kph?
b. proceed together at 5kph?
c. remain stationary relative to the earth?
d. turn into a black hole?
- a. accelerate to 40kph?
b. proceed onward at 10kph?
c. proceed onward at 5kph?
d. proceed onward at 2.5kph?
e. remain stationary relative to the earth?
Uba,
The velocities in your questions are too low. This leads you to conclude that the sum of the mass of the products is equal to the sum of the mass of the initial objects. The unchanging mass and the complete inelasticity of the collision means you can't use conservation of energy, or at least not the simple one that that uses kinetic energy only.
Trippy is able to keep track of all the energy because he's using conservation of energy in a fully relativistic sense, i.e. keeping track of kinetic energy and the energy in the rest mass of the objects.
The fact that the mass isn't invariant in the relativistic collisions seems to be one of your mental stumbling blocks.
The velocities in your questions are too low. This leads you to conclude that the sum of the mass of the products is equal to the sum of the mass of the initial objects. The unchanging mass and the complete inelasticity of the collision means you can't use conservation of energy, or at least not the simple one that that uses kinetic energy only.
Trippy is able to keep track of all the energy because he's using conservation of energy in a fully relativistic sense, i.e. keeping track of kinetic energy and the energy in the rest mass of the objects.
The fact that the mass isn't invariant in the relativistic collisions seems to be one of your mental stumbling blocks.
QUOTE (barakn+Feb 22 2008, 12:01 AM)
Uba,
The velocities in your questions are too low. This leads you to conclude that the sum of the mass of the products is equal to the sum of the mass of the initial objects. The unchanging mass and the complete inelasticity of the collision means you can't use conservation of energy, or at least not the simple one that that uses kinetic energy only.
Trippy is able to keep track of all the energy because he's using conservation of energy in a fully relativistic sense, i.e. keeping track of kinetic energy and the energy in the rest mass of the objects.
The fact that the mass isn't invariant in the relativistic collisions seems to be one of your mental stumbling blocks.
The velocities in your questions are too low. This leads you to conclude that the sum of the mass of the products is equal to the sum of the mass of the initial objects. The unchanging mass and the complete inelasticity of the collision means you can't use conservation of energy, or at least not the simple one that that uses kinetic energy only.
Trippy is able to keep track of all the energy because he's using conservation of energy in a fully relativistic sense, i.e. keeping track of kinetic energy and the energy in the rest mass of the objects.
The fact that the mass isn't invariant in the relativistic collisions seems to be one of your mental stumbling blocks.