QUOTE (buttershug+Oct 22 2008, 08:00 PM)
I just thought of a mind experiment to try to illustrate.
Get a piece of paper. Draw a box. Move your finger in that box.
Which frame are you not counting, the box or your finger?
both are frames of reference.
I'm sorry but I'm sure that analogy is not relevant.
It is also a little patronising, have I not demonstrated that I know what a frame of reference is? This discussion is a little beyond the definition of a frame, I thought.
Get a piece of paper. Draw a box. Move your finger in that box.
Which frame are you not counting, the box or your finger?
both are frames of reference.
I'm sorry but I'm sure that analogy is not relevant.
It is also a little patronising, have I not demonstrated that I know what a frame of reference is? This discussion is a little beyond the definition of a frame, I thought.
QUOTE (Sapo+Oct 22 2008, 08:19 PM)
You have the reasoning and the patience, sir, but do you have the club handy for the time your patience and reason run short? 
In my first discussion with buttershug, I misunderstood his stance and made an assumption; I ended up looking pretty silly!
I am willing to argue my point until he convinces me I'm wrong or 'an authority'* steps in to settle the dispute
(AN, Trout, Mr Homm, RPenner etc..)
In my first discussion with buttershug, I misunderstood his stance and made an assumption; I ended up looking pretty silly!
I am willing to argue my point until he convinces me I'm wrong or 'an authority'* steps in to settle the dispute
(AN, Trout, Mr Homm, RPenner etc..)
QUOTE (bm1957+Oct 22 2008, 07:26 PM)
Velocity being relative means that you can give any frame any velocity, so long as you choose correctly the frame with which you are measuring that velocity relative to.
This means you need two frames to put a value on velocity.
it is inertial then we cannot give it a velocity (we can't even say v=0, I think this is your fundamental error). If it is non-inertial, then we can calculate the acceleration (or equivalent gravitational field) that it is experiencing, no second frame required.
But that's what I'm saying. But Omnibus is saying he has acceleration with only one frame.
And it does not have v but it has v/t? Is that what you are saying?
There is no v but it is changing?
If you were in a box how would you know if you were moving?
acceleration feels exactly like gravity.
It is impossible to observe acceleration in one frame.
I really really don't understand when you say "It doesn't have to accelerate relative to anything except itself." Acceleration is the how fast the rate of moving relative to something else is changing.
Oh and you do need two frames to experience gravity. The observer and whatever is causing the gravity.
This means you need two frames to put a value on velocity.
it is inertial then we cannot give it a velocity (we can't even say v=0, I think this is your fundamental error). If it is non-inertial, then we can calculate the acceleration (or equivalent gravitational field) that it is experiencing, no second frame required.
But that's what I'm saying. But Omnibus is saying he has acceleration with only one frame.
And it does not have v but it has v/t? Is that what you are saying?
There is no v but it is changing?
If you were in a box how would you know if you were moving?
acceleration feels exactly like gravity.
It is impossible to observe acceleration in one frame.
I really really don't understand when you say "It doesn't have to accelerate relative to anything except itself." Acceleration is the how fast the rate of moving relative to something else is changing.
Oh and you do need two frames to experience gravity. The observer and whatever is causing the gravity.
Hi Omnibus,
I have looked at paragraph 10 of Einstein's paper and agree that it is confusing. But if you read carefully, you will see that it is just Einstein's "ponderous style" which causes the confusion. What he starts off saying is that in any reference frame (also the one with coordinates x,y,z) relative to which the speed of a particle with mass m (that is being accelerated) is small, one can use Newton's law F=ma. This is true whether the particle is moving with slow speed relative to reference frame k (using coordinates x,y,z) or with a slow speed relative to reference frame K (using Greek coordinates).
Einstein then considered such a slow-moving particle within the system K and transforms its motion into system k which is moving with a high speed v relative to K. He then derives that F=m(beta^3)a. Thus where-as F=ma is observed by an observer within K, an observer in k moving with a high speed v relative to K will observe F=m(beta^3)a. Thus, the two expressions are NOT relative to the same reference frame. What should happen is that when the relative speed between the two frames becomes small, the two expressions must approach each other. This is exactly the case; since beta then becomes unity.
I hope this is of help.
Regards
I have looked at paragraph 10 of Einstein's paper and agree that it is confusing. But if you read carefully, you will see that it is just Einstein's "ponderous style" which causes the confusion. What he starts off saying is that in any reference frame (also the one with coordinates x,y,z) relative to which the speed of a particle with mass m (that is being accelerated) is small, one can use Newton's law F=ma. This is true whether the particle is moving with slow speed relative to reference frame k (using coordinates x,y,z) or with a slow speed relative to reference frame K (using Greek coordinates).
Einstein then considered such a slow-moving particle within the system K and transforms its motion into system k which is moving with a high speed v relative to K. He then derives that F=m(beta^3)a. Thus where-as F=ma is observed by an observer within K, an observer in k moving with a high speed v relative to K will observe F=m(beta^3)a. Thus, the two expressions are NOT relative to the same reference frame. What should happen is that when the relative speed between the two frames becomes small, the two expressions must approach each other. This is exactly the case; since beta then becomes unity.
I hope this is of help.
Regards
QUOTE (johanfprins+Oct 20 2008, 06:28 PM)
The experiments which had been done by flying clocks on an aeroplane are very controversial and WRONG.
And your evidence is? They aren't controversial, they are seen as the first example of how relativity even has an effect at everyday conditions.
And your evidence is? They aren't controversial, they are seen as the first example of how relativity even has an effect at everyday conditions.
QUOTE (johanfprins+Oct 20 2008, 06:28 PM)
The fact that the clocks have to be adjusted on satellites does NOT prove that time actually ticks slower on the satellite, It only proves that from the a framework attached to Earth it "seems' as if the clock is ticking slower.
If it walks like a duck, quacks like a duck and swims like a duck and is every way indistinguishable from a duck, we call it a duck.
Your logic is akin to saying "Yes, grass may seem green to everyone but it's actually blue, we just can never measure it like that".
Your logic is akin to saying "Yes, grass may seem green to everyone but it's actually blue, we just can never measure it like that".
QUOTE (johanfprins+Oct 20 2008, 06:28 PM)
Hi "real physicist", I challenge you to an open debate in which I will prove by rigid mathematics, using the Lorentz transformations, that the clocks actually do tick at the same rate. No Wikipedia involved!
Given the mathematics of special relativity is a consistent system and not particularly complicated (compared to many other sections of physics), your claims are incorrect.
Of course if you're so right and it's so easy why don't you get your work published in a reputable journal? After all, when mainstream proponents (ie myself) offer to do such things as a public debate with unpublished ignorant cranks all it does is give the impression your view is something we feel must be publically addressed. I'm thinking of the example of MIT students debating Kent Hovind on the topic of evolution. If you're unable to get your work published in a reputable journal you have failed to show it's worth anything scientifically.
Of course if you're so right and it's so easy why don't you get your work published in a reputable journal? After all, when mainstream proponents (ie myself) offer to do such things as a public debate with unpublished ignorant cranks all it does is give the impression your view is something we feel must be publically addressed. I'm thinking of the example of MIT students debating Kent Hovind on the topic of evolution. If you're unable to get your work published in a reputable journal you have failed to show it's worth anything scientifically.
QUOTE (johanfprins+Oct 20 2008, 06:28 PM)
I am not attacking the special theory of relativity: All I am claiming is that the derivation of time-dilation does not imply that the clocks ACTUALLY tick at different rates. Furthermore, Einstein's derivation of length-contraction violates the Lorentz transformation.
You're not attacking special relativity, just its main results and derivations? 
QUOTE (johanfprins+Oct 20 2008, 06:28 PM)
I am willing to meet you face-to face "actual physicist"!!! If you are REALLY ONE!!!
Get yourself published, proving you're not just another loud mouth crank who thinks he's right (there's a dozen a month which go through here, many with opposing views of even one another, you can't all be right!). PM me when you've done that and we'll sort something out 
QUOTE (AlphaNumeric+Oct 23 2008, 08:42 AM)
If it walks like a duck, quacks like a duck and swims like a duck and is every way indistinguishable from a duck, we call it a duck. If its a swine with a Grey Afro with big snout for boot, we should call it a eeeeeeew.
There is no curvature of space. The Hubble telescopes optics cant but could pick it up if it were true.
Agree with me(eyeque) on this THERE IS NO SPACE CURVATURE!!!!
There are professors around the world been teaching BS in Einsteins fantasies
If it walks like a duck, quacks like a duck and swims like a duck and is every way indistinguishable from a duck, we call it a duck. If its a swine with a Grey Afro with big snout for boot, we should call it a eeeeeeew.
There is no curvature of space. The Hubble telescopes optics cant but could pick it up if it were true.
Agree with me(eyeque) on this THERE IS NO SPACE CURVATURE!!!!
There are professors around the world been teaching BS in Einsteins fantasies
There is no curvature of space. The Hubble telescopes optics cant but could pick it up if it were true.
Agree with me(eyeque) on this THERE IS NO SPACE CURVATURE!!!!
There are professors around the world been teaching BS in Einsteins fantasies
If it walks like a duck, quacks like a duck and swims like a duck and is every way indistinguishable from a duck, we call it a duck. If its a swine with a Grey Afro with big snout for boot, we should call it a eeeeeeew.
There is no curvature of space. The Hubble telescopes optics cant but could pick it up if it were true.
Agree with me(eyeque) on this THERE IS NO SPACE CURVATURE!!!!
There are professors around the world been teaching BS in Einsteins fantasies
Obviusly the internet is not keeping up with the quantum documentaries on cable however:
http://www.geocities.com/recycling_universe/ed_spacetime.htm
Alphanumeric, your words are laced with Einstien slober
http://www.geocities.com/recycling_universe/ed_spacetime.htm
Alphanumeric, your words are laced with Einstien slober
QUOTE (AlphaNumeric+Oct 23 2008, 08:42 AM)
Given the mathematics of special relativity is a consistent system and not particularly complicated (compared to many other sections of physics), your claims are incorrect.
Of course if you're so right and it's so easy why don't you get your work published in a reputable journal? After all, when mainstream proponents (ie myself) offer to do such things as a public debate with unpublished ignorant cranks all it does is give the impression your view is something we feel must be publically addressed.
I have tried to copy the arguments on time dilation but the equations did not come out: However, I will try an type them In:
Time-dilation:
Although Einstein claimed that he had used “the first and fourth expressions of the Lorentz-transformation” he did not take the effect of the first expression for the position-coordinate fully into account as he should have. Consider a primary event occurring at the origin x=0 of K at a time t. The coordinates of this event relative to K(/) are given by the Lorentz transformation as:
x(/)=-(beta) vt and t(/)=(beta)t
These two expressions can be combined to give that:
0=x(/)+vt(/)
x(/)=-vt(/) is the distance along the x/-axis of the origin of K from the origin of K(/).
Consider now the symmetric situation where a primary event occurs at a time t(/) at the origin x(/)=0 of K/. The coordinates of this event relative to K follows from the reverse Lorentz transformation as:
x=(beta)vt(/) and t=(beta)t(/)
These two expressions can also be combined to give that:
0=x-vt
x=vt is the distance that the origin of K/ is from the origin of K.
Both x and -x(/) are equal to the actual distance between the origins of K and K/: and must thus be identical: i.e. one must thus have for the synchronized clocks at any instant in time that:
t=t(/)
It seems to me that the mainstream proponents (like yourself) have been INCREDIBLY STUPID FOR MORE THAN 100 YEARS. this is why I cannot get it published in a "reputable journal" . The mainstream IDIOTS like yourself block it. You do not want to be exposed for what you really are.
Of course if you're so right and it's so easy why don't you get your work published in a reputable journal? After all, when mainstream proponents (ie myself) offer to do such things as a public debate with unpublished ignorant cranks all it does is give the impression your view is something we feel must be publically addressed.
I have tried to copy the arguments on time dilation but the equations did not come out: However, I will try an type them In:
Time-dilation:
Although Einstein claimed that he had used “the first and fourth expressions of the Lorentz-transformation” he did not take the effect of the first expression for the position-coordinate fully into account as he should have. Consider a primary event occurring at the origin x=0 of K at a time t. The coordinates of this event relative to K(/) are given by the Lorentz transformation as:
x(/)=-(beta) vt and t(/)=(beta)t
These two expressions can be combined to give that:
0=x(/)+vt(/)
x(/)=-vt(/) is the distance along the x/-axis of the origin of K from the origin of K(/).
Consider now the symmetric situation where a primary event occurs at a time t(/) at the origin x(/)=0 of K/. The coordinates of this event relative to K follows from the reverse Lorentz transformation as:
x=(beta)vt(/) and t=(beta)t(/)
These two expressions can also be combined to give that:
0=x-vt
x=vt is the distance that the origin of K/ is from the origin of K.
Both x and -x(/) are equal to the actual distance between the origins of K and K/: and must thus be identical: i.e. one must thus have for the synchronized clocks at any instant in time that:
t=t(/)
It seems to me that the mainstream proponents (like yourself) have been INCREDIBLY STUPID FOR MORE THAN 100 YEARS. this is why I cannot get it published in a "reputable journal" . The mainstream IDIOTS like yourself block it. You do not want to be exposed for what you really are.
QUOTE (buttershug+Oct 22 2008, 09:56 PM)
But that's what I'm saying. But Omnibus is saying he has acceleration with only one frame.
In that case I don't think I have explained myself adequately.
I guess that is what I am saying. Not quite that it does not have v, but it has no absolute v, so v cannot be defined. But dv/dt can be defined, all you need to know is the speed at time t2, relative to the speed at time t1. No speed relative to any other frame, but a relative speed, all the same.
I guess that is what I am saying. Not quite that it does not have v, but it has no absolute v, so v cannot be defined. But dv/dt can be defined, all you need to know is the speed at time t2, relative to the speed at time t1. No speed relative to any other frame, but a relative speed, all the same.
If you were in a box how would you know if you were moving?
acceleration feels exactly like gravity.
It is impossible to observe acceleration in one frame.
If your argument below that you need a frame for gravity to originate from is correct, then I concede and would have to agree with you.
Hopefully I explained myself above, but let me try an analogy:
If you are at the end of a line, but you don't know how long the line is, and the line is extended; you can never give the line a definite length, but you can still observe and measure the change in length. The same as you can observe and measure the change in velocity, but not the velocity itself.
Hopefully I explained myself above, but let me try an analogy:
If you are at the end of a line, but you don't know how long the line is, and the line is extended; you can never give the line a definite length, but you can still observe and measure the change in length. The same as you can observe and measure the change in velocity, but not the velocity itself.
Oh and you do need two frames to experience gravity. The observer and whatever is causing the gravity.
Like I said above, if you can back that up, I will have to concede and agree with you. I have never really considered the possibility that gravity may have to originate from a frame at all, I just assumed that a field could be applied to a single frame.
Hi all.
bm1957....will this do?.....The only reason someone in the lab can 'feel' acceleration due to gravity is is if the lab FLOOR 'resists' because the lab does NOT accelerate WITH that someone....and hence is NOT 'co-moving' or 'same frame' with the person in the lab 'feeling' it. And to feel inertial acceleration the ROOF/walls/floor must be accelerating and 'pressing' you against it because of YOUR inertial 'resistance'. So in either case, either you oer the lab do the 'resisting' and hence are not in the same fram at all until no more gravity/inertial acceleration force is acting on the you/lab.....and then you and the lab once again co-move/co-fall TOGETHER as a single 'merged' frame.
Have I explained it clearly enough in my usual rushed way?
I'm afraid that won't do, you seem to have a bad misunderstanding there. Let me explain:
We assume a scientist inside a closed lab, where no information can travel between the inside/outside of the lab. This is our frame.
This frame can be subjected to an acceleration, and the scientist will feel a pseudo-force in the opposite direction to the acceleration.
This frame could be subjected to a gravitational field, and the scientist will feel a force in the direction of the gravitational force.
In both situations, the scientist will be able to measure the force and tell which direction it is in, and he will infer an acceleration in units of ms^-2
However, he will never know whether this force is caused by an acceleration in the opposite direction of the force, or an applied gravitational field.
This is a layman's description of one of the principles of equivalence, the equivalence of a gravitational field and acceleration. I hope it helps to clear up my argument for you.
In that case I don't think I have explained myself adequately.
QUOTE
And it does not have v but it has v/t? Is that what you are saying?
There is no v but it is changing?
There is no v but it is changing?
I guess that is what I am saying. Not quite that it does not have v, but it has no absolute v, so v cannot be defined. But dv/dt can be defined, all you need to know is the speed at time t2, relative to the speed at time t1. No speed relative to any other frame, but a relative speed, all the same.
QUOTE (->
| QUOTE |
| And it does not have v but it has v/t? Is that what you are saying? There is no v but it is changing? |
I guess that is what I am saying. Not quite that it does not have v, but it has no absolute v, so v cannot be defined. But dv/dt can be defined, all you need to know is the speed at time t2, relative to the speed at time t1. No speed relative to any other frame, but a relative speed, all the same.
If you were in a box how would you know if you were moving?
acceleration feels exactly like gravity.
It is impossible to observe acceleration in one frame.
If your argument below that you need a frame for gravity to originate from is correct, then I concede and would have to agree with you.
QUOTE
I really really don't understand when you say "It doesn't have to accelerate relative to anything except itself." Acceleration is the how fast the rate of moving relative to something else is changing.
Hopefully I explained myself above, but let me try an analogy:
If you are at the end of a line, but you don't know how long the line is, and the line is extended; you can never give the line a definite length, but you can still observe and measure the change in length. The same as you can observe and measure the change in velocity, but not the velocity itself.
QUOTE (->
| QUOTE |
| I really really don't understand when you say "It doesn't have to accelerate relative to anything except itself." Acceleration is the how fast the rate of moving relative to something else is changing. |
Hopefully I explained myself above, but let me try an analogy:
If you are at the end of a line, but you don't know how long the line is, and the line is extended; you can never give the line a definite length, but you can still observe and measure the change in length. The same as you can observe and measure the change in velocity, but not the velocity itself.
Oh and you do need two frames to experience gravity. The observer and whatever is causing the gravity.
Like I said above, if you can back that up, I will have to concede and agree with you. I have never really considered the possibility that gravity may have to originate from a frame at all, I just assumed that a field could be applied to a single frame.
QUOTE (Fairy+Oct 23 2008, 09:41 AM)
lol thanks
Hi Fairy,
Are you unable to contribute any physics. It is boring to read lol thanks
Hi Fairy,
Are you unable to contribute any physics. It is boring to read lol thanks
QUOTE (eyeque+Oct 23 2008, 08:50 AM)
If it walks like a duck, quacks like a duck and swims like a duck and is every way indistinguishable from a duck, we call it a duck. If its a swine with a Grey Afro with big snout for boot, we should call it a eeeeeeew.
There is no curvature of space. The Hubble telescopes optics cant but could pick it up if it were true.
Agree with me(eyeque) on this THERE IS NO SPACE CURVATURE!!!!
There are professors around the world been teaching BS in Einsteins fantasies
The curvature of space was picked up long before the Hubble telescope. Einstien's following grew greatly when he was able to show it. During an eclipse a star was not where astromnomers thought it would be. It was exactly where Einstien said it would be.
and have you never heard about gravitational lenses? The Hubble telescope also uses those.
There is no curvature of space. The Hubble telescopes optics cant but could pick it up if it were true.
Agree with me(eyeque) on this THERE IS NO SPACE CURVATURE!!!!
There are professors around the world been teaching BS in Einsteins fantasies
The curvature of space was picked up long before the Hubble telescope. Einstien's following grew greatly when he was able to show it. During an eclipse a star was not where astromnomers thought it would be. It was exactly where Einstien said it would be.
and have you never heard about gravitational lenses? The Hubble telescope also uses those.
QUOTE (RealityCheck+Oct 23 2008, 03:11 AM)
Hi all.
bm1957....will this do?.....The only reason someone in the lab can 'feel' acceleration due to gravity is is if the lab FLOOR 'resists' because the lab does NOT accelerate WITH that someone....and hence is NOT 'co-moving' or 'same frame' with the person in the lab 'feeling' it. And to feel inertial acceleration the ROOF/walls/floor must be accelerating and 'pressing' you against it because of YOUR inertial 'resistance'. So in either case, either you oer the lab do the 'resisting' and hence are not in the same fram at all until no more gravity/inertial acceleration force is acting on the you/lab.....and then you and the lab once again co-move/co-fall TOGETHER as a single 'merged' frame.
Have I explained it clearly enough in my usual rushed way?
I'm afraid that won't do, you seem to have a bad misunderstanding there. Let me explain:
We assume a scientist inside a closed lab, where no information can travel between the inside/outside of the lab. This is our frame.
This frame can be subjected to an acceleration, and the scientist will feel a pseudo-force in the opposite direction to the acceleration.
This frame could be subjected to a gravitational field, and the scientist will feel a force in the direction of the gravitational force.
In both situations, the scientist will be able to measure the force and tell which direction it is in, and he will infer an acceleration in units of ms^-2
However, he will never know whether this force is caused by an acceleration in the opposite direction of the force, or an applied gravitational field.
This is a layman's description of one of the principles of equivalence, the equivalence of a gravitational field and acceleration. I hope it helps to clear up my argument for you.
QUOTE (johanfprins+Oct 23 2008, 08:48 AM)
Hi Omnibus,
I have looked at paragraph 10 of Einstein's paper and agree that it is confusing. But if you read carefully, you will see that it is just Einstein's "ponderous style" which causes the confusion. What he starts off saying is that in any reference frame (also the one with coordinates x,y,z) relative to which the speed of a particle with mass m (that is being accelerated) is small, one can use Newton's law F=ma. This is true whether the particle is moving with slow speed relative to reference frame k (using coordinates x,y,z) or with a slow speed relative to reference frame K (using Greek coordinates).
Einstein then considered such a slow-moving particle within the system K and transforms its motion into system k which is moving with a high speed v relative to K. He then derives that F=m(beta^3)a. Thus where-as F=ma is observed by an observer within K, an observer in k moving with a high speed v relative to K will observe F=m(beta^3)a. Thus, the two expressions are NOT relative to the same reference frame. What should happen is that when the relative speed between the two frames becomes small, the two expressions must approach each other. This is exactly the case; since beta then becomes unity.
I hope this is of help.
Regards
I have looked at paragraph 10 of Einstein's paper and agree that it is confusing. But if you read carefully, you will see that it is just Einstein's "ponderous style" which causes the confusion. What he starts off saying is that in any reference frame (also the one with coordinates x,y,z) relative to which the speed of a particle with mass m (that is being accelerated) is small, one can use Newton's law F=ma. This is true whether the particle is moving with slow speed relative to reference frame k (using coordinates x,y,z) or with a slow speed relative to reference frame K (using Greek coordinates).
Einstein then considered such a slow-moving particle within the system K and transforms its motion into system k which is moving with a high speed v relative to K. He then derives that F=m(beta^3)a. Thus where-as F=ma is observed by an observer within K, an observer in k moving with a high speed v relative to K will observe F=m(beta^3)a. Thus, the two expressions are NOT relative to the same reference frame. What should happen is that when the relative speed between the two frames becomes small, the two expressions must approach each other. This is exactly the case; since beta then becomes unity.
I hope this is of help.
Regards
Well I know all about gravitational lenses like our Sun bending radio frequencies to a focus to be received by radio-telescopic equipment. However I'm talking about the WHOLE universe. I stand by the documentary i watched recently. It said: It space-time was curved it would be like a triangular corn chip curved. but it isn't and they cant detect anything to suggest there is a center with the universe curved around it.
QUOTE (bm1957+Oct 23 2008, 09:55 AM)
QUOTE (bm1957+Oct 23 2008, 09:45 AM)
If your argument below that you need a frame for gravity to originate from is correct, then I concede and would have to agree with you.
Hopefully I explained myself above, but let me try an analogy:
If you are at the end of a line, but you don't know how long the line is, and the line is extended; you can never give the line a definite length, but you can still observe and measure the change in length. The same as you can observe and measure the change in velocity, but not the velocity itself.
If you have a line. Both end points are a frame of reference. You don't have to know where they are. And if the line is growing they are moving relative to each other. Even the one that is not moving.
But more to your analogy just because you can't measure it and don't know it does not mean it does not have a definite length.
About the elevator. Your frame of reference is inside the elevator. The other frame of reference is outside the elevator. It is extremely vague but it doesn't have to be precise for there to be a frame of reference.
Just because you don't know what numbers to put into a formula doesn't mean it's not valid.
And why has Omnibus stopped posting?
Hopefully I explained myself above, but let me try an analogy:
If you are at the end of a line, but you don't know how long the line is, and the line is extended; you can never give the line a definite length, but you can still observe and measure the change in length. The same as you can observe and measure the change in velocity, but not the velocity itself.
If you have a line. Both end points are a frame of reference. You don't have to know where they are. And if the line is growing they are moving relative to each other. Even the one that is not moving.
But more to your analogy just because you can't measure it and don't know it does not mean it does not have a definite length.
About the elevator. Your frame of reference is inside the elevator. The other frame of reference is outside the elevator. It is extremely vague but it doesn't have to be precise for there to be a frame of reference.
Just because you don't know what numbers to put into a formula doesn't mean it's not valid.
And why has Omnibus stopped posting?
QUOTE (bm1957+Oct 23 2008, 09:52 AM)
I'm afraid that won't do, you seem to have a bad misunderstanding there. Let me explain:
We assume a scientist inside a closed lab, where no information can travel between the inside/outside of the lab. This is our frame.
inside and outside are two frames.
We assume a scientist inside a closed lab, where no information can travel between the inside/outside of the lab. This is our frame.
inside and outside are two frames.
QUOTE (buttershug+Oct 23 2008, 11:08 AM)
If you have a line. Both end points are a frame of reference. You don't have to know where they are. And if the line is growing they are moving relative to each other. Even the one that is not moving.
But more to your analogy just because you can't measure it and don't know it does not mean it does not have a definite length.
Sorry, you've taken my analogy too far. I was just trying to demonstrate that a value (in our case, v) can have a definite dv without ever having a definite v.
I appreciate that.
Can I clarify your argument... that a non-inertial frame cannot exist in isolation? Would it require an inertial frame to be acceptable? Or would be 2 co-existing non-inertial frames be acceptable? Does being able to measure the acceleration/gravity from within the frame not have any relevance to whether a frame can accelerate background-independently?
I appreciate that.
Can I clarify your argument... that a non-inertial frame cannot exist in isolation? Would it require an inertial frame to be acceptable? Or would be 2 co-existing non-inertial frames be acceptable? Does being able to measure the acceleration/gravity from within the frame not have any relevance to whether a frame can accelerate background-independently?
Just because you don't know what numbers to put into a formula doesn't mean it's not valid.
What formula are you talking about now? I haven't claimed that any formula is invalid, regardless of whether or not you can put the numbers in.
Was that an accidental strawman or have implied something I didn't realise?
My guess... because he is a troll and demonstrably incorrect. But he did provoke an interesting conversation!
But more to your analogy just because you can't measure it and don't know it does not mean it does not have a definite length.
Sorry, you've taken my analogy too far. I was just trying to demonstrate that a value (in our case, v) can have a definite dv without ever having a definite v.
QUOTE
About the elevator. Your frame of reference is inside the elevator. The other frame of reference is outside the elevator. It is extremely vague but it doesn't have to be precise for there to be a frame of reference.
I appreciate that.
Can I clarify your argument... that a non-inertial frame cannot exist in isolation? Would it require an inertial frame to be acceptable? Or would be 2 co-existing non-inertial frames be acceptable? Does being able to measure the acceleration/gravity from within the frame not have any relevance to whether a frame can accelerate background-independently?
QUOTE (->
| QUOTE |
| About the elevator. Your frame of reference is inside the elevator. The other frame of reference is outside the elevator. It is extremely vague but it doesn't have to be precise for there to be a frame of reference. |
I appreciate that.
Can I clarify your argument... that a non-inertial frame cannot exist in isolation? Would it require an inertial frame to be acceptable? Or would be 2 co-existing non-inertial frames be acceptable? Does being able to measure the acceleration/gravity from within the frame not have any relevance to whether a frame can accelerate background-independently?
Just because you don't know what numbers to put into a formula doesn't mean it's not valid.
What formula are you talking about now? I haven't claimed that any formula is invalid, regardless of whether or not you can put the numbers in.
Was that an accidental strawman or have implied something I didn't realise?
QUOTE
And why has Omnibus stopped posting?
My guess... because he is a troll and demonstrably incorrect. But he did provoke an interesting conversation!
The good thing is that everything that physics said by now is not in the universal theory. Thats really a good thing.
QUOTE (bm1957+Oct 23 2008, 10:23 AM)
Sorry, you've taken my analogy too far. I was just trying to demonstrate that a value (in our case, v) can have a definite dv without ever having a definite v.
I appreciate that.
Can I clarify your argument... that a non-inertial frame cannot exist in isolation? Would it require an inertial frame to be acceptable? Or would be 2 co-existing non-inertial frames be acceptable? Does being able to measure the acceleration/gravity from within the frame not have any relevance to whether a frame can accelerate background-independently?
What formula are you talking about now? I haven't claimed that any formula is invalid, regardless of whether or not you can put the numbers in.
Was that an accidental strawman or have implied something I didn't realise?
My guess... because he is a troll and demonstrably incorrect. But he did provoke an interesting conversation!
It doesn't matter if the v is definite or not. You need v to have a and to have v you need two frames, ergo you need two frames to have a. It doesnt' matter if they are definite or not. It doesn't matter if you know them or not.
And I was the one that said the formula was was invalid for a single frame. The OP gave two formulas and said they contradicted each other in a single frame. Later I said that for a single frame the "a" in F=ma would equal zero. I say you can't have motion in a single frame.
And there is no "background". I think that is the part you don't understand. Any "background" you would put there would be another frame.
That's what I meant when I said it's called the theory of relativity for a reason. Everything is relative to everything else and there is no "background".
I appreciate that.
Can I clarify your argument... that a non-inertial frame cannot exist in isolation? Would it require an inertial frame to be acceptable? Or would be 2 co-existing non-inertial frames be acceptable? Does being able to measure the acceleration/gravity from within the frame not have any relevance to whether a frame can accelerate background-independently?
What formula are you talking about now? I haven't claimed that any formula is invalid, regardless of whether or not you can put the numbers in.
Was that an accidental strawman or have implied something I didn't realise?
My guess... because he is a troll and demonstrably incorrect. But he did provoke an interesting conversation!
It doesn't matter if the v is definite or not. You need v to have a and to have v you need two frames, ergo you need two frames to have a. It doesnt' matter if they are definite or not. It doesn't matter if you know them or not.
And I was the one that said the formula was was invalid for a single frame. The OP gave two formulas and said they contradicted each other in a single frame. Later I said that for a single frame the "a" in F=ma would equal zero. I say you can't have motion in a single frame.
And there is no "background". I think that is the part you don't understand. Any "background" you would put there would be another frame.
That's what I meant when I said it's called the theory of relativity for a reason. Everything is relative to everything else and there is no "background".
QUOTE (buttershug+Oct 23 2008, 12:56 PM)
It doesn't matter if the v is definite or not. You need v to have a and to have v you need two frames, ergo you need two frames to have a. It doesnt' matter if they are definite or not. It doesn't matter if you know them or not.
That seems to be logically consistent.
Is this in reply to me asking which formula you were talking about? Because I don't see how your argument about a formula being valid even if you can't put a value on the variables relates to it. I agree that a formula should be valid regardless of whether the variables can be given a value, so your previous argument about that was a strawman, I'm assuming you didn't intend it to be.
So we are back to, "Can a non-inertial frame exist in isolation?". If you can provide reference that it can't, I concede and agree you are right. If you can provide reference for gravity having to originate from a 2nd frame, I concede and agree you are right. Can you back either of these up? In fact, providing reference that applying any force requires a 2nd frame from which the force originates will imply that for an isolated frame, F=0 and a=0 in F=ma
Come on, that's 3 ways for you to shut me up. I will be happy to concede and will have learnt something if you can provide reference or concurrence from an authority (see previous post) on any of these points.
Is this in reply to me asking which formula you were talking about? Because I don't see how your argument about a formula being valid even if you can't put a value on the variables relates to it. I agree that a formula should be valid regardless of whether the variables can be given a value, so your previous argument about that was a strawman, I'm assuming you didn't intend it to be.
So we are back to, "Can a non-inertial frame exist in isolation?". If you can provide reference that it can't, I concede and agree you are right. If you can provide reference for gravity having to originate from a 2nd frame, I concede and agree you are right. Can you back either of these up? In fact, providing reference that applying any force requires a 2nd frame from which the force originates will imply that for an isolated frame, F=0 and a=0 in F=ma
Come on, that's 3 ways for you to shut me up. I will be happy to concede and will have learnt something if you can provide reference or concurrence from an authority (see previous post) on any of these points.
And there is no "background". I think that is the part you don't understand. Any "background" you would put there would be another frame.
That's why I said background-independent. If a frame cannot accelerate background-independently, then the background is required and therefore a 2nd frame is required, you win. (By background, I mean any 2nd frame to use as a reference.)
That's a 4th way, provide reference for the inability of background-independent acceleration. My stance insists that it is possible, if it is not, I am wrong.
I hope that I have demonstrated an understanding of why it is called relativity already, and an understanding of what a 'background' infers (a 2nd frame).
That seems to be logically consistent.
QUOTE
And I was the one that said the formula was was invalid for a single frame. The OP gave two formulas and said they contradicted each other in a single frame. Later I said that for a single frame the "a" in F=ma would equal zero. I say you can't have motion in a single frame.
Is this in reply to me asking which formula you were talking about? Because I don't see how your argument about a formula being valid even if you can't put a value on the variables relates to it. I agree that a formula should be valid regardless of whether the variables can be given a value, so your previous argument about that was a strawman, I'm assuming you didn't intend it to be.
So we are back to, "Can a non-inertial frame exist in isolation?". If you can provide reference that it can't, I concede and agree you are right. If you can provide reference for gravity having to originate from a 2nd frame, I concede and agree you are right. Can you back either of these up? In fact, providing reference that applying any force requires a 2nd frame from which the force originates will imply that for an isolated frame, F=0 and a=0 in F=ma
Come on, that's 3 ways for you to shut me up. I will be happy to concede and will have learnt something if you can provide reference or concurrence from an authority (see previous post) on any of these points.
QUOTE (->
| QUOTE |
| And I was the one that said the formula was was invalid for a single frame. The OP gave two formulas and said they contradicted each other in a single frame. Later I said that for a single frame the "a" in F=ma would equal zero. I say you can't have motion in a single frame. |
Is this in reply to me asking which formula you were talking about? Because I don't see how your argument about a formula being valid even if you can't put a value on the variables relates to it. I agree that a formula should be valid regardless of whether the variables can be given a value, so your previous argument about that was a strawman, I'm assuming you didn't intend it to be.
So we are back to, "Can a non-inertial frame exist in isolation?". If you can provide reference that it can't, I concede and agree you are right. If you can provide reference for gravity having to originate from a 2nd frame, I concede and agree you are right. Can you back either of these up? In fact, providing reference that applying any force requires a 2nd frame from which the force originates will imply that for an isolated frame, F=0 and a=0 in F=ma
Come on, that's 3 ways for you to shut me up. I will be happy to concede and will have learnt something if you can provide reference or concurrence from an authority (see previous post) on any of these points.
And there is no "background". I think that is the part you don't understand. Any "background" you would put there would be another frame.
That's why I said background-independent. If a frame cannot accelerate background-independently, then the background is required and therefore a 2nd frame is required, you win. (By background, I mean any 2nd frame to use as a reference.)
That's a 4th way, provide reference for the inability of background-independent acceleration. My stance insists that it is possible, if it is not, I am wrong.
QUOTE
That's what I meant when I said it's called the theory of relativity for a reason. Everything is relative to everything else and there is no "background".
I hope that I have demonstrated an understanding of why it is called relativity already, and an understanding of what a 'background' infers (a 2nd frame).
johanfprins, Newton’s second law in K is F = ma for any velocity v, not just for small v, correct? However, F = ma differs from F=m(beta^3)a proposed by Einstein to be the law in K for the same particle. This crucial contradiction can only be resolved by having beta = 1, that is when “theory” of relativity becomes invalid.
QUOTE (Omnibus+Oct 23 2008, 12:38 PM)
johanfprins, Newton’s second law in K is F = ma for any velocity v, not just for small v, correct? However, F = ma differs from F=m(beta^3)a proposed by Einstein to be the law in K for the same particle. This crucial contradiction can only be resolved by having beta = 1, that is when “theory” of relativity becomes invalid.
But Newton was wrong.
F=ma is only 100% accurate when a=0.
It's not even accurate enough for the GPS system to work properly.
But Newton was wrong.
F=ma is only 100% accurate when a=0.
It's not even accurate enough for the GPS system to work properly.
QUOTE (buttershug+Oct 23 2008, 12:48 PM)
But Newton was wrong.
F=ma is only 100% accurate when a=0.
It's not even accurate enough for the GPS system to work properly.
You gotta be kidding me.
F=ma is only 100% accurate when a=0.
It's not even accurate enough for the GPS system to work properly.
You gotta be kidding me.
QUOTE (buttershug+Oct 23 2008, 01:48 PM)
But Newton was wrong.
F=ma is only 100% accurate when a=0.
Also correct for any observer which is stationary with respect to the 'm' as I understand. i.e. has a relative velocity of 0. Is that correct?
But essentially Omnibus, bh is right, the k is there to correct the perceived mass when you don't have the simple situation of low speed and/or stationary observer.
F=ma is only an approximation which is valid in most everyday situations. F=m(beta^3)a approaches F=ma as beta approaches 1. beta is close enough to 1 in everyday life that F=ma looks pretty good, but it is still only an approximation.
F=ma is only 100% accurate when a=0.
Also correct for any observer which is stationary with respect to the 'm' as I understand. i.e. has a relative velocity of 0. Is that correct?
But essentially Omnibus, bh is right, the k is there to correct the perceived mass when you don't have the simple situation of low speed and/or stationary observer.
F=ma is only an approximation which is valid in most everyday situations. F=m(beta^3)a approaches F=ma as beta approaches 1. beta is close enough to 1 in everyday life that F=ma looks pretty good, but it is still only an approximation.
BM1957
Can you have a gravity source without a position for that source? I don't mean can you have a gravity source without knowing it's location. I"m asking can a gravity source exist without having a location?
If it does have a location then that is a frame of reference.
I think "background-independent" is meaningless in relativity. I think it's like saying "unicorn idependent".
You have explained how you can have acceleration without knowing velocity but haven't explained how you can have accelleration without velocity. And if you have velocity then you have change in position. If you have change in position then you have more than one frame of reference. The starting position and the new position at least.
Can you have a gravity source without a position for that source? I don't mean can you have a gravity source without knowing it's location. I"m asking can a gravity source exist without having a location?
If it does have a location then that is a frame of reference.
I think "background-independent" is meaningless in relativity. I think it's like saying "unicorn idependent".
You have explained how you can have acceleration without knowing velocity but haven't explained how you can have accelleration without velocity. And if you have velocity then you have change in position. If you have change in position then you have more than one frame of reference. The starting position and the new position at least.
QUOTE (buttershug+Oct 23 2008, 02:21 PM)
BM1957
Can you have a gravity source without a position for that source? I don't mean can you have a gravity source without knowing it's location. I"m asking can a gravity source exist without having a location?
If it does have a location then that is a frame of reference.
I don't care about a gravity source, I am talking about a gravitational field. I am not interested in what causes that field, it is irrelevant.
Do you have any source or reference for your argument? Because I don't think reasoning is going to convince me.
[EDIT][SPECULATION]In GR, gravity is considered as curved space. Empty space is locally flat but unless the universe is steady state, then it is not flat at larger scales. So yes, gravity may arguably exist without requiring a source with an associated position.[/SPECULATION][/EDIT]
And I defined for you what I meant by a background. Any 2nd frame of reference. Maybe background is the wrong word, but I have explained what I mean clearly enough, you are trying to prove me wrong with semantics now, that won't work either.
And I defined for you what I meant by a background. Any 2nd frame of reference. Maybe background is the wrong word, but I have explained what I mean clearly enough, you are trying to prove me wrong with semantics now, that won't work either.
You have explained how you can have acceleration without knowing velocity but haven't explained how you can have accelleration without velocity. And if you have velocity then you have change in position. If you have change in position then you have more than one frame of reference. The starting position and the new position at least.
I don't think that acceleration without velocity is required.
I think that a single frame cannot be given any particular velocity, because there is nothing to relate that velocity to. That includes v=0. You cannot have v=0 for an isolated frame, because what are you suggesting it is at rest relative to?
So, my argument is that it is valid to give an isolated frame a velocity v=v(1), with the acknowledgement that a value can never be put on v(1), neither an absolute value (obviously) nor a relative value. If the frame is non-inertial, then at time t(2), it will have a velocity v(2). Again, we may never know what v(2) is, because we have nothing to relate it to. But, we can quite simply find delta(v), or v(2)-v(1) from within the frame, and we can say how the velocity has changed. No other frames or observers are required to make the calculations and I think that is a key point.
You still have a number of options for sources/references which will force me to concede my position. Do you have any?
Can you have a gravity source without a position for that source? I don't mean can you have a gravity source without knowing it's location. I"m asking can a gravity source exist without having a location?
If it does have a location then that is a frame of reference.
I don't care about a gravity source, I am talking about a gravitational field. I am not interested in what causes that field, it is irrelevant.
Do you have any source or reference for your argument? Because I don't think reasoning is going to convince me.
[EDIT][SPECULATION]In GR, gravity is considered as curved space. Empty space is locally flat but unless the universe is steady state, then it is not flat at larger scales. So yes, gravity may arguably exist without requiring a source with an associated position.[/SPECULATION][/EDIT]
QUOTE
I think "background-independent" is meaningless in relativity. I think it's like saying "unicorn idependent".
And I defined for you what I meant by a background. Any 2nd frame of reference. Maybe background is the wrong word, but I have explained what I mean clearly enough, you are trying to prove me wrong with semantics now, that won't work either.
QUOTE (->
| QUOTE |
| I think "background-independent" is meaningless in relativity. I think it's like saying "unicorn idependent". |
And I defined for you what I meant by a background. Any 2nd frame of reference. Maybe background is the wrong word, but I have explained what I mean clearly enough, you are trying to prove me wrong with semantics now, that won't work either.
You have explained how you can have acceleration without knowing velocity but haven't explained how you can have accelleration without velocity. And if you have velocity then you have change in position. If you have change in position then you have more than one frame of reference. The starting position and the new position at least.
I don't think that acceleration without velocity is required.
I think that a single frame cannot be given any particular velocity, because there is nothing to relate that velocity to. That includes v=0. You cannot have v=0 for an isolated frame, because what are you suggesting it is at rest relative to?
So, my argument is that it is valid to give an isolated frame a velocity v=v(1), with the acknowledgement that a value can never be put on v(1), neither an absolute value (obviously) nor a relative value. If the frame is non-inertial, then at time t(2), it will have a velocity v(2). Again, we may never know what v(2) is, because we have nothing to relate it to. But, we can quite simply find delta(v), or v(2)-v(1) from within the frame, and we can say how the velocity has changed. No other frames or observers are required to make the calculations and I think that is a key point.
You still have a number of options for sources/references which will force me to concede my position. Do you have any?
QUOTE (Omnibus+Oct 23 2008, 12:38 PM)
johanfprins, Newton’s second law in K is F = ma for any velocity v, not just for small v, correct? However, F = ma differs from F=m(beta^3)a proposed by Einstein to be the law in K for the same particle. This crucial contradiction can only be resolved by having beta = 1, that is when “theory” of relativity becomes invalid.
No, Newton's law is only valid when the particle being accelerated is moving at low speed relative to the observer: i.e. when Galileo's transformations are valid. Only in this case acceleration is the same relative to all moving reference frames. Use Galileo's transformation and confirm it (it is easy). And in this case you find a kinetic energy equal to (1/2)mv^2. Relativistic corrections are not required. At high speed, differentiation of the speed requires Lorentz's transformations. It is easy to show that the acceleration is then not the same in all reference frames. Einstein took the special case of small acceleration at low speed relative to one observer and then deduced by using the Lorentz transformations what an observer will observe when a particle being slowly accelerated moves fast relative to this observer. It led to the correct alternative formula for kinetic energy which has been confirmed over and over experimentally.
It has all along been a problem to define acceleration within the framework of special relativity: It is believed that this has been corrected in the general theory of relativity by the principle of equivalence.
No, Newton's law is only valid when the particle being accelerated is moving at low speed relative to the observer: i.e. when Galileo's transformations are valid. Only in this case acceleration is the same relative to all moving reference frames. Use Galileo's transformation and confirm it (it is easy). And in this case you find a kinetic energy equal to (1/2)mv^2. Relativistic corrections are not required. At high speed, differentiation of the speed requires Lorentz's transformations. It is easy to show that the acceleration is then not the same in all reference frames. Einstein took the special case of small acceleration at low speed relative to one observer and then deduced by using the Lorentz transformations what an observer will observe when a particle being slowly accelerated moves fast relative to this observer. It led to the correct alternative formula for kinetic energy which has been confirmed over and over experimentally.
It has all along been a problem to define acceleration within the framework of special relativity: It is believed that this has been corrected in the general theory of relativity by the principle of equivalence.
That is incorrect. In K the valid mechanics is Newtonian mechanics which Einstein also doesn’t deny (cf. §1 of the 1905 paper).
Besides, if we accept, just for the sake of the discussion, that Newton’s law of motion in K has a new form, namely, F = beta^3ma, then the first postulate requires this to be represented in k as F’ = beta^3ma’. As can be seen in §10 of the 1905 paper it is not the form F’ = ma’ Einstein claims the law has in k. To avoid this contradiction in k the value of beta must be 1, that is, Lorentz transformations must not be valid.
Besides, if we accept, just for the sake of the discussion, that Newton’s law of motion in K has a new form, namely, F = beta^3ma, then the first postulate requires this to be represented in k as F’ = beta^3ma’. As can be seen in §10 of the 1905 paper it is not the form F’ = ma’ Einstein claims the law has in k. To avoid this contradiction in k the value of beta must be 1, that is, Lorentz transformations must not be valid.
QUOTE (Omnibus+Oct 23 2008, 02:22 PM)
. As can be seen in §10 of the 1905 paper it is not the form F’ = ma’ Einstein claims the law has in k.
He claims this is close enough. That is all he is claiming for it.
i.e. F' is close enough to ma' to not be bothered with the difference.
But says F'=Ma' is close enough.
He claims this is close enough. That is all he is claiming for it.
i.e. F' is close enough to ma' to not be bothered with the difference.
But says F'=Ma' is close enough.
QUOTE (Omnibus+Oct 23 2008, 02:22 PM)
That is incorrect. In K the valid mechanics is Newtonian mechanics which Einstein also doesn’t deny (cf. §1 of the 1905 paper).
Einstein, although ponderously, clearly stated that he is transforming from a reference frame within which Newton's laws are valid because the body being accelerated is moving slowly relative to this reference frame. In other words when making a relativistic transformation from the framework in which the moving body finds itself momentarily stationary, to the reference frame relative to which it is SLOWLY moving (or vice versa) the Galilean transformations suffice. Therefore Newton's law F=ma is valid. He then transformed this movement from the reference frame of the first observer (relative to which the body being accelerated is moving SLOWLY) to the reference frame of another observer which is moving at a high speed relative to the reference frame of the first observer; CORRECTLY USING THE lORENTZ TRANSFORMATIONS. Thus F=ma is valid in the one reference frame while F=beta^3ma is what is observed within a reference frame moving at high speed relative to the first reference frame.
Einstein, although ponderously, clearly stated that he is transforming from a reference frame within which Newton's laws are valid because the body being accelerated is moving slowly relative to this reference frame. In other words when making a relativistic transformation from the framework in which the moving body finds itself momentarily stationary, to the reference frame relative to which it is SLOWLY moving (or vice versa) the Galilean transformations suffice. Therefore Newton's law F=ma is valid. He then transformed this movement from the reference frame of the first observer (relative to which the body being accelerated is moving SLOWLY) to the reference frame of another observer which is moving at a high speed relative to the reference frame of the first observer; CORRECTLY USING THE lORENTZ TRANSFORMATIONS. Thus F=ma is valid in the one reference frame while F=beta^3ma is what is observed within a reference frame moving at high speed relative to the first reference frame.
QUOTE (bm1957+Oct 23 2008, 01:50 PM)
You still have a number of options for sources/references which will force me to concede my position. Do you have any?
You first.
All you have to do is convince me you can have a gravity field outside the elevator without an "outside the elevator".
That you can have accelleration without v, but not knowing what the v is does not mean there is no v. You can have velocity without knowing what it is.
That you can have movement without changing position.
I trust we both agree that F=ma when a = 0?
I mean 0=m*0.
You first.
All you have to do is convince me you can have a gravity field outside the elevator without an "outside the elevator".
That you can have accelleration without v, but not knowing what the v is does not mean there is no v. You can have velocity without knowing what it is.
That you can have movement without changing position.
I trust we both agree that F=ma when a = 0?
I mean 0=m*0.
QUOTE (johanfprins+Oct 23 2008, 02:48 PM)
Therefore Newton's law F=ma is valid.
I want to make sure I understand. When you say Newton's law is valid. You mean "Acceptable", "it works", or "it's good enough to be used". Or even "it's the proper equation to use at low speeds". Am I correct in saying that?
By valid, you don't mean "exactly" or "precisely", correct?
BTW how fast does something have to move for us to be able to measure a deviation from F=ma?
I want to make sure I understand. When you say Newton's law is valid. You mean "Acceptable", "it works", or "it's good enough to be used". Or even "it's the proper equation to use at low speeds". Am I correct in saying that?
By valid, you don't mean "exactly" or "precisely", correct?
BTW how fast does something have to move for us to be able to measure a deviation from F=ma?
QUOTE (buttershug+Oct 23 2008, 03:51 PM)
You first.
I can't. That's why I'm open to the possibility that I am wrong. I'm aware what my stance implies but your arguments against my stance are less than convincing.
I never said anything about an 'outside the elevator'. Let's assume a gravitational field that is only present where the elevator exists. Is that a fallacious assumption? I don't know. Is it necessary for the gravitational field to originate 'somewhere'? I don't know. If it is necessary, then I am wrong.
I never said anything about an 'outside the elevator'. Let's assume a gravitational field that is only present where the elevator exists. Is that a fallacious assumption? I don't know. Is it necessary for the gravitational field to originate 'somewhere'? I don't know. If it is necessary, then I am wrong.
That you can have accelleration without v, but not knowing what the v is does not mean there is no v. You can have velocity without knowing what it is.
That's another strawman, I never said there was no v, you did. I said that you could put no value on v. If the velocity of an isolated frame is restricted to being 0, then you are right, and I will concede. I believe that putting that restriction on requires a 2nd frame to say what the first frame is stationary with respect to.
That's another strawman, and very vague. I never argued that you could have movement without changing position. Without defining what you mean by movement and defining how you are measuring position, the statement seems absurd to me.
That's another strawman, and very vague. I never argued that you could have movement without changing position. Without defining what you mean by movement and defining how you are measuring position, the statement seems absurd to me.
I trust we both agree that F=ma when a = 0?
I mean 0=m*0.
I don't understand what you mean.
I agree that a=0 implies F=0
I agree that F=0 implies a=0
Is that what you mean?
That's another strawman, and very vague. I never argued that you could have movement without changing position. Without defining what you mean by movement and defining how you are measuring position, the statement seems absurd to me.
Ok then, I say if you change position, you have at least two frames. The initial location and the final location.
F’ = ma’
in k as
F = beta^3ma
where beta = 1/(sqr(1 – v^2/c^2))
This is wrong. In SR force transforms to F = m gamma^3 (v.a)v - m gamma a
At no point is mass ever assumed to vary, since it does not as I said right at the start of this thread. How can anyone trust you when you can't get the formula you're using to try and prove your point right. I know RPenner has corrected you as well
Therefore, one single electron at one single state in the single system K ends up, due to the first postulate on the one hand and the Lorentz transformations on the other, with two different values of the mass, namely m and beta^3m. This is only possible iff beta = 1, that is, if the Lorentz transformations are invalid.
That you can make this conclusion shows you are in no position at all to understand scientific papers. F = ma is the low velocity limit of the relativistic formula, as any undergrad who's done a course on SR in their first year could show. There is no contradiction since you are comparing a system at v = 0 and transformed to some high v. Of course the equations change. Thats the point - the universe is not Newtonian.
Hi bm1957.
I think you missed the point, and that's why you suggest I have "a bad misunderstanding there". :-)
On re-reading I see what you were trying to say, but don't think it changes much. Remove the '2nd frames' that you suggest exist and you still have equivalence; it is just not as obvious, because both instances will feel like free-fall (i.e., nothing).
I can't. That's why I'm open to the possibility that I am wrong. I'm aware what my stance implies but your arguments against my stance are less than convincing.
QUOTE
All you have to do is convince me you can have a gravity field outside the elevator without an "outside the elevator".
I never said anything about an 'outside the elevator'. Let's assume a gravitational field that is only present where the elevator exists. Is that a fallacious assumption? I don't know. Is it necessary for the gravitational field to originate 'somewhere'? I don't know. If it is necessary, then I am wrong.
QUOTE (->
| QUOTE |
| All you have to do is convince me you can have a gravity field outside the elevator without an "outside the elevator". |
I never said anything about an 'outside the elevator'. Let's assume a gravitational field that is only present where the elevator exists. Is that a fallacious assumption? I don't know. Is it necessary for the gravitational field to originate 'somewhere'? I don't know. If it is necessary, then I am wrong.
That you can have accelleration without v, but not knowing what the v is does not mean there is no v. You can have velocity without knowing what it is.
That's another strawman, I never said there was no v, you did. I said that you could put no value on v. If the velocity of an isolated frame is restricted to being 0, then you are right, and I will concede. I believe that putting that restriction on requires a 2nd frame to say what the first frame is stationary with respect to.
QUOTE
That you can have movement without changing position.
That's another strawman, and very vague. I never argued that you could have movement without changing position. Without defining what you mean by movement and defining how you are measuring position, the statement seems absurd to me.
QUOTE (->
| QUOTE |
| That you can have movement without changing position. |
That's another strawman, and very vague. I never argued that you could have movement without changing position. Without defining what you mean by movement and defining how you are measuring position, the statement seems absurd to me.
I trust we both agree that F=ma when a = 0?
I mean 0=m*0.
I don't understand what you mean.
I agree that a=0 implies F=0
I agree that F=0 implies a=0
Is that what you mean?
QUOTE (buttershug+Oct 23 2008, 02:57 PM)
I want to make sure I understand. When you say Newton's law is valid. You mean "Acceptable", "it works", or "it's good enough to be used". Or even "it's the proper equation to use at low speeds". Am I correct in saying that?
By valid, you don't mean "exactly" or "precisely", correct?
BTW how fast does something have to move for us to be able to measure a deviation from F=ma?
By stating that Newton's law is valid, I should have said as originally derived by assuming a constant mass m: i.e. within the regime where acceleration causes purely an increase in speed without an increase in inertia. At very high speeds acceleration cannot keep on just increasing the speed since there is an upper speed limit (the speed of light c). The increase in energy done by the force then increases the mass-inertia. Newton's law can still be considered as still valid but with mass not being a constant anymore.
It is, however, a large speed-interval over which one can take m to be "accurately" a constant. By expanding the relativistic equation for kinetic energy one will be safe to take m as constant up to about 0.75 to 0.6 time light speed. Obviously, if one uses an infinite number of decimal places one can state that Newtons' law for constant mass is only approximately constant.
By valid, you don't mean "exactly" or "precisely", correct?
BTW how fast does something have to move for us to be able to measure a deviation from F=ma?
By stating that Newton's law is valid, I should have said as originally derived by assuming a constant mass m: i.e. within the regime where acceleration causes purely an increase in speed without an increase in inertia. At very high speeds acceleration cannot keep on just increasing the speed since there is an upper speed limit (the speed of light c). The increase in energy done by the force then increases the mass-inertia. Newton's law can still be considered as still valid but with mass not being a constant anymore.
It is, however, a large speed-interval over which one can take m to be "accurately" a constant. By expanding the relativistic equation for kinetic energy one will be safe to take m as constant up to about 0.75 to 0.6 time light speed. Obviously, if one uses an infinite number of decimal places one can state that Newtons' law for constant mass is only approximately constant.
QUOTE (bm1957+Oct 23 2008, 03:03 PM)
That's another strawman, and very vague. I never argued that you could have movement without changing position. Without defining what you mean by movement and defining how you are measuring position, the statement seems absurd to me.
Ok then, I say if you change position, you have at least two frames. The initial location and the final location.
In the stationary system K the valid mechanics is Newtonian for any velocity. That must not be disputed let alone that Einstein himself accepts it too. Thus, the second Newton's law in K is exactly F = ma and in no way is F = m(beta^3)a, unless beta = 1 and Lorentz transformations are invalid.
QUOTE (Omnibus+Oct 23 2008, 03:15 PM)
In the stationary system K the valid mechanics is Newtonian for any velocity.
That is where you are wrong: Consider kinetic energy. When a body moves at a low velocity its kinetic energy is (1/2)mv^2. At a high velocity it is (M-m)c^2 where M-m is the increase in inertia at high velocities. At low velocities (M-m)c^2 becomes (1/2)mv^2. This fact has been experimentally confirmed. So how on Earth can you state that Newtonian mechanics is valid for any velocity?
That is where you are wrong: Consider kinetic energy. When a body moves at a low velocity its kinetic energy is (1/2)mv^2. At a high velocity it is (M-m)c^2 where M-m is the increase in inertia at high velocities. At low velocities (M-m)c^2 becomes (1/2)mv^2. This fact has been experimentally confirmed. So how on Earth can you state that Newtonian mechanics is valid for any velocity?
This isn't my invention but is Einstein premise on which he builds his "theory" and then tries to overthrow the initial premise. The fact that there are experimental data demonstrating whatever is not a proof that Einstein's "theory" derives it.
As a matter of fact, assuming the velocity dependence of m (that is, assuming exactly the thing that has to be proven) and then “proving” it, thus committing the crucial error in reasoning called “petitio principii” (begging the question; circular argument), is the error in another failed exercise by Einstein to derive mass-energy relationship—that in Einstein A., “Ist die Tragheit eines Korpers von seinem Energiegehalt abhangig, Ann.Phys., 17, 1905.
Note again, the theory must derive what is known experimentally and not assume it first as initial condition and then happily “derive” it.
Note again, the theory must derive what is known experimentally and not assume it first as initial condition and then happily “derive” it.
QUOTE (Omnibus+Oct 23 2008, 03:53 PM)
As a matter of fact, assuming the velocity dependence of m
Note again, the theory must derive what is known experimentally and not assume it first as initial condition and then happily “derive” it.
I have NOT done what you are accusing me of. Whether Einstein's special relativity is correct or not, I pointed out that it is experimentally known that Newton's mechanics is NOT valid at all speeds. Your statement was that "Newton's mechanics IS valid at ALL speeds". Experiment is the only arbitrator; is it not?
Note again, the theory must derive what is known experimentally and not assume it first as initial condition and then happily “derive” it.
I have NOT done what you are accusing me of. Whether Einstein's special relativity is correct or not, I pointed out that it is experimentally known that Newton's mechanics is NOT valid at all speeds. Your statement was that "Newton's mechanics IS valid at ALL speeds". Experiment is the only arbitrator; is it not?
You're doing it right now. Don't tell me what experiment shows. Show me a theory that can derive what experiment shows.
QUOTE (Omnibus+Oct 23 2008, 03:53 PM)
Note again, the theory must derive what is known experimentally and not assume it first as initial condition and then happily “derive” it.
But the star was in the sky where Einstien's theory said it would be not where Newton's theory said it would be.
Doesn't that give at least some credence to Einstien's theory?
But the star was in the sky where Einstien's theory said it would be not where Newton's theory said it would be.
Doesn't that give at least some credence to Einstien's theory?
QUOTE (buttershug+Oct 23 2008, 04:10 PM)
But the star was in the sky where Einstien's theory said it would be not where Newton's theory said it would be.
Doesn't that give at least some credence to Einstien's theory?
None whatsoever. That's pure propaganda and has nothing to do with Science. Einstein's "theory", as already demonstrated, is not in a position to predict any experimental result.
Doesn't that give at least some credence to Einstien's theory?
None whatsoever. That's pure propaganda and has nothing to do with Science. Einstein's "theory", as already demonstrated, is not in a position to predict any experimental result.
QUOTE (Omnibus+Oct 23 2008, 04:00 PM)
You're doing it right now. Don't tell me what experiment shows. Show me a theory that can derive what experiment shows.
That is exactly what Einstein's theory does! If you not want to see what is there, I cannot help you. But do not feel too badly about it: There are "expert" scientists who have won Nobel Prizes who react exactly in the samr way as you are reacting!! I will be back tomorrow.
That is exactly what Einstein's theory does! If you not want to see what is there, I cannot help you. But do not feel too badly about it: There are "expert" scientists who have won Nobel Prizes who react exactly in the samr way as you are reacting!! I will be back tomorrow.
No, it doesn't. I already demonstrated it.
QUOTE (Omnibus+Oct 23 2008, 04:22 PM)
None whatsoever. That's pure propaganda and has nothing to do with Science. Einstein's "theory", as already demonstrated, is not in a position to predict any experimental result.
But he did make a predicition.
And the prediction turned out to be accurate.
And F=ma only if you don't look closely enough.
But he did make a predicition.
And the prediction turned out to be accurate.
And F=ma only if you don't look closely enough.
QUOTE (buttershug+Oct 23 2008, 04:56 PM)
But he did make a predicition.
And the prediction turned out to be accurate.
And F=ma only if you don't look closely enough.
What prediction? No prediction can be made, based on his "theory". I've already demonstrated it.
And the prediction turned out to be accurate.
And F=ma only if you don't look closely enough.
What prediction? No prediction can be made, based on his "theory". I've already demonstrated it.
QUOTE (buttershug+Oct 23 2008, 04:12 PM)
Ok then, I say if you change position, you have at least two frames. The initial location and the final location.
That's false.
Take an inertial frame travelling at v=v(1) relative to a second frame. At least one frame is 'moving', but we only have two frames. Ergo, movement does not imply a different frame for every spatial position.
If what you suggest is true, then as soon as you introduce v<>0 (relative to some other frame), then you would find yourself with an infinite number of frames.
That's false.
Take an inertial frame travelling at v=v(1) relative to a second frame. At least one frame is 'moving', but we only have two frames. Ergo, movement does not imply a different frame for every spatial position.
If what you suggest is true, then as soon as you introduce v<>0 (relative to some other frame), then you would find yourself with an infinite number of frames.
QUOTE (Omnibus+Oct 23 2008, 04:15 PM)
In the stationary system K the valid mechanics is Newtonian for any velocity. That must not be disputed let alone that Einstein himself accepts it too. Thus, the second Newton's law in K is exactly F = ma and in no way is F = m(beta^3)a, unless beta = 1 and Lorentz transformations are invalid.
But surely in a stationary system, beta=1? Which satisfies your condition for everything to be self-consistent.
If not, I am not understanding what beta is. Please will you write what beta is equal to?
But surely in a stationary system, beta=1? Which satisfies your condition for everything to be self-consistent.
If not, I am not understanding what beta is. Please will you write what beta is equal to?
QUOTE (Omnibus+Oct 23 2008, 05:01 PM)
What prediction? No prediction can be made, based on his "theory". I've already demonstrated it.
Ask the survivors of Hiroshima and Nagasaki if they agree with you!!
Ask the survivors of Hiroshima and Nagasaki if they agree with you!!
Here’s how one can avoid the doubts, as unsubstantiated as they are, raised, that F = ma may not be the law of motion in K for v =/= 0 and therefore Einstein’s reasoning may be saved:
The first postulate transforms Newton’s second law
F’ = ma’
in k as
F = ma
in K.
On the other hand, Lorentz transformations transform that same Newton’s second law
F’ = ma’
in k as
F = beta^3ma
where beta = 1/(sqr(1 – v^2/c^2))
Therefore, one single electron at one single state in the single system K ends up, due to the first postulate on the one hand and the Lorentz transformations on the other, with two different values of the mass, namely m and beta^3m. This is only possible iff beta = 1, that is, if the Lorentz transformations are invalid.
We need not even address the “theory” of relativity here because it has already failed still in §1 and §2 of the 1905 paper due its own internal contradictions.
This wraps up the technical part of the thread with the definitive conclusion that the “theory” of relativity as well as the Lorentz transformations (which have nothing to do with said “theory”) are invalid in their entirety with no potential for correction, developments and/or possibility for experimental verification whatsoever. The fact that the “theory” of relativity and the Lorentz transformations, which traditionally are used to accompany it, are invalid makes any area of study connected with it such as the string theories, cosmology etc. also invalid. It is advisable that those who are occupying themselves with such studies take heed of the demonstrated invalidity of the discussed “theory” because ignoring it is only a sign of intellectual laxity or even worse, intellectual dishonesty.
The first postulate transforms Newton’s second law
F’ = ma’
in k as
F = ma
in K.
On the other hand, Lorentz transformations transform that same Newton’s second law
F’ = ma’
in k as
F = beta^3ma
where beta = 1/(sqr(1 – v^2/c^2))
Therefore, one single electron at one single state in the single system K ends up, due to the first postulate on the one hand and the Lorentz transformations on the other, with two different values of the mass, namely m and beta^3m. This is only possible iff beta = 1, that is, if the Lorentz transformations are invalid.
We need not even address the “theory” of relativity here because it has already failed still in §1 and §2 of the 1905 paper due its own internal contradictions.
This wraps up the technical part of the thread with the definitive conclusion that the “theory” of relativity as well as the Lorentz transformations (which have nothing to do with said “theory”) are invalid in their entirety with no potential for correction, developments and/or possibility for experimental verification whatsoever. The fact that the “theory” of relativity and the Lorentz transformations, which traditionally are used to accompany it, are invalid makes any area of study connected with it such as the string theories, cosmology etc. also invalid. It is advisable that those who are occupying themselves with such studies take heed of the demonstrated invalidity of the discussed “theory” because ignoring it is only a sign of intellectual laxity or even worse, intellectual dishonesty.
QUOTE (Omnibus+Oct 24 2008, 02:20 AM)
F’ = ma’
in k as
F = beta^3ma
where beta = 1/(sqr(1 – v^2/c^2))
This is wrong. In SR force transforms to F = m gamma^3 (v.a)v - m gamma a
At no point is mass ever assumed to vary, since it does not as I said right at the start of this thread. How can anyone trust you when you can't get the formula you're using to try and prove your point right. I know RPenner has corrected you as well
QUOTE (Omnibus+Oct 24 2008, 02:20 AM)
Therefore, one single electron at one single state in the single system K ends up, due to the first postulate on the one hand and the Lorentz transformations on the other, with two different values of the mass, namely m and beta^3m. This is only possible iff beta = 1, that is, if the Lorentz transformations are invalid.
That you can make this conclusion shows you are in no position at all to understand scientific papers. F = ma is the low velocity limit of the relativistic formula, as any undergrad who's done a course on SR in their first year could show. There is no contradiction since you are comparing a system at v = 0 and transformed to some high v. Of course the equations change. Thats the point - the universe is not Newtonian.
Omnibus,
I am starting to think that your main purpose is not to argue physics but to engage, like Johannes Stark did, in "Einstein bashing".
I am starting to think that your main purpose is not to argue physics but to engage, like Johannes Stark did, in "Einstein bashing".
QUOTE (RealityCheck+Oct 24 2008, 02:38 AM)
Hi bm1957.
I think you missed the point, and that's why you suggest I have "a bad misunderstanding there". :-)
On re-reading I see what you were trying to say, but don't think it changes much. Remove the '2nd frames' that you suggest exist and you still have equivalence; it is just not as obvious, because both instances will feel like free-fall (i.e., nothing).
Going from a frame S, with coordinates (t,x), to a frame S', coordinates (t',x') where S' is moving at v relative to S, you have the transformations :
t' = gamma*(t-vx)
x' = gamma*(x-vt)
To convert back you just swap the position of the primes and change the sign of v, ie
t = gamma*(t'+vx')
x = gamma*(x'+vt')
If you substitute these into one another you find that t = t*(1-v^2)*gamma^2 so gamma = sqrt(1/(1-v^2)), which is true. All entirely consistent and you see that neither frame sees the clock of the other frame tick at the same rate.
You write down things like t = beta t' AND t=t' and somehow think you've been consistent? The only way that's consistent is if beta = 1, ie v=0 and if the frames aren't moving relative to one another of course their clocks tick at the same rate!
It's an experimental fact they tick at different rates due to relative motion. After relative motion it's found atomic clocks are no longer in sync and not by some random amount due to mechanical problems, by an amount in perfect agreement with predictions. Sure, some other interpretation of the results can be given, so as to not accept relativity but the fact the clocks always go out of sync by a precisely predictable amount means the rate of ticking must change.
That's incorrect. First postulate ("Principle of Relativity") transforms F' = ma' in k as F = ma in K.
That's incorrect. First postulate ("Principle of Relativity") transforms F' = ma' in k as F = ma in K.
That you can make this conclusion shows you are in no position at all to understand scientific papers. F = ma is the low velocity limit of the relativistic formula, as any undergrad who's done a course on SR in their first year could show. There is no contradiction since you are comparing a system at v = 0 and transformed to some high v. Of course the equations change. Thats the point - the universe is not Newtonian.
You are cavalierly accepting the above but you must not, if you care for the validity of the theory. What is wrong in this paper is the fact that when transforming the equation describing the single state of a single particle in one frame in forward direction by the first postulate and then transforming it back by the Lorentz transformations the same particle which should be in the same state, because that’s what the point of these transformations is, ends up in a different state. This sole fact is a crucial discrepancy which invalidates said “theory” in its entirety.
As I showed in my last posting you can see this contradiction in the "theory" also from the fact that one single electron in one single state, described by F' = ma' ends up being in two different states, described by both F = ma and F = beta^3ma. This discrepancy is due to the application of two ways of transformation which are in contradiction with each other. Because of this crucial contradiction the “theory” at hand must go.
That's incorrect. First postulate ("Principle of Relativity") transforms F' = ma' in k as F = ma in K.
This is a complete misunderstanding of the first postulate: The first postulate states that if you do the identical mechanical experiment within two different inertial reference frames moving with any speed relative to each other you will get the same result. Thus if you accelerate a mass which is moving with a slow speed relative to reference frame k (F=ma): If you do the same relative to reference frame K: i.e. also accelerate a mass which is moving slowly (but this time relative to reference frame K) you will also get the same result (F=ma).
However, when you look at an experiment being done within a reference frame (say k) while moving past at a high speed, you will not get the same result . In this case F will not be equal to ma since the mass is NOT moving slowly relative to your reference frame K.
This is the essence of all relativity theories since Galileo first pointed out that a mechanical experiment done on a moving ship will give the same result when the identical experiment is done on the shore. However, an observer looking from the shore at the experiment being done on the ship WILL NOT see the same result as the experimenter on the ship.
t' = gamma*(t-vx)
x' = gamma*(x-vt)
To convert back you just swap the position of the primes and change the sign of v, ie
t = gamma*(t'+vx')
x = gamma*(x'+vt')
If you substitute these into one another you find that t = t*(1-v^2)*gamma^2 so gamma = sqrt(1/(1-v^2)), which is true. All entirely consistent and you see that neither frame sees the clock of the other frame tick at the same rate.
You write down things like t = beta t' AND t=t' and somehow think you've been consistent? The only way that's consistent is if beta = 1, ie v=0 and if the frames aren't moving relative to one another of course their clocks tick at the same rate!
It's an experimental fact they tick at different rates due to relative motion. After relative motion it's found atomic clocks are no longer in sync and not by some random amount due to mechanical problems, by an amount in perfect agreement with predictions. Sure, some other interpretation of the results can be given, so as to not accept relativity but the fact the clocks always go out of sync by a precisely predictable amount means the rate of ticking must change.
QUOTE (johanfprins+Oct 23 2008, 10:38 AM)
It seems to me that the mainstream proponents (like yourself) have been INCREDIBLY STUPID FOR MORE THAN 100 YEARS. this is why I cannot get it published in a "reputable journal" . The mainstream IDIOTS like yourself block it. You do not want to be exposed for what you really are.
Mainstream views are knocked over all the time. In 1980 we thought neutrinos were massless and we knew what the universe was made of. Now we know they are not massless, which the Standard Model has trouble handling, and we're not sure what more than 90% of the universe is.
You take a hyperbolic view of things because you don't like being told you're wrong or uninformed. And so like all cranks, you become bitter and resentful to people who have spent time learning and understanding material you haven't and so you whine on internet forums about how you've got all the answers but there's noone listening to you. There's tons of other cranks like that here and on other forums, most of whom contradict one another but they ban together because they call one another "free thinkers" and not "educated robots", unlike people like myself, who haven't been brainwashed by education and professors. Funny how none of them ever get anywhere with their 'work' isn't it.....
You take a hyperbolic view of things because you don't like being told you're wrong or uninformed. And so like all cranks, you become bitter and resentful to people who have spent time learning and understanding material you haven't and so you whine on internet forums about how you've got all the answers but there's noone listening to you. There's tons of other cranks like that here and on other forums, most of whom contradict one another but they ban together because they call one another "free thinkers" and not "educated robots", unlike people like myself, who haven't been brainwashed by education and professors. Funny how none of them ever get anywhere with their 'work' isn't it.....
QUOTE (AlphaNumeric+Oct 24 2008, 08:38 AM)
And so like all cranks, you become bitter and resentful to people who have spent time learning and understanding material you haven't and so you whine on internet forums about how you've got all the answers but there's noone listening to you. There's tons of other cranks like that here and on other forums, most of whom contradict one another but they ban together because they call one another "free thinkers" and not "educated robots", unlike people like myself, who haven't been brainwashed by education and professors. Funny how none of them ever get anywhere with their 'work' isn't it.....
Have YOU got anywhere with the BS you are calculating? I am willing anytime to compare my citation-index with yours: If you have one! You are representative of the bigots who have taken over physics since the late 1920's and has led us through the looking glass to visit Alice.
Obviously, when the times are referenced to other positions than the origins of the two reference frames you will get what you just derived. The fact that you are so stupid to think that your derivation proves me wrong, just shows how incompetent the professors are who brainwashed you into NOT being able to think for yourself.
You must compare synchronized clocks and according to the Lorentz-transformations the clocks at the origins have been synchronized when the passed each other. Therefore to compare the time on two clocks held by twins which synchronized their clocks when the passed each other at time t=t(/)=0, you must transform between the origins as I have done. you will then obtain for t>0 and t(/)>0 that t=t(/). Obviously, when comparing clocks at positions x and x(/) not zero you will obtain different results. They have not been synchronized!
Maybe physics is not for you. Have you considered the priesthood: It is in this vocation where "mainstream dogma" must be defended at all costs and where other ideas are being rejected as blasphemy by cranks!
Have YOU got anywhere with the BS you are calculating? I am willing anytime to compare my citation-index with yours: If you have one! You are representative of the bigots who have taken over physics since the late 1920's and has led us through the looking glass to visit Alice.
Obviously, when the times are referenced to other positions than the origins of the two reference frames you will get what you just derived. The fact that you are so stupid to think that your derivation proves me wrong, just shows how incompetent the professors are who brainwashed you into NOT being able to think for yourself.
You must compare synchronized clocks and according to the Lorentz-transformations the clocks at the origins have been synchronized when the passed each other. Therefore to compare the time on two clocks held by twins which synchronized their clocks when the passed each other at time t=t(/)=0, you must transform between the origins as I have done. you will then obtain for t>0 and t(/)>0 that t=t(/). Obviously, when comparing clocks at positions x and x(/) not zero you will obtain different results. They have not been synchronized!
Maybe physics is not for you. Have you considered the priesthood: It is in this vocation where "mainstream dogma" must be defended at all costs and where other ideas are being rejected as blasphemy by cranks!
QUOTE (Omnibus+Oct 23 2008, 05:01 PM)
What prediction? No prediction can be made, based on his "theory". I've already demonstrated it.
He predicted that during a solar eclipse a certain star would appear at a certain position. Astromers predicted a slightly different location.
They all waited for the solar eclipse to see who's prediction would be accurate.
Einstien's was.
All you have demonstrated is that you don't understand that F=ma is usefull only at relatively low speeds. And even then there is a minute error.
And even wrong basesly crazy theories can make predictions.
Like the guy from the Farmer's Almanac says; " It's easy to predict the weather, it's getting it right that's difficult."
He predicted that during a solar eclipse a certain star would appear at a certain position. Astromers predicted a slightly different location.
They all waited for the solar eclipse to see who's prediction would be accurate.
Einstien's was.
All you have demonstrated is that you don't understand that F=ma is usefull only at relatively low speeds. And even then there is a minute error.
And even wrong basesly crazy theories can make predictions.
Like the guy from the Farmer's Almanac says; " It's easy to predict the weather, it's getting it right that's difficult."
QUOTE (bm1957+Oct 23 2008, 05:26 PM)
That's false.
Take an inertial frame travelling at v=v(1) relative to a second frame. At least one frame is 'moving', but we only have two frames. Ergo, movement does not imply a different frame for every spatial position.
If what you suggest is true, then as soon as you introduce v<>0 (relative to some other frame), then you would find yourself with an infinite number of frames.
Are you saying that there is a second frame if you are moving?
And that you can't move without a second frame?
That's what I was getting at. The OP was talking about one frame. I was saying if there is movement then their must be at least two frames.
And if there is no movement then F=ma only has the trivial solution of 0.
And I guess I should call the other frames potential frames.
Take an inertial frame travelling at v=v(1) relative to a second frame. At least one frame is 'moving', but we only have two frames. Ergo, movement does not imply a different frame for every spatial position.
If what you suggest is true, then as soon as you introduce v<>0 (relative to some other frame), then you would find yourself with an infinite number of frames.
Are you saying that there is a second frame if you are moving?
And that you can't move without a second frame?
That's what I was getting at. The OP was talking about one frame. I was saying if there is movement then their must be at least two frames.
And if there is no movement then F=ma only has the trivial solution of 0.
And I guess I should call the other frames potential frames.
QUOTE (buttershug+Oct 24 2008, 12:02 PM)
Are you saying that there is a second frame if you are moving?
And that you can't move without a second frame?
That's what I was getting at. The OP was talking about one frame. I was saying if there is movement then their must be at least two frames.
And if there is no movement then F=ma only has the trivial solution of 0.
And I guess I should call the other frames potential frames.
No. Am I really that bad at explaining myself? You seem to have misunderstood me many times in this thread.
I am saying that you are implying that you can't move without creating another frame for every position you occupy, which I believe is false.
I am saying that a second frame is required to confirm movement, an observatory frame. But if frames A and B exist, and frame A observes frame B to move with a relative velocity v, then this does not imply that the movement of B is producing frames as it moves, which is what you suggested when you said that moving from a to b requires at least 2 frames; one at a and one at b.
I have tried to reduce our disagreement to a single question with a number of possible answers. Please tell me if you agree that it would resolve our difference and I will try to ask for an authoritative answer:
Does an isolated frame have a velocity?
a) Yes, v=0
b) Yes, v<>0
c) No defined velocity, but dv is valid
d) No, velocity is meaningless in reference to an isolated frame
My stance is that c) is the correct answer. You have switched between a) and d). b) is trivially false.
If c) is correct, I am right. If a) or d) is correct, you are right (I believe a) to be trivially false also)
Do you agree that an authoritative answer to that question would resolve our dfference? Would you like to add/remove/change any of the potential answers?
And that you can't move without a second frame?
That's what I was getting at. The OP was talking about one frame. I was saying if there is movement then their must be at least two frames.
And if there is no movement then F=ma only has the trivial solution of 0.
And I guess I should call the other frames potential frames.
No. Am I really that bad at explaining myself? You seem to have misunderstood me many times in this thread.
I am saying that you are implying that you can't move without creating another frame for every position you occupy, which I believe is false.
I am saying that a second frame is required to confirm movement, an observatory frame. But if frames A and B exist, and frame A observes frame B to move with a relative velocity v, then this does not imply that the movement of B is producing frames as it moves, which is what you suggested when you said that moving from a to b requires at least 2 frames; one at a and one at b.
I have tried to reduce our disagreement to a single question with a number of possible answers. Please tell me if you agree that it would resolve our difference and I will try to ask for an authoritative answer:
Does an isolated frame have a velocity?
a) Yes, v=0
b) Yes, v<>0
c) No defined velocity, but dv is valid
d) No, velocity is meaningless in reference to an isolated frame
My stance is that c) is the correct answer. You have switched between a) and d). b) is trivially false.
If c) is correct, I am right. If a) or d) is correct, you are right (I believe a) to be trivially false also)
Do you agree that an authoritative answer to that question would resolve our dfference? Would you like to add/remove/change any of the potential answers?
None of the above? why? I just like being an a-ss hol-e lol
QUOTE
This is wrong. In SR force transforms to F = m gamma^3 (v.a)v - m gamma a
That's incorrect. First postulate ("Principle of Relativity") transforms F' = ma' in k as F = ma in K.
QUOTE (->
| QUOTE |
| This is wrong. In SR force transforms to F = m gamma^3 (v.a)v - m gamma a |
That's incorrect. First postulate ("Principle of Relativity") transforms F' = ma' in k as F = ma in K.
That you can make this conclusion shows you are in no position at all to understand scientific papers. F = ma is the low velocity limit of the relativistic formula, as any undergrad who's done a course on SR in their first year could show. There is no contradiction since you are comparing a system at v = 0 and transformed to some high v. Of course the equations change. Thats the point - the universe is not Newtonian.
You are cavalierly accepting the above but you must not, if you care for the validity of the theory. What is wrong in this paper is the fact that when transforming the equation describing the single state of a single particle in one frame in forward direction by the first postulate and then transforming it back by the Lorentz transformations the same particle which should be in the same state, because that’s what the point of these transformations is, ends up in a different state. This sole fact is a crucial discrepancy which invalidates said “theory” in its entirety.
As I showed in my last posting you can see this contradiction in the "theory" also from the fact that one single electron in one single state, described by F' = ma' ends up being in two different states, described by both F = ma and F = beta^3ma. This discrepancy is due to the application of two ways of transformation which are in contradiction with each other. Because of this crucial contradiction the “theory” at hand must go.
QUOTE (Omnibus+Oct 24 2008, 03:25 PM)
That's incorrect. First postulate ("Principle of Relativity") transforms F' = ma' in k as F = ma in K.
This is a complete misunderstanding of the first postulate: The first postulate states that if you do the identical mechanical experiment within two different inertial reference frames moving with any speed relative to each other you will get the same result. Thus if you accelerate a mass which is moving with a slow speed relative to reference frame k (F=ma): If you do the same relative to reference frame K: i.e. also accelerate a mass which is moving slowly (but this time relative to reference frame K) you will also get the same result (F=ma).
However, when you look at an experiment being done within a reference frame (say k) while moving past at a high speed, you will not get the same result . In this case F will not be equal to ma since the mass is NOT moving slowly relative to your reference frame K.
This is the essence of all relativity theories since Galileo first pointed out that a mechanical experiment done on a moving ship will give the same result when the identical experiment is done on the shore. However, an observer looking from the shore at the experiment being done on the ship WILL NOT see the same result as the experimenter on the ship.
QUOTE (johanfprins+Oct 24 2008, 09:10 AM)
This is a complete misunderstanding of the first postulate: The first postulate states that if you do the identical mechanical experiment within two different inertial reference frames moving with any speed relative to each other you will get the same result. Thus if you accelerate a mass which is moving with a slow speed relative to reference frame k (F=ma): If you do the same relative to reference frame K: i.e. also accelerate a mass which is moving slowly (but this time relative to reference frame K) you will also get the same result (F=ma).
However, when you look at an experiment being done within a reference frame (say k) while moving past at a high speed, you will not get the same result . In this case F will not be equal to ma since the mass is NOT moving slowly relative to your reference frame K.
This is the essence of all relativity theories since Galileo first pointed out that a mechanical experiment done on a moving ship will give the same result when the identical experiment is done on the shore. However, an observer looking from the shore at the experiment being done on the ship WILL NOT see the same result as the experimenter on the ship.
See, the thing is, I don't follow your understanding of the first postulate but Einstein's.
However, when you look at an experiment being done within a reference frame (say k) while moving past at a high speed, you will not get the same result . In this case F will not be equal to ma since the mass is NOT moving slowly relative to your reference frame K.
This is the essence of all relativity theories since Galileo first pointed out that a mechanical experiment done on a moving ship will give the same result when the identical experiment is done on the shore. However, an observer looking from the shore at the experiment being done on the ship WILL NOT see the same result as the experimenter on the ship.
See, the thing is, I don't follow your understanding of the first postulate but Einstein's.
Oh what the F are you smarts assing about? have you got a bigger IQ score
?
That's fifty league IQ measurement recently
Good luck nerd, I'll curdle you in the Uni exams, have a go, try me!!!!lol
?
That's fifty league IQ measurement recently
Good luck nerd, I'll curdle you in the Uni exams, have a go, try me!!!!lol
QUOTE (Omnibus+Oct 24 2008, 04:21 PM)
See, the thing is, I don't follow your understanding of the first postulate but Einstein's.
OK, now we are getting somewhere! When the Galilean transformations can be used, there is a possibility that there might be a unique stationary reference frame within the Universe. However the Galilean transformation is also valid if there is not such a reference frame.
To incorporate light speed Einstein had to postulate that there is no unique stationary reference frame in the Universe. This leads to the statement that all reference frames are equivalent. There is no known reason why we can discriminate between them. In other words there is no special reference frame. This does not mean that a mechanical experiment being done within one reference frame will be seen identically from another reference frame moving relative to the first. It still only means that the same experiment done identically within any reference frame will give the same results that will be obtained when the identical experiment is done within another reference frame moving relative to the first reference frame. But not that when looking into a reference frame (where the experiment is being done) from another reference frame moving relative to the first reference frame (where the experiment is being done) will be observed identically.
OK, now we are getting somewhere! When the Galilean transformations can be used, there is a possibility that there might be a unique stationary reference frame within the Universe. However the Galilean transformation is also valid if there is not such a reference frame.
To incorporate light speed Einstein had to postulate that there is no unique stationary reference frame in the Universe. This leads to the statement that all reference frames are equivalent. There is no known reason why we can discriminate between them. In other words there is no special reference frame. This does not mean that a mechanical experiment being done within one reference frame will be seen identically from another reference frame moving relative to the first. It still only means that the same experiment done identically within any reference frame will give the same results that will be obtained when the identical experiment is done within another reference frame moving relative to the first reference frame. But not that when looking into a reference frame (where the experiment is being done) from another reference frame moving relative to the first reference frame (where the experiment is being done) will be observed identically.
QUOTE (johanfprins+Oct 24 2008, 09:41 AM)
OK, now we are getting somewhere! When the Galilean transformations can be used, there is a possibility that there might be a unique stationary reference frame within the Universe. However the Galilean transformation is also valid if there is not such a reference frame.
To incorporate light speed Einstein had to postulate that there is no unique stationary reference frame in the Universe. This leads to the statement that all reference frames are equivalent. There is no known reason why we can discriminate between them. In other words there is no special reference frame. This does not mean that a mechanical experiment being done within one reference frame will be seen identically from another reference frame moving relative to the first. It still only means that the same experiment done identically within any reference frame will give the same results that will be obtained when the identical experiment is done within another reference frame moving relative to the first reference frame. But not that when looking into a reference frame (where the experiment is being done) from another reference frame moving relative to the first reference frame (where the experiment is being done) will be observed identically.
That is true but that's not what Einstein has in mind in formulating his first postulate ("Principle of Relativity"). That's why, according to Einstein (not according to what is correct) F' = ma' in k transforms as F = ma in K. This is a postulate which, as such, we must take as stated. The problem is that this particular postulate, even if we take it as true, is in conflict with the Lorentz transformations, as shown, which invalidates the "theory" in question without a trace.
To incorporate light speed Einstein had to postulate that there is no unique stationary reference frame in the Universe. This leads to the statement that all reference frames are equivalent. There is no known reason why we can discriminate between them. In other words there is no special reference frame. This does not mean that a mechanical experiment being done within one reference frame will be seen identically from another reference frame moving relative to the first. It still only means that the same experiment done identically within any reference frame will give the same results that will be obtained when the identical experiment is done within another reference frame moving relative to the first reference frame. But not that when looking into a reference frame (where the experiment is being done) from another reference frame moving relative to the first reference frame (where the experiment is being done) will be observed identically.
That is true but that's not what Einstein has in mind in formulating his first postulate ("Principle of Relativity"). That's why, according to Einstein (not according to what is correct) F' = ma' in k transforms as F = ma in K. This is a postulate which, as such, we must take as stated. The problem is that this particular postulate, even if we take it as true, is in conflict with the Lorentz transformations, as shown, which invalidates the "theory" in question without a trace.
QUOTE (Omnibus+Oct 24 2008, 02:21 PM)
See, the thing is, I don't follow your understanding of the first postulate but Einstein's.
Einstien's postulate was "I can't see the difference, can you see the difference".
He wasn't saying there was no difference (between F and ma). He was only saying that at low speeds (if you call half the speed of light slow) you can't measure the difference.
Einstien's postulate was "I can't see the difference, can you see the difference".
He wasn't saying there was no difference (between F and ma). He was only saying that at low speeds (if you call half the speed of light slow) you can't measure the difference.
QUOTE (buttershug+Oct 24 2008, 10:33 AM)
Einstien's postulate was "I can't see the difference, can you see the difference".
He wasn't saying there was no difference (between F and ma). He was only saying that at low speeds (if you call half the speed of light slow) you can't measure the difference.
Einstein is not saying there is a difference but we clearly see there is. F = ma is different from F = beta^3ma but it shouldn't because both originate from one and the same state, described by F' = ma', of one and the same electron. That crucial discrepancy invalidates the whole theory in its entirety.
He wasn't saying there was no difference (between F and ma). He was only saying that at low speeds (if you call half the speed of light slow) you can't measure the difference.
Einstein is not saying there is a difference but we clearly see there is. F = ma is different from F = beta^3ma but it shouldn't because both originate from one and the same state, described by F' = ma', of one and the same electron. That crucial discrepancy invalidates the whole theory in its entirety.
QUOTE (Omnibus+Oct 24 2008, 03:46 PM)
Einstein is not saying there is a difference but we clearly see there is. F = ma is different from F = beta^3ma but it shouldn't because both originate from one and the same state, described by F' = ma', of one and the same electron. That crucial discrepancy invalidates the whole theory in its entirety.
At low speeds you can't detect the difference.
In a Newtonian world F=ma.
I think Einstien is saying you can't see the difference. And for all intents and purposes there is no differnce. I think he was explaining the old explaination to contrast it to the new explaination.
Your interpretation does not work.
Johanfprins does work.
At low speeds you can't detect the difference.
In a Newtonian world F=ma.
I think Einstien is saying you can't see the difference. And for all intents and purposes there is no differnce. I think he was explaining the old explaination to contrast it to the new explaination.
Your interpretation does not work.
Johanfprins does work.
QUOTE (buttershug+Oct 24 2008, 11:33 AM)
At low speeds you can't detect the difference.
In a Newtonian world F=ma.
I think Einstien is saying you can't see the difference. And for all intents and purposes there is no differnce. I think he was explaining the old explaination to contrast it to the new explaination.
Your interpretation does not work.
Johanfprins does work.
Why are you continuing to insist that there isn't a difference between F = ma and F = beta^3ma since anyone who cares to look at this can see there is a difference? A single electron that has one state in k cannot have two states in K. Any theory that would give such contradictory result is invalid.
In a Newtonian world F=ma.
I think Einstien is saying you can't see the difference. And for all intents and purposes there is no differnce. I think he was explaining the old explaination to contrast it to the new explaination.
Your interpretation does not work.
Johanfprins does work.
Why are you continuing to insist that there isn't a difference between F = ma and F = beta^3ma since anyone who cares to look at this can see there is a difference? A single electron that has one state in k cannot have two states in K. Any theory that would give such contradictory result is invalid.
QUOTE (Omnibus+Oct 24 2008, 06:38 PM)
Why are you continuing to insist that there isn't a difference between F = ma and F = beta^3ma since anyone who cares to look at this can see there is a difference? A single electron that has one state in k cannot have two states in K. Any theory that would give such contradictory result is invalid.
Dear Omnibus,
The only difference between the two formulas arises when beta is not unity. When the particle being accelerated is moving with a slow speed relative to the observer one can approximate v/c with zero and therefore one can approximate beta with being unity: Thus F=ma. If the particle is moving at a high speed relative to the observer v/c cannot be approximated with zero and therefore beta cannot be approximated with unity. Thus F=ma is the approximate expression for F=(beta)^3ma when the speed is small. To test it yourself just put beta equal to unity within the last expression. Really it is NOT that difficult to put beta equal to unity in the second expression and then to see that you get the first expression!!!
Dear Omnibus,
The only difference between the two formulas arises when beta is not unity. When the particle being accelerated is moving with a slow speed relative to the observer one can approximate v/c with zero and therefore one can approximate beta with being unity: Thus F=ma. If the particle is moving at a high speed relative to the observer v/c cannot be approximated with zero and therefore beta cannot be approximated with unity. Thus F=ma is the approximate expression for F=(beta)^3ma when the speed is small. To test it yourself just put beta equal to unity within the last expression. Really it is NOT that difficult to put beta equal to unity in the second expression and then to see that you get the first expression!!!
QUOTE (Omnibus+Oct 24 2008, 06:38 PM)
A single electron that has one state in k cannot have two states in K. Any theory that would give such contradictory result is invalid.
Of course it can and does: if you move with the electron its kinetic energy is zero. When the electron moves relative to you its kinetic energy is NOT zero> So it has two different energy states relative to the two inertial reference frames!!
Of course it can and does: if you move with the electron its kinetic energy is zero. When the electron moves relative to you its kinetic energy is NOT zero> So it has two different energy states relative to the two inertial reference frames!!
QUOTE (johanfprins+Oct 24 2008, 12:01 PM)
Of course it can and does: if you move with the electron its kinetic energy is zero. When the electron moves relative to you its kinetic energy is NOT zero> So it has two different energy states relative to the two inertial reference frames!!
No, it doesn't. A single electron in one given state in one system is represented by exactly one state in another system. That's the point of the transformations from one frame into another.
No, it doesn't. A single electron in one given state in one system is represented by exactly one state in another system. That's the point of the transformations from one frame into another.
QUOTE (Omnibus+Oct 24 2008, 07:07 PM)
No, it doesn't. A single electron in one given state in one system is represented by exactly one state in another system. That's the point of the transformations from one frame into another.
Then why can it have zero kinetic energy in one system but non-zero kinetic energy within all other systems moving relative to the system in which it has zero kinetic energy?
Do two different energies define the same state? If it does we will not be able to measure spectral lines from excited electrons will we? In order for an electron which moves relative to you to lose its kinetic energy it has to be decelerated. It will then emit radiation! This proves that the two states ARE NOT THE SAME!!
Then why can it have zero kinetic energy in one system but non-zero kinetic energy within all other systems moving relative to the system in which it has zero kinetic energy?
Do two different energies define the same state? If it does we will not be able to measure spectral lines from excited electrons will we? In order for an electron which moves relative to you to lose its kinetic energy it has to be decelerated. It will then emit radiation! This proves that the two states ARE NOT THE SAME!!
Sorry I quoted the wrong posting! What i wanted to quote was Omnibus claiming that the electron has "two" states within the same reference frame. This is not the case: It has one energy state within one reference frame and another within another reference frame moving relative to the first!!
If the particle has zero kinetic energy in a given system no transformations can make its kinetic energy to become both zero and non-zero in that given system. This is what Einstein is trying to do with his unsuccessful exercise.
QUOTE (Omnibus+Oct 24 2008, 07:20 PM)
If the particle has zero kinetic energy in a given system no transformations can make its kinetic energy to become both zero and non-zero in that given system. This is what Einstein is trying to do with his unsuccessful exercise.
Wrong: Einstein's transformation is from one system to another system moving relative to the first. If the electron has zero kinetic energy within the first system it is not possible that it can have zero kinetic energy within a reference frame moving relative to the first reference frame within which it has zero kinetic energy! This so simple that even my goldfish can understand it: Sheesh
Wrong: Einstein's transformation is from one system to another system moving relative to the first. If the electron has zero kinetic energy within the first system it is not possible that it can have zero kinetic energy within a reference frame moving relative to the first reference frame within which it has zero kinetic energy! This so simple that even my goldfish can understand it: Sheesh
Read the paper carefully again to convince yourself that Einstein first applies the "Principle of Relativity" to transform F = ma from K to k, obtaining F' = ma' in k. Then he transforms F' = ma' in k back into K but this time applying Lorentz transformations and obtains F = beta^3ma for the same electron. Thus, transforming the single F' = ma' from k into K in two different ways, he obtains two different states in K for that same electron in K.That's unacceptable. One state, described by F' = ma' of one electron in one system (k in this case) cannot be represented in another system (K in this case) by two different states, described namely by F = ma and F = beta^3ma. I already explained that.
QUOTE (Omnibus+Oct 24 2008, 07:34 PM)
Read the paper carefully again to convince yourself that Einstein first applies the "Principle of Relativity" to transform F = ma from K to k, obtaining F' = ma' in k. Then he transforms F' = ma' in k back into K but this time applying Lorentz transformations and obtains F = beta^3ma for the same electron. Thus, transforming the single F' = ma' from k into K in two different ways, he obtains two different states in K for that same electron in K.That's unacceptable. One state, described by F' = ma' of one electron in one system (k in this case) cannot be represented in another system (K in this case) by two different states, described namely by F = ma and F = beta^3ma. I already explained that.
As I have written before: Einstein's style was ponderous and it will thus seem confusing. BUT WHAT YOU WRITE IS NOT WHAT EINSTEIN DID: He started off with a slow-moving particle in one reference frame and then transformed the experiment into another reference frame moving at a high speed relative to the first. I agree that Einstein could have explained what he did better. But what he did is what I just wrote. So there is nothing wrong with his theory: Maybe with his ability to be a good teacher!!
I need some rest: Will be back tomorrow!
As I have written before: Einstein's style was ponderous and it will thus seem confusing. BUT WHAT YOU WRITE IS NOT WHAT EINSTEIN DID: He started off with a slow-moving particle in one reference frame and then transformed the experiment into another reference frame moving at a high speed relative to the first. I agree that Einstein could have explained what he did better. But what he did is what I just wrote. So there is nothing wrong with his theory: Maybe with his ability to be a good teacher!!
I need some rest: Will be back tomorrow!
Einstein did exactly what I explained.
May I interject?
I developed an equation for kinetic energy that gives EXACTLY the same result as Einstein's -- therefore, it is equally valid.
Einstein's equation is basically nmc^2. There is no velocity here except v in the Lorentz transform and c^2.
My equation is derived from Newton and is (where R = Lorentz transformation)
E = mv^2/ (R+R^2).
Notice there is v in the numerator, and as E -->infinity, V --> infinity (as m is constant).
Also note that (R+R^2) is actually 1/(R+R^2) and that when v is very slow, R=!.
And the fraction is 1/2. Thus at low velocities we have E=mv^2 x 1/2
This is only one confirmation that super-c velocity is possible.
This and much more is in my ebook. A Diagnosis of Special Relativity.
More info on my blog: quantum-RelativityBkolni
(I have 5 articles on relativity available for any website that wants them.
vertvergon@gmail.com )
I developed an equation for kinetic energy that gives EXACTLY the same result as Einstein's -- therefore, it is equally valid.
Einstein's equation is basically nmc^2. There is no velocity here except v in the Lorentz transform and c^2.
My equation is derived from Newton and is (where R = Lorentz transformation)
E = mv^2/ (R+R^2).
Notice there is v in the numerator, and as E -->infinity, V --> infinity (as m is constant).
Also note that (R+R^2) is actually 1/(R+R^2) and that when v is very slow, R=!.
And the fraction is 1/2. Thus at low velocities we have E=mv^2 x 1/2
This is only one confirmation that super-c velocity is possible.
This and much more is in my ebook. A Diagnosis of Special Relativity.
More info on my blog: quantum-RelativityBkolni
(I have 5 articles on relativity available for any website that wants them.
vertvergon@gmail.com )
QUOTE (Omnibus+Oct 24 2008, 08:05 PM)
Einstein did exactly what I explained.
I sneaked back because I will not be available at all tomorrow: We are celebrating my Mother-in-Law's 90th birthday. Wow! You thought you had worries? She is on her way to break 100 years. I will probably not be there when that happens! Just think what it means to be outlived by your Mother-in-Law
Nonetheless, I have to confirm that Omnibus is deducing conclusions from Einstein's paper which are not in there. I have just now had the same problem that two referees made deductions from a manuscript that I have submitted on superconduction, which are NOT in the manuscript. Usually this happens when people have ulterior motives: In my case the re frees block it because most, if not all the publications in "reputable journals" on superconduction will be wrong if I am correct: The problem is that I am correct: Just as Einstein was correct in his 1905 publication.
Think Omnibus think: Why do you want at all costs to believe that Einstein was wrong? He did make mistakes, but not the mistake you attribute to him. He is still the greatest physicist who lived during then 20th century. There is not a single person after him who surpassed him!
I sneaked back because I will not be available at all tomorrow: We are celebrating my Mother-in-Law's 90th birthday. Wow! You thought you had worries? She is on her way to break 100 years. I will probably not be there when that happens! Just think what it means to be outlived by your Mother-in-Law
Nonetheless, I have to confirm that Omnibus is deducing conclusions from Einstein's paper which are not in there. I have just now had the same problem that two referees made deductions from a manuscript that I have submitted on superconduction, which are NOT in the manuscript. Usually this happens when people have ulterior motives: In my case the re frees block it because most, if not all the publications in "reputable journals" on superconduction will be wrong if I am correct: The problem is that I am correct: Just as Einstein was correct in his 1905 publication.
Think Omnibus think: Why do you want at all costs to believe that Einstein was wrong? He did make mistakes, but not the mistake you attribute to him. He is still the greatest physicist who lived during then 20th century. There is not a single person after him who surpassed him!
God bless your Mother-in-Law. May she reach well beyond 100. You too.
As for Einstein, what he really did is exactly what I explained. If I were your referee I won't agree with your frivolous interpretations as well.
As for Einstein, what he really did is exactly what I explained. If I were your referee I won't agree with your frivolous interpretations as well.
QUOTE (Omnibus+Oct 24 2008, 05:20 PM)
If the particle has zero kinetic energy in a given system no transformations can make its kinetic energy to become both zero and non-zero in that given system. This is what Einstein is trying to do with his unsuccessful exercise.
No he is showing it the old way and the new way to demonstrate you get different anwers.
No he is showing it the old way and the new way to demonstrate you get different anwers.
The Modified General Relativity!
I have made my page for the Modified Einstein Field Equation.
And I wrote the New Metric.
http://www.geocities.jp/imyfujita/galaxy/galaxy01.html
Now I would like to talk about The Modified General Relativity.
Iori Fujita
I have made my page for the Modified Einstein Field Equation.
And I wrote the New Metric.
http://www.geocities.jp/imyfujita/galaxy/galaxy01.html
Now I would like to talk about The Modified General Relativity.
Iori Fujita
QUOTE (buttershug+Oct 24 2008, 04:40 PM)
No he is showing it the old way and the new way to demonstrate you get different anwers.
What on earth does that mean?
What on earth does that mean?
wankerbus didn't i question your IQ compared to 141?
buttershug, carefully read and try to understand what Einstein had done in paragraph 6 of the discussed 1905 paper (where there is the mentioned by you 'Old manner' and 'New manner' of expression) and try to apply it in the discussed here example from paragraph 10 of the same paper. In agreement with the method in paragraph 6 you will find out that what comes out from transforming F' = ma' from k to K by using the "Principle of Relativity" (that is, the first postulate) should be exactly the same as the outcome from transforming that F' = ma' from k to K by using Lorentz transformations. That is, F = ma (the 'Old manner of expression') and F = beta^3ma (the 'New manner of expression') "must express exactly the same thing". They will express exactly the same thing iff beta = 1, that is, only if the Lorentz transformations are invalid.
I wrote the above text as a favor to you, in an effort to help you somehow understand it in your terms. It is not at all necessary to present my argument in that way. The main thing you have to understand and remember is that a unique particle in a unique state in a given system cannot be presented in another system to have two different states in that other system. If that happens, the theory, such as the discussed here "theory", leading to such discrepancy is invalid.
I wrote the above text as a favor to you, in an effort to help you somehow understand it in your terms. It is not at all necessary to present my argument in that way. The main thing you have to understand and remember is that a unique particle in a unique state in a given system cannot be presented in another system to have two different states in that other system. If that happens, the theory, such as the discussed here "theory", leading to such discrepancy is invalid.
Single experiment which can prove the relativity: Two clocks are synchronized on a satellite, then one clock was sent on the Earth, then the clock have returned on the satellite and have shown smaller time. Was such experiment?
Omnibus, you are absolutely correct -- but for the wrong reason.
The invalidation of SR goes back before your discovery.
Let me explain.
Sometimes analogies are terrific tools to explain a difficult concept.
We have one here: Take a bottle bottom. The reason we choose
is because it is a piece of glass the planes of which are very irregular.
Now we have an observer place the piece up to his eye and observe
the world through it. As we expect, the world appears very distorted.
The word "appears" is the operative word. The world appears distorted
but we know it isn't.
The same is true of Einstein's theory of relativity. According to
this theory an observer on a stationary platform views the conditions
on a moving platform His bottle bottom is replaced by the Lorentz
transformation. It distorts the observations such that a meter rod
contracts in length, the mass of objects increase, time slows down,
nothing can exceed the speed of light and 2 + 2 does not equal 4..
As with the bottle bottom, we realize the distortions are observations only and that which is observed does not change. Thus the rod does not contract ,the mass of an object does not increase, time does not slow down, it is possible to exceed the speed of light, and 2 + 2 equals 4.
Einstein's error was that he believed the distortions were real in the
observed system. No more true than if the world seen through a bottle
bottom actually became distorted as observed,
The details of this revelation can be found in my e-book, "A diagnosis of Special Relativity"
There, Einstein's own words are used to illustrate the point.
So you see, all the notoriety and super fame of the STR are due to a massive error.
And that's why the theory has been attacked so incessantly from day one.
Everybody argues using the nonexistent parameters Einstein generated. They are all barking up the wrong tree.
The parameters as Einsein describes them are OBSERVATIONS ONLY and do not exist in reality.
Length does not contract, mass does not increase, clocks do not slow down, 2+2=4, and the limit of velocity is infinity.
As to the experiments that supposedly confirm SR, there are other explanations for them.
A complete analysis -- and new developments can be found in my e-book
"A Diagnosis of Special Relativity"
More information on my blog Quantum-RelativityBkolni.
The invalidation of SR goes back before your discovery.
Let me explain.
Sometimes analogies are terrific tools to explain a difficult concept.
We have one here: Take a bottle bottom. The reason we choose
is because it is a piece of glass the planes of which are very irregular.
Now we have an observer place the piece up to his eye and observe
the world through it. As we expect, the world appears very distorted.
The word "appears" is the operative word. The world appears distorted
but we know it isn't.
The same is true of Einstein's theory of relativity. According to
this theory an observer on a stationary platform views the conditions
on a moving platform His bottle bottom is replaced by the Lorentz
transformation. It distorts the observations such that a meter rod
contracts in length, the mass of objects increase, time slows down,
nothing can exceed the speed of light and 2 + 2 does not equal 4..
As with the bottle bottom, we realize the distortions are observations only and that which is observed does not change. Thus the rod does not contract ,the mass of an object does not increase, time does not slow down, it is possible to exceed the speed of light, and 2 + 2 equals 4.
Einstein's error was that he believed the distortions were real in the
observed system. No more true than if the world seen through a bottle
bottom actually became distorted as observed,
The details of this revelation can be found in my e-book, "A diagnosis of Special Relativity"
There, Einstein's own words are used to illustrate the point.
So you see, all the notoriety and super fame of the STR are due to a massive error.
And that's why the theory has been attacked so incessantly from day one.
Everybody argues using the nonexistent parameters Einstein generated. They are all barking up the wrong tree.
The parameters as Einsein describes them are OBSERVATIONS ONLY and do not exist in reality.
Length does not contract, mass does not increase, clocks do not slow down, 2+2=4, and the limit of velocity is infinity.
As to the experiments that supposedly confirm SR, there are other explanations for them.
A complete analysis -- and new developments can be found in my e-book
"A Diagnosis of Special Relativity"
More information on my blog Quantum-RelativityBkolni.
QUOTE (Omnibus+Oct 24 2008, 11:33 PM)
God bless your Mother-in-Law. May she reach well beyond 100. You too.
As for Einstein, what he really did is exactly what I explained. If I were your referee I won't agree with your frivolous interpretations as well.
Thanks Omnibus, I will convey your good wishes.
By the way I did counter the referees with impeccable scientifically based logic: Just as I have countered your lack of logic in this thread.
As for Einstein, what he really did is exactly what I explained. If I were your referee I won't agree with your frivolous interpretations as well.
Thanks Omnibus, I will convey your good wishes.
By the way I did counter the referees with impeccable scientifically based logic: Just as I have countered your lack of logic in this thread.
QUOTE (dimazin+Oct 25 2008, 05:23 AM)
Single experiment which can prove the relativity: Two clocks are synchronized on a satellite, then one clock was sent on the Earth, then the clock have returned on the satellite and have shown smaller time. Was such experiment?
This experiment requires acceleration of the clocks relative to each other. This is a different kettle of fish than in the case where no acceleration is present. Therefore all experiments like sending a clock by airplane around the world (Haefele and Keating) must take the accelerations and decelerations of the airplane into account. In the experiment reported by Haefele and Keating this was not done. They only took the average speed relative to the rotating Earth into account. Most airplane flights are continuously accelerating and then continuously decelerating until landing. This saves fuel!
This experiment requires acceleration of the clocks relative to each other. This is a different kettle of fish than in the case where no acceleration is present. Therefore all experiments like sending a clock by airplane around the world (Haefele and Keating) must take the accelerations and decelerations of the airplane into account. In the experiment reported by Haefele and Keating this was not done. They only took the average speed relative to the rotating Earth into account. Most airplane flights are continuously accelerating and then continuously decelerating until landing. This saves fuel!
QUOTE (johanfprins+Oct 25 2008, 12:32 AM)
Thanks Omnibus, I will convey your good wishes.
By the way I did counter the referees with impeccable scientifically based logic: Just as I have countered your lack of logic in this thread.
I disagree.
By the way I did counter the referees with impeccable scientifically based logic: Just as I have countered your lack of logic in this thread.
I disagree.
QUOTE (Omnibus+Oct 25 2008, 07:51 AM)
I disagree.
You are a free person. So I respect your right to disagree even though you are wrong!
Regards,
Johan
You are a free person. So I respect your right to disagree even though you are wrong!
Regards,
Johan
QUOTE (johanfprins+Oct 25 2008, 01:12 AM)
You are a free person. So I respect your right to disagree even though you are wrong!
Regards,
Johan
Can you tell where I'm wrong?
Regards,
Johan
Can you tell where I'm wrong?
QUOTE (Omnibus+Oct 25 2008, 08:14 AM)
Can you tell where I'm wrong?
Yes: Read all my posts above!
OK I am off to celebrate 90 years!
Yes: Read all my posts above!
OK I am off to celebrate 90 years!
I've read them and they are a mosaic of correct statements and statements which are wanting in scientific rigor, let alone that there are gaps in understanding the basics of scientific methodology. Case in point, you have remained with the impression that because there is a change of mass with velocity experimentally established, that would be the reason to pronounce a theory claiming such change a correct theory. This is a fundamental misunderstanding of methodology of science. A theory has to predict such change without presuming it. This is with regard to mass but you are demonstrating similar methodological weakness in other points during the discussion at that not only with me. This weakness makes you very easy target for zealous defenders of the status quo which converts you from a critic into the most desired straw man for such zealots.
QUOTE (johanfprins+Oct 24 2008, 10:07 AM)
Have YOU got anywhere with the BS you are calculating? I am willing anytime to compare my citation-index with yours: If you have one! You are representative of the bigots who have taken over physics since the late 1920's and has led us through the looking glass to visit Alice.
Yeah. one of those 'bigots' who helped build the transistor and the GPS system.
And whose predictions match all experiments. All you say is "They tick the same" and then ignore experiments which show otherwise as 'misunderstood'. Simply denying an experiment doesn't mean it hasn't been done.
Yeah. one of those 'bigots' who helped build the transistor and the GPS system.
And whose predictions match all experiments. All you say is "They tick the same" and then ignore experiments which show otherwise as 'misunderstood'. Simply denying an experiment doesn't mean it hasn't been done. QUOTE (johanfprins+Oct 24 2008, 10:07 AM)
Obviously, when the times are referenced to other positions than the origins of the two reference frames you will get what you just derived. The fact that you are so stupid to think that your derivation proves me wrong, just shows how incompetent the professors are who brainwashed you into NOT being able to think for yourself.
You must compare synchronized clocks and according to the Lorentz-transformations the clocks at the origins have been synchronized when the passed each other. Therefore to compare the time on two clocks held by twins which synchronized their clocks when the passed each other at time t=t(/)=0, you must transform between the origins as I have done. you will then obtain for t>0 and t(/)>0 that t=t(/). Obviously, when comparing clocks at positions x and x(/) not zero you will obtain different results. They have not been synchronized!
You must compare synchronized clocks and according to the Lorentz-transformations the clocks at the origins have been synchronized when the passed each other. Therefore to compare the time on two clocks held by twins which synchronized their clocks when the passed each other at time t=t(/)=0, you must transform between the origins as I have done. you will then obtain for t>0 and t(/)>0 that t=t(/). Obviously, when comparing clocks at positions x and x(/) not zero you will obtain different results. They have not been synchronized!
If x=0 at t=0 and x'=0 at t'=0 and they are initially set to have t=t'=0, this doesn't mean that t'=t for all t or t' (which ever you are taking as the independent variable). For an instant their clocks will agree and then they will not. The gamma factor tells you that immediately. It's like saying :
"Let T = 2(t-t0). To set origins equal we put t0=0, so initially, at t=0 T=0. Therefore t=T for all t."
Obviously false but it's what you're doing.
You do realise these are exercises given to school children? I help teach 'Relativity and Motion' at my university and time dilation is included. Of course they don't do it with scalar equations, they aren't up to speed on using tensor calculus yet, they are only 1st years.
"Let T = 2(t-t0). To set origins equal we put t0=0, so initially, at t=0 T=0. Therefore t=T for all t."
Obviously false but it's what you're doing.
You do realise these are exercises given to school children? I help teach 'Relativity and Motion' at my university and time dilation is included. Of course they don't do it with scalar equations, they aren't up to speed on using tensor calculus yet, they are only 1st years.
QUOTE (johanfprins+Oct 24 2008, 10:07 AM)
Maybe physics is not for you. Have you considered the priesthood: It is in this vocation where "mainstream dogma" must be defended at all costs and where other ideas are being rejected as blasphemy by cranks! mad.gif
Ouch, got an insult which doesn't suck?
Are you a physicist? I doubt it. And given the extremely vitriolic attitude you have towards them, or rather us, it would seem you failed to be any good at it when it came to crunch time and passing exams. Now, like all cranks, you wonder forums whining about your greatness and how terrible it is mainstream physics doesn't listen to you.
Like I said, if it's so easy and obvious you're right, get published.
And for the record, I have a degree in mathematics, a masters in theoretical physics and am in my 3rd year of a string theory PhD.
Are you a physicist? I doubt it. And given the extremely vitriolic attitude you have towards them, or rather us, it would seem you failed to be any good at it when it came to crunch time and passing exams. Now, like all cranks, you wonder forums whining about your greatness and how terrible it is mainstream physics doesn't listen to you.
Like I said, if it's so easy and obvious you're right, get published. And for the record, I have a degree in mathematics, a masters in theoretical physics and am in my 3rd year of a string theory PhD.
QUOTE (johanfprins+Oct 25 2008, 05:40 AM)
This experiment requires acceleration of the clocks relative to each other. This is a different kettle of fish than in the case where no acceleration is present. Therefore all experiments like sending a clock by airplane around the world (Haefele and Keating) must take the accelerations and decelerations of the airplane into account. In the experiment reported by Haefele and Keating this was not done. They only took the average speed relative to the rotating Earth into account. Most airplane flights are continuously accelerating and then continuously decelerating until landing. This saves fuel!
Idiots of the relativity have not checked up the theory from the point of view of the motionless observer on the satellite. Kinematic expansion of time depends on speed. Acceleration only changes speed.
Idiots of the relativity have not checked up the theory from the point of view of the motionless observer on the satellite. Kinematic expansion of time depends on speed. Acceleration only changes speed.
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