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AirForceAcademyboy
Here is a new topic is the gravitational forces from a black hole capable of tearing into Subspace, and what happens if a black hole keeps growing that it becomes to large will it colapes or will something else more terrifying happen though these black hole have existed for a very long time but if it kept growing what would happen.
AlphaNumeric
Black holes are a general relativitic result. 'Hyperspace' is a result from Heim theory, which has little to no experimental validation in the area of gravity. Hence, asking what a black hole can do with hyperspace isn't a valid question, they are different theories.

What would a supermassive black hole collapse into? It's already a zero sized point mass, it can't get any smaller! A black hole will last as long as it absorbed the same or more matter/energy than it emits. Eventually a black hole, no matter it's size, will become warmer than the universe (since the universe is slowly cooling) and so will begin radiating energy faster than it absorbs it. This is a run away effect (since black holes get hotter as they give shrink) and the black hole will evaporate in a flash of high energy radiation, at least according to Hawking, Penrose and most relativity people.

The time required for this to happen though is billions and billions of times longer than the universe has been around so far!
AJ
[I] [COLOR=gray] i have heard that black holes even suck light up and rip it apart. If this is true, would we freeze in time or if it get's torn would it distort and make us move faster then it, and if thats true ... by a 2nd party observer we would be moving faster then light??
AJ
sorry... in the last line I made a mistake on what I was asking ... by a 2nd party observer would we be moving back in time ... sorry for the mess up
Nick
Black holes don't exist.
Harry Costas
Hello All

Its just amazing how people describe the so called blackhole.

Thats right, they do not exist, the closest thing would be a MECO object. Ultra dense plasma matter that does not allow light to escape.

As for zero piont mass, thats funny.

To have any idea of what a so called blackhole is. We need to look at how they can be made.

We look at compact star cores, neutron star cores, quark star cores and preon star cores and the formation of black holes from the supernova of a large star.

The progressive increase in density of the star core eventually reaches a critical mass where light cannot escape.

The movies have descibed black holes differently. As some form of singularity, a whirlppol of electromagnetic gravity waves.

This link may help with some info. Just because I post links does not mean I agree with them 100%.
http://casa.colorado.edu/~ajsh/schw.shtml
AlphaNumeric
QUOTE (AJ+Nov 26 2006, 03:27 AM)
If this is true, would we freeze in time or if it get's torn would it distort and make us move faster then it, and if thats true ... by a 2nd party observer we would be moving faster then light??

From an observers point of view, an infalling person will freeze in time on the event horizon, though from their own point of view they continue to experience time normally, even when in the black hole.

Again, from an observers point of view, the infalling person hits the speed of light when they get to the event horizon, but from their own point of view, they don't. The rule "You can't move faster than light" is a local one, space-time curvature and warping means that person A can't really measure person B's local motion compared to that of light, because they can't measure someone else's time and length properly.
QUOTE (Nick+Nov 26 2006, 06:07 AM)
Black holes don't exist.
Another wonderful contribution from the guy who don't know a thing about relativity.
QUOTE (Harry Costas+Nov 26 2006, 06:07 AM)
Thats right, they do not exist, the closest thing would be a MECO object. Ultra dense plasma matter that does not allow light to escape.
The problem is that if you've a cold object with sufficent mass to form an event horizon, it is unable to support itself against it's own mass. The plasma couldn't just remain plasma, it would be crushed into a point because the electrons and other fermions couldn't support themselves via degeneracy.
Harry Costas
Hello All


Mr Alpha you maybe right. But not in my books.

I do not think you know what ultra dense plasma matter is.

When you make comments please address it to the subject and not to the persons knowledge.

When I speak of plasm ultra dense. I speak of plasma where the atoms break down to the basic particals such as Neutrons in neutrons star cores, quarks in quarks star cores and the theoretical preon particals that produce the most denses core except for the so called black holes.

Search the net and you will find the info.

Also MECO objects

kaneda
Nick. What do you mean, that black holes don't exist? Surely we have evidence?
StevenA
Here's a paradox - if gravity propagates/travels at the speed of light, but light can't escape a black hole, then how can gravity go somehow faster than light and still "escape" in order to pull more light in?

If gravity warps a fabric of space that light travels through then the modes of travel are different and don't need have the same delays just as a shockwave travelling through a pipe doesn't travel at the same speed as water inside it.

If gravity was a repulsive force pushing inward from outside, then no problem would exist, as you could still have "crushing" gravitational forces present from outside a black hole.

Now consider this:

If you were close to the event horizon (as seen by a distant observer) of a black hole, the event horizon would be moving outward, past you at light speed and you'd need to travel at light speed away from it, just to maintain a constant distance from it (again, as witnessed by a distant observer).

So how can you be moving away from an object at light speed, yet have it appear to someone at a distance as if you were stationary? (Truly they couldn't see you anyway as if you were travelling at light speed, you'd be light and and not yet witnessed, but we're working with hypotheticals here)

Well, consider if two black holes were to "look" at each other - both expanding localling at light speed. Would the relative size of one black hole to the other change? Not directly, and each could see the other as occupying a similar volume of space as itself (each being its own "observer" and expanding outward, the other would). Depending on the time or distance between them, each would witness the other as progressively smaller, the further back in time, or equivalently, the further away in space they were from each other, and of course over time, they'd expand and merge .... just like a mysterious gravitational force. (Space could expand in a similar manner, and diffuse into mass, though it would seem it couldn't expand at the same rate - tidal forces would seem to require this flow or diffusion have a gradient/differential in space)

So what would the force experienced standing on the surface of the Earth be, if this expansion caused the gravitational force? It would seem a differential pressure between the expansion of space, primarily outward from the Earth ... so you're being accelerated upward by an expanding mass beneath you. In this case all forces might be unifiable as a single repulsive force and relative observations made from different scales of expansion could make these appear as attractive in comparison - so the entire universe could be expanding, along with everything inside it, and "gravity" makes space appear warped because not all objects diffuse at the same rate.

If you were inside a perfectly spherical shell of mass, you'd experience no gravitational force toward the surface - you could on one hand see this as a delicate and concidental balancing of 1/d^2 attractive forces, or you could see it as a symmetrical expansion outward by mass as it expands uniformly outward (of course keeping it from buckling inward would be tricky).

user posted image

Here's a graphic of Conway's Game of Life. Each cell follows simple rules when updating. (There a Java simulator there)

If you watch the cells over time, you'll find chaotic centers of action, very analogous to mass and little "gliders" that move at a constant rate through space, very similar to light. Depending upon the update rules, the chaotic centers can grow and shift around. (It's been shown that Conway's Game of Life is yet another system capable of "universal computation", and that's just in 2 dimensions)

Here's another couple examples of simple automata systems, but in 3 dimensions:

User posted image

Now imagine if the bottom, spherical example was a black hole expanding at light speed and that distant objects appear small, not directly because of any parallax effect, but because what you see of a distant star is both diffused over a large space (low intensity) and actually was smaller at the time ... you aren't seeing what's "there" now but instead what you're seeing a diffused energy "here" from a prior compressed state of "then". So does the moon look larger when you get near it because you're seeing it from a less diffused and more spacially expanded state, or is that just Euclidean geometry and vanishing points, or are they part of the same thing?

Consider that local time would likely appear to be slower in an expanded/diffused state versus a compressed state, so variations in differential rates of expansion in some areas of space could make internal times appear different as internal changes would occur slower in a diffused state.

Also, light would appear frozen in time, because having a maximum forward rate of motion, leaves no additional cycles available for lateral expansion, whereas a larger chaotic mass would tend to retain spherical traits and expand slower but equally in all directions.

Disclaimer: You can ignore most all my mumbo-jumbo and hand waving, but I still don't see how you can both have black holes, where light can't escape and gravity travelling at light speed, (Hawking Radiation does seem quite possible though) and I also don't believe relativity looks closely enough at relative scales of distances, instead of just relative locations or motions - an "observer" needs references to compare things with, and if space does expand, then relative measurements made from an expanded scale would seem to cancel out much of the expansion and leave differential rates of expansion as being observed as likely close to a warping of space, over time. If the evolution of a system over time is closely correlated to an expanding diffusion of interactions (the nice thing about this is that mass doesn't need to be self contained or have internal binding forces, as variable rates of expansion can give the appearance of this), then having two systems varying in their rate of diffusion could affect how rates of times, velocities and distances are perceived/interpreted.
Harry Costas
Hello All

Some people think that gravity goes much faster than light.
The Speed of Gravity What the Experiments Say

Tom Van Flandern, Meta Research [as published in Physics Letters A 250:1-11 (1998)]

http://metaresearch.org/cosmology/speed_of_gravity.asp

================================================

The Speed of Gravity - Repeal of the Speed Limit

By Tom Van Flandern, Meta Research
http://metaresearch.org/cosmology/gravity/speed_limit.asp

Note there are some links that dispute this.
Nick
QUOTE (kaneda+Nov 27 2006, 02:59 PM)
Nick. What do you mean, that black holes don't exist? Surely we have evidence?

We have evidence of the effect of extreme gravity not of black holes. biggrin.gif

Mitch Raemsch -- Light Falls --
kjw
Nick - a region of space with an "extreme" gravitational field is described as black hole.

by your logic, you are saying that we have evidence of a celestial body of hot gases that radiates energy derived from thermonuclear reactions in the interior but we have no evidence that stars exist.
AlphaNumeric
A black hole is an object from which light can't escape and this is acheived if you have a certain amount of mass within a certain volume. It does not require a singularity, but any physically viable model at present ends up with a singularity unless you start a system in a very precise way (like having all the stars in our galaxy in a region about 0.2 light years in radius and them NOT crashing into one another!).

So even if a star doesn't collapse into a pointlike singularity, if it has it's mass M within a radius R such that R < 2GM/c^2, it will not be able to emit light, it'll be a black hole, despite not having a singularity.
Nick
QUOTE (kjw+Nov 30 2006, 08:39 PM)
Nick - a region of space with an "extreme" gravitational field is described as black hole.

The extreme of gravity is not a black hole. Light can always escape.

The requirement for a black hole is an event horizon.
You are wrong kjw. Black holes don't exist. tongue.gif

Mitch Raemsch -- Light Falls --
AlphaNumeric
QUOTE (Nick+Dec 1 2006, 04:53 AM)
The extreme of gravity is not a black hole. Light can always escape.

If a star's gravitational field gets any more powerful than a neutron star's then it does become a black hole, even if it's not a singularity (as I outlined in my previous post). You can't just keep increasing gravitational field strength without limit without getting a black hole, eventually you hit the point where light can't escape.
QUOTE (Nick+Dec 1 2006, 04:53 AM)
The requirement for a black hole is an event horizon.
Yes and given 'extreme enough gravity' you'll get one.
QUOTE (Nick+Dec 1 2006, 04:53 AM)
You are wrong kjw. Black holes don't exist
Wow, what amazing logic. rolleyes.gif
kaneda
Nick. Escape velocity of Earth = 7 mps. Escape velocity of sun = 27 mps. Escape velocity of neutron star = 120,000 mps. Escape velocity of black hole is more than 186,282 mps.

Since light cannot travel faster than light, how can it escape from a black hole? Since light cannot escape, you cannot see beyond the point where an escape velocity greater than C is needed (event horizon) so the area is without light (ie: black).

Unless you can explain how light can travel faster than light, you must accept black holes exist.
Nick
Escape velocity applies ONLY to matter; NOT LIGHT. Matter slows down when escaping. Light doesn't. It can always escape. tongue.gif tongue.gif tongue.gif

There are no black holes kaneda!

Mitch Raemsch -- Light Falls --
AlphaNumeric
You can still relate the strength of gravity to space-time curvature and when the space-time curvature is high enough, no null geodesics (ie light paths) will point out from the star, they will all point inwards, meaning light can't escape.

Nick, you don't know anything about relativity and your naive comments aren't going to turn it over. Why not just learn some instead?
Nick
What about light emitted outward?

MITCH RAEMSCH -- LIGHT FELL --
AlphaNumeric
An object which has formed an event horizon, be it a singularity or be it an object (or collection of objects) somehow supporting itself against the crushing gravity will not be able to emit ANY light in the manner our Sun goes. ALL the energy created via thermonuclear reaction would be forced, via space-time curvature, to move towards the centre of mass of the entity.

This is because within the event horizon it is the radial coordinate, r, which is 'time-like' and the time coordinate is 'space-like' because the sign of the metric changes from -+++ to +-++. Since ALL objects which are physical must move move through the time-like coordinate, this means that any object is forces to decrease it's r by the gravity and it's motion through space-time. Of course you understood all that Nick, being so amazing at relativity wink.gif laugh.gif

No star which is able to undergo thermonuclear fusion will be smaller than it's Schwarzchild radius anyway because the heat and light pressure generated by the fusion counteracts the gravitational inwards pressure and prevents collapse to the point where an event horizon is formed.

Any object which is crushed below it's Schwarzchild radius won't be 'a star' like our Sun because they can support themselves against collapse.
Nick
QUOTE (AlphaNumeric+Dec 1 2006, 07:46 PM)
An object which has formed an event horizon, be it a singularity or be it an object (or collection of objects) somehow supporting itself against the crushing gravity will not be able to emit ANY light in the manner our Sun goes. ALL the energy created via thermonuclear reaction would be forced, via space-time curvature, to move towards the centre of mass of the entity.

This is because within the event horizon it is the radial coordinate, r, which is 'time-like' and the time coordinate is 'space-like' because the sign of the metric changes from -+++ to +-++. Since ALL objects which are physical must move move through the time-like coordinate, this means that any object is forces to decrease it's r by the gravity and it's motion through space-time. Of course you understood all that Nick, being so amazing at relativity wink.gif laugh.gif

No star which is able to undergo thermonuclear fusion will be smaller than it's Schwarzchild radius anyway because the heat and light pressure generated by the fusion counteracts the gravitational inwards pressure and prevents collapse to the point where an event horizon is formed.

Any object which is crushed below it's Schwarzchild radius won't be 'a star' like our Sun because they can support themselves against collapse.

YOU HAVEN'T ANSWERED THE QUESTION ALPHANUMM.

Mitch Raemsch -- Light Fell --
AlphaNumeric
Yes I did Nick, I said what happens to light in my first paragraph. I then went on to explain why a. that occurs if such a star could form and b. why such a star doesn't work.

Clearly the concepts of time and space-like coordinates went over your heard. So much for you understanding relativity wink.gif
rpenner
QUOTE (Nick+ Dec 1 2006, 09:13 PM)
YOU HAVEN'T ANSWERED THE QUESTION ALPHANUMM.


Yes, he did answer it. It's just not a question you want to hear because to understand it you need to comprehend what curvature of four-dimensional space-time means.

He said it's a ill-posed question, because "outward" is not a direction you can point to on the inside of a black hole exactly like "to the past" is not a direction you can point to here. If the rules of the court room applied, he would object to your question because it presumes facts which are not in evidence.

You cannot presume that there are outward null geodesics without doing the math and making that assumption is why the Schwarzschild coordinates go singular at R=2GM/c^2 even though, locally, nothing appears non-Lorentzian at that point.

QUOTE
No stationary frames inside the Schwarzschild radius
According to the Schwarzschild metric, at the Schwarzschild radius r_s, proper radial distance intervals become infinite, and proper time passes infinitely slowly. Inside the Schwarzschild radius, proper radial distances and proper times appear to become imaginary (that is, the square root of a negative number).

Historically, it took decades before this strange behaviour was understood properly (see again Kip Thorne's book Black Holes and Time Warps for an account). The problem with the Schwarzschild metric is that it describes the geometry as measured by observers at rest. It is now realized that once inside the Schwarzschild radius, there can be no observers at rest: everything plunges inevitably to the central singularity. In effect, the very fabric of spacetime falls to the singularity, carrying everything with it. No pressure can withstand the inexorable collapse.

To paraphrase Misner, Thorne & Wheeler (1973, Gravitation, p. 823), that [the] same unseen power of the world which impels everyone from age 20 to 40, and from 40 to 80, impels objects inside the horizon irresistably towards the singularity.


That is, if you are talking about black holes in GR, which is what everyone else means, seeing as there are no laboratory black holes yet, and the objects in the sky which are called black holes are called so due to their properties matching the predictions of GR.

If you have in mind some other theory of GR, you have to develop the math yourself. If you have some other definition of black hole other than the conventional one, you have to explain it yourself. But if you are talking about GR's black holes, your question:
QUOTE (->
QUOTE
No stationary frames inside the Schwarzschild radius
According to the Schwarzschild metric, at the Schwarzschild radius r_s, proper radial distance intervals become infinite, and proper time passes infinitely slowly. Inside the Schwarzschild radius, proper radial distances and proper times appear to become imaginary (that is, the square root of a negative number).

Historically, it took decades before this strange behaviour was understood properly (see again Kip Thorne's book Black Holes and Time Warps for an account). The problem with the Schwarzschild metric is that it describes the geometry as measured by observers at rest. It is now realized that once inside the Schwarzschild radius, there can be no observers at rest: everything plunges inevitably to the central singularity. In effect, the very fabric of spacetime falls to the singularity, carrying everything with it. No pressure can withstand the inexorable collapse.

To paraphrase Misner, Thorne & Wheeler (1973, Gravitation, p. 823), that [the] same unseen power of the world which impels everyone from age 20 to 40, and from 40 to 80, impels objects inside the horizon irresistably towards the singularity.


That is, if you are talking about black holes in GR, which is what everyone else means, seeing as there are no laboratory black holes yet, and the objects in the sky which are called black holes are called so due to their properties matching the predictions of GR.

If you have in mind some other theory of GR, you have to develop the math yourself. If you have some other definition of black hole other than the conventional one, you have to explain it yourself. But if you are talking about GR's black holes, your question:
What about light emitted outward?

makes as much sense as "What's the weather like 10 billion kilometers north of London?"
Nick
NO. YOU DIDN'T.


WHAT ABOUT LIGHT EMITTED OUTWARD alphanummer !

Mitch Raemsch -- Light FELL! -
AlphaNumeric
QUOTE (Nick+Dec 1 2006, 11:20 PM)
NO. YOU DIDN'T.


WHAT ABOUT LIGHT EMITTED OUTWARD alphanummer !

*sigh* Obviously you know you have been shown to be wrong, because you're having to resort to CAPITALISING ALL YOUR WORDS, a sign of trying to replace substance with form.

As I, and now Rpenner, have explained, to say "what about the light emitted outwards" is meaningless in this case. Pick a point of space-time within the black hole and imagine yourself there. Every single direction you can move in takes you towards the singularity. It is impossible for any object with mass, or even objects without mass such as light, to increase their distance from the singularity.

Therefore there is no outwardly emitted light. Any atom in the star emitting light emits it inwards, always inwards, due to space-time curvature.

This is a topic you and I have discussed before, several months ago when I had to explain to lack of 'paths of escape' from inside a black hole, that all the time-like and null geodesics (ie paths you or light can move along) take you to the singularity.

Simply repeating a question two people have explained is poorly defined and also why it's poorly defined doesn't make your question somehow valid. Once again, you, despite your claims of profound knowledge, fail to grasp tenants of relativity that any decent student in university would be expected to understand. Now this in itself would not a bad thing (we all start somewhere) were it not for the fact you've had this things explained to you before, long enough ago for you to have gone away and learnt them properly. But instead you remained ignorant and now ask the same questions again.

You aren't coming up with profound questions which show problems with relativity, you're coming up with questions which show problems with your own understanding and your own willingness to learn or rather, lack of.
Nick
SO WHAT HAPPENS TO OUTGOING LIGHT alpha#? ph34r.gif
AlphaNumeric
Once again, my reply clearly goes over your head. As I have now explained twice, along with Rpenner, there is no outgoing light. As you know I've said before 'all roads lead to Rome', the singularity being Rome in this case. Any and all light emitted by particles within the event horizon is inward emitted because all null geodesics point at the singularity, the light is always descreasing it's distance from the singularity, so by definition was emitted inwards.

There is no outwardly emitted light. If you ask the same question again or a similar rewording you're either just trolling or you're profoundly stupid.
Nick
QUOTE (AlphaNumeric+Dec 1 2006, 11:01 PM)
There is no outwardly emitted light.

And why is that?
AlphaNumeric
Did you bother reading the rest of my post or my previous ones or did you just read that one sentence?

Space-time curvature within a black hole causes all light ray geodesics (along with all time-like geodesics, the ones matter follows) to bend inwards. I did outline the specific reason in my post mentioning the sign change in the metric coefficents. The metric goes from -+++ to +-++, where the - refers to the coefficent of the dr^2 term in the metric. This means that it is now the time-like coordinate and therefore any physical object, including light, is forces to be in constant motion in it and that motion is decreasing r, moving towards the singularity.

Such explainations are available in detail at your local library or university. You might want to look them up some time wink.gif
Nick
If all roads lead to Rome then you can also leave Rome by those same roads.

WHY CAN'T MATTER EMITT LIGHT OUTWARD AS IT FALLS? What is preventing it?
AlphaNumeric
The roads are all one way wink.gif

As I just said several times space-time curvature prevents them 'leaving'. No matter what direction the light is emitted, it's on a path to the singularity, not outwards. There is no 'outward road' to chose.

Above the event horizon there are paths of escape. At the event horizon no longer is there a path of escape, though light can just about keep itself from falling into the black hole, but it can't escape either. Anything further in and light is forced inwards.

Why do you addamently refuse to learn any relativity? There's tons of websites and books aimed at giving everything from beginners introductions to postgrad research topics in relativity. Why do you not use such resources instead of asking the same questions again and again and again?
Nick
QUOTE (AlphaNumeric+Dec 1 2006, 11:33 PM)
The roads are all one way wink.gif


Prove it. There is no inward without an outward. No up without a down. tongue.gif

>> There is no 'outward road' to chose.

What prevents a photon from being emitted outward? There is space so why not? If light is emitted outward how can it change its direction?



MITCH RAEMSCH ----- LIGHT FALLS -----
AlphaNumeric
The proof is found in the maths of relativity. All causal geodesics point inwards due to space-time curvature. You are trying to use your 3d Euclidean physical intuition on a 4d non-Euclidean space-time.

If you are so amazing at relativity, you should be able to demonstrate to yourself that all lightcones point inwards by chaning from the coordinates (t,r,theta,phi) to (u,r,theta,phi) near the event horizon using the transformation v = t+r+2M ln|r-2M|.

This then gives you the metric form

ds^2 = -(1- 2M/r)dv^2 + 2dvdr + r^2 dP^2

dP^2 is S^2 sphere metric.

You can now see that at r=2M, all the metric coefficents are finite and therefore no problem exists at r=2M in these coordinates. You can also sketch the light cones (supressing the dP^2) to give you this diagram :

User posted image

The time axis points up and increasing r points to the right. You can see on the right hand side of the diagram, the two sets of lines/curves are almost perpendicular to one another, this is to be expected since it's almost flat space-time. As you get closer to the r=2M line (the vertical line to just to left of centre) one of the sets of lines begins to warp anbd bend the other way. This gets so bad that for r=< 2M, there is no line which points outwards, they all point to the left (to the singularity).

If there's no line pointing to the right, there's no path of escape, light and any other particle is forced inwards.

The paths light takes inwards are not altered, you can see that there's a set of lines which remain parallel, that's the paths of light fired straight in to the black hole. But the opposite paths, the ones which would normally take the photon out of the black hole are bent inwards, there is no escape, despite being a way in.

If you don't think this is right, show a flaw in the maths which derived it. Yes, it's hard to understand conceptually at first, but none the less, that is the result.
amrit
according to Newton gravity decreases by going towards the centre of stellar objects

if so how it is possible that massive stars collapse into black hole

might be space is most curve in the center of massive star and gravity as well
this way formation of black holes can be understood better
AlphaNumeric
QUOTE (amrit+Dec 5 2006, 10:22 PM)
if so how it is possible that massive stars collapse into black hole

Why do you ask this question when I just told you the answer in another thread?

Gravity creates enormous pressures within stars. Yes, even with the fact that gravitational acceleation descreases as you get closer to the centre, this doesn't mean that pressure does. Infact, pressure increases as you get closer to the centre. Just as the pressure of a swimming pool is higher at the bottom than at the surface, the same is true for any star.

This means that when no more energy is available via fusion to fight off gravity the crushing pressure does just that, it crushes the star into a white dwarf, then a neutron star and then a black hole (assuming enough mass).

Thus, even using Newton's ideas (well, their relativistic extension but the principle is the same, gravitational acceleration decreases closer to the centre of a star) you still end up with neutron stars and black holes.
amrit
Gravity creates enormous pressures within stars. Yes, even with the fact that gravitational acceleation descreases as you get closer to the centre, this doesn't mean that pressure does. Infact, pressure increases as you get closer to the centre. Just as the pressure of a swimming pool is higher at the bottom than at the surface, the same is true for any star.

alfa i might start calling you idiot too

no pressure possible by going closer to the centre if gravity would be not there
do you have some common sense ?
pressure without gravity,what could make it ?
Wholly ghost ?
Nick
Light can always move against the curve. It need not follow space curvature.

NO BLACK HOLES

Light can always get out. tongue.gif

MITCH RAEMSCH -- LIGHT FALLS ALL THE WAY --
AlphaNumeric
QUOTE (amrit+Dec 5 2006, 11:02 PM)
no pressure possible by going closer to the centre if gravity would be not there

Wrong. Learn some basic mechanics.
QUOTE (amrit+Dec 5 2006, 11:02 PM)
do you have some common sense ?
pressure without gravity,what could make it ?
The gravity elsewhere in the star squeezes the centre. Obviously there is gravity elsewhere in the star, even according to Newton. It's this inward force which increases pressure.

I have common sense and I'm using my brain. You obviously aren't! You aren't even trying to understand what I'm saying to you.
QUOTE (Nick+Dec 5 2006, 11:06 PM)
Light can always move against the curve. It need not follow space curvature.
Nope, the light MUST move along the null geodesics, it is the definition of possible light paths. Light can only move on null geodesics. It's the definition of such geodesics!

You're just doing the same as Armit, you repeat the same nonsense again and again as if it somehow validates your argument. Prove there's something wrong with the metric transformation I posted before and that the resultant geodesics arne't valid. Otherwise, you're just posting your opinion again and again in the face of logic and derivation.
amrit
no pressure possible by going closer to the centre if gravity would be not there

Wrong. Learn some basic mechanics.


than basic mechanic is wrong
i do not have time tolearn wrong things

do an experiment
imagine that gravity is strongest at the centre
you will see
all comes together well


AlphaNumeric
QUOTE (amrit+Dec 5 2006, 11:15 PM)
than basic mechanic is wrong

Despite the fact it's all experimentally verified to high precision rolleyes.gif
QUOTE (amrit+Dec 5 2006, 11:15 PM)
i do not have time tolearn wrong things
You don't have time to learn anything it would seem....
amrit
space is curved according to the gravitational centre of stellar object
Nick
QUOTE (AlphaNumeric+Dec 5 2006, 10:12 PM)
Nope, the light MUST move along the null geodesics, it is the definition of possible light paths. Light can only move on null geodesics. It's the definition of such geodesics!

IF A ROAD LEADS TO ROME IT ALSO LEADS AWAY FROM ROME. tongue.gif

MITCH RAEMSCH -- LIGHT FELL ON YOUR HEAD PUPA --
AlphaNumeric
QUOTE (Nick+Dec 6 2006, 09:33 PM)
IF A ROAD LEADS TO ROME IT ALSO LEADS AWAY FROM ROME. tongue.gif

Remember, analogies are not literal truths, they are examples to help understanding.

In the case of light and black holes, all the roads are one way wink.gif So perhaps I should say "All roads are one way and they all lead to Rome".

If you think that's wrong, feel free to rewrite relativity and send it to Hawking wink.gif
AirForceAcademyboy
QUOTE (Nick+Dec 1 2006, 03:53 AM)
The extreme of gravity is not a black hole. Light can always escape.

The requirement for a black hole is an event horizon.
You are wrong kjw. Black holes don't exist. tongue.gif

Mitch Raemsch -- Light Falls --

WE DO HAVE EVIDENCE THAT THEY DO EXIST HERE IS A SITE THAT I THINK YOU ALL SHOULD LOOK AT ABOUT BLACK HOLES http://en.wikipedia.org/wiki/Black_holes#Evidence
philip347
The term black hole refers to the massdensity of the leftover constituencies, of a once large star.

That is all this is, is a term.

these densities however must nestle in space, which is manifolded and has many accessways in and out of it.

So therefore, by deduction laws, if black holes exist in outerspace, they must not only bend time and space, but also must allow manifolds, in and around them.

They are known to possess relative north and south poles.

If these objects have north and south poles, then they also must manifold space, so that if say a black hole were near a larger object, such as a flotsam trail of stellar debris, or a sun, then these line of force, would be indicated as being expressed around these objects?
AirForceAcademyboy
QUOTE (philip347+Dec 9 2006, 01:24 AM)
The term black hole refers to the massdensity of the leftover constituencies, of a once large star.

That is all this is, is a term.

these densities however must nestle in space, which is manifolded and has many accessways in and out of it.

So therefore, by deduction laws, if black holes exist in outerspace, they must not only bend time and space, but also must allow manifolds, in and around them.

They are known to possess relative north and south poles.

If these objects have north and south poles, then they also must manifold space, so that if say a black hole were near a larger object, such as a flotsam trail of stellar debris, or a sun, then these line of force, would be indicated as being expressed around these objects?

bullcrap, bullcrap, bullcrap, heres an idea show some proof to back up what it is you are saying Like the phrase says "talk is cheap" well sonny you do a lot of talking. But however i did prove what i am saying if you read the website it says that It is possible that a black holes do produce very strong magnetic feilds to the point of 10^8 teslas........... prove yourself!!!!!
Nick
There are NO BLACK HOLES. That theory has effectively been disproven. tongue.gif

Nick -- LIGHT FALLS --

Nick
QUOTE (AirForceAcademyboy+Dec 9 2006, 01:06 AM)
WE DO HAVE EVIDENCE THAT THEY DO EXIST HERE IS A SITE THAT I THINK YOU ALL SHOULD LOOK AT ABOUT BLACK HOLES http://en.wikipedia.org/wiki/Black_holes#Evidence

The only evidence there is the result of the flaw in the theory. I can disprove it AFA. tongue.gif

MITCH RAEMSCH



Nick
QUOTE (AlphaNumeric+Dec 2 2006, 09:55 AM)
The proof is found in the maths of relativity. All causal geodesics point inwards due to space-time curvature. You are trying to use your 3d Euclidean physical intuition on a 4d non-Euclidean space-time.

If you are so amazing at relativity, you should be able to demonstrate to yourself that all lightcones point inwards by chaning from the coordinates (t,r,theta,phi) to (u,r,theta,phi) near the event horizon using the transformation v = t+r+2M ln|r-2M|.

This then gives you the metric form

ds^2 = -(1- 2M/r)dv^2 + 2dvdr + r^2 dP^2

dP^2 is S^2 sphere metric.

You can now see that at r=2M, all the metric coefficents are finite and therefore no problem exists at r=2M in these coordinates. You can also sketch the light cones (supressing the dP^2) to give you this diagram :

User posted image

The time axis points up and increasing r points to the right. You can see on the right hand side of the diagram, the two sets of lines/curves are almost perpendicular to one another, this is to be expected since it's almost flat space-time. As you get closer to the r=2M line (the vertical line to just to left of centre) one of the sets of lines begins to warp anbd bend the other way. This gets so bad that for r=< 2M, there is no line which points outwards, they all point to the left (to the singularity).

If there's no line pointing to the right, there's no path of escape, light and any other particle is forced inwards.

The paths light takes inwards are not altered, you can see that there's a set of lines which remain parallel, that's the paths of light fired straight in to the black hole. But the opposite paths, the ones which would normally take the photon out of the black hole are bent inwards, there is no escape, despite being a way in.

If you don't think this is right, show a flaw in the maths which derived it. Yes, it's hard to understand conceptually at first, but none the less, that is the result.

Your diagram has led you down the wrong road.

Your Black Hole roads seem to bend alfy. But any geodesics would have to be straight. They would always go straight toward the singularity. Therefor light emitted straight outward would not bend and can always get out.

Show me where I am wrong Alfalfa. biggrin.gif

What mechanism is going to drag the light backwards?

MITCH RAEMSCH -- LIGHT FALLS --
philip347
Air Force Boy, if I'm reading what you have said right, then your attacking yourself.

Try the primary Hawking BH investigative group?
Shemi
Newtonian physics, relativity, and most accepted candidate TOEs allow for BHs

A Black hole in newtonian physics is not necessarily completely black, black holes only become truely black when one considers relativity and the constant speed of light.

Due to the speed of light being constant, and the inability of anything that travels faster than it (other than perhaps Tachyons which are generally considered unreal), nothing can escape a BH due to the fact that nothing can reach escape velocity inside of the event horizon because the event horizon is the radius at which the escape velocity is the speed of light.

Photons may not have mass, however, they do have momentum and gravity acts on this momentum producing gravitational lensing.

All that is needed to prove BHs exist is the detection of a sufficiently large mass that is in a small enough volume. Since neutron stars exist there doesn't seem to be any reason why BHs can't (especially given the evidence that indicates the prescence of BHs such as the supermassive ones in the central regions of most galaxies).
AirForceAcademyboy
QUOTE (philip347+Dec 10 2006, 02:33 AM)
Air Force Boy, if I'm reading what you have said right, then your attacking yourself.

Try the primary Hawking BH investigative group?

no that is not what i am saying there is a site that you click on it says Magnetic feilds ok ill just give you the direct site http://en.wikipedia.org/wiki/Magnetic_field no scroll down to the bottom of the article it will say that white dwarfs. nuetron stars, and black holes produce a very large magnetic feild. Now going back to my first theory that is it possible for a black hole to somehow trigger hyperspace seeing as how the hyperspace thoery say that a strong enough magnetic force can open hyperspace but even though the black hole puts out 10^8 teslas of magnetic force it still not strong enough to do so but when you add the gravitational force and the pressure of a black hole then it just might be able to do so. With that amount of Magnetic force it can slow a SUNS rotation to 100- 1000 times given it takes millions of years the fact that it can do that is still quite powerful thats more then a magnetic star can do...
AlphaNumeric
QUOTE (philip347+Dec 9 2006, 02:24 AM)
these densities however must nestle in space, which is manifolded and has many accessways in and out of it.

You cannot 'manifold' something. There is no verb 'to manifold'.
QUOTE (Nick+Dec 10 2006, 03:31 AM)
Your diagram has led you down the wrong road.

Your Black Hole roads seem to bend alfy. But any geodesics would have to be straight. They would always go straight toward the singularity. Therefor light emitted straight outward would not bend and can always get out.

Show me where I am wrong Alfalfa.  biggrin.gif
The geodesics ARE straight in space, but I'm plotting a space-time diagram, so that any accelerating object (and objects falling into a black hole do accelerate) will follow a curved path in most coordinate systems. The objects falling inwards in this coordinate system are accelerated and hence the line is curved.

Nick, I even provided the maths with which I (or rather, people like Eddington) derived the geodesic paths. If you're going to say I'm wrong, you can't just wave your arms, you're going to have to prove my maths wrong. It's not just physical intution here which is giving me these results, it's cold hard logic!
QUOTE (Nick+Dec 10 2006, 03:31 AM)
What mechanism is going to drag the light backwards?
Curved space-time! It doesn't drag light backwards, it means that within an event horizon, every path light can possibly follow points inwards. Yes, it's hard to get your head around, hence why the maths is important. You'd understand why r must descrease if you understood what time and space-like vectors were. Within an event horizon the radial coordinate r is time-like!
Nick
QUOTE (AlphaNumeric+Dec 10 2006, 08:51 PM)
The geodesics ARE straight in space, but I'm plotting a space-time diagram, so that any accelerating object (and objects falling into a black hole do accelerate) will follow a curved path in most coordinate systems. The objects falling inwards in this coordinate system are accelerated and hence the line is curved.

Geodesics are space and TIME. Please show how straight geodesics can drag light backward Alfalfa. biggrin.gif

Also to what degree are objects accelerated inside a black hole? tongue.gif

MITCH RAEMSCH -- LIGHT FELL ON YOUR HEAD ALF --
AlphaNumeric
QUOTE (Nick+Dec 10 2006, 10:11 PM)
Geodesics are space and TIME. Please show how straight geodesics can drag light backward Alfalfa.  biggrin.gif

Geodesics are lines in space and time, not literally space and time. The geodesics only appear straight in 3d Euclidean space, in 4d Minkowski space-time they curve, meaning light moves into the singularity.

Nick, you are asking me these questions as if to say "Ah ha, I've got you! Here's something which doesn't make sense in your theory" but that isn't what you're actually doing at all. You are demonstrating that you're attempting to grasp a theory which you've neglected to learn the basics of. You are trying to use an inadequate understanding to somehow show what I'm saying is wrong. Believe me, much clever and more educated people than you have tried and they didn't manage it.
QUOTE (Nick+Dec 10 2006, 10:11 PM)
Also to what degree are objects accelerated inside a black hole?
I suggest learning basuc Schwarzchild geometry in general relativity, thus allowing you to compute such things as r' and t' in and around the black hole. Again, you will not understanding it without more basic learning because such notions are time and space-like vectors mean nothing to you.
QUOTE (Nick+Dec 10 2006, 10:11 PM)
LIGHT FELL ON YOUR HEAD ALF
Nope, just my explaination and general relativity seems to be going over your head.

Don't kid yourself Nick, you aren't shaking the foundations of relativity here, you're just making yourself appear like the smug, over confident but actually underinformed and ignorant guy you are.
Nick
QUOTE (AlphaNumeric+Dec 10 2006, 09:36 PM)
Geodesics are lines in space and time, not literally space and time. The geodesics only appear straight in 3d Euclidean space, in 4d Minkowski space-time they curve, meaning light moves into the singularity.


Caught you alfalfa. Minkowski space-time doesn't curve Alfy.

Geodesics are space and TIME. If they go straight in they also go straight out.
PLEASE SHOW how light emitted backwards along these geodesics gets dragged back.


Also please explain to what degree are objects accelerated inside a black hole tongue.gif

MITCH RAEMSCH -- LIGHT FELL ON YOUR HEAD ALF --
AlphaNumeric
QUOTE (Nick+Dec 10 2006, 10:53 PM)
Caught you alfalfa. Minkowski space-time doesn't curve Alfy.

That's true, perhaps I should have said 'Lorentzian'. I was referring to the fact that distance isn't measured as ds = dx + dy + dz but as ds = dt- dx + dy + dz, so you can't use your usual notions of distance. It's bad enough in Minkowski space-time, in extremely curved space-time like that of a black hole, you're going to get it wrong. And you do.
QUOTE (Nick+Dec 10 2006, 10:53 PM)
PLEASE SHOW how light emitted backwards along these geodesics gets dragged back.
I have, I even gave you the maths for the derivation. The fact you're deliberatelly ignorant of it doesn't mean the derivation and logic isn't there, you're just blind to it. This is taught to hundreds of thousands of people a year, across the planet, and huge quantities of research are done on it. As I said, cleverer and more educated people than you have tried to find flaws in this work and failed. You're not doing any better, you're just making an idiot of yourself.
QUOTE (Nick+Dec 10 2006, 10:53 PM)
Also please explain to what degree are objects accelerated inside a black hole
As explained, it's complicated and probably beyond you. Actually, screw that, it IS beyond you. I've given you the expression for the space-time interval, surely with all you're knowledge of relativity you know how to work it out wink.gif

You still haven't given a flaw in my mathematical derivation. All you're doing is flailing your arms laugh.gif What's the matter, the maths over your head ? wink.gif
Nick
TRy to be smart and Answer the following questions alfalfa:

What is the speed of freefall at the event horizon?
What is the curved timerate there Alf?

MITCH RAEMSCH -- LIGHT FELL ON YOUR HEAD --

AlphaNumeric
QUOTE (Nick+Dec 10 2006, 11:18 PM)
What is the speed of freefall at the event horizon?
What is the curved timerate there Alf?

Both are ill defined questions. From whose perspective? From the person falling into the black hole the timerate is the same timerate you are experiencing now and the velocity of freefall can be taken to be almost anything you want, depending on whose perspective and what frame you're in.

It is meaningless to ask such a question from the point of view of a distant observer because they can never know the motion of something at the moment of crossing the event horizon or later, because the information cannot reach them.

Again, it's not hard to give quanative answers Nick and as someone you proclaimed himself to be someone who understood relativity deeply, it should be within your grasp.

This thread on www.physicsforums.com might help you understand, particularly the notion that light speed violation is a local fact, not a global one (as the universe's expansion demonstrates).

Nick, if you want to contribute something to relativity rather than just asking somewhat innane questions constantly, enroll at your local college and learn some vector calculus, linear algebra , special relativity and then general relativity. You'll be infinitely more competent at relativity then and won't be asking these questions again and again and again. It's for your benefit, not mine. I don't care if you live the rest of your life in ignorance.
Nick
What is the speed of freefall at the event horizon?
What is the curved timerate there Alf?

QUOTE (AlphaNumeric+Dec 10 2006, 10:43 PM)
Both are ill defined questions. From whose perspective?


What is vague about freefall speed?
From whose perspective Alf?
The singularities. The destiny.
What is the speed of freefall?

What is vague about curved timerate?

Einstein said that there would be a curved timerate that matter would exeprience anywhere in curved space-time. What is matter's curved timerate there at the event horizon Alf?

MITCH RAEMSCH -- LIGHT FALLS --
AlphaNumeric
Time and speed are relative Nick, so of course you have to define from whose perspective they are measured. Hence asking "How fast does an object move?" or "What is the rate of time's passage?" is ill defined in relativity without "from the perspective of observer..." and then defining the position and motion of an observer.

A second problem also arises in general relativity (the above problem applies to both special and general relativity) in that some notions are only local notions.

As I illuded to before, the expansion of the universe allows for some things to move away from one another 'faster than light'. The objects 15 billion light years from us are being carried away from us faster than light due to the expansion of space-time. It would seem the statement 'Nothing moves faster than light from an observers pespective' isn't applying?! But why? Because it's a local statement and the expansion of space-time is not a local phenomena, it's more a global statement, while the light speed limit is a local one.

This effect, though slightly less intuitive, is what makes an object experience infinite redshift, gravity is making it appear an object moving across the event horizon is moving at the speed of light, from the point of view of a distant observer, but that's a global situation. Locally, the person falling in is moving slower than light. Similarly with the passage of time.

Much of black hole mechanics suffers from such complications and you, Nick, have fallen foul of them a number of time, such as geodesic paths and time passage on or in the event horizon. You gave me a negative feedback comment saying I'm a 'maths nut' but it's through maths that I'm able to understand these things, that allow me to plot those space-time curves, that allow me to see that some 'paradoxes' with black holes are not literally paradoxes (just as the Twin Paradox isn't actually a paradox in special relativity). You, like Zephir, rely too much on misguided intuition and not enough on logic. After all, intuition is another way of saying "I expect all new things to behave like things I've seen before". I think you'll agree that's a poor way of learning new things because some things are totally different from past experience. Black holes and their details are just such an example.

As I said, you'd do yourself a favour if you learnt some of this formally, not just flailing your arms and thinking yourself a genius. You're not. You're very misguided and almost always wrong in your understanding of relativity. Frankly I don't see why you think being ignorant of a subject makes you an expert at it, that's utterly flawed logic. It's no skin off my nose if you don't learn relativity, my ability and understanding in it are not contingent on you knowing any of it and noone like me will give a **** what you say because we know you're just an ignorant crank while you remain in your current state of knowledge. I don't see what half the members of this forum's problem with learning it! Even in spare time, it's not too hard to learn enough maths and physics to be 'competent' at the basics of general relativity. Someone with decent physics ability could do it in under a year if they put their mind to it!
Nick
What is the curved timerate for matter when at the event horizon?
What is the speed of freefall of that matter relative to the singularity?

MITCH RAEMSCH -- LIGHT FELL ON YOUR HEAD ALFALFA --

AlphaNumeric
The timerate for an infalling object from it's own point of view is constant, just as you experience time in a 'normal' way right now. That's the whole point of proper time.

As for the speed, asking about it from the singularity's perspective is like asking for the photon's point of view of distance or time, it's meaningless, given it's the point where physical quantities are no longer meaningul, hence why it's a proper singularity, unlike the event horizon which has meaningful concepts of time and space in some coordinates.

Once again, you ask innane questions because you don't understand. You try a cheap shot at me by saying " LIGHT FELL ON YOUR HEAD ALFALFA" but that's just because you've not actually got anything else but cheap shots.

A bit of time spent learning the required maths and physics and you'd be able to answer these questions (or realise their illdefined nature) yourself. This is pretty basic on the scale of general relativity.
Nick
QUOTE (AlphaNumeric+Dec 10 2006, 11:44 PM)
The timerate for an infalling object from it's own point of view is constant,
As for the speed, asking about it from the singularity's perspective is like asking for the photon's point of view of distance or time, it's meaningless,

The curved timerate for matter is continously changing during freefall. Time is not constant in gravity. The definition of EVENT HORIZON is where time ends. If time ends in gravity then there is no time to be observed. There is NO PROPER TIME THERE ALFALFA.

Why can't there be a motion of freefalling matter with regards to the singularity? After all it will arive there at a certain speed. What speed is that Alfalfa?

MITCH RAEMSCH -- LIGHT FALLS ON ALFALFA'S HEAD --


AlphaNumeric
QUOTE (Nick+Dec 11 2006, 01:02 AM)
The curved timerate for matter is continously changing during freefall. Time is not constant in gravity.

The time rate from the point of view of a distant observer is not constant, but for the person falling it it is constant. It's everyone else who isn't!
QUOTE (Nick+Dec 11 2006, 01:02 AM)
he definition of EVENT HORIZON is where time ends.
No it isn't, it's more complicated and well defined than that. Such things as null Killing vectors probably don't mean anything to you. You are making up your own definition then claiming there's a problem. You're right, you're making it up! laugh.gif
QUOTE (Nick+Dec 11 2006, 01:02 AM)
There is NO PROPER TIME THERE ALFALFA
Just saying it isn't making it true. Kruskall coordinates clearly demonstrate otherwise.
QUOTE (Nick+Dec 11 2006, 01:02 AM)
Why can't there be a motion of freefalling matter with regards to the singularity? After all it will arive there at a certain speed. What speed is that Alfalfa?
*sigh* Once again, you neglect to say from whose perspective. There is motion of freealling matter, you just can't describe it from the point of view of the singularity (ie how fast it sees things fall towards it). You have to define a series of points within the event horizon and then you can only define the velocity of the infalling object when it's above those points.

Still just grasping at straws there Nick. I'm still waiting for you to find a flaw in my maths derivation. What's the matter, too complicated for you? If you can't refute that, all your armwaving is meaningless because all my comments derive from the logical results of the maths.
Nick
QUOTE (ALFY+)
The time rate from the point of view of a distant observer is not constant, but for the person falling it it is constant. It's everyone else who isn't!


You got it wrong. Propertime is not a timerate at all but mearly an appearence ALF.


QUOTE (ALFY+)
You are making up your own definition then claiming there's a problem. You're right, you're making it up!


Look at the text ALF. The definition of AN EVENT HORIZON is an end to time.


QUOTE (ALFY+)
There is motion of freealling matter, you just can't describe it from the point of view of the singularity


And Why not ALFALFA?


If curved time for matter ends there is NO MORE TIME WHATSOEVER.

MITCH RAEMSCH -- LIGHT FELL ON YOUR HEAD ALF --
AlphaNumeric
QUOTE (Nick+Dec 11 2006, 02:08 AM)
You got it wrong. Propertime is not a timerate at all but mearly an appearence ALF.

You might want to open a textbook on relativity and look up what the 'tau' coordinate represents.
QUOTE (Nick+Dec 11 2006, 02:08 AM)
Look at the text ALF. The definition of AN EVENT HORIZON is an end to time.
Actually, it's an event horizon is the Killing horizon of an asymptotically flat space-time, a theorem proved by Hawking and which appears in a copy of some Cambridge lecture notes based on lectures by Hawking, but what would he know eh? wink.gif

Just making up your own misguided definition and then proclaiming it's flawed doesn't prove anything. Me saying "2+2=5, therefore maths is wrong!!" doesn't prove anything wink.gif
QUOTE (Nick+Dec 11 2006, 02:08 AM)
And Why not ALFALFA?
Because, as I said, it's a physical singularity. You cannot give a notion of time from it's point of view. You can however at the event horizon because it's only a coordinate singularity, in that by a good choice of coordinates, everything works fine. Spherical coordinates don't work everywhere, there's no reason to expect them to, they don't work everywhere even in flat space-time! Kruskal coordinates work fine on the event horizon and give time a well defined meaning there. I've told you this already so you're just making yourself appear stupid by repeating the same questions as if repetition backs up your argument.
QUOTE (Nick+Dec 11 2006, 02:08 AM)
If curved time for matter ends there is NO MORE TIME WHATSOEVER.
Hence why you can't get the singularity's point of view of time, it doesn't exist properly there. The event horizon isn't a physical singularity though, as has been explained to you probably more than a dozen times now.
Nick
You're smart ALFi but you don't have the truth.
If the event horizon is not an end to time alfalfa explain the redshift there.

QUOTE (ALF+)
You might want to open a textbook on relativity and look up what the 'tau' coordinate represents.


If Tau represents time it is an appearence.
QUOTE (ALFALFA+)
Actually, it's an event horizon is the Killing horizon of an asymptotically flat space-time,


I see you say flat space-time at the event horizon. You've been Had.

QUOTE (ALFi+)
as I said, it's a physical singularity. You cannot give a notion of time from it's point of view. You can however at the event horizon because it's only a coordinate singularity,


Why can't there be relative motion towards it?

QUOTE (ALFI+)
The event horizon isn't a physical singularity.


You have MATH FUDGE then ALF.
Again, If the event horizon is not an end to time alfalfa explain the redshift there.
AlphaNumeric
QUOTE (Nick+Dec 11 2006, 02:38 AM)
If the event horizon is not an end to time alfalfa explain the redshift there.

Because it is the surface at which the paths of light rays cannot escape outwards, thus as an infalling object emits light, the light has to fight against stronger and stronger gravitational effects, causing more and more redshift.

You are using the WRONG definition for the event horizon, not to mention you are still attatched to the incorrect nothing that time stops from everyone's perspective on the event horizon!! You must detatch yourself from the idea that everyone sees the same thing in relativity!
QUOTE (Nick+Dec 11 2006, 02:38am)
If Tau represents time it is an appearence
That's meaningless unless you define what 'an appearance' is.
QUOTE (Nick+Dec 11 2006, 02:38am)
I see you say flat space-time at the event horizon. You've been Had.
No, I said asymptotically flat. Obviously you're not familiar with such a critical concept in physics. Asymptotically flat space-time means that as you move AWAY from the black hole to a very large distance, the space-time tends to flat space-time. You can see in the Schwarzchild metric that as r-> infinity, ds^2 -> dt^2 - dx^2 - dy^2 - dz^2, that of Minkowski space-time. This is what 'asymptotically flat' means. There's other notions like 'asymptotically Anti deSitter', which is the metric you'd want in a universe which has a cosmological constant like ours appears to.

You think 'I've been had' because you are unfamiliar with relativity. It's your ignorance which is 'had', not me.
QUOTE (Nick+Dec 11 2006, 02:38 AM)
Why can't there be relative motion towards it?
Read what I'm saying. You asked about motion from the perspective of the singularity, not about relative motion towards the singularity. You cannot measure motion from the point of view of the singularity, it's a physical singularity, it has no well defined concept of time or distance because it's a physical singularity. You can however measure motion from other points within the event horizon's perspective.
QUOTE (Nick+Dec 11 2006, 02:38 AM)
You have MATH FUDGE then ALF.
You mean Hawking, Penrose, Einstein, Schwarzchild, Kerr, Newman, Riessner, Nordstrom and every other general relativity researcher for the last 90 years has a fudge. This isn't my working, it's theirs!

You think that the event horizon is defined as 'time stops', it isn't. It's defined by Killing factors which are a special type of 4 vector in differential geometry. In some coordinates they are not nice on the event horizon, but in others they were perfectly well defined. Because of this, your definition isn't the right one. Instead you have to use 'coordinate free' definitions, just as the one I quoted last time that Hawking uses!

As I said, you make up a definition and then complain it's wrong! Do you not see the false logic in that?
Nick
SHOW THE MATH FOR TIRED LIGHT.

QUOTE (ALFALFA+)
Because it is the surface at which the paths of light rays cannot escape outwards, thus as an infalling object emits light, the light has to fight against stronger and stronger gravitational effects, causing more and more redshift.


NO ALFALFA THERE IS NO TIRED LIGHT. THAT IS NOT GENERAL RELATIVITY. THAT IS YOUR EXCUSE ALONE.

QUOTE (ALFALFA+)
No, I said asymptotically flat.

SO WHICH IS IT FLAT OR CURVED?

QUOTE (ALFALFA+)
You can however measure motion from other points within the event horizon's perspective.


BY WHAT MEASURING DEVICE?
HOW FAST ARE THINGS FALLING AND WHAT IS THEIR CURVED TIMERATE?

QUOTE (ALFALFA+)
Because of this, your definition isn't the right one. Instead you have to use 'coordinate free' definitions, just as the one I quoted last time that Hawking uses!


IF TIME DOESN'T STOP THEN HOW FAST IS IT GOING?

QUOTE (ALFALFA+)
As I said, you make up a definition and then complain it's wrong! Do you not see the false logic in that?


IF TIME DOESN'T END AT THE EVENT HORIZON THEN WHY THE INFINITE REDSHIFT THERE ALFALFA?

SHOW ME WHERE I AM WRONG. SHOW ME THE MATH FOR TIRED LIGHT. biggrin.gif


MITCH RAEMSCH -- LIGHT FALLS --
AlphaNumeric
Nick, I'm perfectly capable of reading your posts without you USING YOUR CAPS LOCK ALL THE TIME. You don't make your argument more valid, you make yourself look like a 5 year old who is upset.
QUOTE (Nick+Dec 11 2006, 09:14 PM)
NO ALFALFA THERE IS NO TIRED LIGHT. THAT IS NOT GENERAL RELATIVITY. THAT IS YOUR EXCUSE ALONE.

Light is redshifted when it moves out of a gravitational well, that's an experimental fact (see the Pound Rebka experiment in the 50s). That isn't 'my excuse', it's a fact of the universe and one of the experimental predictions Einstein gave for general relativity.
QUOTE (Nick+Dec 11 2006, 09:14 PM)
SO WHICH IS IT FLAT OR CURVED?
This is another case of you showing your ignorance and thinking that you have trapped me. Look up what 'asymptotically flat' means.

Since I know you're too lazy or stubborn for that, I'll tell you. It means at large distances, the space-time tends to flatness, but close to the object with mass, it's curved. All black hole solutions to the Einstein Equations are asymptotically flat because for r-> infinity they go to flat space-time. You obviously ignored my explaination in my last post.
QUOTE (Nick+Dec 11 2006, 09:14 PM)
BY WHAT MEASURING DEVICE?
HOW FAST ARE THINGS FALLING AND WHAT IS THEIR CURVED TIMERATE?
For about the 10th time, it depends on where you are and from whose point of view. Why don't you learn the maths of relativity so you can answer this yourself?
QUOTE (Nick+Dec 11 2006, 09:14 PM)
IF TIME DOESN'T STOP THEN HOW FAST IS IT GOING?
From the point of view of observers inside the event horizon, time on the event horizon passes normally.
QUOTE (Nick+Dec 11 2006, 09:14 PM)
IF TIME DOESN'T END AT THE EVENT HORIZON THEN WHY THE INFINITE REDSHIFT THERE ALFALFA?
Because the event horizon isn't defined as 'the surface where time ends', it's a Killing surface, where the Killing vectors are null (Killing is someone's surname, it doesn't imply the verb 'to kill').

The ever increasing redshift is due to the photon having to climb out of more and more powerful gravity potentials. Simple energy conservation tells you the light gets redshifted. As I said, it's an experimental fact.

And before you ask, there is no photon which is 'infinitely redshifted', it's just the amount of redshift is not bounded above. Remember, light emission is a discrete process. I know I've discussed this with you before and explained why a discrete emission process means no 'infinitely tired light' but you obviously either don't remember, didn't understand or don't care.

For repetition sake, here's the explaination why no photon is infinitely redshifted :

An object is falling into a black hole and giving off light. The emission process is discrete because if it was continous, the object would be emitting infinitely many photons in any unit of time. Therefore, there is a non-zero amount of time, T, between photon emissions (T can be variable, but it's sufficent to consider it constant in this case). As the object falls inwards, the light it's giving off becomes more and more redshifted by gravity, from the point of view of a distant observer. Consider the few moments before it crosses the event horizon. The last but one photon to be emitted before it hits the EH is red shifted an enormous but finite amount. This is because T>0, so the distant the object is above the event horizon, d, is greater than 0 too. No problems with any infinities. The next photon is emitted when the object is on the event horizon. Light can only move on null geodesics. If it's emitted directly away from the event horizon it moves along a geodesic which has constant distance from the event horizon! Since redshifting is related to the change in radial distance and there is no change, the photon isn't redshifted at all! No problems with any infinities (0 is a lot nicer than infinity). The next photon to be emitted will be within the event horizon and by space-time curvature will end up in the singularity.

No problems at all. No photon is redshifted an infinite amount, so there's no 'energyless light', all the photons are either redshifted a finite amount, not redshifted at all or blue shifted.

I know I've said that to you before, so perhaps you'll remember it this time!
QUOTE (Nick+Dec 11 2006, 09:14 PM)
SHOW ME WHERE I AM WRONG. SHOW ME THE MATH FOR TIRED LIGHT.
I've already explained to you why your definition of the event horizon isn't correct. I've explained why the event horizon isn't 'the end of time'. I've explained how light paths behave around the event horizon.
Nick
QUOTE (ALFALFA+)
Light is red shifted when it moves out of a gravitational well,


PLEASE SHOW THE MATH FOR TIRED LIGHT.
ACCORDING TO EINSTEIN LIGHT STARTS OFF RED SHIFTED AT EMISSION.
AND SHOW THE MATH FOR THE GRAVITATIONAL WELL. THERE ARE NO "HOLES" IN SPACE-TIME CURVATURE.

MITCH RAEMSCH -- LIGHT FELL ON YOUR HEAD ZEPH --
AlphaNumeric
QUOTE (Nick+Dec 11 2006, 10:08 PM)
ACCORDING TO EINSTEIN LIGHT STARTS OFF RED SHIFTED AT EMISSION.

No it doesn't. The redshifting happens as it moves away from the object with mass. It doesn't start redshifted, it redshifts over it's journey.

http://en.wikipedia.org/wiki/Pound-Rebka-Snider_experiment

Have a read.
QUOTE (Nick+Dec 11 2006, 10:08 PM)
AND SHOW THE MATH FOR THE GRAVITATIONAL WELL. THERE ARE NO "HOLES" IN SPACE-TIME CURVATURE.
It isn't a hole in space-time. A 'gravity well' is a way of saying 'a region with a lot of mass and thus has strong gravity'. Saying 'the photon climbs out of the gravity well' just means 'the photon moves away from the mass, towards a region of weaker gravity'.

Obviously you aren't familiar with common phrases in physics.
Nick
Light doesn't climb out of a "HOLE." And it doesn't change energy. It starts out that way.

The common phrases are WRONG. Show me where I AM WRONG. biggrin.gif

MITCH RAEMSCH
AlphaNumeric
QUOTE (Nick+Dec 12 2006, 12:46 AM)
Light doesn't climb out of a "HOLE."

It's a turn of phrase. It's related to such geometric analogies as representing a mass's gravity as a 'dent' in the fabric of space-time. It's not a literal hole. You seem to be making a habit of jumping to the wrong understanding then complaining that it's the theory at fault.
QUOTE (Nick+Dec 12 2006, 12:46 AM)
And it doesn't change energy. It starts out that way.
This is another discussion we're repeating. Your view violates conservation of energy. Your view has the energy (and hence relativistic mass) of the photon moving through a gravitational field without a change in energy. Just as a ball thrown upwards slows, in line with energy conservation, a photon fired upwards redshifts gradually in line with energy conservation.

If it started instantly redshifted, there are two problems. Firstly, energy conservation and secondly it would mean the photon knows when and where it's going to be absorbed. Are you saying that not only is energy not conserved but photons can tell the future? laugh.gif
QUOTE (Nick+Dec 12 2006, 12:46 AM)
The common phrases are WRONG. Show me where I AM WRONG.
Reread this thread, I've done it plenty of times.
Nick
THE REDSHIFT TO LIGHT AT EMISSION WAS NAMED AFTER EINSTEIN.
PROVE THAT THE EINSTEIN SHIFT IS WRONG ALFALFA.


QUOTE (ALFALFA+)
It's a turn of phrase. It's related to such geometric analogies as representing a mass's gravity as a 'dent' in the fabric of space-time. It's not a literal hole.


THEN THERE IS NO GRAVITY WELL; NO HOLE IN SPACETIME CURVATURE.

QUOTE (Alf+)
Your view violates conservation of energy.


NO. THAT IS YOUR PROBLEM. IF LIGHT IS REDSHIFTED IT STARTS OFF AND STAYS THAT WAY. SHOW THE MATH BEHIND TIRED LIGHT.

QUOTE (ALFALFA+)
If it started instantly redshifted, there are two problems. Firstly, energy conservation and secondly it would mean the photon knows when and where it's going to be absorbed. Are you saying that not only is energy not conserved but photons can tell the future?


ENERGY CONSERVATION IS YOUR PROBLEM. IT DOESN'T CHANGE.

MITCH RAEMSCH


AlphaNumeric
QUOTE (Nick+Dec 12 2006, 01:15 AM)
THE REDSHIFT TO LIGHT AT EMISSION WAS NAMED AFTER EINSTEIN.
PROVE THAT THE EINSTEIN SHIFT IS WRONG ALFALFA.

No, the redshift due to gravity (ie gravitational redshift) was named after Einstein. It wasn't an effect which occured at emission, but during the journey of the photon. This is all in relativity textbooks. Chapter 5, page 121 of Schutz's "A First Course in General Relativity" specifically talks about this effect.
QUOTE (Schutz+)
We predict that a photon climbing in Earth's gravitational field will lose energy (not suprisingly) and will consequently be redshifted
Emphasis added by me.

So I totally agree with the Einstein redshifting, it's just you aren't understanding it properly. It's a process which occurs as the photon moves against gravity.
QUOTE (Nick+Dec 12 2006, 01:15 AM)
THEN THERE IS NO GRAVITY WELL; NO HOLE IN SPACETIME CURVATURE.
I never said it was. You seem to be arguing again your own misunderstanding here Nick. The more you attempt to complain, the more you just argue with yourself and your lack of comprehension.
QUOTE (Nick+Dec 12 2006, 01:15 AM)
NO. THAT IS YOUR PROBLEM. IF LIGHT IS REDSHIFTED IT STARTS OFF AND STAYS THAT WAY. SHOW THE MATH BEHIND TIRED LIGHT.
It's not 'my problem', it's a fact of relativity.

The 'math' is that the ratio of energies of the emitted to absorbed photon is

E'/E = hv'/hv = 1 - gh/c^2 + O(1/c^4)

You'll find it in plenty of textbooks. Why don't you show your proof that light is instantly redshifted rather than the more logical point of view relativity has, where energy is conserved and so is causality. How do you explain the photon knows it's own future, ie where and when it will be absorbed?
QUOTE (Nick+Dec 12 2006, 01:15 AM)
ENERGY CONSERVATION IS YOUR PROBLEM. IT DOESN'T CHANGE.
No, energy conservation is your problem because your view point claims energy isn't conserved. How do you explain that away? Where does the energy go? Why isn't it conserved? Not only are you arguing against relativity, which gives a coherent, logic and experimentally verified description of this effect, you're arguing against pretty much the most fundamental premise of modern science!

Let's see. In a single post you managed to misinterpret gravitational redshifting, misunderstand a comment term in relativity (and end up arguing with your own misunderstanding), claim a problem your point of view has is actually problem and then claim one of the most experimentally verified, fundamental premises of science is wrong.

Wow, even for you that's doing pretty damn poorly Nick. I expect your next post will be similar to your last few, the caps lock stuck on, a bunch of ill thought out comments and a poor attempt to bring down relativity which is actually just your terrible understanding and, I would say I'm justified in saying this for once, your ignorance. You demand I produce the maths for this stuff when it's on hundreds of websites and in thousands of books.

http://en.wikipedia.org/wiki/Redshift#Gravitational_redshift would be a good place to start (you'll notice that page uses the phrase 'gravity well' too, but obviously not to mean a literal hole in space-time, you just aren't bothering to think about it).
Montec
Hello all

If we look at gravity as a right angle resultant force similar to the force that affects moving charged particles in a magnetic field then there is no need for gravity to escape a black hole.

smile.gif

Nick
This is all in relativity textbooks. Chapter 5, page 121 of Schutz's "A First Course in General Relativity" specifically talks about this effect.

QUOTE (Schutz)
We predict that a photon climbing in Earth's gravitational field will lose energy (not suprisingly) and will consequently be redshifted

WE? THAT WAS NOT EINSTEIN. THE EINSTEIN SHIFT OCCURES AT EMISSION. LIGHT STARTS OFF LESS ENERGETIC AND STAYS THAT WAY.

GO BACK TO YOUR BOOKS ALF. IT AINT EINSTEIN.

The 'math' is that the ratio of energies of the emitted to absorbed photon is
E'/E = hv'/hv = 1 - gh/c^2 + O(1/c^4)

WRONG. IT IS THE DIFFERENCE IN CURVED TIMERATE AT EMISSION THAT CREATES THE REDSHIFT. THAT IS THE EINSTEIN.

MITCH RAEMSCH
AlphaNumeric
QUOTE (Nick+Dec 12 2006, 01:58 AM)
WE? THAT WAS NOT EINSTEIN. THE EINSTEIN SHIFT OCCURES AT EMISSION. LIGHT STARTS OFF LESS ENERGETIC AND STAYS THAT WAY.

I'm afraid not Nick. Gravitational redshift in the way I have outlined it was one of Einstein's specific predictions about how to verify relativity was superior to Newtonian theory.

Can you find me a quote of Einstein where he specifically says that the redshift occurs the instant of emission and nothing changes after that? Somehow I doubt it, considering all the things I've pointed you to demonstrating relativity specifically says otherwise.
QUOTE (Nick+Dec 12 2006, 01:58 AM)
GO BACK TO YOUR BOOKS ALF. IT AINT EINSTEIN.
Considering both relativity textbooks I have sitting on my desk both say the same thing, as does Wikipedia, it would seem that I'm fairly justified in what I've been saying. Where's your evidence that relativity or Einstein say otherwise?
QUOTE (Nick+Dec 12 2006, 01:58 AM)
WRONG. IT IS THE DIFFERENCE IN CURVED TIMERATE AT EMISSION THAT CREATES THE REDSHIFT. THAT IS THE EINSTEIN.
You might be thinking of the difference in the g_00 values of the metric for the point of emission and the point of absorption. That is how you work out the change in frequency yes (it gives the same equation I just gave), but that doesn't mean the effect is instant at emission and never changes, it's a continously changing redshift whose total change you compute by comparing g_00 metric values. That is what Einstein said, he said nothing about the effect being done totally at point of emission and never changing again.

If you'd done a bit of relativity, you'd see what I'm say, but instead it would seem you've picked up a tiny snippet of information about relativity and due to a lack of understanding about the maths of relativity and a lack of proper physical understanding for the maths' implications, you've latched onto the initial guess you had as to the maths meaning and are not willing to change your mind.

Nick, you keeping saying "That's not Einstein" or "That's not what relativity says", but everytime you say that, you're wrong. I've got 2 textbooks to hand andseveral lecture course notes (along with the fact I actually went to those lectures) on relativity and they all contradict what you're saying, as do sites like Wikipedia.

Why don't you open a book on relativity and learn it, so that you do know what Einstein actually said and what his theory does or doesn't say? Besides, Einstein's comments and understanding (sometimes even about relativity) aren't always perfect. It took a while before he accepted the notion of black holes, but eventually he realised the results of people like Schwarzchild were correct. Why don't you do the same, actually learn something.

Oh, and stop typing in caps lock. It doesn't do you any favours, it just makes you appear more irrational.
Nick
THE REDSHIFT TO LIGHT IS NOT A RELATIVE. THE EINSTEIN SHIFT IS DUE TO SLOWER TIME AT EMISSION. AND CURVED TIMERATE IS AN ABSOLUTE.

GO BACK TO YOUR BOOKS ALF. I AM SORRY BUT YOU FAIL HERE.

PLEASE SHOW EINSTEIN'S WORDS ON TIRED LIGHT.
THERE ARE NONE.

LIGHT IS REDSHIFTED AT EMISSION. THAT IS HOW TO UNDERSTAND SPECTRUM SHIFT OF THE SUN.

THE SUN'S SLOWER TIME CAUSES THE EINSTEIN SHIFT TO LIGHT AT EMISSION.

YOU HAVE YET TO SHOW EINSTEIN'S WORDS FOR YOUR CASE.
I AM NOT AFRAID TO HERE THEM. BUT ALAS YOU WILL BE ARE UNABLE TO PRODUCE THEM FOR YOUR CASE. HE NEVER USED THE TERM.

MITCH RAEMSCH -- LIGHT FALLS --
N O M
I think you're finally getting through to Nick, Alphanumeric. He is having to resort to all capitals to try and make his point. No doubt foaming at the mouth too.
philip347
Nick does not understand the topolgy of black holes.
Nick
I UNDERSTAND THAT BLACK HOLES DON'T EXIST.

SUCH REASONING AS TIRED LIGHT IS SIMPLY WRONG.

BLACK HOLE PHYSICS FAILS AT THE EVENT HORIZON. I CAN SHOW WHY.

IT IS SIMPLY DEMONSTRATED.
AlphaNumeric
QUOTE (Nick+Dec 12 2006, 03:33 AM)
THE REDSHIFT TO LIGHT IS NOT A RELATIVE. THE EINSTEIN SHIFT IS DUE TO SLOWER TIME AT EMISSION. AND CURVED TIMERATE IS AN ABSOLUTE.

Yes it is due to the different values of the g_00 metric entry at point of emission compared to point of absorption. However, that doesn't say that the photon is redshifted instantly at emission and never changes.

Infact, since you can consider the point of absorption as a variable, you can see that redshift varies. To have it occur all at once breaks energy conservation and causality. You still haven't explained what happens to the energy or how the photon can know it's future if your point of view was right Nick.
QUOTE (Nick+Dec 12 2006, 03:33 AM)
PLEASE SHOW EINSTEIN'S WORDS ON TIRED LIGHT. THERE ARE NONE.
He didn't call it 'tired light', he called it gravitational redshift. You yourself said it's called 'the Einstein redshift', so obviously he had something to say on the matter, why else call it after him?

All the material I've given you is work Einstein did. As I said, the gravitational redshift of photons on Earth was an experiment he devised to test relativity over Newtonian theory and it was done in the 50s. It was the last one of the experiments Einstein devised to be done.
QUOTE (Nick+Dec 12 2006, 03:33 AM)
LIGHT IS REDSHIFTED AT EMISSION. THAT IS HOW TO UNDERSTAND SPECTRUM SHIFT OF THE SUN.
No, the shift is a continous effect which happens as the photon moves away from the sun. It's well understood and well modelled. It also is logically consistent with energy conservation and causality. Your claims aren't.
QUOTE (Nick+Dec 12 2006, 03:33 AM)
THE SUN'S SLOWER TIME CAUSES THE EINSTEIN SHIFT TO LIGHT AT EMISSION.
Nope, it's a continous effect as the photon moves away. This is what relativity predicts. You are claiming relativity says something it doesn't.

Please quote Einstein or a relativity textbook saying that the effect happens at the point of emission and not over the journey. I've quoted a textbook backing up my claims, now you do it. You'll be hard pressed to do it though.
QUOTE (Nick+Dec 12 2006, 03:33 AM)
YOU HAVE YET TO SHOW EINSTEIN'S WORDS FOR YOUR CASE.
I AM NOT AFRAID TO HERE THEM. BUT ALAS YOU WILL BE ARE UNABLE TO PRODUCE THEM FOR YOUR CASE. HE NEVER USED THE TERM.
You're right, he probably never did use the term 'tired light', but then I never said he did. You have been using the term 'tired light', not me, so it's you who is in the wrong in that case. Einstein obviously talked about gravitational redshift and his predictions were the ones I've been explaining to you.

For instance, http://archive.ncsa.uiuc.edu/Cyberia/NumRe...nsteinTest.html explains the various tests of relativity Einstein gave and mentions the redshift. It explains the photon loses energy as it climbs up against gravity, it's a continous process during it's journey. It also gives a picture of the 'gravity well' example. Remember, it's not a literal meaning, it's just good for understanding (well, for most people, obviously not you).

So far you've claimed that "Einstein said..." or "That's not what Einstein said..." and never backed up a single comment. I've given you links to places outlining Einstein's work and predictions and given you quotes from textbooks. You haven't demonstrated anything other than you don't actually know what Einstein said, you just make up what he said in your head.
QUOTE (Nick+Dec 12 2006, 03:33 AM)
SUCH REASONING AS TIRED LIGHT IS SIMPLY WRONG.
And I have explained to you why there is no infinitely redshifted photons. Partly redshifted photons are an experimental fact. Hence, there's no problem with photons losing energy as they move away from a planet, star or black hole. It's simple conservation of energy!
QUOTE (Nick+Dec 12 2006, 03:33 AM)
BLACK HOLE PHYSICS FAILS AT THE EVENT HORIZON. I CAN SHOW WHY.

IT IS SIMPLY DEMONSTRATED.
No, you demonstrate that you are utterly unfamiliar with the notions of multiple charts in any manifold based theory (of which relativity is one). As I outlined to you, even the flat 2d Euclidean plane can need 2 coordinate systems to describe it, because some coordinate systems are not universally valid (polar coordinates at r=0). It's nothing to do with the space itself, just your choice of description. Similarly, the event horizon is just where your choice of description breaks down so you change your description method.

Saying the event horizon is where black hole physics breaks down is akin to saying that because there's no Ancient Greek word for 'computer', computers don't exist. No, it's just you've choosen an inadequate method of description, so change languages, say to English, and then you can describe a computer. Similarly with black holes at the event horizon, you change coordinates (Eddington-Finklestein or Kruskal coordinate work perfectly well at the event horizon) and physics is perfectly well defined there.

You have a problem with it because you don't open your mind to the possibility you might be wrong, that you have misunderstand a rather complex, but well established, nothing of physics/maths.

The problems are all in your understanding here Nick. All your questions do is just uncover more and more of your stubborn intent never to learn anything. Asking questions is fine, we all had to ask things like this at some point to be able to understand such material, but when you ask the same ones again and again and again it just shows you don't want to learn, you want to preach. The problem is, you're demonstratably wrong.
mott.carl
alphanumeric

read my topic..on the violation of pt.

i go send you by e-mail,the mathematics of thaose my thoughts.be patient.

congratulation mott

i think that the BH,if possible a observer is near of the singulaty.that does the warp
of the all the curvatures of space-time,and does all the space-time of all the the
universe twisted within of a "point",where the "past" and "future" already would
be seen by these observers.
Nick
NO. THE EINSTEIN SHIFT IS NOT CAUSED BY LIGHT LOOSING ENERGY AS IT STRUGGLES AGAINST GRAVITY. IF IT HAS LESS ENERGY IT IS BECAUSE IT STARTS OFF THAT WAY.

MITCH RAEMSCH
AlphaNumeric
Where is your derivation of this claim? I've given numerous references and links which show otherwise, that it's actually a continous shift.

How do you explain your view contradicts the conservation of energy? What about causality?

You're going to have to give some evidence for your claims because you're claiming that some very well established results are false.
Nick
NO.THERE IS NO CONFLICT IN THE RESULTS. YOUR TIRED LIGHT INTERPRETATION IS NOT WHAT THE EINSTEIN SHIFT IS ABOUT.

MITCH RAEMSCH -- LIGHT FELL --
AlphaNumeric
QUOTE (Nick+Dec 12 2006, 10:44 PM)
NO.THERE IS NO CONFLICT IN THE RESULTS.

Aside from violating causlity and conservation of energy you mean?
QUOTE (Nick+Dec 12 2006, 10:44 PM)
YOUR TIRED LIGHT INTERPRETATION IS NOT WHAT THE EINSTEIN SHIFT IS ABOUT.
I've provided links and citations from books to demonstrate it is what the Einstein shift is about. I asked you before but you ignored me, where's your evidence Einstein said otherwise? That it's not a continous process?

That's right, you have none laugh.gif
Nick
Your books are wrong alf. The Einstein shift isn't Tired Light.

Please show Einstein's derivation.

Mitch Raemsch -- LIGHT FALLS --
RMC
Nick -
QUOTE
Black holes don't exist.


Saying that something does or doesn't exist when little is known about it is an act of stupidity. In this case there is logic behind black holes which makes your statement sound even worse.
Nick
I CAN SHOW WHY RMC.
*vanadesse
QUOTE (Nick+Dec 12 2006, 09:03 PM)
I CAN SHOW WHY RMC.

Well why don't you get on with it??
Nick
I HAVE DEMONSTRATED IT VAN.

WHERE TIME SLOWS LIGHT SLOWS. AS LIGHT WERE TO APPROACH THE SURFACE OF A BLACK HOLE IT WOULD SLOW DOWN AND SLOW ALL THE WAY DOWN AND PARK ITSELF ON THE EDGE.

IF RMC WANTS MORE I WILL WAIT.

*vanadesse
QUOTE (Nick+Dec 12 2006, 09:21 PM)
WHERE TIME SLOWS LIGHT SLOWS.

Actually it doesn't. If it did, we would measure the speed of light to be different.
QUOTE
AS LIGHT WERE TO APPROACH THE SURFACE OF A BLACK HOLE IT WOULD SLOW DOWN AND SLOW ALL THE WAY DOWN AND PARK ITSELF ON THE EDGE.

No it wouldn't - do you have any common sense at all?
Nick
THIS IS THE REASON FOR THE SHAPIRO EFFECT.
*vanadesse
QUOTE (Nick+Dec 12 2006, 09:35 PM)
THIS IS THE REASON FOR THE SHAPIRO EFFECT.

dry.gif

You are correct, but only relative to a separate reference frame. If you have a black hole, light doesn't just slow down. It looks like it slows down to an outside observer. But in reference to itself or the black hole, it is still moving at the speed of light into the black hole. There is no paradox.
Nick
LIGHT MOVES SLOWER IN A SLOWER TIME.
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