High speeds - becoming a black hole

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Do you see that actually you are not responding to some questions I ask, but instead you do something else?
 

Hans de Vries

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aaroman said:
Do you see that actually you are not responding to some questions I ask, but instead you do something else?
My interpretation of:

aaroman said:
Probably a lot of things go against special relativity when there are things that have gravitation, isn't it?
Was that your were just casting doubt on my reply. If you instead want to
change the subject of the discussion then maybe you should make that clearer.


Regards, Hans
 
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Ok, here was the last question with no reply:
Ok, so the light will be Doppler shifted. That means more energetic photons. Does that mean stronger gravity from them? If that is true, just accelerating enough we could obtain gravity as strong as we want (in theory).
 

pervect

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aaroman said:
Well, I think that is already established that what one sees from one reference system is not necessarily what one sees from another.
I don't expect "to see us crushed", I expect to not see us at all.
You are apparently grossly misunderstanding the answer that several different people (and the written FAQ written specifically to answer the question you just asked) are all trying to give you.

Physical reality is independent of the coordinates used to describe it. If person A shoots person B on a moving train, person B is dead in all frames of reference.

He is not alive in some frames, and dead in others. He is dead in all frames.

It does not matter if person C is watching on the train, or person D is watching from the station - person B is dead. (C and D may disagree on the time of death, but they won't disagree that B dies).
 
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Physical reality is independent of the coordinates used to describe it. If person A shoots person B on a moving train, person B is dead in all frames of reference.
Not at the same time, though. Fore some is dead, for some is still living. For some is still unborn, so no, person B is not dead in all frames.
Going back to gravity from this analogy, in one frame the "force" felt could be very weak, in another one could be very strong. One frame just has to accelerate enough :)
I'm not questioning physical reality, but the appearance. If I'm not mistaken, for an outside observer, a falling object will never pass through the event horizon of a black hole. That is not true for the object that falls in. It passes through the event horizon in a finite time. So in some sort of way, for one observer B never "dies", for the one that falls though it's completely another matter.

PS I just looked over some papers: either they are wrong, or the affirmation that a black hole is a black hole in all reference frames which was presented on this thread is wrong. Talking abaout the Schwartzchild solution, is says something like that:
one can discover different coordinate frames in terms of which no singularity is seen. ... Introduce new coordinates ("Kruskal coordinates")
Well, I think I gave enough. It's from "Introduction to General Relativity" by G. Hooft. So what's the truth here? Is it a black hole or not in all coordinates?
 
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Physics Monkey

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aaroman said:
PS I am your friend :) But when somebody tells me that things look the same in two frames of refference, when obviously do not, I don't believe.
Wonderful! However, you have taken the mantra "everything is relative" too far. Black holes are not relative (at least not in the sense that I think you want them to be). No matter how fast you move, the components of the curvature are always finite if there is no black hole, and the components are always divergent at some point if there is a black hole. Also, like prevect said, no matter how fast you move, light beams that escape in one frame will still escape in every other frame hence no black hole in one frame means no black hole in every frame.

Regarding your question about the doppler shifted photons, the gravitational field is the same! What do I mean by this? If you make a transformation of coordinates corresponding to a Lorentz transformation then all that happens is that the components of [tex] R_{\alpha \beta \gamma \delta} [/tex] (or any other tensor) get scrambled up. The curvature invariant that I defined earlier is the same in any frame. Geodesics still deviate in the same way as before, so no true new gravitational effects have been produced (although, of course, the coordinates you assign to geodesic and its tangent vector change).
 
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no black hole in one frame means no black hole in every frame
Well, I just edited previous post, I found a paper that says that a singularity can be no singularity in another coordinate frame, so they seem relative if that is true.
Regarding your question about the doppler shifted photons, the gravitational field is the same!
Energy and momentum is different. Why is the gravitational result the same?
The curvature invariant that I defined earlier is the same in any frame
The curvature is invariant, but the force one feels might be different, or else no matter of the reference frame, all would feel the same gravity, which is not true. I tested this experimentally :)
 

Physics Monkey

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aaroman said:
Well, I just edited previous post, I found a paper that says that a singularity can be no singularity in another coordinate frame, so they seem relative if that is true.
You have confused a so called coordinate singularity with a true space time singularity. The traditional Schwarzschild metric is
[tex]
ds^2 = - \left(1 - \frac{2 M}{r}\right) dt^2 + \frac{1}{\left(1 - \frac{2 M}{r}\right)}dr^2 + r^2 d\Omega^2
[/tex]
and there is clearly a coordinate singularity at [tex] r = 2 M [/tex], the event horizon. However, the components of the curvature are perfectly well defined here so there is no actual singularity. The apparent singularity is an artifact of the coordinate system. The Kruskal-Szekeres coordinate system is one way of removing this artifical coordinate singularity. However, the singularity at [tex] r = 0 [/tex] is a true spacetime singularity and cannot be removed by a clever choice of coordinates. The curvature diverges at [tex] r = 0 [/tex].
 
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Now I might have a problem with points in physical reality, but for the rest I get it.
 

pervect

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aaroman said:
Not at the same time, though. Fore some is dead, for some is still living. For some is still unborn, so no, person B is not dead in all frames.
Going back to gravity from this analogy, in one frame the "force" felt could be very weak, in another one could be very strong. One frame just has to accelerate enough :)

I'm not questioning physical reality, but the appearance. If I'm not mistaken, for an outside observer, a falling object will never pass through the event horizon of a black hole. That is not true for the object that falls in. It passes through the event horizon in a finite time. So in some sort of way, for one observer B never "dies", for the one that falls though it's completely another matter.
You are correct that a falling object will pass through the event horizon at t=infinity in the coordinate system of the outside observer, and that it will pass through the event horizon in a finite proper time as measured by its own clock. This happens because the coordinate transforms are singular - specifically, it is the Schwarzschild coordinates that are singular in this cae.

If you look at the more detailed post I gave earlier, response #12

https://www.physicsforums.com/showpost.php?p=825259&postcount=12

I mentioned that it was important that the coordinate transforms were not singular.

The result of this is that because the coordinate transforms aren't singular, a beam of light that escapes to infinity in the rest frame of the object still escapes to infinity in "the" coordinate system in which the object is moving.
In flat space-time, you are probably familiar with time dilation for a moving object. The time dilation factor can be very large, but will always be finite. This finite time dilation means that there are no singularities in the coordinate transform for a moving observer - the Lorentz transform - any event t occuring on the moving train at a finite time occurs at a finite time t' on the non-moving station, and vica-versa.

PS I just looked over some papers: either they are wrong, or the affirmation that a black hole is a black hole in all reference frames which was presented on this thread is wrong. Talking abaout the Schwartzchild solution, is says something like that:
one can discover different coordinate frames in terms of which no singularity is seen. ... Introduce new coordinates ("Kruskal coordinates")
Well, I think I gave enough. It's from "Introduction to General Relativity" by G. Hooft. So what's the truth here? Is it a black hole or not in all coordinates?
This quote is about the "bad behavior" of the Schwarzschild coordinates at the event horizon that was mentioned earlier. A black hole in Kruskal coordinates remains a black hole - the Kruskal coordinates simply remove the "bad behavior" that the Schwarzschild coordinates exhibit at the event horizon.
 

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