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Black hole moving at the SOL. |
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| Aug12-12, 01:46 PM | #1 |
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Black hole moving at the SOL.
I know that in scenarios where QM and GR are both applicable the answerers come out ridiculous. I believe this may be one of those scenarios. It is also possible that I have some misunderstanding that leads to a ridiculous answer. My question is which of these is the case.
A photon is traveling through empty space. A space ship is traveling in the opposite directions at such a velocity that the photon is extremely blue shifted. It is shifted to such a high frequency that in the ships frame of reference, it has enough energy to form an event horizon. The ship and the photon collide. Does the crew observe their ship being hit by a microscopic black hole traveling at the speed of light? How do they reconcile this with the idea that no massive object can travel at C? An observer wittinesses the collision from a nearby planet. In his frame of reference the photon is just a high energy photon. What does he observe and how can it be reconciled with the observations made by the crew of the ship? |
| Aug12-12, 02:26 PM | #2 |
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This is a classical issue, not a quantum-mechanical one. The reasoning is the same as if the photon was replaced by a baseball. |
| Aug12-12, 03:03 PM | #3 |
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I understand in the baseball scenario that the relativistic mass is a misnomer, in that it is not equivalent to gravitational mass. Therefore, no event horizon forms as "relativistic mass" increases.
My proposed scenario seems different to me in that the object in question is a photon, and therefore, has no rest mass. It must have gravitational mass because it has momentum and can be deflected by the gravitational field of a massive body. Conservation of momentum dictates that the massive body must also change momentum. If the energy of a photon increases as it's wavelength decreases then it's momentum, and gravitational influence, should also increase. Why is this not the case? |
| Aug12-12, 03:54 PM | #4 |
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Black hole moving at the SOL.
The source of the gravitational field is the stress-energy tensor, not the mass or the energy-momentum vector.
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| Sep15-12, 08:13 PM | #5 |
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Yes, I know this thread is over a month old but I'm still stumped on the same question. I did come across this recently...
http://www.youtube.com/watch?v=rYQTA...feature=relmfu Around time stamp 40:00 Mr. Susskind talks about energy gravitating. I believe he's a credible source. Back to my original question... Given that energy gravitates And The energy of a photon is frame dependent Can a photon be so energetic in a particular frame of reference that it gravitates enough to form an event horizon. If not, why? |
| Sep15-12, 08:33 PM | #6 |
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| Sep15-12, 08:34 PM | #7 |
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Recognitions:
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The very short simple-speak version above would go something like "energy gravitates, but so does momentum, and pressure - but while they gravitate, they don't do so in a way you can just plug simply into Newton's law to get the answer." If you want to understand it more precisely, I think you'll need to understand the stress energy tensor, and the Riemann curvature tensor, for starters. But you should be able to understand that the answer to your question is "no", even if you don't understand the details. A very hyper-relativistic mass would NOT be a black hole. It may or may not be helpful for you to know that the name of the solution that DOES describe a rapidly moving object (one that would have a low mass at rest) is the Aichelburg–Sexl ultraboost, http://en.wikipedia.org/w/index.php?...ldid=426066275 http://arxiv.org/abs/gr-qc/0110032 The second paper (the arxiv paper) more-or-less completely describes the gravitational field of such a boosted particle in the way that's perfectly natural for general relativity - i.e. through the Riemann curvature tensor. Unfortunately, it's unclear how much use this will be to you, even if we do explain that the Riemann curvature tensor is very much like the tidal force in Newtonian gravity. As the second paper mentions, the gravitational field of the ultraboost is analogous to the electric field of a rapidly moving charge, which is also an impulsive wave. Anyway, maybe I've gone on too much. The main thing to remember is that the answer to your original question is still "No, it doesn't form a black hole", just as it was before. |
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