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Q & A: length contraction and black holes

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Most recent answer: 07/01/2015
Q:
Perceived Black Holes in the Special Theory I am aware that the following is in all probability the result of a flawed understanding of the Special Theory for which I expect I will soon be slaughtered for. This idea focuses on motion in one direction which is where part of the problem lies. This may sound absurd or idiotic. An observer in motion relative to a massive rod will measure its length as the proper length multiplied by the inverse of the Lorentz factor. The length the observer measures will contract as the velocity between the two systems approaches the speed of light. Eventually, the length the observer measures will be short enough so that a photon emitted from the rod would be unable to escape the gravitational field of the rod. To the observer the rod and all of its mass would appear to be concentrated into a small enough volume to form a black hole. Take a rod with a specific mass M: L=L_o √(1-v^2/c^2 )R_S=√(2GM/c^2 )Set the observed length equal to the Schwarzschild radius: 2GM/c^2 =L_o √(1-v^2/c^2 )(4G^2 M^2)/(L_o^2 c^4 )=1-v^2/c^2 v^2/c^2 =1-(4G^2 M^2)/(L_o^2 c^4 )v^2=c^2-(4G^2 M^2)/(L_o^2 c^2 )v=√(c^2-(4G^2 M^2)/(L_o^2 c^2 ))Thus, the observer would perceive the massive rod as a small black hole. Why is this notion wrong?
- Jack
New Jersey
A:

That's a very thoughtful question. You'd think that between the inertial mass increase and the length contraction that an object has when viewed in a frame in which it's moving that it would form a black hole. Yet that doesn't make sense because in it's own frame light can escape easily. There must be some mistake. 

Without really knowing General Relativity, I do know that its gravitational source term must include not just energy (inertial mass) but the full relativistic four-vector, for which the other three components are the vector momentum. So the source of gravity is very different for that moving object with its momentum than it would be for something with that same energy but no momentum. I guess that makes the solutions of the GR equations come out with no black hole, unlike for the zero-momentum object.

Mike W.


(published on 07/01/2015)

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