Rindler Observer: Unruh Effect & Spacetime Geometry

[pythagoras88]

Hi,

I read sean carroll book, "spacetime and geometry" and in the last chapter where he tries to derive unruh effect, he introduced the concept of Rindler coordinate.

There is this part where he considers an accelerating observer in minkowski metric and introduce the trajectory blah blah. My question is, since the observer is accelerating, there must be some sort of fuel or energy propelling, so this energy by itself will actually result in curvature in the spacetime manifold. So isn't the consideration of accelerating observer in minkowski space invalid??

Thanks in advance for any response.

 

[Bill_K]

The observer is supposed to be a 'test particle,' i.e. one who's mass is small enough that its own gravitational field is negligible.

 

[pythagoras88]

oh, ok... haha... i guess it is a trivial question. So it is juest an accelerating particle that does not affect the spacetime.

Initially i was thinking, because an accelerating observer will have an rindler horizon given by x=t and x=-t. Then is it possible that the energy used to accelerate the particle cause a curvature in spacetime such that it create an event horizon. Then maybe this event horizon may actually coincide with the rindler horizon.

Anyway, Thanks for the reply.

 

[pervect]

The Rindler space-time isn't curved in the sense of having a non-zero curvature tensor.

My rather vague understanding of the Unruh effect is that, although the Rindler and standard space-times are both flat, the coordinates are different. And while they both have time translation symmetries, the time translation symmetry represented by the t coordinate in the Minkowski space is different from the time translation symmetry represented by the coordinate t' in the Rindler space, because the coordinates are different. This results in concepts of energy (which result from Noether's theorem - every time translation symmetry corresponds to a conserved energy) which are different in the two spaces.