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kent davidge

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- #1

kent davidge

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- #2

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p=0 is a coordinate singularity, usually called the Rindler horzion, because the metric coefficient of dn^2 vanishes. I don't believe that the null geodesics should be able to cross the Rindler horizon, and in your diagram they don't.

- #3

kent davidge

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So it seems like I have obtained the correct diagram from wrong equations.Wiki has a derivation of the null geodesics in <<link>>

It seems that I should have considered the geodesic equations first.

- #4

PeterDonis

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I don't believe that the null geodesics should be able to cross the Rindler horizon

"Outgoing" ones (the ones going in the same direction as the Rindler observers' proper acceleration) can't, but "ingoing" ones (the ones going in the opposite direction) certainly can. However, this won't be visible in Rindler coordinates because they don't cover the Rindler horizon; instead, an ingoing null geodesic will look like it asymptotically approaches ##p = 0## as ##n \rightarrow \infty## but never reaches it.

Also, the above is assuming that we are looking at where geodesics go, but we also need to look at where they come from. That's just the time reverse of the above: outgoing null geodesics can come from below the Rindler horizon while ingoing null geodesics can't. An outgoing null geodesic will asymptotically approach ##p = 0## in Rindler coordinates but never reach it as ##n \rightarrow - \infty##.

- #5

PeterDonis

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So it seems like I have obtained the correct diagram from wrong equations.

It seems that I should have considered the geodesic equations first.

Your equations aren't wrong. What you did, for this simple case, is equivalent to solving the geodesic equations for null geodesics. In the Wikipedia article, that corresponds to setting ##P = Q = 0## and ##y = z = 0##.

In the more general case where we put back the other two spatial dimensions, what you did is equivalent to solving the restricted set of null geodesic equations that only apply to "radial" geodesics--geodesics that only move in the ##n - p## plane, not in the other two coordinate directions.

- #6

kent davidge

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Ah, ok. What seems weird to me is that if you let ##n \rightarrow \infty## the past light cone of the particle will cover only half of the space. Is that right? So it seems that there are events that will never cross the particles world line.Your equations aren't wrong. What you did, for this simple case, is equivalent to solving the geodesic equations for null geodesics. In the Wikipedia article, that corresponds to setting ##P = Q = 0## and ##y = z = 0##.

In the more general case where we put back the other two spatial dimensions, what you did is equivalent to solving the restricted set of null geodesic equations that only apply to "radial" geodesics--geodesics that only move in the ##n - p## plane, not in the other two coordinate directions.

- #7

Nugatory

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Yes, there are events that will never be in the past light cone of the accelerating observer. That’s the Rindler horizon at work.So it seems that there are events that will never cross the particles world line.

[Edit: “past light cone of the accelerated observer” is sloppy wording. “Past light cone of any event on the worldline of the accelerated observer” might be better]

Compare with the event horizon around a Schwarzschild black hole: events on the inside of the horizon will never be in the past light cone of an observer outside.

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