Poynting vector and Rindler flux under time inversion

Question

This question is about some reply by John Baez on sci.physics.research

the post is this: https://groups.google.com/d/msg/sci.physics.research/F6x5GkFt0ic/fxsfuNl9d8gJ

the article he is talking about is this one: http://xxx.lanl.gov/abs/gr-qc/0106075

the question itself is pretty simple. In that reply, which i'll quote for completeness, reads:

Unruh says that Haisch and Rueda's calculations are wrong, and
that a correct calculation shows a uniformly accelerating observer
zipping through the vacuum state of a quantized electromagnetic 
field on Minkowski spacetime sees a *perfectly thermalized* bath 
of photons.

In particular, this means such an observer will see no "Rindler flux" - 
i.e., the expectation value of the Poynting vector is zero.  Or in
less fancy language: there will be, on average, no net flux of momentum 
in the photons seen by the accelerating observer.

He gives a very simple argument showing that the expectation value 
of the Poynting vector *must* be zero: the whole situation is 
time-symmetric, and time reversal flips the direction of the 
Poynting vector!  

Now, i don't understand at all that last symmetry argument. Yes, time inversion will reverse direction of real vectors, so? an accelerated particle has a definite acceleration vector, that is reversed under time inversion, and is not zero. Why would any Poynting vector be zero?

Answer 1

Your mistake is simple. The vector of acceleration is clearly even (unchanged), not odd (flipping sign), under the time reversal because it is the second derivative of the position, $d^2 \vec x/ dt^2$, and "second" is derived from two, an even number. Two minus signs cancel.

On the other hand, the Poynting vector is odd because it is essentially the density of energy times its velocity and velocity is the first derivative. So they can't be proportional to one another.