Poynting vector and Rindler flux under time inversion

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

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?

-

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.