In article , (Mitchell) writes:
But there is energy lost from the photons. Do the math. That's where
the Doppler shift matters.

That's just not true. I'm a senior year physics undergrad at Cornell,
and I was in a group of 6 people who got to sit down and talk with
Gold about his "perfect mirror and violations of conservation of
energy/momentum" paper before it was published. Among the arguments we
attempted to refute is theory was the idea of Doppler shift. He
correctly countered that a Doppler shift does not affect the total
energy, only the power.

:-)))

Consider an elastic collision of two particles. For convenience and
simplicity we'll make it a 1D problem. Two particles coming one
towards the other, colliding and flying away. Now, lets consider it
relative to three different reference frames.

1) In the center of mass frame, both particles approach (with the
same momentum), collide and recede, still with the same momentum and
same velocity (for each one) as before the collision (remember, the
collision is elastic). Total energy is preserved and the energy of
each of the particles individually is preserved as well.

2) In the initial reference frame of A, we see prior to the collision
A stationary and B approaching, after the collision both are moving.
Total energy still preserved but A gained some energy and B lost some.

3) In the (initial) frame of B, the situation is just reversed.
Total energy still preserved but as a result of the collision A lost
some energy and B gained some.

In all (infinity of) other reference frames you'll get other,
intermediate results. Mind you, we're not talking different processes.
We're talking about *same process* as viewed from different reference
frames. The question of who gains and who loses energy in a collision
is not "inherent" to the process but a matter of accounting, depending on
the reference frame.

So, if you take an elastic collision of a photon with the mirror, in
their CM reference frame, indeed, nobody gains or loses energy, the
photon bounces back with the same energy with which it hit. But,
you're not observing from the CM reference frame but from "our" frame,
stationed on the Sun. In this frame, the photon bounces back with
different energy. And the energy difference can be represented
through the Doppler shift between the CM reference frame and our.
Note, you've to apply the shift twice, first transforming from our
frame to the CM, then back.

As I said, do the math. Will do you good.

Mati Meron | "When you argue with a fool,
| chances are he is doing just the same"