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This question is a continuation from this one.

A material disk have two sides, one that is reflective and another absorptive of electromagnetic radiation in the range where the background cosmic radiation spectrum is significative.

It is pretty clear that the disk will not accelerate any further after thermalizing with the background radiation.

Question: what is the condition for the net exchange of momentum between the background radiation and the disk to become zero?

will the disk accelerate until the momentum of the blueshifted radiation absorbed at the front equates the momentum of redshifted photons being reflected at the back side?

Will the reflective side stop reflecting efficiently after the disk reaches the same temperature as the background radiation?

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If you give your disk enough ideal properties, such as being infinitely thin and conducting heat perfectly, so that there is no temperature difference between both sides, so that its own radiation has no net effect on its movement, and being able to reflect and absorb radiation perfectly no matter what the wavelength, then methinks it will be the redshift/blueshift thing that will eventually stop it from accelerating.

If the disk is moving with velocity $v$ then the radiation hitting the absorptive side will have its frequency modified (blueshift) by a factor $f$, while the reflective side will be (redshift) by a factor $f'$. Since radiation pressure is proportional to frequency, the thrust on the disk will be proportional to $2f'-f$.

With the classical approach to Doppler effect, we would get

$$2f'-f = 2\frac{c-v}{c}-\frac{c+v}{c} = \frac{c-3v}{c},$$

and there would be acceleration until the disk reaches 1/3 of the speed of light.

With the relativistic approach, and $\beta = v/c$, we would have

$$2f'-f = 2\sqrt{\frac{1-\beta}{1+\beta}}-\sqrt{\frac{1+\beta}{1-\beta}}= \frac{1-3\beta}{\sqrt{1-\beta^2}}.$$

The numerator of that expression has a root at $\beta=\frac{1}{3}= 0.33$, so acceleration would stop at that fraction of the speed of light.

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the numerator in the relativistic version is wrong: is actually $1 - 3 \beta$ – lurscher Jan 3 '13 at 18:50
Thanks for spotting the error! Funny that the result is the same whether in a classical or relativistic setting... – Jaime Jan 3 '13 at 19:09

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