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If you send a light beam to a surface which reflects the light perfectly I would say that on the surface is put a force of the light so the surface moves backwards. But where does that energy comes from? If it perfectly reflects the light than the frequency shouldn't be changed and the speed of light wouldn't change either. So were does the moving surface gets his energy from? Or isn't it possible to maintain their frequency or perhaps the surface is not moving at all?

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The latter - energy is conserved, so the recoil of an atom (as a simple example) translates to a lowering of energy (lower frequency) of the reflected photon.

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Let's treat a one-dimensional example, of an elastic collision of a single photon with a mirror. Conserving energy (and cheating slightly by using newtonian kinetic energy, assuming $v \ll c$), we have:

$$hf_{0} = hf_{f} + \frac{1}{2}mv^{2}$$

While, conserving momentum, we have:

$$hf_{0}/c = mv - hf_{f}/c$$

from the second equation, we get:

$$hf_{f} = mvc - hf_{0}$$

so, we have:

$$\begin{align} 2 hf_{0} = mvc - \frac{1}{2}mv^{2}&\\ v^{2} - 2 vc + 4 \frac{hf_{0}}{m} &= 0\\ v &= c \pm \sqrt{c^{2}-4 \frac{hf}{m}} \end{align}$$

the positive root is obviously unphysical, and in the limit $m \gg hf_{0}$ we have $v = 0$. So perfect reflection and conserved energy are completely consistent.

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Electromagnetic radiation does exert a radiation pressure and so there must be momentum transfer between the incident radiation and the surface. So the reflector exerts a force on the incoming radiation. In the case of light being reflected from a mirror this force is not very large and so the recoil of the relevtor is not noticeable. The analysis can also be done in terms of photons which carry momentum being "reflected".

Although in the case of light and a mirror the radiation pressure force is comparatively small it is radiation pressure produced by photons trying to escape from the centre of the Sun which balances the inward gravitational attractive force.

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