# light falling on a completely reflecting friction less mirror

If you had a perfectly reflective mirror whose face was perpendicular to the ground and was kept on a friction less track on the ground, and you shined a flashlight on it, what would happen?

• Would the mirror move?

• Would the light reflect back?

• Would the mirror need to be fixed for it to reflect back because all energy from the light would be going into the mirror's motion?

NOTE: I feel that some energy of incident light would make mirror move and some would be of reflected light but that would imply the wavelength of reflected light (and hence its color) would be different. Am I wrong? Please explain. I am only a high school student so I don't know anything about relativity.

Light does indeed bear momentum. If you have a quantity $E$ joules of light in a plane wave beam, it has linear momentum $\frac{E}{c}$ in the direction of propagation. So if it bounces off a mirror, the impulse transferred to the mirror is $2\frac{E}{c}$, in the direction of propagation, at least at first.

Most certainly, the light bounces back. If the mirror was on a frictionless cart as described, it would begin to move - to translate - in the direction of the beam. The force is the impulse transfer per unit time, namely $2\frac{P}{c}$, where $P$ is the beam's power. And you are absolutely, altogether right: the wavelength does indeed change. There is a redshift: the reflected light is Doppler shifted to longer wavelengths. A reflected beam like this actually undergoes "twice" the Doppler shift of a source fixed to the mirror: I mean that you use the normal Doppler scaling factor squared.

As the mirror gets moving, so that it is moving at $v\ll c$, then the reflexional Doppler scale factor is, to first order approximation, $1-2\,\frac{v}{c}$. If you assume the same number of photons come back as are incident on the mirror, then this implies a power loss of $2\frac{v\,P}{c}$.

But look at our force equation. It is $2\frac{P}{c}$. A force $F$ working on something moving at speed $v$ works at a rate of $F\,v$. So, just thinking classically and not thinking about photons at all, we also conclude that there is a power loss of $2\frac{v\,P}{c}$! This, of course, is where the kinetic energy of your mirror on your cart comes from.

So you've shown some fantastic intuition in your question and your reasoning is exactly right. You should be very pleased with yourself and I hope that seeing the numbers come out exactly in keeping with what you foretold by your own reasoning is satisfying for you.

Maybe you should study physics later?

• yes i am definitely studying physics(relativity) later after completing my BTech from India(i live here). then go to US for higher studies. Is the dropler effect you are talking same as in case of sound because i have studied that in school.Can you please explain the Dropler shift in your answer – ragvri Feb 6 '15 at 11:26
• @rjmessibarca The Doppler for light is very like that for sound, i.e. you can think of it as being a variation in the rate at which you cross wavefronts depending on your speed relative to the source. The equations one uses to calculate it are slightly different, though: you need to take special relativity into account for light Doppler. To first order, though, as I've calculated here, the equations are the same. – WetSavannaAnimal Feb 6 '15 at 11:31
• @rjmessibarca :Please note ,its Doppler effect not Dropler effect. – Paul Feb 6 '15 at 12:27