# Light (Photon) reflecting from an accelerated mirror

Is there well know formulas (or some articles) from General Relativity how photon (light) is changing (how frequency is changing) while reflecting from the accelerated mirror?

I suspect it should be different from the reflecting from just moving mirror with constant speed because accelerated mirror is Non-inertial reference frame and general relativity is applicable in that case.

UPDATE 1

After some comments, I want to clarify the question. What will happen in the picture below? Let's say I have a harmonically oscillating mirror x(t) = Sin(t) on my desk. The blue line is the mirror, red/orange circle is a photon. I put values of coordinate x, speed v and acceleration a on top of the image.

At the point x=1, v=0 and a=1. It looks strange, how is it possible v=0, but a is not. But it is the same if we throw a stone up. Acceleration is always g, but at the top point, speed is zero. At this point where x=1, v=0, a=1 photon is falling on the mirror.

The question is what will happen with photon frequency from my side (if this oscillating mirror is on my desk)? What will I see?

I change colour from red to orange (blueshift), because I assumed, that acceleration direction is from right to left, that means fictitious force will be in opposite direction, which is equivalent to gravitational force from left to right and photon is reflecting to the source of gravitational force (to the source of the gravity). Or it is not correct? Maybe I will not see any changes and all changes will be visible only for the observer on the mirror?

Also what if the photon hits the mirror when speed is not zero and acceleration is not zero? Will I see only Doppler shift or something different?

Of course, I understand, that in real life with the real oscillating mirror on the desk with any achievable speed any difference in frequency will be undetectable, but I just want to know theoretically with the assumption, that I can detect very very small changes in frequency. If we assume the plane of the mirror is perpendicular to the acceleration, the situation is the same as the situation where the mirror is (horizontally) at rest in a gravitational field. When a photon (moving downwards) approaches the mirror its frequency increases. The moment the photon hits the mirror its (vertical) direction is reversed but its frequency doesn't change. After the reflection, the photon's frequency decreases while travelling upwards. These three processes, in which I divided up the total process, each conserve the energy ($(\frac{hc}{\lambda})^2$) of the photon.

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Mar 16 '17 at 1:25

I answer to your question for when the mirror is uniformly accelerating parallel to its normal away from any gravitational field from the viewpoint of the observer in the lab frame of reference. Assume that when the EM wave hits the mirror, the mirror has an instantaneous velocity $$v$$, receding with a uniform acceleration $$a$$ from the lab observer. And, assume that the incident wavelength is $$\lambda$$, and the reflected one is $$\lambda^{\prime}$$ as measured by the lab observer.

If we behave every single wave as a rod with a length $$\lambda$$, by the time the right end of the incident wave (the blue rod) hits the mirror, the left end of the reflected wave (the red rod) emerges from the mirror. Moreover, as the left end of the incident wave is absorbed by the mirror, the right end of the reflected wave is anticipated to be detached from the mirror abruptly. [See the attached figure.]

The important point with this illustration is that the time $$t_i$$ for the incident wave to disappear inside the mirror must exactly be equal to the time $$t_r$$ for the reflected wave to completely leave the mirror as measured by the lab observer. Considering the fact that the mirror recedes from the left end of the incident wave, and that the mirror also recedes from the left end of the reflected wave, we can write for the incident wave: [See the attached figure.]

$$\lambda+\frac{a}{2}{t_i}^2+vt_i=ct_i\rightarrow \frac{a}{2}{t_i}^2-(c-v)t_i+\lambda=0\rightarrow$$ $$t_i=\frac{c-v\pm \sqrt{(c-v)^2-2a\lambda}}{a} \space.\space\space\space\space (1)$$

For the reflected wave, we have:

$$\lambda^\prime=ct_r+\frac{a}{2}{t_r}^2+vt_r=ct_r+ct_i-\lambda\space.$$

In that $$t_i=t_r$$, it yeilds:

$$\lambda^\prime=2ct_i-\lambda \space.\space\space\space\space (2)$$

Substituting Eq. (1) into Eq. (2), and choosing the minus sign, we have:

$$\lambda^\prime={\frac{2c}{a}} [\space c-v-\sqrt{(c-v)^2-2a\lambda}\space]-\lambda \space.$$

If the acceleration tends to zero, for a mirror with a constant speed $$v$$, we obtain:

$$\lim_{a\to 0}\lambda^\prime=\frac{c+v}{c-v}\lambda$$

This result is correct. 