# Einstein's equivalence principle applied to a constant accelerated system comprising two mirrors

Let us consider a system comprising two mirrors with an observer placed in between. The distance from the observer to each of the mirrors is $$d$$. At $$t=0$$ the observer emits a light signal in both directions. If the system is at rest, the observer should register both reflected light signals at $$t=2d/c$$. However, if the system accelerates with constant acceleration $$a$$, the observer in the center should measure a time delay in between the registration of the light reflected by the left mirror and the light reflected by the right mirror. We want to derive an expression for the said delay.

We assume that we work in the non-relativistic regime $$at\ll c$$. We can think of acoustic instead of light waves if it helps.

For starters, we write down the position of the mirrors and the observer. We define the observers position to be, $$x_0(t)=\frac{a}{2}t^2,$$ thereafter, the positions of the left and right mirrors follow to be, \begin{align}y_-(t)=\frac{a}{2}t^2-d, &&y_+(t)=\frac{a}{2}t^2+d.\end{align} The time when the light reaches the left mirror $$M_-$$ can be found from the quadratic equation, $$y_-(t)=-ct$$. There are two times that solve the quadratic equation. The (positive) time is given by, $$t_1=\frac{c}{a}\left(-1+\sqrt{1+2\frac{ad}{c^2}}\right)\approx\frac{d}{c}.$$ At $$t_1$$ the light is reflected by $$M_-$$ and travels back towards the observer. The equation to solve is, $$y_-(t_1)+c (t-t_1)=-d_1+ct=x_0(t),$$ with, $$t_2=\frac{c}{a}\left(1-\sqrt{1-2\frac{a d_1}{c^2}}\right)\approx\frac{d_1}{c}\approx2\frac{d}{c}-\frac{ad^2}{2c^3},$$ being the time of interest. We identify $$2d/c$$ to be the time of the system at rest and the second term to be the acceleration dependent correction to that.

Repeating the analysis for the right mirror $$M_+$$, we first solve $$c t=\frac{a}{2}t^2+d$$, yielding, $$t_3=\frac{c}{a}\left(1-\sqrt{1-2\frac{ad}{c^2}}\right)\approx\frac{d}{c}.$$ Consequently, light is reflectd at $$t_3$$ from $$M_+$$ back to the observer. We solve, $$y_+(t_3)-c(t-t_3)=d_2-ct=x_0(t),$$ and find, $$t_4=\frac{c}{a}\left(-1+\sqrt{1+2\frac{a d_2}{c^2}}\right)\approx\frac{d_2}{c}\approx2\frac{d}{c}+\frac{ad^2}{2c^3}.$$ Finally, we predict the delay between the registration of the reflected signals to be, $$\Delta t:=t_4-t_2\approx\frac{ad^2}{2c^3}.$$

Invoking Einstein's equivalence principle, we should measure the same delay $$\Delta t$$ for the mirror system being at rest and perpendicular to the surface of the earth with $$a=g$$ being the gravitational acceleration on the surface of the earth.

Questions:

1. I don't understand why the correction term for the light reflected from the left mirror $$M_-$$ has a negative sign. I would expect the light to take more time to travel back to the observer. The reason is that the observer has picked up additional velocity in the time where the light has traveled to the mirror because of the constant acceleration. The light now has to "overtake" the said additional velocity which would manifest through an increase in travel time.
2. Is it correct that if we would conduct the same experiment in the solar system, we would have to consider a position-dependent acceleration as the timelike component of the Schwarzschild metric depends on the radius $$g_{00}=-(1-\frac{GM}{r})$$ in order to apply Einstein's equivalence principle correctly?
• For your first question: I think it is negative because the light should arrive before $2d/c$ from $M_{-}$ because it is moving towards the observer. Similarly from $M_{+}$ it should be later than $2d/c$ so the correction is positive. Also I think if you Taylor expand $t_2$ you should get the same. – Manvendra Somvanshi Feb 15 at 11:13
• @ManvendraSomvanshi Can you elaborate on your argument? It is unclear why the light should arrive earlier than $2d/c$. PS: I updated my argument related to the first question. – bodokaiser Feb 15 at 11:16