# What is path of light in the accelerating elevator?

1. Mathematically, (by mathematically I means by equations) what is path of light in the accelerating elevator?

2. What is the difference between an ordinary derivative and covariant derivative (which is used in curved geodesic)?

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You are asking two different questions here.. Perhaps you should create two different threads. –  John M Apr 21 '13 at 20:46

By the equivalence principle, the uniformly accelerated frame of the elevator than be treated as a spacetime with a uniform gravitional field. The metric for a spacetime in which there is a uniform gravitational field in the $z$-direction is (with $c=1$) $$ds^2 = -\left(1+ gz\right)^2dt^2 + dx^2 + dy^2 + dz^2$$ Since this metric is invariant under translations in $t,x,y$, we immediately get three killing vectors $\partial_t, \partial_x, \partial_y$ and three corresponding conserved quantities along a geodesic $\gamma^\mu(\lambda) = (t(\lambda), x(\lambda), y(\lambda), z(\lambda))$; \begin{align} c_t &= \dot \gamma \cdot \partial_t = -\left(1+ gz\right)^2 \dot t \\ c_x &= \dot \gamma \cdot \partial_x = \dot x \\ c_y &= \dot \gamma\cdot \partial_y = \dot y \end{align} where here overdots mean derivative with respect to affine parameter $\lambda$. Light travels along null geodesics which satisfy $\dot \gamma^2 = 0$ which gives the equation $$-\left(1+ gz\right)^2\dot t^2 + \dot x^2 + \dot y^2 + \dot z^2 = 0$$ Combining these results gives $$-\left(1+ gz\right)^{-2}c_t^2 + c_x^2+c_y^2+\dot z^2 = 0$$ and therefore all in all we have $$\ddot x = 0, \qquad \ddot y = 0, \qquad \ddot z = \frac{c_t^2g}{(1+g z)^3}$$ If we choose $\dot t = 1$, so that the affine parameter corresponds to time, then we get the following $$\ddot x = 0, \qquad \ddot y = 0, \qquad \ddot z = -\frac{g}{(1-g z)^3}$$ In which case we see that the motion of light is such that it experiences no acceleration in the $x$ and $y$ directions and a position-dependent acceleration in the $z$-direction. In fact, if we taylor expand the right hand side of the $z$ equation of motion with respect to the parameter $g$ then we find $$\ddot z = -g + 3z g^2 + \mathcal O(g^3)$$ So for small accelerations $g$, the light just experiences the acceleration of the elevator downward, plus higher order corrections.