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  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
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2 Answers 2

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.

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As long as you are asking about covariant derivative, I assume that you are interested in a relativistic treatment of the light trajectories. For this, a defenition of uniformly accelerating frame is needed. Take a look at Rindler coordinates. The article contains some information about geodesics, but it is hard to follow. The easiest way is to solve for geodesics in some inertial frame, where they are straight lines, and then pass to the accelerated system. I think that once I thought that they can be characterized as ellipses orthogonal to Rindler horizont, but I am not sure. You can google about Rindler Coodinates if Wiki is not enough.

As for the covariant derivative, the difference between a covariant derivative and an ordinary one is just some summ of Christoffel symbols :) Taken seriously -- the ordinary derivative just don't make any sense for non-scalar objects. General Relativity, and basically all Physics are not defined in a concrete coordinate system (here I mean that all theories are defined not by coordinates, but by some structures -- Riemann metric on a 4-manifold, Lorentz metric on a flat space-time, symplectic structre, a principal bundle, conformal structure, etc.). This means that components of your vectors should transform nicely under coordinate transformations. Ordinary derivative of a vector does not transform nicely, while the covariant does. Mathematically, ordinary derivative just cannot be defined without picking a specific coodinate system, while the covariant can. See the wiki article.

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