Question: Show that the Euler-Lagrange equation for the action principle (27.8) is equivalent to the geodesic equation for a photon in the static spacetime metric g

Show that the Euler-Lagrange equation for the action principle (27.8) is equivalent to the geodesic equation for a photon in the static spacetime metric g00(xk), gij(xk). Specifically, do the following.

(a) The action (27.8) is the same as that for a geodesic in a 3-dimensional space with metric γjk and with t playing the role of proper distance traveled [Eq. (25.19) converted to a positive-definite, 3-dimensional metric]. Therefore, the Euler- Lagrange equation for Eq. (27.8) is the geodesic equation in that (fictitious) space [Eq. (25.14) with t the affine parameter]. Using Eq. (24.38c) for the connection coefficients, show that the geodesic equation can be written in the form

dxk Yjk dt 1 - -/- (Yjk,l + Y jl,k - Ykl,

(b) Take the geodesic equation (25.14) for the light ray in the real spacetime, with spacetime affine parameter ζ, and change parameters to coordinate time t. Thereby obtain

j) = 0. dxk dx dt 2 dt (27.12a)

(c) Insert the second of these equations into the first, and write the connection coefficients in terms of derivatives of the spacetime metric. With a little algebra, bring your result into the form Eq. (27.12a) of the Fermat-principle Euler-Lagrange equation.


Equations.

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dxk Yjk dt 1 - -/- (Yjk,l + Y jl,k - Ykl, j) = 0. dxk dx dt 2 dt (27.12a)

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