Geodesic (general relativity)/Proofs

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[edit] Proof 1

$\nabla_{\vec U} \vec U = 0$ ,

$U^\alpha \nabla_\alpha \vec U = 0$ ,

U α U β;α = 0

U α (U β,α + U σ Γ β σα) = 0

U α U β,α + Γ β σα U α U σ = 0

$\ddot x^\beta + \Gamma^\beta {}_{\sigma \alpha} \dot x^\sigma \dot x^\alpha = 0. \$

(return to article)

[edit] Proof 2

The goal being to extremize the value of

$l = \int d\tau = \int {d\tau \over d\phi} \, d\phi = \int \sqrt{{(d\tau)^2 \over (d\phi)^2}} \, d\phi = \int \sqrt{{-g_{\mu \nu} dx^\mu dx^\nu \over d\phi \, d\phi}} \, d\phi = \int f \, d\phi$

where

$f = \sqrt{-g_{\mu \nu} \dot x^\mu \dot x^\nu}$

such goal can be accomplished by calculating the Euler-Lagrange equation for f, which is

${d \over d\tau} {\partial f \over \partial \dot x^\lambda} = {\partial f \over \partial x^\lambda}$ .

Substituting the expression of f into the Euler-Lagrange equation (which extremizes the value of the integral l), gives

${d \over d\tau} {\partial \sqrt{-g_{\mu \nu} \dot x^\mu \dot x^\nu} \over \partial \dot x^\lambda} = {\partial \sqrt{-g_{\mu \nu} \dot x^\mu \dot x^\nu} \over \partial x^\lambda}$

Now calculate the derivatives: ${d \over d\tau} \left( {-g_{\mu \nu} {\partial \dot x^\mu \over \partial \dot x^\lambda} \dot x^\nu - g_{\mu \nu} \dot x^\mu {\partial \dot x^\nu \over \partial \dot x^\lambda} \over 2 \sqrt{-g_{\mu \nu} \dot x^\mu \dot x^\nu}} \right) = {-g_{\mu \nu, \lambda} \dot x^\mu \dot x^\nu \over 2 \sqrt{-g_{\mu \nu} \dot x^\mu \dot x^\nu}} \qquad \qquad (1)$

${d \over d\tau} \left( {g_{\mu \nu} \delta^\mu {}_\lambda \dot x^\nu + g_{\mu \nu} \dot x^\mu \delta^\nu {}_\lambda \over 2 \sqrt{-g_{\mu \nu} \dot x^\mu \dot x^\nu}} \right) = {g_{\mu \nu , \lambda} \dot x^\mu \dot x^\nu \over 2 \sqrt{-g_{\mu \nu} \dot x^\mu \dot x^\nu}} \qquad \qquad (2)$

${d \over d\tau} \left( {g_{\lambda \nu} \dot x^\nu + g_{\mu \lambda} \dot x^\mu \over \sqrt{-g_{\mu \nu} \dot x^\mu \dot x^\nu}} \right) = {g_{\mu \nu , \lambda} \dot x^\mu \dot x^\nu \over \sqrt{-g_{\mu \nu} \dot x^\mu \dot x^\nu}} \qquad \qquad (3)$

${\sqrt{-g_{\mu \nu} \dot x^\mu \dot x^\nu} {d \over d\tau} (g_{\lambda \nu} \dot x^\nu + g_{\mu \lambda} \dot x^\mu) - (g_{\lambda \nu} \dot x^\nu + g_{\mu \lambda} \dot x^\mu) {d \over d\tau} \sqrt{-g_{\mu \nu} \dot x^\mu \dot x^\nu} \over -g_{\mu \nu} \dot x^\mu \dot x^\nu} = {g_{\mu \nu , \lambda} \dot x^\mu \dot x^\nu \over \sqrt{-g_{\mu \nu} \dot x^\mu \dot x^\nu}} \qquad \qquad (4)$

${(-g_{\mu \nu} \dot x^\mu \dot x^\nu) {d \over d\tau} (g_{\lambda \nu} \dot x^\nu + g_{\mu \lambda} \dot x^\mu) + {1 \over 2} (g_{\lambda \nu} \dot x^\nu + g_{\mu \lambda} \dot x^\mu) {d \over d\tau} (g_{\mu \nu} \dot x^\mu \dot x^\nu) \over -g_{\mu \nu} \dot x^\mu \dot x^\nu} = g_{\mu \nu ,\lambda} \dot x^\mu \dot x^\nu \qquad \qquad (5)$

$(g_{\mu \nu} \dot x^\mu \dot x^\nu) (g_{\lambda \nu ,\mu} \dot x^\nu \dot x^\mu + g_{\mu \lambda ,\nu} \dot x^\mu \dot x^\nu + g_{\lambda \nu} \ddot x^\nu + g_{\lambda \mu} \ddot x^\mu)$

$= (g_{\mu \nu ,\lambda} \dot x^\mu \dot x^\nu) (g_{\alpha \beta} \dot x^\alpha \dot x^\beta) + {1 \over 2} (g_{\lambda \nu} \dot x^\nu + g_{\lambda \mu} \dot x^\mu) {d \over d\tau} (g_{\mu \nu} \dot x^\mu \dot x^\nu) \qquad \qquad (6)$

$g_{\lambda \nu ,\mu} \dot x^\mu \dot x^\nu + g_{\lambda \mu ,\nu} \dot x^\mu \dot x^\nu - g_{\mu \nu ,\lambda} \dot x^\mu \dot x^\nu + 2 g_{\lambda \mu} \ddot x^\mu = {\dot x_\lambda {d \over d\tau} (g_{\mu \nu} \dot x^\mu \dot x^\nu) \over g_{\alpha \beta} \dot x^\alpha \dot x^\beta} \qquad \qquad (7)$

$2(\Gamma_{\lambda \mu \nu} \dot x^\mu \dot x^\nu + \ddot x_\lambda) = {\dot x_\lambda {d \over d\tau} (\dot x_\nu \dot x^\nu) \over \dot x_\beta \dot x^\beta} = {U_\lambda {d \over d\tau} (U_\nu U^\nu) \over U_\beta U^\beta} = U_\lambda {d \over d\tau} \ln |U_\nu U^\nu| \qquad \qquad (8)$