Geodesic (general relativity)/Proofs

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This mathematics article is devoted entirely to providing mathematical proofs and support for claims and statements made in the article geodesic (general relativity). This article is currently an experimental vehicle to see how we might be able to provide proofs and details for math articles without cluttering up the main article itself. See [WikiProject Mathematics/Proofs] for current discussion. This article is "experimental" in that it is a proposal for one way that we might be able to deal with expressing proofs.

Proof 1

[ \nabla_ \vec U = 0 ],

[ U^\alpha \nabla_\alpha \vec U = 0 ],

[ U^\alpha U^\beta _ = 0 ],

[ U^\alpha (U^\beta _ + U^\sigma \Gamma^\beta _) = 0 ],

[ U^\alpha U^\beta _ + \Gamma^\beta _ U^\alpha U^\sigma = 0 ],

[ \ddot x^\beta + \Gamma^\beta _ \dot x^\sigma \dot x^\alpha = 0. \ ]

(return to article)

Proof 2

The goal being to extremize the value of

[ l = \int d\tau = \int \, d\phi = \int \sqrt \, d\phi = \int \sqrt \, d\phi = \int f \, d\phi]

where

[ f = \sqrt \dot x^\mu \dot x^\nu} ]

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

[ = ].

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

[ \dot x^\mu \dot x^\nu} \over \partial \dot x^\lambda} = \dot x^\mu \dot x^\nu} \over \partial x^\lambda} ]

Now calculate the derivatives: [ \left( \dot x^\nu - g_ \dot x^\mu \over 2 \sqrt \dot x^\mu \dot x^\nu}} \right) = \dot x^\mu \dot x^\nu \over 2 \sqrt \dot x^\mu \dot x^\nu}} \qquad \qquad (1)]

[ \left( \delta^\mu _\lambda \dot x^\nu + g_ \dot x^\mu \delta^\nu _\lambda \over 2 \sqrt \dot x^\mu \dot x^\nu}} \right) = \dot x^\mu \dot x^\nu \over 2 \sqrt \dot x^\mu \dot x^\nu}} \qquad \qquad (2) ]

[ \left( \dot x^\nu + g_ \dot x^\mu \over \sqrt \dot x^\mu \dot x^\nu}} \right) = \dot x^\mu \dot x^\nu \over \sqrt \dot x^\mu \dot x^\nu}} \qquad \qquad (3) ]

[ \dot x^\mu \dot x^\nu} (g_ \dot x^\nu + g_ \dot x^\mu) - (g_ \dot x^\nu + g_ \dot x^\mu) \sqrt \dot x^\mu \dot x^\nu} \over -g_ \dot x^\mu \dot x^\nu} = \dot x^\mu \dot x^\nu \over \sqrt \dot x^\mu \dot x^\nu}} \qquad \qquad (4) ]

[ \dot x^\mu \dot x^\nu) (g_ \dot x^\nu + g_ \dot x^\mu) + (g_ \dot x^\nu + g_ \dot x^\mu) (g_ \dot x^\mu \dot x^\nu) \over -g_ \dot x^\mu \dot x^\nu} = g_ \dot x^\mu \dot x^\nu \qquad \qquad (5) ]

[ (g_ \dot x^\mu \dot x^\nu) (g_ \dot x^\nu \dot x^\mu + g_ \dot x^\mu \dot x^\nu + g_ \ddot x^\nu + g_ \ddot x^\mu) ]

[= (g_ \dot x^\mu \dot x^\nu) (g_ \dot x^\alpha \dot x^\beta) + (g_ \dot x^\nu + g_ \dot x^\mu) (g_ \dot x^\mu \dot x^\nu) \qquad \qquad (6) ]

[ g_ \dot x^\mu \dot x^\nu + g_ \dot x^\mu \dot x^\nu - g_ \dot x^\mu \dot x^\nu + 2 g_ \ddot x^\mu = (g_ \dot x^\mu \dot x^\nu) \over g_ \dot x^\alpha \dot x^\beta} \qquad \qquad (7) ]

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

This is just one step away from the geodesic equation. (return to article)

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