Alternate form of geodesic equation

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SUMMARY

The discussion focuses on the transformation of the geodesic equation, specifically demonstrating that the equation \(\frac{du^{a}}{dt} +\Gamma^{a}_{bc}u^{b}u^{c}=0\) can be reformulated as \(\frac{du_{a}}{dt}=\frac{1}{2}(\partial_{a}g_{cd})u^{c}u^{d}\). The relevant Christoffel symbol is defined as \(\Gamma^{a}_{bc}=\frac{1}{2}g^{ad}(\partial_{b}g_{dc}+\partial_{c}g_{bd}-\partial_{d}g_{bc})\). The original poster encountered difficulties when attempting to contract the geodesic equation with \(g_{ab}\), leading to confusion with Kronecker deltas. They inquired about potential simplifying assumptions or approximations that could clarify the relationship.

PREREQUISITES
  • Understanding of differential geometry concepts, particularly geodesics.
  • Familiarity with Christoffel symbols and their derivation.
  • Knowledge of tensor calculus, specifically contraction of tensors.
  • Basic principles of general relativity and metric tensors.
NEXT STEPS
  • Study the derivation and applications of Christoffel symbols in general relativity.
  • Explore the implications of the Newtonian limit in the context of geodesic equations.
  • Learn about tensor contraction techniques and their significance in differential geometry.
  • Investigate the role of metric tensors in defining geodesics and their properties.
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This discussion is beneficial for students and researchers in mathematics and physics, particularly those focusing on general relativity, differential geometry, and the mathematical foundations of geodesic equations.

Alexrey
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Homework Statement


We're asked to show that the geodesic equation \frac{du^{a}}{dt} +\Gamma^{a}_{bc}u^{b}u^{c}=0 can be written in the form \frac{du_{a}}{dt}=\frac{1}{2}(\partial_{a}g_{cd})u^{c}u^{d}




Homework Equations


\Gamma^{a}_{bc}=\frac{1}{2}g^{ad}(\partial_{b}g_{dc}+\partial_{c}g_{bd}-\partial_{d}g_{bc})



The Attempt at a Solution


I tried contracting the geodesic equation with g_{ab} but came out with some Kronecker deltas which stumped me a bit.
 
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It can't be true in general. (pun!)

Is there some kind of simplifying assumption or approximation that makes 2 of the terms go away? Perhaps you're studying the Newtonian limit?
 

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