Geodesic Eq: Deriving 2nd Term on RHS

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SUMMARY

The discussion focuses on deriving the second term on the right-hand side of the geodesic equation when expressed in terms of the time coordinate parameter \( t \). The participant struggles to derive the term \( \frac{1}{2} \frac{d}{dt} \left[ \ln \gamma_{\alpha\beta} \left( \frac{dx^\alpha}{dt} \frac{dx^\beta}{dt} \right) \right] \). It is advised to avoid working directly with the metric and instead utilize variable substitution to parameterize the curve with \( t \) rather than \( s \). The transformation of the geodesic equation under this change is crucial for obtaining the desired term.

PREREQUISITES
  • Understanding of geodesic equations in differential geometry
  • Familiarity with the Christoffel symbols, specifically \( \Gamma^0_{\alpha\beta} \)
  • Knowledge of parameterization techniques in calculus
  • Basic concepts of tensor calculus and metric tensors
NEXT STEPS
  • Study the derivation of geodesic equations in general relativity
  • Learn about variable substitution techniques in differential equations
  • Explore the role of Christoffel symbols in geodesic motion
  • Investigate the transformation properties of tensors under coordinate changes
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Students and researchers in theoretical physics, particularly those focusing on general relativity and differential geometry, will benefit from this discussion.

peterpang1994
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As the geodesic equation in a form of
ed2e808ce2b6aa1859eb947f21f23ec0.png

is quite familiar for me. But I still cannot derive it in terms of time coordinate parameter;
a82eae864b04bc27b468fc0becfabe9d.png

I can't get the second term on the right hand side
what I can get is
½{d[lngαβ(dxα/dt)(dxβ/dt)]/dt}dxμ/dt

How can I obtain that term?
 
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I suggest you do not work in terms of the metric. The equation follows directly from the variable substitution to parameterise the curve with ##t## instead of ##s## and looking at how the geodesic equation transforms under this change. Keep in mind that
$$
\frac{d^2 t}{ds^2} = - \Gamma^0_{\alpha\beta} \dot x^\alpha \dot x^\beta .
$$
 
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