Undergrad Variation of Four-Velocity Vector w/ Respect to Metric Tensor

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SUMMARY

The variation of the four-velocity vector, defined as Uμ=dxμ/dτ, with respect to the metric tensor gαβ is expressed as δUμ=1/2 UμδgαβUαUβ. To derive this, one must first prove the equation δg^{\alpha\beta}U_{\alpha}U_{\beta} = 2 U_{\mu}δU^{\mu}. By utilizing the identity U_{\mu}U^{\mu} = 1, the left-hand side can be rewritten, leading to the desired result through comparison of the derived equations.

PREREQUISITES
  • Understanding of four-velocity in general relativity
  • Familiarity with metric tensors and their variations
  • Knowledge of LaTeX for mathematical expressions
  • Basic principles of differential geometry
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  • Study the derivation of variations in tensor calculus
  • Learn about the properties of metric tensors in general relativity
  • Explore the application of LaTeX for typesetting mathematical equations
  • Investigate the implications of the four-velocity in relativistic physics
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Hubble_92
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Hi everyone! I'm having some difficulty showing that the variation of the four-velocity,

Uμ=dxμ/dτ

with respect the metric tensor gαβ is

δUμ=1/2 UμδgαβUαUβ

Does anyone have any suggestion?

Cheers,
Rafael.

PD: Thanks in advances for your answers; this is my first post! I think ill be active sharing and discussing in other Physics/ Astrophysics topics ;)
 
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Hubble_92 said:
Hi everyone! [...] PD: Thanks in advances for your answers; this is my first post! I think ill be active sharing and discussing in other Physics/ Astrophysics topics ;)
Just a suggestion for the future. Please, use the LaTex code. This website has MathJax implemented, so that the equations are made to look great. Just search here for a tutorial on how to write with the simple code.
 
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Likes martinbn, vanhees71 and Hubble_92
First prove that
\delta g^{\alpha\beta}U_{\alpha}U_{\beta} = 2 U_{\mu}\delta U^{\mu}. \ \ \ \ \ \ \ (1) Then, using U_{\mu}U^{\mu} = 1, you can write the left-hand-side of (1) as
\delta g^{\alpha \beta}U_{\alpha}U_{\beta} \equiv 2 U_{\mu} \left( \frac{1}{2}U^{\mu} \delta g^{\alpha \beta}U_{\alpha}U_{\beta}\right) . \ \ \ \ \ (2)
The result follows by comparing (1) with (2).
 
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Likes anuttarasammyak, vanhees71, PeroK and 1 other person
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