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

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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From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
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