I would like to ask you a question - maybe simple - but bothering me.

We have two four-position vectors product in curvilinear coordinates given by

$(1) \quad X^{\alpha}g_{\alpha \beta}X^{\beta} = \tau^2$

and taking proper-time derivative I expected to get

$(2) \quad 2\cdot U^{\alpha}g_{\alpha \beta}X^{\beta} = 2\tau$

But if we would decompose it into partial derivatives and proper-time derivatives of separate vectors it would be:

$(3) \quad \frac{dX^{\alpha}}{d\tau}g_{\alpha \beta}X^{\beta} + X^{\alpha}g_{\alpha \beta}\frac{dX^{\beta}}{d\tau} + X^{\alpha}X^{\beta}\frac{dg_{\alpha \beta}}{d\tau} = 2\cdot U^{\alpha}g_{\alpha \beta}X^{\beta} +X^{\alpha}X^{\beta}\frac{dg_{\alpha \beta}}{d\tau} \neq 2\tau$

$(4) \quad U^{\gamma} \nabla_{\gamma} X^{\alpha}X^{\beta}g_{\alpha \beta}= U^{\gamma} g_{\gamma}\,^{ \alpha} X^{\beta}g_{\alpha \beta} + U^{\gamma} g_{\gamma}\,^{\beta} X^{\alpha}g_{\alpha \beta} + X^{\alpha}X^{\beta} U^{\gamma} \nabla_{\gamma} \, g_{\alpha \beta} = 2\cdot U^{\alpha}g_{\alpha \beta}X^{\beta} +X^{\alpha}X^{\beta}U^{\gamma} \,[??] \neq 2\tau $

What is bothering me is that:

A) $Eq (2) \neq Eq (3) \neq Eq (3)$

B) If equation (2) is correct, then why? Because metric tensor may be changing over time as I suppose.

C) What it is $[??]=\nabla_{\gamma} \, g_{\alpha \beta} $ in Eq. (4)? - As far as I know, metric tensor should be invariant under covariant four-gradient, and therefore I have no idea what is the result of such derivative.

I would appreciate if you find a while to answer.

I suppose for you it will be a minute, but I have really doubt about above formulas.


1 Answer 1


The source of your problem seems to be C). The Christoffel symbols satisfy $$\Gamma_{abc} + \Gamma_{bac} = g_{ab,c} $$ So (assuming no silly typos) $$g_{ab;i} = g_{ab,i} - \Gamma^c_{bi}g_{ac} - \Gamma^c_{ai}g_{cb}= g_{ab,i} - \Gamma_{abi} - \Gamma_{bai} =0 $$ Thus, the metric behaves as a constant with respect to covariant differentiation; this simply states that vectors have constant magnitude under parallel displacement.

  • $\begingroup$ Thank you Charles! $\endgroup$
    – pog
    Commented Apr 8, 2020 at 22:48

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