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In the process of getting the formula of Christoffel symbols in the terms of metric tensors, we get

$$\partial_n g_{rm}+\partial_m g_{rn}-\partial_r g_{mn}=2\Gamma ^t_{mn} g_{rt}. $$

The book I have read goes ' multiply both sides of the equation by the inverse metric tensor' and yields

$$\Gamma ^t_{mn}={1\over 2} g^{rt}(\partial_n g_{rm}+\partial_m g_{rn}-\partial_r g_{mn}).$$

But I am confused with dummy variables. In the first equation, r was the real variable and t was a dummy variable, but in the second one, the opposite seems to be true. If we write the right hand side of the first equation without summation, it'll be

$$2(\Gamma ^0_{mn} g_{r0}+\Gamma ^1_{mn} g_{r1}+\Gamma ^2_{mn} g_{r2}+\Gamma ^3_{mn} g_{r3})$$

I will multiply this with $g^{rt}$. In this case it's for t=0, 1, 2, 3, so the left hand side is $$(\Gamma ^0_{mn} g_{r0}+\Gamma ^1_{mn} g_{r1}+\Gamma ^2_{mn} g_{r2}+\Gamma ^3_{mn} g_{r3})(g^{r0}+g^{r1}+g^{r2}+g^{r3}) \\ = \Gamma ^0_{mn}+\Gamma ^1_{mn}+\Gamma ^2_{mn}+\Gamma ^3_{mn}$$ since $g_{rm}g^{rn}=\delta ^n_m$. (Let's divide both sides with 2)

The left hand side will be $${1\over 2} (g^{r0}+g^{r1}+g^{r2}+g^{r3})(\partial_n g_{rm}+\partial_m g_{rn}-\partial_r g_{mn})$$ and it is not a summation with r because r is not a dummy variable.

So I don't know the way to get to the right result from the last equation. And how the real and dummy variables changed during the calculation.

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  • $\begingroup$ Because t is dummy variable you can change it with any letter you want, for example change t in first equation with an other letter like f and try solution $\endgroup$
    – Sancol.
    Feb 27 at 14:55

2 Answers 2

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No, it is not working like that.

As one disposes of ( as one has 3 different indices r, m, n which are not contracted of which each runs from 0 to 3 -- so 4 different values) $4\times 4\times 4$ equations (some of these might be identical due to symmetry relations) one can combines these equations upon multiplication with the metric tensor $g^{rs}$ --- i.e. the multiplication with the metric tensor is not just multiplication but also implicates summation:

$$g^{rs}(\partial_n g_{rm} + \partial_m g_{rn} -\partial_r g_{nm})= 2g^{rs}g_{rt}\Gamma^t_{mn}$$

which actually means: $$\sum\limits_{r=0,\ldots, 3} g^{rs}(\partial_n g_{rm} + \partial_m g_{rn} -\partial_r g_{nm})=\sum\limits_{r=0,\ldots, 3} 2 g^{rs}g_{rt} \Gamma^t_{mn}$$

This means that one takes from the $4\times 4\times 4$ equations $4\times 4$ eqns (for m and n) but with fixed $r=0$ multiplying it with $g^{0s}$, adds it to the next set of equations for $r=1$ multiplying that one with $g^{1s}$ and so forth, i.e. also adds the set of eqns for $r=2$ weighted by $g^{2s}$ and for $r=3$ weighted by $g^{3s}$. And of course one does it for each $r$ on LHS and RHS.

Then one gets: $$ \sum\limits_{r=0,\ldots, 3} g^{rs}\frac{1}{2}(\partial_n g_{rm} + \partial_m g_{rn} -\partial_r g_{nm})= \delta^s_t\Gamma^t_{mn} = \Gamma^s_{mn}$$

because

$$ \sum\limits_{r=0,\ldots, 3} g^{rs}g_{rt} = \delta^s_t$$

where $\delta^s_t$ is the Kronecker symbol. Or applying Einstein's summation convention the summation symbol can be omitted.

$$g^{rs}\frac{1}{2}(\partial_n g_{rm} + \partial_m g_{rn} -\partial_r g_{nm})= \delta^s_t\Gamma^t_{mn} = \Gamma^s_{mn}$$

As the very last step one can replace the index $s$ by $t$ again. But the name of an index can be chosen (almost) arbitrarily --- better do not use indices which are already used in the equation.

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Act on your first equation with $g^{ar}$. Then the action on the right-hand side will produce $2\Gamma^t_{mn} \delta^a_t$. Contracting with the Kronecker delta and rearranging yields: $$ \Gamma^a_{mn} = \frac{1}{2} g^{ar} \left( \partial_n g_{rm} + \partial_m g_{rn} - \partial_r g_{mn}\right) $$ This is identical to the expression in your second line (with the free index $t$ replaced with $a$, and $g^{ab} = g^{ba}$).

You are right about what is free and what is dummy. When you contract with the "inverse metric" you should use a new free index, $a$, and the old free index $r$. The book then, confusingly, renames the new free index $t$ again, even though it was summed over in the calculation.

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