Dirac's book *General Theory of Relativity*: Doesn't this show the partial derivative of the metric tensor is zero?

Question

See the bold text for my question.

This question regards Dirac's General Theory of Relativity page 13. In his demonstration that the length of a vector is unchanged by parallel displacement Dirac does the following:

Expand the differential of the scalar product of the vector with itself

$$d\left[a_{\alpha}a_{\beta}g^{\alpha\beta}\right]=\left(a_{\beta}da_{\alpha}+a_{\alpha}da_{\beta}\right)g^{\alpha\beta}+a_{\alpha}a_{\beta}g^{\alpha\beta}{}_{,\gamma}dx^{\gamma}.$$

$$=a^{\alpha}da_{\alpha}+a^{\beta}da_{\beta}+a_{\alpha}a_{\beta}g^{\alpha\beta}{}_{,\gamma}dx^{\gamma}$$

$$=a^{\alpha}\Gamma_{\beta\alpha\gamma}a^{\beta}dx^{\gamma}+a^{\beta}\Gamma_{\alpha\beta\gamma}a^{\alpha}dx^{\gamma}+a_{\alpha}a_{\beta}g^{\alpha\beta}{}_{,\gamma}dx^{\gamma}$$

$$=\left(\Gamma_{\beta\alpha\gamma}+\Gamma_{\alpha\beta\gamma}\right)a^{\alpha}a^{\beta}dx^{\gamma}+a_{\alpha}a_{\beta}g^{\alpha\beta}{}_{,\gamma}dx^{\gamma}$$

$$=g_{\alpha\beta,\gamma}a^{\alpha}a^{\beta}dx^{\gamma}+a_{\alpha}a_{\beta}g^{\alpha\beta}{}_{,\gamma}dx^{\gamma}.$$

Next we differentiate the Kronecker delta

$$0=d\left[\delta_{\nu}^{\mu}\right]=d\left[g^{\mu\alpha}g_{\nu\alpha}\right]=\left(g^{\mu\alpha}{}_{,\beta}g_{\nu\alpha}+g^{\mu\alpha}g_{\nu\alpha,\beta}\right)dx^{\beta}.$$

Then we contract the parenthetic expression with the metric tensor $g_{\mu\delta}$ to obtain

$$g_{\mu\delta}g_{\nu\alpha}g^{\mu\alpha}{}_{,\beta}=-g_{\nu\delta,\beta}.$$

At this point Dirac says:

This is a useful formula giving us the derivative of $g^{\alpha\beta}$ in terms of the derivative $g_{\mu\nu}.$ It allows us to infer $$g_{\alpha\beta,\gamma}a^{\alpha}a^{\beta}=-a_{\alpha}a_{\beta}g^{\alpha\beta}{}_{,\gamma}$$ and so the expression $d\left[a_{\alpha}a_{\beta}g^{\alpha\beta}\right]$ vanishes.

I don't dispute Dirac's conclusion, but it appears he omitted a more fundamental and important result of

$$g_{\mu\delta}g_{\nu\alpha}g^{\mu\alpha}{}_{,\beta}=-g_{\nu\delta,\beta}.$$

According to chapter 4 Nontensors, given any set of indexed symbols defined at a point we may use the metric tensor to raise and lower indices. Applying that to the last equation gives us

$$g_{\nu\delta,\beta}=-g_{\nu\delta,\beta}.$$

Which can only hold if $g_{\nu\delta,\beta}=0$. Is this correct?

I believe it is. So I really don't understand why Dirac didn't simply use that fact to draw his conclusion that the differential of $a_{\alpha}a_{\beta}g^{\alpha\beta}$ vanishes.

This is what Dirac says in chapter 4:

We can have a quantity $N^{\mu}{}_{\nu\rho\dots}$ with various up and down suffixes, which is not a tensor. If it is a tensor, it must transform under a change of coordinate system according to the law exemplified by (3.6). With any other law it is a nontensor. A tensor has the property that if all the components vanish in one system of coordinates, they vanish in every system of coordinates. This may not hold for nontensors.

For a nontensor we can raise and lower suffixes by the same rules as for a tensor. Thus, for example,

$$g^{\alpha\nu}N^{\mu}{}_{\nu\rho}=N^{\mu\alpha}{}_{\rho}.$$

The consistency of these rules is quite independent of the transformation laws to a different system of coordinates. Similarly we can contract by putting an upper and lower suffix equal.

We may have tensors and nontensors appearing together in the same equation. The rules for balancing suffixes apply equally to tensors and nontensors.

Answer 1

Be careful on what you apply the lowering of operators.

\begin{eqnarray} g_{\nu\delta, \beta} &=& \partial_{\beta}(g^{\mu\alpha}g_{\mu\delta} g_{\nu\alpha})\\ &=& g^{\mu\alpha}_{,\beta}g_{\mu\delta} g_{\nu\alpha} + g^{\mu\alpha}g_{\mu\delta,\beta} g_{\nu\alpha} +g^{\mu\alpha}g_{\mu\delta} g_{\nu\alpha,\beta} \end{eqnarray}

It is not the same as $ g^{\mu\alpha}_{,\beta}g_{\mu\delta} g_{\nu\alpha}$