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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.

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  • $\begingroup$ That is wrong. The partial of a tensor is NOT a tensor, you cannot raise and lower indices through a derivative, only through a covariant derivative. $\endgroup$ – ggcg Dec 20 '18 at 11:45
  • $\begingroup$ Steven, Dirac's famous book is...not this one :-) Seriously, I know you have read or are reading MTW, so why bother with Dirac? If there is an interesting physical situation described in this book, try to get the main point and redo the calculations using your MTW (or even more modern) knowledge/language/techniques $\endgroup$ – magma Dec 21 '18 at 11:27
  • $\begingroup$ By the way, you are a Mathematica user. Are you aware of the powerful (and free) xAct package www.xact.es? $\endgroup$ – magma Dec 21 '18 at 11:29
  • $\begingroup$ @magma Dirac's book on GR is incredible in it's deft conciseness. He goes directly to the essentials in a way that I can (usually) follow. MTW treats the Bianci identities, beginning on page 364. Dirac derives them on page 23. $\endgroup$ – Steven Thomas Hatton Dec 25 '18 at 0:37
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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}$

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  • $\begingroup$ That's not my claim. I claim $g^{\mu\alpha}{}_{,\beta}g_{\mu\delta}g_{\nu\alpha}=g_{\nu\delta,\beta}=-g_{\mu\delta}g^{\mu\alpha}g_{\nu\alpha,\beta}=-g_{\nu\delta,\beta}.$ $\endgroup$ – Steven Thomas Hatton Dec 20 '18 at 11:01
  • $\begingroup$ @StevenHatton The point here is that even the first step is wrong. You can't just pull a metric through a derivative. You can do that in special relativity where the components of the metric are constant, but not here. $\endgroup$ – knzhou Dec 20 '18 at 11:06
  • $\begingroup$ No, that is exactly your claim. You are not doing tensor analysis on manifolds correctly $\endgroup$ – ggcg Dec 20 '18 at 11:46
  • $\begingroup$ OK. Now I see the point of the answer. What I don't understand is what Dirac says in chapter 4. It gives the impression that one can raise and lower indices on any indexed point function. I am going to add the quoted text to my original question. $\endgroup$ – Steven Thomas Hatton Dec 20 '18 at 11:55
  • $\begingroup$ @knzhou It is true in Riemann normal coordinates. See sections 11.6 and 13.3 of Misner Thorne and Wheeler. $\endgroup$ – Steven Thomas Hatton Dec 20 '18 at 12:37

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