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This is a follow on to my previous question: Dirac's book *General Theory of Relativity*: Doesn't this show the partial derivative of the metric tensor is zero?

I should not have made that mistake. I "know" better. For the sake of my own ego I plead:

Wheeler’s First Moral Principle: Never make a calculation until you know the answer. Make an estimate before every calculation, try a simple physical argument (symmetry! invariance! conservation!) before every derivation, guess the answer to every paradox and puzzle. Courage: No one else needs to know what the guess is. Therefore make it quickly, by instinct. A right guess reinforces this instinct. A wrong guess brings the refreshment of surprise. In either case life as a spacetime expert, however long, is more fun!

I concede that Dr. Wheeler didn't recommend posting my guess to p.se. Nonetheless, I did gain from the mistake. There are a few reasons why I guessed wrongly regarding the vanishing of $g_{\alpha\beta,\gamma}$. One is that I knew, at least under certain circumstances $g_{\alpha\beta,\gamma}$ can be made to vanish, at least locally. As I now recall, that requires a Riemann normal local coordinate system. I also knew that, either the partial or the covariant derivative of the metric tensor always vanishes. But more significantly, I was following what Dirac apparently prescribes in section 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.

When I first read that, it made me uneasy. I know the affine connection coefficients (Christoffel symbols) have the characteristics of "nontensors" described by Dirac. I simply guessed, and guessed wrong, that his rule was applicable to the partial derivative of the metric tensor. I observe that, in section 10 Dirac calls the partial derivative of a vector in covariant form a "nontensor", which suggests incorrectly that the above prescription is applicable. So here is my question:

When and why can we apply Dirac's prescription for raising and lowering indices on "nontensors"? In other words, why does it work for Christoffel symbols, but not for partial derivatives of tensors? My best guess is that, in the case of Christoffel symbols, there is no preexisting definition for the Chrstoffel symbol of the second kind, so we can define it, at will.

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Dirac's remark should be use with diligence. E.g. by raising the lower $\nu_1$-index on the nontensor $$N_{\lambda_1\ldots\lambda_t}{}^{\mu_1\ldots\mu_r}{}_{\nu_1\ldots\nu_s}~:=~\partial_{\lambda_1}\ldots \partial_{\lambda_t}T^{\mu_1\ldots\mu_r}{}_{\nu_1\ldots\nu_s},\tag{A}$$ Dirac means the nontensor $$N_{\lambda_1\ldots\lambda_t}{}^{\mu_1\ldots\mu_r\alpha}{}_{\nu_2\ldots\nu_s}~:=~g^{\alpha\nu_1}N_{\lambda_1\ldots\lambda_t}{}^{\mu_1\ldots\mu_r}{}_{\nu_1\ldots\nu_s}\tag{B}$$ rather than the nontensor $$\partial_{\lambda_1}\ldots \partial_{\lambda_t}T^{\mu_1\ldots\mu_r\alpha}{}_{\nu_2\ldots\nu_s} ~:=~\partial_{\lambda_1}\ldots \partial_{\lambda_t}(g^{\alpha\nu_1}T^{\mu_1\ldots\mu_r}{}_{\nu_1\ldots\nu_s}).\tag{C}$$ Eqs. (B) & (C) are generically not the same.

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  • $\begingroup$ OK, I think I see the problem. Dirac's rule applies to any nontensor. My error was in interpreting the result. If I start with the nontensor $N^{\mu}{}_{\nu\rho}=T^{\mu}{}_{\nu,\rho},$ I cannot raise the index on $N^{\mu}{}_{\nu\rho}$ and put the comma back in. This is clearer when written using $\partial$ symbols rather than commas. $\endgroup$ – Steven Thomas Hatton Dec 27 '18 at 21:07

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