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I have been trying to express $\eta^{\mu\nu}$ in terms of $\eta_{\mu\nu}$ and I have stumble upon the following relation:

$\eta^{\mu\nu} = \eta^{\mu\alpha}\eta^{\nu\beta}\eta_{\alpha\beta}$

I can see how the indices $\alpha$ and $\beta$ contract to form the contravariant tensor $\eta^{\mu\nu}$, but I do not understand why the ordering of the tensors $\eta^{\mu\alpha}$, $\eta^{\nu\beta}$ and $\eta_{\alpha\beta}$ does not matter. If I am correct, only the inner upper and lower indices can contract, so that the ordering $\eta^{\mu\alpha}\eta_{\alpha\beta}\eta^{\nu\beta}$ is more appropriate. So, $\eta^{\mu\alpha}\eta_{\alpha\beta}\eta^{\nu\beta} = \eta^{\mu}_{\beta}\eta^{\nu\beta}$, so that inner upper and lower index $\alpha$ contracts. Then, $\eta^{\mu}_{\beta}\eta^{\nu\beta} = \eta^{\mu}_{\beta}\eta^{\beta\nu}$, because the metric tensor is symmetric. Next, $\eta^{\mu}_{\beta}\eta^{\beta\nu} = \eta^{\mu\nu}$ and we end up with the desired form.

Do you think ordering of the tensors in a tensor product matters? I say this, because matrix multiplication is non-commutative. Therefore, shouldn't tensor multiplication also be non-commutative?

Also, I could simplify the RHS into LHS only because the metric tensor is symmetric? What if the tensor were not symmetric? How could we then have managed to simplify?

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Well they are real numbers... if you like you can express to another physicist your relation just as well like so:

Suppose $\eta_{\mu \nu}$ is a metric in $N$ dimensions with components who vary contravariantly and $\bar{\eta}_{\mu \nu}$ the corresponding covariant components. Then $\eta_{\mu\nu} = \sum_{\alpha=1}^N\sum_{\beta=1}^N\eta_{\mu\alpha}\eta_{\nu\beta}\bar{\eta}_{\alpha\beta}$

the summand is just a product of real numbers, and products of real numbers can be arranged however you please: $abc=cab=bca=cba=\cdots$.

You'll see this kind of phrasing in books just before they introduce the Einstein summation convention! "whose components vary contravariantly" is replaced with a superscript and "whose components vary covariantly" is replaced with a subscript and the sums are dropped, so you can write the whole thing more quickly. But you're always talking about sums and your components are actual numbers.

In matrix multiplication $AB\neq BA$ can occur. This equation has no indices. It is still the case that $\sum_j A_{ij}B_{jk}=\sum_j B_{jk}A_{ij}$, because these are just real numbers.

You really should take a second look at the definition of the Einstein tensor notation. There's absolutely no magic to it (excluding covariance/contravariance!)

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  • $\begingroup$ Thanks for the kind reply. I was also wondering why $\eta^{\mu\alpha}\eta^{\mu\beta} = 1$. Is it because the metric tensor is symmetric, so that we can rewrite the product as $\eta^{\alpha\mu}\eta^{\mu\beta}$, and then we sum over the index $\mu$ to form the product of 1? $\endgroup$ – nightmarish Mar 12 '15 at 6:03
  • $\begingroup$ watch your upper and lower indices! It's also not true it equals $1$. See property 18 on mathworld.wolfram.com/MetricTensor.html $\endgroup$ – user12029 Mar 12 '15 at 6:37

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