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In maths that is

## Question

Verify the consequences of metric compatibility: If\begin{align}

{\mathrm{\nabla }}_{\sigma}g_{\mu \nu }=0 & \phantom {10000}(1) \\

\end{align}then (a)\begin{align}

{\mathrm{\nabla }}_{\sigma}g^{\mu \nu }=0 & \phantom {10000}(2) \\

\end{align}and (b)\begin{align}

{\mathrm{\nabla }}_{\lambda}{\varepsilon }_{\mu \nu \sigma \rho }=0 & \phantom {10000}(3) \\

\end{align} I am not sure if we are assuming ##{\mathrm{\Gamma }}^{\tau }_{\lambda \mu }={\mathrm{\Gamma }}^{\tau }_{ \mu \lambda }## or not.

## Answer

Part (a) was quite simple but I struggled with the part (b) until 23 March and had to give up. Along the way I had lots of practice at index manipulation, I reacquainted myself with Cramer's rule for solving simultaneous equations, proved (b) on the surface of a sphere, found the 'dynamite' version of Carroll's streamlined matrix determinant equation (2.66) and added some equation shortcut keys to my keyboard. The time was not wasted.We make frequent use here of the fact that ##g_{\mu \nu }g^{\mu \rho }\mathrm{=}{\delta}^{\rho }_{\nu }## and the indexing effect of the Kronecker delta: ##{\delta}^{\lambda }_{\beta }\mathrm{\Gamma }^{\mu }_{\sigma \lambda }={\mathrm{\Gamma }}^{\mu }_{\sigma \beta }## because we are summing over ##\lambda ## and the only non-zero term is when ##\beta =\lambda ##. In this case ##\mathrm{\Gamma }## can be replaced by any symbol or tensor of any rank.

Here is the full effort Ex 3.01 Consequences of metric compatibility.pdf (7 pages of which 4 might be worth looking at).

**The comment left by JSBach1801 solved this problem very easily as I fully realised in Jan 2021. Thanks! That answer is right at the end of the pdf.**

Check out this link for some really clever solutions to the porblem from Carroll's book. The Part B is so easier than you realize.

ReplyDeletehttps://indexguy.wordpress.com/2007/07/13/some-identities-regarding-metric-compatibility/

The proof for part a is great. But I don't follow the proof for part b. It is true that the determinant will only depend on the metric entries, but why would that mean that the covariant derivative will vanish?

Deletethe fact that the covariant derivative vanishes follows directly from part (a) XD one way of thinking about it is that g is 'constant' wrt covariant derivatives. Other than that I suppose that since the Levi-Civita symbol is not a tensor the covariant derivative of the Levi-Civita symbol is just a partial derivative, so everything vanishes. I wasn't quite sure about this argument so I did things a bit more explicitly.

DeleteMy fully typed-up solution can be found here: https://drive.google.com/file/d/13dBHxlh8CIBFfUIrf18mFPERDQ7IWkHJ/view?usp=sharing

The solution to (b) in the link provided by JSBach1801 is simply wrong. It is very important to keep in mind that covariant derivatives are meaningful only when operating on a tensor field. (A tensor field may be a field of scalars, vectors, one-forms, etc.) |g|, or its square root, does not form a scalar field, because its value at each point is coordinate dependent. Idem for the Levi-Civita symbol.

ReplyDeleteHere's [my solution to the exercise](https://physics.stackexchange.com/a/743720/269697). It's too long to be posted as a comment.

ReplyDelete