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In lecture 3 (about 97 min into the lecture) of Leonard Susskind's general relativity course, he suggests finding the Riemann curvature tensor in terms of the Christoffel symbols as an exercise.

I started from $$D_r D_s V_n - D_s D_r V_n = R_{rs}^t{_n} V_t$$ as a definition of the Riemann curvature tensor.

The covariant derivative is defined by $$D_sV_n = \partial_sV_n - \Gamma_s^u{_n}V_u$$ and the Christoffel symbols are defined in terms of the metric by $$\Gamma_s^u{_n} = \dfrac12 g^{ut}\left( \partial_sg_{tn} + \partial_n g_{st} - \partial_tg_{sn}\right).$$

I'm supposed to find $$R_{rs}^t{_n} = -\partial_r\Gamma_s^t{_n} + \partial_s\Gamma_r^t{_n} + \Gamma_r^w{_n}\Gamma_s^t{_w} - \Gamma_s^w{_n}\Gamma_r^t{_w}.$$

I get that this works, but only if $$\Gamma_r^w{_n}\partial_sV_w = \Gamma_s^w{_n}\partial_rV_w.$$

I'm stuck showing this. I tried expanding the Christoffel symbols out into the metric, but I wind up trying to show things like $$\partial_r g_{tn}\partial_s = \partial_s g_{tn}\partial_r,$$ which I don't see how to justify.

(It may also be that I'm making algebra errors because I'm unfamiliar with tensor calculus, so any advice you have would be helpful here.)

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    $\begingroup$ The identity is false. Try e.g. acting with it on a function that only depends on one of the coordinates, say, $s$, with $s\neq r$. The r.h.s. vanishes, while the l.h.s. does not. $\endgroup$ – AccidentalFourierTransform Jul 20 '19 at 15:57
  • $\begingroup$ So is the setup of the calculation right? $\endgroup$ – Mark Eichenlaub Jul 20 '19 at 16:04
  • $\begingroup$ I think so, but I didn't double check all the indices. The derivation is pretty standard though, so you should be able to find it online. $\endgroup$ – AccidentalFourierTransform Jul 20 '19 at 16:09
  • $\begingroup$ Okay, I found this page: einsteinrelativelyeasy.com/index.php/general-relativity/… But it seems to say that the identity I asked about is true. (near the bottom, just below the first blue box) $\endgroup$ – Mark Eichenlaub Jul 20 '19 at 16:53
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    $\begingroup$ I'm not sure I follow; shouldn't I be able to choose $v^i$ arbitrarily to show that if the sum is zero, each individual operator is also zero? I don't know which identity you mean by "the quoted identity". Are you saying that the page I linked above is wrong? $\endgroup$ – Mark Eichenlaub Jul 20 '19 at 18:04
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I'm over a year late to answer this question, but here goes. I think you must have misapplied some identities or got lost in the weeds during your derivation. This has to do with the torsion-freeness of the connection. We have, using the formulas for the covariant derivative of a rank 2 tensor:

\begin{align*} D_rD_sV_n=&D_r(\partial_s V_n-\Gamma^u_{sn}V_u)\\ &=\partial_r \partial_s V_n-(\partial_r \Gamma^u_{sn})V_u-\Gamma^u_{sn}\partial_r V_u\\ &-\Gamma^a_{rs}(\partial_a V_n-\Gamma^u_{an}V_u)\\ &-\Gamma^a_{rn}(\partial_s V_a-\Gamma^u_{sa}V_u) \end{align*}

We want to antisymmetrize with respect to r and s, and I want to use bracket notation $V_{[a} U_{b]}\equiv \frac{1}{2}(V_a U_b-V_b U_a)$ for brevity. To do this without ambiguity, we should try to rearrange all our indices to put r and s next to each other. Using the symmetry of the Christoffel symbol in the lower two indices (a consequence of torsion freeness)...

\begin{align*} D_rD_sV_n=&-(\partial_r \Gamma^u_{sn})V_u-\Gamma^u_{ns}\partial_r V_u\\ &-\Gamma^a_{nr}\partial_s V_a+\Gamma^a_{nr}\Gamma^u_{sa}V_u\\ &+\text{symmetric stuff} \end{align*} Then we have: $$ D_{[r}D_{s]}V_n=-(\partial_{[r} \Gamma^u_{s]n})V_u-\Gamma^u_{n[s}\partial_{r]} V_u-\Gamma^a_{n[r}\partial_{s]} V_a+\Gamma^a_{n[r}\Gamma^u_{s]a}V_u$$

The first and last terms give the Riemann tensor up to factors of two. The middle two terms add up to zero because by definition, $\Gamma^u_{n[s}\partial_{r]}=-\Gamma^u_{n[r}\partial_{s]}$.

What did you do wrong?

You might have forgot to properly distribute the partial derivative across $\partial_r(\Gamma^u_{sn}V_u)$. If you work with the convention that $\partial_r$ acts on everything to the right of it, you have to be very careful about it. If you find yourself making mistakes or overusing parentheses, you can use the symbol $\overrightarrow\partial$ to mean "acts on everything to the right" and the symbol $\partial$ to mean "acts on the thing immediately to the right". So as an operator equation $\overrightarrow\partial_x f=\partial_xf+f\overrightarrow\partial_x$.

You also might have misapplied the covariant derivative of a 2-tensor. We have:

$$D_a (V_{bc})=\partial_a V_{bc}-\Gamma^u_{ab}V_{uc}-\Gamma^u_{ac}V_{bu}$$

There are also plenty of other bookkeeping / index mistakes that could be made here.

Other References

pg. 82 / equation 3.66 of Sean Carroll's lecture notes covers this, making no assumptions about torsion.

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