I'm doing an exercise where I needed to find, for an arbitary $4$-vector $V^\mu$ $$ V_{r;\theta;\phi}-V_{r;\phi;\theta}=\ ???$$ for the space-time metric $$ ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}=\left(1-\frac{r_g}{r}\right)c^2dt^2-\left(1+\frac{r_g}{r}\right)dr^2-r^2g_\Omega $$ Note that here $;$ denote the covariant differential so that $$V_{r;\theta;\phi}=\frac{D }{Dx^\phi}\left(\frac{D V_r}{Dx^\theta}\right)$$

I'm able to show that the required quantity turn out to be

$$V_{r;\theta;\phi}-V_{r;\phi;\theta}=\frac{1}{r}\left(\frac{\partial V_\phi}{\partial x^\theta}-\frac{\partial V_\theta}{\partial x^\phi}\right)$$

It's asked to interpret this result which I'm not able to do. I can understand that the right-hand side is a radial component of the curl. But I'm not able to see any geometrical interpretation why this should be equal to left hand side. Please help me with this.


You certainly have an interesting question, YoungKidachi.

As soon as I saw the indices in the vectors $V$, I immediately recognized a commutator in the covariant derivatives, and you probably did too. The thing is tah I link a commutator of covariant derivatives directly to the Riemann curvature tensor, and in fact it is the right way to go; take a look here Riemman curvature tensor. What you are considering is just a component of:

$ V_\beta R^\beta \,_{\nu \sigma \rho} = \nabla_{[\rho}\nabla_{\sigma]} V_\nu $

So my take at it is that you are measuring the component $ r $ (radial distance) of the vector that tells you what the difference is between transporting $ V_\nu $ in the $ \theta $ then in the $ \phi $ direction vs. transporting it first in the $ \phi $ direction and then in the $ \theta $ direction.

I hope this helps you, or at least that it gets you on the right track!

  • $\begingroup$ In the link, take a look in the "coordinate expression" section, to see what I'm talking about. $\endgroup$ Nov 28 '21 at 13:23
  • $\begingroup$ Notice that as we take the limit $r\rightarrow \infty$, We find that the commutator vanishes. What's the reason for this? Can we see it geometrically or with intuition? $\endgroup$ Nov 28 '21 at 13:46
  • 1
    $\begingroup$ Good question, it actually helps a lot with the intuition. Think what effects taking $ r \rightarrow \infty $ has on the spacetime. The Schwarzschild metric reduces to the minkowski metric, and the connection vanishes. Because the space is flat, there is no preference in the order of parallel transports (kind of the definition of a flat space). This means that the vector V should be zero in all of the components, as the order of transportations don't matter. $\endgroup$ Nov 28 '21 at 13:52

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