0
$\begingroup$

We have a covariant derivative of a covariant tensor: $$ A_{\mu ; \nu} = A_{\mu , \nu} - \Gamma^{\alpha}_{\mu \nu} A_{\alpha} $$ The covariant derivative of a contravariant tensor is: $$ A^{\mu}_{;\nu} = A^{\mu}_{,\nu} + \Gamma^{\mu}_{\nu \alpha} A^{\alpha} $$

I am trying to use $A^{\mu}_{;\nu} = (g^{\mu \sigma} A_{\sigma})_{;\nu}$ to derive the second equation, given the first.

My attempt is as follows: $$ \begin{align} A^{\mu}_{;\nu} &= (g^{\mu \sigma} A_{\sigma})_{,\nu} - g^{\mu \sigma} \Gamma^{\alpha}_{\sigma \nu}A_{\alpha} \\ &= A^{\mu}_{,\nu} - \Gamma^{\mu \alpha}_{\nu}A_{\alpha} \\ &= A^{\mu}_{,\nu} - (g^{\alpha \epsilon}\Gamma^{\mu}_{\nu \epsilon})A_{\alpha} \\ &= A^{\mu}_{,\nu} - \Gamma^{\mu}_{\nu \epsilon}(g^{\alpha \epsilon}A_{\alpha}) \\ &= A^{\mu}_{,\nu} - \Gamma^{\mu}_{\nu \epsilon}A^{\epsilon} \end{align} $$ Relabelling $\epsilon$ as $\alpha$ we have: $$ A^{\mu}_{;\nu}= A^{\mu}_{,\nu} - \Gamma^{\mu}_{\nu \alpha}A^{\alpha} $$

Clearly there is an error in my method here, as I have a minus where I ought to have a plus. Have I missed a step, or am I going about this the completely wrong way?

Thanks, Sean.

$\endgroup$
  • $\begingroup$ Is the Christoffel symbol symmetric in the upper two indices? $\endgroup$ – GodotMisogi May 13 '16 at 16:06
  • $\begingroup$ This is the first that I've ever encountered a Christoffel with two up indices. I'm assuming that if it is antisymmetric in the upper indices, then the sign would change between my 2nd and 3rd line of working? $\endgroup$ – Vielbein May 13 '16 at 16:15
  • $\begingroup$ Covariant derivative of the metric tensor is exactly 0, so you can freely move it in or out the derivative expression. $\endgroup$ – Alexander May 13 '16 at 16:16
2
$\begingroup$

Explicitly working with the components of $g$ and $\Gamma$ can be messy. Here is a nicer way to derive the coordinate expression for the covariant derivative of a vector (or any tensor) without raising or lowering indices.

Consider vector fields $A^\mu, B_\mu$. The covariant derivative of a scalar is the same as the coordinate derivative (in any coordinate system), so $$ \nabla_\nu(A^\mu B_\mu)=\partial_\nu(A^\mu B_\mu)= (\partial_\nu A^\mu)B_\mu+A^\mu\partial_\nu B_\mu \ . $$ On the other hand using Leibniz's rule for $\nabla$ and your expression for $\nabla_\nu B_\mu$ $$ \nabla_\nu(A^\mu B_\mu)=(\nabla_\nu A^\mu)B_\mu+A^\mu\nabla_\nu B_\mu =(\nabla_\nu A^\mu)B_\mu+A^\mu\partial_\nu B_\mu-A^\mu \Gamma^\alpha_{\mu\nu}B_\alpha \ . $$ Taking the difference of the two equations and relabelling the dumb indices in the Christoffel symbol term we obtain $$ B_\mu(\partial_\nu A^\mu+ \Gamma^\mu_{\sigma\nu}A^\sigma-\nabla_\nu A^\mu)=0. $$ Since this must be true for any one-form $B_\mu$ the term in brackets vanishes, and the components of the covariant derivative of a vector in some coordinates are given by

$$\nabla_\nu A^\mu=\partial_\nu A^\mu+ \Gamma^\mu_{\sigma\nu}A^\sigma \ ,$$

as you wanted to show. In a similar way you can now derive the coordinate expression for the covariant derivative of any higher order tensor.

$\endgroup$
1
$\begingroup$

$$ \nabla_\mu A^\nu = g^{\nu\alpha} \nabla_\mu A_\alpha = g^{\nu\alpha} ( \partial_\mu A_\alpha - \Gamma^\lambda_{\mu\alpha} A_\lambda ) = g^{\nu\alpha} \partial_\mu ( g_{\alpha\beta} A^\beta ) - g^{\nu\alpha} \Gamma^\lambda_{\mu\alpha} g_{\lambda\beta} A^\beta $$ This simplifies to $$ \nabla_\mu A^\nu = \partial_\mu A^\nu + ( g^{\nu\alpha} \partial_\mu g_{\alpha\beta} - g^{\nu\alpha} \Gamma^\lambda_{\mu\alpha} g_{\lambda\beta} ) A^\beta $$ Finally, note that $$ g^{\nu\alpha} \partial_\mu g_{\alpha\beta} - g^{\nu\alpha} g_{\lambda\beta}\Gamma^\lambda_{\mu\alpha} = g^{\nu\alpha} \partial_\mu g_{\alpha\beta} - \frac{1}{2} g^{\nu\alpha} g_{\lambda\beta} g^{\lambda \rho} ( \partial_\mu g_{\rho\alpha} + \partial_\alpha g_{\rho\mu} - \partial_\rho g_{\mu\alpha} ) = g^{\nu\alpha} \partial_\mu g_{\alpha\beta} - \frac{1}{2} g^{\nu\alpha} ( \partial_\mu g_{\beta\alpha} + \partial_\alpha g_{\beta\mu} - \partial_\beta g_{\mu\alpha} ) = \frac{1}{2} g^{\nu\alpha} ( \partial_\mu g_{\beta\alpha} + \partial_\beta g_{\mu\alpha} - \partial_\alpha g_{\beta\mu} ) = \Gamma^\nu_{\mu\beta} $$ Thus, we find $$ \nabla_\mu A^\nu = \partial_\mu A^\nu + \Gamma^\nu_{\mu\beta} A^\beta $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.