# Riemann tensor and covariant derivative

• A
hi, I tried to take the covariant derivative of riemann tensor using christoffel symbols, but it is such a long equation that I have always been mixing up something. So, Could you share the entire solution, pdf file, or links with me????? ((( I know this is the long way to derive the einstein tensor and using bianchi identities shorten the way, but I really want to obtain the same solution using the long way......)))

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Nevertheless, you can share a book which includes the solution of covariant derivative of riemann tensor I really appreciate if you help me ))

I have not received a answer for a long time., and still waiting for your answer. I know it is a long derivation, but I would like you to remember that you do not have to make a derivation on this forum, I just ask you to share pdf, book, link.... which includes the proof. I hope your valuable return...

stevendaryl
Staff Emeritus
I'm not exactly sure what you are looking for. The Riemann tensor looks like this:

$R^\mu_{\alpha \beta \lambda}$

when you take covariant derivatives, you get an extra factor of a connection coefficient for each index (with a minus sign in the case of lowered indices). So:

$\nabla_\nu R^\mu_{\alpha \beta \lambda} = \partial_\nu R^\mu_{\alpha \beta \lambda} + \Gamma^\mu_{\nu \sigma} R^\sigma_{\alpha \beta \lambda} - \Gamma^\tau_{\nu \alpha} R^\mu_{\tau \beta \lambda} - \Gamma^\tau_{\nu \beta} R^\mu_{\alpha \tau \lambda} - \Gamma^\tau_{\nu \lambda} R^\mu_{\alpha \beta \tau}$

What else are you wanting to know about it? Did you want to write out $R$ in terms of $\Gamma$? That's an enormous mess.

I'm not exactly sure what you are looking for. The Riemann tensor looks like this:

$R^\mu_{\alpha \beta \lambda}$

when you take covariant derivatives, you get an extra factor of a connection coefficient for each index (with a minus sign in the case of lowered indices). So:

$\nabla_\nu R^\mu_{\alpha \beta \lambda} = \partial_\nu R^\mu_{\alpha \beta \lambda} + \Gamma^\mu_{\nu \sigma} R^\sigma_{\alpha \beta \lambda} - \Gamma^\tau_{\nu \alpha} R^\mu_{\tau \beta \lambda} - \Gamma^\tau_{\nu \beta} R^\mu_{\alpha \tau \lambda} - \Gamma^\tau_{\nu \lambda} R^\mu_{\alpha \beta \tau}$

What else are you wanting to know about it? Did you want to write out $R$ in terms of $\Gamma$? That's an enormous mess.
I'm not exactly sure what you are looking for. The Riemann tensor looks like this:

$R^\mu_{\alpha \beta \lambda}$

when you take covariant derivatives, you get an extra factor of a connection coefficient for each index (with a minus sign in the case of lowered indices). So:

$\nabla_\nu R^\mu_{\alpha \beta \lambda} = \partial_\nu R^\mu_{\alpha \beta \lambda} + \Gamma^\mu_{\nu \sigma} R^\sigma_{\alpha \beta \lambda} - \Gamma^\tau_{\nu \alpha} R^\mu_{\tau \beta \lambda} - \Gamma^\tau_{\nu \beta} R^\mu_{\alpha \tau \lambda} - \Gamma^\tau_{\nu \lambda} R^\mu_{\alpha \beta \tau}$

What else are you wanting to know about it? Did you want to write out $R$ in terms of $\Gamma$? That's an enormous mess.
I started with what you had written down in your post, but as you said it is a mess, and I am mixing up something due to enormous mess. I just want to see the full solution and I do not want you to be exhausted to derive the solution on this forum, so that is why I am asking you to share a file or other things which include this enormous solution) If you have, I will be very pleased

But why would you want to write ##R^\mu_{\quad \alpha\beta\lambda}## in terms of the connection?
It's a useless exercise if you ask me.

Other than that you could use mathematica but it will still take a bit of writing, especially paying attention that you have a valid expression.