Why Do These Riemann Tensor Terms Cancel Each Other Out?

[ProfDawgstein]

I was working on the derivation of the riemann tensor and got this

(1) $\Gamma^{\lambda}_{\ \alpha\mu} \partial_\beta A_\lambda$

and this

(2) $\Gamma^{\lambda}_{\ \beta\mu} \partial_\alpha A_\lambda$

How do I see that they cancel (1 - 2)?

$\Gamma^{\lambda}_{\ \alpha\mu} \partial_\beta A_\lambda - \Gamma^{\lambda}_{\ \beta\mu} \partial_\alpha A_\lambda = 0$

The only difference is $\alpha \leftrightarrow \beta$

First step was $\left[ D_\alpha, D_\beta \right] A_\mu = D_\alpha (D_\beta A_\mu) - D_\beta (D_\alpha A_\mu)$

then

$D_\beta A_\mu = \partial_\beta A_\mu - \Gamma^{\lambda}_{\mu\beta} A_\lambda = A_{\mu ;\beta} => V_{\mu\beta}$

then another covariant derivative

$D_\alpha V_{\mu\beta} = \partial_\alpha V_{\mu\beta} - \Gamma^{\lambda}_{\ \alpha\mu} V_{\lambda\beta} - \Gamma^{\lambda}_{\ \alpha\beta} V_{\mu\lambda}$

then plug in

$D_\alpha (D_\beta A_\mu) = \partial_\alpha (\partial_\beta A_\mu - \Gamma^{\sigma}_{\ \mu\beta} A_\sigma) - \Gamma^{\lambda}_{\ \alpha \mu} (\partial_\beta A_\lambda - \Gamma^{\sigma}_{\ \lambda \beta} A_{\sigma}) - \Gamma^{\lambda}_{\ \alpha \beta} (\partial_\lambda A_\mu - \Gamma^{\sigma}_{\ \mu\lambda} A_\sigma)$

And later

$-\Gamma^{\lambda}_{\ \alpha \mu} (\partial_\beta A_\lambda - \Gamma^{\sigma}_{\ \lambda \beta} A_{\sigma})$

which is

$-\Gamma^{\lambda}_{\ \alpha \mu} \partial_\beta A_\lambda + \Gamma^{\lambda}_{\ \alpha \mu} \Gamma^{\sigma}_{\ \lambda \beta} A_{\sigma}$

the 2nd term cancels later, but the 1st one does not (see above)

Fleisch (Students Guide to Vectors and Tensors) also does this derivation, but he never had two terms like this.

 

[ProfDawgstein]

updated first post a few hours ago.

Why can't I edit it now?

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the 2nd calculation ($D_\beta D_\alpha$) should be the same, except that $\alpha \leftrightarrow \beta$

could it be that I forgot the product rule for the 2nd term in $( ... )$?

I am so stupid :(

$\partial_\alpha (\partial_\beta A_\mu - \Gamma^{\sigma}_{\ \mu\beta} A_\sigma)$

$= \partial_\alpha \partial_\beta A_\mu - \partial_\alpha (\Gamma^{\sigma}_{\ \mu\beta} A_\sigma)$

using the product rule on the 2nd term

$= \partial_\alpha \Gamma^{\sigma}_{\ \mu\beta} A_\sigma - \Gamma^{\sigma}_{\ \mu\beta} \partial_\alpha A_\sigma$

doing $\alpha \leftrightarrow \beta$ for the 2nd commutator term

$= \partial_\beta \Gamma^{\sigma}_{\ \mu\alpha} A_\sigma - \Gamma^{\sigma}_{\ \mu\alpha} \partial_\beta A_\sigma$

which just produces the terms I need to cancel the ones from post #1 :)

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Thanks for not posting the answer.

Sometimes it is hard to see the obvious...

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The full derivation now is

$A_{\mu ;\beta \alpha} = \partial_\alpha \partial_\beta A_\mu - \partial_\alpha \Gamma^{\sigma}_{\ \mu\beta} A_\sigma - \Gamma^{\sigma}_{\ \mu\beta} \partial_\alpha A_\sigma - \Gamma^{\lambda}_{\ \alpha\mu} \partial_\beta A_\lambda + \Gamma^{\lambda}_{\alpha\mu} \Gamma^{\sigma}_{\ \lambda\beta} A_\sigma - \Gamma^{\lambda}_{\ \alpha\beta} \partial_\lambda A_\mu + \Gamma^{\lambda}_{\ \alpha\beta} \Gamma^{\sigma}_{\ \mu\lambda} A_\sigma$

and

$A_{\mu ;\alpha \beta} = \partial_\beta \partial_\alpha A_\mu - \partial_\beta \Gamma^{\sigma}_{\ \mu\alpha} A_\sigma - \Gamma^{\sigma}_{\ \mu\alpha} \partial_\beta A_\sigma - \Gamma^{\lambda}_{\ \beta\mu} \partial_\alpha A_\lambda + \Gamma^{\lambda}_{\beta\mu} \Gamma^{\sigma}_{\ \lambda\alpha} A_\sigma - \Gamma^{\lambda}_{\ \beta\alpha} \partial_\lambda A_\mu + \Gamma^{\lambda}_{\ \beta\alpha} \Gamma^{\sigma}_{\ \mu\lambda} A_\sigma$

subtracting both

$A_{\mu ;\beta \alpha} - A_{\mu ;\alpha \beta}$

using symmetry of the christoffel symbols and $\partial_\alpha \partial_\beta = \partial_\beta \partial_\alpha$ and moving the minus sign out of $( ... )$ we get

$\left[ D_\alpha, D_\beta \right] A_\mu = A_{\mu ;\beta \alpha} - A_{\mu ;\alpha \beta} = - R^{\sigma}_{\ \mu\alpha\beta} A_\sigma$

where

$R^{\sigma}_{\ \mu\alpha\beta} = \partial_\alpha \Gamma^{\sigma}_{\ \mu\beta} - \partial_\beta \Gamma^{\sigma}_{\ \mu\alpha} + \Gamma^{\lambda}_{\ \beta\mu} \Gamma^{\sigma}_{\ \lambda\alpha} - \Gamma^{\lambda}_{\ \alpha\mu} \Gamma^{\sigma}_{\ \lambda\beta}$

 

[zn5252]

You may also see the derivation in Dirac's book : General theory of Relativity under equation 11.1 if I'm not wrong

 

[ProfDawgstein]

Everything is solved now.

After some messy messing around and remembering the product rule ( LOL , Thanks Newton ;) ) I got it.

Can be closed.

 

[sardar]

I can see that these two terms cancel out because they are related by the exchange of $\alpha$ and $\beta$. In other words, they are symmetric with respect to these indices. This is a common occurrence in tensor calculations, where we can simplify expressions by taking advantage of the symmetry of certain terms. In this case, the fact that $\alpha$ and $\beta$ can be interchanged means that the terms are equivalent and can be subtracted from each other without changing the overall value of the expression.

Furthermore, the presence of the Christoffel symbols $\Gamma^{\lambda}_{\ \alpha \beta}$ in both terms indicates that these expressions are related to the covariant derivative, which is a fundamental concept in differential geometry. This also supports the idea that these two terms should cancel out, as the covariant derivative is a symmetric operation with respect to its indices.

It is important to carefully consider the properties and relationships of tensors when performing calculations, as it allows us to simplify and make sense of complex expressions. In this case, the symmetry of the terms and the use of the covariant derivative lead us to the conclusion that they cancel out.