Why Do These Riemann Tensor Terms Cancel Each Other Out?

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The discussion focuses on the cancellation of two terms in the derivation of the Riemann tensor, specifically how the expressions involving the Christoffel symbols and partial derivatives lead to a zero result when subtracted. The key step involves recognizing that the only difference between the two terms is the interchange of indices, which allows for their cancellation. The participants also highlight the importance of applying the product rule correctly during the derivation process. Ultimately, the derivation confirms the relationship between the commutator of covariant derivatives and the Riemann curvature tensor. The discussion concludes with the realization that understanding these cancellations is crucial for grasping the underlying geometry of the tensor calculus.
ProfDawgstein
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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.
 
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updated first post a few hours ago.

Why can't I edit it now?

----------------------------------------

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 :)

-------------------------------------------------------

Thanks for not posting the answer.

Sometimes it is hard to see the obvious...

-------------------------------------------------------

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}##
 
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You may also see the derivation in Dirac's book : General theory of Relativity under equation 11.1 if I'm not wrong
 
Everything is solved now.

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

Can be closed.
 
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In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

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