- #1

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- TL;DR Summary
- I constructed an attempted proof of the Contracted Bianchi Identity, and cannot identify any errors myself. However, obviously, this does not mean that the proof is watertight. For any errors present, I am hoping to get details and explanations on where I went wrong, how to avoid the mistake in the future, etc.

Criticism welcome!

__My Attempted Proof__

##R^{mn}_{;n} = \frac {1} {2} g^{mn} R_{;n}##

So, we want ##2 R^{mn}_{;n} = g^{mn} R_{;n} ##

Start w/ 2nd Bianchi Identity ##R_{abmn;l} + R_{ablm;n} + R_{abnl;m} = 0##

Sum w/ inverse metric tensor twice ##g^{bn} g^{am} (R_{abmn;l} + R_{ablm;n} + R_{abnl;m}) = g^{bn} g^{am} (0)##

Distributing ##g^{bn} (g^{am} R_{abmn;l} + g^{am} R_{ablm;n} + g^{am} R_{abnl;m}) = 0##

Index manipulating ##g^{bn} (R^{m}_{~~~bmn;l} + R^{m}_{~~~blm;n} + R^{m}_{~~~bnl;m}) = 0##

Contracting ##g^{bn} (R_{bn;l} - R_{bl;n} + R^{m}_{~~~bnl;m}) = 0##

Distributing and manipulating indices ## R^n_{n;l} - R^n_{l;n} + g^{bn} R^{m}_{~~~bnl;m} = 0##

Summing ##R_{;l} - R^n_{l;n} + g^{bn} R^{m}_{~~~bnl;m} = 0##

Sum W/ ##g^{lp}##: ##g^{lp} (R_{;l} - R^n_{l;n} + g^{bn} R^{m}_{~~~bnl;m} = 0##

Distributing ##g^{lp} R_{;l} - g^{lp} R^n_{l;n} + g^{lp} g^{bn} R^{m}_{~~~bnl;m} = 0##

Summing ##g^{lp} R_{;l} - R^{np}_{;n} + g^{lp} g^{bn} R^{m}_{~~~bnl;m} = 0##

Ricci Tensor (downstairs indices is symmetric, so I THINK it is valid to flip the 2 top indices here ##g^{lp} R_{;l} - R^{pn}_{;n} + g^{lp} g^{bn} R^{m}_{~~~bnl;m} = 0##

##g^{lp} R_{;l} + g^{lp} g^{bn} R^{m}_{~~~bnl;m} = R^{pn}_{;n}## becomes ##R^{pn}_{;n} = g^{lp} R_{;l} + g^{lp} g^{bn} R^{m}_{~~~bnl;m}##

##R^{pn}_{;n} = g^{lp} R_{;l} - g^{lp} g^{bn} R^{m}_{~~~bln;m}##

##R^{pn}_{;n} = g^{lp} R_{;l} - R^{mnp}_{~~~~~~~~n;m}##

##R^{pn}_{;n} + R^{mnp}_{~~~~~~~~n;m} = g^{lp} R_{;l} ##

The metric tensor and its inverse are symmetric ##R^{pn}_{;n} + R^{mnp}_{~~~~~~~~n;m} = g^{pl} R_{;l} ##

Summing over ##n##, ##R^{pn}_{;n} + R^{mp}_{;m} = g^{pl} R_{;l} ##

I've seen others just replace indices like I am about to, so I am about 60% sure the following operation is valid. However, I am not sure. Any insight and corrections would be much appreciated here especially! ##R^{mn}_{;n} + R^{mn}_{;m} = g^{mn} R_{;n} ##

Combining like terms ##2 R^{mn}_{;n} = g^{mn} R_{;n} ##, which is just the result we're looking for, assuming I did it right.

Any help is much appreciated!

P.S., I'm not always great at articulating my thoughts, so my apologies if this question isn't clear. Also, I know this isn't high school material, but I am currently in high school, which is why I made my level "Basic/high school level."

Lastly, I know that this isn't the typical way one would go about proving this identity, but I was wondering if this was still a valid avenue to do so (I did it from scratch).