R computation from 1 independent Riemann tensor component

  • Thread starter binbagsss
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  • #1
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We have ##R^{1}_{212}## as the single independent Riemann tensor component, and I'm after ##R##. From symmetry properties and contracting we can attain the other non-zero components.

The solution then states that ##R_{11}=R^{1}_{111} + R^{2}_{121}=R^{2}_{121}## .

I thought it would have been
##R_{11}=R^{1}_{111} + R^{2}_{121}+R^{2}_{112}+R^{2}_{211}##

All I can think of then is that the upper indicie can only contract with the middle indice? I've never heard this before though, is that what's going on here?

Similarlly for ##R_{22}=R^{2}_{222} + R^{1}_{212}=R^{2}_{121}## is the solution.

The components ##R^{2}_{211}, R^{2}_{112}, R^{1}_{221}, R^{1}_{122}, ##are the unused components with two 2s, two 1s - if the upper indice only contracts with the middle, these would not come into the formula for any ricci vector...

If someone could let me know if I'm on the right or wrong track here,
cheers !
 

Answers and Replies

  • #3
1,232
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Ok. so it looks like only the middle indice contracts. But doesnt this differ then from when we contract vectors with the metric, here the position of the indices isn't important is it? When we contract with the metric all that matters is upper and lower? Thanks.
 
  • #4
Matterwave
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Ok. so it looks like only the middle indice contracts. But doesnt this differ then from when we contract vectors with the metric, here the position of the indices isn't important is it? When we contract with the metric all that matters is upper and lower? Thanks.
The Ricci tensor is defined as ##R_{ab}\equiv R^c_{~acb}## the contraction is in the middle index.
 

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