- #1
ProfDawgstein
- 80
- 1
Hey,
I have been doing a few proofs and stumbled across this little problem.
Trying to show the symmetry of the Ricci tensor by using the Riemann tensor definition
##R^m_{\ ikp} = \partial_k \Gamma^m_{\ ip} - \partial_p \Gamma^m_{\ ki} + \Gamma^a_{\ ip} \Gamma^m_{\ ak} - \Gamma^a_{\ ik} \Gamma^m_{\ ap}##
Now set m = k
##R^m_{\ \,imp} = R_{ip} = \partial_m \Gamma^m_{\ ip} - \partial_p \Gamma^m_{\ mi} + \Gamma^a_{\ ip} \Gamma^m_{\ am} - \Gamma^a_{\ im} \Gamma^m_{\ ap}##
Checking every term for symmetry (i <-> p)
1. symmetric because ##\Gamma^{m}_{\ ip} = \Gamma^{m}_{\ pi}##
2. see below
3. symmetric because ##\Gamma^{m}_{\ ip} = \Gamma^{m}_{\ pi}##
4. symmetric because ##\Gamma^a_{\ im} \Gamma^m_{\ ap} = \Gamma^m_{\ pa} \Gamma^a_{\ mi}## (interchanging a & m, and just rotate) and ##\Gamma^{a}_{\ im} = \Gamma^{a}_{\ mi}##
Now for the 2nd term
##\partial_p \Gamma^m_{\ mi} == \partial_i \Gamma^m_{\ mp}##
This is a contracted Christoffel symbol...
using ##\Gamma^k_{ij} = \frac{1}{2} g^{kl} \left(\partial_i g_{jl} + \partial_j g_{il} - \partial_l g_{ij}\right)##
leads to
##\Gamma^{m}_{\ mi} = \frac{1}{2} g^{ml} \frac{\partial g_{ml}}{\partial {x^i}} ##
Now the derivative
##\partial_p \Gamma^m_{\ mi} = \frac{1}{2} \left( \frac{\partial g^{ml}}{\partial x^p} \frac{\partial g_{ml}}{\partial {x^i}} + g^{ml} \frac{\partial g_{ml}}{\partial {x^i} \partial {x^p}}\right)##
The 2nd term in this expression is symmetric (i <-> p), because order of partial differentiation doesn't matter.
For the first term I am not sure though.
Is ##\frac{\partial g^{ml}}{\partial x^p} \frac{\partial g_{ml}}{\partial {x^i}} == \frac{\partial g^{ml}}{\partial x^i} \frac{\partial g_{ml}}{\partial {x^p}}##?
The inverse is something totally different, but you also contract it, this is the point where I am slightly confused.
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I know there are other ways to "derive" the symmetry of the Ricci tensor, but I wanted to try this one :)
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Thank you in advance!
It's probably so obvious that I don't see it![Roll Eyes :rolleyes: :rolleyes:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Hope you guys had a nice Christmas ;)
I have been doing a few proofs and stumbled across this little problem.
Trying to show the symmetry of the Ricci tensor by using the Riemann tensor definition
##R^m_{\ ikp} = \partial_k \Gamma^m_{\ ip} - \partial_p \Gamma^m_{\ ki} + \Gamma^a_{\ ip} \Gamma^m_{\ ak} - \Gamma^a_{\ ik} \Gamma^m_{\ ap}##
Now set m = k
##R^m_{\ \,imp} = R_{ip} = \partial_m \Gamma^m_{\ ip} - \partial_p \Gamma^m_{\ mi} + \Gamma^a_{\ ip} \Gamma^m_{\ am} - \Gamma^a_{\ im} \Gamma^m_{\ ap}##
Checking every term for symmetry (i <-> p)
1. symmetric because ##\Gamma^{m}_{\ ip} = \Gamma^{m}_{\ pi}##
2. see below
3. symmetric because ##\Gamma^{m}_{\ ip} = \Gamma^{m}_{\ pi}##
4. symmetric because ##\Gamma^a_{\ im} \Gamma^m_{\ ap} = \Gamma^m_{\ pa} \Gamma^a_{\ mi}## (interchanging a & m, and just rotate) and ##\Gamma^{a}_{\ im} = \Gamma^{a}_{\ mi}##
Now for the 2nd term
##\partial_p \Gamma^m_{\ mi} == \partial_i \Gamma^m_{\ mp}##
This is a contracted Christoffel symbol...
using ##\Gamma^k_{ij} = \frac{1}{2} g^{kl} \left(\partial_i g_{jl} + \partial_j g_{il} - \partial_l g_{ij}\right)##
leads to
##\Gamma^{m}_{\ mi} = \frac{1}{2} g^{ml} \frac{\partial g_{ml}}{\partial {x^i}} ##
Now the derivative
##\partial_p \Gamma^m_{\ mi} = \frac{1}{2} \left( \frac{\partial g^{ml}}{\partial x^p} \frac{\partial g_{ml}}{\partial {x^i}} + g^{ml} \frac{\partial g_{ml}}{\partial {x^i} \partial {x^p}}\right)##
The 2nd term in this expression is symmetric (i <-> p), because order of partial differentiation doesn't matter.
For the first term I am not sure though.
Is ##\frac{\partial g^{ml}}{\partial x^p} \frac{\partial g_{ml}}{\partial {x^i}} == \frac{\partial g^{ml}}{\partial x^i} \frac{\partial g_{ml}}{\partial {x^p}}##?
The inverse is something totally different, but you also contract it, this is the point where I am slightly confused.
--------------
I know there are other ways to "derive" the symmetry of the Ricci tensor, but I wanted to try this one :)
--------------
Thank you in advance!
It's probably so obvious that I don't see it
Hope you guys had a nice Christmas ;)