Calculating Christoffel Symbols from Metric Tensor

Click For Summary
SUMMARY

This discussion focuses on the calculation of Christoffel symbols from the metric tensor, specifically using the equation 0=\frac{\partial g_{ik}}{\partial x^l}-g_{mk}\Gamma^m_{il}-g_{im}\Gamma^m_{kl}. Participants clarify the process of deriving the Christoffel symbols, emphasizing the symmetry in lower indices and the manipulation of indices through the metric tensor. The final expression for the Christoffel symbols is given as \Gamma^i_{kl}=\frac{1}{2}g^{im}(\frac{\partial g_{mk}}{\partial x^l}+\frac{\partial g_{ml}}{\partial x^k}-\frac{\partial g_{kl}}{\partial x^m). Participants also discuss the implications of raising and lowering indices using the metric.

PREREQUISITES
  • Understanding of differential geometry concepts, particularly metric tensors.
  • Familiarity with Christoffel symbols and their properties.
  • Knowledge of tensor notation and index manipulation.
  • Basic grasp of how metrics operate in raising and lowering indices.
NEXT STEPS
  • Study the derivation of Christoffel symbols from metric tensors in "Introduction to Riemannian Manifolds" by John M. Lee.
  • Learn about the properties of symmetric tensors and their implications in differential geometry.
  • Explore the role of the metric tensor in general relativity and its applications in physics.
  • Investigate advanced tensor calculus techniques, particularly in the context of curved spaces.
USEFUL FOR

Mathematicians, physicists, and students of differential geometry who are looking to deepen their understanding of Christoffel symbols and their derivation from metric tensors.

Qyzren
Messages
41
Reaction score
0
http://en.wikipedia.org/wiki/Christoffel_symbols#Definition

start with 0=\frac{\partial g_{ik}}{\partial x^l}-g_{mk}\Gamma^m_{il}-g_{im}\Gamma^m_{kl}

in wiki it said "By permuting the indices, and resumming, one can solve explicitly for the Christoffel symbols as a function of the metric tensor:"
\Gamma^i_{kl}=\frac{1}{2}g^{im}(\frac{\partial g_{mk}}{\partial x^l}+\frac{\partial g_{ml}}{\partial x^k}-\frac{\partial g_{kl}}{\partial x^m})

but i don't see how they did this step. Can someone please show me?
Thanks you
 
Physics news on Phys.org
well g_{mk} \Gamma^m_{il}=\Gamma_{kil} so you get:

0=\frac{\partial{g_ik}}{\partial{x^l}}-\Gamma_{kil}-\Gamma_{ikl}

i think that wwhat they mean by permuting the indices is just that you can relabel them to create two other equations:
0=\frac{\partial{g_lk}}{\partial{x^i}}-\Gamma_{kli}-\Gamma_{kil}
and
0=\frac{\partial{g_il}}{\partial{x^k}}-\Gamma_{lki}-\Gamma_{kli}

now you know the Christoffel symbol \Gamma^a_{bc} is symmetric in its' lower indices. If you add the first two of those three equations and subtract the last one you get:

2 \Gamma_{kil}=\frac{\partial{g_ik}}{\partial{x^l}}+\frac{\partial{g_lk}}{\partial{x^i}}-\frac{\partial{g_il}}{\partial{x^k}}

and now just bring the 2 across and pull a g^{im} out the front.

this isn't quite right as I've rushed it through... the ideas are right though so try doing it yourself with all the working and seeing where i went wrong.
 
I do not understand this
"well g_{mk} \Gamma^m_{il}=\Gamma_{kil}"
specifically, how you can have a christoffel symbol with 3 lower indices, how does that work?

should the 2nd equation not be
0=\frac{\partial{g_il}}{\partial{x^k}}-\Gamma_{lik}-\Gamma_{ilk}
instead of
0=\frac{\partial{g_il}}{\partial{x^k}}-\Gamma_{lki}-\Gamma_{kli}

and finally how do you pull out a g^{im} factor at the end?
 
do you know how the metric g^{ab} operates?

It acts as follows:

g^{\mu \nu} X_{\nu} = X^{\mu} for a vector X_{\mu}
and
g_{\mu \nu} X^{\nu}=X_{\mu} for a X^{\mu} \in \Lambda^1 i.e. in the space of one forms.

and so we can manipulate the Christoffel symbol to have as many up and down indices as we like simply by acting on it with a series of metrics.

i pulled out the g^{im} at the end simply so it would be in the same form as the one you quoted in your first post. remember however that when you pull out that factor it will alter the up and down indices accordingly.

hope this helps a bit.
 
Thanks,
I just did not know it was possible to raise or lower indices for the christoffel symbol.

I've figured it out now though
 

Similar threads

Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 19 ·
Replies
19
Views
2K