How to contract the christoffel symbol

In summary, the conversation discusses a problem with calculating the Christoffel symbols and contracting the indices in a given formula. The suggested approach is to transform to a coordinate system where the metric is diagonal and work out the derivatives in terms of the components of the metric. The end goal is to find an equation that can be used in any coordinate system.
  • #1
254
5
http://www.thephysicsforum.com/vlatex/pics/92_db3a451c7d105432675bef473582556e.png [Broken]

Anyone can help me? I am really stuck at here.
 
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  • #2
TimeRip496 said:
http://www.thephysicsforum.com/vlatex/pics/92_db3a451c7d105432675bef473582556e.png [Broken]

Anyone can help me? I am really stuck at here.
Well, you should at least show your work so far, and exactly where you get "stuck". I.e., like in the homework forums.
 
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  • #3
You might start with writing out how the Christoffel symbols are defined, and then contracting the two indices to see what you get.
 
  • #4
strangerep said:
Well, you should at least show your work so far, and exactly where you get "stuck". I.e., like in the homework forums.
I get (0.5)(d/dxμ)(gσv)(gσv) . I don't even know to get that ln.

Sorry for my untidiness as I am still new to this.
 
  • #5
If you don't show your intermediate steps, we can't really tell where you went wrong. By the way ##g_{\sigma\nu}g^{\sigma\nu}=\delta_\sigma^{~~\sigma}=n## where ##n## is the dimension of the manifold. So you definitely went wrong somewhere as your expression is the derivative of a constant so will equal 0.
 
  • #6
Matterwave said:
If you don't show your intermediate steps, we can't really tell where you went wrong. By the way ##g_{\sigma\nu}g^{\sigma\nu}=\delta_\sigma^{~~\sigma}=n## where ##n## is the dimension of the manifold. So you definitely went wrong somewhere as your expression is the derivative of a constant so will equal 0.
IMG_20150125_142314-1.jpg

I did that by cancelling out the indices.
 
  • #7
Ok, but your final result ##\Gamma^\sigma_{\sigma\mu}=\frac{1}{2}g^{\sigma\nu}\partial_\mu g_{\sigma\nu}## is very different than the one you wrote in post #4.

You are pretty close to the answer. Perhaps the easiest way to move forward now is to expand the original statement ##\Gamma^\sigma_{\sigma\mu}=\partial_\mu (\ln\sqrt{g})## out to see what that looks like in terms of components ##g_{\mu\nu}## (it will be pretty hard to reverse-engineer this I think). The easiest way to work out the equation in the OP is to transform to a coordinate system in which the metric is diagonal (which can always be done), work out the derivatives in terms of components of the metric in that coordinate system, and then produce an equation which will work in any coordinate system. :)
 
  • #8
Matterwave said:
Ok, but your final result ##\Gamma^\sigma_{\sigma\mu}=\frac{1}{2}g^{\sigma\nu}\partial_\mu g_{\sigma\nu}## is very different than the one you wrote in post #4.

You are pretty close to the answer. Perhaps the easiest way to move forward now is to expand the original statement ##\Gamma^\sigma_{\sigma\mu}=\partial_\mu (\ln\sqrt{g})## out to see what that looks like in terms of components ##g_{\mu\nu}## (it will be pretty hard to reverse-engineer this I think). The easiest way to work out the equation in the OP is to transform to a coordinate system in which the metric is diagonal (which can always be done), work out the derivatives in terms of components of the metric in that coordinate system, and then produce an equation which will work in any coordinate system. :)
Thanks! I did finally manage to contract it. However, I have difficulty contracting Γσμμ. Is contracting it going to give us the same formula as shown above? I search online but I can't find any contraction of that christoffel symbol given above.
 
  • #9
TimeRip496 said:
Thanks! I did finally manage to contract it. However, I have difficulty contracting Γσμμ. Is contracting it going to give us the same formula as shown above? I search online but I can't find any contraction of that christoffel symbol given above.

You can't "contract" two indices which are both lower indices. You can only contract one upper index with a lower index. So what you want to look for is actually ##g^{\mu\nu}\Gamma^\rho_{\mu\nu}##. Try writing that out and seeing what it turns out to be.
 

1. What is the purpose of the Christoffel symbol in mathematics?

The Christoffel symbol is used in mathematics, specifically in differential geometry, to represent the components of the connection between tangent spaces.

2. How is the Christoffel symbol calculated?

The Christoffel symbol is calculated by taking the derivative of the metric tensor and then solving for the Christoffel symbol components using the Einstein summation convention.

3. What is the significance of contracting the Christoffel symbol?

Contracting the Christoffel symbol involves summing over the indices of the symbol. This is significant because it allows for the simplification of calculations and the reduction of equations.

4. How is the Christoffel symbol used in general relativity?

In general relativity, the Christoffel symbol is used to calculate the geodesic equation, which describes the path of a free-falling particle in curved spacetime. It is also used to calculate the Riemann curvature tensor, which represents the curvature of spacetime.

5. Can the Christoffel symbol be contracted in any coordinate system?

Yes, the Christoffel symbol can be contracted in any coordinate system as long as it follows the rules of tensor calculus. However, the resulting equations may differ depending on the chosen coordinate system.

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