luc Posted May 12, 2005 Posted May 12, 2005 I'm trying to weave my way through General Relativity, but I find difficult to learn the concept of Christoffel symbols of the second kind. They are defined as [tex] \Gamma^l_{ki} = \frac{1}{2} g^{lj} (\partial_k g_{ij} + \partial_i g_{jk} - \partial_j g_{ki}) [/tex] However I really don't know how to do the partial derivative of the metric tensor; I do know how to do partial derivarives of 0-rank tensor fields and 1-rank tensor fields, but I'd be very happy if someone could explain me how to do partial derivatives of 2-rank tensor fields. A link to a page explaining how to do this kind of derivative would also be welcome Edit:It seems that latex is down, but I hope you know what I'm talking about...
timo Posted May 12, 2005 Posted May 12, 2005 It´s just a partial derivative. [math] \partial_i g_{jk} [/math] has 4*16 = 64 entries ijk, each corresponding to the derivative [math] \partial_i [/math] of the metric´s entry [math] g_{jk} [/math]. EDIT: nice TeX errors. Ascii version: It´s just a partial derivative. \partial_i g_{jk} has 4*16 = 64 entries ijk, each corresponding to the derivative \partial_i of the metric´s entry g_{jk}. EDIT2: Oh, and the keyword for TeX-code in this forum is "math", not "tex".
luc Posted May 13, 2005 Author Posted May 13, 2005 I guess that you say that I have to do the partial derivative of each of the 16 entries of the metric tensor. Doing the partial derivative with respect to the 3 spatial coordinates and with respect to time gives 4 results for each entry of the metric tensor, so 16*4=64 are the entries of the resulting 3-tensor field. Is that right?
fuhrerkeebs Posted May 13, 2005 Posted May 13, 2005 Just take the partial of each entry...say you had \partial_i g^{ij}, where g^{1,0}=x^0, g^{0,1}=x^1, and g^{ij}=0 when i != j, then you would just do the sum and take the partials, like this: A^j = \partial_i g^{ij} = \partial_=0 g^{0j} + \partial_1 g^{1j} So, A^0=2, A^1=0. Note that I ranged my indices from 0 to 1, and that I used the summation convention (in case that wasn't implied).
timo Posted May 16, 2005 Posted May 16, 2005 No, it´s not a tensor of rank 3 - it´s not a tensor at all. Just a bunch of 64 entries. That´s because the result does not transform tensorial. You´ll learn this when you come to the covariant derivative.
□h=-16πT Posted May 21, 2005 Posted May 21, 2005 Bare in mind that that definiton is for manifolds whose connection is torsion free.
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