 Quote by Marin
Hi all!
I read about tensor analysis and came about following expressions, where also a questions arose which I cannot explain to me. Perhaps you could help me:
I: Consider the following expressions:
[tex]d\vec v=dc^k e^{(k)}[/tex]
[tex]d\vec v=dc^k e_{(k)}[/tex]
where:
[tex]dc^k=dv^k+v^t\Gamma_{wt}^k dx^w[/tex]
[tex]dc_k=dv_k-v_t\Gamma_{wk}^t dx^w[/tex]
Now, consider the covariant derivatives:
[tex]\frac{\partial c^k}{\partial x^q}=\frac{\partial v^k}{\partial x^q}+v^t\Gamma_{wt}^k \frac{\partial x^w}{\partial x^q}=\frac{\partial v^k}{\partial x^q}+v^t\Gamma_{wt}^k \delta^w_q=\frac{\partial v^k}{\partial x^q}+v^t\Gamma_{qt}^k[/tex]
analagous:
(1)[tex]\frac{\partial c_k}{\partial x^q}=\frac{\partial v_k}{\partial x^q}-v_t\Gamma_{qk}^t [/tex]
So far so good, here I start transforming:
[tex]\frac{\partial c_k}{\partial x^q}=\frac{g_{kl}\partial c^l}{\partial x^q}=g_{kl}\frac{\partial c^l}{\partial x^q}=\frac{g_{kl}\partial v^l}{\partial x^q}+v^t\Gamma_{qt}^l g_{kl}=\frac{\partial v_l}{\partial x^q}+v^t\Gamma_{qtk}[/tex]
As the second term looks different from the one above we continue transforming it:
[tex]v^t\Gamma_{qtk}=v^t\Gamma_{qk}^s g_{ts}=(t\rightarrow s, g_{ts}=g_{st})=v^s\Gamma_{qk}^t g_{ts}=v_t\Gamma_{qk}^t[/tex]
so, we finally get:
(2)[tex]\frac{\partial c_k}{\partial x^q}=\frac{\partial v_l}{\partial x^q}+v_t\Gamma_{qk}^t[/tex]
By comparing (1) and (2) I miss a minus sign!
I suspect that the Christoffel symbol of first kind is antisymmetric and indices permute just like they do in the epsilon tensor and thereby generate a minus but I am not sure...
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Be careful how you "transform" the Christoffel symbols. They are not tensors and do not transform as tensors.
It is true that [itex]\Gamma^{qk}_t= -\Gamma{tk}_q[/itex] and that [itex]\Gamma^{qk}_t= \Gamma^{kq}_t[/itex].
[qote]II: In both of the above daces of derivatives one uses dx^q as differential which is contravariant. Does it make sence to also use a covariant dx_q? Is in general differentiation of covariant vectors with respect to a covariant variable defined? (I suppose it must be, since you also differentiate a contravariantvector w.r.t. a contravariant variable)[/quote]
Yes, you can use covariant [itex]dx_q[/itex] if you also lower the indices on the Christoffel symbols. [itex]\Gamma_{ij, t}= g_{ki} g_{jq}\Gamma^{kq}_t[/itex].
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III: And another question: Is the Kronecker delta symmetric in non-orthogonal coordinates
[tex]\delta^i_j=\delta^j_i[/tex] ???
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The Kronecker delta is definded by [itex]\delta_{ij}= 1[/itex] if i= j, 0 if [itex]i\ne j[/itex], independent of the coordinate system, so, yes, it is always symmetric. (The metric tensor is, although dependent, of course, on the coordinate system, is also symmetric in all coordinate systems.)
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If not, then which one of the two definitions is correct: (Iīve seen both in the net)
[tex]e^{i}e_j=\delta^j_i[/tex]
or
[tex]e^je_i=\delta^j_i[/tex]
I have also seen two types in which you define covariant vectors:
[tex]\vec v=v_ke^k[/tex] and [tex]\vec v=v_ke_k[/tex]
Which one is correct, or do they just represent the same covariant vector once in the covariant and in the contravariant basis?
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They are really the same thing although the second would not make sense in the
standard "Einstein summation convention" that we sum when the same index appears both as a superscript and a subscipt.
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IV: And the last one: I havenīt seen a classification of the Christoffel symbol of this kind:
[tex]\Gamma^{kl}_m[/tex] Is it also symmetric in the upper indices?
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Yes, it is. That should be clear from the definition of the Christoffel symbols (of the first kind) in terms of derivatives of the metric tensor. What definition are you using?
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Thanks a lot, I really appreciate your help!
marin
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You are welcome.