Thread: Christoffel symbols View Single Post
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 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: $$d\vec v=dc^k e^{(k)}$$ $$d\vec v=dc^k e_{(k)}$$ where: $$dc^k=dv^k+v^t\Gamma_{wt}^k dx^w$$ $$dc_k=dv_k-v_t\Gamma_{wk}^t dx^w$$ Now, consider the covariant derivatives: $$\frac{\partial c^k}{\partial x^q}=\frac{\partial v^k}{\partial x^q}+v^t\Gamma_{wt}^k \frac{\partial x^w}{\partial x^q}=\frac{\partial v^k}{\partial x^q}+v^t\Gamma_{wt}^k \delta^w_q=\frac{\partial v^k}{\partial x^q}+v^t\Gamma_{qt}^k$$ analagous: (1)$$\frac{\partial c_k}{\partial x^q}=\frac{\partial v_k}{\partial x^q}-v_t\Gamma_{qk}^t$$ So far so good, here I start transforming: $$\frac{\partial c_k}{\partial x^q}=\frac{g_{kl}\partial c^l}{\partial x^q}=g_{kl}\frac{\partial c^l}{\partial x^q}=\frac{g_{kl}\partial v^l}{\partial x^q}+v^t\Gamma_{qt}^l g_{kl}=\frac{\partial v_l}{\partial x^q}+v^t\Gamma_{qtk}$$ As the second term looks different from the one above we continue transforming it: $$v^t\Gamma_{qtk}=v^t\Gamma_{qk}^s g_{ts}=(t\rightarrow s, g_{ts}=g_{st})=v^s\Gamma_{qk}^t g_{ts}=v_t\Gamma_{qk}^t$$ so, we finally get: (2)$$\frac{\partial c_k}{\partial x^q}=\frac{\partial v_l}{\partial x^q}+v_t\Gamma_{qk}^t$$ 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...
Be careful how you "transform" the Christoffel symbols. They are not tensors and do not transform as tensors.

It is true that $\Gamma^{qk}_t= -\Gamma{tk}_q$ and that $\Gamma^{qk}_t= \Gamma^{kq}_t$.

[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 $dx_q$ if you also lower the indices on the Christoffel symbols. $\Gamma_{ij, t}= g_{ki} g_{jq}\Gamma^{kq}_t$.

 III: And another question: Is the Kronecker delta symmetric in non-orthogonal coordinates $$\delta^i_j=\delta^j_i$$ ???
The Kronecker delta is definded by $\delta_{ij}= 1$ if i= j, 0 if $i\ne j$, 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.)

 If not, then which one of the two definitions is correct: (Iīve seen both in the net) $$e^{i}e_j=\delta^j_i$$ or $$e^je_i=\delta^j_i$$ I have also seen two types in which you define covariant vectors: $$\vec v=v_ke^k$$ and $$\vec v=v_ke_k$$ Which one is correct, or do they just represent the same covariant vector once in the covariant and in the contravariant basis?
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.

 IV: And the last one: I havenīt seen a classification of the Christoffel symbol of this kind: $$\Gamma^{kl}_m$$ Is it also symmetric in the upper indices?
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?

 Thanks a lot, I really appreciate your help! marin
You are welcome.