# Christoffel symbols

1. Feb 21, 2009

### 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)$$\frac{\partial c_k}{\partial x^q}=\frac{\partial v_k}{\partial x^q}+v_t\Gamma_{qk}^t$$

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...

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)

III: And another question: Is the Kronecker delta symmetric in non-orthogonal coordinates

$$\delta^i_j=\delta^j_i$$ ???

If not, then which one of the two definitions is correct: (I´ve 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?

IV: And the last one: I haven´t seen a classification of the Christoffel symbol of this kind:

$$\Gamma^{kl}_m$$ Is it also symmetric in the upper indices?

Thanks a lot, I really appreciate your help!

marin

Last edited: Feb 21, 2009
2. Feb 21, 2009

### HallsofIvy

Staff Emeritus
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. $\Gamma_{ij, t}= g_{ki} g_{jq}\Gamma^{kq}_t$.

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.)

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.

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?

You are welcome.

3. Feb 21, 2009

### Marin

ok, now let me try and see:

$$\Gamma^l_{qt}=\Gamma^l_{tq}=-\Gamma^q_{lt}=-\Gamma^q_{tl}=\Gamma^t_{lq}$$

and also:

$$\Gamma^l_{qt}=-\Gamma^t_{ql}=-\Gamma^t_{lq}$$

seems to me like a contradiction..

Can you please point out the wrong equalities, so that I get a better understanding?

thanks

4. Feb 21, 2009

### HallsofIvy

Staff Emeritus
You are right. I wrote too fast $\Gamma^l_{qt}= \Gamma^l_{tq}$ but $\Gamma^t_{lq}\ne -\Gamma^q_{lt}$. What is true is that $\Gamma^l_{qt}+ \Gamma^q_{tl}+ \Gamma^t_{lq}= 0$. Notice that the three indices are "rotated"- the three even permutations of tlq.

I asked before what definition of the Krisstofel symbols you are using. The ones I am familiar with are
$$\Gamma_{ij,k}= \frac{1}{2}\left(\frac{\partial g_ik}{\partial x^j}+ \frac{\partial g_{jk}}{\partial x^i}- \frac{\partial g_{ij}}{\partial x^k}\right)$$

$$\Gamma^{i}_{jk}= g^{im}\Gamma_{jk,m}$$
The symmetry rules follow from that.

5. Feb 22, 2009

### Marin

I also "use" the definition you stated, but haven´t studied it thoroughly yet, perhaps I had to, before I try to take on expressions.

Anyway, I´ll go on trying to understand this stuff and some new and will pose some questions again.

Thanks once again for the help!

6. Feb 22, 2009

### Marin

ok, I tried the following transformation:

$$\Gamma^w_{bt}=\Gamma^w_{bt}\delta^b_b=\Gamma^w_{bt}\delta^b_w\delta^w_b=\Gamma^w_{bt}\delta^b_w\delta^b_w=(\Gamma^w_{bt}\delta^b_w)\delta^b_w=\Gamma^w_{wt}\delta^b_w=\Gamma^b_{wt}=\Gamma^b_{tw}$$

But this implies that nothing changes with the Christoffel symbol if I permute the indices evenly.

But then, HallsofIvy, the equation you posted does not hold any more..