# Transformation law for Christoffel's first kind

I don't understand this particular part in this image. I am following schaum's series book on "vector analysis". I didn't find any explanation for it. I also tried searching in Internet and somewhere i found an explanation that because p and q are symmetric. I am not satisfied with the answer because there were no futher explanation about it. Could anybody help me with it? I am stuck with it.

• I assume you have looked at en.wikipedia.org/wiki/Christoffel_symbols and en.wikipedia.org/wiki/Metric_tensor at the very least? – honeste_vivere Feb 27 '18 at 14:23
• I don't understand how it is derived from the previous line. – Rima Feb 27 '18 at 14:28
• I gathered as much. If you look at the definition of the metric tensor and the definition of $\Gamma_{ijk}$ in terms of the metric tensor, you should be able to slog through the symbol gymnastics pretty quickly. – honeste_vivere Feb 27 '18 at 14:30
• I don't have any problem with the symbols. What i don't understand is how did the remaining terms get removed. I hope you understand what i mean to say. – Rima Feb 27 '18 at 14:38

When you subtract equation (1) from the sum of equations (2) and (3), the term $$\frac{\partial x^p}{\partial \bar{x}^j}\frac{\partial^2 x^q}{\partial \bar{x}^m \partial \bar{x}^k}g_{pq}$$ cancels out with the term $$\frac{\partial^2 x^r}{\partial \bar{x}^k \partial \bar{x}^m}\frac{\partial x^p}{\partial \bar{x}^j}g_{rp}$$ because those two terms are in fact identical: \begin{align}\frac{\partial x^p}{\partial \bar{x}^j}\frac{\partial^2 x^q}{\partial \bar{x}^m \partial \bar{x}^k}g_{pq}&=\frac{\partial^2 x^q}{\partial \bar{x}^m \partial \bar{x}^k}\frac{\partial x^p}{\partial \bar{x}^j}g_{pq}\\ &=\frac{\partial^2 x^q}{\partial \bar{x}^m \partial \bar{x}^k}\frac{\partial x^p}{\partial \bar{x}^j}g_{qp}\\ &=\frac{\partial^2 x^q}{\partial \bar{x}^k \partial \bar{x}^m}\frac{\partial x^p}{\partial \bar{x}^j}g_{qp}\\ &=\frac{\partial^2 x^r}{\partial \bar{x}^k \partial \bar{x}^m}\frac{\partial x^p}{\partial \bar{x}^j}g_{rp}\ \ . \end{align} In the above, the second equality is due to the metric tensor always being symmetric, $g_{pq}=g_{qp}$, the third equality is due to the order of partial differentiation not mattering, and the fourth equality is due to it not mattering what you name a summation index.

Similarly, $$\frac{\partial x^q}{\partial \bar{x}^k}\frac{\partial^2 x^r}{\partial \bar{x}^j \partial \bar{x}^m}g_{qr}=\frac{\partial^2 x^p}{\partial \bar{x}^m \partial \bar{x}^j}\frac{\partial x^q}{\partial \bar{x}^k}g_{pq}\ \ ,$$ so those two terms also cancel out.

Finally, $$\frac{\partial x^r}{\partial \bar{x}^m}\frac{\partial^2 x^p}{\partial \bar{x}^k \partial \bar{x}^j}g_{rp}=\frac{\partial^2 x^q}{\partial \bar{x}^j \partial \bar{x}^k}\frac{\partial x^r}{\partial \bar{x}^m}g_{qr}\ \ ,$$ so when you add two copies of that term and multiply by $1/2$, you just get $$\frac{\partial^2 x^q}{\partial \bar{x}^j \partial \bar{x}^k}\frac{\partial x^r}{\partial \bar{x}^m}g_{qr}=\frac{\partial^2 x^p}{\partial \bar{x}^j \partial \bar{x}^k}\frac{\partial x^r}{\partial \bar{x}^m}g_{pr}=\frac{\partial^2 x^p}{\partial \bar{x}^j \partial \bar{x}^k}\frac{\partial x^q}{\partial \bar{x}^m}g_{pq}\ \ ,$$ where the first equality is just the renaming of a summation index $q\to p$, and the second equality is similarly just the renaming $r\to q$.

It sounds like you understand the terms that don't involve a second partial derivative, so I won't comment on those terms.

• Since $\{g}_{pq}$ is symnetrical with $\{g}_{qp}$ can we say that double derivative of r cancels with q instead of p as you mentioned above because in the final answer double derivative of p remains instead of q as you mentioned. Is it just the same? – Rima Feb 28 '18 at 3:07
• @Rima I'm afraid I don't understand your follow-up question. Can you express what you're asking as an equality between terms? – Red Act Feb 28 '18 at 12:51
• In your final answer when you have added two copies of that term,it has double derivative of x^q while in the image i have posted has x^p instead. If you understand what i mean. I am unable to use latex form in my comment right now. Sorry for that. – Rima Feb 28 '18 at 12:58
• @Rima I edited my answer to spell out why $\frac{\partial^2 x^q}{\partial \bar{x}^j \partial \bar{x}^k}\frac{\partial x^r}{\partial \bar{x}^m}g_{qr}$ is the same as the circled term in the image. The equality doesn't actually rely on the symmetry of $g$. – Red Act Feb 28 '18 at 13:23
• Just a little doubt to clear. At first you said that those two terms for p and r cancel each other because they have identical terms. Finally you have renamed the indices. But doesn't that mean that they are symmetrical? Is "identical" and "symmetrical" different?you said the equality doesn't rely on the symmetry of g. – Rima Feb 28 '18 at 13:28