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I have to derive the transformation law for the Christoffel symbols:

Let

$\Gamma_{bc}^{a} = \frac{1}{2}g^{ad}\big(\partial_{b}g_{dc}+\partial_{c}g_{bd}-\partial_{d}g_{bc}\big)$

be the Chritoffel symbols in a basis denoted by $\{x^{i}\}$ and

$\overline{\Gamma}_{\beta\gamma}^{\alpha} = \frac{1}{2}\overline{g}^{\alpha\delta}\big(\overline{\partial}_{\beta}\overline{g}_{\delta\gamma}+\overline{\partial}_{\gamma}\overline{g}_{\beta\delta}-\overline{\partial}_{\delta}\overline{g}_{\beta\gamma}\big)$

be the Chritoffel symbols in a basis denoted by $\{\overline{x}^{i}\}$.

Using the facts:

(1) $\overline{g}^{\alpha\delta} = \frac{\partial\overline{x}^{\alpha}}{\partial x^{a}}\frac{\partial\overline{x}^{\delta}}{\partial x^{d}}g^{ad}$

(2) $\frac{\partial}{\partial \overline{x}^{\beta}}\overline{g}_{\delta\gamma} = g_{dc}\frac{\partial}{\partial \overline{x}^{\beta}}\big (\frac{\partial x^{d}}{\partial \overline{x}^{\delta}}\frac{\partial x^{c}}{\partial \overline{x}^{\gamma}}\big ) + \frac{\partial x^{d}}{\partial\overline{x}^{\delta}}\frac{\partial x^{c}}{\partial \overline{x}^{\gamma}}\frac{\partial x^{f}}{\partial \overline{x}^{\beta}}\frac{\partial}{\partial x^{f}}g_{dc}$

(3) $\frac{\partial}{\partial \overline{x}^{\gamma}}\overline{g}_{\beta\delta} = g_{bd}\frac{\partial}{\partial \overline{x}^{\gamma}}\big (\frac{\partial x^{b}}{\partial \overline{x}^{\beta}}\frac{\partial x^{d}}{\partial \overline{x}^{\delta}}\big ) + \frac{\partial x^{b}}{\partial\overline{x}^{\beta}}\frac{\partial x^{d}}{\partial \overline{x}^{\delta}}\frac{\partial x^{g}}{\partial \overline{x}^{\gamma}}\frac{\partial}{\partial x^{g}}g_{bd}$

(4) $\frac{\partial}{\partial \overline{x}^{\delta}}\overline{g}_{\beta\gamma} = g_{bc}\frac{\partial}{\partial \overline{x}^{\delta}}\big (\frac{\partial x^{b}}{\partial \overline{x}^{\beta}}\frac{\partial x^{c}}{\partial \overline{x}^{\gamma}}\big ) + \frac{\partial x^{b}}{\partial\overline{x}^{\beta}}\frac{\partial x^{c}}{\partial \overline{x}^{\gamma}}\frac{\partial x^{h}}{\partial \overline{x}^{\delta}}\frac{\partial}{\partial x^{h}}g_{bc}$

and rename $f\to b$, $g\to c$ and $h\to d$ I get the result:

$\overline{\Gamma}_{\beta\gamma}^{\alpha} = \frac{\partial \overline{x}^{\alpha}}{\partial x^{a}}\frac{\partial x^{b}}{\partial \overline{x}^{\beta}}\frac{\partial x^{c}}{\partial \overline{x}^{\gamma}}\Gamma_{bc}^{a} + N$,

where N is the cumbersome expression

$N = \frac{1}{2}\frac{\partial\overline{x}^{\alpha}}{\partial x^{a}}\frac{\partial\overline{x}^{\delta}}{\partial x^{d}}g^{ad}\bigg [g_{dc}\frac{\partial}{\partial \overline{x}^{\beta}}\big (\frac{\partial x^{d}}{\partial \overline{x}^{\delta}}\frac{\partial x^{c}}{\partial \overline{x}^{\gamma}}\big ) + g_{bd}\frac{\partial}{\partial \overline{x}^{\gamma}}\big (\frac{\partial x^{b}}{\partial \overline{x}^{\beta}}\frac{\partial x^{d}}{\partial \overline{x}^{\delta}}\big ) - g_{bc}\frac{\partial}{\partial \overline{x}^{\delta}}\big (\frac{\partial x^{b}}{\partial \overline{x}^{\beta}}\frac{\partial x^{c}}{\partial \overline{x}^{\gamma}}\big )\bigg ]$

According to the solution, this term should be reducible to

$N = \frac{\partial x^{a}}{\partial \overline{x}^{\gamma}}\frac{\partial x^{b}}{\partial \overline{x}^{\beta}}\frac{\partial^{2} \overline{x}^{\alpha}}{\partial x^{a}\partial x^{b}}$.

I tried it many times but I failed allways....Can anyone give me a hint? Especially, what should I do with the term $g^{ad}g_{bc}$, which is appearing in $N$?

Thanks!

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1 Answer 1

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Write$$\begin{align}\Gamma_{\beta\gamma}^\alpha&=\frac12g^{ad}\left(\frac{\partial\overline{x}^\alpha}{\partial x^a}\frac{\partial\overline{x}^\delta}{\partial x^d}\frac{\partial}{\partial\overline{x}^\beta}\left(\frac{\partial x^e}{\partial\overline{x}^\delta}\frac{\partial x^c}{\partial\overline{x}^\gamma}g_{ec}\right)+\beta\leftrightarrow\gamma-\frac{\partial\overline{x}^\alpha}{\partial x^a}\frac{\partial\overline{x}^\delta}{\partial x^d}\frac{\partial}{\partial\overline{x}^\delta}\left(\frac{\partial x^b}{\partial\overline{x}^\beta}\frac{\partial x^c}{\partial\overline{x}^\gamma}g_{bc}\right)\right)\\&=\color{limegreen}{\frac12g^{ad}\left(\frac{\partial\overline{x}^\alpha}{\partial x^a}\frac{\partial\overline{x}^\delta}{\partial x^d}\frac{\partial x^e}{\partial\overline{x}^\delta}\frac{\partial x^c}{\partial\overline{x}^\gamma}\frac{\partial}{\partial\overline{x}^\beta}g_{ec}+\beta\leftrightarrow\gamma-\frac{\partial\overline{x}^\alpha}{\partial x^a}\frac{\partial\overline{x}^\delta}{\partial x^d}\frac{\partial x^b}{\partial\overline{x}^\beta}\frac{\partial x^c}{\partial\overline{x}^\gamma}\frac{\partial}{\partial\overline{x}^\delta}g_{bc}\right)}\\&+\color{red}{\frac12g^{ad}\left(\frac{\partial\overline{x}^\alpha}{\partial x^a}\frac{\partial\overline{x}^\delta}{\partial x^d}g_{ec}\frac{\partial}{\partial\overline{x}^\beta}\left(\frac{\partial x^e}{\partial\overline{x}^\delta}\frac{\partial x^c}{\partial\overline{x}^\gamma}\right)+\beta\leftrightarrow\gamma-\frac{\partial\overline{x}^\alpha}{\partial x^a}\frac{\partial\overline{x}^\delta}{\partial x^d}g_{bc}\frac{\partial}{\partial\overline{x}^\delta}\left(\frac{\partial x^b}{\partial\overline{x}^\beta}\frac{\partial x^c}{\partial\overline{x}^\gamma}\right)\right)},\end{align}$$where the green part looks like a tensor's transformation while the red part doesn't. You mistakenly used $d$ for two sets of contracted indices: notice my use of $e$ for one of them.

The green part is$$\begin{align}&\frac12g^{ad}\left(\frac{\partial\overline{x}^\alpha}{\partial x^a}\frac{\partial x^c}{\partial\overline{x}^\gamma}\frac{\partial}{\partial\overline{x}^\beta}g_{dc}+\beta\leftrightarrow\gamma-\frac{\partial\overline{x}^\alpha}{\partial x^a}\frac{\partial x^b}{\partial\overline{x}^\beta}\frac{\partial x^c}{\partial\overline{x}^\gamma}\frac{\partial}{\partial x^d}g_{bc}\right)\\&=\frac{\partial\overline{x}^\alpha}{\partial x^a}\frac{\partial x^b}{\partial\overline{x}^\beta}\frac{\partial x^c}{\partial\overline{x}^\gamma}\frac12g^{ad}\left(\frac{\partial}{\partial x^b}g_{dc}+b\leftrightarrow c-\frac{\partial}{\partial x^d}g_{bc}\right)=\frac{\partial\overline{x}^\alpha}{\partial x^a}\frac{\partial x^b}{\partial\overline{x}^\beta}\frac{\partial x^c}{\partial\overline{x}^\gamma}\Gamma^a_{bc},\end{align}$$as expected. The red part is$$\frac12g^{ad}\left(\left[\frac{\partial\overline{x}^\alpha}{\partial x^a}g_{dc}\frac{\partial^2x^c}{\partial\overline{x}^\beta\partial\overline{x}^\gamma}+\frac{\partial\overline{x}^\alpha}{\partial x^a}\frac{\partial\overline{x}^\delta}{\partial x^d}g_{ec}\frac{\partial x^c}{\partial\overline{x}^\gamma}\frac{\partial^2x^e}{\partial\overline{x}^\beta\partial\overline{x}^\delta}\right]+\beta\leftrightarrow\gamma\\-\frac{\partial\overline{x}^\alpha}{\partial x^a}\frac{\partial\overline{x}^\delta}{\partial x^d}g_{bc}\frac{\partial x^b}{\partial\overline{x}^\beta}\frac{\partial^2x^c}{\partial\overline{x}^\gamma\partial\overline{x}^\delta}-\frac{\partial\overline{x}^\alpha}{\partial x^a}\frac{\partial\overline{x}^\delta}{\partial x^d}g_{bc}\frac{\partial x^c}{\partial\overline{x}^\gamma}\frac{\partial^2x^b}{\partial\overline{x}^\beta\partial\overline{x}^\delta}\right),$$but only one term doesn't cancel, namely $\frac{\partial\overline{x}^\alpha}{\partial x^c}\frac{\partial^2x^c}{\partial\overline{x}^\beta\partial\overline{x}^\gamma}$. That's almost what your source had, so I think it contains a typo. (Wikipedia agrees with me.)

A perhaps easier way to transform Christoffel symbols is to use $\nabla_{a}V_{b}=\partial_{a}V_{b}-\Gamma_{ab}^{c}V_{c}$ for a vector $V$, so$$\begin{align}\Gamma_{\beta\gamma}^{\alpha}V_{\alpha} &=\left(\partial_{\gamma}-\nabla_{\gamma}\right)V_{\beta} \\&=\left(\partial_{\gamma}-\nabla_{\gamma}\right)\left(\frac{\partial x^{b}}{\partial\overline{x}^{\beta}}V_{b}\right) \\&=\color{limegreen}{\frac{\partial x^{b}}{\partial\overline{x}^{\beta}}\left(\partial_{\gamma}-\nabla_{\gamma}\right)V_{b}}+\color{red}{V_{b}\left(\partial_{\gamma}-\nabla_{\gamma}\right)\frac{\partial x^{b}}{\partial\overline{x}^{\beta}}} \\&=\color{limegreen}{\frac{\partial\overline{x}^{\alpha}}{\partial x^{a}}\frac{\partial x^{b}}{\partial\overline{x}^{\beta}}\frac{\partial x^{c}}{\partial\overline{x}^{\gamma}}\Gamma_{bc}^{a}V_{\alpha}}+\color{red}{\frac{\partial\overline{x}^{\alpha}}{\partial x^{b}}\frac{\partial x^{b}}{\partial\overline{x}^{\beta}\partial\overline{x}^{\gamma}}V_{\alpha}},\end{align}$$whence$$\Gamma_{\alpha\beta}^{\gamma}=\color{limegreen}{\frac{\partial\overline{x}^{\alpha}}{\partial x^{a}}\frac{\partial x^{b}}{\partial\overline{x}^{\beta}}\frac{\partial x^{c}}{\partial\overline{x}^{\gamma}}\Gamma_{bc}^{a}}+\color{red}{\frac{\partial\overline{x}^{\alpha}}{\partial x^{b}}\frac{\partial x^{b}}{\partial\overline{x}^{\beta}\partial\overline{x}^{\gamma}}}$$as before.

Incidentally, we can make these calculations a lot more succinct by using subscripts to denote partial derivatives, so the result is $\Gamma^\alpha_{\beta\gamma}=\color{\limegreen}{\overline{x}^\alpha_ax^b_\beta x^c_\gamma\Gamma^a_{bc}}+\color{red}{\overline{x}^\alpha_bx^b_{\beta\gamma}}$. This is a lot easier to remember and perhaps even to guess, and see as "morally correct" given the rules of tensor indices. But some authors would insist on giving each derivative a comma, to emphasise it's partial, not covariant. Some would also want the Greek indices to be barred or primed. (Mind you, if you go that route they don't need to be Greek at all.)

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  • $\begingroup$ In the second proof, in the third step, shouldn't the covariant derivative with respect to γ of the jacobian b->β just be the partial derivative with respect to γ making the right (red) side disappear ? $\endgroup$ Commented May 6, 2022 at 15:28
  • $\begingroup$ @T4l0n I'm not sure what your reasoning is, but I suspect it involves forgetting which quantities are held constant in each partial derivative's definition. In particular,$$\left(\frac{\partial x^b}{\partial\bar{x}^\beta\partial\bar{x}^\gamma}\right)_{x^a}=\left(\frac{\partial x^c}{\partial\bar{x}^\gamma}\left(\frac{\partial}{\partial x^c}\frac{\partial x^b}{\partial\bar{x}^\beta}\right)_{x^\alpha}\right)_{x^a}=\frac{\partial x^c}{\partial\bar{x}^\gamma}\left(\left(\frac{\partial}{\partial\bar{x}^\beta}\delta_c^b\right)_{x^\alpha}\right)_{x^a}$$doesn't make sense. $\endgroup$
    – J.G.
    Commented May 6, 2022 at 16:07
  • $\begingroup$ I'm probably doing something wrong with this: $$ \nabla_\gamma{\frac{\partial x^b}{\partial x^\beta}} = \frac{\partial x^b}{\partial x^\beta \partial x^\gamma} + \Gamma^b_{\gamma k}\frac{\partial x^k}{\partial x^\beta} - \Gamma^k_{\gamma \beta}\frac{\partial x^b}{\partial x^k}=\frac{\partial x^b}{\partial x^\beta \partial x^\gamma} + \Gamma^b_{\gamma \beta} - \Gamma^b_{\gamma \beta} = \frac{\partial x^b}{\partial x^\beta \partial x^\gamma} $$ $\endgroup$ Commented May 6, 2022 at 17:49
  • $\begingroup$ @T4l0n For starters, you can't mix indices from different coordinate systems in a tensor (or in Christoffel symbols), including the Kronecker delta; similarly, $x^b_\beta$ isn't a tensor at all. $\endgroup$
    – J.G.
    Commented May 6, 2022 at 17:53
  • $\begingroup$ I'm used to different notations but isn't $x^b_\beta = \frac{\partial x^b}{\partial x^\beta}$ the jacobian from $b$ to $\beta$ coordinates ? If so how is that not a tensor ? $\endgroup$ Commented May 7, 2022 at 5:15

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