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I'm trying to calculate a Christoffel symbol, but I'm stuck on showing that $\frac{\partial(e^{2A})}{\partial r}=2e^{2A}\frac{\partial A}{\partial r}\ ? $

Nice easy steps would be much appreciated.

Thank you

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    Your're welcome, Peter. Good luck with the general relativity! (It's not easy!)2011-12-07

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When $A$ is a scalar function of $r$, the formula $\frac{d }{d r}e^{2A}=2e^{2A}\frac{dA}{dr}$ is nothing but the chain rule.

When $A$ is a matrix, one has to watch out for the lack of commutativity. If $A$ and $\frac{dA}{dr}$ commute, everything works as for scalars. Otherwise, the formula breaks down: for example, if $A(r)=\begin{pmatrix}0 & 1 \\ r & 0\end{pmatrix} \tag1$ then $\frac{d}{dr}_{\, \big|r=0} \exp A(r) = \begin{pmatrix}1/2 & 1/6 \\ 1 & 1/2\end{pmatrix}\tag2$ which is not a product of $A'(0)=\begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}$ and $\exp A(0)=\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}$, in either order: $A'(0)\exp A(0) =\begin{pmatrix}0 & 0 \\ 1 & 1\end{pmatrix}\qquad (\exp A(0)) \,A'(0)=\begin{pmatrix}1 & 0 \\ 1 & 0\end{pmatrix}\tag3$ One may observe in passing that the average of two products in (3) is $\frac12 (A'(0)\exp A(0) + (\exp A(0)) \,A'(0)) = \begin{pmatrix}1/2 & 0 \\ 1 & 1/2\end{pmatrix} \tag4$ which does a decent job of approximating (2). This suggests that (4) may be the beginning of a power series that gives the correct value of the derivative of matrix exponential, which (I guess) can be fished out of the Zassenhaus formula.