Can I contract index in this expression?

Question

I'm reading Carrol text on general relativity, on page 96 they arrive to the term

\begin{equation} \frac{\partial x^{\mu}}{\partial x^{\mu '}}\frac{\partial x^{\lambda}}{\partial x^{\lambda '}}\frac{\partial^2 x^{\nu '}}{\partial x^{\mu}\partial x^{\lambda}}.\tag{1} \end{equation}

Can I contract this expression to get

\begin{equation} \frac{\partial^2 x^{\nu '}}{\partial x^{\mu '}\partial x^{\lambda '}}~?\tag{2} \end{equation}

I'm using the chain rule $$\frac{\partial x^{\mu}}{\partial x^{\mu '}} \frac{\partial}{\partial x^{\mu}}=\frac{\partial}{\partial x^{\mu '}}\tag{3}$$ (which I think is correct).

Answer 1

The chain rule (3) is correct, but expression (1) is only 1 out of 2 terms in expression (2)

$$ \frac{\partial^2 x^{\nu ^\prime}}{\partial x^{\mu ^\prime}\partial x^{\lambda ^\prime}}~=~ \frac{\partial x^{\mu}}{\partial x^{\mu ^\prime}} \frac{\partial}{\partial x^{\mu}}\left( \frac{\partial x^{\lambda}}{\partial x^{\lambda ^\prime}}\frac{\partial x^{\nu ^\prime}}{\partial x^{\lambda}}\right)~=~ \frac{\partial x^{\mu}}{\partial x^{\mu ^\prime}}\left( \frac{\partial}{\partial x^{\mu}} \frac{\partial x^{\lambda}}{\partial x^{\lambda ^\prime}}\right)\frac{\partial x^{\nu ^\prime}}{\partial x^{\lambda}}+\frac{\partial x^{\mu}}{\partial x^{\mu ^\prime}}\frac{\partial x^{\lambda}}{\partial x^{\lambda ^\prime}}\frac{\partial^2 x^{\nu ^\prime}}{\partial x^{\mu}\partial x^{\lambda}}, $$

cf. Leibniz rule.

Answer 2

No, because $$\color{red}{\sum_{\mu,\lambda}}\frac{\partial x^{\color{red}{\mu}}}{\partial x^{\mu '}}\frac{\partial x^{\color{red}{\lambda}}}{\partial x^{\lambda '}}\frac{\partial^2 x^{\nu '}}{\partial x^{\color{red}{\mu}}\partial x^{\color{red}{\lambda}}}$$already has a double summation implied in $\mu$ and $\lambda$, so these are not free indices, and hence cannot be contracted with anything.

Answer 3

No, you can't. $$\frac{\partial x^{\mu}}{\partial x^{\mu '}}\frac{\partial x^{\lambda}}{\partial x^{\lambda '}}\frac{\partial^2 x^{\nu '}}{\partial x^{\mu}\partial x^{\lambda}}$$

can be re-written as : $$\frac{\partial x^{\mu}}{\partial x^{\mu '}}\frac{\partial x^{\lambda}}{\partial x^{\lambda '}}\frac{\partial}{\partial x^{\mu}}\left(\frac{\partial x^{\nu '}}{\partial x^{\lambda}}\right)$$

Now it can be clearly seen that $\partial x^{\lambda}$ in last term is part of first-order partial derivative, which needs to be differentiated again against $\partial x^{\mu}$. So you can't simplify things like that, because partial derivatives differentiates a function which depends on multiple variables. And in general you need to get used to the idea that derivatives are not ratios.