Differentiation in general relativity

Question

If we have:

$$ \frac{d\phi^a}{d\tau}= \frac{\partial \phi^a}{\partial x^\mu} \frac{dx^\mu}{d\tau} \tag{1}$$

Differentiating it, we get: $$ \frac{\partial \phi^a}{\partial x^\mu}\frac{d^2x^\mu}{d\tau^2} + \frac{\partial^2\phi^a}{\partial x^\nu \partial x^\lambda} \frac{dx^\nu}{d\tau}\frac{dx^\lambda}{d\tau} \tag{2}$$

I got how we got the first term in equation (2) but what about the second term? Where did the $dx^\lambda , dx^\nu $ come from? Where did the $dx^\mu $ go?

Thank you a lot for taking the time to read this.

Answer 1

I think your equation (2) should read $$ \frac{d^2 \phi^a}{d \tau^2} = \frac{\partial \phi^a}{\partial x^\mu}\frac{d^2x^\mu}{d\tau^2} + \frac{\partial^2\phi^a}{\partial x^\nu \partial x^\lambda} \frac{dx^\nu}{d\tau}\frac{dx^\lambda}{d\tau}$$ This follows from application of the product and chain rule: $$ \frac{d}{d \tau} \left(\frac{\partial \phi^a}{\partial x^\mu} \frac{dx^\mu}{d\tau} \right) = \left(\frac{d}{d \tau} \frac{\partial \phi^a}{\partial x^\mu}\right) \frac{dx^\mu}{d\tau} + \frac{\partial \phi^a}{\partial x^\mu}\left(\frac{d}{d \tau} \frac{dx^\mu}{d\tau} \right) $$ The second term becomes the first term on the RHS of the first equation. The first term can be seen to equal the second term on the RHS of the first equation using the following form of the chain rule: $$ \frac{d}{d \tau} = \frac{d x^\lambda}{d \tau} \frac{\partial}{\partial x^\lambda}$$ where we sum over $\lambda$ --- we could choose to use any letter here instead of $\lambda$, as long as we haven't used it already. The result you've shown then follows by relabelling $\mu \to \nu$, which is unnecessary in my view, but legitimate --- we can label our objects with whatever letters we like. This is so even if our index is not a dummy index --- provided we consistently replace all instances of $\mu$ by $\nu$, our equation hasn't changed. As Shakespeare might say, what's in a name?

Answer 2

The first equation (which is just the chain rule really) tells us generally $$ \frac{\mathrm{d}}{\mathrm{d}\tau} (\cdot) = \frac{\partial(\cdot)}{\partial x^\nu} \frac{\mathrm{d}x^\nu}{\mathrm{d}\tau}, $$ where I can replace the dummy indices $\nu$ with anything I want (as long as they are the same and not repeated elsewhere), and indeed I wrote them as $\nu$ rather than $\mu$ to avoid conflicting with $\mu$ later.

The second part of the product rule is then $$ \frac{\mathrm{d}}{\mathrm{d}\tau} \left(\frac{\partial\phi^a}{\partial x^\mu}\right) \cdot \frac{\mathrm{d}x^\mu}{\mathrm{d}\tau} = \frac{\partial}{\partial x^\nu} \left(\frac{\partial\phi^a}{\partial x^\mu}\right) \frac{\mathrm{d}x^\nu}{\mathrm{d}\tau} \cdot \frac{\mathrm{d}x^\mu}{\mathrm{d}\tau} = \frac{\partial^2\phi^a}{\partial x^\nu\partial x^\mu} \frac{\mathrm{d}x^\nu}{\mathrm{d}\tau} \frac{\mathrm{d}x^\mu}{\mathrm{d}\tau}. $$ You can replace the dummy indices however you see fit, including $\mu \to \lambda$.