Does this derivative make sense?

Question

I know that the following derivative makes sense and is $$ \frac{\partial r}{\partial x^\mu}=\frac{x_\mu}{r}\,, $$ where $r=|x|$. Does the following derivative make sense? $$ \frac{\partial x^\mu}{\partial r} $$ If it does, what is the result?

Answer 1

(note: this post is working with the individual scalars directly and not using the usual index conventions)

No. Partial derivative notation is somewhat misleading. For example, the notation $\frac{\partial r}{\partial x_0}$ doesn't mean "the derivative of $r$ with respect to $x_0$"; it means "the derivative of $r$ with respect to $x_0$ when $x_1$, $x_2$, and $x_3$ are all held constant".

Similarly, the tangent vector $\frac{\partial}{\partial x_0}$ is not "the vector in the direction where $x_0$ increases"; it is "the vector in the direction that $x_0$ increases when $x_1$, $x_2$, and $x_3$ are held constant".

In other words, partial derivative notation only makes sense when you're asking for the derivative with respect to one variable out of a system of coordinates..

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Differentials are, in my opinion, much more natural to work with. There is nothing implicitly hidden in the equation

$$ \mathrm{d}r = \sum_{i=0}^3 \frac{x_i}{r} \mathrm{d}x_i $$

and you can rearrange it to your liking in the obvious ways; e.g. if you wanted, you could could multiply through by $r$ and shuffle the terms around:

$$ x_1 \mathrm{d}x_1 + x_3 \mathrm{d}x_3 = r \mathrm{d}r - x_0 \mathrm{d} x_0 - x_2 \mathrm{d} x_2 $$

and the result would remain true. (no particular reason for this particular form; I just picked something to do more or less at random)