# Coordinate transform of partial derivative

## Homework Statement

How does ∂aAb behave under coordinate transformations in special relativity? Work out ∂'aA'b

## The Attempt at a Solution

I have been given back the solution sheet to this problem, but I don't understand it. This is what I have I get the first line, (13). We have treated the partial as a covariant vector, and written out the usual transform for it. The part within the brackets is the normal transform of a contravariant vector A.

Looking at (14), the text says we have used the product rule here. I am familiar with this in normal calculus, where if k(x) = f(x)g(x) then k'(x) = f'(x)g(x) + f(x)g'(x). If I apply that here, I would expect to see ∂'aAb + ∂aA'b

Written out long hand, this would be similar to (14). However, the solution has an extra ∂Xd/∂X'a in the second term, and that weird ∂∂X'b thing in the first term.

For the second term in (14) we have every item in (13) just multiplied together in a different order; and the first term in (14) is the same, but with two of those items conflated to form the ∂∂X'b strangeness. Are they multiplied together in some sense, or is this now a second order partial derivative term? What on earth is going on? If someone could post up a clear rule for how to approach this type of problem it would be great, as it will keep cropping up in my course and I am totally clueless how else to tackle it.

My tutor advised me to take a look at the wikipedia link for the chain rule, but I can't see how that is relevant to this problem, although I see it is different in higher dimensions to what I am used to. I have also searched to see if the product rule is different in higher dimensions, but it seems to work just the same for vectors as for normal functions.

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vela
Staff Emeritus
Homework Helper
You're starting off with
$$\partial'_a A'^b = \frac{\partial X^d}{\partial X'^a} \frac{\partial}{\partial X^d} \left(\frac{\partial X'^b}{\partial X^c} A^c\right) = \frac{\partial X^d}{\partial X'^a} \left[\frac{\partial}{\partial X^d} \left(\frac{\partial X'^b}{\partial X^c} A^c\right)\right]$$You use the product rule to evaluate the term in the square brackets to get
$$\partial'_a A'^b = \frac{\partial X^d}{\partial X'^a} \left[ \left(\frac{\partial}{\partial X^d} \frac{\partial X'^b}{\partial X^c}\right) A^c + \frac{\partial X'^b}{\partial X^c} \left(\frac{\partial}{\partial X^d} A^c\right)\right]$$

Thanks for that Vela, I can follow your explanation although I am not sure why we do it that way instead of putting the terms in a different order or changing the position of the square brackets - probably some property of the partial operator which I am not familiar with. As long as I know how to do it, that is a victory!

Could you (or someone else) explain what the "double derivative" ∂∂X'b/∂Xd∂Xc means? It looks like it's the rate of change of a derivative, which should be a second-order derivative I think, but it's not written that way.

vela
Staff Emeritus
That term is a second-order derivative. In the "normal" notation, it would correspond go a term like $\frac{\partial^2}{\partial x\partial y}$ if c≠d, and to a term like $\frac{\partial^2}{\partial x^2}$ if c=d.