How to evaluate $\partial_u(x^2 x_v)$?

Question

I need to use some basic tensor calculus but I did not receive any introduction to the topic. Just a couple of statements. I am stumbling with this problem to evaluate: $$\partial_u(x^2 x_v)$$ I know that $$x^2 = x_u x^u = x_u g^{uv} x_v = x^u g_{uv} x^v$$ and $$x_v = [ct, -x, -y, -z]$$ $$x^v = [ct, x, y, z]$$

I mostly don't know how to evaluate $x^2x_v$? My intuition says

$$x^2 = x_vx^v = c^2 t^2 + x^2 + y^2 + z^2$$ which is scalar $$\partial_u (x^2 x_u ) = x^2 \partial_u x_u= x^2 (1-1-1-1)= -2x^2$$

This doesn't make sense, right? How do you actually evaluate this?

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Also would be nice if you can suggest a terse resource to get up to speed on these manipulations and tensor calculus in general.

Answer 1

In general one first writes the term out completely, $$\partial_u (x^2 ~x_v) = \partial_u (\eta_{mn}~\eta_{vt}~x^m~x^n~x^t),$$ then one expands out with the product rule, $$ \begin{align} \partial_u (x^2 ~x_v) =~& (\partial_u \eta_{mn})~\eta_{vt}~x^m~x^n~x^t + \eta_{mn}~(\partial_u \eta_{vt})~x^m~x^n~x^t + ~\\ & \eta_{mn}~\eta_{vt}~(\partial_u x^m)~x^n~x^t + \eta_{mn}~\eta_{vt}~x^m~(\partial_u x^n)~x^t +\eta_{mn}~\eta_{vt}~x^m~x^n~(\partial_u x^t). \end{align}$$Note that in special relativity the first two terms are zero because the metric tensor $\eta$ is constant over spacetime. (In general relativity $\eta$ becomes $g$ and $\partial$ becomes $\nabla$ and these terms still vanish, but that's only true because we choose a specific connection which makes them vanish.)

Aside from $\partial_{\bullet} \eta_{\bullet\bullet} = 0$ we also can simplify the last three terms because we have that $\partial_a x^b = \delta_a^b,$ the Kronecker delta. This forces in the first of the three an identification $u=m$, so one gets $x_u x_v$ for the first term, for example.

Note that your gut intuition, which is that $\partial_u$ is selecting out what to act upon by its index, is wholly incorrect here and needs to be abandoned. $\partial_\bullet$ is a spacetime gradient, full-stop. It applies to all three $x$ terms which all refer to the spacetime position. It happens to have a lower index, because spatial gradients always produce covectors, and the lower index identifies that this is a covector field, not a scalar field.

Answer 2

First we introduce a new dummy variable for $x^2=x^\rho x_\rho$. So we have

$$\partial_\mu(x^2x_\nu)=\partial_\mu(x^\rho x_\rho x_\nu).$$

Then we use $\,\partial_\mu x^\rho=\delta_\mu^\rho\,$ and $\,\partial_\mu x_\rho=\eta_{\mu\rho}\,$ to obtain

\begin{align} \partial_\mu(x^2x_\nu)&=\delta_\mu^\rho x_\rho x_\nu+x^\rho\eta_{\mu\rho}x_\nu+x^\rho x_\rho\eta_{\mu\nu}\\ &=2x_\mu x_\nu+x^2\eta_{\mu\nu}, \end{align}

assuming special relativity so $\,\eta_{\mu\nu}\,$ is constant.