Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose I have something like $ f = g_{\mu \nu} x^{\mu} x^{\nu} $, where the Einstein summation convention is implied. Now suppose I want to to take the derivative $ \partial_{\mu}f = \frac{\partial f}{\partial x^{\mu}} $. How would I go about doing this? I figure it's not just going to be $ \partial_{\mu} f = g_{\mu \nu} x^{\nu} $.

share|cite|improve this question
up vote 10 down vote accepted

Use the product rule: $$ \begin{align*} \partial_\lambda(g_{\mu\nu} x^\mu x^\nu) &= g_{\mu\nu,\lambda} x^\mu x^\nu + g_{\mu\nu} x^\mu_{,\lambda} x^\nu + g_{\mu\nu} x^\mu x^\nu_{,\lambda} \\&= g_{\mu\nu,\lambda} x^\mu x^\nu + g_{\mu\nu} \delta^\mu_\lambda x^\nu + g_{\mu\nu} x^\mu \delta^\nu_\lambda \\&= g_{\mu\nu,\lambda} x^\mu x^\nu + g_{\lambda\nu} x^\nu + g_{\mu\lambda} x^\mu \\&= g_{\mu\nu,\lambda} x^\mu x^\nu + 2 g_{\lambda\nu} x^\nu \end{align*} $$ where I've used the convention $T^{\mu...}_{\nu...,\lambda} \equiv \partial_\lambda T^{\mu...}_{\nu...}$, the fact that $x^\mu_{,\lambda}=\delta^\mu_\lambda$ and the symmetry of $g_{\mu\nu}$.

share|cite|improve this answer

This is a little expansion on the accepted answer. The Einstein summation convention works because summation commutes with all linear operators including other summations because of distributivity. Therefore it does not matter where you put the summation sign as long as its scope includes all the occurrences of the summed index and you can do all calculations as if the summations were not there. You just need a way to keep track what is being summed over. This avoids a lot of trivial steps that just move summations within expressions. So you can proceed with differentiation as if the summation was not there. There is one subtlety though and it comes up in your question: contraction effectively binds and hides variables being contracted like quantifiers do in logic and $\lambda$ does in the $\lambda$-calculus and as do most indexed operators like $\sum_i$, $\prod_i$, $\bigcup_i$ and $\bigcap_i$. This means that if you define $f = g_{\mu\nu} x^\mu x^\nu$ then it is somewhat misleading to use $\mu$ again in $\partial_\mu f$ since the $\mu$ in the definition of $f$ is really bound so the index in the differentiation should be fresh. That is why you should write $\partial_\lambda f = \partial_\lambda (g_{\mu\nu} x^\mu x^\nu)$ and then proceed with ordinary algebraic manipulation as in Christoph's answer.

share|cite|improve this answer

Assuming that $g_{\mu \nu}$ is the Minkowski metric. One can say that

$ \partial_{\mu}x^{\alpha} = \frac{\partial x^{\alpha}}{\partial x^{\mu}} =\delta_{\mu}^{\alpha}$

Therefore $ \partial_{\mu} f = 2 g_{\mu \nu} x^{\nu} $

If $g_{\mu \nu}$ is not a constant there is another term containing derivative of $g_{\mu \nu}$.

share|cite|improve this answer
you're missing a factor of 2 – Christoph Oct 6 '12 at 12:59
yes I noticed it – Prathyush Oct 6 '12 at 13:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.