Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have a term I want to vary by a field, $\phi$. $$ `S' = \frac{-1}{2}\,\sqrt{-g}\,g^{\mu\,\nu}\,\delta\left[h(\phi)\,\partial_{\mu}\phi\,\partial_{\nu}\phi \right]. $$

Is it correct to get this?

(I am unsure about the derivatives)

let $$a:=\frac{-1}{2}\,\sqrt{-g}\,g^{\mu\,\nu}$$

Basically I want to know how to do this:

$$ a\,h(\phi)\,\delta\left[ \partial_{\mu}\phi\,\partial_{\nu}\phi\right] $$

I should also say, that unlike the usual case, I am only interested in varying the field, not the coordinates. So I think $\partial_a\phi \mapsto \partial_a\phi+\partial_a(\delta\,\phi)$....

If I take Taylor series would then the variation be, $$ \left( \partial_{\mu}\phi + \frac{1}{2}\,\partial_{\mu'}\partial_{\mu}\phi\,(\delta\phi) \right)\,\left( \partial_{\nu}\phi + \frac{1}{2}\,\partial_{\nu'}\partial_{\nu}\phi\,(\delta\phi) \right) = $$

$$ \cdots = \partial_{\mu}\phi\,\frac{1}{2}\,\partial_{\nu'}\partial_{\nu}\phi\,(\delta\phi) + \partial_{\nu}\phi\,\frac{1}{2}\,\partial_{\mu'}\partial_{\mu}\phi\,(\delta\phi) $$ and then collect on $\delta\phi$? I have 4 index values... I think $\mu' \neq \nu$ and $\nu'\neq \mu$, as there seems to be no reason why they would be.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

I'm not really sure what you're asking, but I think it's about how to deal with varying the term $\partial_\mu \phi \partial_\nu \phi$. I'll work in Minkowski space + Cartesian coordinates so we don't have to worry about the metric determinant.

Say you have the action:

$$S=\int d^4x~ g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi $$

Varying this action yields:

$$\delta S=\int d^4x~ g^{\mu \nu} \delta (\partial_\mu \phi \partial_\nu \phi)$$

Now we can use the product rule to expand this out:

$$\int d^4x~ g^{\mu \nu} \delta (\partial_\mu \phi \partial_\nu \phi)=\int d^4x~ g^{\mu \nu} [\partial_\nu \phi \delta (\partial_\mu \phi)+\partial_\mu \phi \delta (\partial_\nu \phi)]$$

Since $\partial_\mu \phi \delta (\partial_\nu \phi)$ is clearly symmetric, we can simplify this to:

$$\delta S=2\int d^4x~ g^{\mu \nu} \partial_\mu \phi \delta (\partial_\nu \phi)$$

Now comes the part I believe you're confused about. We integrate by parts, i.e. we use the fact that:

$$\int d^4x~ a~\partial_\mu b=\int d^4x~ \partial_\mu (ab) -\int d^4x~ b~\partial_\mu a$$

We also use the fact that the gradient and the functional derivative commute: $\partial_\mu \delta\phi =\delta (\partial_\mu \phi)$. So, we get:

$$\delta S=2\int d^4x~ \partial_\nu (g^{\mu \nu} \partial_\mu \phi \delta \phi)-2\int d^4x~ g^{\mu \nu} \partial_\nu \partial_\mu \phi ~\delta \phi$$

Now all we have to do is notice that the first integral is of a total derivative, i.e. something of the form $\partial_\mu V^\mu$, so it can be converted into a surface term by the 4D divergence theorem. Since this is a variational problem we know that $\delta \phi=0$ on the surface, and therefore the whole first integral must be zero. What we're left with is:

$$\delta S=-2\int d^4x~ g^{\mu \nu} \partial_\nu \partial_\mu \phi ~\delta \phi$$

This, of course (by the stationary action principle), tells us that the equations of motion for our field are:

$$g^{\mu \nu} \partial_\nu \partial_\mu \phi =\square \phi =0$$.

share|improve this answer
Thank you! Well I'm familar w/ this so far. I asked poorly I suppose (trying to do a MWE), but I have a term $g^{ab}\,\partial_a\phi\,\partial_b\phi$ and I need to get it to $(1/\phi)\,\phi^{;c}\,\phi_{;c}$. Ugg. Its the Brans-Dicke term if that rings a bell. :) –  nate Apr 19 '13 at 4:47
I guess I have a mistake further upstream. Your results seem to confirm that what I am asking of this first term is not possible... Thanks much! –  nate Apr 19 '13 at 4:52

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.