# proper variation of action term

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.

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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:

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

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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