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I'm having problem with calculating the functional derivative of $F$ with respect to $\phi(x)$ while

$$F = \int d^{4}x \phi^2 \partial_{\mu}\phi\partial^{\mu}\phi.$$

I want to obtain $\frac{\delta F}{\delta \phi}$ by making form of $$\delta F = \int d^{4}x \frac{\delta F}{\delta \phi}\delta \phi.$$

But, I am confused how could I compute $\delta F$ because of $\partial_{\mu}\phi$.

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Beginning with $$ F[\phi] = \int d^{4}x \ \phi^2 \ \partial_{\mu}\phi \ \partial^{\mu}\phi $$

we replace $\phi$ with $\phi+\lambda\eta$: $$ F[\phi+\lambda\eta] = \int d^{4}x \ (\phi+\lambda\eta)^2 \ \partial_{\mu}(\phi+\lambda\eta) \ \partial^{\mu}(\phi+\lambda\eta) $$

Then we take the derivative w.r.t. $\lambda$ at $\lambda=0,$ and use integration by parts to remove any derivatives of $\eta$: $$ \frac{d}{d\lambda} \left. F[\phi+\lambda\eta] \right|_{\lambda=0} = \int d^{4}x \ \left( 2\phi\eta \ \partial_{\mu}\phi\ \partial^{\mu}\phi + \phi^2 \ \partial_{\mu}\eta \ \partial^{\mu}\phi + \phi^2 \ \partial_{\mu}\phi \ \partial^{\mu}\eta \right) \\ = \int d^{4}x \ \left( 2\phi\eta \ \partial_{\mu}\phi\ \partial^{\mu}\phi - \partial_{\mu}(\phi^2\ \partial^{\mu}\phi) \eta - \partial^{\mu}(\phi^2\ \partial_{\mu}\phi) \eta \right) \\ = \int d^{4}x \ \left( 2\phi \ \partial_{\mu}\phi\ \partial^{\mu}\phi - \partial_{\mu}(\phi^2\ \partial^{\mu}\phi) - \partial^{\mu}(\phi^2\ \partial_{\mu}\phi) \right) \eta \\ $$

The $\eta$-independent factor in front of $\eta$ is now the functional derivative: $$ \frac{\delta F}{\delta\phi} = 2\phi \ \partial_{\mu}\phi\ \partial^{\mu}\phi - \partial_{\mu}(\phi^2\ \partial^{\mu}\phi) - \partial^{\mu}(\phi^2\ \partial_{\mu}\phi) \\ = 2\phi\ \partial_{\mu}\phi\ \partial^{\mu}\phi - (2\phi\ \partial_{\mu}\phi\ \partial^{\mu}\phi + \phi^2 \partial_{\mu}\partial^{\mu}\phi) - (2\phi\ \partial^{\mu}\phi\ \partial_{\mu}\phi + \phi^2\ \partial^{\mu}\partial_{\mu}\phi) \\ = -2\phi\ \partial^{\mu}\phi\ \partial_{\mu}\phi - \phi^2\ \partial^{\mu}\partial_{\mu}\phi $$

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  • $\begingroup$ Thank you. I solved it. You gave me a key : integration by parts :) $\endgroup$ – Seal Jan 19 at 15:11
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HINT: When you're doing this variation, you're implicitly replacing $\phi \to \phi + \delta \phi$ and keeping all the terms linear in $\delta \phi$. So you have $$ \delta F = \int d^4x \left[ (2 \phi \, \delta \phi) \partial_\mu \phi \partial^\mu \phi + \phi^2 (\partial_\mu \delta \phi) (\partial^\mu \phi) \right]. $$ The first term is of the form $(\text{something}) \times \delta \phi$, but the second one isn't. So you need to get rid of the derivative on the $\delta \phi$ in the second term, somehow.

Do you know of any mathematical techniques where you can "shift" the derivative on one term in an integrand onto a different term in the integrand?

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  • $\begingroup$ That is exactly what I want to know. Thanks, I solved. $\endgroup$ – Seal Jan 19 at 15:10

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