# Variational technique

1. May 11, 2008

### arroy_0205

Suppose I try to find variation of the term \phi^2 under variation of \phi i.e., (\delta \phi). Then I take derivative of \phi^2 with respect to \phi and multiply by \delta \phi. In case of more complicated objects containing derivative of \phi what is the procedure? for example:
$$\delta(\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi})$$
what is the variation of this quantity under variation of \phi?

2. May 11, 2008

### arroy_0205

Is the rule here
$$\delta(\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi})=\frac{\partial(\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi})}{\partial \phi}\delta(\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi)$$
But then the derivative part creates problem. The part under square root depends on derivative of \phi and not on \phi itself, so the result is zero. I am confused.

Last edited: May 11, 2008
3. May 11, 2008

### lbrits

Always go back to the definition of the derivative:

$$\delta F[\phi] = F[\phi + \delta \phi] - F[\phi]$$.

In any case, in your case the chain rule applies:
$$\delta(\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi})= \frac{1}{ 2\sqrt{\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi} } \delta(\eta^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi)$$