# Variation of a function with respect to the metric

I was reading this paper and I think that I find a mistake, may be I'm wrong, but I want to be sure.

They take the variation with respect to the metric $$g_{\alpha\beta}$$ of this function

$$S(\delta \Omega)=\int_{\delta \Omega}n_{\nu}s^{\nu}\sqrt{h}d^{d-1}x$$

With some fixed boundary conditions $$g_{\alpha\beta}(\delta \Omega)=g_{\alpha\beta}^{\delta \Omega}$$. $$s^{\nu}$$ is a function that depends of the metric, $$\delta\Omega$$ is a Jordan Orientable surface with normal $$n_{\nu}$$.

They define a family of metrics

$$g_{\alpha\beta}(x^{\mu})=g*_{\alpha\beta}(\mu)+\delta_{\epsilon}(g_{\alpha\beta})x^{\mu}$$

Where $$g*_{\alpha\beta}$$ is the metric that extremize $$S(\delta\Omega)$$, $$\epsilon\in R$$. $$\delta_{\epsilon}(g_{\alpha\beta})$$ satisfices the boundary condition $$\delta_{\epsilon}(g_{\alpha\beta})(\delta\Omega)=0$$ and $$\lim_{\epsilon \rightarrow 0} \delta_{\epsilon}(g_{\alpha\beta})(x^{\mu})=0$$.

The the variation with respect to the metric of the first equation is

$$\lim_{\epsilon \rightarrow 0} \frac{\delta_{\epsilon(S)(\delta \Omega)}}{\epsilon}=0$$ I agree with this equation.

However I don't agree with this equation

$$\lim_{\epsilon \rightarrow 0} \frac{\delta_{\epsilon(S)(\delta \Omega)}}{\epsilon}=\int_{\delta \Omega}n_{\nu}\lim_{\epsilon \rightarrow 0}\frac{\delta_{\epsilon}(s^{\nu})}{\epsilon}\sqrt{h}d^{d-1}x=0$$

I think that we also have to take the variation with respect to the metric of the normal $$n_{\nu}$$ and then we get something like this $$\lim_{\epsilon \rightarrow 0} \frac{\delta_{\epsilon(S)(\delta \Omega)}}{\epsilon}=\int_{\delta \Omega}\lim_{\epsilon \rightarrow 0}\frac{\delta_{\epsilon}(n_{\nu}s^{\nu})}{\epsilon}\sqrt{h}d^{d-1}x=0$$

Since

$$n_{\alpha}=\frac{\partial_{\alpha}f}{\sqrt{|g^{\alpha \beta}\partial_{\alpha}f \partial_{\beta}f | }}$$

\ $$\textbf{EDIT}$$ I'm not looking for the complete way of taking the variation of this function. I'm looking for and answer that say's if I have to do something like this $$\delta_{\epsilon}n_{\nu}(s^{\nu})$$ or like this $$\delta_{\epsilon}(n_{\nu}s^{\nu})$$ in the variation.

• The normal is defined on the boundary and the variation of the metric on the boundary vanishes? – chichi Apr 21 '20 at 7:00
• That is what i understand. – Nothing Apr 21 '20 at 15:24

## 1 Answer

In general you would be correct: we require the normal vector to be normalized, and that clearly depends on the metric. But notice that we require the metric on the surface $$\partial \Omega$$ to stay constant, so $$n_\mu$$ doesn't change. By the same reasoning, $$\sqrt{h}$$ is constant too. In fact, since the integral is on the surface, almost nothing changes. The only possible variation is if $$s^\mu$$ depends on derivatives of the metric, which are not fixed on the surface.

• Thanks! I'm going to wait to another answer to decide if a give you the status of correct aswer. But I really like what you say. – Nothing Apr 24 '20 at 18:52
• Why the normal vector needs to be normalized? – Nothing Apr 24 '20 at 18:56
• @Cruz your definition gives a normalized vector; as you noticed, it involves the metric. You can use a non-normalized vector, but then that factor shows up elsewhere in the integral. – Javier Apr 24 '20 at 19:10
• @Cruz Javier is correct :) The same thing happens when variation of the York boundary term is taken into consideration for standard gravitational action. – A. Ok Apr 25 '20 at 21:52