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


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

\ $$\textbf{EDIT}$$ 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.

  • 1
    $\begingroup$ The normal is defined on the boundary and the variation of the metric on the boundary vanishes? $\endgroup$
    – chichi
    Commented Apr 21, 2020 at 7:00
  • $\begingroup$ That is what i understand. $\endgroup$
    – Nothing
    Commented Apr 21, 2020 at 15:24

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Nothing
    Commented Apr 24, 2020 at 18:52
  • $\begingroup$ Why the normal vector needs to be normalized? $\endgroup$
    – Nothing
    Commented Apr 24, 2020 at 18:56
  • $\begingroup$ @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. $\endgroup$
    – Javier
    Commented Apr 24, 2020 at 19:10
  • 1
    $\begingroup$ @Cruz Javier is correct :) The same thing happens when variation of the York boundary term is taken into consideration for standard gravitational action. $\endgroup$
    – A. Ok
    Commented Apr 25, 2020 at 21:52

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