How can I take the derivative?

[etotheipi]

The action considered is \begin{align*}
S[\Phi] = \int_M dt d^3 x \sqrt{-g} \left( -\frac{1}{2}g^{ab} \partial_a \Phi \partial_b \Phi - \frac{1}{2}m^2 \Phi^2\right)
\end{align*}I can "see" unrigorously that variation with respect to $\partial_t \Phi(x)$ it is going to be\begin{align*}
\frac{\delta S}{\delta(\partial_t \Phi(x))} &= -\frac{1}{2}\sqrt{-g} g^{ab} \frac{\delta}{\delta(\partial_t \Phi(x))} \left( \partial_a \Phi \partial_b \Phi \right) \\

&= -\frac{1}{2}\sqrt{-g} g^{ab} \cdot 2 \partial_a \Phi \delta^t_b \\

&= -\sqrt{-g} g^{at} \partial_a \Phi \\

&= \sqrt{h} n^{a} \partial_a \Phi
\end{align*}How do I do this calculation properly? For instance to find the functional variation with respect to $\Phi$ I could consider $\lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon}\left( S[\Phi + \epsilon \alpha] - S[\Phi] \right)$. But how do I write it when considering the functional derivative of $S$ with respect to $\partial_t \Phi$; do I let the action have a second argument $S = S[\Phi, \partial_a \Phi]$ and proceed from there?

 

[Gaussian97]

Why do you need such a strange derivative? I don't really see how such a derivative would be defined, maybe if you give me some more context I can help

 

[etotheipi]

$(M, g_{ab})$ is a globally hyperbolic spacetime and the metric is here written $ds^2 = -N^2 dt^2 + h_{ij} (dx^i + N^i dt)(dx^j + N^j dt)$ with $t$ a global time coordinate labelling the cauchy surfaces $\Sigma_t$ of normal $-N(dt)_a$ and induced metric $h_{ab}$, such that $\sqrt{-g} = N\sqrt{h}$.

The aim is to find the canonical momentum $\Pi(x)$ conjugate to $\Phi$ by varying $S$ with respect to $\partial_t \Phi(x)$

 

[Gaussian97]

Ok, usually when one has an action of the form
$$S = \int \mathscr{L}(\Phi, \partial_\mu \Phi, t) d^4 x$$
the canonical momentum is defined as
$$\Pi(x)=\frac{\partial \mathscr{L}(\Phi, \partial_\mu \Phi, t)}{\partial(\partial_0\Phi)}$$
This one is almost immediate to compute and indeed gives the answer you have in #1. But, otherwise, I don't know how one is supposed to define something like
$$\frac{\delta S[\Phi]}{\delta (\partial_0 \Phi)}$$
At first, something like that seems nonsensical to me, but maybe there's a way to properly define such a thing...

 

[etotheipi]

Yeah, it's a bit strange. Landau & Lifshitz explained similar sorts of things, e.g. varying the action with respect to the co-ordinates (i.e. considering only admissible trajectories between $a \longrightarrow b$ and keeping $(\delta x^i)_a = 0$ but $(\delta x^i)_b$ variable) then you can show that$$\delta S = -mcu_i \delta x^i \implies p_i = - \frac{\partial S}{\partial x^i}$$so the time-component of the momentum four-vector would just be $p_t = - \partial S / \partial t$.

I figured the motivation for $\Pi(x) = \delta S / \delta (\partial_t \Phi)$ in the OP would be analogous to that but I don't know how to formalise that at all. I don't know any field theory, really

 

[Gaussian97]

Indeed, now that I think about it, since functional derivatives have the chain rule, maybe we can write it like
$$\frac{\delta S[\Phi]}{\delta (\partial_0 \Phi)} = \frac{\delta S[\Phi]}{\delta \Phi}\frac{\delta \Phi}{\delta (\partial_0 \Phi)}$$
Then the first derivative is the usual one that is used to derive E-L equations, the problem is the second term.
We know that $$\frac{\delta (\partial_0 \Phi)(y)}{\delta \Phi(x)}=-\partial_0 \delta^{(4)}(x-y),$$ but I don't know if there exists a property for functional derivatives that allow us to invert this, similar to normal derivatives.