Deriving equation 9.91 in Carroll's Book "Spacetime and Geometry"

[silverwhale]

Homework Statement

Derive the following formula in page 395 in Sean Carrolls Book "Spacetime and Geometry"

Relevant Equations

$\pi = \sqrt{-g} \nabla_0 \phi.$

Hello dear PhysicsForums attendees!

I tried to solve for somebody the aforementioned problem. But I am not sure if my attempt is correct. So I am writing down what I suggested.

Looking at eq 2.46 in Carrolls book; The metric is Lorentzian in General Relativity so that $g^{\mu \nu} = diag(-1,+1, ...,+1)$. This is the canonical form of the metric, where the signatures are, as written, -1 for the time coordinate and +1, ...,+1 for the spatial coordinates and where the metric is diagonal.

Next, we say that at any point p element of the manifold M there exists a coordinate system where the metric is canonical and further the following equation holds (eq 2.47 in Carrolls book):
$$g_{\hat{\mu} \hat{\nu}}(p) = \eta_{\hat{\mu} \hat{\nu}}.$$
We say the coordinate system at p is locally intertial. This is very important because we are assuming that locally curved space looks like flat space. An assumption which, in General Relativity, must be fulfilled.

Following equation 3.66 in Carrolls book, we have:
$g_{\hat{\mu} \hat{\nu}} = g(\hat{e}_{(\hat{\mu})},\hat{e}_{(\hat{\nu})})$, and by using formulae 3.64, 3.65 and 3.66 we have an actual algorithm on how to rewrite the metric into one defined on a coordinate system that is locally inertial. It should be noted that the $\hat{e}_{(\hat{\mu})}$ are basis vectors of the tangent space $T_p$.

Now, by taking equation 2.47, it is easy to derive eq 9.91 in Carrolls book:

$$\pi = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{\mu\nu} \nabla_\mu \phi \nabla_\nu \phi - \cdots \})$$
$$ = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{00} \nabla_0 \phi \nabla_0 \phi + g^{0i} \nabla_0 \phi \nabla_i \phi + g^{i0} \nabla_i \phi \nabla_0 \phi + g^{ij} \nabla_i \phi \nabla_j \phi - \cdots \})$$
by only looking at the first term, as the rest does not depend on $\nabla_0 \phi$, we further derive:
$$ \frac{\partial}{\partial(\nabla_0 \phi)} (g^{00} \nabla_0 \phi \nabla_0 \phi) = 2 g^{00} \nabla_0 \phi.$$
Just use the Leibniz rule. Then put this into the starting equation to get:
$$\pi = \sqrt{-g} \nabla_0 \phi.$$

We are actually done. But by assuming that the metric is Minkowskian locally we can calculate easily its determinant, which should be -1. Such that the $\sqrt{-g}$ term is equal to 1. So finally we get:
$$\pi = \nabla_0 \phi.$$

Please note that $g^{\mu\nu}$ is defined here to be $g^{\hat{\mu} \hat{\nu}}$ which is just a shorthand for $g^{\hat{\mu} \hat{\nu}}(p)$. So to make the computation clear.

As a notice, it is not clear why the author wrote the term $\sqrt{-g}$ explicitly in equation 9.91 ..

 

[etotheipi]

\begin{align*}\pi &= \frac{\partial \mathcal{L}}{\partial(\nabla_0 \phi)} \\

&= -\frac{1}{2}\sqrt{-g} g^{\mu \nu} \cdot 2 \delta^0_{\nu} \partial_{\mu} \phi \\

&= -\sqrt{-g} g^{\mu 0} \nabla_{\mu} \phi \\

&= -\sqrt{-g} \nabla^{0} \phi \\

&= \sqrt{-g} \nabla_0 \phi

\end{align*}