# Conjugate momentum for scalar field in curved spacetime

The Lagrange density $$\mathcal{L}$$ of a scalar field $$\phi$$ in curved spacetime is $$\mathcal{L}=\sqrt{-g}(-\frac{1}{2} g^{\mu\nu}\nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2-\xi R\phi^2)\tag{9.87}$$ where $$g$$ is metric determinant, $$\xi$$ is a constant and $$R$$ is the curvature scalar, and the signature convention is $$(-,+,+,+)$$.

I read from Sean Carroll's Spacetime and Geometry book, pg 394-395, that the conjugate momentum $$\pi$$ is given by $$\pi=\frac{\partial\mathcal{L}}{\partial(\nabla_0\phi)}\tag{9.90}$$ and $$\pi=\sqrt{-g}\nabla_0\phi.\tag{9.91}$$

How can one show (9.91) without knowing the $$g^{\mu\nu}$$ components?

OP has a point: Eq. (9.91) should read

$$\pi~=~-\sqrt{-g}g^{0\mu}\nabla_{\mu}\phi~=~-\sqrt{-g}\nabla^0\phi,\tag{9.91'}$$ i.e. the 0-index should be upstairs.

To prove equation 9.91 globally we will first prove it locally, the generalization is then straightforward.

We take a point $$p$$ of the spacetime-manifold $$M$$, at which a tangent space $$T_p$$ is defined. The scalar fields $$\phi(x^{\mu})$$ are then defined with respect to a coordinate system constructed from the basis vectors of $$T_p$$.

Now, in general relativity, curved spacetime looks locally like Minkowski space + a gravitational force (Equivalence Principle). So we can construct Riemann Normal Coordinates $$x^{\hat{\mu}}(p)$$ satisfying:

$$g_{\hat{\mu} \hat{\nu}}(p) = \eta_{\hat{\mu} \hat{\nu}}, \partial_\hat{\sigma} g_{\hat{\mu} \hat{\nu}}(p) = 0.$$

These coordinates are called locally inertial coordinates (see eq 2.47 in Carroll's book).

Next relabel $$g_{\hat{\mu} \hat{\nu}}$$ to $$g_{\mu\nu}$$ to avoid confusion. Then, we show that equation 9.91 holds in these coordinates:

$$\pi = \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{\mu\nu} \nabla_\mu \phi \nabla_\nu \phi - \cdots \})$$ $$= \frac{\partial}{\partial(\nabla_0 \phi)} ( \sqrt{-g} \{-\frac{1}{2} g^{00} \nabla_0 \phi \nabla_0 \phi + g^{0i} \nabla_0 \phi \nabla_i \phi + g^{i0} \nabla_i \phi \nabla_0 \phi + g^{ij} \nabla_i \phi \nabla_j \phi - \cdots \})$$ 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 to prove this statement. Then: $$\pi = \sqrt{-g} \nabla_0 \phi.$$ Where the metric is put in its canonical form $$g_{\mu \nu} = diag(-1, +1, +1, +1)$$ as we are using locally inertial coordinates.

Finally, as $$\pi = \sqrt{-g} \nabla_0 \phi$$ is a tensorial equation (as the covariant derivative of a scalar field is independent of the used coordinate system) so 9.91 is globally true.

It should be noted that $$g^{00} = g_{00} = -1$$ and by this equation 9.91 is true.

• Hello, welcome to Physics SE! Please note that this space is strictly for answers, this kind of comment should be posted as such. Nov 29 '20 at 10:31
• Hello Mauro! I actually thought that I answered the question, didn‘t I? Nov 29 '20 at 12:40