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?


2 Answers 2


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.

  • 1
    $\begingroup$ Hello, welcome to Physics SE! Please note that this space is strictly for answers, this kind of comment should be posted as such. $\endgroup$ Nov 29, 2020 at 10:31
  • $\begingroup$ Hello Mauro! I actually thought that I answered the question, didn‘t I? $\endgroup$ Nov 29, 2020 at 12:40
  • $\begingroup$ I edited the answer according to your comments. $\endgroup$ Nov 30, 2020 at 6:07
  • $\begingroup$ I did some more reasearch. I now think that Carroll, just took 9.50 and rewrote it covariantly, assuming one can write "any" physics equation in covariant form.. But I do not know if this really is right or not.. $\endgroup$ Nov 30, 2020 at 6:33

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