# Canonical momentum density vs. energy-momentum tensor

Suppose we have a scalar field $\varphi$ with Lagrangian

$$\mathcal{L} = \frac{1}{2} \kappa \left( \frac{\partial \varphi}{\partial x} \right)^2 + \frac{1}{2} \rho \left( \frac{\partial \varphi}{\partial t} \right)^2 \,.$$

Then the canonical momentum density is

$$\pi = \frac{\partial \mathcal{L}}{\partial \dot{\varphi}} = \rho \dot{\varphi} \,.$$

Whereas the energy-momentum tensor:

$$T^\mu{}_\nu =\frac{ \partial \mathcal{L}}{\partial (\partial_\mu \varphi)} \partial_\nu \varphi - \delta^\mu_\nu \mathcal{L}$$

Has a 01 component whose interpretation (I believe) is momentum density,

$$T^0{}_1 = -\rho \dot{\varphi}\varphi' \,.$$

These two quantities don't correspond; what is going on here?

Thank you.

## 2 Answers

The two quantities don't correspond because they are conserved quantities corresponding to different symmetries. One is a symmetry from shifting your field, the other from shifting space-time itself. Here is what is going on precisely:

Let us do a simpler case first: In a particle mechanics system, let's say a free particle with $L = \frac{1}{2}m\dot{x}^2$, the "field" is simply $x(t)$. However $L$ does not explicitly depend on $x$, so we may shift our system $x$ to $x + \epsilon$ and have our overall action be unchanged. Following the Noether procedure, we make $\epsilon$ time dependent, do the variation again, and recover the conserved "canonical" momentum, the usual $m\dot{x}$. Also note that the system does not explicitly depend on time either. Varying $t$ according to Noether gives us $H$.

In the field theoretic case, recovering the canonical momentum is exactly analogous. In this case, our field is $\varphi$, so we if take $\varphi$ to $\varphi + \epsilon \psi$, i.e $\delta \varphi = \epsilon \psi$ to be our variation, we can recover the canonical momentum. The symmetry that gives you the stress-energy tensor, on the other hand, is if you shift your space-time variables. Take $x^{\mu}$ to $x^{\mu} + \epsilon^{\mu}$. This is equivalent to letting $\delta \varphi$ = $\epsilon^{\mu}\partial_{\mu}\varphi$. Proceeding with Noether gives you the stress energy tensor (this computation can be found in standard field theory textbooks).

The canonical momentum $\pi$ is not the same as certain components of the energy momentum tensor $T$. This can be seen by going over to the Hamiltonian description. gj255's action gives the Hamiltonian, $$H=\frac{1}{2}\int dx ( \rho\phi_{,0}\phi_{,0}-\kappa\phi_{,1}\phi_{,1}) \ .$$ Here the coords of the Hamiltonian formalism are $q^{i}(t)\rightarrow q^{x}(t)\rightarrow \phi(t,x)$. In the Hamiltonian formalism, the canonical momenta $p_{i}$ are invariant under canonical transformations generated by the $p^{j}$ themselves; this is clear from the Poisson bracket, $$\frac{dp_{i}}{d\epsilon}=[p_{i},p_{j}]_{PB}=0 \ .$$ In the field theory, the momenta $p_{i}(t)\rightarrow p_{x}(t)\rightarrow \pi(t,x)$ and so the $\pi(t,x)$ is invariant under canonical transformations generated by $\pi(t,y)$. $$\frac{d\pi(t,x)}{d\epsilon}=[\pi(t,x),\pi(t,y)]_{PB}=0$$ This is completely different to the energy momentum tensor $T$ which is derived from the fact that the Lagrangian density $\mathcal{L}=\mathcal{L}(\phi,\phi_{,\mu})$ has no explicit dependence on $x$ and $t$.

This answer is essentially the same as that of dayareishq but it uses the Hamiltonian formalism in place of Noether's.