3
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.