Conservation of Momentum in Classical Field Theory

[fayled]

In classical field theory, translational (in space and time) symmetry leads the derivation of the energy-momentum tensor using Noether's theorem.

From this it is possible to derive four conserved charges. The first turns out to be the Hamiltonian, and thus we have energy conservation.

The remaining three turn out to be
Pⁱ=-∫d³xπ^a∂_iφ_a
where φ_a are fields and π^a are their conjugate momentum densities. It is often stated that these are the three (physical) momentum components, and so we have conservation of momentum. Is there an easy way to see that this is true?

 

[samalkhaiat]

...
Is there an easy way to see that this is true?

If you are asking to prove that \frac{d}{dt}P^{i} = 0 , then the answer to your question is an easy yes. Translation invariance means that the Lagrangian, \mathcal{L}(\varphi_{a},\dot{\varphi}_{a},\partial_{k}\varphi_{a}) , has no explicit x^{k}-dependence, i.e., \partial^{i}\mathcal{L} = 0 . This implies that \int d^{3}x \ \partial^{i}\mathcal{L} = 0 .

The rest is just algebra: Calculate \partial^{i}\mathcal{L} and use the equation of motion to obtain

0 = \int d^{3}x \ \partial^{i}\mathcal{L} = \frac{d}{dt} \int d^{3}x \ \frac{\partial \mathcal{L}}{\partial \dot{\varphi_{a}}} \ \partial^{i}\varphi_{a} + \int d^{3}x \ \partial_{k}\left( \frac{\partial \mathcal{L}}{\partial \partial_{k} \varphi_{a}} \ \partial^{i}\varphi_{a}\right) . The second term vanishes by the divergence theorem, and the first term is just \frac{d}{dt}P^{i}.

 

[fayled]

If you are asking to prove that \frac{d}{dt}P^{i} = 0 , then the answer to your question is an easy yes. Translation invariance means that the Lagrangian, \mathcal{L}(\varphi_{a},\dot{\varphi}_{a},\partial_{k}\varphi_{a}) , has no explicit x^{k}-dependence, i.e., \partial^{i}\mathcal{L} = 0 . This implies that \int d^{3}x \ \partial^{i}\mathcal{L} = 0 .

The rest is just algebra: Calculate \partial^{i}\mathcal{L} and use the equation of motion to obtain

0 = \int d^{3}x \ \partial^{i}\mathcal{L} = \frac{d}{dt} \int d^{3}x \ \frac{\partial \mathcal{L}}{\partial \dot{\varphi_{a}}} \ \partial^{i}\varphi_{a} + \int d^{3}x \ \partial_{k}\left( \frac{\partial \mathcal{L}}{\partial \partial_{k} \varphi_{a}} \ \partial^{i}\varphi_{a}\right) . The second term vanishes by the divergence theorem, and the first term is just \frac{d}{dt}P^{i}.

I'm happy that this quantity is conserved - It just isn't very clear that it is in fact the momentum.

 

[samalkhaiat]

Then you should consider some concrete Lagrangians (for example the Lagrangian of electromagnetic, long elastic rod, sound wave, etc.). Or you can use the argument of relativistic covariance.

 

[vanhees71]

Well, it's momentum by definition, because the conserved quantity that's conserved because of spatial translations is called momentum (as for temporal translations it's called energy).