What is the difference between kinetic momentum $p=mv$ and canonical momentum?

What is the difference, if any, between kinetic momentum $p=mv$ and canonical momentum? Why is canonical momentum important (specifically to classical field theory)?

• Can you be more specific what exactly you want to know? If you know these two names you should be able to see the difference between the definitions for yourself, what is unclear/confusing or what do you think is missing? – ACuriousMind Dec 27 '16 at 22:40
• @ACuriousMind Why and how is canonical momentum used in classical field theory; and does it have a different physical meaning than regular momentum? – Stoby Dec 27 '16 at 22:46
• Doesn't e.g. the corresponding section of the Wikipedia article sufficiently answer that? – ACuriousMind Dec 27 '16 at 22:47
• @ACuriousMind it does, thank you for pointing me in that direction – Stoby Dec 27 '16 at 22:51

The Lagrangian for a charged particle in an arbitrary Electric and Magnetic field can be written as: $$L = \frac{1}{2}m |\dot{\vec{q}}|^2 - e \phi + e \vec{A}\cdot \dot{\vec{q}}.$$
Of course the kinematic momentum is just $m \frac{d\vec{q}}{dt}$. The canonical momentum is $\frac{\partial L}{\partial\dot{\vec{q}}}$ which is equal to $\vec{p}=m \dot{\vec{q}}+e\vec{A}$. We use this to write the hamiltonian: $$H = \frac{1}{2m}|{\vec{p}}-e\vec{A}|^2 - e \phi.$$ Notice that the magnetic field thus effectively contributes nothing to the energy (Hamiltonian) which is good because the magnetic field does no work on charged particles.
$$L = -\frac{1}{4}F^{\mu \nu} F_{\mu \nu},$$ where F is the electromagnetic field tensor. Go ahead and try to find the momentum densities and construct the hamiltonian as practice!