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

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  • $\begingroup$ 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? $\endgroup$ – ACuriousMind Dec 27 '16 at 22:40
  • $\begingroup$ @ACuriousMind Why and how is canonical momentum used in classical field theory; and does it have a different physical meaning than regular momentum? $\endgroup$ – Stoby Dec 27 '16 at 22:46
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    $\begingroup$ Doesn't e.g. the corresponding section of the Wikipedia article sufficiently answer that? $\endgroup$ – ACuriousMind Dec 27 '16 at 22:47
  • $\begingroup$ @ACuriousMind it does, thank you for pointing me in that direction $\endgroup$ – Stoby Dec 27 '16 at 22:51
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Regular momentum (or "kinematic momentum") is the mass times the actual velocity of the particle (for the particle case of course). Canonical momentum is that which is conjugate to a spatial degree of freedom (i.e. it is that derivative of the lagrangian with respect to a velocity term). Canonical momentum is important for a couple of reasons but I would argue that it's greatest use is in constructing the hamiltonian (H = qp-L, p = canonical momentum). My favorite example to highlight the difference in canonical and kinematic momentum is the lagrangian for a charged particle:

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.

Now when discussing classical field theory the generalization is straightforward. Kinematic momentum becomes kinematic momentum density and canonical momentum becomes canonical momentum density. The kinematic momentum density is just the density times the velocity term and the canonical momentum density is the derivative of the lagrangian density with respect to a velocity term. An important example to study here is E&M field lagrangian density:

$$ L = -\frac{1}{4}F^{\mu \nu} F_{\mu \nu}, $$ where F is the electromagnetic field tensor. Go ahead and try to find the momentum densities and construct the hamiltonian as practice!

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