# 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

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 ($$H = \dot qp-L$$, $$p =$$ 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!

• But isn't the kinetic momentum the canonical momentum of a free particle? Is the kinetic momentum anything more than just the canonical momentum? – JGBM Jan 19 at 22:07

This is a question I've been playing with myself a lot lately! But the simplest way to describe it is that they are simply different quantities. The canonical momentum is the derivative of the Lagrangian function with respect to the generalized velocities. The kinetic momentum is essentially just the "normal" momentum of a free particle with no interactions. In classical mechanics the kinetic momentum is just the mass times the velocity. In general, these two quantities do not have to be the same at all. For an example, see Charlie's answer where the Lagrangian includes terms describing electromagnetic interactions. For a completely free particle, however, these two different definitions of momentum are exactly the same. The importance of the canonical momentum is really in how it allows us to connect to electromagnetism, which also lets us include these effects in the Hamiltonian in quantum mechanics. Hope this helps!

Canonical momentum is the momentum. Kinetic momentum is only part of the momentum. The other part is the potential momentum. For example, the magnetic part of the Lorentz force does not conserve kinetic momentum. Only the total, canonical, momentum mv+qA is conserved. A problem could be that the total momentum is not gauge invariant. There are two ways to corner this. You can just accept this, as dynamics is determined by the rate of change of momentum and this is captured by the Lorentz force expression. Alternatively, you can abandon gauge invariance as I propose in my peer reviewed paper.