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What does the canonical momentum $\textbf{p}=m\textbf{v}+e\textbf{A}$ mean? Is it just momentum accounting for electromagnetic effects?

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To clarify, are you looking for physical intuition? For example, the magnitude of the standard, non-relativistic momentum of a particle tells you "how difficult it would be to stop the particle" in a sense that can be made precise using the concept of impulse. Are you looking for something analogous to this but for canonical momentum as opposed to, say, a discussion of how it mathematically arises from a Lagrangian for electrodynamics? –  joshphysics Jul 24 '13 at 22:27
Intuitively, is it the entire momentum of asystem? So magnetic fields have momentum? –  cpc333 Jul 25 '13 at 1:53

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In Lagrangian mechanics, "momentum" is just a conserved quantity, and is the derivative of the Lagrangian with respect to velocity ($\frac{dL}{d\dot{q}}$). For the case of a point charge traveling through a uniform magnetic field $\mathbf{B}$, $\mathbf{p} = m \mathbf{v}$ simply isn't conserved anymore, as the charge travels in a circular path due to the magnetic field, causing its momentum to constantly change direction. A quantity known as the canonical momentum, $\mathbf{P} = m\mathbf{v} + e \mathbf{A}$ ends up being conserved throughout the charged particle's trajectory. (Setting the total time derivative of the canonical momentum equal to zero simply results in $m \mathbf{a} = e \mathbf{v} \times \mathbf{B}$, which is just the expression for magnetic force.) In short, the canonical momentum is simply "the quantity that is conserved" in electromagnetic interactions, while the kinetic momentum is just the product of mass and velocity.

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It is not true that the canonical momentum is "conserved in electromagnetic interactions." It satisfies the Euler-Lagrange equation of motion, $\frac{d}{dt}\frac{dL}{d\dot{q}} - \frac{dL}{dq}$. For instance, in a uniform magnetic field in the Landau gauge, $\mathbf{A} = -By\mathbf{\hat{x}}$ and the Lagrangian is $L = \frac{1}{2}mv^2 - e\Phi(x) + e \mathbf{v}.\mathbf{A}$, so $\frac{dP_y}{dt} = e \dot{x} B \neq 0$. –  Ted Pudlik Jul 2 at 19:55

Yes, it accounts for the effect of the vector potential on a moving charge. But there is also a more fundamental role that it plays: assuming position-independent vector potential, the canonical momentum is a conserved quantity, while the "normal" (or kinetic) momentum (mass times velocity) is not.

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The canonical momentum is not conserved. Check with the Lagrangian $L= \frac{1}{2}m \vec v^2 - q\Phi(x) + q \vec v .\vec A(x)$ and Euler-Lagrange equations –  Trimok Jul 25 '13 at 10:09
I assumed position-independent vector potential. –  Frederic Brünner Jul 25 '13 at 10:35
But this means a zero electric and magnetic field. –  Trimok Jul 25 '13 at 10:38
Magnetic yes, but the electric field is not necessarily zero. –  Frederic Brünner Jul 25 '13 at 10:42
No, you have to choose $\Phi=$Constant in order to have a conserved canonical momentum, so the electric field is zero too. –  Trimok Jul 25 '13 at 10:44

The whole problem starts when you try to do electromagnetism with the Lagrangian because you can't write the magnetic field in terms of a potential. However we CAN write it in terms of a vectorpotential $\vec{A}$:

$\vec{B} = \nabla\times\vec{A}$.

It seems that this is usefull and can be used to derive the appropriate Lagrangian and Hamiltonian which are given and checked here.

It seems (from the calculations given in the link) that to include the magnetic field, we need to replace our momentum with:


By replacing the momentum by this term, you are able to do Lagrangain and Hamiltonian mechanics (which work with potentials) for magnetic fields (which can't be written in terms of a potential).

For electric fields you can still include them by using the electronic potential.

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The canonical (total) momentum is the sum of the kinetic (mechanical) momentum and the potential momentum. Potential momentum occurs only if the potential energy depends explicitly on velocity.

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