# Is the additional term in the canonical momentum exactly equal to the momentum of the electromagnetic field?

The canonical momentum of a particle in an electromagnetic field is given by $$\textbf{P}=m\textbf{v}+q\textbf{A}$$ Is the term $$q \textbf{A}$$ equal to the momentum of the electromagnetic field (which would be $$\mathbf{P}_\text{field} = \int \epsilon_0 \left(\textbf{E}\times\textbf{B}\right) dV$$ ) ?

Otherwise, where is this part of the momentum stored?

When we write the rate of change of momentum in a region in terms of the stress energy tensor, we write: $$\frac{d}{d t}\left(\mathbf{P}_{\text {mech }}+\mathbf{P}_{\text {Field}}\right)_{\mathrm{\alpha}}=\oint_{S} \sum_{\beta} T_{\mathrm{\alpha} \mathrm{\beta}} n_{\mathrm{\beta}} d a$$ (this is equation 6.122 from section 6.7 of Jackson's electrodynamics book) where $$\textbf{P}_{\text {mech }} = m \dot{\textbf{v}}$$.

This equation does not take into account the change of momentum due to change in $$q\textbf{A}$$, unless this term is the momentum in the field itself.

Yes, you are right. $$qA$$ is called the field momentum. And it is indeed equal to the momentum carried by the electromagnetic field. It can be derived from the definition of Poynting vector, where you try to express its contribution to the total electromagnetic field energy.

Actually, it can be expressed using a wide variety of expressions, given by

$$\boxed{\mathbf{P_{EM}= qA = \int_V \rho A \, dV = \epsilon \int (E \times B) \, dV}}$$

As mentioned by @ChiralAnomaly in the comments, this assumes the validity of the Coulomb gauge $$\mathbf{\nabla \cdot A =0}$$. This renders the vector potential to be a bit unrealistic but it is good enough for semi-classical calculations. In general, $$\mathbf{\int_V \rho A \, dV = \epsilon \int (E \times B) - E \,(\nabla \cdot {A}) \, dV}$$.

• Might be worth mentioning that this assumes the gauge condition $\nabla\cdot\mathbf{A}=0$. Commented May 29, 2020 at 16:23
• @ChiralAnomaly Nice point. Commented May 29, 2020 at 17:33
• This is all wrong, see my answer. Commented Aug 31, 2023 at 15:30

The canonical momentum of a particle in an electromagnetic field is given by $$\textbf{P}=m\textbf{v}+q\textbf{A}$$ Is the term $$q \textbf{A}$$ equal to the momentum of the electromagnetic field (which would be $$\mathbf{P}_\text{field} = \int \epsilon_0 \left(\textbf{E}\times\textbf{B}\right) dV$$ ) ?

No. In the Lagrangian and Hamiltonian formulation of equations of motion of a particle moving in external field, $$\mathbf A$$ refers to external field experienced by the particle, that is, field due to sources other than the particle which is being described. Thus it is not the total EM vector field $$\mathbf A_{total}$$.

Thus the product $$q\mathbf A$$ cannot be expressed as

$$q\mathbf A_{total}$$ or $$\int \rho \mathbf A_{total}dV.\tag{*}$$ At best, $$q\mathbf A$$ is a contribution to the integral (*), but it is not equal to it.

The latter integral may sometimes give the same value as EM momentum in the whole space

$$\int\frac{\mathbf E \times \mathbf B}{\mu_0}dV,$$ but that requires that 1) we use the Coulomb potential $$\mathbf A_C$$; 2) integration goes over the whole space, including the sources of the field, and there is no radiation at the infinity; or the integration goes over a region of space which has a special boundary where the electric field or vector potential vanishes (such as when the boundary is entirely in a perfect conductor).

Otherwise, where is this part of the momentum stored?

EM momentum is distributed in space, and we can ascribe it to the system of particles that generates EM field, or separately to the EM field itself.

When we write the rate of change of momentum in a region in terms of the stress energy tensor, we write: $$\frac{d}{d t}\left(\mathbf{P}_{\text {mech }}+\mathbf{P}_{\text {Field}}\right)_{\mathrm{\alpha}}=\oint_{S} \sum_{\beta} T_{\mathrm{\alpha} \mathrm{\beta}} n_{\mathrm{\beta}} d a$$ (this is equation 6.122 from section 6.7 of Jackson's electrodynamics book) where $$\textbf{P}_{\text {mech }} = m \dot{\textbf{v}}$$.

This equation does not take into account the change of momentum due to change in $$q\textbf{A}$$, unless this term is the momentum in the field itself.

Change of $$q\mathbf A$$ does not imply momentum is changing. For example, consider charged particle moving outside a very long solenoid in which constant current flows, so the solenoid creates external field $$\mathbf A$$ outside the solenoid. As the particle moves away from the solenoid, it experiences lower and lower value of $$\mathbf A$$, yet it experiences zero EM force, because both magnetic and electric field vanishes outside the solenoid. Both momentum of the particle and momentum of the field are thus constant in time.

• $\vec{A}$ refers to external field experienced by the particle, that is, field due to sources other than the particle which is being described. Is it really necessary to distinguish between the sources of $\vec{A}$? At the fundamental level (in QFT), the Lagrangian has a term ${j}_\mu {A}^\mu$, which does not distinguish between which part of $A^\mu$ is external, and which part is locally produced by $j_\mu$. Commented Aug 31, 2023 at 16:21
• @ArchismanPanigrahi It is if we want to have Lagrangian/Hamiltonian for point particles, which is the context of the question. They are two different fields - $\mathbf A$ is finite and continuous where the particle is, so terms like $\mathbf v\cdot \mathbf A$ are defined. Field $\mathbf A_{total}$, on the other hand, is singular where the particle is, and thus $\mathbf v\cdot \mathbf A_{total}$ is not defined. If sources are continuous without singularities, working with total field $\mathbf A_{total}$ may be sufficient. But such formulation cannot describe systems with point particles. Commented Aug 31, 2023 at 16:29
• @ArchismanPanigrahi Lagrangian term $j_{\mu}A^\mu$ works fine for point particles, only if $A$ is the external field. If $A$ in this expression is the total field, the Lagrangian works only for regular charge and current distributions, "distributed as a fluid". Commented Aug 31, 2023 at 16:36