What is the "momentum density" of an electromagnetic field? How does the Poynting vector relate to the momentum "stored" in such field?

I'm taking introductory, undergraduate-level E&M, for which we're following Griffiths. In his chapter on the conservation laws, he gives the following as the statement for conservation of momentum under electrodynamics:

$$\frac{d\vec{p}_{\textrm{mech}}}{dt} = \oint_S \overleftrightarrow{T}d\vec{a} -\epsilon_0 \mu_0 \frac{d}{dt} \int_V \vec{S} \; d\tau$$

Griffiths claims (without much elaboration) that the term involving Maxwell's stress tensor $$\overleftrightarrow{T}$$ yields the "momentum per unit time flowing through the surface" (which is believable, on a purely intuitive basis), and that the term involving $$\vec{S}$$ (the Poynting vector) yields the "momentum stored in the fields".

While I'm able to follow the mathematical derivation leading to this result without much trouble, I'm struggling to grasp on an intuitive level how an integral over the Poynting vector leads to a measure of some kind of "momentum stored". I'm more accustomed to thinking about the Poynting vector as a measure of energy flux (i.e. energy entering or leaving a region). It also isn't clear to me what a time derivative of such an integral "does".

Some similar questions exist, and several mention the momentum density $$g$$, which involves the Poynting vector $$\vec{S}$$, often pointing to this resource letter by Griffiths. None, however, elaborate on what precisely is meant by "momentum density", nor why this is related in any manner to the Poynting vector, except to say that "it must be" in order for conservation of momentum to work out.

In a more intuitive sense, what is going on here?

• An excellent read on this is §32 of Landau&Lifshitz 'The Classical Theory of Fields'. You can find it online. Commented Jan 19 at 20:18

It may be simpler to shift to the left the integral over the volume which involves the Poynting vector and which represents the momentum associated with the field inside the volume.

The term inside this integral is therefore the momentum density associated with the field.

In the end, the equation simply states that the derivative of the total momentum (particle and field) inside the volume is equal to the force acting on the boundary surface of the system.

Hope it can help and sorry for my poor english.

One way to gain intuition for how the volume integral over the Poynting vector represents momentum is to do dimensional analysis. The volume integral of the Poynting vector

$$\int_V \vec{S} \; d \tau$$

has dimensions: $$\frac{energy}{area \cdot time} \cdot volume = energy \cdot \frac{length}{time} = energy \cdot velocity.$$ Additionallly, $$\mu_0 \epsilon_0$$ has units of $$\frac{1}{(velocity)^2}$$, since $$\mu_0 \epsilon_0 = \frac{1}{c^2}$$. This gives that the units of $$\epsilon_0 \mu_0 \int_V \vec{S} \; d \tau$$ are $$\frac{energy}{velocity}.$$ These are the same units as momentum, which can be seen by considering that kinetic energy ($$\frac{1}{2} m v^2$$) divided by velocity has the same units as linear momentum, $$p=mv$$. Dimensionally, then, it is clear that $$\epsilon_0 \mu_0 \int_V \vec{S} \; d \tau$$ represents momentum, and its time derivative represents force.

In other words, the volume integral of the Poynting vector gives the energy of an EM wave times its velocity, which is $$c$$ in free space. The prefactor of $$1/c^2$$ converts the quantity to the energy of the EM wave divided by its velocity, which is the same as its momentum. The momentum density $$\vec{g}$$ is then defined as the momentum per volume, which is simply the quantity above before the volume integral: $$\vec{g} = \frac{1}{c^2}\vec{S}.$$

The term $$\frac{1}{c^2}\frac{d}{dt}\int_V \vec{S} \; d \tau = \frac{d\vec{p}_{\mathrm{field}}}{dt}$$ thus represents the rate of change of the momentum of the fields within the volume V. The full equation can be rewritten as:

$$\frac{d}{dt} (\vec{p}_{\mathrm{mech}} + \vec{p}_{\mathrm{field}}) = \oint_S \mathbf{T} \cdot d\vec{a}.$$

The surface integral of the Maxwell stress tensor gives the force on the surface, which is, as Griffiths writes, the momentum per time flowing in through the surface. $$\frac{d}{dt} (\vec{p}_{\mathrm{mech}} + \vec{p}_{\mathrm{field}}) = \frac{d}{dt} \vec{p}_{\mathrm{in}}.$$

So overall, the equation says that, within some volume V, the rate of change of the momentum of the charges inside plus the rate of change of the momentum stored in the EM fields, is equal to the rate of change of the momentum flowing into the volume. If there are no charges in the volume, the equation simplifies to:

$$\frac{d \vec{p}_{\mathrm{field}}}{dt} = \frac{d \vec{p}_{\mathrm{in}}}{dt}$$

which says that the rate of change of the momentum stored in the fields within the volume is equal to the rate of change of the momentum flowing into the surface. If the fields are traveling out of the volume, the flow of momentum is out of the surface, so $$\frac{d}{dt} \vec{p}_{in}$$ is negative, as expected.