Electromagnetic field or direct interactions between charged particles?

Question

Consider a small distribution of charged particles enclosed by an arbitrary volume $V$ with boundary $S$.

It can be shown that the total mechanical momentum of the particles, $\mathbf{P_{mech}}$, obeys the equation:

$$\frac{\partial}{\partial t}\left(\mathbf{P_{mech}+\mathbf{P_{em}}}\right)=\int_S \mathbf{\overset{\leftrightarrow}{T}}\cdot d\mathbf{S}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

where $\mathbf{\overset{\leftrightarrow}{T}}$ is the Maxwell stress tensor which is a $3\times 3$ matrix with elements defined by:

$$T_{ij} = \epsilon_0\left(E_iE_j-\frac{1}{2}\delta_{ij}E^2\right)+\frac{1}{\mu_0}\left(B_iB_j-\frac{1}{2}\delta_{ij}B^2\right)$$ and $\mathbf{P_{em}}$ is the total electromagnetic field momentum in the volume $V$ given by:

$$\mathbf{P_{em}}=\epsilon_0 \int_V \mathbf{E}\times \mathbf{B}\ dV$$

Let us explore the meaning of equation (1) by varying the arbitrary volume $V$. Let us make the reasonable assumption that we only have a finite distribution of charged particles. In that case the components of the $\mathbf{E}$ and $\mathbf{B}$ fields must drop to zero at large distances. Therefore if we extend the enclosing volume $V$ so that its boundary goes to infinity then the surface integral of the Maxwell stress tensor must go to zero. Therefore equation (1) simply becomes:

$$\frac{\partial}{\partial t}\left(\mathbf{P_{mech}+\mathbf{P_{em}}}\right)=0$$

Thus the total momentum of the particles and the total momentum of the electromagnetic field, evaluated over all space, is conserved. In other words the state of the electromagnetic field simply mirrors the state of the particles. Also we have a global momentum conservation condition rather than a local one. Perhaps this does not represent a "fundamental" description of the system?

Now let us shrink the volume $V$ right down to the size of the distribution of charged particles. Since the volume $V$ is small then the total electromagnetic momentum in that volume will be small. Thus equation (1) now becomes:

$$\frac{\partial}{\partial t}\mathbf{P_{mech}}=\int_S \mathbf{\overset{\leftrightarrow}{T}}\cdot d\mathbf{S}$$

Thus the total force on the particles, $\mathbf{F}=\partial \mathbf{P_{mech}}/\partial t$, is provided by the surface integral of the Maxwell stress tensor in the region of the particles. This is equivalent to the local electromagnetic field acting on the particle charges through the Lorentz force law.

Now my question is this: if we take the second volume definition how is momentum conserved locally?

The Maxwell stress tensor supplies mechanical momentum to the particles but where does that momentum come from? It cannot come from the electromagnetic field itself as our second definition of the volume $V$ is so small that there is very little electromagnetic field momentum inside it.

I think the momentum supplied to the particles must come directly from the particles themselves. This makes sense as the system is closed - there are no external forces acting on the particles.

Thus we have:

$$\frac{\partial}{\partial t}\mathbf{P_{mech}}=\int_S \mathbf{\overset{\leftrightarrow}{T}}\cdot d\mathbf{S}=0$$

Thus in this description of the system, which is perhaps more fundamental as momentum is conserved locally, only the particles themselves carry that momentum. The local electromagnetic fields transfer energy-momentum between the particles but do not contain energy-momentum themselves. It could be that for each transfer of energy-momentum from particle A to B there is a simultaneous transfer of negative energy-momentum from B to A. Thus the net energy-momentum in the local field is zero.

So perhaps electromagnetic fields, as such, do not exist and instead there are only direct electromagnetic interactions between charged particles?

Answer 1

The problem here is in the way you take the volume to be small.

You assert that $$ \int \mathbf E \times \mathbf B \mathrm dV \to 0$$ while $$ \int_S \mathbf T \cdot \mathrm d\mathbf S \neq 0.$$ But this limit is not well-defined.

$\mathbf S$ is the surface of the volume over which you integrate and will vanish as well in your limit $V \to 0$. If you shrink the volume so much that the momentum of the EM field is negligible, the force it exerts on your particle is negligible too.

Answer 2

You assert that total electromagnetic momentum is small in a small region. Even when this is so, it is not right either morally or mathematically to then throw away a term like $\frac{d}{dt} \vec{P}_{em}$ because something can have a very large time derivative even if it is quite small. A stiff spring oscillating with a small amplitude can have large velocities.

Adding it to the mechanical momentum won't change this. Even when the field momentum is small compared to the mechanical momentum, the field momentum could be changing much more quickly. In general you have to be careful about approximating before you subtract, and a derivative involves a subtraction.

So $\frac{d}{d t}\int_V\mathscr{\vec P}_\text{mech}+\vec{\mathscr{P}}_\text{em} dxdydz=\int_S \overset{\leftrightarrow}{T}\cdot d\vec{S},$ is correct, where $\mathscr{\vec P}$ is momentum density and the volume is stationary, whereas $\frac{d}{d t}\int_V\mathscr{\vec P}_\text{mech} dxdydz\neq \int_S \overset{\leftrightarrow}{T}\cdot d\vec{S},$ no matter how small the volume.

Keep in mind that you can also have a large flux of field momentum even when the field momentum is small (or zero). This comes up in your next question.

The Maxwell stress tensor supplies mechanical momentum to the particles but where does that momentum come from?

The stress tensor measures the flux of momentum. Recall how $\frac{\partial \rho}{\partial t}=-\vec\nabla \cdot \vec J.$ The $\vec J$ measures the flux of charge, the per unit area per unit time flow of charge, it is pretty much unrelated to charge density in the sense that tif you had a charge distribution $\rho_1(x,y,z,t)$ with a certain current density $\vec J$ then the charge distribution $\rho_2(x,y,z,t)=A+\rho_1(x,y,z,t)$ could have the exact same current density $\vec J.$ So the charge density could be positive,negative, large or small or zero and it tells you nothing about the current.

Now, the same thing happens for momentum. Instead of $\rho$ (a density) having $\vec J=(J_x,J_y,J_z)$, a current density (flux), that satisfies

$$\frac{\partial \rho}{\partial t}+\frac{\partial J_{x}}{\partial x}+\frac{\partial J_{y}}{\partial y}+\frac{\partial J_{z}}{\partial z}=0,$$

you instead have a density like $\mathscr P_x$ with has a corresponding flux $(T_{xx},T_{yx},T_{zx})$ that satisfies

$$\frac{\partial \mathscr P_x}{\partial t}+\frac{\partial T_{xx}}{\partial x}+\frac{\partial T_{yx}}{\partial y}+\frac{\partial T_{zx}}{\partial z}=0,$$ just like for charge density and its flux. And similarly

$$\frac{\partial \mathscr P_y}{\partial t}+\frac{\partial T_{xy}}{\partial x}+\frac{\partial T_{yy}}{\partial y}+\frac{\partial T_{zy}}{\partial z}=0$$

$$\frac{\partial \mathscr P_z}{\partial t}+\frac{\partial T_{xz}}{\partial x}+\frac{\partial T_{yz}}{\partial y}+\frac{\partial T_{zz}}{\partial z}=0$$

So you can have large fluxes (stresses) even when the density (momentum) is small, just like you can have large currents even when the charge density is small.

It cannot come from the electromagnetic field itself as our second definition of the volume V is so small that there is very little electromagnetic field momentum inside it.

This is false, firstly the momentum density might not be small even in a small region, but secondly even if it were small, the flux could still be large. In the exact same way that the electric current density can be large even when the electric charge density is small (or even zero). Flux is a flow, if just as much momentum is flowing into a region as flowing out then the amount of momentum in the region remains the same, so if it was small then it stays small. if it was negative, then it stays negative. You are allowed to have a flux that extends for a long region. For instance with electric charge you could have current rushing from the edge of a disk to the center where in each region in between just as much current flows in as out, so the charge density in between can be zero. Momentum can flow the same way. There might be a region where it builds up and it might build up because it flowed into that region but the regions that adjoin it it might simply have an equal flow in as flow out and might have little or no momentum themselves. In fact momentum can go negative and negative momentum just points in the opposite direction it is just as sensible as positive momentum. Just think of three scalar densities $\mathscr P_x$, $\mathscr P_y$, and $\mathscr P_z$, that act very similar to $\rho$, that flow in a conserved way.

The only real tricky part is that electromagnetic momentum and mechanical momentum are not by themselves conserved, so it is only the total momentum that flows in a conserved way (like electric charge does). But in regions without electric charge or electric current electromagnetic momentum is conserved, so there it flows just fine. And total momentum density is a vector, that has three components $\mathscr P_x$, $\mathscr P_y$, and $\mathscr P_z$, that act very similar to $\rho$, and that do flow in a conserved way.

So perhaps electromagnetic fields, as such, do not exist and instead there are only direct electromagnetic interactions between charged particles?

Without the contribution of the electromagnetic momentum density to the total momentum density, then the total momentum density does not flow in a conserved way and momentum is not conserved. There is no stronger way to say the momentum of fields are real, and thus fields are real. As real as anything else.

Obviously, making some mathematical errors and not understanding flux and how it relates to conserved flows of conserved quantities can make fields seem less real, but now the details are there. Fields are real if you want momentum to be conserved.