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From reading Griffiths, I understand that the total EM force on a set of charges in volume $\mathcal{V}$ can be found as $$ \textbf{F}=\oint_{\mathcal{S}} \overleftrightarrow{\textbf{T}} \cdot d\textbf{a}-\epsilon_0\mu_0\frac{d}{dt}\int_{\mathcal{V}}\textbf{S}\ d\tau. $$ where $\mathcal{S}$ is the boundary of $\mathcal{V}$. $\overleftrightarrow{\textbf{T}}$ is the Maxwell stress tensor, and $\textbf{S}$ is the Poynting vector. I have the following questions:

  1. The stress tensor physically represents the force per unit area acting on the surface. Thus, its appearance here is understandable. But, how can I interpret the second term? It seems to be the rate at which energy flows out of the volume, but what does that have to do with force?
  2. How is it that the EM force on charges is the same as the EM force on the surface of any volume that encloses the charges? How can I understand this intuitively without the equations? Someone asked a similar question here (Maxwell's Stress Tensor) but I didn't quite understand the answers.
  3. To me, it seems like you consider both the $\textbf{E}$ and $\textbf{B}$ fields that pass through the surface $\mathcal{S}$. Looking at problem 8.3 in Griffiths (a spherical shell of charge density $\sigma$ rotating at $\omega$, we want to know the magnetic force on the upper hemisphere), I would use a volume that encloses the upper bowl and the base disk. In this case, I have magnetic fields inside and outside, and an electric field outside that I have to care for. However, when I look at the solutions, Griffiths only considers the inside and outside $\textbf{B}$ fields, ignoring $\textbf{E}$. Is this because he asked for the magnetic force, and not the total force? Am I right in thinking that the $\textbf{E}$ should too be accounted for if one is to find the total force on the upper hemisphere?
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  • $\begingroup$ For Q1, my guess is that force acting on charged particles is a result of (1): static EM field (Maxwell stress tensor) and (2): dynamically, the rate of change of momentum of EM field (Poynting vector term) $\endgroup$
    – K_inverse
    Sep 12, 2018 at 3:37
  • $\begingroup$ Ok I think that makes sense. Does this mean that in a static case, the force calculated using the tensor is the same as the force calculated using Columbs law? If yes, then this brings me to Q2: how is it that the integral over surface gives me a quantity that I can find from considering the direct interaction between charges. $\endgroup$
    – Ptheguy
    Sep 12, 2018 at 3:47
  • $\begingroup$ It is the result of divergence theorem. Maybe it is better to review the derivation en.wikipedia.org/wiki/Maxwell_stress_tensor#Motivation. $\endgroup$
    – K_inverse
    Sep 12, 2018 at 3:56
  • $\begingroup$ Integral of poyting vector is total momentum. The time derivative of momentum is force (think about newtons law in this way) $\endgroup$
    – lalala
    Sep 12, 2018 at 4:52
  • $\begingroup$ Yes, but isn't the Poynting vector energy per unit time per unit area? Why integrate over volume? $\endgroup$
    – Ptheguy
    Sep 13, 2018 at 5:04

2 Answers 2

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Field momentum:

$\mu_0 \epsilon_0 \vec{S}$ is the momentum density of the field. Integrating this about a volume, finds the TOTAL field momentum in that volume. The rate of change of this, is the rate of change of field momentum within a volume.

Let's look closely, how can there be a change in field momentum within a volume?

Mechanical momentum:

f - this by definition is the rate at which mechanical momentum increases. If there is an increase in mechanical momentum, the total field momentum decreases ( hence the minus sign)

Momentum entering through the surface boundary:

The other term is $\iint T \cdot \vec{da}$

As you pointed out, this represents the force acting on the surface

Another way of putting it, is that this term represents the rate at which momentum is carried into that volume, by the fields. Much like in poyntings theorem, the poynting vector represents the rate at which energy flows out of that volume.

This makes sense, the stress tensor is dotted with the da vector, analogous to flow directly through of that volume.

It is also pleasing that the rate at which momentum flows through the surface, IS BY DEFINITION, the force acting on the surface.

When there is no change in mechanical momentum:

$$\textbf{F} = 0$$

$$\oint_{\mathcal{S}} \overleftrightarrow{\textbf{T}} \cdot d\textbf{a}=\epsilon_0\mu_0\frac{d}{dt}\int_{\mathcal{V}}\textbf{S}\ d\tau. $$

$$ \int_{\mathcal{V}} \nabla \cdot \overleftrightarrow{\textbf{T}} d\tau=\epsilon_0\mu_0\frac{d}{dt}\int_{\mathcal{V}}\textbf{S}\ d\tau. $$

$$ \nabla \cdot \overleftrightarrow{\textbf{T}} = \epsilon_0\mu_0\frac{d\textbf{S}}{dt}\ $$

Which is a continuity equation for field momentum.

Also your point (2) is incorrect, you are ignoring the actual change of field momentum. There are 2 terms determining the force on charges, not 1

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  1. But, how can I interpret the second term? It seems to be the rate at which energy flows out of the volume, but what does that have to do with force?

You are right, the Poynting vector $$\mathbf{S}=\frac{1}{\mu_0}\mathbf{E}\times\mathbf{B}$$ is the energy flow density (i.e. energy per time and area) of the EM field. But incidentally, the vector $$\mathbf{g}=\epsilon_0\mu_0\mathbf{S}=\epsilon_0\mathbf{E}\times\mathbf{B}$$ is also the momentum density (i.e. momentum per volume) of the EM field. See for example "Resource Letter EM-1: Electromagnetic Momentum" by D.J. Griffiths. Hence the term $$-\epsilon_0\mu_0\frac{d}{dt}\int_{\mathcal{V}}\textbf{S}\ d\tau$$ or $$-\frac{d}{dt}\int_{\mathcal{V}}\textbf{g}\ d\tau$$ is the decrease rate of the momentum contained in the EM field in the volume. And this relates well to the force $\mathbf{F}$.

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  • $\begingroup$ Are you sure your conclusion statement is correct? This term represents the rate at which field momentum changes within a volume. This doesn't represent the rate at which field momentum flows out of the volume. Consider the field momentum decreasing as a result of mechanical momentum increasing within that volume. I am fairly confident the stress tensor term is analogous to the flow of field momentum out of that volume. $\endgroup$ Apr 30, 2022 at 15:39
  • $\begingroup$ It is also pleasing that the stress tensor has a dot product with the da vector, analogous to a flow out of that volume. $\endgroup$ Apr 30, 2022 at 15:41
  • $\begingroup$ This can also be seen as setting f to zero. This is a statement about the conservation of field momentum. Which yields a familiar continuity equation of field momentum. Where the stress tensor term is momentum flow out of that volume. $\endgroup$ Apr 30, 2022 at 15:42
  • $\begingroup$ The stress tensor is also flow into not out, look at the signs. $\endgroup$ Apr 30, 2022 at 16:32
  • $\begingroup$ @jensenpaull I have improved the wording at the end. $\endgroup$ Apr 30, 2022 at 16:47

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