# Maxwell's Stress Tensor

What really is the Maxwell Stress Tensor? I understand that it's derived from $$\mathbf {F} = \int _V ( \mathbf E + \mathbf v \times \mathbf B )\rho \ d \tau$$

Griffiths describes this as "total EM force on the charges in the volume $\mathcal V$".

$$T_{i j} = \epsilon_0 \left(E_i E_j - \frac{1}{2} \delta_{ij} E^2\right) + \frac{1}{\mu_0} \left(B_i B_j - \frac{1}{2} \delta_{ij} B^2\right)$$

This leads us the the stress tensor, but there's something that I don't understand. The description given is

Physically, $T$ is the force per unit area acting on the surface.

What surface are we speaking about here? An arbitrary surface? In the case of example 8.2 (net force on a uniformly charge sphere's upper hemisphere), the surface in question is clearly the boundary of the upper hemisphere and it's "disk" separating the two hemispheres. In other cases, such as problem 8.4, where we have two point charges separated by a distance, we must integrate over a particular surface. For such a problem, we must "integrate the stress tensor over the plane equidistant from the two charges", but why? How would summing up the force on the plane separating the two point charges equal the force on each charge?

How could there be "force per unit area" across an empty plane?

The Maxwell stress tensor is introduced as analogue to the stress tensor in continuum mechanics, and its form is derived from the equation $$\frac{d}{dt} \int_V \boldsymbol{\mathscr{p}} + \boldsymbol{\mathscr{g}} \,dV = \oint_{S} d\mathbf S \cdot \mathbf T$$ where $\boldsymbol{\mathscr{p}}$ is density of momentum of matter and $\boldsymbol{\mathscr{g}}$ is density of momentum of the EM field. The right-hand side is a surface integral over $S$, the boundary of the region $V$ and it looks like total surface force in continuum mechanics. The region and thus also its surface is arbitrary, the equation is valid for any choice. If the boundary is in free space, obviously there is no "surface force" in the original sense, but the equation has the same form as if there was - as in continuum mechanics.

Alternatively, one can interpret $d\mathbf S\cdot\mathbf T$ on the right-hand side not as a surface force per unit area, but as EM momentum that enters the region $V$ through $dS$ per unit time. This is perhaps better as we do not have to talk about "force of tension" acting in free space (on what? is a good question). But both views are commonly used.

EDIT: In case everything is static, force on regular charge distribution $\rho(\mathbf x)$ (first charged particle) that is contained within region $V$ is $$\mathbf F = \int_V \rho(\mathbf x)\mathbf E(\mathbf x) d^3\mathbf x$$ This can be transformed using Maxwell's equations and vector calculus into $$\mathbf F = \oint_S d\mathbf S \cdot \mathbf T$$ where $S$ is boundary surface of $V$. This only signifies that electrostatic force on a charged body can be calculated either as sum of elementary forces acting "locally" or "in bulk" as "volume forces", or one can alternatively calculate the same force from the field at distance surface that encloses the body as a sum of fictitious surface forces. The latter case reminds of continuum stress force, but the connection is only mathematical - physically, there is no force on the surface since there is no matter there.

Sidenote: All this is usually derived only for regular distributions, where $\rho$ is bounded. The derivation does not work for point charges, because for them the left-hand integral does not have sense - they are a special case that leads to different theory. However, incidentally the right-hand side is still valid for point particles. This is because there is similar theorem that can be derived for point particles, with slightly different EM stress tensor, which somewhat surprisingly gives the same surface integral. This is easily seen from the fact that the force between two charged bodies does not depend on whether the bodies are points or say, uniformly charged spheres.

Back to the OP's questions, all this equivalence between the two ways to express the force breaks down once the fields cease to be static; then the presence of EM momentum in $V$ cannot be neglected. Then the force acting on a charged body is not given solely by the right-hand type of integral, but it is believed that the original equation for the force - the first equation of the OP (integral of a local expression) is still correct for extended bodies.

• While I understand what you're saying, it leaves me somewhat unsatisfied. It still doesn't explain why the force between two point charges can be worked out by integrating the stress tensor across the plane between them. – Astrum May 29 '14 at 10:03
• @Astrum, I've edited the answer to explain why the stress tensor can be used to find the force. – Ján Lalinský May 29 '14 at 18:53
• I'm not sure why the left hand side no sense for say a dirac delta distribution. You have the distribution act on the identity function to give one, so the integral makes sense (as all integrals of distributions make sense as the distribution acting on the identify function). If by different theory you just mean distributions, then I follow you. But you talked about the non point particles as bounded distributions so I'm not sure if you just meant 3-form by distribution. Could you provide a name of the different theory or a citation? – Timaeus May 7 '15 at 2:06
• What I meant is the integral $$\int_V \boldsymbol{\mathscr{g}} \,dV$$ has no sense for point particle. This is because point particle has EM field that varies as $~\frac{1}{r^2}$ in the neighborhood of the particle. Quantity $\mathscr{g} = k\mathbf E\times \mathbf B$ is then not integrable at the point where the particle is. – Ján Lalinský May 7 '15 at 19:07

A stress tensor has nine components at every point in space. If you group them into three sets of three, you can imagine it as three vector fields. Do so. Each of those vector fields has a divergence, and that would be three scalar fields. You could combine those three scalar fields together to get one vector field. What if that vector field were the force density (force per unit volume)?

The force per unit volume tells you the rate at which momentum changes (per volume) in a that volume. So this doesn't tell us what the stress is, only what it's divergence is. But you can apply the divergence theorem to any nice bounded region to say that that the flux of the stress tensor out of the region it the total force acting on the volume inside region. The flux through the surface equals the divergence integrated over the region, so they are both insensitive to the same things.

In relation to all this here is another issue: in an electrostatic field $\mathbf{g}=0$ while $\mathbf{T}\ne 0$ so we have flux of momentum while the density of momentum is zero. In my opinion $T$ should be interpreted as a combination of momentum flux plus stress just as in continuumm mechanics the anolog of $T$ is (using indicial notation) $$\sigma_{ik}-\rho V_iV_k$$ The second tem in the sum gives flux of momentum owed to matter motion while the first term gives the forces acting on a given portion of matter

• How would you divide the entries of the matrix $\mathbf T$ into such two components? – Ján Lalinský May 6 '15 at 22:48
• I don't Know but if we want to accept things like the one I said(flux of momentum$\ne0$ while momentum density=0) me must accept that $\mathbf{T}$ involves stress as well as momentum flux – facenian May 8 '15 at 10:45
• May be it is not that you can divide the entries of the matrix $\mathbf{T}$ but how you interpret them according to the physical situation – facenian May 8 '15 at 10:54
• In the macroscopic theory of electric current conduction, it is common to assume there is net electric current while the charge density vanishes. Similarly, I do not think there is a real problem with having non-vanishing EM momentum flux with density of EM momentum vanishing. These are even more abstract mathematical quantities than charge density and current density. – Ján Lalinský May 10 '15 at 22:29
• The example of macroscopic conduction would be ok if we had $\rho\mathbf{V}=\mathbf{J}\ne 0$ while $\rho=0$ however this is not case, the $\rho$ in the expression for $\mathbf{J}$ is not the total charge density is only the density in motion – facenian May 11 '15 at 13:51

## protected by Qmechanic♦Sep 12 '15 at 7:52

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