Electromagnetic stress tensor from pressure and tension

[Glenn Rowe]

I'm puzzling over Exercise 1.14 in Thorne & Blandford's Modern Classical Physics. We are given that an electric field $\boldsymbol{E}$ exerts a pressure $\epsilon_{0}\boldsymbol{E}^{2}/2$ orthogonal to itself and a tension of the same magnitude along itself. (The magnetic field does the same, but for simplicity I'll assume we have only an electric field.) We're then asked to verify that the stress tensor embodies these stresses. The tensor (with $\boldsymbol{B}=0$) is $$

\mathsf{T}=\frac{\epsilon_{0}}{2}\left[\boldsymbol{E}^{2}\mathsf{g}-2\boldsymbol{E}\otimes\boldsymbol{E}\right]$$
where $\mathsf{g}$ is the metric tensor (we're working in Newtonian space, so $\mathsf{g}=\delta_{ij}$ is 3-d, not 4-d as in relativity).
In components, the tensor is $$

T_{ij}=\frac{\epsilon_{0}}{2}\left(\boldsymbol{E}^{2}\delta_{ij}-2E_{i}E_{j}\right)$$ which agrees with definitions I've seen in other books.
Now suppose we take $\boldsymbol{E}=E\hat{\boldsymbol{x}}$ so that the electric field is along the x axis. Then we get $T_{xx}=-\frac{\epsilon_{0}}{2}E^{2}$ and $T_{yy}=T_{zz}=\frac{\epsilon_{0}}{2}E^{2}$, with all off-diagonal elements equal to zero, which seems to match the requirement, since we have a negative 'pressure' (i.e. a tension) along the x-axis and positive pressures orthogonal to the x axis.
My problem is trying to match this with Thorne & Blandford's equation 1.32, which states that if we have a force $\boldsymbol{F}$ acting across a directed area element $\boldsymbol{\Sigma}$, then the force can be written in terms of the stress tensor and the area as (sum over $j$): $$

F_{i}=T_{ij}\Sigma_{j}$$
If we take
$\boldsymbol{\Sigma}$ to be an area perpendicular to the x axis, then $\Sigma=A\hat{\boldsymbol{x}}=[A,0,0]$ and, with the tensor worked out above, we have $$

\boldsymbol{F}=\left[T_{xx}\Sigma_{x},T_{yx}\Sigma_{x},T_{zx}\Sigma_{x}\right]=\left[-\frac{\epsilon_{0}}{2}AE^{2},0,0\right]$$ That is, the pressure force perpendicular to the x-axis has disappeared, and we have only the tension force along the x axis.
I feel that I'm missing something either obvious or fundamental here, but if anyone could explain where my reasoning is wrong, I'd be grateful.

 

[vanhees71]

But that's simply, because the stress tensor is diagonal in the basis chosen, i.e., your basis vector are along the principal axis of the stress tensor and you put also a surface with a normal vector in the direction of one of these principle axis. Then there are only stresses along this axis. In other words you work in a basis system of eigenvectors of the stress tensor. The stress tensor maps the surface normal vector to the stress, and if the surface normal vector is an eigenvector of the stress tensor the stress is only along this axis.

 

[Glenn Rowe]

That explains why the stress tensor is diagonal, but if the electric field exerts a pressure orthogonal to itself, there should be a force in the y and z directions, as well as the tension along the x axis, shouldn't there? In other words, is eqn 1.32 in Thorne & Blandford (above) the right way to get the force from the stress tensor?

 

[vanhees71]

Why shouldn't this be correct?

You get the Maxwell stress tensor from Maxwell's equations and an analysis of the conservation laws (most elegantly using Noether's theorem to the Lagrangian for the electromagnetic field coupled to charge and current distributions). In other words it follows from the momentum balance of a system consisting of the em. field and charged particles.

To find the effect of the other diagonal elements of course you need a surface not perpendicular to the field. For simplicity take a surface with normal vector in the $y$ direction as an example.

 

[Glenn Rowe]

I guess what's confusing me is that according to the equation in Thorne & Blandford, the force is given by
$$

F_{i}=T_{ij}\Sigma_{j}$$
Assuming that the stress tensor is fixed, then according to this formula, the force depends on the orientation of the surface. (Of course it also depends on the magnitude of the area, but that's just because the stress tensor has units of force per unit area.) The way it's written in the textbook, it gives the impression that it is *the* force vector which is given by this equation, but clearly that can't be true if it depends on the surface orientation. It would seem to me that if you consider some infinitesimal volume, and write out the actual physical forces that act on it and add them up vectorially, that you end up with a unique force vector acting on that volume element. How does that unique force vector relate to the one calculated by the above formula?

Sorry to be kind of thick on this point, but I've never encountered stress tensors in this context before and I'm finding the concepts hard to grasp. Thanks for your time.

 

[vanhees71]

Of course the force depends on the orientation of the surface. Take an ideal fluid in non-relativistic physics. Its stress tensor is $T_{ij}=-p \delta_{ij}$. It's also intuitively clear that the force on a surface due to this pressure is in direction of the normal of the surface.

A stress tensor always has this meaning: It's describing the forces on a surface. By definition the force on an infinitesimal surface is $\mathrm{d} F_i = T_{ij} \mathrm{d}^2 \Sigma_j$.

 

[Glenn Rowe]

Thanks for your help.