Trying to understand how to apply Maxwell stress tensor to calculate forces

Question

I'm struggling to understand how to use Maxwell's stress tensor to compute electromagnetic forces acting on surfaces. I'll take problem 8.7 from Griffths Introduction to Electrodynamics as an example.

An infinite parallel-plate capacitor, with the lower plate (at $z = −d/2$) carrying surface charge density $−σ$, and the upper plate (at $z = +d/2$) carrying charge density $+σ$, is considered. The Maxwell's stress tensor is defined as (neglecting magnetic fields) \begin{equation} \overleftrightarrow{T} = \epsilon_0[\mathbf{E}\mathbf{E}-\frac{1}{2}|\mathbf{E}|^2I]\,, \end{equation} where $\mathbf{E}$ is the electric field and $I$ is the identity tensor. In my opinion, being $\mathbf{E} = -E\mathbf{e}_z$, the expression for $\overleftrightarrow{T}$ in this case would be $$ \overleftrightarrow{T} = \epsilon_0\left[ \begin{matrix} -E^2/2 & 0 & 0 \\ 0 & -E^2/2 & 0 \\ 0 & 0 & E^2/2 \\ \end{matrix} \right] \,. $$ Now, in order to calculate, for instance, the force acting on the upper plate, being the Lorentz force per unit volume $\mathbf{f}$ \begin{equation} \mathbf{f} = \nabla \cdot \overleftrightarrow{T}\,, \end{equation} I would integrate this equation over volume $V$ enclosed by the parallel plates and then apply the divergence theorem in order to compute the total Lorentz force $\mathbf{F}$ as \begin{equation} \mathbf{F} = \int_{S_{u}} \overleftrightarrow{T} \cdot \mathbf{n}_u \, \text{dS} + \int_{S_{l}} \overleftrightarrow{T} \cdot \mathbf{n}_l \, \text{dS}\,. \end{equation} Here, the normal unit vectors must be outwardly-pointing, so $\mathbf{n}_u = [0,0,1]$ for the upper plate with surface $S_u$ and $\mathbf{n}_l = [0,0,-1]$ for the lower plate with surface $S_l$. My question is how do I extract from this previous expression the forces acting on surfaces $S_u$ and $S_l$. Are the two terms on the right hand side already the forces acting respectively on the upper and lower plate? If this was the case I would be surprised since, for example, the force acting on $S_u$ would push the upper plate in the positive $z$, while from a physical point of view I would expect it to be in the negative $z$ direction.

Answer 1

To find the net force $\mathbf{F}$ on the charges within a particular volume $\mathcal{V}$, you need to perform the integral $$ \int_{\partial \mathcal{V}} \overleftrightarrow{\mathbf{T}} \cdot \mathbf{n} \, \text{d}S $$ where $\partial \mathcal{V}$ is the boundary of the volume $\mathcal{V}$ and $\mathbf{n}$ is the outward-pointing normal. This is the strangeness/power of this technique: you don't need to know where the charges are to figure out the forces on them, you just need to know the fields on some surface containing them. As Griffiths himself puts it (p. 366):

Take a moment to appreciate what happened. We were calculating the force on a solid object, but instead of doing a volume integral, as you might expect, Eq. 8.21 allowed us to set it up as a surface integral; somehow the stress tensor sniffs out what is going on inside.

So to find the force on the upper plate $S_u$, you don't integrate over the surface $S_u$ itself; you integrate over the boundary of a volume containing $S_u$. For example, you could integrate over a "pillbox" with one end between the plates and one end above the upper plate. Note on the end of the pillbox where $\overleftrightarrow{\mathbf{T}} \neq 0$ (between the plates), the surface normal is downward-pointing, which obviates the sign problem you were having.

If you think about it, the idea of integrating over $S_u$ and $S_d$ themselves doesn't make a lot of sense, since $\mathbf{E}$ is discontinuous at both sheets (it's zero on one side and non-zero on the other. So $\overleftrightarrow{\mathbf{T}}$ is ill-defined on these surfaces as well.