# Maxwell Stress Tensor at material boundaries

I am trying to grasp the meaning of the Maxwell Stress tensor $T_i^j$ at material boundaries. Concretely, I am trying to calculate the force between two waveguides. The results are given in an article by Povinelli et al.

The problem is, I get the same results when I evaluate the stress tensor IN the material and integrate this over the boundary of the waveguide. $$F_i=\int n_j T_{i,in}^j dS$$ However, since there is a discontinuity of the Maxwell tensor at the boundary, I would expect an extra surface force $f_i=n_j (T_{i,out}^j-T_{i,in}^j)$. This would essentially mean that I would have to calculate: $$F_i=\int n_j T_{i,out}^j dS$$

What is wrong with this reasoning?

• What is this funny notation of yours, with the double bar? – Danu Aug 6 '15 at 9:31
• It's just the symbol for tensor, but it isn't displayed very beautiful... – Fork2 Aug 6 '15 at 9:34
• Perhaps you could convert to index notation? I find it to be much clearer, and it is more widely used. – Danu Aug 6 '15 at 9:35
• I'm not that acquainted with tensor notation, but I hope this is alright. – Fork2 Aug 6 '15 at 9:43

I'm not sure discontinuity has anything to do with it.

If you have two parallel waveguides running in the x direction separated by a distance d with the electricomagnetic field having momentum inside the guides pointing entirely in the x direction then for every little piece of matter inside the waveguide the force density for components other than the x component is given by the divergence of the stress tensor (other than the x component of the divergence).

Thus the total force on waveguide 1 in directions other than x is the surface integral $$\int_{in1} n_jT^j_{i,in1} dS,$$ for $i\neq x.$

In the x direction the stress tensor tells us only how the field and mechanical momentum have their sum change, and so there isn't as simple a relationship between stress and mechanical momentum (since there is field momentum in the x direction which is possibly changing).

Now we basically found the force on the guide by looking at just the force on the body (inside) of the guide. That is not always OK, so let's check if that was OK.

Well, if there is surface charge or surface current then we need to consider not just force density but also force per area (and if we have line charge then we need even more) but if our waveguide has no surface current or surface charge then we are done. This is because the mechanical momentum changes because of the Lorentz Force and hence we need the charge and current if these are volume charges and volume currents we get all the charge and current by considering little volumes and adding them up.

• Yes, but a change in dielectric constant means a change in polarization. And this can be viewed as a surface polarization charge $-\vec{n} \cdot (\vec{P}_1-\vec{P}_2)$. So isn't there always some sort of a charge at a material boundary? – Fork2 Aug 10 '15 at 13:15
• @Fork2 Polarization is a measure of dipole density per unit volume. So just because that volume density is discontinuous does not mean there is a surface charge. There is a bound surface charge, but the force is on all charges not just on bound or just on free. The D field itself is ambiguous and so therefore so is P and thus bound and free charge is outright ambiguous. The electric field is real (though frame dependent) and a discontinuity of E would contain some real surface charge. I agree: if the stress tensor itself is discontinuous that there can be real surface charge, I'll go edit – Timaeus Aug 10 '15 at 16:14