In 'Introduction to Electrodynamics' by D. Griffiths, shortly after introducing the Maxwell stress tensor there is a paragraph concerning the physical interpretation of the stress tensor $\boldsymbol{T}$

Physically, $\boldsymbol{T}$ is the force per unit area (or stress) acting on the surface. More precisely, $T_{ij}$ is the force (per unit area) in the $i$th direction acting on an element of the surface oriented in the $j$th direction - "diagonal" elements ($T_{xx}$, $T_{yy}$, $T_{zz}$) represent pressures, and "off-diagonal" elements ($T_{xy}$,$T_{xz}$, etc.) are shears.

I understand where all this comes from mathematically, but I fail to grasp how this translates into an actual force, and specifically what is meant by "an element of the surface oriented in the $j$th direction". Any clarification would be greatly appreciated.


It means a surface element whose tangent plane has a normal in the $j$th direction. For a flat surface, we can shorten that to the normal to the surface pointing in the $j$th direction. For a curved surface, each infinitesimal patch has its own tangent plane.

  • $\begingroup$ It is implied that the surface element is slightly charged; otherwise there is no force to apply to ;-) $\endgroup$ Jun 1 '18 at 19:21
  • $\begingroup$ @VladimirKalitvianski I was replying to the OP's question of what the surface element comment meant. The $i$th component of the force has a clear interpretation. $\endgroup$
    – J.G.
    Jun 1 '18 at 19:56

Another way to think of it is, take a surface with area element $d{\vec A}$. The force felt on that surface has components given by:

$$F^{i} = \int\sum_{j} T^{ij}dA^{j}$$

The point of the off-diagonal components is that if they are present, this force will be in a different direction than that given by $d{\vec A}$, even for the case of the area element being a flat surface aligned with one of the axes.

  • $\begingroup$ Force can be in a different direction even for diagonal T; it suffices that the diagonal elements are not the same. $\endgroup$ Jun 1 '18 at 18:56
  • $\begingroup$ @JánLalinský yes, of course, I'm sorry, I was thinking for the case of a flat area element aligned with one of the axes. Will edit answer. $\endgroup$ Jun 1 '18 at 19:04
  • $\begingroup$ wouldn't we need to say something about the charge / current density at that area element to deduce the force? $\endgroup$ Jun 1 '18 at 19:29
  • $\begingroup$ @creillyucla : From the fact that we're doign this in three dimensions, I thought we were talking about the ordinary stress tensor, which is a concept from non-electrodynamic classical mechanics, of which the electromagnetic one is a generalization. $\endgroup$ Jun 1 '18 at 19:32
  • 1
    $\begingroup$ @JerrySchirmer that is true; $j_b$ are distributions that characterize charge and current density of some body in some reference frame. The force expression $\int dV\, F^{ab}j_b$ can be rewritten (under some conditions) as $\int dV\, \partial_b T^{ba}$, using the EM stress-energy tensor $T$ and without using $j^b$. $\endgroup$ Jun 2 '18 at 18:05

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