# Electromagnetic Stress Tensor

1. Aug 3, 2017

### Vrbic

1. The problem statement, all variables and given/known data
An electric field E exerts (in Gaussian cgs units) a pressure E2/8π orthogonal to itself and a tension of this same magnitude along itself. Similarly, a magnetic field B exerts a pressure B2/8π orthogonal to itself and a tension of this same magnitude along itself. Verify that the following stress tensor embodies these stresses:
T = 1/ 8π ((E2 + B2 )g − 2(EE + B B))

2. Relevant equations
$T_{ii}=P$

3. The attempt at a solution
$T_{xx}=P=1/ 8π(\textbf{E}^2 + \textbf{B}^2-2(E_x^2 + B_x^2))=1/ 8π(-E_x^2- B_x^2 + E_y^2+ B_y^2+ E_z^2 + B_z^2)$
$T_{xy}=2/ 8π(-E_xE_y- B_xB_y )$
Why I didn't get a right result? I probably don't understand "a pressure orthogonal to itself". In particular, it means that if E=(Ex,0,0) so in which direction is pressure? I would say in same but there is written orthogonal.

2. Aug 3, 2017

### Orodruin

Staff Emeritus
$T_{xy}$ is not a pressure, it is a shear stress component. The correct question to ask is what $T_{xx}$ is when the electric (or magnetic) field is in the $x$-direction versus what it is when the electric (or magnetic) field is in the $y$-direction.

3. Aug 3, 2017

### Vrbic

Ok. So to your question (only in electric field):
1) $\textbf{E}=(E1,0,0)$: $T_{xx}=\textbf{E}^2-2E1^2=-E1^2$

2) $\textbf{E}=(0,E1,0)$: $T_{xx}=\textbf{E}^2=E1^2$

Second thing is whether I understand stress tensor. So I try to explain what it is $T_{xx}$: We have defined unit area $\textbf{a}$ as unit vector perpendicular on this area. Let's say $\textbf{a}=(1,0,0)$ ($y-z$ plane). Then $T_{xx}$ is force in $x$ direction on unit area $\textbf{a}$ (pressure).
From that Im confused from results which I have got. Minus pressure? What is wrong?

4. Aug 3, 2017

### Orodruin

Staff Emeritus
Nothing is wrong. Strains can be both positive and negative. It depends on the force across the area.

If you think about the stress tensor in regular solid mechanics, it would correspond to pulling vs pushing the material. (The EM stress tensor is just the same!)