Why Isn't My EM Stress Tensor Calculation Giving the Expected Result?

[Vrbic]

Homework Statement

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(E ⊗ E + B ⊗ B))

Homework Equations

$T_{ii}=P$

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.
Please advise.

 

[Orodruin]

$T_{xy}$ 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.

 

[Vrbic]

$T_{xy}$ 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.

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 I am confused from results which I have got. Minus pressure? What is wrong?

 

[Orodruin]

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!)