Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the physical meaning of the Maxwell Stress tensor symmetry?

share|cite|improve this question
Nothing. The Maxwell stress tensor was designed to write the Maxwell Equations in a symmetric way. – centralcharge Jun 19 '13 at 10:26
It's antisymmetric. That's a way to pack two vectors (E and B) into a second rank tensor. – jinawee Jun 19 '13 at 11:03
It's not always symmetric. The symmetry of the stress tensor is related to the conservation of angular momentum. In the vacuum, you would wish that not only linear, but also angular momentum, of the electromagnetic field alone would be conserved. This is ensured by trying to create an symmetric stress tensor. In material media, this is not necessarily true. – Hydro Guy Jun 19 '13 at 12:08
@jinawee and dimension10: The question is referring to the Maxwell stress tensor, not the electromagnetic tensor. – Ben Crowell Jun 19 '13 at 13:37
@user23873 thank you! Could you explain me a little better how its symmetry is connected with the conservation of angular momentum? – Caos Jun 19 '13 at 14:28
up vote 1 down vote accepted

No, it's not symmetric. Let me explain:

Say for instance that you only take the magnetic part of the Maxwell stress tensor (let's ignore the electric part). Then you would have the outer product $BB$ + (diagonal tensor). A lot of textbooks usually write it as 1/$\mu BB+$ (diagonal tensor), which is wrong and misleading, since it assumes that the material has a linear behavior $B = \mu * H $.

The right expression is $BH+$ (diagonal tensor), where $B = \mu_0 (M + H)$ Therefore if M is not colinear with $H$ you will get a non-symmetric tensor. However if M is colinear with H then you will get a symmetric one. This colinearity between M and $H$ holds true when the magnetization can be described by a linear and isotropic relationship ... that is $M = $some_physical_constant * $H$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.