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The physical meaning of Maxwell stress tensor

  1. Jan 21, 2008 #1
    Maxwell stress tensor:

    [tex] T_{ij} = \epsilon_0 \left( E_i E_j - \frac{1}{2} \delta_{ij} E^2 \right) + \frac{1}{\mu_0} \left( B_i B_j - \frac{1}{2} \delta_{ij} E^2 \right)[/tex]

    We can interpret T as the force per unit area acting on the surface. But what surprises me is, [tex]T_{ij} = T_{ji}[/tex], i.e. the electromagnetic force (per unit area) in the ith direction acting on an element of surface oriented in the jth direction is equal to the electromagnetic force (per unit area) in the jth direction acting on an element of surface oriented in the ith direction !!

    It seems like a result of mathematical derivation, but could we get the same result by simple physical interpretation without math??

    Anyone have idea?? Thanks a lot!!
     
  2. jcsd
  3. Jan 21, 2008 #2
    The physical meaning is quite simple. Stress by definition means the forces acting per unit area in the internal surfaces of a body. In the Maxwell stress tensor, the diagonal elements [itex]T_{xx},T_{yy},T_{zz}[/itex] represent pressures and the non-diagonal elements [itex]T_{xy},T_{yz},T_{zx}[/itex] etc. represent shears. However, the Maxwell tensor has a special property due to the Kronecker delta, [itex]\delta_{ij}[/itex] i.e. all the non-diagonal elements vanish.
    Yes, you can derive a similar result using Gauss's law and Faraday's law but the equation will be complicated. The tensor is a much comprehensive and general representation.
     
  4. Jan 21, 2008 #3
    Why do the non-diagonal vanishes:confused:

    Despite the Kronecker delta term which will vanishes, there are still have terms that will not vanish.( [tex] \epsilon_0 E_i E_j + \frac{1}{\mu _0} B_i B_j [/tex] ).

    I think I could understand what the stress tensor means.
    But what I'm curious about is why [tex]T_{ij} = T_{ji}[/tex](which is not generally zero, i think :uhh: )

    Of course it can be derived mathematically. But can we have some intuition that can tell us the same thing qualitatively??
    (Maybe it's a consequence of some basic property of E and B ?? )
     
  5. Jan 21, 2008 #4

    Ben Niehoff

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    I think what Reshma means is that you can always choose a local coordinate system in which T is diagonal.

    The diagonal elements of T represent compressive/tensile stresses, while the off-diagonal elements represent shear stresses. So the above statement amounts to saying that shear stresses can always be decomposed into pure compressive/tensile stresses by rotating the coordinate system.

    The basic properties of E and B that imply this consequence are precisely the Maxwell equations themselves.
     
  6. Jan 21, 2008 #5
    Could you sketch the idea/method how to get the diagonal tensor?

    This consequence is due to Maxwell equations, in fact, we use maxwell equations and some vector calculus to derive it. But this consequence are so "simple" that I think it might have some simple way that can comes to the same conclusion. But after 2 days of thought, I still can't find other ways. It seems like we can only get the conclusion by math XDDD
     
  7. Jan 21, 2008 #6
    Hy,
    i think, guys, that if you think in a CM way... and following Noether theorem you can reach easly the symmettry on T(mu/nu)... assuming only uniformity and isotropy of your system... i think there can be cases in wich T is not zero.

    regards
    marco
     
  8. Jan 22, 2008 #7
    Hi psi-string
    I think this link might provide a nice explanation for you answer :smile:
    http://www.nhn.ou.edu/~shaferry/41832005_files/l4.pdf
     
  9. Jan 22, 2008 #8
    Thanks! It does provide plenty of help!!
     
  10. Feb 8, 2008 #9

    MarcusAgrippa

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    You say, in connection with the Maxwell stress tensor that "Stress by definition means the forces acting per unit area in the internal surfaces of a body." Yet the Maxwell stress tensor is defined throughout all of space. What is being stressed when we are not inside a material body but in a vacuum, say?
     
    Last edited: Feb 8, 2008
  11. Feb 8, 2008 #10

    pam

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    Several textbooks imply this, but it is not true.
    The derivation shows that the integral [tex]\oint{\vec dS}\cdot[{\bf T}][/tex]
    equals the force on charges and currents within a closed surface.
    But that is only true for the integral, and T is not the stress at any one point on the surface.
     
    Last edited: Feb 8, 2008
  12. Feb 9, 2008 #11

    MarcusAgrippa

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    You say, "The derivation shows that the integral ... equals the force on charges and currents within a closed surface. But that is only true for the integral, and T is not the stress at any one point on the surface."

    So, does the Maxwell tensor itself, without performing any integrations, have any physical meaning? What exactly (if anything) does it, viewed as a field, describe?

    Since it occurs in a momentum conservation law, and in that conservation law it supposedly represents momentum current density, is it possible that the "force" it represents is the rate of transfer of mometum per unit area by the field across a surface?
     
  13. Feb 9, 2008 #12

    pam

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    Only its integral over a closed surface has physical meaning.

    The MST does not represent momentum current density. It represents the rate of flow of momentum into a volume when integrated over the surface enclosing that volume.
     
  14. Feb 19, 2008 #13

    MarcusAgrippa

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    Re the necessity of the Maxwell stress tensor being symmetric, you may find the following helpful: E N Parker, Cosmical Magnetic fields, Clarendon press, Oxford, 1979, p 52, "The stress tensor ... is necessarily symmetric in its two indices, because any non-vanishing anti-symmetric part represents an unbalanced torque on each infinitesimal element of volume." Incidentally, Parker's view of the stress tensor does not agree with Pam's. He does not regard it as essential for the stress tensor to be integrated over a closed surface before one can interpret it. His discussion of this point in Chapter 5 of Cosmical Magnetic Fields is interesting. Griffiths too disagrees with Pam: see his Electromagnetism, Chapter 8, particularly in the worked examples but also the solutions in his Solutions Manual. Technically, I suppose, Pam is correct in her insistence on integrating over a closed surface before interpreting. The definition of the stress tensor and of the Poynting vector are uncertain up to terms that vanish on integration over a closed boundary. In practice, however, all practitioners of the theory seem to regard these tensors as having their obvious pointwise significance without insisting on their integration on a closed surface.
     
  15. Feb 19, 2008 #14

    pam

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    A demonstration that the MST gives the wrong answer for the stress on a dielectric surface surface is given in the graduate text "Classical Electromagnetism", J. Franklin,
    Addison-Wesley.
    That text also shows that if epsilon or mu is non-isotropic, the derivation of the MST breaks down, so there is no valid derivation of a non-symmetric MST.
     
  16. Feb 20, 2008 #15

    MarcusAgrippa

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    Thank you Pam. I don't know Franklin's text but will source it in the next few days. There seem to be very few useful discussions on these issues in older texts. Perhaps you might consider writing an article for the AJP which irons out these thorny issues. I for one would like to see some focused discussion dedicated to this topic that collects together both conceptions and misconceptions and that guides one into the proper way to think about stress, energy and momentum in the EM field.
     
  17. Feb 21, 2008 #16

    pam

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    A simple way to see that the MST does not give the right stress at a point on the surface of a dielectric is to consider a point charge outside a dielectric. The stress at a point on the surface of the dielectric is given by the field of the charge times the bound surface charge density. The BSC is proportional to (epsilon-1), so the correct stress is proportional to (epsilon-1). The MST depends on E^2, and so will have terms with epsilon^2. These terms give the wrong stress at a point, but integrate to zero over a closed surface.
     
  18. Apr 26, 2011 #17
    I have some question and will be happy if you help me.
    when we study electromagnetic ,it is easy to find force of a magnetic field on particle witch has electric charge (for example electron and ...) but what is force of magnetic field on ferromagnetic materials? for example if we put a cube of soft iron with a*a*a dimentions near a magnetic field B or H what is the force witch attract it?
     
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