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How is the electromagnetic tensor derived?

  1. Jul 10, 2012 #1
    I understand that the EM tensor is a way of expressing the electromagnetic field in a frame invariant way, but how is it derived? Please use the (-+++) convention as I mostly use that.
  2. jcsd
  3. Jul 10, 2012 #2
    I'd be interested in this question as well--if someone has some historical insight into how the leap was made from E and B to the full-on Faraday tensor.
  4. Jul 10, 2012 #3


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  5. Jul 10, 2012 #4
    Interesting. I think that gets 90% of the way there, but the small subtlety of needing to construct the four-potential and using the relations between that and the EM fields is something they seem to gloss over. On the one hand, it seems fairly obvious, but on the other hand, the exact reasoning there that justifies it seems elusive.
  6. Jul 10, 2012 #5
    I hate it when they use SI units in relativity, annoying c or c^2 factors
  7. Jul 10, 2012 #6
    This page shows how we can derive it, but doesn't explain the reasoning for doing things the way they are done.
  8. Jul 10, 2012 #7
    In the end, I think some amount of argument concerning symmetry has to be invoked. If the magnetic field is really a plane field and not a vector field, the object that it is a part of has to be at least a plane as well. The Faraday field is what results through a simple analogy of the field being a derivative of a potential. There is some amount of, yes, by magic it works. Trying to build up to 3+1D from 3D is, well, hard. Trying to go the other way is much easier--and, in my opinion, much cleaner from a conceptual standpoint, though it does have drawbacks.

    Were I to teach EM, I might be tempted to start with the Faraday field and SR and show that the E and B fields in 3D come about from looking at particular components that are of relevant interest to 4-currents with various properties.
  9. Jul 10, 2012 #8
    Im looking for a more qualitative explanation for this, so far all articles have ignored the logic and jumped to the maths.
  10. Jul 10, 2012 #9
    The electric field and the magnetic field can be expresed in terms of the Coulomb potential [tex]\Phi[/tex] and magnetic vector potential a as

    [tex]E = -\nabla \Phi - \frac{\partial \Phi}{\partial t}[/tex]

    [tex]B = \nabla × a[/tex]

    TGhe 4-potential is defined as A = ([tex]\Phi/c[/tex], a)

    The Faraday Tensor is then defined as

    [tex]F^{\alpha\beta} = ∂^{\alpha}A^{\beta} - ∂^{\beta}A^{\alpha}[/tex]

    (Sorry. I don't know why the equation is broken up that way.)
  11. Jul 10, 2012 #10


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    I'm not sure that there is any qualitative explanation. You have a mathematical object, you prove that it is a tensor, you use it to write an equation that expands to Maxwells equations, therefore it is the field tensor.
  12. Jul 10, 2012 #11
    For any action, if one performs an (artificial) variation of the metric tensor [itex]\delta g_{\mu \nu}[/itex], then, whatever comes as a coefficient in the expression for the variation of the action:
    \delta S = -\frac{1}{2 c} \, \int{T^{\mu \nu} \, \delta g_{\mu \nu} \, \sqrt{-g} \, d^4 x}
    is identified as the stress-energy tensor.
    Last edited: Jul 10, 2012
  13. Jul 14, 2012 #12
    Then the question is... is there really no other object which could be Lorentz invariant and sum up the two fields of classical theory? Is it the only possible tensor combination?
  14. Jul 14, 2012 #13
    I believe so. When you look at the differential equations for various ranks of field and source tensors, you should see that the field tensor must be rank two, and the source tensor must be rank one or three (they're related by duality). Anything else is fundamentally different.
  15. Jul 14, 2012 #14


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    One way to derive it is to use a covariant Lagrangian with minimal coupling to the electromagnetic potential [itex]A^{\mu}[/itex]:

    [itex]L = \dfrac{m}{2} U^{\mu} U_{\mu} - q A_{\mu} U^{\mu}[/itex]

    where [itex]U^{\mu}[/itex] is the 4-velocity. Then the lagrangian equations of motion give:

    [itex]\dfrac{d}{d \tau} \dfrac{\partial L}{\partial U^\mu} = \dfrac{\partial L}{\partial X^\mu} [/itex]

    which gives:

    [itex]m \dfrac{d U_\mu}{d \tau} - q \dfrac{d A_\mu}{d \tau}= - q \dfrac{\partial A_{\nu}}{\partial X^\mu} U^{\nu}[/itex]

    Then we use [itex]\dfrac{d A_\mu}{d \tau}= U^{\nu} \dfrac{\partial A_{\mu}}{\partial X^\nu}[/itex] to get:

    [itex]m \dfrac{d U_\mu}{d \tau} = q U^{\nu} (\dfrac{\partial A_{\mu}}{\partial X^\nu} - \dfrac{\partial A_{\nu}}{\partial X^\mu}) [/itex]

    So we just define [itex]F_{\mu \nu}[/itex] to be [itex]\dfrac{\partial A_{\mu}}{\partial X^\nu} - \dfrac{\partial A_{\nu}}{\partial X^\mu}[/itex] to get

    [itex]m \dfrac{d U_\mu}{d \tau} = q U^{\nu} F_{\mu \nu}[/itex]

    Then you just look at the spatial components:

    [itex]m \dfrac{d U_j}{d \tau} = q U^{\nu} F_{j \nu}[/itex]
    [itex]= q U^{t} F_{j t} + q U^{i} F_{j i}[/itex]

    In the non-relativistic limit, this becomes:

    [itex]m \dfrac{d v_j}{d t} = q c F_{j t} + q v^{i} F_{j i}[/itex]

    The right-hand side is just [itex] q E_j + q (v \times B)_j[/itex], provided that you identify:

    [itex]E_j = c F_{j t}[/itex]
    [itex]B_j = \epsilon_{i j k} F_{j i}[/itex]
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