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4 potential

  1. Feb 1, 2012 #1
    How to show that the electromagnetic 4 potential is a 4-vector?
     
  2. jcsd
  3. Feb 1, 2012 #2

    tiny-tim

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    hi quantum123! :smile:

    how are you defining the electromagnetic 4 potential?
     
  4. Feb 1, 2012 #3

    clem

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    In order for the continuity equation [tex]\nabla\cdot{\bf j}+\partial_t\rho=0[/tex]
    to be an invariant, [tex]\j^\mu=[\rho,{\bf j}][/tex] must be a four-vector.
    Then [tex]\partial^2_t A^\mu-\nabla^2 A^\mu=4\pi j^\mu[/tex] shows that [tex]A^\mu[/tex] must be a four-vector.
     
    Last edited: Feb 1, 2012
  5. Feb 1, 2012 #4
    I see, so local conservation of charge leads to the 4 potentials being a 4 vector.
    I guess it works the other way round too.
     
  6. Feb 2, 2012 #5

    clem

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    It doesn't go the other way.
    I don't know of another good argument for A^\mu to be a 4-vector.
     
  7. Feb 2, 2012 #6
    You can just list it as a postulate. In QM, we say the states are Hilbert space vectors. So in electroynamics, we say that the field potentials are 4 vectors.
     
  8. Feb 2, 2012 #7

    Meir Achuz

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    Then why did you ask your original question? When you say "we", don't include us.
     
  9. Feb 2, 2012 #8
    I just want to ascertain that whether there is a link between the conservation of charge and the tensorial nature of 4 potentials, and then which is more fundamental.

    Physics is tough!
     
  10. Feb 2, 2012 #9

    Mentz114

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    The property that makes a vector a 4-vector is that its magnitude is invariant under a change of basis.

    A2 = vava = (L.v)a (L.v)a = va'va'

    where L is the matrix dxa/dxa'. The 4-potential doesn't need this property because it is never contracted.

    It looks as if the 4-potential is more like a position vector, from which we can get a 4-vector by differentiation, or by subtraction to get an invariant interval.

    The relativistic invariant of EM is F = FabFab which is expressed in derivatives of the potential. If the potential is transformed F is not invariant. It is also necessary to boost the differentials.

    I'm not sure about this but I take the above to indicate that the 4-potential is not a tensor.
     
    Last edited: Feb 2, 2012
  11. Feb 2, 2012 #10

    Matterwave

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    I'd like to start from the Faraday tensor. One can define the Faraday tensor [itex]F_{\mu\nu}[/itex] in such a way that it is obviously a tensor (i.e. by postulating a Covariant form of the Lorentz force law by contracting that tensor with the 4-velocity). By requiring that "tubes of Faraday never end" (i.e. that there are no magnetic monopoles), we can see that its exterior derivative is 0, and therefore it is a closed 2-form. A closed 2 form is also an exact two form (under some topological requirements - something mathematicians may worry about when you carry over to GR perhaps), so we can deduce F=dA for some one form A. Thus, A is a one-form, and since our space has a metric defined on it, we can (trivially) associate a vector potential with this one form.
     
  12. Feb 2, 2012 #11

    Mentz114

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    Matterwave, I understand what you're saying. But I still think the 4-potential does not have to obey the tensorial rule. If A is a 1-form we can find a vector by raising its index with the metric, but it isn't the vector we differentiate to get the field tensor.

    Can't I write down any A I like and get a field tensor from it ?

    [Edit] I've been overlooking the fact that the length of any vector (t,x,y,z) remains the same under coordinate transformations. This might affect my logic a bit.
     
    Last edited: Feb 2, 2012
  13. Feb 2, 2012 #12

    Meir Achuz

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    Conservation of charge is a fundamental starting point. I showed the link to A^\mu in my first post.
     
  14. Feb 2, 2012 #13

    Matterwave

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    I'm not sure what your point is exactly. We take the exterior derivative of the 1 form potential to get the 2 form field strength. This trivially generalizes to "we take the anti-symmetric derivative of the vector potential to get the field strength tensor".

    See:

    [tex]F_{\mu\nu}=\partial_\mu A_\nu -\partial_\nu A_\mu[/tex]

    [tex]F^{\mu\nu}=g^{\mu\rho}g^{\nu\tau}F_{\mu\nu}=g^{\mu\rho}g^{\nu\tau}(\partial_\rho A_\tau -\partial_\tau A_\rho)=(\partial^\mu A^\nu -\partial^\nu A^\mu)[/tex]

    You can't just write down "any" vector potential. The symmetry is simply the gauge. You can write any "any" vector potential related to each other by gauge transformations.
     
    Last edited: Feb 2, 2012
  15. Feb 2, 2012 #14
    It need not be you see, for 2 reasons:-
    1) It can be derived as a mathematical consequence of Maxwell's equations
    2) It has got to be proven experimentally. And I know experiments have been done to verify it even in the 20th century.
     
  16. Feb 3, 2012 #15

    clem

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    You have 1) backwards. Maxwell got his displacement current term by using the continuity equation.
    We're going in circles. You're on your own now.
     
  17. Feb 3, 2012 #16
    Actually, electromagnetic potential is a principal connection
     
  18. Feb 4, 2012 #17

    samalkhaiat

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    You cannot show that because the gauge potential IS NOT a 4-vector. If you think it is a 4-vector, then you didn't understand the quantum theory of the electromagnetic field! Generally, massless fields can not be described by vector fields! The gauge potential we use to describe the EM field transforms inhomogenuosly under Lorentz transformation. However, the Field strength does transform as antisymmetric tensor field.
     
  19. Feb 5, 2012 #18
    How about in the case of the classical theory of electromagnetic fields?
     
  20. Feb 5, 2012 #19

    samalkhaiat

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    What I said about the transformation property of [itex]A_{\mu}[/itex] hold true in classical EM as well as in QFT. The issue, however, in the classical theory is not as important.
     
  21. Feb 5, 2012 #20
    Strange:
    David Griffiths, Introduction to electrodynamics 3rd edition, page 541,
    "V and A together constitutes a 4-vector".
     
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