Prove that 4 vector potential is really a 4 vector?

  1. Prove that 4 vector potential does really a 4 vector? Most of the textbooks I found only mention that divergence of 4 vector potential equals to zero and the d'Alembertian of it is a four vector current and therefore it 'should be' a four vector. However I do not see there is any tensor theorem to get this conclusion. Could anyone prove it for me, thank you.
     
    Last edited: May 2, 2008
  2. jcsd
  3. The covariance of the Lorenz gauge, i.e. div A +(1/c)@(phi)/@t = 0 (where @ = partial sign), implies that Av = (phi,A) is a (contravariant) 4-vector. This follows from the quotient theorem which states that if O is a 4-operator and OA = scalar then A must be a 4-vector.

    Pete
     
  4. Reply to pmb phy, to the best of my knowledge, quotient theorem does not apply to operator acting on a tensor. Are you sure you are right about this?
     
  5. If you already believe that [tex]j^\mu[/tex] is a 4-vector (from the invariance of charge), then [tex]\partial_\nu\partial^\nu A_\mu=4\pi j^\mu[/tex] proves that [tex]A^\mu[/tex] is a 4-vector.
     
    Last edited: May 2, 2008
  6. I can't seem to get the tex right. It should be 4\pi j^\mu.
     
  7. Reply to pam, I do not think you are right. What you said is just like x^2=1, then x must be equal to 1. But it is not true, x can be -1. This is just A-level logic tells us that a proposition is true does not mean its converse is true.
     
  8. Yes. If you recall the derivation of the quotient theorem you will note that the derivation only requires the transformation properties of a quantity and since the vector and operator transform as a covariant vector and a contravariant vector it follows that the quotient theorem also applies to operators.

    Pete
     
  9. Reply to pmb phy, I am still skeptic about this. It is because when I refer to the derivation of quotient theorem, I found that even though it only requires the transformation properties of the quantity acting on the tensor, but that quantity must be arbitrary. Am I wrong about this? Anyway, I found a way to prove that 4-vector potential is indeed a 4-vector. Thank you for your help sincerely.
     
  10. Here I have to thanks those who have attempted to help me solve my problems, expecially to pmb phy and pam.
     
  11. Your interpretation about arbitray is a bit wrong. What is arbitrary is that the 4-vector (in this case 4-potential) is an arbitraty 4-vector and the derivative of that 4-vector is not zero in general. With these facts it follows that the 4-potential is a 4-vector. Jackson's EM text explains this in the same exact way, but my explanation is clearer (Jackson does not mention the quotient theorem but does state that the invariance implies that the 4-potential is a 4-vector).

    Pete
     
  12. That's not what I said. In my post, the LHS is a scalar times a vector. I am saying that is a vector. My upper and lower \mu got a bit mixed up, but I couldn't correct it.
     
  13. Hans de Vries

    Hans de Vries 1,135
    Science Advisor



    If you start with the assumption that:

    1) charge is invariant under Lorentz transform then.

    2) The charge/current vector must transform like [itex](1,\beta_x,\beta_y,\beta_z)[/itex]

    3) The charge/current density [itex]j^\mu[/itex] must transform like [itex](\gamma,~\beta_x\gamma,~\beta_y\gamma,~\beta_z\gamma)[/itex] because of Lorentz contraction

    4) [itex]A^\mu[/itex] must also transform like [itex](\gamma,~\beta_x\gamma,~\beta_y\gamma,~\beta_z\gamma)[/itex] because of [itex]\Box~A^\mu=j^\mu[/itex]


    The d'Alembertian [itex]\partial_\mu\partial^\mu[/itex] is a Lorentz invariant contraction, it has one raised and
    one lowered index. The result of the operator transforms the same as the operand.
    See for instance Jackson at the end of section 11.6


    Regards, Hans.

    (Edit, pmb_phy is right, but it is because of the d'Alembertian)
     
    Last edited: May 7, 2008
  14. samalkhaiat

    samalkhaiat 1,055
    Science Advisor

     
    Last edited: May 9, 2008
  15.  
    Last edited: May 10, 2008
  16. Hans de Vries

    Hans de Vries 1,135
    Science Advisor

     
    Last edited: May 16, 2008
  17. samalkhaiat

    samalkhaiat 1,055
    Science Advisor

     
  18. Shhh! No yelling please. Some of us may have a hangover. :biggrin:
    Please show me the source of this expression. I don't recall ever seeing it before.

    Pete
     
  19. samalkhaiat

    samalkhaiat 1,055
    Science Advisor

     
  20. samalkhaiat

    samalkhaiat 1,055
    Science Advisor

     
    Last edited: May 17, 2008
  21. Hans de Vries

    Hans de Vries 1,135
    Science Advisor

    In the Coulomb gauge the electric potential propagates with infinite speed while the
    magnetic vector potential propagates with c. The final results of Bjorken & Drell are
    only Lorentz invariant simply because V is set to zero and only [itex]\vec{A}[/itex] is used...

    The Coulomb gauge does violate Special Relativity. Changes in V which propagate
    instantaneously (dV/dx) give observable violations. Therefor it is not physically
    meaningful.


    True, the Lorentz gauge does not determine the potentials uniquely. The Lorentz
    gauge does contain the Liènard Wiechert potentials which are uniquely determined
    and the Liènard-Wiechert potential [itex]A^{\mu}[/itex] does transform like a 4-vector.

    Using other potentials as the Liènard-Wiechert potentials can give rise to all kinds
    of problems like Lorentz violations, space being not "simply-connected" and erroneous
    results for the spin density of the electromagnetic field in vacuum:

    [tex]
    {\cal S}^\mu\ \ =\ \ \frac{e^2}{4\pi^2}\ \mbox{\large $\varepsilon$}^{\mu\alpha\beta\gamma} A_\alpha\partial_\beta A_\gamma\ \ =\ -\frac{e^2}{4\pi^2}\ \mbox{\large $\varepsilon$}^{\mu\alpha\beta\gamma}\ F^{\mu\alpha}A_\alpha
    [/tex]

    Which depends explicitly on the absolute value of the potentials.


    Regards, Hans
     
    Last edited: May 17, 2008
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