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Scalar Potential Problem

  1. Aug 6, 2005 #1
    There´s an assertion that the electric field depends only on Vector potential in radiation region. But I couldn´t see clearly why the contribution of scalar potential could be comparatively ignored. Could anyone give me some explanations? Thanks!
  2. jcsd
  3. Aug 6, 2005 #2
    well, radiation is nothing more than oscillating electric and magnetic fields, so let's look at maxwell's equations in vacuo...

    div E = 0
    div B = 0
    curl E = -dB/dt
    curl B = mu_0 epsilon_0 dE/dt

    since the divergence of the electric field is zero (in vacuum), it is "solenoidal," that is, it's purely rotational, and in that situation, you can't use scalar potential--you have to use vector potential.

    now it's in the same boat as the B-field! it's divergence is always zero (so long as we don't find any magnetic monopoles!), so it can't have an associated scalar potential.

    ...did that help?

    now...i'm unsure of how the situation changes once charges are introduced. ...is that your question? :redface:
  4. Aug 6, 2005 #3
    Just to point what seemed to me as an incorrect reasoning:

    Uniform electric field has zero divergence and is consequence of a well defined scalar potential V(X,Y,Z) = kX, for instance.

    Best regards,

  5. Aug 6, 2005 #4
    Aditionally, it seems that in the space betweem charges (which can be put very far apart) the field is neither uniform, nor can only be described by vector potentials.
  6. Aug 6, 2005 #5

    right! my mistake. :redface:
  7. Aug 6, 2005 #6

    from pg. 43 of arfken and weber...

    "If we have the special case of the divergence of a vector vanishing, the vector...is said to be solenoidal.... When a vector is solenoidal it may be written as the curl of another vector known as the vector potential."
  8. Aug 7, 2005 #7

    Hans de Vries

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    Science Advisor
    Gold Member

    This is not true. It does depend on the scalar potential as well.

    The separation of E in dV/dr and dA/dt is unfortunately often not given.
    (Jackson doesn't show it for instance in Chapter 14 on Radiation by moving charges)

    You can find the formulas online here:


    see formula 15 for dA/dt and formula 16 for dV/dx. The dotted beta (v/c) is
    the accelaration (a/c) which gives the radiation terms. You can simplify the
    formula by setting beta itself to zero (v<<c).

    You also should be able to find them here:


    At least when rochester.edu is back up again (maintanance?) at page 28.

    Regards, Hans
    Last edited: Aug 7, 2005
  9. Aug 7, 2005 #8
    I would like to introduce in this discussion the concept of "gauge". One seems to have some liberty of expressing the same physical situation with different formal expressions.

    I once used the radiation gauge and, as I understand it, it represents one possible way of expressing field situations. V and A concepts provide some amount of redundancy, and this gauge choice is a way of fixing this redundant system of language.

    Best Regards,

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