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Covariant formulation of Coulomb's Law

  1. Jun 5, 2010 #1
    I was noodling about with 4-vectors and Maxwell's Equations when I took it upon myself to figure out how to write the field of a point charge in terms of 4-vectors, since I'd never seen it before (if anyone knows a book that has this I'd be happy to see it). It's actually relatively simple, which I thought was pretty cool.

    So let's take a point particle with charge [tex]q[/tex] and four-velocity [tex]U^{\mu}[/tex] that passes through the origin at [tex]t = 0[/tex]. Let [tex]X^{\mu}[/tex] be the coordinate where we want to evaluate the potential [tex]\Phi^{\mu}[/tex] and fields [tex]F^{\mu\nu}[/tex]. The speed of light and the Coulomb constant are both 1. I'm using metric signature (+, -, -, -), because I like timelike vectors to have positive norms, as God intended.

    So after mucking around a bit I got this:

    [tex]\Phi^{\mu} = \frac{qU^{\mu}}{\sqrt{(X^{\alpha}U_{\alpha})^{2} - X^{\alpha}X_{\alpha}}}[/tex]

    [tex]F^{\mu\nu} = \frac{q(X^{\mu}U^{\nu} - X^{\nu}U^{\mu})}{((X^{\alpha}U_{\alpha})^{2} - X^{\alpha}X_{\alpha})^{3/2}}[/tex]

    The [tex](X^{\alpha}U_{\alpha})^{2} - X^{\alpha}X_{\alpha}[/tex] expression in the denominator of both equations works out to just [tex]r^{2}[/tex] if we're in a frame where the particle is at rest ([tex]U^{\mu} \rightarrow (1, 0, 0, 0)[/tex]), so you get that the potential drops off as [tex]1/r[/tex] and the fields as [tex]1/r^{2}[/tex] as they should. If the particle is moving that expression gives the "length contracted" [tex]r^{2}[/tex] that we expect from Special Relativity. Note that this expression is also (up to a constant and a sign) equal to the norm of the [tex]X^{\mu}U^{\nu} - X^{\nu}U^{\mu}[/tex] expression in the numerator of [tex]F^{\mu\nu}[/tex]. I'm not sure if that means anything but I thought it was kind of interesting.

    Okay, so these are pretty neat and tidy formulas. But then I noticed that they kind of give you CT symmetry! If we flip the sign of the particle's charge and it's four-velocity, the potential and fields are exactly the same. So here we have a purely classical justification for the whole "a positron is an electron moving backwards in time" idea in QED. Which was new to me.

    Thought this was cool and felt like sharing.
     
  2. jcsd
  3. Jun 6, 2010 #2

    Meir Achuz

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    [tex]F^{\mu\nu}[/tex] for a charge moving with constant velocity is done in most EM textbooks by Lorentz transforming [tex]F^{\mu\nu}[/tex] from rest.
    You have given the covariant form of these expressions, which are usually given in vector notation in terms of \phi, A, E, and B.
     
    Last edited: Jun 6, 2010
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