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Relativistic Coulomb potential - how to understand the equation

  1. Jul 16, 2012 #1

    I need a formulation of the equation for Coulomb's potential. It needs to be an integral that applies to densities (so no delta functions). (I think the relevant densities are charge densities?) Also, it needs to be relativistic.

    So far I have:

    ? = [itex]\int[/itex][itex]\frac{ρ(r', f(r'))}{|?-r'|}[/itex]d3r'

    What exactly do I write instead of the two question marks?
    And what is the function f?

    Anyone who can help gets a big cyberhug.
  2. jcsd
  3. Jul 17, 2012 #2


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    Gauge-fixing electrodynamics in Coulomb gauge or in A°=0 gauge + solving Gauss law on flat Minkowski spacetime, one piece of the interaction reads

    [tex]\int d^3x\,d^3y\,\frac{\rho(x)\,\rho(y)}{|x-y|}[/tex]

    In addition you have transversal photons coupling to the current density. Using QED including fermion fields one can show that the full theory preserves Poincare invariance.

    I haven't done the calculation in GR, so I don't know this generalizes to space-time curvature.
    Last edited: Jul 17, 2012
  4. Jul 17, 2012 #3
    Thanks Tom!

    Although I think your numerator has two densities - I'm looking for the equation for the field (or Coulomb potential) generated by a single charge density.

    Also, I should have been more clear, that while I wish to stay away from GR for the time being, I'm after the relativistic equation in the SR sense. Now, I think this means we need to introduce times into the equation, and that these are somehow a part the function f(r') that I mentiond, and am hoping to flesh out (something about keeping track of the density's past light cone). So any help with this would be great.
  5. Jul 17, 2012 #4


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    What I presented is the Coulomb potential.

    For fields you have to use the so-called retarded potentials; they depend on the gauge, so I present the Lorentz gauge version (you can find similar expressions for Coulomb gauge and other gauges as well)

    [tex]A_\mu(x,t) = \frac{1}{4\pi}\int d^3y \frac{j_\mu(y,t_-)}{|x-y|}[/tex]

    with the retarded time

    [tex]t_- = t - \frac{|x-y|}{c}[/tex]

    This is classical electrodynamics and no longer valid in QED!
  6. Jul 17, 2012 #5
    Thanks, Tom. I think you're showing me that I know much less about this than I thought, for example, I didn't think there was a difference between a potential and a field. For example, in Newtonian gravitation, a mass density determines values (g(x)) in accordance with the standard law, and these values can either be thought of as field values or gravitational potentials. Either I've got that wrong too or the concepts come apart in electromagnetism?

    Anyway, the equation you've given looks like a nice clarification of the incomplete equation I gave above. Particularly if we phrase your equation with a similar structure to the way I phrased mine:

    [tex]A_\mu(x,t) = \frac{1}{4\pi}\int [ \frac{j_\mu(y,t - \frac{|x-y|}{c})}{|x-y|}] d^3y

    ...it looks as though my f(r') is in fact the retarded time you spelled out for me. Furthermore, my ρ is your jμ which I assume is the charge density?

    The major difference between the two equations is that yours adds a further constant: the one over four pi. I actually thought that wasn't needed because of something to do with the units in which we have chosen to write the equation down in. But since you seem to really know what your talking about, I'm happy to take your word that this constant is required.

    One question I have is, what is A? (I guess this is connected with my confusion above about the difference between fields and potentials?) I also wonder why you use μ subscripts and do not invoke ρ, which I thought was the standard symbol for densities.

    Thanks this is very helpful!!
  7. Jul 17, 2012 #6


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    OK; I see that you are not familiar with 4-vectors

    [tex]x_\mu = (t, -\vec{x});\;\mu = 0 \ldots 3[/tex]
    [tex]j_\mu = (\rho, -\vec{j})[/tex]
    [tex]A_\mu = (\Phi, -\vec{A})[/tex]

    Therefore my equation holds for the charge density (and the electric potential) as well as for the current density (and the vector potential from which you may calulate the B-fields).

    The difference between fields and potentials becomes important once you want to study the dynamics not only of the charge in a given potential but of the fields themselves. Of course you can fix a Coulomb field and calculate how it affects the charge. But now you want to do it the other way round, you want to calculate the fields that are generated by (moving) charges. That means you have to use field theory and Maxwell equations instead of of point-particle mechanics in a given background field.

    Restricting yourself to point particles, i.e. delta functions for mass and current density you will find equations looking more familiar to you.
    Last edited: Jul 17, 2012
  8. Jul 17, 2012 #7
    That's helpful that it applies to both charge and current densities. Perhaps point particle equations are more familiar to me in one sense, but the sort of equation you have given is exactly what I'm after, given what I'm trying to prove. Perhaps it will help to give some background.

    Basically, what I'm trying to do is show that a specific kind of algebraic transformation that obtains in pre-relativistic electric field equations, also obtain in the relativistic equations. The algebraic transformation is relatively simple; we start with the equation that determines the field for a single pre-relativistic charge density:

    E(x) = [itex]\frac{1}{4\pi\in_{0}} [/itex][itex]\int[/itex][[itex]\frac{ρ_{1}(x-y)}{|x-y|^{3}} [/itex] ] d3y

    That's for a single density. For multiple (pre-relativistic) densities we just add a sum (superposition principle for interactions):

    E(x) = [itex]\frac{1}{4\pi\in_{0}}[/itex][[itex]\sum^{N}_{i=1}[/itex][[itex]\int[/itex] [itex]\frac{ρ_{i}(x-y)}{|x-y|^{3}}[/itex]]d3y]

    OK, now for the algebraic transformation that interests me. What I want to do is transform this last equation back into the single density equation, by swapping the sum and the integral:

    E(x) = [itex]\frac{1}{4\pi\in_{0}}[/itex] [itex]\int[/itex][ [itex]\frac{(x-y)}{|x-y|^{3}}[/itex] [itex]\sum^{N}_{i=1}[/itex] ρi ] d3y

    This is supposed to show that charge densities are additive (the charge of the whole is the sum of the charges of the parts). For to get the right field, the composite density distribution that must be inserted in the single density equation has to be the sum of its parts, as the final equation shows.

    OK so my goal is just to show that similar arguments apply in the relativistic context. And, at least if the superposition principle applies, then it looks as though I can run the same argument with your equation? In other words, does it make sense to say that the field determined by multiple relativistic charge densities is given by:

    [tex]A_\mu(x,t) = \frac{1}{4\pi} [\sum^{N}_{i=1}[ \int [ \frac{j_i\mu(y,t - \frac{|x-y|}{c})}{|x-y|}] d^3y]

    And can we again just swap the sum and the integral so as to recover an instance of the single density equation you gave that determines the same field as the above?
  9. Jul 17, 2012 #8


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    In electrodynamics (linear field theory with superposition principle) non-overlapping charge densities and fields are additive; interchanging sum and integral is usually OK, sometimes you will have to check for convergence explicitly.

    In order to calculate the field strength you have to calculate the gradient of the integrals I posted; make sure that you get the retared time correctly; it depends on the coordinates and has to be taken into account for the gradient! In addition for non-stationary densities E- and B field become time-dependent (see Maxwell's equations) and you may have to take time derivatives and rotation into account.

    Therefore you better study the simple equations for the potential first.
  10. Jul 18, 2012 #9
    OK great, so I take from your discussion that firstly, this:

    [tex]A_\mu(x,t) = \frac{1}{4\pi} [ \int [ \frac{(y,t - \frac{|x-y|}{c})}{|x-y|}\sum^{N}_{i=1}j_i\mu] d^3y]

    Is a valid expression, that is equivalent to the previous equation, the only difference being that the sum and the integral has been swapped--or at least if our fields and densities do not overlap, this is valid; whereas if there is overlap, we will have to be aware of this in our calculations and do what ever is required to account for them.

    Secondly, for simplicity, I should stipulate that I only intend to prove what I want to prove for stationary densities, because non-stationary densities give rise to further complications and potential adjustments to the equation.

    And thirdly, to properly underdstand what is going on in this equation, I need to study four vectors (which is where \mu comes in), and also, how to calculate the gradient of an interval, and that to study these properly, I am well advised to begin with the simplest equations for the potential first?

    If I'm right that these are the upshots of our discussion then this has been very helpful!
  11. Jul 18, 2012 #10


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    OK, let's start with stationary densities.

    For the E-field you have to use the (static) j0 = ρ the calculate A0; then the E-field is just E = grad A0

    For the B-field you have to use the (static) vector j the calculate the vector potential A; then the B-field is just B = rot A

    The simplest case is the static electric charge with an E-field, so we start with this case - and postpone the B-field

    [tex]A_0(x,t) = \Phi(x,t) = \frac{1}{4\pi}\int d^3y \frac{\rho(y,t_-)}{|x-y|}[/tex]

    with the retarded time

    [tex]t_- = t - \frac{|x-y|}{c}[/tex]

    Your expression became wrong in the meantime; you should first check the argument of j; it must read

    [tex]\rho(y,t_-) = \sum_i \rho(y,t_-)[/tex]

    and therefore

    [tex]A_0(x,t) = \Phi(x,t) = \frac{1}{4\pi}\int d^3y \left[\frac{1}{|x-y|}\sum_n \rho_n(y,t_-)\right][/tex]

    Now you can calculate the E-field:

    [tex]\nabla A_0(x,t) = \nabla \Phi(x,t) \to \partial_i \Phi(x,t) = \frac{\partial}{\partial x^i} \Phi(x,t) \to \frac{1}{4\pi}\int d^3y \left[\frac{\partial}{\partial x^i}\frac{1}{|x-y|}\sum_n \rho_n(y,t_-)\right][/tex]

    In many cases will be be simpler differentiate after integration, but this depends on your specific problem.

    How does a charge distribution you are interested in look like?

    Pointlike charges at different locations? Then I would propose the following:
    - calculate A0 for one pointlike charge
    - calculate E for one pointlike charge
    - sum over all charges
    Last edited: Jul 18, 2012
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