# Maxwell equations in Lorenz gauge

1. Apr 13, 2012

### jjustinn

In the Lorenz gauge, the Maxwell equations reduce to four inhomogenous wave equations, with the charge density acting as the source for V, and the current density for A.

For now, just take a static charge distribution -- say, a point charge at the origin.

It is well known that a static charge distribution leads to the electrostatic field; for our point charge at the origin, V(r, t=infinity) = 1/r.

However, if the charge is static, V would be constantly increasing...wouldn't it? So at t=infinity, V would be infinity everywhere? Now, running a simulation (because I suck at PDEs), it appears that while this appears to be true, E = gradV does appear to stay constant...so perhaps this is a red herring, but I would feel much more confident if I could find an analytical solution (or a logical explanation).

Any takers?

2. Apr 14, 2012

### fzero

If the charge distribution is static, then $\dot{\rho}(\mathbf{x})=0$. The scalar potential is

$$V = \int \frac{\rho(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|} d\mathbf{x}',$$

so $\dot{V}=0$ as well. So $V$ is static. Note that when the potential is static, the inhomogenous wave equation reduces to the Poisson equation of electrostatics. For the point charge at the origin, $V\sim 1/r$ at all times. The only divergence is at the origin.

3. Apr 14, 2012

### jjustinn

Right - but that assumes that it was always static, and therefore you're assuming what I want to derive: that's the green function for the Laplacian, not the Wave equation.

Put another way, you're assuming it was static for all time; But if the point charge was turned on any time after t=-infinity, the static approximation won't hold (except possibly at t=infinity, but it seems that at any large but finite time, V will be much larger than 1/r...

So to restate the problem (all scalar factors set to 1)

V,tt = V,xx + V,yy + V,zz + p

p(t<0) = 0

p(t>=0) = point charge at origin

I'm wondering if one trick might be putting a boundary condition so V(r=infinity) = 0; since it only falls off as 1/r that would (I think?) be an extra assumption with no apparent justification (other than the fact it might give the right answer ;) ).

This sort of reminds me of how I understand the process of renormalization in QFT, which of course is exactly that sort of unjustifiable hack, but again, I could be way off base here.

Thanks,
Justin

4. Apr 14, 2012

### vanhees71

Take the retarded Green's function of the wave equation and use it to calculate the electromagnetic field for a static charge distribution. On the one hand that's a weird way to calculate the Coulomb potential, on the other it furthers the understanding of the matter quite well :-).

5. Apr 14, 2012

### fzero

I assumed that it was static because that's what you said it was. If we introduce a time-dependent charge distribution, we have to use the retarded Green function to solve the wave equation. The solution is now

$$V(\mathbf{x},t) = \int \frac{\delta(t'-t + |\mathbf{x}-\mathbf{x}'|/c)}{|\mathbf{x}-\mathbf{x}'|} \rho(\mathbf{x}',t') d\mathbf{x}' dt' .$$

This is derived in, for example, Ch 6 of Jackson.

There's no reason to introduce a boundary condition, since the initial condition is enough. We can do the integral above to find

$$V(\mathbf{x},t) = \begin{cases} 0, & t<|\mathbf{x}|/c, \\ \frac{\rho_0}{|\mathbf{x}|}, & t\geq |\mathbf{x}|/c. \end{cases}$$

At a given distance from the origin, the potential is zero until late enough times that a signal traveling at light speed can reach the point. After that, we have the ordinary static potential. This is as expected. By causality, we can't measure the potential until the signal has reached us. Once it does, the charge is effectively static, so we should measure the static field.

Renormalization is actually conceptually deeper than just removing infinities. It is the realization that physics can depend crucially on the energy scale that we use to probe a system. You can find some context here: http://en.wikipedia.org/wiki/Renormalization_group This Wilsonian viewpoint generally isn't covered until grad-level statistical mechanics or an advanced QFT course.

6. Apr 14, 2012

### vanhees71

This cannot be since then you have an electric field popping out of nothing instantaneously and instantly at any point in space. Of course this comes from the totally unphysical assumption of "jjustinn" concerning his charge distribution: You cannot pop up a point charge out of nothing at $t=0$. This violates not only causality but also the necessary condition of charge conservation and thus gauge invariance. What you can have, is a charge sitting at the origin forever (which is of course also an unrealistic assumption, but it's not in contradiction with any physical laws). Then you get immediately the Coulomb field as it shoud be (here written in its form using non-rationalized Gaussian units).

7. Apr 14, 2012

### jjustinn

First, apologies for my poor phrasing: I didn't mean to suggest you should have read my mind.

I'm familiar with the derivation of the Wave equation green function, and that's another thing that's puzzled me -- the Green function seems to say, as you point out, that the potential at point r goes from 0 to 1/r, then stays at 1/r for all time -- which not only sounds reasonable (since the charge appearing at t=0 is jut as discontinuous), but agrees with observations.

So perhaps my problems are stemming from my lattice-based simulation, where the charge is acting as a constant source, adding to the V around it, and since there is no dissipation / damping, it never goes away.

Or, it might also have something to do with the fact I'm only updating it in the plane, and the 2D Green function isn't a spreading delta function, but IIRC a log function of some sort.

To give some background, I originally thought my simulation was broken, since V was monotonically increasing everywhere (while E stabilized to a constant value) -- but I started reading this (http://psych.colorado.edu/~oreilly/papers/OReilly05_md.pdf), and he mentions adding a empirically-determined damping factor at the (artificial, Sommerfield-condition-ed) boundary "to achieve the 1/r dependence"; before I'd seen this, I'd found that adding a damping term at/near the boundary had a similar effect, but rejected it outright as non-physical.

Anyway, thanks for taking the time to respond. I'll play with this some more and will undoubtedly be back with more dumb questions with crucial unstated assumptions ;)

8. Apr 14, 2012

### jjustinn

Ah hah! I think this explains my problem: (from http://ieeexplore.ieee.org/xpl/logi...l5/97/5487427/05492190.pdf?arnumber=5492190):

Of course, you need a subscription to see the paper >:/

So...this actually brings up anther question, for a new thread: shouldn't our point charge at origin generate the same waves in the XY plane (or any plane through the origin) as a 2D charge would, via radial symmetry?

9. Apr 14, 2012

### jjustinn

Hey vanhees -- I missed yr replies earlier, sandwiched between the other ones. I've also been concerned with the non-physicality of the situation, but it seemed like mathematically it should have been fine.

However, I did find something interesting when trying to make it more physical: by adding an equal/opposite negative charge, its outward monotonically-decreasing negative wave cancels the monotonically-increasing wave from the positive source, making the average value stable, rather than increasing to infinity...but it's not too useful to have a simulation where the net charge has to be zero. This is what initially made me think of "renormalization" (used in the lay 'removing infinities' sense) -- it wasn't until later I found that gradV actually stayed constant.

In any case, would you care to expand on that last comment? If you are uncomfortable with a single charge coming out of nowhere, how about a positron/electron pair pulled out of the vacuum? Or since this is explicitly classical, how about if I separate the charges on a glass rod and a bit of fur at t=0, then fire the glass rod from a cannon into the distance at t= 1?