Electromagnetic Waves: Solving Wave Equation for Point Charge

[Bobhawke]

How does, for example, the field of a single point charge arise as the solution of a wave equation? I tried solving the wave equation that arises from Maxwell's equations with a source, that is rho = q delta(0) (I can't get Latex to work), but it got too complicated for me. I think it might be necessary to approximate the charge distribution of a single charge as a gaussian and then take the limit in which it becomes a delta function at some point to solve this. Anyways, when you solve the equation in a vacuum you get perpendicular E and B fields oscillating in phase with one another. My question is, do you get the same thing when you solve in the presence of a source, and if so how does this lead to the 1/r^2 coulomb field?

 

[Phrak]

How does, for example, the field of a single point charge arise as the solution of a wave equation?

Point charges don't show up in Maxwell's equations, but for a region with a charge density you use Gauss Law,

$$q/ \epsilon = \nabla \cdot E$$

to find the static electric fields.

 

[Troels]

Anyways, when you solve the equation in a vacuum you get perpendicular E and B fields oscillating in phase with one another. My question is, do you get the same thing when you solve in the presence of a source, and if so how does this lead to the 1/r^2 coulomb field?

It doesn't - the static 1/r^2 coulomb field doesn't radiate and neither does the 1/r^2 amperian magnetic field. The radiation fields (those who goes as 1/r) are - in a sense - completely separate fields that only show up when your source is accelerating.

It is a very complicated business, even David J. Griffiths warns that the derivation is "rough going - but the answer is well worth the effort"

That being said, "Introduction to electrodynamics" by griffiths has a very beatiful coverage of both fields from arbitrary sources and how those field lead to electromagnetic waves

 

[rbj]

dunno if i agree with Troels (no 1/r with a point charge, as far as i understand).

the first Maxwell Equation:

$$\nabla \cdot E = \frac{\rho}{\epsilon_0}$$

is a differential (infinitesimally small) version of Gauss's Law. and Gauss's Law follows from the inverse-square law and a couple other very reasonable axioms.

 

[Bobhawke]

It doesn't - the static 1/r^2 coulomb field doesn't radiate and neither does the 1/r^2 amperian magnetic field. The radiation fields (those who goes as 1/r) are - in a sense - completely separate fields that only show up when your source is accelerating.

It is a very complicated business, even David J. Griffiths warns that the derivation is "rough going - but the answer is well worth the effort"

That being said, "Introduction to electrodynamics" by griffiths has a very beatiful coverage of both fields from arbitrary sources and how those field lead to electromagnetic waves

For sure the poynting vector is zero for a static charge so there is no energy flow, but still the field propagates through space with speed c. I want to know what the solutions of the wave equation look like in such a case, and how this ties in with the 1/r^2 coulomb field.

I will try and have a look at Griffiths and see if it can enlighten me some more.

 

[jtbell]

For sure the poynting vector is zero for a static charge so there is no energy flow, but still the field propagates through space with speed c.

For a static charge I don't think there is any useful way in you can say that the field produced by it "propagates through space with speed c". It simply "is".

Changes in the charge distribution (which is no longer a static situation, of course) produce changes in the field, which propagate with speed c.

 

[schroder]

For a static charge I don't think there is any useful way in you can say that the field produced by it "propagates through space with speed c". It simply "is".

Changes in the charge distribution (which is no longer a static situation, of course) produce changes in the field, which propagate with speed c.

Hmmmm. The static field simply is throughout all of space? But how does it get there? In setting up or creating a static field, the setting up produces a change, and that change propagates at c through space. After that change is propagated, the static field exists. So it seems correct to say that the static field is propagated through space. Or is that just semantics?

 

[Ben Niehoff]

Hmmmm. The static field simply is throughout all of space? But how does it get there? In setting up or creating a static field, the setting up produces a change, and that change propagates at c through space. After that change is propagated, the static field exists. So it seems correct to say that the static field is propagated through space. Or is that just semantics?

If you have no charge before time t=0, and then you have a point charge afterward, then there is a field that propagates through space. I think in this case, the Poynting vector will be a delta function confined to the forward light cone from the point at which the charge popped into existence.

Since charges don't generally pop into existence, we don't generally consider such problems. When you have a static point charge, you generally consider that the point charge has existed at that particular point in space for all eternity. That's what electrostatics is.

If you then accelerate the charge, you will get radiation fields that propagate at finite speed.

 

[Count Iblis]

$$\mathit V\left(\mathbf r ,\mathit t\right) = \frac{1}{4\pi\epsilon_0}\int \frac{\rho \left(\mathbf r' , \mathit t - \frac{\left|\mathbf r - \mathbf r'\right|}{\mathit c}\right)}{\left|\mathbf r - \mathbf r'\right|}\, d^3 r'$$

 

[jtbell]

If you have no charge before time t=0, and then you have a point charge afterward,

...which violates the charge continuity equation, and therefore also violates Maxwell's equations, unless there is a current flowing "into" that point:

$$\frac{\partial \rho}{\partial t} = - \vec \nabla \cdot \vec J$$

 

[atyy]

For sure the poynting vector is zero for a static charge so there is no energy flow, but still the field propagates through space with speed c. I want to know what the solutions of the wave equation look like in such a case, and how this ties in with the 1/r^2 coulomb field.

I will try and have a look at Griffiths and see if it can enlighten me some more.

This link has a section about EM waves which may be useful: "So in electromagnetism, knowing all about the sources isn't enough to specify the fields." http://math.ucr.edu/home/baez/gr/ricci.weyl.html

Feynman discusses what happens when you move an infinite plane of charge - I don't remember if that produces a plane wave, though that would be my guess.

The radiation field produced by a point charge is given by the Lamor formula, Griffith should have it. Some places online:

Dan Schroeder's "Purcell Simplified" http://physics.weber.edu/schroeder/mrr/MRRhandout.pdf.

Xiaochao Zheng's "Radiation reaction and electron's self energy -- an unsolved problem" http://www.jlab.org/~xiaochao/teaching/PHYS343/tex/chap11-6.pdf

 

[Phrak]

You can ask about the electric field about a small charged spherical region (a point charge will result in infinite field strength). Imagine the charge at rest, displaced some distance x at some velocity v, then once again at rest.

There will be jag in the electric field with an accompanying magnetic field. This jag in the field will propagate outward in a spherical shell at the velocity c.


The senerio of a charge appearing out of nowhere would be nice, but the change in the electric field must be accompanied by perpendicular magnetic field. There is no physically unambiguous solution for the magnetic field! There's nowhere for it to go within the surface of a sphere. So even without invoking the charge continuity equation, it doesn't work.

 

[schroder]

If you have no charge before time t=0, and then you have a point charge afterward, then there is a field that propagates through space. I think in this case, the Poynting vector will be a delta function confined to the forward light cone from the point at which the charge popped into existence.

Since charges don't generally pop into existence, we don't generally consider such problems. When you have a static point charge, you generally consider that the point charge has existed at that particular point in space for all eternity. That's what electrostatics is.

If you then accelerate the charge, you will get radiation fields that propagate at finite speed.

Then, from the conservation of electric charge, all of the electric charge that ever existed in the Universe existed at time zero, and presumably a static field existed then also. From that point on, only changes are super positioned/propagated throughout that field. Is that correct? It sounds to me something like the old idea of ether. What I question is the significance of a “static” field at all, since none has existed since time zero; everything since then is dynamic. And it would seem to me that the gravity field must be treated in a similar way. Is this current theory?