# Moving electrical charges and Maxwell's equations

1. Mar 25, 2010

### closet mathemetician

1) does an electrical charge (say, an electron) traveling with a constant velocity (say, in the x-direction) travel as an electromagnetic wave?

I'm thinking of an analogy with flowing mass. Suppose you have massive particles, evenly distributed through some volume, traveling with constant velocity, say the x-direction. The flux across the (y,z) surface of a volume is calculated as the integral of the current density, vector $$\vec{J}$$ over the (y,z) surface area.

where

$$\vec{J}=\rho\cdot\vec{v}$$

and
$$\int\int_A\vec{J}\cdot dA$$

The result is the flow of mass across the (y,z) surface area per unit time, with units of kg/s. There is no curl in this vector field.

Now, imagine that instead of flowing mass, we have flowing electric charge in the x-direction across the (y,z) surface of a volume. Assume the electrons are not being accelerated as they flow across the (y,z) surface.

With this substitution the flux of mass across the surface becomes the flux of electrons across the surface with units of q/s, which is electrical current.

Does each electron rotatate as it travels in a straight line, producing an "infintesimal" rotation, which, when integrated over the surface area (y,z), produces an overall rotation of the vector field (Stokes theorem).

Does any of this cause the electrons to radiate? I don't see any light radiating when current travels through a metal wire.

Maybe I'm confusing electromagnetic waves traveling along the electric field lines with electrons traveling through space.

2. Mar 27, 2010

### uzair_ha91

MY knowledge (which still remains untested) is that an accelerating electric charge produces a changing electric field. This changing electric field then produces a changing magnetic field which again sets up a changing electric field. This continuous process is known as an electromagnetic wave. Thus electromagnetic wave is actually caused by oscillations of electric and magnetic fields which arise due to an accelerating charge not a charge moving with constant velocity.
A charge moving with constant velocity only creates a constant magnetic field.

3. Mar 27, 2010

### kcdodd

The simple answer is no. If a particle is moving at a constant (non-zero) velocity, that means there exists an inertial frame in which the particle has velocity of zero. Clearly there would be no EM waves in this frame since you simply have E = kq/r^2 (EM waves cannot be "gotten rid of" in this way). The other way to know it is not a EM wave is that EM waves travel at the speed of light (in free space). However, this "wave" simply travels at the speed of the particle.

As a side note, if you know electric field in this "rest frame", then you can perform a certain type of transformation (called a Lorentz transformation) to know the electric and magnetic fields in the original frame.

4. Mar 30, 2010

### closet mathemetician

Thanks for the responses. I guess the key is that to radiate the charge must be accelerating, but must the acceleration always be in the form of an oscillation? What if you accelerated a charge in a straight line? Or, maybe if you tried that the charge would always oscillate anyway?

Maybe what you are saying is that the very occurrence of the mutual induction of E and B fields causes the oscillation, so any time you accelerate a charge it will oscillate?

And yes, kcdodd, I know about Lorentz transformations (got lots of questions about that too, but I'll save for another thread).

5. Mar 30, 2010

### elect_eng

An oscillating charge will create a steady wave with a fixed frequency. The oscillation can be charge flowing around a loop, or a straight line back and forth motion.

A straight line accelleration in one direction still creates radiation, but the wave is a transient pulse with a broad spectrum.

6. Mar 30, 2010

### kcdodd

If I get too technical just tell me. The definition of radiation has to do with integrating the flux of power, or pointing vector (S = ExB), over a closed surface around the particle. For a single particle, the flux across this surface is proportional to the acceleration of the particle. Contrary to the last post, a constant acceleration will create a constant radiation field in a direction perpendicular to the path of travel. Clearly it also has no period, but it is clearly not a burst.

In fact, a particle which is rotating is the one which gives bursts of radiation (which, at least at one time, was the explanation for pulsars). Since the perpendicular direction to the path also rotates (except along the axis of rotation), you get an outward spiral of radiation. You actually call the "bursts" as waves, but depending on the exact parameters they may not look like waves (aka, broad spectrum of waves as the poster mentioned). So in short, radiation is not necessarily the same thing as waves.

Last edited: Mar 30, 2010
7. Mar 31, 2010

### closet mathemetician

you're good, kcdodd, I'm getting it, little, by little.

8. Mar 31, 2010

### elect_eng

So, according to you, a charge accellerates by you and continues traveling on out to an infinite distance. When the the charge is 100 million light years away, you will still be experiencing a constant radiation field. Also, you will be seeing photons with infinite frequency (implying infinite energy) since you say they have no period (I assume no period means 0 period, hence f=1/T=infinity).

9. Mar 31, 2010

### elect_eng

I didn't say rotating, I said traveling in a loop. A charge traveling in a circle can be viewed as a sinusoidal oscillation in both the x and y directions. Photons are generated at the frequency of the charge circulation (cyclic frequency). If the angular velocity is constant and this exists for a long time, then you have a steady state solution, which is what I mean by a steady wave with fixed frequency.

10. Apr 1, 2010

### kcdodd

By no period, I mean it is not periodic. And I also meant travelling in a loop.

11. Apr 1, 2010

### bjacoby

As pointed out the answer is "no". But that has to do with waves that are electromagnetic as you are thinking of. In that case acceleration is needed to create propagating waves.

But there is more to this than that. Apparently waves of some type do exist even for electrons traveling at constant velocity. And the evidence for that is that electrons, even single electrons exhibit wave behavior such as diffraction at a slit. So if one has an electron beam traveling at constant velocity through space one has to wonder just what it is that causes the statistical paths of electrons to be on average solutions to a wave equation? It's not clear at all why such at thing is true. But it is. But these waves are apparently not electromagnetic in any sense.

12. Apr 1, 2010

### elect_eng

If we are talking about outward radiation of real power (not reactive power in the near field), then this energy transmission must take the form of photons. Each photon will have a quantized energy and a corresponding frequency. How can photons have no period (or frequency)? If we restrict the interpretation to classical EM theory, how can a traveling electromagnetic wave not be periodic? Doesn't it need to be a solution of the wave equation?

13. Apr 1, 2010

### Born2bwire

I too am at a lost on kcdodd's views here. Charges that have a regular oscillation that elect_eng described previously give rise to electromagnetic waves. All electromagnetic radiation are waves, I am not sure why kcdodd would state to the contrary. The only exception I would give is if we are talking about alpha or beta radiation which involves particles. Synchrotron radiation is just an example of highly directed electromagnetic wave excitation.

14. Apr 1, 2010

### kcdodd

I am simply stating a definition of radiation as the Poynting vector. If you integrate this flux around a closed surface, and is non-zero, then energy is moving out from whatever is inside the surface. I have said nothing about waves. The poynting vector of a classical em plane wave is periodic in magnitude, but always in the direction of motion indicating the movement of energy. If you accelerate a charge with constant acceleration, then you must be doing work on EM field at a rate proportional to v*a (which is proportional to the total flux of energy out). Nowhere here is a period.

How this relates to photons I cannot answer you.

15. Apr 1, 2010

### elect_eng

I disagree with you. There can indeed be a period. Also, any explanation must be related to photons. This is what I tried to do in my first post, yet you chose to shoot it down even though you can't correlate your explanation with simple physics.

I'm OK with your explanation of the Poynting vector, but I disagree with your statement that my description of a radiation pulse is not correct. Imagine that you are a stationary observer and you see a charge accellerating by you. What will you observe from the "Poynting vector"? You will see a radiation pulse as the charge accellerates by you. The charge is initially far away and you see a very small intensity, however when the charge is going by near you, maximum intensity is observed. Then as the charge fades in the distance you see small intensity again. Now think of Fourier analysis. How can a pulse have a period? Well, it can't have one period, but it can be an integration of a spectrum of sinusoidal waves. This is how the solution obeys the wave equation. Superposition applies and the summation of an infinite number of sine waves forms a pulse in the time domain. This is the classical explanation in terms of Maxwells equations. However, in the quantum world, this radiation is interpreted as photons. As an observer, you will see the emission of photons with a broad energy spectrum.

At least that is my opinion. This is the only way I can make sense of your description of the Poynting vector radiating power away (which I agree with), and the well known fact that the energy must disappear as energy carried away by photons (which you should agree with).

Last edited: Apr 1, 2010
16. Apr 1, 2010

### kcdodd

Firstly, photons are not "simple physics". If you cannot explain classical events with classical physics, then adding photons and quantum mechanics is not going to help. How do you "simply" explain the electric force on a charged particle with photons? That is out of the scope of this forum.

Second, you do not see radiation of a particle coming directly toward you, only the perpendicular component.(thinking now about this it is hard to believe, but look at a dipole antenna radiation pattern). Granted, if a particle passes you it's distance will go something like tan(angle), and you will get a magnitude change due to the distance effect, but that is not radiation. You would see that even from a particle with a constant velocity that is just passing by.

I suppose you might call that the induction definition of radiation. If a charged particle passes by, you will see a changing E and B, and you can use a "probe" to extract energy, and therefore conclude that radiation exists. However, it is really the probe which is doing work on the field, and so the flux through a surface surrounding the probe will be non-zero. But the flux through the surface around the particle of interest still is zero. So this is not a good definition of radiation.

The magnitude of the poynting vector, like I said already, is proportional to the rate of work done on the particle, which is also proportional to v*a. If acceleration is constant, you will get something like a^2*t (where i just subbed in the velocity as a function of time). If you wish to do Fourier analysis on this function that's fine, but I just don't think it illuminates very much.

edit: I realized there is something wrong with the power relation I have stated here. see my later post.

Last edited: Apr 2, 2010
17. Apr 1, 2010

### elect_eng

I did give a conceptual explanation in terms of classical physics. By bringing in the notion of photons, I'm just trying to show an additional point of view. I could avoid any further discussion of photons if it is viewed as out of the scope of this forum, or this thread. It's not a critical part of my view on what is happening here. Still, why ignore the well-known wave-particle duality? It is a useful concept.

In any event, you do make a good point about "explaining classical events with classical physics". I should present a mathematical derivation in terms of Maxwells equations in the classical context. If I don't do this, then I'm just giving an opinion without proof. The weekend is coming up, so I can spare an hour or two to write something up.

If you are inclined, you can also present a more detailed explanations with equations and derivations. It's hard to follow some of your points without more details. Some things are confusing. For example you said "The magnitude of the poynting vector is proportional to the rate of work done on the particle", but some of the work done is used to generate kinetic energy, since any charged particle has mass and velocity is continually increasing. This may just be a wording issue, but equations will make it more clear and leave no ambiguity. There are other examples of confusion I have in interpreting, but I don't want to be nit-picky. -I'm just saying that sometimes equations leave less wiggle-room for misinterpretation.

18. Apr 1, 2010

### kcdodd

That sounds good.

By work done on the particle I did mean by the em field. It would probably be better to say the work done on the field by the particle, since the field energy is going up.

That leads to a misunderstanding I realized I have when I posted earlier, and that is the acceleration dependence of radiation power. Apparently it is the second power of a (acceleration), not the first. I only had an intuitive idea that the radiation reaction force on the particle was simply an induction like effect, resisting the change of velocity. Basically, an induced electric field in the opposite direction as acceleration. And so you get work done against this efield as you accelerate putting energy into the field.

In math terms

$$\nabla \cdot \vec{S}= - \vec{J}\cdot\vec{E}$$

where, now we have for a particle;

$$\nabla \cdot \vec{S} = - q\vec{v}\cdot\vec{E}$$

at the position of the particle, and zero everywhere else.

The electric field on the right hand side, in my mind, was simply proportional to the first time derivative of velocity in the way I already mentioned (so that v and E oppose each other). I am missing where the second power of acceleration comes from. If you go to the power of radiation formula you also loose the velocity dependence, which doesn't seem logical because that seems to indicate you could get radiation power out with no mechanical power in (when v = 0). So, I need to do some homework on where the disconnect is.

Last edited: Apr 2, 2010
19. Apr 3, 2010

### elect_eng

I ran across this article which seems interesting and relevant to what you are saying here.

http://www.mathpages.com/home/kmath528/kmath528.htm

It seems there is some controversy about the case of a uniformly accelerating charge. One of the paradoxes seems tied to the equivalence principle. If a uniformly accelerating charge radiates, then a stationary charge in a gravitational field should radiate also. These types of questions are beyond my expertise. For this reason, anything I say should apply to a charge accelerating in one direction, but not necessarily a constant acceleration. The above link makes reference to the first and third derivatives of position. It's also an interesting discussion.

I'm still putting together a derivation and explanation for why an observer would experience a radiation burst with a broad spectrum if a charge accelerates while traveling by in one direction. I've mapped out the approach and it is straightforward and based on the standard derivation of radiation from a short dipole. I'm trying to keep the explanation on the lowest possible level, and also make the explanation clear. Hence, it takes some time to put this together. The approach I'll show allows someone familiar with basic antenna theory to follow.

My basic approach considers that the radiation seen at any particular time is the result of the current from the accelerating charge at one particular section of it's path. This small section can be viewed as a short antenna with a current that is an impulse function. Fourier analysis can be used to represent this current pulse as a broad spectrum (that is, an integral of sinewaves over a frequency band). The full effect of what an observer sees over time is then the integration over the entire path of the charged particle. The overall effect is a pulse of radiation due to distance effect (low radiation intensity when particle is far away), combined with the broad spectrum generated by a current pulse in each short section of the particle's path.

Last edited: Apr 3, 2010
20. Apr 3, 2010

### kcdodd

Yes I encountered that "paradox" a while back, but it is not really the issue here. The short answer is uniform acceleration does radiate. But if acceleration is constant, then magnitude of poynting flux is constant (assuming literature is correct), which has a spectrum of a single frequency of zero. I am not really sure how this frequency domain relates to photon frequency domain. Interested to see what you get with a different approach.

Last edited: Apr 3, 2010