Accelerating Charges: Classical Theory and Light Radiation

[pmb_phy]

Hi Reilly

Thanks for stopping by. Always great to see you here!

Radiation means energy flux through the surface at infinity, so the radiation fields must go as 1/r. In the L-W potential, given in one of dextercioby's prior posts,the 1/r term is proportional to the acceleration. No acceleration, no radiation.

Now see? This is something I don't get. A plane wave is a solution of Maxwell's equation and it does not decrease with distance traveled. Let me guess - It doesn't appear in nature so we don't talk about it?

Pete

 

[reilly]

Pete -- Not really. The basic idea is that, locally, a spherical wave front far from its source will look like a plane wave. And, macroscopic light bouncing off mirrors comes pretty much in plane waves. In such a case, the plane waves are superpositions of lots of sources. There's lots of math to go with all of this.
Plane waves, spherical waves are all alive and well. But, indeed plane E&M waves are a bit of an abstraction, but a useful one at that, particularly in QED.

Thanks & regards,
Reilly

 

[pmb_phy]

But, indeed plane E&M waves are a bit of an abstraction, but a useful one at that, particularly in QED.

Yep. That's what I was referring to by not occurring in nature. Thank God I don't have to teach this point in a class. I think the students would razz me.

Pete

 

[pmb_phy]

It just occurred to me how simple the answer is to the question regarding the Poynting vector. Simply provide a simple example of the Poynting vector and then see if the wave equation is satisfied. E.g. Let an uncharged current carrying wire sit on the x-axis. Here there is only a B-field. Now transform to a frame moving parallel to the wire. Now there is both an E field and a B field. These fields are static and therefore do not satisfy the wave equation and therefore there is no radiation in the new frame. This can be placed in the covariant form

\Box A^{\alpha} = 0

which is the wave equation in the absense of charges and current as is the case above outside the wire. So no radiation in one inertial frame -> no radiation in all inertial frames. If you're wondering about the case A^{\alpha} = constant it represents a wave with infinite wavelength. At least this is what sounds right. I'm not 100% sure. Hopefully Reilly or Rob will chime in and make sure.

Actually "to radiate" is different than "there is radiation" but it should be convincing about the absolute existence of detecting radiation in an inertial frame. But it does address the question in an indirect way.

Pete

 

[Borogoves]

Creating an electromagnetic field with nonzero Poynting vector means emitting radiation...An electric charge moving with constant velocity emits radiation.

The end.

Daniel.

Oh don't be so arrogant young man. The Poynting vector is locally invariant.

_________________________________________________________________

 

[Meir Achuz]

Creating an electromagnetic field with nonzero Poynting vector means emitting radiation...An electric charge moving with constant velocity emits radiation.

The end.

Daniel.

I just came across this thread. Each statement above is completely wrong.
It is contradicted in several posts in this thread, and by all textbooks.
Radiation must fall off no faster than 1/r^2, and this only occurs if there is acceleration.

 

[Pwantar]

Alright,Pëte.Computing the Poyting vector for a Liénard-Wiechert field generated by a constant velocity moving electric charge,one gets a nonzero result.Ergo,radiation is being emitted.

I'm aware people always ponder about Larmor's formula,but this case must not be excluded.

Daniel.

Yes, but did you integrate the Poynting vector over a sphere containing the charge? This gives you the power. A non-zero result there means it radiates... a non-zero Poynting vector does not. That is what it means for something to radiate... To carry energy out to infinity.

What we are looking for, is a pointing vector whose radial component is inversely proportionate to r^2, so that the limit as r approaches infinity of the Poynting vector integrated over the sphere containing the point charge is not equal to 0. (Note, the area element in the radial direction is given by r^2*sin(t)*dt*dp, where t=theta, p=phi... it makes sense that an inversely proportionate r^2 dependence would give a non-zero result.)

And if you look closely, the Poynting vector given by the E and B fields given by the Lienard-Weichart formula has an inverse dependence on r^4... Thus, the Poynting vector integrated over a sphere containing the charge tends to 0 as r approaches infinity. Thus, it does not radiate.

As for the original question of the thread, maybe it's an if and only if thing. We already know the answer in one direction.. Can we show it the other? I'm unsure, and will certainly ponder over it.

By the way, everyone interested in this stuff, check out chapter 10 of Griffith's Introduction to Electrodynamics... it's great for a high-level introduction. I'm actually studying for an exam on this stuff now... well, back to stuff that will actually be on my exam.

-Tina the Bunny

 

[PhyDude]

Hi Everyone

Is it possible to determine the force that a positive test charge experiences when it is released near a stationery/fixed positive charge by simply using Coulomb's law? The positive test charge obviously accelerates non-uniformly but can Coulombs law really be used to determine the force that the stationery point charge exerts on the positive test charge?

Any help would be appreciated and I know that this is kinda a homework question and not really a radiation question but its is related to the topic of this thread.

 

[jason12345]

Check out the Liénard-Wiechert potentials and fields in any serious electrodynamics book,J.D.Jackson being the first to turn to.

Daniel.

It's far too complicated in Jackson. Giffiths' Introduction to Electrodynamics gives a far easier to understand derivation