Falling Electrons

Not long ago I wrote about one of the conceptual problems between intro mechanics and intro E&M from the freshman physics standpoint: developing a sense of the size of units between the two subjects. For instance, accelerating a spacecraft to escape velocity is no easy feat, but accelerating an electron to escape velocity only requires a tiny fraction of a volt. Let's do a similar calculation with power radiated from an accelerating point source.

As we know, a particle with an electric charge produces an electric field. A moving charged particle also generates a magnetic field. And though we won't worry about the details yet, it turns out that an accelerating charge actually causes its own electric and magnetic fields to interact in such a way as to cause electromagnetic waves to ripple out in the form of light. Electromagnetic waves carry energy, and so accelerating a charged particle will require a bit of additional force in order to supply that power. The actual amount of power is given by the Larmor formula:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

As usual, a is acceleration, q is charge, c is the speed of light, and μ0 is the magnetic constant. We can use this to answer a conceptual question of the type that beginning students can use to develop a sense of scale. This is a classic question, which I first saw in Griffiths' book.

An electron is released from rest and falls under the influence of gravity. How much power does it radiate?

We can look up the charge of the electron and plug it into the Larmor formula with the acceleration equal to the usual 9.8 m/s2. I get:

i-88516ca505d84f969b7ab15a965c3fb3-2.png

This is not a lot of power. Even if you get a macroscopic glob of electrons (say, Avogadro's number of them) it's still an absolutely tiny total radiated power.

In retrospect it had to be tiny. Hairbrushes don't burst into glow when dropped. But as always, having a good order-of-magnitude grasp of the general concept can be the difference between understanding and confusion.

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Reconcile the local observer's and external observers' point of view.

An electron and its local observer vacuum free fall into the sun, a = 273.95 m/sec^2 (27.9 gees) as seen by an external observer. Does the electron radiate? (Do it at night to reduce background noise.)

An electron and its local observer ride a rocket, a = 273.95 m/sec^2 (27.9 gees) as seen by an external observer. Does the electron radiate?

An electron and its local observer ride the inner rim of a Beckman Optima 10^6 gee ultracentrifuge as seen by an external observer. Does the electron radiate?

A conducting loop has a magnetic field. Do its accelerating (direction) electrons radiate? A superconducting loop has a magnetic field. Do its accelerating (direction) electrons radiate? A permanent magnet's field is generated by aligned orbital angular momenta (e.g., 37% of Sm_2Co_17 magnetic field). Do its participating electrons radiate?

@ Matt,
On what scale is the radiation from protons in a particle accelerator such as the Tevatron or the LHC?

@ Uncle Al,
Re your first question: could you please define "night" at the sun :D

It's pretty hefty. Hyperphysics has a good summary of it. At the relativistic speeds of particle accelerators, the effect is much more pronounced. That's the reason the Tevatron and the LHC use protons; the increased mass means you gan get them to higher energies at a smaller velocity.

An electron is released from rest and falls under the influence of gravity. How much power does it radiate?

Supposed to be EXACTLY the same as how much power is radiated by a positron released from rest and falling under the influence of gravity.

But this has not been well-tested, nor the crucial test of falling anti-hydrogen atoms, behaving identically to falling hydrogen atoms, which Michael Salamon, head of all Astrophysics missions at NASA HQ, would dearly like to see experimentally tested.

Is the Cherenkov effect caused by the deceleration of a charged particule, or is it unrelated ?

It shares some similarities with the Cherenkov effect since both are a result of acceleration. However, the Cherenkov effect specifically requires the electron to be decelerating in an insulator while traveling at a speed greater than c/n, where n is the refractive index of the material. That creates sort of a shock wave effect.

Cherenkov light has to do with the fact that the speed of light in a medium is c/n (where n is the index of refraction). When a charged particle enters a medium moving faster than c/n, it creates radiation similar to a sonic boom.

By creeky belly (not verified) on 06 Aug 2008 #permalink

Whoops, you were quicker on the draw Matt. Sorry for the echo chamber.

By creeky belly (not verified) on 06 Aug 2008 #permalink

If an observer and an electron fall together in a graviational field, the observer will cinclude that the electron is at rest and radiates no energy.

If the electron is stationary, the observer will conclude that the electron is accelerating upward, and does radiate energy.

When I charge up my van-de-graf generator in my garage, how come it doesn't glow from all the electrons accelerating upward against gravity?

By Paul Murray (not verified) on 06 Aug 2008 #permalink

Crickey Belly, it is always a good sign to receive answers from 2 different sources, and both match.

Thanks for the responses, guys.

I'd still like an answer to mine.

By Paul Murray (not verified) on 07 Aug 2008 #permalink

The van de Graff question is easy: the electrons on the belt aren't accelerating in any inertial frame, and thus won't radiate.

But the guy falling along beside the electron? I have to be honest, I don't know. I'm pretty sure the answer is going to involve general relativity, which at this point I haven't yet had occasion to learn in any detail. It's an excellent question, and when I find out I'll definitely write about it.

"the electrons on the belt aren't accelerating in any inertial frame"

Well - I thought they were. My understanding of GR was that standing stationary on the earth's surface is not an "inertial frame" because of the effect of gravity: there is no way to tell the difference between standing on the ground and being accelerated upward.

So why don't my generator glow?

By Paul Murray (not verified) on 10 Aug 2008 #permalink

Hmm. That sounds right, but after thinking about it I can't figure out how to reconcile the various reference frames. I've been taught special relativistic electrodynamics but I just don't know GR and I can't say anything definitive. I have checked out a GR textbook from the library and I'm going to try to figure this out before I'm back at school for the fall semester in a week and a half. Failing that I've got an NAS specialist in quantum optics in my department that I can ask.

By Matt Springer (not verified) on 10 Aug 2008 #permalink

The electrons in a van de graff don't really accelerate very much. Remember, it doesn't take a lot of electrons to produce a strong charge, the belt in a van de graff doesn't go too fast, and the electrons aren't being accelerated for most of the trip. Also, I'd suspect that the charge is carried in the form of some ionization of the belt material, in which case the electrons are in bound states and (I think) don't radiate for quantum mechanical reasons.

The questions about gravity being the same as acceleration is interesting. When I got my physics degree we used the same book (Griffiths) for electrodynamics, so I've seen the falling electron question. However, we never tied gravity and electrodynamics together in any fundamental sense. I suspect, as Matt said, that this involves GR and is far from trivial.

It's been a while since I got my physics degree, so I could be wrong, but thats my guess.