6

The Larmor Formula suggests that the power radiated by an accelerating charge is non zero. But we know that a uniformly charged non-conducting ring rotating about its central axis does not radiate. Why is this so?

This is somehow very strange to me. For example, suppose I charge a very small portion of the non conducting ring with charge q. Now when I rotate this ring with a constant angular velocity, there should be radiation. Now, all I have to do is spray some charge uniformly on the rotating ring, and the radiation just ceases! Isn't this strange? We can negate radiation simply by spraying charge carefully?

Please explain why there is no radiation in this case.

sammy gerbil
  • 27,277
Lelouch
  • 3,616
  • Some differences of opinion in another forum. https://www.researchgate.net/post/An_accelerated_charge_produces_electromagnetic_waves_and_a_circular_motion_is_accelerated_Why_does_a_steady_current_in_a_coil_not_produce_waves – Farcher Aug 05 '16 at 14:42
  • 1
    I think this is about interference. I mean if you have +Q and -Q charges on top of each other, you don't expect that they will radiate. However, due to linearity of Maxwell equations, one can solve corresponding Maxwell equations using accelerating +Q and -Q as source term separately. They both radiate, but with 180 phase difference. Then one can sum these, and radiation vanishes. In your case, the problem is more subtle however. But in principle, one should do $S=(E_1+E_2) \times (B_1 + B_2)$ instead of $S = S_1 + S_2 = E_1 \times B_1 + E_2 \times B_2$. – Mikael Kuisma Aug 05 '16 at 15:44
  • @Farcher I read a the first few answers in that forum, and I think they are all either wrong or missing the point. For example, an oscillating dipole is not required for radiation. – garyp Aug 05 '16 at 15:50
  • 3
    I am not familiar with the fact that a rotating charged ring does not radiate. Can you provide a reference to that? – garyp Aug 05 '16 at 15:51
  • @garyp , see the first answer here : http://physics.stackexchange.com/questions/13416/is-it-true-that-any-system-of-accelerating-charges-will-radiate – Lelouch Aug 05 '16 at 17:58
  • Paraphrasing the question: A charged particle moving in a circle will radiate, but a continuous charge distribution will not. What's the essential feature responsible for the difference? How does synchrotron radiation fit into the picture? – garyp Aug 05 '16 at 18:49
  • The question would be improved if it was explicit that the axis of rotation was the symmetry axis (which seems to assumed by comments and answers but does not seem to be stated anywhere). – dmckee --- ex-moderator kitten Aug 11 '16 at 14:35
  • @dmckee, I have mentioned 'central axis' in the 1st paragraph. ALthough i should've been more explicit, but i'm not changing anything since everyone understood what i meant. – Lelouch Aug 11 '16 at 15:32

5 Answers5

8

To radiate, you must have a fluctuation with time

If you take a small portion of the ring, charge it, and then make it rotate, the distribution of charges does change with time.

If the ring is evenly charged, with no additional currents, and you make the ring rotate, then in this (ideal) situation the distribution of charges between two times is always the same. There would be no way to differentiate the situation at $t_1$ and $t_2$, so no radiation. I assume that if you compute the EM fields radiated by each charge, and sum them over the ring, they cancel out.

Albert Beton
  • 281
  • Im talking about an essentially classical situation here – Lelouch Aug 05 '16 at 19:02
  • I have checked with multiple sources, all confirming that no power is radiated in this case. I simply don't understand why. I have read in griffiths about the poynting vector and how in the limit r tends to infinity, the flux of S through a sphere of radius r, is taken as the power radiated. But i could not prove it in this case, as the vector forms are not so easily manipulated , to get the poynting vector. – Lelouch Aug 06 '16 at 09:16
  • http://www.commonsensescience.org/pdf/articles/spinning_charged_ring_model_of_electron_yields_new.pdf – Lelouch Aug 06 '16 at 14:36
  • please see pg:2, last line, contd. in page 3. – Lelouch Aug 06 '16 at 14:37
  • 1
    @Lelouch I think I had misunderstood your original question then. But as the article explains, to radiate, you must have a fluctuation with time, in your situation the distribution of charges between two times is always the same, so no radiation. I assume that if you compute the EM fields radiated by each charge, and sum them over the ring, they cancel out. – Albert Beton Aug 06 '16 at 14:46
  • I think that is true. I read up on some other things as well. Anyway thank you. – Lelouch Aug 06 '16 at 14:55
4

Photons are their own anti-particle and can cancel out. Or put another way, light is a wave and waves can destructively interfere. With the rotating ring of charge, there are many radiating fields, but the symmetry of the problem lets them cancel destructively so there is no net wave and no power lost to radiation. Radiation here is just an EM wave carrying power.

Think of a somewhat similar situation from electrostatics: If you add electrical charge to a perfect conductor, the charges arrange themselves on the surface and the electric field is 0 within the conductor. There are many charges with many fields, but they all cancel out. This is easiest to calculate for a sphere, in case you want to try it.

EL_DON
  • 2,624
  • 1
    let me ask something a little unrelated. The fact that electrons arrange themselves on a conductor so as to make it am equipotential, is also because, in terms of total field energy, this is the minimum energy configuration(thomsons theorem i think this is) . If we also want to make a least energy configuration out of this rotating ring, and give it a chance to radiate away its energy and acquire a lower field energy state, why wont it go for it? – Lelouch Aug 11 '16 at 00:14
  • @Lelouch: that is an excellent question. I think in order to begin radiating, the charges would first have to move to an asymmetric, higher energy state. But what if they start in the asymmetric state? The power lost to cyclotron radiation would not be very large, and I can easily imagine that energy could be lost faster by rearranging to be symmetric. – EL_DON Aug 11 '16 at 00:29
1

I address this question in my answer Why don't loop currents produce light?. As you divide the single charged particle into more and more pieces while keeping the total charge and average current fixed, the charge distribution varies less and less over time and the radiated power smoothly decreases to zero.

Community
  • 1
tparker
  • 47,418
0

If we were to describe emission from radiation from Maxwell equations, then what couples to electromagnetic radiation is really current, not the charge itself (although there may be some complicated situations, depending on the gauge chosen.) A rotating homogeneous ring or a standing wave do not correspond to current flows changing in time, even though particles constituting them do move.

Another way to see it is by disentangling Lagrangian and Eulerian descriptions of the current and charges: the particles do move (Lagrangian view), but the charge distribution and the current distribution remain unchanged in every point of space.

Roger V.
  • 58,522
0

When a single electron turns in a circle it emits radiation to help conserve momentum along the axis it has just turned from . But a ring of many electrons replaces the momentum of the turning electron by another electron which takes its former position.