13
$\begingroup$

Does an accelerated charge always radiate? For example the current electrons in an electric circuit when moving through a turn they are accelerated, do they radiate because of that acceleration?

  1. If the answer is no, then why not?
  2. If the answer is yes, then how small the radiation is?
Improve this question
$\endgroup$
1
  • $\begingroup$ Can you clarify that you're talking about synchrotron or cyclotron radiation? $\endgroup$
    – endolith
    Commented Aug 12, 2011 at 21:50

2 Answers 2

Reset to default
13
$\begingroup$

An individual electron going around the loop would of course radiate. However, if we have a constant current going around the loop, there are many electrons and their individual radiations cancel each other, so there is no net radiation. They balance each other out. We can show there is no net radiation by solving Maxwell's equations on the net current, which is constant. See, e.g., McDonald's Why Doesn’t a Steady Current Loop Radiate? (pdf).

There would be some transient radiation when you change the current. Also, in the real world, there would be some infinitesimal radiation due to the fact that the current won't truly be constant at the microscopic level, i.e. because of Johnson noise etc.

Improve this answer
$\endgroup$
12
  • 2
    $\begingroup$ I do not understand how individual radiations of many electrons would cancel each other. How can we show this using Maxwell's equations, is this a standard calculation or something? Pointing out some textbook in which this is explained (or even listed as an exercise) would be greatly appreciated and helpful. $\endgroup$
    – Revo
    Commented Aug 10, 2011 at 19:20
  • 1
    $\begingroup$ Well, you can formally prove it using vector potentials and Green functions, if you are comfortable with those. However for a more intuitive take, consider a single electron going around in a loop. His total vector acceleration averaged over the loop must be zero. Now imagine many identical electrons spread around the loop; the total vector acceleration for all electrons at any instant must be the same as the loop-average for a single electron, i.e. zero. Since radiation is proportional the charge times acceleration, they cancel. $\endgroup$
    – user1631
    Commented Aug 10, 2011 at 19:58
  • 2
    $\begingroup$ Hmmm, that means a current in a superconducting ring will fade away by radiation? $\endgroup$
    – Revo
    Commented Aug 10, 2011 at 20:11
  • 1
    $\begingroup$ A "shut up and calculate" way to prove that a ring with current produces no electromagnetic waves is to solve Maxwell's equations for this configuration of electric currents. Thankfully, the solution for a loop with current in infinite space is readily available. It can be found in any physics textbook under "magnetostatics", along with a sketch of magnetic field lines curling around the wire. This solution is unique. No time varying magnetic or electric fields in this solution. Therefore, no electromagnetic waves. $\endgroup$
    – drlemon
    Commented Aug 10, 2011 at 20:35
  • 2
    $\begingroup$ Without appealing to Green's functions and other esoterica, an easy way to see that it doesn't radiate is that since there is no time-variation in the currents, there is no parameter that would set the frequency of the waves. You might think that the period would equal the time required for an electron to go once around the circuit, but that would be different for different types of wire because of the different drift velocities. Maxwell's equations don't have velocity in them, only current. (The time to drift around the circuit is also ~1 s, so the frequency would be ~1 Hz!) $\endgroup$
    – user4552
    Commented Aug 10, 2011 at 20:56
1
$\begingroup$

Generally speaking, accelerated charges radiate. For the case of a current in a wire, the acceleration is so small that the radiation is hard or impossible to detect.

For example, consider 1 amp traveling down a wire of radius 1mm. How fast do the electrons move? To solve this we need to know the number of mobile charge carriers per unit volume and a few constants. The answer is $v = $ 2 meters per day. You can convert this speed into an acceleration using the formula $a = v^2/R$ where $R$ is the radius of curvature of the corner. It is a very small number.

For comparison, the radiation that makes it so difficult to speed up electrons in a collider experiment comes from electrons that are traveling essentially at the speed of light.

Improve this answer
$\endgroup$
9
  • $\begingroup$ Yeah that is the intuitive answer I had in mind, electrons in a current loop radiate but the radiation is infinitesimally small because the drift velocity is infinitesimally small. $\endgroup$
    – Revo
    Commented Aug 10, 2011 at 22:20
  • 1
    $\begingroup$ The answer by user1631 is also correct (I believe), but only applies to constant currents. In a synchrotron, the currents are not constant but instead come as pulses. $\endgroup$
    – Carl Brannen
    Commented Aug 10, 2011 at 22:30
  • 2
    $\begingroup$ There are a couple of problems with this answer. One is that the drift velocity is not universal, as you seem to be claiming; it depends on the material. The other is that it is possible to make the acceleration as large as desired simply by making v large and/or making R small. Even with large a, there is no radiation, but your answer makes it sound as though there is. $\endgroup$
    – user4552
    Commented Aug 11, 2011 at 14:50
  • $\begingroup$ @Ben; I keep going back and forth on this. First, I never said drift velocity is universal; I did the calculation with copper because this is common for circuits. Second, if you make v large enough you end up with synchrotron radiation. But I agree that if the - charges of the electrons are exactly cancelled by the + charges of the metal then there is no radiation of this sort. But an exact cancellation is not compatible with a change in the current; for the usual metals, one must have voltage to have current and this implies that the + and - charges did not cancel exactly. $\endgroup$
    – Carl Brannen
    Commented Aug 11, 2011 at 16:33
  • $\begingroup$ I'm feeling really doubtful of my arguments here and would appreciate a critique so that I can go ahead and delete the answer. $\endgroup$
    – Carl Brannen
    Commented Aug 12, 2011 at 17:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.