Let's consider it case by case:

Case 1: Charged particle is at rest. It has an electric field around it. No problem. That is its property.

Case 2: Charged particle started moving (it's accelerating). We were told that it starts radiating EM radiation. Why? What happened to it? What made it do this?

Follow up question: Suppose a charged particle is placed in uniform electric field. It accelerates because of electric force it experiences. Then work done by the electric field should not be equal to change in its kinetic energy, right? It should be equal to change in K.E + energy it has radiated in the form of EM Waves. But then, why don't we take energy radiated into consideration when solving problems? (I'm tutoring grade 12 students. I never encountered a problem in which energy radiated is considered.)

How do moving charges produce magnetic field?

  • $\begingroup$ A question that comes to my mind right now is, given an acceleration of magnitude $a$, what is the resulting frequency of the EM wave? I am almost a physicist now, I did not continue because the amazing world of programming has captured me so I am working right now! But someday I will go back and continue studying physics, and it also was mentioned to us but I haven't seen any mathematical derivation. $\endgroup$ – iharob Nov 11 '15 at 0:08
  • $\begingroup$ Only an oscillating charge would radiate at a definite frequency, its frequency of oscillation. Otherwise you get pulses with a wide-band spectrum. $\endgroup$ – Pieter Nov 29 '16 at 23:19
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    $\begingroup$ physics.stackexchange.com/a/349367/47511 $\endgroup$ – Count Iblis Sep 3 '17 at 20:14

A diagram may help:

enter image description here

Here, the charged particle was initially stationary, uniformly accelerated for a short period of time, and then stopped accelerating.

The electric field outside the imaginary outer ring is still in the configuration of the stationary charge.

The electric field inside the imaginary inner ring is in the configuration of the uniformly moving charge.

Within the inner and outer ring, the electric field lines, which cannot break, must transition from the inner configuration to the outer configuration.

This transition region propagates outward at the speed of light and, as you can see from the diagram, the electric field lines in the transition region are (more or less) transverse to the direction of propagation.

Also, see this Wolfram demonstration: Radiation Pulse from an Accelerated Point Charge

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    $\begingroup$ Could you kindly comment on my "Follow up question" $\endgroup$ – claws May 21 '13 at 12:55
  • $\begingroup$ @claws, my only comment for now (I'll save more for later when I have time) is to refer you to this on the question of "Does a uniformly accelerated charge radiate?": mathpages.com/home/kmath528/kmath528.htm $\endgroup$ – Alfred Centauri May 21 '13 at 13:13
  • $\begingroup$ @Alfred Centauri: Purcell also followed the same procedure. Check this. $\endgroup$ – user36790 Jan 11 '16 at 3:14
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    $\begingroup$ This is such a beautiful explanation! $\endgroup$ – Betelgeuse Feb 9 '18 at 15:33
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    $\begingroup$ Wouldn't the same argument apply to a an electron in uniform motion? $\endgroup$ – gardenhead Jan 7 at 22:14


The electrons intrinsic properties are their electric charge and their magnetic dipole moment. So the electron has two fields around it. The magnetic field is observable if one put a magnetizable material into an external magnetic field. Often the magnetization of the material holds for a while, which is explained by the alignment of the magnetic dipole moments of the subatomic constituents.


If one observe an electron beam in a vacuum chamber hardly one will observe that the electrons slow down (except the change of velocity and direction from the earths gravitation). Since there is no decrease of the speed of a constant moving electron there wouldn't be any loss of energy, hence the electron does not radiate. So you are right that only particles under acceleration radiate.

How an why do accelerating charges radiate electromagnetic radiation?

Accelerated charges radiate and they do this in portions, in the past called by Einstein quanta and later called photons. Every photon - as well as the emitting particle - has an electric field component and a magnetic field component and that is why such radiation is called EM radiation.

Why EM radiation occurs?

Suppose you have to slow down a car. Not having EM radiation you would be able to stop your care only by transferring your kinetic energy to another body, be this another massive body or a rotating disc for example. To our luck the loss of energy in every energy transfer happens at any case. So for a why question the answer has to be because nature works this way. The better questions are how something happens. The answer how would be an explanation on a more detailed level (including new hows) as the observation level.

How EM radiation occurs?

There is a phenomenon in nature called the Lorentz force. As soon as an electron moves inside a magnetic field and if the electrons direction of movement is not parallel to the north-south direction of the magnetic field then the electron gets deflection in the direction perpendicular to both directions of the electrons movement and the magnetic field.

An external constant magnetic field does not contribute (put in) energy for the deflection of the electron. Means, one can let through the magnet device electrons as long as he wants, the magnetic device does not weaken. So the reason for the deflection and the escorting EM radiation from the electron has to lie in the electron and its kinetic energy (an electron in rest to the external magnetic field won't be deflected.)

I started with the statement that an electron has a magnetic dipole moment. Coming into an external magnetic field the electrons magnetic field gets aligned to this external field. At the same time the photon emission happens. If we suppose that during the alignment process the radiation of the photon happens, this will disbalance the alignment again And - because the photon has a momentum - the electron gets pushed against the direction of the photon emission which is in accordance with the observation radially outwards directed.

Now we have an effective chain: alignment - photon emissio - deflection - again alignement - ... By this the electron lose kinetic energy and moves in a spiral path until it stops. In detail the spiral path is a path of tangerine slices.

  • $\begingroup$ But how about when electron is not in external magnetic field. What causes it's acceleration to produce photons and slow it down? $\endgroup$ – MaDrung May 31 '17 at 10:51
  • $\begingroup$ Nothing. As long as an electron isn't under the influence of a field or electromagnetic radiation it doesn't emit photons nor slows down. Why this question occurs to you? $\endgroup$ – HolgerFiedler Jun 3 '17 at 13:06
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    $\begingroup$ I thought accelerating electrons emit photons and slow down. Let's say there is electric field pointing in the opposite direction. The photon Will be accelerating. Will he not be emiting light and slowing down? $\endgroup$ – MaDrung Jun 5 '17 at 5:54
  • $\begingroup$ @MaDrung Where is the contradiction of what you comment right to my comment? $\endgroup$ – HolgerFiedler Jun 5 '17 at 6:52

The second problem is quite tough. J. D. Jackson comments, in the introductory remarks of his chapter on 'Radiation Damping, Classical Models of Charged Particles', that we know how to solve classical electrodynamics problems in two ideal conditions - a) given charge and current densities, how to compute the fields and b) given the fields, how to find the motion of charged particles in their presence. When charged particles accelerate, they do produce radiation which in turn affects the motion of all other charged particles. However, that problem is, Jackson says, still unsolved.

Coming to the first problem, if you calculate $\vec{E}$ and $\vec{B}$ for a moving charged particle, you will see that they depend on the acceleration $\vec{a}$ of the charged particle. Now calculate the Poynting vector $\vec{S}$. You will observe that $\vec{S}$, depends on acceleration but not velocity. Integrating it to get power radiated gives the famous Larmor formula. You may want to refer to Griffiths' chapter on 'Electromagnetic Radiation'.

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    $\begingroup$ For Jackson, the radiation also affects the particle itself. For Griffiths, the Poynting vector does have terms that contain velocity, they just fall off faster than 1/r and so don't contribute a finite energy to an infinitely distant surface. But they are carrying power away, just not to super far away. So back to Jackson, that energy loss also affects the motion of the charge. $\endgroup$ – Timaeus Nov 12 '15 at 1:40

Accelerating charges don't have to radiate. Look at an electron at rest on the earth (or constantly accelerating for a long time). It will not radiate. The radiation acceleration formula such as Lamor's only hold for particles with changing acceleration - like a sinusoidal movement.

See for instance Feynman: From http://www.mathpages.com/home/kmath528/kmath528.htm

For example, in Feynman's "Lectures on Gravitation" he says "we have inherited a prejudice that an accelerating charge should radiate", and then he goes on to argue that the usual formula giving the power radiated by an accelerating charge as proportional to the square of the acceleration "has led us astray" because it applies only to cyclic or bounded motions.

  • $\begingroup$ The derivation of Larmor's formula does not require that the charged particle be in a cyclic or bounded motion. According to the classical electromagnetic theory (Maxwell's equations), accelerating charged body is associated with radiation EM field - a field component that decays with distance as $1/r$. In this sense, accelerated charges do radiate, no exceptions. The reason why lots of people are confused by this is because they tend to assume that just because there is this radiation field, it has to transmit energy away from the charged body. But that is a different, independent question. $\endgroup$ – Ján Lalinský Sep 3 '17 at 21:54
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    $\begingroup$ "In General Relativity there is what is called The Equivalence Principle. This is the assertion that there should be no difference between a body at rest experiencing a uniform gravitational field and a body experiencing uniform acceleration. There is no reason to expect a charged particle resting in a uniform gravitational field to be emitting radiation. Therefore, according to the Equivalence Principle there should be no radiation from a charged particle experiencing uniform acceleration. But the Larmor formula and its derivation are there, so something must be wrong." - I side with GR $\endgroup$ – Tom Andersen Sep 3 '17 at 23:07
  • $\begingroup$ Maybe there should be no radiation according to the reasoning based on the Equivalence principle, but there has to be radiation if EM field is to obey Maxwell's equations. When it comes to EM radiation, I side with electromagnetic laws, not theory of gravity. There is a puzzle to be solved, sure, but I doubt its resolution lies in invalidation of Maxwell's equations. Perhaps the two situations you mention are not as equivalent as would be required to apply the principle of equivalence. $\endgroup$ – Ján Lalinský Sep 4 '17 at 22:21
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    $\begingroup$ Or perhaps there can be radiation field around the particle in the freely falling frame while there is none in the frame of the body that is source of the gravity field. $\endgroup$ – Ján Lalinský Sep 4 '17 at 22:26
  • $\begingroup$ Agree with Ján Lalinský. Think of an electron sitting still in Earth's gravity field. It should not radiate at this moment. Now approach it with a ball made of neutrons so the electron experiences changing gravity (thus changing equivalent acceleration according to Einstein). Does it start to radiate? To my intuition it should not. $\endgroup$ – verdelite Jan 23 at 17:07
