What does it mean that charged particles in orbits undergoing acceleration would emit electromagnetic radiation?

One of the drawbacks of Rutherford's atomic model was that electron would emit electromagnetic radiation as it is continuously changing direction and therefore undergoing acceleration, in what form would this radiation be emitted?

And what does it mean that the electron is undergoing acceleration, does it mean that while at turns of the orbit the electron is slowing down and on straight paths it is going fast,or is it just a case of change in direction.

If it is just a case of change in direction then why should the electron release energy(electromagnetic radiation) as its circular motion around the nucleus is just due to the electrostatic forces in the first place.

• It’s electromagnetic radiation, so what do you mean by “in what form would this radiation be emitted”? – G. Smith Jun 18 '19 at 4:39
• Why do you think there would be any “straight paths”? – G. Smith Jun 18 '19 at 4:41
• why don't you read some textbook? physics.usask.ca/~hirose/p812/notes/Ch8.pdf – anna v Jun 18 '19 at 5:41
• “Why should the electron release energy?” Because Maxwell’s equations say it does when it accelerates. Moving in a circle counts as accelerating. The reason why it is moving in a circle (electrostatic attraction to the nucleus) is completely irrelevant. – G. Smith Jun 18 '19 at 5:47
• @G.smith I assume the radiation would be in the form of photons which means that the electron would lose energy ,but where does that energy comes from? As acceleration doesn't provide with any energy as it is just due to the change in direction. – Fardeen Khan Jun 18 '19 at 11:59

3 Answers

How can we illustrate that an oscillating charge radiates energy? Let's build up our physical picture:

1. First, let's locate an electron at different positions in the x-y plane. For simplicity, let us choose these positions of the electron such that they lie on a straight line. For each of these positions we calculate the electric field at $$(x_0,y_0,z_0) = (0,0,1)$$ -- we call this position the observer. Since the distance between the electron and the observer changes, the electric field strength at $$(x_0,y_0,z_0)$$ changes.

2. Next, let us include the "movement" of the charge. So, for example, we choose the position of the electron as $$(x_e(t), y_e(t), z_e(t)) = (sin(t/10), 0, 0)$$. However, instead of assuming that the $$t$$ variable is continuous, we take a static view: We take $$t$$ to be an integer, $$t = 0, 1, 2, \ldots$$ Hence, we obtain the same interpretation as in item 1. However, now we already have an oscillating electric field.

So far, we have not used Maxwell's equations, but only electro-statics. Nevertheless, we obtained an oscillating electric field and the step to the propagating electro-magnetic wave is not really surprising anymore. By using Maxwell's eq, i.e. $$\frac{dE}{dt} \propto B$$ and $$\frac{dB}{dt} \propto E$$ the propagating electro-magnetic wave seems to be the "natural" conclusion -- from the perspective of todays knowledge. An other way of realising that a electro-magnetic wave must be emitted in situation 2 is the fact, that we just described the setup of an Hertzian dipole.

1. So far we considered only a 1D oscillation of the electron. However, if the electron follows a circular path in the $$x-y$$ plane (=it rotates) this movements can be described as a superposition of two oscillations: one in the $$x$$-direction and one in the $$y$$-direction. Hence, we immediately conclude that the electron emits an electro-magnetic wave also in this configuration [Sidemark: Now the wave would be circular polarised. In contrast situation 2 results in a linearly polarised wave.]

To your second paragraph: Newton's law states, that a body travels in a uniform motion (= constant velocity vector), if no force acs upon it. Therefore, a rotating electron is accelerated.

In conclusion: The classical physics predicts that an accelerating charge (note: circular motion is an accelerating motion) emits radiation. A deviation from this classical picture is worth noting.

Rutherford's model itself has no such problem, because it is based on electrostatic law of force. There is no radiation in electrostatics and thus no radiation implied by the Rutherford's model alone.

But the problem arises when we try to reformulate Rutherford's model in relativistic electromagnetic theory, where changes in EM field propagate with finite speed. We have to replace the electrostatic Coulomb law by Maxwell's equations or similar relativistic laws, which imply that the force acting on the electron is no longer a central conservative force.

Two particles moving in circles around each other means both particles are accelerating. Any particle's acceleration points from the particle to the center of the circle. It is called centripetal acceleration, its value is given by $$v^2/R$$ where $$v$$ speed of the particle and $$R$$ is radius of the circle.

So we have two charged bodies circling around common center point, they produce time-dependent fields and each experiences force due to the other particle; these forces are correlated in time.

In the retarded field variant of EM theory, motion of such system was analyzed by Synge [1]. He found that above decribed motion of both particles implies the distance between the particles decreases in time. Using Frenkel's formulation of EM theory of point particles [2], which Synge's work is consistent with (although it seems Synge was unaware of it), it is easy to show that for such motion electromagnetic energy is being lost from the region because of EM radiation passing through the region boundary.

If the advanced field variant of EM theory was used instead, the opposite result would have been obtained; electromagnetic energy comes from infinity into the region through the boundary and electromagnetic energy of the system inside this region increases.

The dominant view is that EM fields of such a system should be given by the retarded solution to Maxwell's equations, the advanced solution is usually discarded as unphysical. So such system is expected to lose EM energy, unless something else can resupply it. This something can be, for example, background radiation due to other, external bodies.

[1] J. L. Synge, On the electromagnetic two–body problem., Proc. Roy. Soc. A 177 118–39 (1940)

https://royalsocietypublishing.org/doi/pdf/10.1098/rspa.1940.0114

[2] J. Frenkel, Zur Elektrodynamik punktförmiger Elektronen, Zeits. f. Phys., 32, (1925), p. 518-534. http://dx.doi.org/10.1007/BF01331692

1) If we force a charged particle to spiral around in a large circle, the form of electromagnetic radiation it emits is called synchrotron radiation if its velocity is relativistic and cyclotron radiation if it is not. See Jan Lalinsky's comments below for more detail.

2) Any object that is forced to move in a circle will experience an acceleration towards the center of the circle because the direction of its velocity vector is continually changing direction.

3) It is a fundamental fact of electromagnetism that accelerating a charged object causes it to emit electromagnetic radiation.

• Only relativistic electron produces synchrotron radiation, its characteristic is that it is radiated into a narrow solid angle around the velocity vector and has a broad spectrum. If the electron is not relativistic, the radiation is called cyclotron radiation, has angular distribution of dipole oscillator aligned with radius vector and has comb-like many-peak spectrum with fundamental frequency equal to the cyclotron frequency. – Ján Lalinský Jun 18 '19 at 19:55
• Will edit my answer. – niels nielsen Jun 18 '19 at 20:15