Why doesn't a charged particle moving with constant velocity produce electromagnetic waves?

Question

A charged particle moving with an acceleration produces electromagnetic waves. Why doesn't a charged particle moving with a constant velocity produce electromagnetic waves? As far I understand, the electric and magnetic fields in space will still be time-dependent, if a charged particle is moving with constant velocity, so they could have given rise to electromagnetic waves, but they don't.

Also, why do accelerating charged particles produce electromagnetic waves? What is Nature's intention behind this phenomena?

Answer 1

Riemannium's answer tackles why you need acceleration to form EM waves. I will hit from a different way that I think gets at your question title as to why charges moving at a constant velocity do not produce EM waves. In the subsequent discussion all mentioned reference frames are inertial reference frames.

The easiest way to reason that charges moving at a constant velocity relative to us will not emit radiation is to observe that we can always boost to a frame moving along with the charge. Then we will just see a stationary charge with just a constant electric field.

Now, it wouldn't make sense that we don't see an EM wave in our frame, but someone moving by at some us would. If an EM wave exists in one inertial frame it must exist in all inertial frames. Therefore, it must be that a charge moving at a constant velocity (in some inertial reference frame) cannot produce an EM wave.

Answer 2

Okay, I'll try with a poor but "intuitive" explanation.

According to relativity theories, "it is impossible to tell if you're at rest or moving with constant velocity".

We know that a charge at rest does not emit any wave.

If you were moving at constant velocity and you saw a static charge emitting a wave, you'd think "this charge is not actually static because it emitting waves, so I'm seing static because I'm moving with the same velocity as it, so I am not at rest".

That would violate one of the most basic principles of physics: you cannot tell if the train is moving forward, or the landscape is moving backwards, provided that $\vec{v}$ is constant.

Check that the opposite would lead to the existence of "priviledged observers", or "observers who are at absolute rest". This doesn't make sense.

So we must discard the idea of charges moving at constant velocity emitting waves. It must be accelerated charges, only because there isn't any other option.

Answer 3

Classically: you need an acceleration of a charge to produce electromagnetic radiation. The electromagnetic tensor couples to charge, so if you look at Newton second law: $$\dot{P^\mu}=QF^{\mu\nu}\dot{X_{\nu}}$$ So, in order to have an electromagnetic wave, a solution of $$\square^2 A^\mu=\mu_0J^\mu$$ with $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ in form of, e.g., a plane wave, you need a non-null field strength $F_{\mu\nu}$ and thus a varying momentum, and thus an acceleration. Quantumly (much more complicated answer, but I will simplify it without quantum electrodynamics or electroweak theory): in order to have an interacting field and to have "waves" by excitations of the vacuum of the theory, you need something that vibrates...And perturbations are impossible without some time variation of the generalized momentum. For a spinor field, you also requires a proper "wave field". They fill up the complete continuum space-time.

A good loophole to my above argument should be about what causes gravitational waves, since momentum is conserved...Then, what causes gravitational waves? Perturbation of the vacuum of the theory of gravity (the metric field itself)! The source of gravity are the connection fields and the local spacetime variations, since you have energy-momentum tensor conservation, you need something else...Indeed, there are another momenta...Dipoles are sources of electromagnetic fields, you need two charges to oscillate or a single charge oscillating rapidly to produce a electromagnetic wave, you need asymmetric masses moving to get gravitational waves...And, at least, a non-null quadrupole momentum variation in order to get gravitational waves...

Answer 4

Ok so if you take maxwell equations and manipulate them a little you can get $$\begin{aligned} \frac{\partial B}{\partial t} \quad & = -\quad \nabla\times E, \\[5pt] \frac{\partial E}{\partial t} \quad & = \frac{1}{c^2} \nabla\times B - \frac{1}{\epsilon_0}J \\[5pt] . \end{aligned}$$ You see the left hand side guarantees you that a varying magnetic field will generate an electric field, and a varying (means it changes in time) electric field will produce a magnetic field. The definition of electric field is given by the force a test charge feels from another charge. more exactly, $$ E = k q_1 \frac{\vec{x} - \vec{x_1}}{|\vec{x} - \vec{x_1}|^3} $$ where q1 is the charge that gives you the electric field in point x, and this q1 charge is situated at the point x1.

nvm, ignore everything...

Intuitive approach: You sit on an electron, you see Electric field spreading around you but nothing else. The thing that Newton taught us is that you can't tell the difference between standing still or moving with constant velocity. Thus if you can't tell you're moving at all, from Maxwell eq, you can't have B, which is the magnetic field. If you can't have B, you can't have varying B, you can't have varying E, thus you cant have EM field, you will only detect that electrostatic field if you sit on an electron (that moves with constant velocity). fml I'm the worse explainer ever.