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Why Accelerating Charges Radiate

A charge at rest sits quietly in its Coulomb field; a charge in uniform motion just drags that field along. Only when you shake it does energy break free and race to infinity. Build the physical picture of radiation before touching a single integral.

Three states of a charge

A charge sitting still carries a static electric field — the Coulomb field, falling off as 1/r^2. It stores energy but sends none away. Set the charge moving at constant velocity and it merely carries that field along; by relativity a uniformly moving charge is just a charge at rest seen from another inertial frame, so it cannot radiate either. But accelerate it and something genuinely new appears: a disturbance detaches from the charge and runs outward forever. That escaping disturbance is electromagnetic radiation.

The deep reason is causality together with a finite speed of light. Field lines cannot rearrange themselves instantaneously across all space; news of a change in the source can only spread at c. That delay is what lets a piece of the field peel off and become independent of its parent charge.

The kink model of radiation

Picture the field lines of a charge that sits still, jerks sideways briefly, then stops. News of the jerk travels outward at c. Inside a sphere of radius ct the lines already point at the new position; outside, they still point at the old one. On the thin shell between them the lines must kink to connect — and that transverse kink, sweeping outward, is the radiation pulse.

E_{\text{rad}} = \frac{\mu_0\, q\, a \sin\theta}{4\pi r} \;\propto\; \frac{1}{r}

The transverse radiation field of an accelerating charge: proportional to the acceleration a, to \sin\theta (angle from the acceleration axis), and to $1/r$.

What comes out is a wave

Far from the source the kink is, locally, a plane EM wave: the electric and magnetic fields are perpendicular to each other and to the propagation direction \mathbf{k}, they oscillate in phase, and their magnitudes are locked by B = E/c. Maxwell's equations force this — take their curl in empty space and each field obeys a wave equation travelling at c = 1/\sqrt{\mu_0\varepsilon_0}.

c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = f\lambda, \qquad B = \frac{E}{c}

The wave speed is set by the vacuum constants; frequency and wavelength satisfy c=f\lambda; the fields carry equal energy since E=cB.

Drag the wavelength slider: a plane EM wave with \mathbf{E}\perp\mathbf{B}\perp\mathbf{k}, oscillating in phase and travelling at c=f\lambda. Notice E and B peak together — this is a far-field radiation wave, not a static field.

The energy streaming outward is measured by the Poynting vector \mathbf{S}=\tfrac{1}{\mu_0}\mathbf{E}\times\mathbf{B}, which points along \mathbf{k}. Time-averaged, the intensity of a wave of amplitude E_0 is \tfrac{1}{2}c\varepsilon_0 E_0^2. Because E=cB in a radiation field, the electric and magnetic contributions to the energy density are exactly equal.

\mathbf{S} = \frac{1}{\mu_0}\,\mathbf{E}\times\mathbf{B}, \qquad \langle I\rangle = \tfrac{1}{2}\,c\,\varepsilon_0 E_0^{2}

Energy flow and the time-averaged intensity carried by a plane wave.

The road ahead

To make all this quantitative we will build, in order: (2) potentials that respect the finite speed of light — the retarded potentials and the Liénard-Wiechert fields of a moving charge; (3) how much power radiates and in what shape — the Larmor formula and dipole radiation; (4) the realization that \mathbf{E} and \mathbf{B} are one relativistic object — the covariant formulation; and (5) relativistic radiators — synchrotron light, beaming, and radiation reaction.