Radiation sourceConsider a charge moving at a constant velocity in a vacuum. As we know, the electric field lines of any charge extend radially outward. What we have not considered, however, is the time it takes for this electric field to propagate. At time $t=0$ let the charge be at the origin moving with some constant velocity $\dot x$. The level curves of the charges electric field propagate outward at the speed of light. At some time $t=t'$ later the charge has moved some distance. Meanwhile, the electric field of the charge from time $t=0$ has propagated a distance $ct'$ outward from $x(0)$. The figure below shows the electric field lines of the charge at time $t=t'$. The imaginary outer circle, having radius $r=ct'$, has just received news of the electric field from $t=0$. Generally we can imagine a circle of some radius $ct_1$ centered at $x(t_1)$, which has just received news of the electric field from $t=t_1$.
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In this nonrelativistic model we assume $\dot x \ll c$. These figures are greatly exagerated, in reality the particles change in position is much less than the radius of the circle. Now consider a charge moving with some acceleration. The electric field lines are no longer straight because the distance the charge travels is no longer proportional to the distance the electric field propagates. The figure below shows the field lines of a particle accelerating from rest at $t=0$ to $t=t'$. As above, the outer circle represents the information that has been received at $t'$.
We see that an accelerating charge creates an electric field varying in time and space, which is radiation! Let's look at a more specific case. A charge $q$ is initially at rest at point $A$ at $t=0$, then accelerates for a short time $\Delta t$ to the point $A'$. It then continues at a constant velocity $v$ for $t'$, ending at point $B$ at time $t=t'+\Delta t$. An observer is far away at an angle $\theta$ from the horizontal.
At $t=t'+\Delta t$ news that the charge has accelerated has not yet reached region 1, so the Observer still sees the electric field line from the stationary charge (shown in $\color{orange} \text{orange}$). In region 2, observers already know that the charge accelerated and is moving at constant velocity. Because it's moving at a constant velocity, the electric field lines are straight. The electric field line seen in region 2 at angle $\theta$ is shown in $\color{blue}\text{blue}$. The electric field line at angle $\theta$ must be continuous (and indeed, as seen in figure 2, the electric field lines of an accelerating charge are continuous), so the field line in region 3 ($\color{red}\text{red}$) must connect the lines in regions 1 and 2. This gives us a bend in the electric field that moves outward at the speed of light–radiation! In reality the figure above is an exageration. We assume that $\Delta t \ll t'$, so $A' \approx A$. Consider a simpler diagram:
Where $\mathbf E$ is the bent electric field. Note that since we assume $v \ll c$, $v_\perp t' \ll c t'$, so the red triangle is right. The red triangle above is similar to the triangle formed by the vectors $(\mathbf E, \mathbf E_\perp, \mathbf E_\parallel)$. $\mathbf E_\parallel$ is the component in the $A$Observer line (the normal electric field we find using Gauss's law) and $\mathbf E_\perp$ is the radiative component caused by the charge accelerating. From this similarity we can say $\begin{align*} \frac{E_\perp}{E_\parallel} &= \frac{v_\perp t'}{c\Delta t} = \frac{a_\perp \Delta t t'}{c\Delta t} \\ &= \frac{a_\perp t'}{c} = \frac{a_\perp r}{c^2} \end{align*}$where $r=ct'$ is the distance from the charge to the bend. This gives us $E_\perp = \frac{a_\perp r}{c^2} E_\parallel.$We find $E_\parallel=q/(4\pi \epsilon_0 r^2)$ by Gauss's law, giving us the Larmor formula, $E_\perp = \frac{q a_\perp}{4 \pi \epsilon_0 c^2 r}.$A few things to note:
