= Radiation fields
The consistent <Liénard–Wiechert potential> obeys $\mathbf A=\mathbf u\varphi/c^2$. To first order in $\boldsymbol\beta=\mathbf u/c$, its large-distance form is
$$
\varphi\simeq\frac{q}{4\pi\epsilon_0r}[1+\mathbf n\cdot\boldsymbol\beta(t_r)],\qquad
\mathbf A\simeq\frac{\mu_0q}{4\pi r}\mathbf u(t_r),\qquad
t_r\simeq t-\frac rc.
$$
Terms varying as $1/r^2$ give near fields and are negligible for the radiated <power> at large $r$. Keep the first-order velocity term in $\varphi$: although small, its retarded-time derivative contributes a leading <acceleration> field. To the required order, $\partial t_r/\partial t\simeq1$ and $\nabla t_r\simeq-\mathbf n/c$. Holding the observation direction fixed at leading order gives
$$
-\nabla\varphi\big|_{1/r}=\frac{q}{4\pi\epsilon_0cr}\mathbf n(\mathbf n\cdot\dot{\boldsymbol\beta}),\qquad
-\frac{\partial\mathbf A}{\partial t}\bigg|_{1/r}=-\frac{q}{4\pi\epsilon_0cr}\dot{\boldsymbol\beta}.
$$
Thus the outgoing <electric field> is transverse:
$$
\boxed{\mathbf E_{\rm rad}(\mathbf r,t)=\frac{q}{4\pi\epsilon_0c r}
\left[\mathbf n\times(\mathbf n\times\dot{\boldsymbol\beta})\right]_{t_r}
=\frac{q}{4\pi\epsilon_0c^2r}
\left[\mathbf n\times(\mathbf n\times\mathbf a)\right]_{t_r}.}
$$
Here $\mathbf a=d\mathbf u/dt_r$. The <magnetic field> follows from $\nabla\times\mathbf A$, or from the outgoing <plane electromagnetic wave> relation:
$$
\boxed{\mathbf B_{\rm rad}=-\frac{\mu_0q}{4\pi r}
[\mathbf n\times\dot{\boldsymbol\beta}]_{t_r}
=\frac1c\mathbf n\times\mathbf E_{\rm rad}.}
$$
The minus sign is essential. Both fields fall as $1/r$, are mutually orthogonal and obey $|\mathbf E|=c|\mathbf B|$. The corresponding <Liénard-Wiechert field> also contains nonradiative terms, omitted in this leading far-field calculation.
Back to article page
