06 May 2025
Retarded Potentials and Liénard-Wiechert Fields
Causal electromagnetic potentials and the exact electric and magnetic fields of a point charge in uniform or arbitrary motion.
In the Lorenz gauge,
\[\nabla\cdot\mathbf A+\frac{1}{c^2}\frac{\partial\phi}{\partial t}=0,\]Maxwell’s equations reduce to wave equations,
\[\left(\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\phi =-\frac{\rho}{\epsilon_0}, \qquad \left(\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\mathbf A =-\mu_0\mathbf J.\]The retarded Green function is
\[G_{\rm ret}(\mathbf r,t;\mathbf r',t') =\frac{\delta\!\left(t-t'-\lvert\mathbf r-\mathbf r'\rvert/c\right)} {4\pi\lvert\mathbf r-\mathbf r'\rvert}.\]It vanishes unless a signal emitted at $(\mathbf r’,t’)$ can reach $(\mathbf r,t)$ at speed $c$. Convolution with the sources gives
\[\boxed{ \phi(\mathbf r,t)=\frac{1}{4\pi\epsilon_0} \int\frac{\rho(\mathbf r',t-R/c)}{R}\,d^3r', }\] \[\boxed{ \mathbf A(\mathbf r,t)=\frac{\mu_0}{4\pi} \int\frac{\mathbf J(\mathbf r',t-R/c)}{R}\,d^3r', } \qquad R=\lvert\mathbf r-\mathbf r'\rvert.\]These are the retarded potentials. Their SI units are volts and tesla-metres, respectively.
Point charge and retarded geometry
Let a charge $q$ follow $\mathbf r_q(t)$. Its density and current are
\[\rho(\mathbf r,t)=q\,\delta^3[\mathbf r-\mathbf r_q(t)], \qquad \mathbf J(\mathbf r,t)=q\mathbf v(t)\, \delta^3[\mathbf r-\mathbf r_q(t)].\]For a fixed observation event $(\mathbf r,t)$, define
\[t_r=t-\frac{R(t_r)}{c},\qquad \mathbf R=\mathbf r-\mathbf r_q(t_r),\qquad R=\lvert\mathbf R\rvert,\qquad \mathbf n=\frac{\mathbf R}{R},\] \[\boldsymbol\beta=\frac{\mathbf v(t_r)}{c}, \qquad \kappa=1-\mathbf n\cdot\boldsymbol\beta.\]To make the schematic trajectory equation-generated, use the dimensionless path parameter $\tau=(t’-t_r)/(t-t_r)$ and diagram coordinates
\[x_d(\tau)=-2.2+3.2\tau, \qquad y_d(\tau)=-0.8\tau^2(1-\tau), \qquad 0\le\tau\le1.\]Thus the two source events are the plotted endpoints at $\tau=0$ and $1$; the field derivation itself retains an arbitrary $\mathbf r_q(t)$.
When the delta function is integrated over source time, its argument $g(t’)=t-t’-R(t’)/c$ contributes the Jacobian
\[\left\lvert\frac{dg}{dt'}\right\rvert_{t_r} =\left\lvert-1+\frac{\mathbf n\cdot\mathbf v}{c}\right\rvert_{t_r} =\kappa.\]Therefore the Liénard-Wiechert potentials are
\[\boxed{ \phi(\mathbf r,t)=\left[\frac{q}{4\pi\epsilon_0\kappa R}\right]_{t_r}, \qquad \mathbf A(\mathbf r,t)=\left[\frac{\mu_0q\mathbf v}{4\pi\kappa R}\right]_{t_r} =\frac{\boldsymbol\beta}{c}\phi. }\]Differentiating retarded quantities
The implicit definition of $t_r$ must be differentiated before computing $\mathbf E=-\nabla\phi-\partial_t\mathbf A$. At fixed $\mathbf r$,
\[\frac{dt_r}{dt} =1-\frac{1}{c}\frac{dR}{dt_r}\frac{dt_r}{dt}, \qquad \frac{dR}{dt_r}=-\mathbf n\cdot\mathbf v,\]so
\[\boxed{\frac{dt_r}{dt}=\frac{1}{\kappa}}.\]At fixed $t$, take a spatial gradient of $t-t_r-R(t_r)/c=0$:
\[-\nabla t_r-\frac{1}{c} \left(\mathbf n+\frac{dR}{dt_r}\nabla t_r\right)=0,\]which gives
\[\boxed{\nabla t_r=-\frac{\mathbf n}{c\kappa}}.\]For any source quantity $Q$ whose observation-point dependence occurs only through $t_r$,
\[\frac{\partial Q(t_r)}{\partial t}=\frac{\dot Q}{\kappa}, \qquad \nabla Q(t_r)=-\frac{\mathbf n\dot Q}{c\kappa}.\]The remaining geometric derivatives follow before any field differentiation:
\[\frac{\partial R}{\partial t} =\frac{dR}{dt_r}\frac{dt_r}{dt} =-\frac{c\,\mathbf n\cdot\boldsymbol\beta}{\kappa}, \qquad \nabla R=\mathbf n+\frac{dR}{dt_r}\nabla t_r =\frac{\mathbf n}{\kappa},\] \[\frac{\partial\mathbf n}{\partial t} =\frac{c[(\mathbf n\cdot\boldsymbol\beta)\mathbf n-\boldsymbol\beta]} {\kappa R}, \qquad \frac{d\kappa}{dt_r} =\frac{c[\beta^2-(\mathbf n\cdot\boldsymbol\beta)^2]}R -\mathbf n\cdot\dot{\boldsymbol\beta}.\]The spatial dependence of $\mathbf n$ must also be retained. In components,
\[\frac{\partial n_i}{\partial r_j} =\frac1R\left[\delta_{ij} +\frac{(\beta_i-n_i)n_j}{\kappa}\right].\]Consequently, with $b=\mathbf n\cdot\boldsymbol\beta$,
\[\boxed{ \nabla\kappa =-\frac{\boldsymbol\beta}{R} -\frac{\beta^2-b}{\kappa R}\mathbf n +\frac{\mathbf n\cdot\dot{\boldsymbol\beta}}{c\kappa}\mathbf n. }\]Put $C=q/(4\pi\epsilon_0)$. Since $\partial_t\kappa=\dot\kappa/\kappa$, direct differentiation of $\phi=C/(\kappa R)$ and $\mathbf A=\boldsymbol\beta\phi/c$ gives
\[\nabla\phi=-C\left[ \frac{\nabla\kappa}{\kappa^2R} +\frac{\mathbf n}{\kappa^2R^2}\right],\] \[\frac{\partial\mathbf A}{\partial t} =\frac Cc\left[ \frac{\dot{\boldsymbol\beta}}{\kappa^2R} +\boldsymbol\beta\left( \frac{cb}{\kappa^2R^2}-\frac{\dot\kappa}{\kappa^3R} \right)\right].\]Substitute the displayed formulas for $\nabla\kappa$ and $\dot\kappa$ into $\mathbf E=-\nabla\phi-\partial_t\mathbf A$. Separating inverse powers of $R$ gives
\[\begin{aligned} \frac{\mathbf E}{C} ={}&\frac1{R^2}\left[ \frac{\mathbf n-(1+b)\boldsymbol\beta}{\kappa^2} -\frac{(\beta^2-b)\mathbf n}{\kappa^3} +\frac{(\beta^2-b^2)\boldsymbol\beta}{\kappa^3} \right]\\ &+\frac1{c\kappa^3R}\left[ (\mathbf n\cdot\dot{\boldsymbol\beta}) (\mathbf n-\boldsymbol\beta) -\kappa\dot{\boldsymbol\beta}\right]. \end{aligned}\]Using $\kappa=1-b$, the $R^{-2}$ coefficients reduce as
\[\kappa-(\beta^2-b)=1-\beta^2,\] \[-(1+b)\kappa+(\beta^2-b^2)=-(1-\beta^2).\]The vector triple-product identity gives
\[(\mathbf n\cdot\dot{\boldsymbol\beta}) (\mathbf n-\boldsymbol\beta)-\kappa\dot{\boldsymbol\beta} =\mathbf n\times[(\mathbf n-\boldsymbol\beta) \times\dot{\boldsymbol\beta}].\]Therefore
\[\boxed{ \mathbf E(\mathbf r,t)=\frac{q}{4\pi\epsilon_0} \left[ \frac{(1-\beta^2)(\mathbf n-\boldsymbol\beta)}{\kappa^3R^2} +\frac{\mathbf n\times[(\mathbf n-\boldsymbol\beta) \times\dot{\boldsymbol\beta}]}{c\kappa^3R} \right]_{t_r}, }\]For the magnetic field,
\[\nabla\times\boldsymbol\beta =\nabla t_r\times\dot{\boldsymbol\beta} =-\frac{\mathbf n\times\dot{\boldsymbol\beta}}{c\kappa},\]so
\[\mathbf B=\nabla\times\mathbf A =\frac1c\left(\nabla\phi\times\boldsymbol\beta -\frac{\phi}{c\kappa}\mathbf n\times\dot{\boldsymbol\beta}\right).\]Inserting $\nabla\phi$ and collecting its $R^{-2}$ and $R^{-1}$ terms gives
\[\boxed{\mathbf B(\mathbf r,t)=\frac{1}{c}\mathbf n\times\mathbf E.}\]Here $\dot{\boldsymbol\beta}=d\boldsymbol\beta/dt_r$. The first term is the velocity field, proportional to $R^{-2}$; the second is the acceleration field, proportional to $R^{-1}$. Dimensional checking gives $q/(\epsilon_0R^2)$ for the first term and $q\dot\beta/(\epsilon_0cR)$ for the second, both in volts per metre.
Uniform rectilinear motion
For $\mathbf v=$ constant, $\dot{\boldsymbol\beta}=0$. Express the field in terms of the vector from the present charge position to the observer,
\[\mathbf s=\mathbf r-\mathbf r_q(t),\qquad \mathbf s=\mathbf s_\parallel+\mathbf s_\perp,\]where the subscripts are relative to $\mathbf v$. For uniform motion, $\mathbf r_q(t)=\mathbf r_q(t_r)+\mathbf v(t-t_r)$ and $c(t-t_r)=R$, so
\[\mathbf s=\mathbf R-\boldsymbol\beta R =R(\mathbf n-\boldsymbol\beta).\]Its parallel and perpendicular components obey
\[s_\parallel=R(\mathbf n\cdot\hat{\mathbf v}-\beta), \qquad s_\perp=R\sqrt{1-(\mathbf n\cdot\hat{\mathbf v})^2}.\]A direct expansion then gives
\[s_\parallel^2+(1-\beta^2)s_\perp^2 =R^2(1-\mathbf n\cdot\boldsymbol\beta)^2 =R^2\kappa^2.\]Since $\mathbf n-\boldsymbol\beta=\mathbf s/R$, the velocity field becomes
\[\boxed{ \mathbf E=\frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)\mathbf s} {\left[s_\parallel^2+(1-\beta^2)s_\perp^2\right]^{3/2}}, \qquad \mathbf B=\frac{1}{c^2}\mathbf v\times\mathbf E. }\]On the line of motion, $\mathbf s_\perp=0$ and $E=q/(4\pi\epsilon_0\gamma^2s^2)$. In the transverse plane, $s_\parallel=0$ and $E=\gamma q/(4\pi\epsilon_0s^2)$. Thus increasing $\gamma=(1-\beta^2)^{-1/2}$ suppresses the longitudinal field and compresses the field into transverse directions. The arbitrary-motion formula above reduces continuously to this result when the acceleration is set to zero.
The retarded-time identities and the algebraic reduction of the uniform-motion field are checked in the Maxima worksheet; every printed residual is zero.
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