Retarded Potentials & Liénard–Wiechert Fields

Step 1: Gauss's law in terms of potentials

$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} \quad \Rightarrow \quad \nabla \cdot \left(-\nabla V - \frac{\partial \mathbf{A}}{\partial t}\right) = \frac{\rho}{\epsilon_0}$$

$$-\nabla^2 V - \frac{\partial}{\partial t}(\nabla \cdot \mathbf{A}) = \frac{\rho}{\epsilon_0}$$

Step 2: Ampere-Maxwell law in terms of potentials

$$\nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0\epsilon_0\frac{\partial \mathbf{E}}{\partial t} \quad \Rightarrow \quad \nabla \times (\nabla \times \mathbf{A}) = \mu_0\mathbf{J} + \mu_0\epsilon_0\frac{\partial}{\partial t}\left(-\nabla V - \frac{\partial \mathbf{A}}{\partial t}\right)$$

Step 3: Expand the double curl

$$\nabla(\nabla \cdot \mathbf{A}) - \nabla^2\mathbf{A} = \mu_0\mathbf{J} - \mu_0\epsilon_0\nabla\frac{\partial V}{\partial t} - \mu_0\epsilon_0\frac{\partial^2\mathbf{A}}{\partial t^2}$$

Step 4: Impose the Lorenz gauge

The potentials $V$ and $\mathbf{A}$ are not unique — we can perform gauge transformations $\mathbf{A} \to \mathbf{A} + \nabla\lambda$, $V \to V - \partial\lambda/\partial t$. Choose the Lorenz gauge:

$$\boxed{\nabla \cdot \mathbf{A} + \mu_0\epsilon_0\frac{\partial V}{\partial t} = 0}$$

Step 5: Connection to charge conservation

Take $\partial/\partial t$ of Gauss's law and $\nabla \cdot$ of Ampere's law. The continuity equation $\nabla \cdot \mathbf{J} + \partial\rho/\partial t = 0$ guarantees that if the Lorenz condition holds at $t = 0$, it holds for all time:

$$\frac{\partial}{\partial t}\left(\nabla \cdot \mathbf{A} + \mu_0\epsilon_0\frac{\partial V}{\partial t}\right) = 0 \quad \Leftarrow \quad \nabla \cdot \mathbf{J} + \frac{\partial \rho}{\partial t} = 0$$

The Lorenz gauge is thus the natural gauge compatible with charge conservation and Lorentz invariance.

Step 1: Gauss's law equation with Lorenz gauge

From Step 1 above: $-\nabla^2 V - \partial(\nabla \cdot \mathbf{A})/\partial t = \rho/\epsilon_0$. Substituting $\nabla \cdot \mathbf{A} = -\mu_0\epsilon_0 \partial V/\partial t$:

$$-\nabla^2 V + \mu_0\epsilon_0\frac{\partial^2 V}{\partial t^2} = \frac{\rho}{\epsilon_0}$$

Step 2: Write as a d'Alembertian

Define $\Box^2 \equiv \nabla^2 - \mu_0\epsilon_0\frac{\partial^2}{\partial t^2} = \nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}$. Then:

$$\boxed{\Box^2 V = -\frac{\rho}{\epsilon_0}}$$

Step 3: Ampere-Maxwell equation with Lorenz gauge

From Step 3 above, with $\nabla \cdot \mathbf{A} = -\mu_0\epsilon_0 \partial V/\partial t$:

$$-\mu_0\epsilon_0\nabla\frac{\partial V}{\partial t} - \nabla^2\mathbf{A} = \mu_0\mathbf{J} - \mu_0\epsilon_0\nabla\frac{\partial V}{\partial t} - \mu_0\epsilon_0\frac{\partial^2\mathbf{A}}{\partial t^2}$$

Step 4: Cancel and simplify

The $\nabla(\partial V/\partial t)$ terms cancel on both sides, leaving:

$$-\nabla^2\mathbf{A} = \mu_0\mathbf{J} - \mu_0\epsilon_0\frac{\partial^2\mathbf{A}}{\partial t^2}$$

$$\boxed{\Box^2 \mathbf{A} = -\mu_0\mathbf{J}}$$

Step 5: Summary — four decoupled wave equations

In Lorenz gauge, Maxwell's equations reduce to four decoupled inhomogeneous wave equations:

$$\nabla^2 V - \frac{1}{c^2}\frac{\partial^2 V}{\partial t^2} = -\frac{\rho}{\epsilon_0}, \qquad \nabla^2 \mathbf{A} - \frac{1}{c^2}\frac{\partial^2 \mathbf{A}}{\partial t^2} = -\mu_0\mathbf{J}$$

Each component satisfies the same wave equation with a source term, solvable by the retarded Green's function.

Step 1: Green's function of the wave equation

The Green's function satisfies $\Box^2 G(\mathbf{r},t;\mathbf{r}',t') = -\delta^3(\mathbf{r}-\mathbf{r}')\delta(t-t')$. By causality we want the retarded solution: the field at $(\mathbf{r},t)$ depends only on the source at earlier times.

Step 2: Fourier transform to find G

Fourier transforming in time, the wave equation becomes the Helmholtz equation. Solving in spherical coordinates about $\mathbf{r}'$ and requiring outgoing waves:

$$\tilde{G}(\mathbf{r},\omega;\mathbf{r}') = \frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{4\pi|\mathbf{r}-\mathbf{r}'|}, \qquad k = \omega/c$$

Step 3: Inverse Fourier transform

Transforming back to the time domain with the retarded (causal) boundary condition:

$$G_{\rm ret}(\mathbf{r},t;\mathbf{r}',t') = \frac{1}{4\pi|\mathbf{r}-\mathbf{r}'|}\,\delta\!\left(t' - \left[t - \frac{|\mathbf{r}-\mathbf{r}'|}{c}\right]\right)$$

The delta function enforces that only the source at the retarded time $t_r = t - |\mathbf{r}-\mathbf{r}'|/c$ contributes.

Step 4: Convolve with source

$$V(\mathbf{r},t) = \frac{1}{\epsilon_0}\int G_{\rm ret}(\mathbf{r},t;\mathbf{r}',t')\,\rho(\mathbf{r}',t')\,d\tau'\,dt'$$

Performing the $t'$ integration using the delta function:

$$V(\mathbf{r},t) = \frac{1}{4\pi\epsilon_0}\int\frac{\rho(\mathbf{r}',t_r)}{|\mathbf{r}-\mathbf{r}'|}\,d\tau'$$

Step 5: Identically for the vector potential

The same Green's function solves $\Box^2\mathbf{A} = -\mu_0\mathbf{J}$:

$$\boxed{\mathbf{A}(\mathbf{r},t) = \frac{\mu_0}{4\pi}\int\frac{\mathbf{J}(\mathbf{r}',t_r)}{|\mathbf{r}-\mathbf{r}'|}\,d\tau'}, \qquad t_r = t - \frac{|\mathbf{r}-\mathbf{r}'|}{c}$$

Each source point contributes with a time delay proportional to its distance from the field point — this is retardation.

Step 1: Define the retarded time condition

The signal from the charge at position $\mathbf{w}(t_r)$ reaches the field point $\mathbf{r}$ at time $t$ if it travels at speed $c$:

$$c(t - t_r) = |\mathbf{r} - \mathbf{w}(t_r)| \equiv \mathscr{r}(t_r)$$

Step 2: Geometric interpretation

Define $f(t_r) = c(t - t_r) - |\mathbf{r} - \mathbf{w}(t_r)|$. We seek roots of $f(t_r) = 0$. The left side $c(t - t_r)$ is a line decreasing with slope $-c$. The right side $|\mathbf{r} - \mathbf{w}(t_r)|$ changes at most at speed $v < c$.

Step 3: Compute the derivative

$$\frac{df}{dt_r} = -c - \frac{d}{dt_r}|\mathbf{r} - \mathbf{w}(t_r)| = -c + \frac{(\mathbf{r} - \mathbf{w})\cdot\mathbf{v}}{|\mathbf{r} - \mathbf{w}|} = -c + \hat{\boldsymbol{\mathscr{r}}}\cdot\mathbf{v}$$

Step 4: Show the derivative is always negative

Since $|\hat{\boldsymbol{\mathscr{r}}}\cdot\mathbf{v}| \leq |\mathbf{v}| = v < c$, we have:

$$\frac{df}{dt_r} = -c + \hat{\boldsymbol{\mathscr{r}}}\cdot\mathbf{v} < -c + c = 0$$

The function $f(t_r)$ is strictly decreasing for all subluminal charges ($v < c$).

Step 5: Uniqueness follows

Since $f(t_r) \to +\infty$ as $t_r \to -\infty$ and $f(t_r) \to -\infty$ as $t_r \to t$, and $f$ is strictly monotonically decreasing, by the intermediate value theorem there is exactly one root:

$$\boxed{t_r = t - \frac{|\mathbf{r} - \mathbf{w}(t_r)|}{c} \quad \text{has a unique solution for } v < c}$$

This is essential: each field event $(\mathbf{r},t)$ is causally connected to exactly one point on the charge's worldline.

Step 1: Key differentiation identity

Let $\boldsymbol{\mathscr{r}} = \mathbf{r} - \mathbf{r}'$, $\mathscr{r} = |\boldsymbol{\mathscr{r}}|$, and $t_r = t - \mathscr{r}/c$. When we differentiate $\rho(\mathbf{r}',t_r)$ with respect to the field point $\mathbf{r}$, the retarded time also depends on $\mathbf{r}$:

$$\nabla t_r = -\frac{1}{c}\nabla\mathscr{r} = -\frac{\hat{\boldsymbol{\mathscr{r}}}}{c}$$

Step 2: Gradient of the retarded scalar potential

$$\nabla\left[\frac{\rho(\mathbf{r}',t_r)}{\mathscr{r}}\right] = \frac{1}{\mathscr{r}}\nabla\rho\big|_{t_r} + \rho\,\nabla\!\left(\frac{1}{\mathscr{r}}\right) = \frac{1}{\mathscr{r}}\dot{\rho}\,\nabla t_r - \frac{\rho}{\mathscr{r}^2}\hat{\boldsymbol{\mathscr{r}}}$$

$$= -\frac{\dot{\rho}}{c\mathscr{r}}\hat{\boldsymbol{\mathscr{r}}} - \frac{\rho}{\mathscr{r}^2}\hat{\boldsymbol{\mathscr{r}}}$$

Step 3: Time derivative of A

Since $t$ appears in $\mathbf{A}$ only through $t_r$ and $\partial t_r/\partial t = 1$:

$$\frac{\partial \mathbf{A}}{\partial t} = \frac{\mu_0}{4\pi}\int\frac{\dot{\mathbf{J}}(\mathbf{r}',t_r)}{\mathscr{r}}\,d\tau'$$

Step 4: Combine for E (Jefimenko)

$$\mathbf{E}(\mathbf{r},t) = -\nabla V - \frac{\partial\mathbf{A}}{\partial t} = \frac{1}{4\pi\epsilon_0}\int\left[\frac{\rho(\mathbf{r}',t_r)}{\mathscr{r}^2}\hat{\boldsymbol{\mathscr{r}}} + \frac{\dot{\rho}(\mathbf{r}',t_r)}{c\mathscr{r}}\hat{\boldsymbol{\mathscr{r}}} - \frac{\dot{\mathbf{J}}(\mathbf{r}',t_r)}{c^2\mathscr{r}}\right]d\tau'$$

Step 5: Curl of A for B

Similarly, computing $\nabla \times \mathbf{A}$ with the chain rule on $t_r$:

$$\nabla \times \left[\frac{\mathbf{J}(\mathbf{r}',t_r)}{\mathscr{r}}\right] = \frac{1}{\mathscr{r}}\dot{\mathbf{J}} \times \nabla t_r + \mathbf{J} \times \nabla\!\left(\frac{1}{\mathscr{r}}\right) = -\frac{\dot{\mathbf{J}}}{c\mathscr{r}} \times \hat{\boldsymbol{\mathscr{r}}} + \frac{\mathbf{J}}{\mathscr{r}^2}\times\hat{\boldsymbol{\mathscr{r}}}$$

Step 6: The Jefimenko equations

$$\boxed{\mathbf{B}(\mathbf{r},t) = \frac{\mu_0}{4\pi}\int\left[\frac{\mathbf{J}(\mathbf{r}',t_r)}{\mathscr{r}^2} + \frac{\dot{\mathbf{J}}(\mathbf{r}',t_r)}{c\mathscr{r}}\right]\times\hat{\boldsymbol{\mathscr{r}}}\,d\tau'}$$

These are the causal generalizations of Coulomb's law and the Biot-Savart law. All quantities are evaluated at the retarded time, and $\dot{\rho}$, $\dot{\mathbf{J}}$ terms capture radiation effects.

Step 1: Point charge sources

For a point charge, the charge and current densities are:

$$\rho(\mathbf{r}',t') = q\,\delta^3(\mathbf{r}' - \mathbf{w}(t')), \qquad \mathbf{J}(\mathbf{r}',t') = q\mathbf{v}(t')\,\delta^3(\mathbf{r}' - \mathbf{w}(t'))$$

Step 2: Substitute into the retarded potential

$$V(\mathbf{r},t) = \frac{1}{4\pi\epsilon_0}\int\frac{q\,\delta^3(\mathbf{r}' - \mathbf{w}(t_r))}{|\mathbf{r}-\mathbf{r}'|}\,d\tau'$$

where $t_r = t - |\mathbf{r}-\mathbf{r}'|/c$. The difficulty: $t_r$ itself depends on $\mathbf{r}'$, so this is not a simple delta function integral.

Step 3: Use the 4D formulation

Rewrite using an explicit time integral with a delta function enforcing the retarded time constraint:

$$V = \frac{q}{4\pi\epsilon_0}\int\!\!\int\frac{\delta^3(\mathbf{r}'-\mathbf{w}(t'))\,\delta(t' - t + |\mathbf{r}-\mathbf{r}'|/c)}{|\mathbf{r}-\mathbf{r}'|}\,d\tau'\,dt'$$

First integrate over $d\tau'$ using the spatial delta function, setting $\mathbf{r}' = \mathbf{w}(t')$:

$$V = \frac{q}{4\pi\epsilon_0}\int\frac{\delta(t' - t + |\mathbf{r}-\mathbf{w}(t')|/c)}{|\mathbf{r}-\mathbf{w}(t')|}\,dt'$$

Step 4: Evaluate the time integral using the delta function identity

Let $g(t') = t' - t + |\mathbf{r}-\mathbf{w}(t')|/c$. By the delta function rule $\delta(g(t')) = \delta(t'-t_r)/|g'(t_r)|$:

$$g'(t') = 1 + \frac{1}{c}\frac{d}{dt'}|\mathbf{r}-\mathbf{w}(t')| = 1 - \frac{\hat{\boldsymbol{\mathscr{r}}}\cdot\mathbf{v}}{c}$$

where $\boldsymbol{\mathscr{r}} = \mathbf{r} - \mathbf{w}(t')$ and we used $\frac{d}{dt'}|\mathbf{r}-\mathbf{w}| = -\hat{\boldsymbol{\mathscr{r}}}\cdot\mathbf{v}$.

Step 5: The Jacobian factor

Performing the $t'$ integration:

$$V = \frac{q}{4\pi\epsilon_0}\frac{1}{\mathscr{r}(1 - \hat{\boldsymbol{\mathscr{r}}}\cdot\mathbf{v}/c)}\Bigg|_{t_r} = \frac{q}{4\pi\epsilon_0}\frac{c}{\mathscr{r}c - \boldsymbol{\mathscr{r}}\cdot\mathbf{v}}\Bigg|_{t_r}$$

The factor $(1 - \hat{\boldsymbol{\mathscr{r}}}\cdot\mathbf{v}/c)^{-1}$ is not a relativistic correction — it arises because the retarded time varies across the volume of the source.

Step 6: The vector potential follows identically

Repeating for $\mathbf{A}$ with the extra factor of $\mathbf{v}$ from $\mathbf{J} = q\mathbf{v}\delta^3$:

$$\boxed{V = \frac{qc}{4\pi\epsilon_0(\mathscr{r}c - \boldsymbol{\mathscr{r}}\cdot\mathbf{v})}\bigg|_{t_r}, \qquad \mathbf{A} = \frac{\mu_0 qc\mathbf{v}}{4\pi(\mathscr{r}c - \boldsymbol{\mathscr{r}}\cdot\mathbf{v})}\bigg|_{t_r} = \frac{\mathbf{v}}{c^2}V}$$

These are the Liénard-Wiechert potentials. The denominator $\mathscr{r}c - \boldsymbol{\mathscr{r}}\cdot\mathbf{v}$ enhances the field in the forward direction (approaching charge) and suppresses it behind.

Step 1: Derivatives of the retarded time

From $c(t - t_r) = \mathscr{r} = |\mathbf{r} - \mathbf{w}(t_r)|$, differentiate with respect to $t$:

$$c\left(1 - \frac{\partial t_r}{\partial t}\right) = \frac{\partial \mathscr{r}}{\partial t} = -\hat{\boldsymbol{\mathscr{r}}}\cdot\mathbf{v}\frac{\partial t_r}{\partial t}$$

$$\frac{\partial t_r}{\partial t} = \frac{c}{c - \hat{\boldsymbol{\mathscr{r}}}\cdot\mathbf{v}} = \frac{\mathscr{r}c}{\mathscr{r}c - \boldsymbol{\mathscr{r}}\cdot\mathbf{v}}$$

Step 2: Gradient of the retarded time

Differentiating the constraint with respect to $\mathbf{r}$:

$$-c\nabla t_r = \nabla\mathscr{r} = \hat{\boldsymbol{\mathscr{r}}} - \hat{\boldsymbol{\mathscr{r}}}\cdot\mathbf{v}\,\nabla t_r \cdot \frac{\boldsymbol{\mathscr{r}}}{\mathscr{r}} \quad \Rightarrow \quad \nabla t_r = -\frac{\hat{\boldsymbol{\mathscr{r}}}}{c - \hat{\boldsymbol{\mathscr{r}}}\cdot\mathbf{v}} = -\frac{\boldsymbol{\mathscr{r}}}{\mathscr{r}c - \boldsymbol{\mathscr{r}}\cdot\mathbf{v}}$$

Step 3: Define the auxiliary variable u

Let $\mathbf{u} \equiv c\hat{\boldsymbol{\mathscr{r}}} - \mathbf{v}$. Then $\boldsymbol{\mathscr{r}}\cdot\mathbf{u} = \mathscr{r}c - \boldsymbol{\mathscr{r}}\cdot\mathbf{v}$ (the L-W denominator). This encodes the Doppler-like geometry of the problem.

Step 4: Compute the gradient of V

Write $V = qc/(4\pi\epsilon_0\,\boldsymbol{\mathscr{r}}\cdot\mathbf{u})$. Taking $\nabla$ requires differentiating $\boldsymbol{\mathscr{r}}$ and $\mathbf{u}$ through $t_r(\mathbf{r})$. After lengthy but systematic calculation using the product and chain rules:

$$\nabla V = -\frac{q}{4\pi\epsilon_0}\frac{1}{(\boldsymbol{\mathscr{r}}\cdot\mathbf{u})^3}\left[\mathscr{r}(c^2 - v^2)\hat{\boldsymbol{\mathscr{r}}} + (\boldsymbol{\mathscr{r}}\cdot\mathbf{u})\mathbf{v} - \mathscr{r}(\boldsymbol{\mathscr{r}}\cdot\mathbf{a})\hat{\boldsymbol{\mathscr{r}}}\right]$$

Step 5: Compute the time derivative of A

Since $\mathbf{A} = (\mathbf{v}/c^2)V$ and the time dependence comes through $t_r$:

$$\frac{\partial\mathbf{A}}{\partial t} = \frac{1}{c^2}\left[\mathbf{a}\frac{\partial t_r}{\partial t}V + \mathbf{v}\frac{\partial V}{\partial t}\right]$$

Using $\partial t_r/\partial t = \mathscr{r}c/(\boldsymbol{\mathscr{r}}\cdot\mathbf{u})$ and a similar lengthy computation for $\partial V/\partial t$.

Step 6: Combine using the BAC-CAB rule

After combining $\mathbf{E} = -\nabla V - \partial\mathbf{A}/\partial t$ and simplifying using $\boldsymbol{\mathscr{r}}\times(\mathbf{u}\times\mathbf{a}) = \mathbf{u}(\boldsymbol{\mathscr{r}}\cdot\mathbf{a}) - \mathbf{a}(\boldsymbol{\mathscr{r}}\cdot\mathbf{u})$:

$$\boxed{\mathbf{E} = \frac{q}{4\pi\epsilon_0}\frac{\mathscr{r}}{(\boldsymbol{\mathscr{r}}\cdot\mathbf{u})^3}\Big[\underbrace{(c^2-v^2)\mathbf{u}}_{\text{velocity field } \sim 1/\mathscr{r}^2} + \underbrace{\boldsymbol{\mathscr{r}}\times(\mathbf{u}\times\mathbf{a})}_{\text{acceleration field } \sim 1/\mathscr{r}}\Big]}$$

Step 7: The magnetic field

A remarkable result: $\mathbf{B}$ is always perpendicular to $\mathbf{E}$ and $\hat{\boldsymbol{\mathscr{r}}}$:

$$\boxed{\mathbf{B} = \frac{1}{c}\hat{\boldsymbol{\mathscr{r}}}\times\mathbf{E}}$$

The velocity field ($\propto 1/\mathscr{r}^2$) is the generalized Coulomb field of a moving charge — it does not radiate. The acceleration field ($\propto 1/\mathscr{r}$) is the radiation field — it carries energy to infinity and exists only when the charge accelerates.