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$\vec{E} = -\nabla V - \partial\vec{A}/\partial t$ and $\vec{B} = \nabla \times \vec{A}$, gauge transformations, the Lorenz gauge, and retarded potentials evaluated at $t - \mathscr{r}/c$.
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By the end of this lesson you will be able to compute fields from potentials, carry out gauge transformations, find induced fields from the vector potential, and apply retardation delays.
You know the electric potential of electrostatics, the vector potential of magnetostatics, and Maxwell's four equations. You know that changing magnetic fields produce curling electric fields, which no scalar potential can describe. This lesson rebuilds the potentials for changing fields and finds that they carry the finite speed of light built in.
| Term | What it means |
|---|---|
| Scalar potential | $V$, which with $\vec{A}$ gives $\vec{E} = -\nabla V - \partial\vec{A}/\partial t$. |
| Vector potential | $\vec{A}$, with $\vec{B} = \nabla \times \vec{A}$. |
| Gauge transformation | $\vec{A} \to \vec{A} + \nabla\lambda$, $V \to V - \partial\lambda/\partial t$, which leaves the fields unchanged. |
| Coulomb gauge | The choice $\nabla \cdot \vec{A} = 0$. |
| Lorenz gauge | The choice $\nabla \cdot \vec{A} = -\mu_0\varepsilon_0\partial V/\partial t$. |
| Retarded time | $t_r = t - \mathscr{r}/c$, when a signal now arriving left its source. |
| Retarded potentials | $V = \frac{1}{4\pi\varepsilon_0}\int\frac{\rho(t_r)}{\mathscr{r}}d\tau$ and the same for $\vec{A}$ with $\vec{J}$. |
Since $\nabla \cdot \vec{B} = 0$ always, $\vec{B} = \nabla \times \vec{A}$ still holds. But Faraday's law now gives $\nabla \times (\vec{E} + \partial\vec{A}/\partial t) = 0$, so it is $\vec{E} + \partial\vec{A}/\partial t$, not $\vec{E}$, that is a gradient:
$$\vec{E} = -\nabla V - \frac{\partial\vec{A}}{\partial t}, \qquad \vec{B} = \nabla \times \vec{A}.$$
Four functions, $V$ and the three components of $\vec{A}$, replace six field components, and two of Maxwell's equations are satisfied automatically. The potentials are not unique: for any function $\lambda$, the gauge transformation $\vec{A}' = \vec{A} + \nabla\lambda$, $V' = V - \partial\lambda/\partial t$ gives exactly the same $\vec{E}$ and $\vec{B}$. That freedom can be used to simplify the equations.
In the Lorenz gauge, $\nabla \cdot \vec{A} = -\mu_0\varepsilon_0\,\partial V/\partial t$, the remaining Maxwell equations become wave equations for each potential, with sources $\rho$ and $\vec{J}$. Their solutions are the retarded potentials,
$$V(\vec{r}, t) = \frac{1}{4\pi\varepsilon_0}\int\frac{\rho(\vec{r}', t_r)}{\mathscr{r}}\,d\tau', \qquad \vec{A}(\vec{r}, t) = \frac{\mu_0}{4\pi}\int\frac{\vec{J}(\vec{r}', t_r)}{\mathscr{r}}\,d\tau', \qquad t_r = t - \frac{\mathscr{r}}{c}.$$
They look like the static formulas, except that each source is evaluated at the retarded time, when a signal traveling at $c$ would have left it. Information about charges and currents spreads outward at the speed of light.
Another way: picture
Picture looking at the night sky. Each star you see is shown as it was when its light left it, years or centuries ago, and farther stars are shown further in the past. The retarded potentials work the same way: at every point in space, each charge and current contributes as it was a time $\mathscr{r}/c$ earlier. The fields you measure now are a mosaic of the past, assembled at the speed of light.
Another way: steps
Checks. In the static limit, $\partial\vec{A}/\partial t = 0$ and the formulas reduce to electrostatics and magnetostatics. A gauge transformation must never change $\vec{E}$ or $\vec{B}$. The electric field from $-\partial\vec{A}/\partial t$ must agree with Faraday's law. And retardation delays must equal distance over $c$.
In statics, $\nabla \times \vec{E} = 0$ and $\vec{E} = -\nabla V$. With changing magnetic fields, $\nabla \times \vec{E} = -\partial\vec{B}/\partial t = -\nabla \times (\partial\vec{A}/\partial t)$, so the combination $\vec{E} + \partial\vec{A}/\partial t$ is curl-free and can be written as $-\nabla V$. The induced electric field of Faraday's law lives entirely in the $-\partial\vec{A}/\partial t$ term.
Inside a solenoid with growing current, $\vec{A}$ circles the axis and grows, and $-\partial\vec{A}/\partial t$ gives exactly the circulating field that Faraday's law predicted, with no scalar potential at all. The two descriptions agree, but the potential form packages the physics more compactly and generalizes to radiation.
Adding the gradient of any function to $\vec{A}$ leaves its curl unchanged, so $\vec{B}$ is unchanged. To keep $\vec{E}$ unchanged too, $V$ must shift by $-\partial\lambda/\partial t$. The potentials are therefore not directly measurable in classical physics; only the fields are. Choosing a gauge is choosing a convenient description.
The Coulomb gauge, $\nabla \cdot \vec{A} = 0$, makes $V$ satisfy Poisson's equation instantaneously, which is convenient for atoms and slowly varying problems. The Lorenz gauge treats $V$ and $\vec{A}$ symmetrically and fits relativity. In the Coulomb gauge $V$ appears to respond instantly to distant charges, but the fields do not: the instantaneous part is cancelled by a matching part of $\vec{A}$, and every measurable effect still travels at $c$.
The wave equations for the potentials have two families of solutions: retarded, in which effects follow causes by $\mathscr{r}/c$, and advanced, in which they would precede them. Nature uses the retarded ones; why is related to the arrow of time. The retarded potentials guarantee that a change at one place cannot be felt elsewhere before light could get there.
For slowly varying sources close by, the delay is negligible and the static formulas work. For a circuit a meter across running at $1$ MHz, the delay is $3$ ns against a period of $1000$ ns, negligible. For the same circuit at $1$ GHz, the delay is three periods, and retardation cannot be ignored: this is where circuits start to radiate and the wires behave as antennas.
Differentiating the retarded potentials gives the fields, Jefimenko's equations. Each has two kinds of terms: ones falling as $1/\mathscr{r}^2$, which for steady sources reduce to Coulomb and Biot–Savart, and new ones falling as $1/\mathscr{r}$, proportional to the rate of change of the sources at the retarded time.
Those $1/\mathscr{r}$ terms are radiation. Their energy flux falls as $1/\mathscr{r}^2$, so the total power through a large sphere stays constant: the energy escapes to infinity. The next two lessons compute that power for oscillating dipoles and for accelerating charges, the two basic sources of all electromagnetic radiation.
In quantum mechanics the potentials, not just the fields, enter the Schrödinger equation: a charged particle's momentum operator becomes $\hat{\vec{p}} - q\vec{A}$, and its energy includes $qV$. Gauge transformations then change the wave function's phase, and physical predictions stay the same only because the phase shifts compensate.
The Aharonov–Bohm effect shows the potentials have direct consequences: electrons passing on either side of a shielded solenoid, through regions where $\vec{B} = 0$ but $\vec{A} \ne 0$, acquire a relative phase equal to $q\Phi/\hbar$, shifting their interference pattern. The gauge invariant quantity is the flux, $\oint\vec{A} \cdot d\vec{l}$ around the loop. Gauge symmetry, extended, is the organizing principle of the entire Standard Model of particle physics.
Retardation is part of daily engineering. A signal crossing a data center's fiber takes about $5$ ns per meter, so the physical layout of servers matters for high-frequency trading, where firms have paid for straighter fiber routes and microwave links between Chicago and New Jersey to shave milliseconds. Satellite phone calls through geostationary satellites, $36{,}000$ km up, suffer a quarter-second round-trip delay.
In space, the delays are minutes. Commands to Mars rovers take between about $4$ and $22$ minutes one way, depending on the planets' positions, so NASA's rovers are given a day's plan at a time and drive autonomously between check-ins. Voyager 1, more than $160$ AU away, hears from Earth nearly a day after each command is sent.
The six components of $\vec{E}$ and $\vec{B}$ are not independent: Maxwell's two source-free equations, $\nabla \cdot \vec{B} = 0$ and Faraday's law, constrain them. The potentials build those constraints in, reducing the problem to four functions, and gauge freedom removes one more. What remains, in empty space, is two independent degrees of freedom — the two polarizations of light.
This counting reappears in quantum electrodynamics, where the photon, massless, has exactly two polarization states, and in the relativistic formulation of the final lesson, where $V/c$ and $\vec{A}$ combine into a single four-vector potential. The potentials are the natural language for relativity, quantum theory and radiation alike.
NASA's Perseverance and Curiosity rovers receive commands from the Deep Space Network's antennas in California, Spain and Australia. The radio signal's fields at Mars are set by the antenna currents on Earth at the retarded time: $4$ to $22$ minutes earlier, depending on where the two planets are in their orbits. A round trip takes twice that.
No one can drive a rover by joystick across such a delay. Instead, engineers at the Jet Propulsion Laboratory send a full day's plan, and the rover's software navigates autonomously, choosing paths around rocks with its own cameras. Near solar conjunction, when Mars passes behind the Sun, the Sun's plasma disrupts the signals and commands are suspended for about two weeks — a reminder that the speed of light, built into the retarded potentials, sets the rhythm of planetary exploration.
Kirchhoff's circuit laws assume that current flows instantaneously around a loop, which is true only when the circuit is much smaller than the wavelength $c/f$. At the $60$ Hz of the power grid the wavelength is $5000$ km, and even continent-spanning lines are only a fraction of it. At the gigahertz frequencies of modern processors and wireless chips, the wavelength is centimeters, comparable to the circuit board.
Then retardation matters: signals take time to travel along traces, different parts of a circuit are out of phase, and wires radiate. Engineers treat fast traces as transmission lines, match their impedances, and shield them. The Federal Communications Commission requires digital devices to be tested for unintentional radiation, because every fast circuit is, to some degree, an antenna — a direct consequence of the retarded potentials.
Coulomb's law, used carelessly, suggests that moving a charge here instantly changes the field on the Moon. Maxwell's equations forbid it: the retarded potentials make every effect arrive a time $\mathscr{r}/c$ after its cause. Static formulas are the limit in which sources change so slowly that the delay does not matter.
A second misconception is that the electric field is always minus the gradient of $V$. That is true only in statics. When magnetic fields change, the $-\partial\vec{A}/\partial t$ term is essential; without it, the induced fields of transformers and generators would be impossible to describe.
A long solenoid has field $B = \mu_0nI$ inside. Use the circulation of $\vec{A}$.
$\oint\vec{A} \cdot d\vec{l} = \displaystyle\int\vec{B} \cdot d\vec{a} = \Phi_B$
From $\vec{B} = \nabla \times \vec{A}$ and Stokes's theorem.
Apply it inside, at radius $s < R$.
$A \cdot 2\pi s = \mu_0nI\,\pi s^2 \quad\Rightarrow\quad A = \tfrac{1}{2}\mu_0nIs$
Circling the axis.
Apply it outside, at $s > R$.
$A \cdot 2\pi s = \mu_0nI\,\pi R^2 \quad\Rightarrow\quad A = \dfrac{\mu_0nIR^2}{2s}$
Nonzero where $\vec{B} = 0$.
Find the induced field when the current changes.
$E = -\dfrac{\partial A}{\partial t} = -\tfrac{1}{2}\mu_0ns\dfrac{dI}{dt} \quad (s < R)$
The Faraday field, from the potential.
Evaluate for $n = 1000$ m⁻¹, $s = 2$ cm, $dI/dt = 100$ A/s.
$E = \tfrac{1}{2} \times 4\pi \times 10^{-7} \times 1000 \times 0.02 \times 100 = 1.26\ \text{mV/m}$
A small circling electric field.
Start with $V = 0$, $\vec{A} = 0$: no fields at all. Choose $\lambda = axt$.
$\lambda = axt$
Any smooth function will do.
Transform the scalar potential.
$V' = 0 - \dfrac{\partial\lambda}{\partial t} = -ax$
A potential that increases toward $-x$.
Transform the vector potential.
$\vec{A}' = 0 + \nabla\lambda = at\,\hat{x}$
Growing in time.
Compute the new electric field.
$\vec{E}' = -\nabla V' - \dfrac{\partial\vec{A}'}{\partial t} = a\hat{x} - a\hat{x} = 0$
The two terms cancel.
Compute the new magnetic field.
$\vec{B}' = \nabla \times (at\,\hat{x}) = 0$
Uniform in space, so no curl.
Draw the conclusion.
$\text{same fields, different potentials}$
Potentials carry redundant information.
A loop of wire $1$ m across carries a current oscillating at frequency $f$. Find the light-travel time across it.
$\Delta t = \dfrac{1}{3.00 \times 10^8} = 3.3\ \text{ns}$
The largest retardation delay within the circuit.
Compare with the period at $60$ Hz.
$T = 16.7\ \text{ms} \gg 3.3\ \text{ns}$
Retardation is utterly negligible.
Compare at $1$ MHz.
$T = 1000\ \text{ns} \gg 3.3\ \text{ns}$
Still negligible: quasi-static circuit theory works.
Compare at $300$ MHz.
$T = 3.3\ \text{ns} = \Delta t$
The circuit is one wavelength across.
State the criterion.
$\text{size} \ll \lambda = \dfrac{c}{f}$
Kirchhoff's laws hold only for circuits small compared with the wavelength.
Describe what happens above it.
$\text{currents at different points are out of phase}$
The circuit radiates, and its wires act as antennas.
Apply to a computer motherboard at $3$ GHz.
$\lambda = 10\ \text{cm}$
Traces several centimeters long must be treated as transmission lines.
Connect to the next lessons.
$\text{radiation} \propto \text{size}/\lambda$
Small, slow circuits barely radiate; large or fast ones radiate strongly.
Write the delay.
$\Delta t = \dfrac{d}{c}$
Distance over the speed of light.
Substitute the values.
$\Delta t = \dfrac{3.6 \times 10^{7}}{3.00 \times 10^8}$
SI units.
Evaluate the delay.
In some region the scalar potential is zero everywhere, while the vector potential is uniform and growing: $\vec{A} = 13t\,\hat{x}$ V s/m per second. What is the electric field?
Complete the worked solution: for $\lambda = 8xt$, find $\partial\lambda/\partial t$ at $x = 7$, $\partial\lambda/\partial x$ at $t = 2$, and the change in $E_x$ the transformation produces.
Differentiate in time.
$\dfrac{\partial\lambda}{\partial t} = 8x =$ p
This shifts $V$ by minus this amount.
Differentiate in space.
$\dfrac{\partial\lambda}{\partial x} = 8t =$ q
This shifts $A_x$ by this amount.
Combine the changes in $E_x$.
$\Delta E_x = \dfrac{\partial^2\lambda}{\partial x\,\partial t} - \dfrac{\partial^2\lambda}{\partial t\,\partial x} =$ r
Mixed partial derivatives are equal: gauge changes never alter the fields.
Match each statement to its expression.
| $-\nabla V - \partial_t\vec{A}$ | $\nabla \times \vec{A}$ | $\nabla \cdot \vec{A} = -\mu_0\varepsilon_0\partial_tV$ | $t - \mathscr{r}/c$ | |
|---|---|---|---|---|
| $\vec{E}$ from potentials | ||||
| $\vec{B}$ from potentials | ||||
| the Lorenz gauge | ||||
| the retarded time |
A source changes suddenly. Fill in how long, in seconds, before the change is felt at distances of $3$ m, $300$ km and $1.5 \times 10^{11}$ m (about $1$ AU).
| delay (s) | |
|---|---|
| $3$ m | |
| $300$ km | |
| $1.5 \times 10^{11}$ m |
Starting from $V = 0$ and $\vec{A} = 0$, apply the gauge transformation with $\lambda = 6xt$ (SI units). What is the new scalar potential at $x = 1$ m, in volts?
Answer: V
A long solenoid with $500$ turns per meter has its current rising at $200$ A/s. Using the vector potential, find the induced electric field inside it $3$ cm from the axis, in mV/m.
Answer: mV/m
Mars is $0.52$ AU from Earth. How long does a command from mission control take to reach a rover there, in minutes? Use $1$ AU $= 1.496 \times 10^{11}$ m and $c = 2.998 \times 10^8$ m/s.
Answer: min
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Complete the worked solution: for $\lambda = 8xt$, find $\partial\lambda/\partial t$ at $x = 6$, $\partial\lambda/\partial x$ at $t = 5$, and the change in $E_x$ the transformation produces.
Differentiate in time.
$\dfrac{\partial\lambda}{\partial t} = 8x =$ p
This shifts $V$ by minus this amount.
Differentiate in space.
$\dfrac{\partial\lambda}{\partial x} = 8t =$ q
This shifts $A_x$ by this amount.
Combine the changes in $E_x$.
$\Delta E_x = \dfrac{\partial^2\lambda}{\partial x\,\partial t} - \dfrac{\partial^2\lambda}{\partial t\,\partial x} =$ r
Mixed partial derivatives are equal: gauge changes never alter the fields.
You can work with electrodynamic potentials. Explain to someone why moving a charge here cannot instantly change the field far away.
19. Your turn: how long does a radio signal take to reach a satellite $36{,}000$ km overhead?, step 3
$\Delta t = 0.12\ \text{s}$
A quarter second for the round trip.