Problem 5.5.5: A pair of potentials that describes nothing

Problem 5.5.5

A pair of potentials that describes nothing

In free space a vector potential is given as

A = A_{0} cos (ω t - k z) a_{z}, ω = k c .

Find the scalar potential $V$ that satisfies the Lorenz gauge (take the solution without a static part).
Show that $A$ and $V$ satisfy the homogeneous wave equations.
Find $E$ and $B$ .
Find a scalar function $ψ$ with $A = \nabla ψ$ and $V = - \partial ψ / \partial t$ . What does this mean?
Repeat (c) for $A = A_{0} cos (ω t - k z) a_{x}$ , $V = 0$ .

Try it yourself first.

Strategy

The Lorenz gauge Eq. (5.5.41) links $\partial V / \partial t$ to $\nabla \cdot A$ , so one time integration gives $V$ . The fields then follow from Eq. (5.5.37). A pair of potentials of the form $(\nabla ψ, - \partial ψ / \partial t)$ is the gauge transformation Eq. (5.5.45) of the pair $(0, 0)$ .

$\nabla \cdot A = \partial A_{z} / \partial z = k A_{0} sin (ω t - k z)$ , and the gauge $\nabla \cdot A + μ_{0} ε_{0} \partial V / \partial t = 0$ gives

\frac{\partial V}{\partial t} = - c^{2} k A_{0} sin (ω t - k z) ⟹ V = \frac{c ^{2} k}{ω} A_{0} cos (ω t - k z) .

Answer

V = c A_{0} cos (ω t - k z)

Both potentials are functions of $ω t - k z = - k (z - c t)$ alone, so they have the d'Alembert form $f (z - c t)$ and satisfy $\nabla^{2} u - c^{- 2} \partial^{2} u / \partial t^{2} = 0$ . Directly: $\nabla^{2} A_{z} = - k^{2} A_{z}$ and $c^{- 2} \partial^{2} A_{z} / \partial t^{2} = - (ω^{2} / c^{2}) A_{z} = - k^{2} A_{z}$ , and the same for $V$ .

$A_{z}$ depends on $z$ and $t$ only, so $B = \nabla \times A = a_{x} \partial A_{z} / \partial y - a_{y} \partial A_{z} / \partial x = 0$ . The electric field has only a $z$ component:

E_{z} = - \frac{\partial V}{\partial z} - \frac{\partial A _{z}}{\partial t} = (- c k A_{0} + ω A_{0}) sin (ω t - k z) = 0.

Answer

E = 0, B = 0

Try $ψ = - (A_{0} / k) sin (ω t - k z)$ :

\nabla ψ - \frac{\partial ψ}{\partial t} = a_{z} \frac{\partial ψ}{\partial z} = A_{0} cos (ω t - k z) a_{z} = A, = \frac{ω}{k} A_{0} cos (ω t - k z) = V .

The pair is a pure gauge transformation of $(A, V) = (0, 0)$ . It looks like a wave, but no field exists. Note that $ψ$ itself satisfies the homogeneous wave equation, as the gauge function between two Lorenz pairs must.

Now $\nabla \cdot A = \partial A_{x} / \partial x = 0$ , so $V = 0$ already satisfies the Lorenz gauge, and

E B = - \frac{\partial A}{\partial t} = ω A_{0} sin (ω t - k z) a_{x}, = \nabla \times A = a_{y} \frac{\partial A _{x}}{\partial z} = k A_{0} sin (ω t - k z) a_{y} .

Answer

E B = ω A_{0} sin (ω t - k z) a_{x}, = k A_{0} sin (ω t - k z) a_{y}

Check

In (e) $E_{x} / B_{y} = ω / k = c$ , and $E \times B$ points along $+ a_{z}$ , the direction of travel. Faraday's law holds: $\nabla \times E = a_{y} \partial E_{x} / \partial z = - ω k A_{0} cos (ω t - k z) a_{y} = - \partial B / \partial t$ .

(a) Longitudinal \vec A, (a) to (d). — (a) Longitudinal $A$ , (a) to (d).

(b) Transverse \vec A, (e). — (a) Longitudinal $A$ , (a) to (d).

Remark

Only the part of $A$ across the direction of travel makes a field. In (e) the result is the uniform plane wave of Section 6.1, obtained here from one potential component. Exercise 5.5.8 shows that a longitudinal $A$ does produce a field when it also varies across the direction of travel, between two plates.