Electromagnetism

Electromagnetic Field of a Uniformly Moving Charge

Derivation from Maxwell's equations using the Lorenz gauge and retarded potentials

Starting from Maxwell's equations in the Lorenz gauge, we derive the electromagnetic field of a uniformly moving point charge. The wave equations for the scalar and vector potentials are solved using Green's functions, leading naturally to the retarded potentials and the Heaviside expressions for the electric and magnetic fields.

Prerequisites

Vector calculus
Maxwell's equations
The Lorenz gauge
Green's functions
Wave equation

I. Introduction
II. Green Function Solution
1. Evaluation of the Delta Function
III. The Potentials
1. The Lorentz Gauge
IV. The Electromagnetic Fields
V. Properties of the Fields

I. Introduction

We derive the electromagnetic field generated by a point charge moving with constant velocity. Rather than starting from the known Lienard--Wiechert potentials, we derive the result directly from Maxwell's equations by introducing the scalar and vector potentials and solving the resulting wave equations. Throughout this section we use Gaussian units. We begin with Maxwell's equations in vacuum,

EQ:MAXWELL

\nabla \cdot E \nabla \cdot B \nabla \times E \nabla \times B = 4 π ρ, = 0, = - \frac{1}{c} \frac{\partial B}{\partial t}, = \frac{1}{c} (4 π J + \frac{\partial E}{\partial t}) .

These equations must be solved for the charge and current densities corresponding to a point charge moving with constant velocity

v

. Assuming that the charge passes through the origin at

t = 0

, the source terms are

Eq:Uniform_motion_sources

ρ (r, t) J (r, t) = q δ^{3} (r - v t), = q v δ^{3} (r - v t) .

A convenient way to solve Maxwell's equations is to introduce a scalar potential

ϕ

and a vector potential

A

defined by

E B = - \nabla ϕ - \frac{1}{c} \frac{\partial A}{\partial t}, = \nabla \times A .

At this stage

ϕ

and

A

are unknown functions that must be determined by solving Maxwell's equations. Their introduction is motivated by the fact that Maxwell's equations acquire a much simpler form when expressed in terms of the electromagnetic potentials. The second Maxwell equation is automatically satisfied since

\nabla \cdot B = \nabla \cdot (\nabla \times A) = 0.

The third Maxwell equation is also identically satisfied. Indeed,

\nabla \times E = \nabla \times (- \nabla ϕ - \frac{1}{c} \frac{\partial A}{\partial t}) = - \nabla \times (\nabla ϕ) - \frac{1}{c} \frac{\partial}{\partial t} (\nabla \times A) = - \frac{1}{c} \frac{\partial B}{\partial t},

since

\nabla \times \nabla ϕ = 0

. Hence, after introducing the scalar and vector potentials, the two homogeneous Maxwell equations are automatically satisfied. We now substitute the potentials into Ampere's law. Using the identity

\nabla \times (\nabla \times A) = \nabla (\nabla \cdot A) - \nabla^{2} A,

we obtain

\nabla (\nabla \cdot A) - \nabla^{2} A = \frac{1}{c} (4 π J + \frac{\partial E}{\partial t}) .

Since

E = - \nabla ϕ - \frac{1}{c} \frac{\partial A}{\partial t},

its time derivative is

\frac{\partial E}{\partial t} = - \nabla \frac{\partial ϕ}{\partial t} - \frac{1}{c} \frac{\partial ^{2} A}{\partial t ^{2}} .

Substituting this expression into the previous equation yields

\nabla (\nabla \cdot A) + \frac{1}{c} \nabla \frac{\partial ϕ}{\partial t} = \nabla^{2} A - \frac{1}{c ^{2}} \frac{\partial ^{2} A}{\partial t ^{2}} + \frac{4 π}{c} J .

Our goal is to obtain an uncoupled wave equation for the vector potential. This is achieved by requiring

A

to satisfy

Eq:Wave_vector_potential

\nabla^{2} A - \frac{1}{c ^{2}} \frac{\partial ^{2} A}{\partial t ^{2}} = - \frac{4 π}{c} J,

which implies that

\nabla (\nabla \cdot A + \frac{1}{c} \frac{\partial ϕ}{\partial t}) = 0 .

The quantity inside the parentheses is therefore independent of position. Since an arbitrary constant may be absorbed into the definition of the potentials, we choose it to vanish and obtain the Lorenz gauge,

Eq:Lorenz_gauge

\nabla \cdot A + \frac{1}{c} \frac{\partial ϕ}{\partial t} = 0.

The wave equation satisfied by the scalar potential follows from Gauss's law,

- \nabla^{2} ϕ - \frac{\partial}{\partial t} (\nabla \cdot A) = 4 π ρ .

Substituting the Lorenz gauge condition,

\nabla \cdot A = - \frac{1}{c} \frac{\partial ϕ}{\partial t},

gives

- \nabla^{2} ϕ + \frac{1}{c ^{2}} \frac{\partial ^{2} ϕ}{\partial t ^{2}} = 4 π ρ,

or equivalently,

Eq:Wave_scalar_potential

\nabla^{2} ϕ - \frac{1}{c ^{2}} \frac{\partial ^{2} ϕ}{\partial t ^{2}} = - 4 π ρ .

We have therefore transformed Maxwell's equations into two wave equations for the scalar and vector potentials,

Eq:Potential_wave_equations

⎩ ⎨ ⎧ \nabla^{2} A - \frac{1}{c ^{2}} \frac{\partial ^{2} A}{\partial t ^{2}} \nabla^{2} ϕ - \frac{1}{c ^{2}} \frac{\partial ^{2} ϕ}{\partial t ^{2}} = - \frac{4 π}{c} J, = - 4 π ρ .

supplemented by the Lorenz gauge condition lorenz gauge , which couples the scalar and vector potentials.

II. Green Function Solution

The two equations

\nabla^{2} ϕ - \frac{1}{c ^{2}} \frac{\partial ^{2} ϕ}{\partial t ^{2}} \nabla^{2} A - \frac{1}{c ^{2}} \frac{\partial ^{2} A}{\partial t ^{2}} = - 4 π ρ, = - \frac{4 π}{c} J

can be solved using the retarded Green function. In the present case, however, it is sufficient to solve only the equation for the scalar potential. Indeed, from uniform motion sources the current density is simply

J = ρ v,

where the velocity

v

is constant. Consequently, if

ϕ

is the solution of the scalar equation, the vector potential is immediately obtained as

A = \frac{v}{c} ϕ .

Hence we proceed to find the solution for the scalar potential from the retarded Green function

Eq:Scalar_potential_green

ϕ (r, t) = \int \frac{ρ ( r ^{'} , t - \frac{∣ r - r ^{'} ∣}{c} )}{∣ r - r ^{'} ∣} d^{3} r^{'} .

Here

r^{'}

is the integration variable. For the uniformly moving point charge,

ρ (r^{'}, t^{'}) = q δ^{3} (r^{'} - v t^{'}),

and therefore

Eq:Scalar_potential_delta

ϕ (r, t) = q \int \frac{δ ^{3} [ r ^{'} - v ( t - \frac{∣ r - r ^{'} ∣}{c} ) ]}{∣ r - r ^{'} ∣} d^{3} r^{'} .

The task is now to evaluate this integral. The difficulty is that the argument of the delta function contains the integration variable

r^{'}

not only explicitly, but also through the distance

∣ r - r^{'} ∣

. Therefore, in order to find the region of space where the argument of the

δ

vanishes, we need to solve the equation

Eq:Source_position_condition

r^{'} - v (t - \frac{∣ r - r ^{'} ∣}{c}) = 0 .

Given the particular form of this equation, it is useful to decompose the position vector into components parallel and perpendicular to the velocity. Let us define the unitary vector in the direction of the particle velocity,

n = \frac{v}{∣ v ∣},

and decompose the position vectors in component parallel and perpendicolar to

n

r = r_{∥} n + r_{⊥} n_{⊥}, ⟨ n_{⊥} ∣ n ⟩ = 0.

Similarly,

r^{'} = r_{∥}^{'} n + r_{⊥}^{'} n_{⊥} .

From source position condition we immediately conclude that

r^{'}

can only have a component parallel to the velocity. Therefore

r_{⊥}^{'} = 0 .

The vector equation source position condition then reduces to the scalar equation

Eq:Source_position_parallel

r_{∥}^{'} - v t - \frac{( r _{∥} - r _{∥}^{'} ) ^{2} + r _{⊥}^{2}}{c} = 0,

We define

s = r_{∥}^{'} - r_{∥}, μ = r_{∥} - v t .

Then equation source position parallel becomes

s + μ + v \frac{s ^{2} + r _{⊥}^{2}}{c} = 0,

or equivalently,

s + μ = - \frac{v}{c} s^{2} + r_{⊥}^{2} .

Since the square root is non-negative, every solution of this equation must satisfy

s + μ \leq 0.

Under this condition the equation may be squared, yielding

s^{2} + μ^{2} + 2 μ s = \frac{v ^{2}}{c ^{2}} (s^{2} + r_{⊥}^{2}) .

Thus

(1 - \frac{v ^{2}}{c ^{2}}) s^{2} + 2 μ s + μ^{2} - \frac{v ^{2}}{c ^{2}} r_{⊥}^{2} = 0.

This equation has the two algebraic solutions

Eq:S_algebraic_solutions

s = \frac{- μ \pm \frac{v}{c} ( 1 - \frac{v ^{2}}{c ^{2}} ) r _{⊥}^{2} + μ ^{2}}{1 - \frac{v ^{2}}{c ^{2}}} .

However, not both algebraic solutions are admissible. We now prove which of the two algebraic solutions is admissible. Let

α = (1 - \frac{v ^{2}}{c ^{2}}) r_{⊥}^{2} + μ^{2} .

Then the two roots can be written as

s = \frac{- μ \pm \frac{v}{c} α}{1 - \frac{v ^{2}}{c ^{2}}} .

Adding

μ

gives

s + μ = \frac{- μ \pm \frac{v}{c} α}{1 - \frac{v ^{2}}{c ^{2}}} + μ = \frac{\pm \frac{v}{c} α - μ \frac{v ^{2}}{c ^{2}}}{1 - \frac{v ^{2}}{c ^{2}}} .

Since

1 - v^{2} / c^{2} > 0

, the admissibility condition is equivalent to

\pm \frac{v}{c} α - μ \frac{v ^{2}}{c ^{2}} < 0.

Equivalently,

\pm α - μ \frac{v}{c} < 0.

For the root with the plus sign this would require

α < μ \frac{v}{c} .

μ < 0

, this is impossible because

α \geq 0

while

μv / c < 0

. If

μ > 0

, squaring would imply

α^{2} < μ^{2} \frac{v ^{2}}{c ^{2}} .

But

α^{2} = (1 - \frac{v ^{2}}{c ^{2}}) r_{⊥}^{2} + μ^{2},

so this would require

(1 - \frac{v ^{2}}{c ^{2}}) r_{⊥}^{2} < μ^{2} (\frac{v ^{2}}{c ^{2}} - 1),

which is impossible, because the left-hand side is non-negative while the right-hand side is negative. Therefore the root with the plus sign is not admissible. For the root with the minus sign, the condition becomes

- α - μ \frac{v}{c} < 0.

μ > 0

, this is immediately satisfied. If

μ < 0

, it is equivalent to

α > - μ \frac{v}{c},

which follows from

α^{2} = (1 - \frac{v ^{2}}{c ^{2}}) r_{⊥}^{2} + μ^{2} > μ^{2} \frac{v ^{2}}{c ^{2}} .

Thus the only admissible solution is the root with the minus sign,

Eq:S_admissible_solution

s = \frac{- μ - \frac{v}{c} ( 1 - \frac{v ^{2}}{c ^{2}} ) r _{⊥}^{2} + μ ^{2}}{1 - \frac{v ^{2}}{c ^{2}}} .

Introducing the standard notation

β = \frac{v}{c},

and expressing the result in terms of the original variables, we obtain

Eq:Source_position_parallel_solution

r_{∥}^{'} = \frac{( 1 - β ^{2} ) μ - β ( 1 - β ^{2} ) r _{⊥}^{2} + μ ^{2}}{1 - β ^{2}} .

1. Evaluation of the Delta Function

We have now determined the values of

r^{'}

for which the argument of the delta function in the Green function vanishes. At this point one might be tempted simply to remove the integral and evaluate the integrand at these values. This, however, is not correct, because the argument of the delta function is itself a function of the integration variable. To evaluate the integral we must therefore use the general identity for the delta function of a vector-valued function,

Eq:Delta_function_vector

δ^{3} (F (r^{'})) = \frac{δ ^{3} ( r ^{'} - r _{0}^{'} )}{∣ det D F ( r _{0}^{'} ) ∣},

where

r_{0}^{'}

is the solution of

F (r^{'}) = 0,

and

D F

denotes the Jacobian matrix of

F

. In the present case,

Eq:Green_function_argument

F (r^{'}) = r^{'} - v (t - \frac{∣ r - r ^{'} ∣}{c}) .

To compute the Jacobian we differentiate each component of

F

with respect to the coordinates of

r^{'}

. Since

\frac{\partial}{\partial r _{i}^{'}} ∣ r - r^{'} ∣ = - \frac{r _{i} - r _{i}^{'}}{∣ r - r ^{'} ∣},

we obtain

\frac{\partial F _{i}}{\partial r _{j}^{'}} = δ_{ij} - \frac{v _{i}}{c} \frac{r _{j} - r _{j}^{'}}{∣ r - r ^{'} ∣} .

Introducing the unit vector

n = \frac{r - r ^{'}}{∣ r - r ^{'} ∣},

the Jacobian matrix can be written as

D F = I - \frac{1}{c} v \otimes n .

Using the identity

det (I + a \otimes b) = 1 + b \cdot a,

it follows immediately that

det D F = 1 - \frac{v \cdot n}{c} .

The scalar potential therefore becomes

ϕ (r, t) = q \int \frac{δ ^{3} ( r ^{'} - r _{0}^{'} )}{( 1 - \frac{v \cdot n}{c} ) ∣ r - r ^{'} ∣} d^{3} r^{'},

where

r_{0}^{'}

is the solution of

F (r^{'}) = 0,

that is, the solution of source position condition . The integration over the delta function can now be carried out immediately, but the resulting expression still depends on the unknown source position

r_{0}^{'}

. Our next task is therefore to eliminate this dependence.

III. The Potentials

The integral can now be evaluated immediately, yielding

ϕ (r, t) = q \frac{1}{∣ r - r ^{'} ∣} \frac{1}{1 - \frac{v \cdot n}{c}},

where it is important to emphasize once again that

r^{'}

is the solution of source position condition , since this equation is precisely the condition under which the argument of the function

F

vanishes. We now eliminate the dependence on the source position

r^{'}

so that the scalar potential is expressed solely in terms of the field point

r

. To this end, we first compute the scalar product between the particle velocity and the unit vector

n

. Since the velocity is parallel to the particle trajectory, only the parallel component contributes, giving

v \cdot n = v \frac{r _{∥} - r _{∥}^{'}}{( r _{∥} - r _{∥}^{'} ) ^{2} + r _{⊥}^{2}} .

the scalar potential becomes

ϕ (r, t) = q \frac{c}{c ∣ r - r ^{'} ∣ - v ( r _{∥} - r _{∥}^{'} )} .

From the solution source position parallel solution , we obtain

r_{∥}^{'} - v (t - \frac{∣ r - r ^{'} ∣}{c}) = 0.

Multiplying through by

v

and rearranging gives

\frac{v ^{2}}{c} ∣ r - r^{'} ∣ . = v (v t - r_{∥}^{'})

Finally, using the condition source position parallel once more, defining

μ = r_{∥} - v t

and after some straightforward algebra, we arrive at the final expression for the scalar potential, which now depends only on the coordinates of the field point,

Eq:Scalar_potential_uniform_motion

ϕ (r, t) = \frac{q}{( 1 - β ^{2} ) r _{⊥}^{2} + μ ^{2}} .

and the vector potential is,

Eq:Vector_potential_uniform_motion

A = v ϕ = v \frac{q}{( 1 - β ^{2} ) r _{⊥}^{2} + μ ^{2}} .

1. The Lorentz Gauge

We have now the expressions for

ϕ

and

A

, but it rests to prove that these solutions fulfill the Lorentz gauge condition. Since

A = \frac{v}{c} ϕ,

and

v

is constant, we have

\nabla \cdot A = \frac{1}{c} v \cdot \nabla ϕ .

Therefore, the Lorenz gauge condition reduces to

Eq:Lorenz_gauge_scalar_potential_condition

v \cdot \nabla ϕ = - \frac{\partial ϕ}{\partial t} .

Since

ϕ (r, t) = q [(1 - β^{2}) r_{⊥}^{2} + μ^{2}]^{- 1/2},

its gradient is

\nabla ϕ = - q [(1 - β^{2}) r_{⊥}^{2} + μ^{2}]^{- 3/2} [(1 - β^{2}) r_{⊥} \nabla r_{⊥} + μ \nabla μ] .

From the decomposition

r = r_{∥} n + r_{⊥} n_{⊥},

it follows immediately that

r_{∥} = r \cdot n .

Taking the gradient of both sides gives

\nabla r_{∥} = n,

and therefore

v \cdot \nabla r_{∥} = v \cdot n = v .

Since

r_{⊥}

is the component of

r

perpendicular to

n

, we also have

v \cdot \nabla r_{⊥} = 0.

From the expression for the gradient,

\nabla ϕ = - q [(1 - β^{2}) r_{⊥}^{2} + μ^{2}]^{- 3/2} [(1 - β^{2}) r_{⊥} \nabla r_{⊥} + μ \nabla μ],

and using

v \cdot \nabla r_{⊥} = 0, v \cdot \nabla μ = v,

we obtain

v \cdot \nabla ϕ = - q [(1 - β^{2}) r_{⊥}^{2} + μ^{2}]^{- 3/2} μv .

On the other hand,

ϕ = q [(1 - β^{2}) r_{⊥}^{2} + μ^{2}]^{- 1/2},

and since

\frac{\partial μ}{\partial t} = - v,

it follows that

\frac{\partial ϕ}{\partial t} = q [(1 - β^{2}) r_{⊥}^{2} + μ^{2}]^{- 3/2} μv .

Hence,

v \cdot \nabla ϕ = - \frac{\partial ϕ}{\partial t},

which is precisely the condition

\nabla \cdot A + \frac{1}{c} \frac{\partial ϕ}{\partial t} = 0.

IV. The Electromagnetic Fields

Having obtained both potentials, we can now derive the electric and magnetic fields. The electric field is obtained from

E = - \nabla ϕ - \frac{1}{c} \frac{\partial A}{\partial t} .

Substituting the expressions for

ϕ

and

A

, and after some long algebra, we obtain

Eq:Electric_field_uniform_motion

E (r, t) = q (1 - β^{2}) \frac{r - v t n}{[ ( 1 - β ^{2} ) r _{⊥}^{2} + μ ^{2} ] ^{3/2}} .

The magnetic field is

B = \nabla \times A

From

A = \frac{v}{c} ϕ,

Since

v

is constant,

\nabla \times A = e_{x} \partial_{x} v_{x} ϕ e_{y} \partial_{y} v_{y} ϕ e_{z} \partial_{z} v_{z} ϕ = \frac{1}{c} \nabla ϕ \times v .

B = \frac{1}{c} \nabla ϕ \times v .

On the hother hand, from the electric field equation above,

E \times v = - \nabla ϕ \times v - \frac{1}{c} \frac{\partial A}{\partial t} \times v = - c B

A

has the same direction of

v

. Hence the final formula for the magnetic field is

Eq:Magnetic_field_uniform_motion

B (r, t) = \frac{1}{c} v \times E

The electromagnetic field of a point charge moving with constant velocity is therefore

Eq:Electromagnetics_fields

E (r, t) B (r, t) = \frac{q ( 1 - β ^{2} ) ( r - v t )}{[ ( r _{∥} - v t ) ^{2} + ( 1 - β ^{2} ) r _{⊥}^{2} ] ^{3/2}}, = \frac{1}{c} v \times E (r, t),

V. Properties of the Fields

We devote this section to discussing a few important properties of the electric and magnetic fields of a uniformly moving charge. From electromagnetics fields it follows immediately that the electric and magnetic fields are orthogonal at every point in space and time:

E (r, t) \cdot B (r, t) = 0

The Poynting vector is given by

S = E \times B = \frac{1}{c} E \times (v \times E) = \frac{1}{c} [v E^{2} - E (E \cdot v)] .

To simplify the algebra, we introduce the notation

E = α (r - v t),

where

α = \frac{q ( 1 - β ^{2} )}{[ ( r _{∥} - v t ) ^{2} + ( 1 - β ^{2} ) r _{⊥}^{2} ] ^{3/2}},

Substituting this expression into the definition of the Poynting vector yields

S = \frac{α ^{2}}{c} [v (r - v t)^{2} - (r - v t) (v \cdot r - v^{2} t)] .

Using the identity

(r - v t)^{2} = (r_{∥} - v t)^{2} + r_{⊥}^{2} .

which leads to the final expression for the Poynting vector,

Eq:Poynting_vector

S = \frac{α ^{2} v r _{⊥}}{c} [r_{⊥} n_{∥} - (r_{∥} - v t) n_{⊥}] .

Equation Poynting vector shows that the energy flux is symmetric about the direction of the particle velocity, since the direction of the unit vector

n_{⊥}

is arbitrary within the plane orthogonal to the velocity.

Contents

I. Introduction

II. Green Function Solution

1. Evaluation of the Delta Function

III. The Potentials

1. The Lorentz Gauge

IV. The Electromagnetic Fields

V. Properties of the Fields