# Lienard-Wiechert potential

*From*: D.McAnally@i'm_a_gnu.uq.net.au (David McAnally)*Date*: Wed, 14 Dec 2005 14:43:04 +0000 (UTC)

There was some confusion a while back about the part that the Lienard-

Wiechert potential of an accelerating charge plays, so I thought that I

would clarify.

Firstly, it should be pointed out that gauge transformations have

absolutely no effect on the physics, so any choice of gauge has no

effect on the physics. Some people seem to have missed that fact.

The Lienard-Wiechert potential is an electromagnetic potential which

satisfies the Lorentz condition (that div A + (1/c^2) @\phi/@t = 0) -

since the physics is unchanged under gauge transformations, then we are

free to take any gauge condition that we like.

Maxwell's Equations are inhomogeneous linear equations, and in vacuo, the

constitutive equations are linear. Maxwell's Equations are:

div B = 0,

curl E = - @B/@t,

div D = \rho,

curl H = J + @D/@t,

where \rho is the charge density and J is the current density. Since we

are working in a Minkowskian spacetime, which is contractible to a point

in the rigorous mathematical definition of that term (see below), then

there exist a vector potential A and a scalar potential \phi such that

B = curl A,

E = - grad \phi - @A/@t.

By the statement that Minkowski spacetime (denoted by M) is contractible

to a point, I mean that there exists a continuous function f mapping

Mx[0,1] to M, and an event x_0 in M, such that

f(x,0) = x_0 for all x in M,

f(x,1) = x for all x in M.

The very fact that such a continuous function exists is enough to

guarantee the existence of the required potentials A and \phi.

In vacuo, the constitutive equations are D = e_0 E and B = m_0 H, where

e_0 is the permittivity of free space and m_0 is the permeability of free

space. It follows that

div E = \rho/e_0,

curl B = m_0 J + (1/c^2) @E/@t,

where c^2 = 1/(m_0 e_0), so that c is a speed, equal to 299 792 458 m/s.

For a given vector potential A and scalar potential \phi which correspond

to the fields E and B, it is possible to solve the equation

del^2 \psi - (1/c^2) @^2 \psi/@t^2 = div A + (1/c^2) @\phi/@t

for \psi, e.g.

\psi(r,t) = (1/(4 \pi)) \int g(r-R,t-|R|/c) |R| sin(\theta) d|R| d\theta

d\Phi

= (1/(4 \pi)) \int g(r-R,t-|R|/c)/|R| dV,

where the integral with respect to |R| is taken from 0 to infinity, the

integral with respect to \theta is taken from 0 to \pi, and the integral

with respect to \Phi is taken from 0 to 2\pi, and

g = div A + (1/c^2) @\phi/@t.

When we take the gauge transformation,

A' = A - grad \psi, \phi' = \phi + @\psi/@t,

then we find that

div A' + (1/c^2) @\phi'/@t = 0.

Since (A',\phi') is related to (A,\phi) by a gauge transformation, then

the physics is unchanged, i.e. A' and \phi' are vector and scalar

potentials for the fields E and B. It follows that for any fields E and

B, there exist a vector potential A and a scalar potential \phi such that

the Lorentz condition, div A + (1/c^2) @\phi/@t = 0, holds.

In vacuo, since div E = \rho/e_0, then

- del^2 \phi - div @A/@t = \rho/e_0,

and so, by the Lorentz condition,

- del^2 \phi + (1/c^2) @^2 \phi/@t^2 = \rho/e_0.

Since curl B = m_0 J + (1/c^2) @E/@t, then

curl curl A + (1/c^2) grad @\phi/@t + (1/c^2) @^2 A/@t^2 = m_0 J.

Since curl curl A = grad div A - del^2 A, then

grad div A - del^2 A + (1/c^2) grad @\phi/@t + (1/c^2) @^2 A/@t^2 = m_0 J,

and so, by the Lorentz condition,

- del^2 A + (1/c^2) @^2 A/@t^2 = m_0 J.

These equations are linear in A and \phi, as is the Lorentz condition.

This means that the most general solution to these equations is equal to

one particular solution plus the general solution of the homogeneous

equations.

The solutions of the homogeneous equations for A and \phi form a vector

space over the real numbers. The solutions of the inhomogeneous equation

are of the form of a specified solution of the inhomogeneous equations

plus an arbitrary solution of the homogeneous equations. The exact

solution of the homogeneous equation that you add to the specific solution

of the inhomogeneous equation is determined by the boundary conditions and

initial conditions

There are nonzero solutions of the homogeneous equations, for example,

A = a cos(k.r - \omega t), \phi = 0,

where \omega = c |k|, and where a is a constant vector such that k.a = 0.

The corresponding EM field is

E = |k| c a sin(k.r - |k| c t),

B = - [k,a] sin(k.r - |k| c t).

This solution is radiative. In fact, if k and \omega are nonzero, then

this is the most general form needed for the term of the type

A = b cos(k.r - \omega t),

\phi = f cos(k.r - \omega t).

If

A = b cos(k.r - \omega t),

\phi = f cos(k.r - \omega t),

for a constant vector b and a constant scalar f, then the Lorentz

condition requires that k.b = f \omega/c^2. When we take

A' = A - grad \psi, \phi' = \phi + @\psi/@t,

where \psi = f/\omega sin(k.r - \omega t), then the map from (A,\phi) to

(A',\phi') is a gauge transformation, so that (A',\phi') and (A,\phi)

correspond to the same EM field, and

A' = (b - f k/\omega) cos (k.r - \omega t), \phi' = 0,

and k.(b - f k/\omega) = k.b - f |k|^2/\omega = k.b - f \omega/c^2 = 0. It

follows that

A = a cos(k.r - \omega t), \phi = 0,

where \omega = c |k|, and k.a = 0, is the most general potential needed of

the form,

A = b cos(k.r - \omega t),

\phi = f cos(k.r - \omega t).

Similarly,

A = a sin{k.r - \omega t), \phi = 0,

where \omega = c |k| and k.a = 0, is the most general potential needed of

the form,

A = b sin(k.r - \omega t),

\phi = f sin(k.r - \omega t).

Since these are solutions of the homogeneous equation, then they are

sourceless solutions, and do not enter into the Lienard-Wiechert

potential, which corresponds to the field generated by the charge. The

solutions discussed above are radiation solutions.

We are specifically interested in the EM field generated by the sources.

An additional EM field which is a solution of the homogeneous equations

could then be added on afterwards.

The EM potential which is generated by the sources is given by

A(r,t) = m_0/(4 \pi) \int J(r-R,t-|R|/c)/|R| dV,

\phi(r,t) = 1/(4 \pi e_0) \int \rho(r-R,t-|R|/c)/|R| dV.

This potential obeys the Lorentz condition.

Instead of integrating with respect to volume, we could integrate with

respect to charge. For an element of charge dq, which is at r-R at time

t-|R|/c, take an element of volume dV = dS d|R|, where dS is an element of

area on the sphere of radius R centred on r. As the sphere shrinks

through the distance of d|R| in time d|R|/c, then the contribution to the

charge is given by

\rho(r-R,t-|R|/c) dS d|R| - \rho(r-R,t-|R|/c) u.R dS dt

= \rho(r-R,t-|R|/c) (1 - u.R/(|R| c)) dS d|R|

= \rho(r-R,t-|R|/c) (1 - u.R/(|R| c)) dV,

where u is the velocity of the charge dq. It follows that

A(r,t) = m_0/(4 \pi) \int u/(|R|-u.R/c) dq,

\phi(r,t) = 1/(4 \pi e_0) \int 1/(|R|-u.R/c) dq,

where R is such that the element dq of charge is at r-R at time t-|R|/c,

and u is the velocity of the element when it is at r-R at time t-|R|/c.

In the case of a point charge, this reduces to

A(r,t) = m_0/(4 \pi) q u/(|R|-u.R/c),

\phi(r,t) = 1/(4 \pi e_0) q/(|R|-u.R/c),

where R is such that the charge is at r-R at time t-|R|/c (since the

charge is travelling at less than the speed of light, then there is at

most one value of R which satisfies this condition). If there is no such

value of R, then A(r,t) = 0 and \phi(r,t) = 0.

Note that u is taken for the charge at r-R at time t-|R|/c.

In the case where the charge is stationary at 0 (the origin of the spatial

coordinate system), R = r and u = 0, so that

A(r,t) = 0,

\phi(r,t) = q/(4 \pi e_0 |r|),

which is indeed the electromagnetic potential associated with Coulomb

force law.

Panofsky and Phillips include a derivation of the EM field corresponding

to the Lienard-Wiechert potential, which is therefore the EM field

generated by the charge. The EM field is given by

E = q/(4 \pi e_0) {(1/s^3) S (1-u^2/c^2) + (1/(c^2 s^3)) [R,[S,du/dt]]},

B = [R,E]/(|R| c),

where R and u are as before, S = R - |R| u/c, s = |R| - u.R/c, du/dt is

the acceleration of the charge when it is at r-R at time t-|R|/c, and for

vectors v and w, [v,w] denotes the cross product of v and w. The first

term within the braces in the expression for E yields the electric field

induced by a uniformly moving charge (and the corresponding term in B in

the full expansion yields the magnetic field induced by a uniformly moving

charge). The first term in the braces makes no contribution to the

radiation. The second term, which depends on the acceleration,

contributes to the EM radiation.

Note that the first term in the braces in the expression for E is directed

along the line joining where the charge would be at time t if it continued

moving at uniform velocity u from r-R at time t-|R|/c. Specifically, the

electric field at (r,t) generated by a uniformly moving charge is always

directed along the line joining the position of the charge at time t with

r.

If (r,t) is such that no value of R exists for which the charge is at r-R

at time t-|R|/c, then the contribution to the EM field from the charge is

E = 0 and B = 0.

In the case of a stationary charge at 0, u = 0, du/dt = 0, R = r, S = r,

and s = |r|, so that

E = q r/(4 \pi e_0 |r|^3),

B = 0,

which is the Coulomb electric field associated with a stationary charge,

and a zero magnetic field (which is the magnetic field of a stationary

charge).

In the case of an instantaneously stationary charge q accelerating with

acceleration a, integration of the long-distance Poynting vector over a

sphere demonstrates that the charge is losing energy to radiation at the

instantaneous rate of q^2 a^2/(6 \pi e_0 c^3) = m_0 q^2 a^2/(6 \pi c),

i.e. the instantaneous power being supplied by the charge to the radiation

is q^2 a^2/(6 \pi e_0 c^3).

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**Follow-Ups**:**Re: Lienard-Wiechert potential***From:*Ken S. Tucker

**Re: Lienard-Wiechert potential***From:*JanPB

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