LienardWiechert 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 LienardWiechert 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(rR,tR/c) R sin(\theta) dR d\theta
d\Phi
= (1/(4 \pi)) \int g(rR,tR/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 LienardWiechert
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(rR,tR/c)/R dV,
\phi(r,t) = 1/(4 \pi e_0) \int \rho(rR,tR/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 rR at time
tR/c, take an element of volume dV = dS dR, where dS is an element of
area on the sphere of radius R centred on r. As the sphere shrinks
through the distance of dR in time dR/c, then the contribution to the
charge is given by
\rho(rR,tR/c) dS dR  \rho(rR,tR/c) u.R dS dt
= \rho(rR,tR/c) (1  u.R/(R c)) dS dR
= \rho(rR,tR/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/(Ru.R/c) dq,
\phi(r,t) = 1/(4 \pi e_0) \int 1/(Ru.R/c) dq,
where R is such that the element dq of charge is at rR at time tR/c,
and u is the velocity of the element when it is at rR at time tR/c.
In the case of a point charge, this reduces to
A(r,t) = m_0/(4 \pi) q u/(Ru.R/c),
\phi(r,t) = 1/(4 \pi e_0) q/(Ru.R/c),
where R is such that the charge is at rR at time tR/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 rR at time tR/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 LienardWiechert 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 (1u^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 rR at time tR/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 rR at time tR/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 rR
at time tR/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 longdistance 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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