Natural Science Forum / Physics / Research / March 2008

Spin is a quantum phenomenon and is not accounted for by the classical
theory of the electron. Thus the conclusion given by Jackson is correct,
though up to quantum corrections which should depend on spin.

Why should you make an assumption? There are models of an electron
coupled to radiation that can give an answer directly (QED a fortiori,
but perhaps also some simplified models). A stationary state, by
definition, implies that there will be no emission of radiation (which
is a dynamical process).

Is your question perhaps whether minimization of U garantees the
identification of the ground state (a state particularly unlikely to
radiate)? Quantum mechanics dictates that the ground state is identified
by minimizing the total energy (the Hamiltonian, to be precise), which
includes *both* the potential and kinetic energies. And yes, this result
has been extensively verified experimentally.

Hope this helps.

Igor

Well, magnetic field doesn't move things I think. Its a change in
orientation. However, I'd ask if electrons radiate while falling down
to earth.

Actually, it's not reasonable. One takes the potential for a point
source to be that obtained by transforming the static potential to a
moving frame. However, an important proviso is that the source point
is to be that where the source WAS located at that time in the past
where its image would have come from.

Thus, if the position the field is taken at is given by the vector r,
and the source lies on a trajectory given by r(t) (also denoted by the
letter r), then the separation from the field point r and the source
point r(t) is given by
  R = r - r(T)
where the time T is t, itself, with the light equivalent of the
distance subtracted out:
 T = t - |R|/c.

All of what I'm about to describe also has a classical analogue and
there, too, there will be radiation terms involving the acceleration.
But the expressions are quite a bit different and I won't bother with
them here.

The potential is given in the language of differential forms as
  A = (A_x dx + A_y dy + A_z dz - phi dt).
I'll use the vectors
  dr = (dx, dy, dz); dS = (dy dz, dz dx, dx dy).
So, the field is given by the 2-form
  F = E_x dx dt + E_y dy dt + E_z dz dt + B_x dy dz + B_y dz dx + B_z
dx dy
  = (E.dr) dt + B.dS.

The usual convention with forms is that dx dy = -dy dx, dx dx = -dx dx
= 0, etc. Thus, one can write
  dS = 1/2 dr X dr
where X denote the vector product.

Thus, one can form products
  (a.dr) (b.dr) = (aXb).dS = -(b.dr) (a.dr)
  (a.dr) (b.dS) = (a.b) dV = (b.dS) (a.dr)
where dV = dx dy dz.

Using the relation F = dA, the field can be calculated in a direct
fashion. Mostly, this comes down to the somewhat routine matter of
expressing the differentials of R, |R| and T in terms of dr and dt.
Indeed, one can write
  dT = dt - 1/c d|R|
  dR = dr - v dT
  d|R| = R.dR/|R|
where ().() denotes the dot product. The velocity is that taken at
time T; v = r'(T).

Solving for dT, dR and d|R|, one obtains
  dT = (|R|cdt - R.dr)/(|R|c - R.v)
  dR = (Rc (dr - v dt) + R X (v X dr))/(|R|c - R.v)
  d|R| = Rc.(dr - v dt)/(|R|c - R.v).

The potential 1-form fitting the description provided above is given
by
  A = (mu_0/4pi) ec (v.dr - c^2 dt)/(|R|c - R.v).

Immediately, you can see that differentiating it will yield something
involving the acceleration. In particular,
  d(v.dr - c^2 dt) = v'(T). dT dr - d(dt) = -a(T).dr dT,
making use of the property d(dt) = 0.

Substituting in dT, one gets
  d(v.dr - c^2 dt) = a.dr (R.dr - |R|c dt)/(|R|c - R.v)
  = ((a X R).dS - |R|ac dr dt)/(|R|c - R.v).
  = a.(R X dS - |R|c dr dt)/(|R|c - R.v).

Working out the algebrain a similar way, one gets
  d(|R|c - R.v) = c d|R| - v.dR - R.a dT.
After substituting for d|R|, dR and dT, and doing a little vector
algebra, ultimately one gets
  d(|R|c - R.v)
  = (c^2 - v^2 + R.a) (R.dr - |R|c dt)/(|R|c - R.v) - (v.dr - c^2
dt).

So, here, too, there is an acceleration involved.

Multiplying on the right by (v.dr - c^2 dt) the last term cancels out.
The product with the first term is obtained via
  (R.dr - |R|cdt)(v.dr - c^2 dt) = R X v.dS  + c(|R|v - Rc).drdt
   = (|R|v - Rc).(Rc dr dt - R X dS).

Combining these results, and using a little more vector algebra, one
gets for the field strength
   F = (mu_0 c e/4 pi |R|) (K/(|R|c - R.v)^3).(|R|c dr dt - R X dS)
where
  K = R X ((Rc - |R|v) X a) + (c^2 - v^2) (Rc - |R|v).

The electric field is extracted out of this,
  E = (mu_0 c^2 e/(4 pi)) K/(|R|c - R.v)^3
  B = R X E/|R|c.

So, there is definitely a dependence on the acceleration. You can
extract from this the (1/r) part and that is the radiation field.

Even in the non-relativistic form of electrodynamics, there will be a
radiation term, as mentioned before.