This morning I started a long and perhaps pointless
calculation to figure out the radiation losses when two electrons collide. I
took as an example the case of two electrons colliding head on with an initial
velocity of 10^6 m/sec each, and rebounding exactly the way they came. I drew a
graph of the collision, and got values for the acceleration and time: namely,
an acceleration of 4x10^21 m/sec^2, and a collision time of 5 x 10^(-16) sec.
Putting these into the Larmor Formula (you can check the numbers yourself) I
get a total power of 4.5 x 10^(-28) joules, of about one part in 10^7 of the
initial energy of each electron.

This isn’t all that surprising after all. The electron
energy is typical for atomic electrons, and the sharpness of spectral linewidths
is typically on the order of 10^7. The proximities, the forces, and the charges
are all the same as we deal with in spectroscopy, except that we have only
one-half of one cycle of oscillation. So the energy radiated is about what it
should be.

But remember, this does not give us the correct answer for
the radiative losses for the overall collision. That’s because the
accelerations of the two electrons are almost mirror images of each other, so
the radiation pattern of one is almost entirely cancelled out by the other. I want
to deal with the question of: just how much is the residue?

A lot of first-year physics students notice something funny
about the formulas for acoustic power: if you have two sources one on top of
the other, the amplitude of the wave doubles, so you get four times as much
power. The other side of the coin is that for oppositie amplitudes, you get
zero power. If this doesn’t bother you when you first encounter it, well…it
ought to.

The paradox starts to make sense when you look at the effect
of proximity of the two sources. If they are close enough to be coherent, you
do indeed get these interference effects. But once the separation of the
sources becomes on the order of one wavelength, the effect disappears. Oh, you
can still have constructive interference…but only in selected directions. In
other directions, you get destructive interference; and mostly, you get random
degrees of interference, positive and negative. It averages out so the power of
the two sources adds up to just the sum of the independent sources.

The case we are dealing with here is two sources of opposite phase which are very close to each other. It’s pretty hard to do the exact
calculation, but it’s not so hard to get a pretty good idea of how the
interference works. If we take the nominal power to be just the sum of the
independent powers of the two sources, the actual power has to equal the
nominal power for anything over one wavelength of separation, and it has to go
to zero as the separation narrows below one wavelength. We can draw a graph of
this:

It’s not too hard to believe that for short separations,
that is, the quadrupole approximation, the power is proportional to the square
of the separation. What does that mean for our present example? Well, we had a
half-cycle time of 500 micro-picoseconds, which corresponds to a wavelength of
300 nano-meters; with an “antenna length” of close to 300 picometers, this
gives us a factor of 10^6 when we square the ratio. So the quadrupole combination
radiates a million times more weakly than the independent electrons. Since we
calculated a “naïve” power loss of 10^(-7), the realistic power loss, taking
into account the quadrupole effect, is only one part in 10^13.

Do the losses ever become appreciable? We are dealing in the
present case with speeds on the order of 1% of the speed of light. The
dimensions of the radiating source are of a similar ratio as compared to the
wavelengths of the emitted radiation. The radiated power actually increases
rapidly with energy, becomes significant, not surprisingly, when the speeds
become relativistic and the geometry of the radiating system becomes a large
fraction of the wavelength.

At least that’s how it works for the classical case of
charged billiard balls. The sad fact is, with all the calculations, I still don’t
know what to make of the quantum mechanical case.

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