Re: On Pair Annihilation and Electromagnetic Field

Anamitra Palit wrote:
Let us consider the annihilation of an electron and a positron into a
pair of photons in relation to Feynman's Diagerams. Carerful
consideration of the energy-momentum conservation is taken into
account by a four-dimensional delta function. We consider each
particle in the field of the four current created by other(A-mu). The
energy of interaction is taken into careful account amongst othert
things. But what about the field energy of the currents created by the
incident particles themselves,I mean, the electric and magnetic fields
surrounding the currents(and extending to a very large distance)What
happens to all this energy when the particles (and hence the source
currents)get annihilated to photons?


You are attempting to discuss an inherently quantum phenomenon
(annihilation), using classical language and concepts (e.g. electric and
magnetic fields, field energy, etc.). Igor gave a response using quantum
concepts; permit me to discuss it using the classical concepts of your
question above.

Let me assume two colliding charged particles with identical masses,
equal but opposite charges, and let me use their center-of-mass frame.

Classically, when a positive and negative charge move toward each other,
there are nonzero E and B fields throughout space which are highly time
dependent (look up Lienard-Wiechert potentials). When they collide and
stick together with zero separation [%], there could be radiation from
the deceleration of their stopping; it's also clear that ultimately the
fields go to zero in a region that expands outward at the speed of light
from their location. If this is modeled as the two charge tracks
intersecting and ending, so that their stopping and sticking together is
instantaneous [%], then there is no actual deceleration and therefore no
actual radiation; at a location a distance L away, the fields smoothly
approach zero at time L/c after their coalescing, and remain zero
thereafter [&].

[%] A classical analog of charge annihilation.

[&] The fields go smoothly to zero and remain zero because
the dipole moment is proportional to their separation,
which goes smoothly to zero and remains zero. All other
moments are zero for point charges. Remember the L-W
potentials use the retarded positions of the charges.


Is it reasonable to assume that field energy moves
out in the form of radiation when annihilation takes place?

No. See above. The CANCELLATION expands outward; there is no radiation.
Classically.


How do we
take care of this particular type of energy changes in high-energy
physics experiments for instance in the colliders?

Particle experiments in some colliders, such as the Tevatron and LEP,
have annihilations occurring when the beams collide [#]. None of these
experiments have detectors sensitive to the E or B fields from the
beams, so the question you asked never comes up [@]. Moreover, for a
given beam crossing only a very small fraction of the beams' charge gets
annihilated, so even if they were measured [*] they would not change
very much (perhaps a few parts in 10^12 per crossing).

[#] The Tevatron collides protons and anti-protons; LEP
collided e+ and e- (it's now dismantled). The LHC collides
protons with protons so no annihilations occur.

[@] These experiments have detectors that measure the
secondary particles that emerge from the interaction
region when an inelastic interaction occurs. They do this
by measuring the electromagnetic interactions of the
particles with the detectors -- not their E and B fields
but rather ionization energy loss. There are no such
detectors located in the beams (they would be destroyed
in less than one beam pulse).

[*] The beams' charges are not measured by the particle
experiments; they are measured directly by accelerator
systems such as beam position monitors and current toroids,
and indirectly by luminosity monitors.


Tom Roberts
.

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