--------------------------------- rubber plate
^ ||||/
| || hand
v
oscillation
We have coined the name "rubber plate model" to describe the elasticity of the electric field. It is easy to imagine in one's mind how an elastic rubber plate behaves when we hold it in our hand at one point and wave it. If we make the rubber plate to oscillate with our hand in a cyclic fashion, the effective mass of the plate is reduced because the far parts of the plate oscillate in a different phase from our hand.
Why QED is able to calculate the classical process correctly - after infinities are removed with regularization?
virtual photon q
~~~~~~~
/ \
e- ---------------------------------------
| virtual
| photon
| p
Z+ ---------------------------------------
In the above diagram the electron scatters from the nucleus Z+, p contains spatial momentum, and q is arbitrary 4-momentum.
The virtual photon q is due to the interaction of the electron with its own elastic electric field. The far field does not have time to react to the acceleration caused by the nucleus Z+. The net effect is that the effective mass of the electron is reduced in its collision with the nucleus.
Let p be fixed.
QED uses propagators to calculate the complex-valued weight of the path for each value of q. It then sums the weights. The sum is infinite, and infinities are cut off in some systematic way - regularized. Miraculously, the final result is correct and agrees with empirical measurements.
How is this procedure related to the elastic electric field of the electron?
QED summarizes the smooth approach and receding phases of the electron into a simple line in the diagram: the virtual photon p.
Before the fly-by, the electron gives a Dirac delta impulse to its own electric field in the first vertex. The propagation of that impulse is calculated with the photon propagator. The electron itself moves under its own propagator.
After the fly-by, the electron absorbs the components of the impulse.
How is this procedure related to the elastic interaction of the electron with its own field?
Why can the smooth interaction with the nucleus Z+ be summarized into a single photon line p?
Classical analysis of the fly-by: the non-relativistic case
y
^
| ● Z+ nucleus
|
| e- -------->
|
-------------------------------> x
Let the electron fly along the x-axis. Let the velocity of the electron be substantially less than the speed of light.
The nucleus first accelerates the electron along the x-axis, then decelerates it. Simultaneously, the nucleus pulls the electron up. The momentum upward becomes p.
The reduced mass of the electron allows the nucleus to accelerate the electron faster. We assume no radiation out. The momentum p is determined exactly by the integral of the upward force over time.
A lighter particle is deflected more. But how does that affect the momentum p? A lighter particle takes a route which comes closer to the nucleus Z+ but the particle moves faster.
Let us have a potential
V = C / r.
According to H. Friedrich (2013), the scattering angle is
Θ(b) = 2 arccos(1 / sqrt(γ^2 + 1)),
where b is the impact parameter,
γ = C / (2 E b),
and E is the particle kinetic energy at the start.
Let γ be small. Then
arccos(1 - γ^2 / 2) = γ.
If we reduce the mass m by 1%, E is 1% less and the deflection angle is 1% more. The momentum p is not affected by the reduced mass, if γ is small. The deflection angle is 2 γ.
Let the particle be very light. Its final upward momentum p cannot be more than m v, where v is its initial velocity. We conclude that if γ is not small, then reducing the mass of the non-relativistic particle reduces p.
If the particle does a head-on collision with the nucleus, then its final momentum q along the x-axis is reduced with a reduced mass because the particle moves faster and has less time to collect momentum.
The effect of mass reduction is to reduce the scattering when the electron flies very close to the proton. Thus, it reduces the cross section for large deflection angles.
The electron fly-by: the relativistic case
What happens if the speed of the electron is close to the light speed? Reduced mass allows it to come closer to the proton but does not affect the speed much. Let us calculate the magnitude on an electron which flies at a distance
r = λ_e / (2 π) = 4 * 10^-13 m
from the nucleus. Let the kinetic energy of the electron be m_e c^2.
The mass reduction is
~ r_e / r = 1/137.
Let us assume that the electric force only affects the electron in an equilateral triangle whose side is r. The acceleration:
a = k e^2 / (r^2 * 2 m_e)
= 10^27 m/s^2.
The distance which the electron has come closer to the proton when it leaves the equilateral triangle:
s = 1/2 a t^2
= 1/2 a r^2 / c^2
= 1/2 k e^2 / (2 m_e c^2)
= 7 * 10^-16 m.
During its journey in the triangle, the electron comes 1/550 closer which adds ~ 1/800 to the momentum.
The mass reduction would increase the acceleration a by 1/137. The effect on the momentum is small and positive, only
1/800 * 1/137 = 10^-5.
How much does reduced mass affect the speed of the electron?
The electron gains
1/2 * 1 / 137 m_e c^2
in kinetic energy while it is in the triangle.
The Lorentz factor is
γ = sqrt(1 / (1 - v^2 / c^2)).
The electron speed is 1/1400 larger in the triangle. Reducing the mass would increase the speed by a ratio
1/1400 * 1/137 = 0.5 * 10^-5.
The effect on the momentum is small and negative. For a relativistic electron, mass reduction seems to have little effect in Coulomb scattering.
Analysis of the Feynman diagram
In the Feynman diagram, we calculate the response to a Dirac delta impulse which the electron makes to its own field before meeting the nucleus.
The impulse contains waves of any 4-momentum q, weighted by the photon propagator.
We know that the Feynman formula should calculate the effect of mass reduction, where the reduction depends on p. Usually, the mass reduction is quite small.
The electron propagators become infinite when q = 0. The Feynman formula favors electrons which are almost on-shell.
The regularization probably subtracts the q = 0 case from a q != 0 case. That may make the integrals finite.
The electron throws part of its energy and momentum (q) away before the approach, and catches them back as it recedes. The reduction (or increase) in the electron package alters scattering probability amplitudes.
Why does it alter them in the way which gives the correct result after regularization?
We sum all packages. This sounds like a path integral for a lagrangian.
The classical path and lagrangian density for the process
package of energy E
and momentum q
~~~~~~~
/ \
momentum k e- --------------------------------->
|
| momentum p
|
● Z+
Let us analyze the problem in classical physics. The electron makes a very complex wave into its own field. In principle, we should take at least a billion volume elements of space and calculate the action for all the "paths" the field can take in those billion elements.
In the Feynman diagram, all this complex process is summarized into a simple process where the electron throws a package of energy E and momentum q to itself before meeting the nucleus.
The Feynman formula for the diagram probably calculates some kind of classical action for the simplified path.
A classical lagrangian is typically of the form
L = T - V,
where T is the kinetic energy and V is the potential energy.
● field
/
/ rubber
/ band
● e-
Let us imagine that the field of the electron is another particle which is attached to the electron via a rubber band. The electron exchanges spatial momentum and energy with the other particle.
It makes sense to describe the process as the electron sending a package of 4-momentum to the other particle and later taking that package back. It is a simplified way to describe the complex process.
What would a suitable lagrangian density be for the path? The first guess is, of course, that the Feynman integral formula calculates the action for the path. However, divergences in the Feynman formula mean that something is wrong with it. The regularized version of the formula probably is the correct way to calculate the classical process.
The Feynman formula includes arbitrary momentum q in the package. That does not look right. The momentum q should be
q = E / E_e * k,
where E_e is the total energy of the electron and k its initial momentum. Maybe wrong momenta cancel each other in the integral?
Momentum p determines which package size is favored. Other sizes should have a small weight. The mass-energy of the electric field is the origin of the packages. That part of the field which does not have time to react to the acceleration of p, is thrown as the package.
We will study this in detail in coming blog postings. There has to be a logical reason why a Feynman diagram calculates classical processes right. What is it?
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