There is nothing which prevents the particles in a Feynman diagram from being extremely heavy, making them essentially classical particles. We could have 1 kg particles doing Coulomb scattering.
In formula (63), Matthew Schwartz (2012) presents the vacuum polarization effect on a spatial momentum transfer. The effective coupling constant becomes smaller when -p^2 is increased:
α_eff = 1 / 137 (1 + 0.00077 ln (-p^2 / m^2),
where m is a fixed energy, for example, m might be 511 keV.
Let us then imagine that we have very heavy particles with very large electric charges. They are the classical limit of Coulomb scattering. We can monitor the paths of the particles very accurately.
Suppose that we then double the mass and the electric charge of our particles. The formula above claims that the coupling constant changes, because the momentum transfer p is double. The change is not small, about 0.1 %.
That does not make sense, since we are working at the classical limit. The Coulomb force should have the same formula for all classical objects.
What is wrong?
Our new way of handling vacuum polarization restores the classical limit.
Recall that a virtual pair typically lives just for 0.1 Compton wavelengths. Here, we define the generalized Compton wavelength for a general 4-momentum as
λ = h c / |k|,
where k is a 4-momentum (E, p), and we define
|k| = sqrt(E^2 + p^2).
The norm || is the euclidean length of the 4-momentum vector. It is not the Minkowski metric length sqrt(-E^2 + p^2).
Let the heavy particles pass each other at a distance L. We add to the usual Feynman vacuum polarization integral an extra coefficient
C(|k|) = 0.1 λ / L = 0.1 h c / (|k| L),
if C(|k|) < 1. If it is bigger than 1, we do not add the coefficient.
That is, we only take into account those pairs which are born at a distance < 0.1 λ from the passing particle.
The coefficient C(|k|) makes the Feynman integral to converge. Or does it? We have to find online a brute force calculation of the Feynman integral with a cutoff |k| < Λ and check that it only diverges logarithmically on Λ.
L. Alvarez-Gaume and M. A. Vazquez-Mozo (2010) have done the calculation (though they did not include it in their paper).
Their formula (372) states:
Π(q^2) = e^2 / (12 π^2) log(q^2 / Λ^2)
+ finite terms.
There, the integral has been taken over all |k| < Λ. The integral diverges only logarithmically when we increase the (large) cutoff Λ.
Our corrective term C(|k|) makes the integral to converge, because it has 1 / |k| in it. But we are also interested in a numerical value.
Suppose that heavy particles pass each other at the distance of 1 meter and exchange momentum q. How does our corrective term C(|k|) affect the value of the integral?
The value of |q| is macroscopic. Its Compton wavelength is microscopic, of the order 10^-34 m. If |k| is of the order |q|, then our corrective term is roughly 10^-34.
In the SI unit system, the 4-momentum of the electron at rest is ~ 10^-21. Our corrective term for it would be ~ 10^-13.
The integrand in formula (365) of the paper for |k| << 10^-21 is roughly
1 / |q|^2,
which is integrated over a 4-volume 10^-84. The value is negligible.
We conclude that the integral is essentially zero when we correct the integrand with our term C(|k|).
The effect of vacuum polarization is essentially zero when macroscopic charges pass each other at a distance 1 meter.
We have restored the classical limit. Heavy particles behave as in the simplest Feynman diagram for Coulomb scattering. Vacuum polarization, which would depend on q, has essentially a zero effect.
Apparently, prior to this blog entry, no one has noticed the classical limit problem in Coulomb scattering vacuum polarization. Breaking the classical limit is a serious error in quantum mechanics.
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