*2.1. S-Matrix Evaluation of the Charged Fermionic Self-Energy within QED*

It is instructive for our purposes to first review in this subsection the standard treatment of the charged fermionic self-energy within QED, ignoring the ECSK gravity. To this end, recall that the canonical evaluation of the electron self-energy was performed by Weisskopf in 1939 [20], which revealed that in general the self-energy of an electron is logarithmically divergent. An S-matrix evaluation for the electromagnetic mass, *δm*, confirms Weisskopf's result [21–23]. In the units of *h*¯ = *c* = 1 and *α* = *e* <sup>2</sup>/4*π*, together with a high-energy cutoff Λ, this electromagnetic mass can be expressed as

$$
\delta m = \frac{3\alpha m}{2\pi} \ln \left(\frac{\Lambda}{m}\right) \text{ (for } \Lambda >> m),
\tag{8}
$$

where "ln" stands for the natural logarithm. Let us investigate this result using simple numerical analysis. What happens, for example, if we use Planck mass for the cutoff? This should be a reasonable assumption since the Planck scale can be considered a limiting scale in the same fashion as *h*¯ and *c* are considered. Then, for an electron we have

$$
\delta m\_{\varepsilon} = \frac{3\alpha}{2\pi} \ln \left(\frac{m\_P}{m\_{\varepsilon}}\right) \approx 0.1795 \, m\_{\varepsilon}. \tag{9}
$$

This suggests that the electromagnetic mass-energy of an electron in this case would be about 18 percent of the total rest mass-energy. That does not seem unreasonable, but there is no way to confirm that it is correct. Moreover, according to some views, if the electromagnetic mass is truly infinite then there should be a compensating negative infinite mechanical mass that produces the observed positive rest mass. On the other hand, Weinberg in Reference [22] gives a renormalised expression for the "complete self-energy function" as

$$\begin{split} \boldsymbol{\Sigma}\_{ordr}^{\*} \boldsymbol{\varepsilon}^{2}(\boldsymbol{p}) &= \boldsymbol{\Sigma}\_{1loop}^{\*} (\boldsymbol{p}) - (\boldsymbol{Z}\_{2} - 1) (i \boldsymbol{p} + m\_{\boldsymbol{x}}) + \boldsymbol{Z}\_{2} \, \delta m\_{\boldsymbol{x}} \\ &= \frac{-2\pi^{2}e^{2}}{(2\pi)^{4}} \int\_{0}^{1} d\mathbf{x} \left\{ [i(1-\mathbf{x})\,\boldsymbol{p} + 2m\_{\boldsymbol{x}}] \ln \left( \frac{m\_{\boldsymbol{x}}^{2}(1-\mathbf{x})}{p^{2}\mathbf{x}(1-\mathbf{x}) + m\_{\boldsymbol{x}}^{2}\mathbf{x}} \right) \right. \\ & \left. - m\_{\boldsymbol{x}}[1+\mathbf{x}] \ln \left( \frac{1-\mathbf{x}}{\mathbf{x}^{2}} \right) \\ & \quad - (i\boldsymbol{p} + m\_{\boldsymbol{x}}) \left[ (1-\mathbf{x})\ln \left( \frac{1-\mathbf{x}}{\mathbf{x}^{2}} \right) - \frac{2(1-\mathbf{x}^{2})}{\mathbf{x}} \right] \right\}. \end{split} \tag{10}$$

The middle term in this integrand represents the renormalised electromagnetic mass so that for an electron it gives

$$\delta m\_{\varepsilon \text{error}} = \frac{-2\pi^2 e^2}{(2\pi)^4} \int\_0^1 dx \left\{-m\_\mathbf{x}[1+\mathbf{x}] \ln\left(\frac{1-\mathbf{x}}{\mathbf{x}^2}\right)\right\} = \frac{3\alpha}{8\pi} \approx 0.0008711 \, m\_\varepsilon. \tag{11}$$

In this case the electromagnetic mass is considerably less than 1 percent of the total rest mass. That indicates that the cutoff should be very close to the rest mass of the electron. Thus we seem to have two conflicting results. The conflict can be resolved by realizing that the mass in the equations leading to these results was plugged in by hand in an ad hoc manner. Moreover, from a quantum field theory and Feynman diagram perspective, it is easy to see that the mass comes in via the fermion propagator. Therefore by definition it is off-mass-shell, since the propagator represents an internal virtual line. It can thus have any value. In other words, it is really a variable. For the time being we will treat it as a variable and use the relation *m* = 1/*r*. Now that the apparent conflict is resolved, in what follows we focus on the renormalisation process.

It is easy to see that, in the rest frame of the particle in question with *i*/*p* = −*m*, the renormalised complete self-energy function in Equation (10) is identically zero. That may seem odd, since one would expect that in the rest frame the self-energy would be equal to the rest mass-energy. It really just means that the propagator is zero, as it should be for the rest frame. Thanks to the Hehl–Datta equation, we have an additional term in the Lagrangian to consider in the renormalisation process. In our view, this additional term represents the mechanical (bare) mass, since it is negative with respects to the observed mass. Thus, in the renormalised self-energy function in Equation (10), the counter terms are superfluous since the torsion term from the Hehl–Data equation is the counter term! Moreover, we will see later that the rest mass is produced in a natural way from energy considerations generated only from the physical constants and geometry.

Using *m* = 1/*r*, we can now put the renormalised electromagnetic mass equation in a more familiar form for electromagnetic energy,

$$
\delta m\_{\text{renorm}} = \frac{3am}{8\pi} = \frac{3a}{8\pi r} = \frac{3\,e^2}{32\pi^2 r}.\tag{12}
$$

*Universe* **2020**, *6*, 112

However, *δm* in this expression still goes to infinity as *r* → 0. Perhaps a Planck length cutoff may be used to tame this infinity. We will soon see that *r* does take on a finite value very close to Planck length. Now we know from experiments that the radius of an electron is likely to be less than 10−<sup>22</sup> m [24]. Substituting that bound for *r* gives

$$
\delta m = \frac{3 \, e^2}{32 \pi^2 \, (10^{-22} \,\text{m})} \gtrsim 1.719 \times 10^3 \,\text{GeV},\tag{13}
$$

which, although finite, is still a very large electromagnetic energy contribution, with the actual value likely to be even greater. Thus, according to this estimate, the electromagnetic mass is going to be very large near the Planck length and will have to be compensated for in order to recover the observed rest mass of a charged fermion. The compensation will have to be negative mass-energy relative to the positive electromagnetic energy:

$$m\_{obs} = -X + \delta m\_R = -X + \frac{3\,\mathrm{a}}{8\pi r\_B} \longrightarrow -\frac{X}{r} + \frac{3\,\mathrm{a}}{8\pi r\_B}.\tag{14}$$

We suspect that the unknown variable *X* might be related to the Hehl–Datta self-interaction term (7) because that term varies as 1/*rx*. Our goal then is to investigate this possibility.

To that end, note that the full second order S-matrix calculation within QED worked out by Milonni [21] gives

$$S\_{fi}^{(2)}(E) = -i(2\pi)^4 \delta^4(p\_f - p\_i) \sqrt{\frac{m^2}{E\_i E\_f}} \frac{1}{r^3} \|(p\_{f'}s\_f)\Sigma(p\_i)u(p\_{i\prime}s\_i) \tag{15}$$

$$\text{where}\quad\Sigma(p\_i) = -ie^2 \int \frac{d^4k}{(2\pi)^4} \frac{g\_{\mu\nu}}{k^2 + i\epsilon} \gamma^\mu \frac{1}{p\_i - \not\!k - m + i\epsilon} \gamma^\nu. \tag{16}$$

Milonni's evaluation of Σ(*pi*) produces the result for *δm* given in Equation (8). However, we doubt this evaluation is appropriate since, as we noted previously, the mass comes in via the propagator and therefore it is off mass shell. We will not use it in our evaluation.
