*2.2. S-Matrix Evaluation for the Charged Fermionic Self-Energy within Einstein–Cartan Gravity*

Building on the results of the previous subsection, we now evaluate the self-energy S-matrix process including the contribution from ECSK gravity. Our calculation will be different from the old way of calculating self-energy in the manner of Weisskopf because in our approach we have the gravitational torsion term as a built-in counter-term. The following calculations will be somewhat unconventional as there is no Feynman diagram for evaluation in the rest frame. We begin with our complete QED Lagrangian for free particles:

$$\frac{\mathfrak{L}\_{\text{QED}}}{\sqrt{-g}} = i\hbar c \,\bar{\Psi}\gamma^{\mu}\partial\_{\mu}\psi + e\,\bar{\Psi}\gamma^{\mu}A\_{\mu}\psi - \frac{1}{4}F\_{\mu\nu}F^{\mu\nu} - mc^{2}\bar{\Psi}\psi - \frac{3\kappa\hbar^{2}c^{2}}{8}(\bar{\Psi}\gamma^{5}\gamma\_{\mu}\psi)(\bar{\Psi}\gamma^{5}\gamma^{\mu}\psi). \tag{17}$$

Here the last term is from the Hehl–Datta equation for gravitational torsion via spin density squared. We take it to be a self-interaction involving intrinsic spin. In other words, the spin interacts with itself much like the charge interacts with the field it creates. For our analysis of a charged lepton in the rest frame, only three terms are applicable because the derivative term is zero and normal gravity is negligible. The self-interaction Hamiltonian density for the first term with *h*¯ = *c* = 1 and (*x*) = (**x**, *t*) is then

$$\mathbf{h}\_I(\mathbf{x})\_1 = e \,\bar{\psi}(\mathbf{x}) \gamma^\mu A\_\mu \psi(\mathbf{x}). \tag{18}$$

The next self-interaction term is

$$\mathbf{h}\_I(\mathbf{x})\_2 = -\frac{3\kappa}{8} (\overline{\psi}(\mathbf{x})\gamma^5\gamma\_k\psi(\mathbf{x})) (\overline{\psi}(\mathbf{x})\gamma^5\gamma^k\psi(\mathbf{x})).\tag{19}$$

For our purposes it is sufficient to evaluate the S-matrix in the rest frame, where the quantities *ψ*¯(*x*)*γ* <sup>5</sup>*γkψ*(*x*) and *ψ*¯(*x*)*γ* 5*γ <sup>k</sup>ψ*(*x*) appearing on the RHS are nonzero even in the rest frame because the coupling between fermions and anti-fermions is an intermediate step in the self-interaction, as we have shown in Equations (A14) and (A15) of Appendix B. Furthermore, as shown in Appendix B, the first order S-matrix terms in the rest frame are now simplified to

$$S\_{fi}^{(1)}(E1) = -i t \frac{\alpha}{r} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} . \tag{20}$$

$$S\_{fi}^{(1)}(E2) = it\frac{3\kappa}{8\pi^3} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} . \tag{21}$$

On the other hand, for *mψψ*¯ in the rest frame, applying Equation (A8) from Appendix B we have

$$S\_{fi}^{(1)}(E3) = im \int d^4x \left< f \mid \overline{\wp}(x)\overline{\wp}(x) \mid i \right> = i \frac{m \, r^3 t}{r^3} \,\overline{u}^f(m) u^i(m) = i \, mt \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} . \tag{22}$$

Equation (5) suggests that the sum of these three terms is equal to zero:

$$-it\frac{\boldsymbol{\alpha}}{r}\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} + it\frac{3\boldsymbol{\kappa}}{8r^3}\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} + i\, int\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} = 0.\tag{23}$$

Consequently, we have

$$
\frac{\mathfrak{a}}{r} - \frac{\mathfrak{Z}\kappa}{8r^3} = m.\tag{24}
$$

Reinstating ¯*h* and *c* we thus arrive at

$$\frac{a\hbar c}{r} - \frac{3\kappa(\hbar c)^2}{8\,r^3} = mc^2. \tag{25}$$

An identical result can be obtained also for the anti-fermion spinor, *v*(*m*), so that this equation remains the same for both fermion and anti-fermion. In what follows we will use this S-matrix result for our numerical approximations. From these results, after solving for *r*, it is evident that the complete process is finite, without divergences. This suggests that the above is the correct S-matrix solution for the fermion self-energy problem, with all other orders of the S-matrix expansion vanishing because the genuine fermionic self-energy must naturally be evaluated only in the rest frame, with all other contributions summing to zero. Any higher loop corrections in squared charge will be automatically compensated by higher loop corrections from the squared spin.

It is also worth noting that without ameliorating the Dirac equation with a cubic term, the Dirac equation would reduce for an electron to *<sup>α</sup>h*¯ /*r<sup>e</sup>* <sup>=</sup> *<sup>m</sup>ec*, giving *<sup>r</sup><sup>e</sup>* <sup>=</sup> *<sup>α</sup>h*¯ /(*mec*) <sup>∼</sup> <sup>10</sup>−<sup>15</sup> m, where *α* = *e* <sup>2</sup>/(4*πhc*¯ ) is the fine structure constant. This is the classical electron radius. Experimental evidence, however, suggests that electron radius is much smaller [24]. As we shall see, our calculations

with the cubic term included predicts the electron radius to be of the order of 10−<sup>34</sup> m, which is closer to the Planck length. This may turn out to be the correct value of the electron radius.

Needless to say, what we have presented above is a derivation of Equation (25) within a theory that may be viewed as a quantum field theory of Dirac fields in a Riemann–Cartan spacetime [4,5]. It can be interpreted also as a theory of gravity-coupled self-interaction within standard general relativity [1,5]. However, any such generalization must necessarily reproduce the Hehl–Datta Equation (5) for single fermions even at reasonably high energies, just as Dirac equation remains valid for single fermions at high energies [18].

Finally, it is important to note here that, despite the appearance of four spinors in the interaction term of Equation (6), it describes the self-interaction of a single fermion, of range <sup>∼</sup>10−<sup>34</sup> m, not mutual interactions among the spins of four distinct fermions. That is to say, it does not describe a "spin field" of some sort as a carrier of a new interaction [5]. If, however, one insists on interpreting the interaction term in Equation (6) as describing interactions among four distinct fermions, then the mass of the corresponding exchange boson would have to exceed 10<sup>15</sup> GeV, which is evidently quite unreasonable. Moreover, as we shall see in Section 3, this energy is a fictitious quantity and therefore there is no justification for assuming some kind of a new torsion interaction between different fermions. What is more, as we shall soon see, within our scheme any corrections due to vacuum polarization are automatically compensated for in the production of electroweak mass-energy, dictated by Equation (25) above.

### *2.3. Evaluation of Charged Fermionic Self-Energy by Dimensional Analysis*

It is instructive to compare the results of the previous two subsections with the evaluation of the charged fermionic self-energy using only dimensional analysis. To this end, we begin with the following physically reasonable assumptions:


Given these assumptions, we ask: What physical mechanism is responsible for the observed rest mass *m<sup>x</sup>* of the elementary charged fermions? To answer this question we express the rest mass energy in CGS units, and assume that it is at least partially<sup>1</sup> electromagnetic in nature, so that it satisfies a relation like

$$
\mu\_{\rm x}c^2 \sim \frac{e^2}{r} + X\_{\prime} \tag{26}
$$

where the dimensionality of *X* is necessarily that of energy. However we already know that the value of *r* < 10−<sup>15</sup> *m* produces an energy greater than *mxc* 2 . Therefore *X* must be negative energy, giving

$$
\varepsilon m\_{\text{x}} c^2 \sim \frac{e^2}{r} + (-X). \tag{27}
$$

<sup>1</sup> As is well known, assuming that the rest mass energy is entirely electromagnetic in nature leads to the classical radius of the electron, which has been ruled out by experiments. Our assumption of it being at least partially electromagnetic in nature is quite reasonable.

Now, since fermions have spin *h*¯ /2, it is reasonable to assume that it is involved in a mechanical-like energy resulting from the spin interacting with itself analogous to charge, so that we may have a relation like

$$-X \sim -\left(\frac{\hbar}{2}\right)^2 \frac{1}{r}.\tag{28}$$

It is evident from this expression that what we have on its RHS is energy × mass × length = *EML*, so we will have to divide out by mass and length to get the dimensions of energy, giving

$$-X \sim -\left(\frac{\hbar}{2r}\right)^2 \frac{1}{M'} \tag{29}$$

which in terms of the gravitational constant *G* ∼ *L* <sup>3</sup>/*MT*<sup>2</sup> can be written as

$$-X \sim -G\left(\frac{\hbar}{2r}\right)^2. \tag{30}$$

The dimensions on the RHS of this expression now give us *E* × *L* <sup>3</sup>/*T* 2 , so we will have to cancel out *L* <sup>3</sup>/*T* 2 . A natural candidate to accomplish that is the speed of light, *c*, giving

$$-X \sim -\frac{G}{c^2 r} \left(\frac{\hbar}{2r}\right)^2 = -\frac{G}{r^3} \left(\frac{\hbar}{2c}\right)^2,\tag{31}$$

provided we cancel out the extra *L* with 1/*r*. Thus, we now have the dimensions of energy on the RHS, so that we finally have

$$
\sigma\_{\rm x}c^2 \sim \frac{e^2}{r\_{\rm x}} - \frac{G}{r\_{\rm x}^3} \left(\frac{\hbar}{2c}\right)^2. \tag{32}
$$

This is the basic form of our central equation (apart from some numerical constants) which we have derived also using two other methods in the previous subsections. Solving this equation for *r<sup>x</sup>* with electron mass for *m<sup>x</sup>* gives a value of the order of 10−<sup>34</sup> m. However, rather surprisingly, we also found a solution for *r<sup>x</sup>* for the classical electron radius.

### **3. Particle Masses via Torsion Energy Contribution**

We start with our S-matrix result for a charged fermion:

$$\frac{a\hbar c}{r} - \frac{3\pi G\hbar^2}{c^2 r^3} = mc^2. \tag{33}$$

The first term in Equation (33) diverges as *r* → 0. If we set *r* to Planck length we obtain

$$\frac{a\hbar c}{l\_P} \approx 8.909 \times 10^{16} \,\text{GeV},\tag{34}$$

which is close to Planck energy. Although finite, this is still an extremely large energy. Such a large energy for charged leptons is never realised in Nature. A natural question then is: Is there a negative mechanical energy that cancels out most of this energy to produce the observed rest mass-energy of leptons? We believe the answer lies in the second term of Equation (33), which—as we saw above—arises from the non-linear amelioration of the Dirac equation within ECSK theory. Indeed, if we again set the Planck length for *r* in the second term of Equation (33), then we obtain

$$-\frac{3\pi \,\mathrm{G}\hbar^2}{2\,\mathrm{c}^2 \,(l\_P)^3} \approx -5.773 \times 10^{19} \,\mathrm{GeV}.\tag{35}$$

Comparing this value with the electrostatic energy at the Planck length estimated in Equation (34) we see at once that the torsion-induced mechanical energy in Equation (35) can indeed counterbalance the huge electrostatic energy. This is a surprising observation, considering the widespread belief that "the numerical differences which arise [between GR and ECSK theories] are normally very small, so that the advantages of including torsion are entirely theoretical" [18].

Moving forward to our goal of numerical estimates, let us note that whenever terms quadratic in spin happen to be negligible, then the ECSK theory is observationally indistinguishable from general relativity. Therefore, for post-general-relativistic effects, the density of spin-squared has to be comparable to the density of mass. The corresponding characteristic length scale, say for a nucleon, is referred to as the Cartan or Einstein–Cartan radius [2,18], defined as

$$r\_{\mathsf{Cart}} \approx (l\_P^2 \lambda\_\mathsf{C})^{\frac{1}{5}},\tag{36}$$

where *λ<sup>C</sup>* is the Compton wavelength of the nucleon. Now it has been noted by Poplawski [6–9,25,26] that quantum field theory based on the Hehl–Datta equation may avert divergent integrals normally encountered in calculating radiative corrections, by self-regulating propagators. Moreover, the multipole expansion applied to Dirac fields within the ECSK theory shows that such fields cannot form singular, point-like configurations because these configurations would violate the conservation law for the spin density, and thus the Bianchi identities. These fields in fact describe non-singular particles whose spatial dimensions are at least of the order of their Cartan radii, defined by the condition

$$
\mathfrak{e} \sim \mathfrak{x} \mathfrak{s}^2,\tag{37}
$$

where <sup>√</sup> *s* <sup>2</sup> ∼ *hc*¯ |*ψ*| 2 is the spin density, *<sup>ǫ</sup>* <sup>∼</sup> *mc*<sup>2</sup> |*ψ*| 2 is the rest energy density and |*ψ*| <sup>2</sup> <sup>∼</sup> 1/*<sup>r</sup>* 3 is the probability density, giving the radius in Equation (36). Consequently, at the least the de Broglie energy associated with the Cartan radius of a fermion (which is approximately 10−<sup>27</sup> m for an electron) may introduce an effective ultraviolet cutoff for it in quantum field theory in the ECKS spacetime. The avoidance of divergences in radiative corrections in quantum fields may thus come from spacetime torsion originating from intrinsic spin. Poplawski and others, however, took *ǫ* to be the mass-energy density of the fermion to arrive at the Cartan radius Equation (36). It is easy to work out from the first term of our Equation (33) that at the Cartan radius the electrostatic energy density for an electron is still extremely large:

$$\frac{a\hbar c}{(10^{-27}m)^4} \approx 1.440 \times 10^{90} \,\text{GeV} \,\text{m}^{-3}.\tag{38}$$

For this reason it is not correct to identify *ǫ* with the rest mass-energy density, which is <sup>≈</sup> 5.1099 <sup>×</sup> <sup>10</sup><sup>77</sup> GeV m−<sup>3</sup> for an electron at the Cartan radius. The electrostatic energy density of an electron is thus about thirteen orders of magnitude higher. Therefore *ǫ* is better identified with the electrostatic energy density provided most of it is cancelled out.

If in Equation (33) we set the electrostatic energy appearing in its first term to be equal to the spin squared energy induced by the self-interaction appearing in its second term and solve for *r*, then we obtain

$$\left| r\_t = \sqrt{\frac{3\pi}{\alpha}} \, l\_P \right|,\tag{39}$$

the value of which works out to be

$$r\_t \approx 5.808 \times 10^{-34} \,\mathrm{m} \,\mathrm{.}\tag{40}$$

Thus, our *r<sup>t</sup>* is about 36 times larger than the Planck length, and, as we can see, it is a remarkably simple constant in terms of the Planck length, *lP*, and the fine structure constant, *α*. According to Equation (33), *r<sup>t</sup>* is the effective radius at which energy density due to spin density should completely compensate the huge electrostatic energy seen in Equation (34). In our view, this is the correct Cartan radius, at least for the charged leptons, and that may still provide a plausible mechanism for averting

singularities, since it is still larger than the Planck length. It is important to note, however, that these huge energy densities never actually occur in Nature, because according to our Equation (33) they are automatically compensated. The physical mechanism described above is simply to enable extraction of the radius *r<sup>x</sup>* for different charged fermions.

We can now use Equation (33) to solve for *r<sup>x</sup>* for the different charged leptons and anti-leptons which leads us to the following formula for our numerical estimates:

$$\frac{a\hbar c}{r\_{\chi}} - \frac{3}{c^2 r\_{\chi}^3} = +m\_{\chi}c^2 \,. \tag{41}$$

As shown in the Appendix A below, we were able to find solutions for *r<sup>x</sup>* for the charged leptons using arbitrary precision in Mathematica. The first in our results listed below is the solution for *r<sup>t</sup>* up to 24 significant figures. Then, using the same precision for comparison, we list the results for *r<sup>e</sup>* for an electron, *r<sup>µ</sup>* for a muon and *r<sup>τ</sup>* for a tauon, along with the anti-fermions:

$$r\_{\ell} = 5.80838808109165274355010 \times 10^{-34} \text{ m} \longrightarrow 0.0 \text{ MeV},$$

$$r\_{\ell-} = r\_{\ell+} = 5.80838808109165274414872 \times 10^{-34} \text{ m} \longrightarrow 0.511 \text{ MeV},$$

$$r\_{\mu-} = r\_{\mu+} = 5.80838808109165286732523 \times 10^{-34} \text{ m} \longrightarrow 106 \text{ MeV},$$

$$r\_{\tau-} = r\_{\tau+} = 5.80838808109165482503366 \times 10^{-34} \text{ m} \longrightarrow 1777 \text{ MeV}.$$

We should note that there is also a positive solution obtained for the reduced Compton wavelength for these fermions in the form of

$$
\sigma\_{\mathcal{X}} = \alpha \frac{\hbar}{m\_{\mathcal{X}}c}.\tag{42}
$$

This is so because the spin density squared term for an electron becomes very small of order 10−<sup>38</sup> MeV when *r<sup>x</sup>* is equal to that result so it can be considered effectively as zero. This solution for fermion radius appears to have been ruled out by scattering experiments. However, while it is true that no structure has been found via scattering experiments, there does seem to be some structure involving the magnetic moment and zitterbewegung. The apparent conflict with the scattering experiments can be resolved if there is a very small object for electric charge near Planck length, as our solution indicates, that is "circulating" about the Compton wavelength. Then the scattering at high energies can be understood as from that point-like object and the scattering at low energies can be understood as from the Compton wavelength "size" due to the Coulomb field, with the Coulomb field "barrier" near the Compton wavelength penetrated in the high energy scatterings.

Evidently, very minute changes in the radii are seen to cause large changes in the observed rest mass-energies of the fermions. As the differences in the radii go larger, the resultant mass-energies go higher, as one would expect. It seems extraordinary that Nature would subscribe to such tiny differences resulting from a large number of significant figures, but that might explain why the underlying relationship between the observed values of the masses of the elementary particles has remained elusive so far. In addition to the possible reasons for this mentioned above, it is not inconceivable that the difference between the spin energy density and the electrostatic energy density radii with respect to *r<sup>t</sup>* arises due to purely geometrical factors. We also suspect that there may possibly be some kind of symmetry breaking mechanism at work similar to the Higgs mechanism, and this symmetry breaking results in the observed mass-energy generation.

As a consistency check, let us verify that the tiny length differences seen above vanish, <sup>∆</sup>*<sup>r</sup>* → 0, as the corresponding rest mass-energy differences tend to zero: <sup>∆</sup>*<sup>E</sup>* → 0. To this end, we recast Equation (41) for arbitrary *r<sup>x</sup>* in a form involving only rest mass-energy on the RHS as:

$$\frac{a\hbar c}{r\_{\chi}} - \frac{3\pi l\_{\rm P}^2 \hbar c}{r\_{\chi}^3} = m\_{\rm x} c^2. \tag{43}$$

If we now set

$$A \equiv a\hbar c \quad \text{and} \quad B \equiv 3\pi \, l\_P^2 \hbar c \,, \tag{44}$$

then, with ∆*E* = *mxc* <sup>2</sup> and setting *r<sup>x</sup>* = *r<sup>t</sup>* as the cancellation radius for which ∆*E* = 0, we obtain

*r<sup>t</sup>* = √ *B*/*A* . (45)

This allows us to derive a general expression for *<sup>r</sup><sup>x</sup>* when <sup>∆</sup>*<sup>E</sup>* 6= 0:

$$\frac{A}{r\_{\chi}} - \frac{B}{r\_{\chi}^3} = \Delta E.\tag{46}$$

From this expression it is now easy to see that

$$\lim\_{\Delta E \to 0} \left\{ \left( \frac{A}{r\_{\chi}} - \frac{B}{r\_{\chi}^3} \right) = \Delta E \right\} = \left( \frac{\sqrt{A}^3}{\sqrt{B}} - \frac{\sqrt{A}^3}{\sqrt{B}} \right) \ln \frac{\sqrt{B/A}}{l\_P} = 0 \implies r\_{\chi} = \sqrt{B/A} = r\_{\mathfrak{l}} \tag{47}$$

and conversely, using Equation (45),

$$\lim\_{r\_X \to r\_Y} \left\{ \left( \frac{A}{r\_X} - \frac{B}{r\_X^3} \right) = \Delta E \right\} \implies \Delta E = 0. \tag{48}$$

Consequently, with <sup>∆</sup>*<sup>r</sup>* ≡ |*r<sup>t</sup>* − *<sup>r</sup>x*|, we see from the above limits that <sup>∆</sup>*<sup>r</sup>* → 0 as <sup>∆</sup>*<sup>E</sup>* → 0, and vice versa.

As a rough estimate, the calculation for the radius *r<sup>q</sup>* of elementary quarks can be performed in a similar manner as that for charged leptons, since at such short distances the strong force reduces to a Coulomb-like force. One must also factor-in the electrostatic energy, so that a relationship like the following must be calculated, say, for the top quark:

$$\frac{9\,\text{a}\hbar c}{4\,r\_{q\text{x}}} + \frac{\alpha\_s \hbar c}{3\,r\_{q\text{x}}} - \frac{3\pi G\hbar^2}{c^2 \,r\_{q\text{x}}^3} = m\_l c^2 \,. \tag{49}$$

Here *α<sup>s</sup>* is the appropriate strong force coupling (we use 0.1). Needless to say, a cancellation radius different from that of the charged leptons should be calculated for comparison, by setting

$$\frac{9}{4} \frac{a\hbar c}{r\_{qt}} + \frac{a\_s \hbar c}{3 \, r\_{qt}} - \frac{3\pi G\hbar^2}{c^2 \, r\_{qt}^3} = 0.\tag{50}$$

A calculation of the radius for the top quark based on Equation (49) can be found in the Appendix A and is <sup>≈</sup>2.594 <sup>×</sup> <sup>10</sup>−34m. We expect it to be only a very rough estimate of the actual value of the radius. Since only one spin density is involved, the above calculation might be able to approximate the behaviour of the quarks. The calculation of the radii *rqx* for the up and down quark will probably be problematic, since their masses are not well known.

With regard to neutrinos, the self-energy function is very similar to that for charged leptons with the Z boson replacing the photon and weak coupling replacing *α*. A rough calculation gives the same order for the radius at <sup>≈</sup>10−<sup>34</sup> m. However, if this does not hold for the neutrino self-energy, then the story is quite different because they would not have self-energy due to electric or colour charge. That means that their rest mass-energy comes entirely from torsion energy due to intrinsic spin. Solving that gives us a radius for electron neutrinos of the order of 10−<sup>26</sup> m. However, the torsion (spin-squared) self-energy is negative relative to the positive rest mass. We suspect that this means that neutrinos could have anti-gravity properties [27]. The anti-gravity effect with normal matter for single neutrinos is practically negligible but the cosmological implications could be large.
