Microscopic considerations and dilatations of $\mathcal{T}$.

As we have discussed if Section 2.2, the choice of the fundamental vector fields that determine the absolute parallelism structure of $\mathcal{S}$ cannot be arbitrary if the minimum time necessary to perform physical operations cannot be neglected. In fact, if we consider the above described theory as a macroscopic approximation of a more complete theory in which the cone $\mathcal{T}^+$ and the constant $\ell$ play a nontrivial role, the fields $A_{\alpha}$ used to define the cone have to be unambiguously specified (up to a transformation of $GL(4, \mathbf{R})$). Then also the field $\phi$ and the exponent $m$ acquire a physical relevance.

The constant $\ell$ that appears in the definition of $\mathcal{T}^+$ has dimension

$$[\ell]= [\omega^i] [\omega^{[ik]}]^{-1} = [L] [\phi] = [L^{1+1/(2-m)} M^{-1/(2-m)}]$$ (6.53)

and in general it is not a length. The acceleration and the angular velocity of a frame are defined in terms of the fields $\tilde A_{[ik]}$ and therefore the maximal acceleration introduced by $\mathcal{T}^+$ is given by $\phi \ell^{-1}$ and may depend on the point.

In a theory with variable couplings, one has to decide which parameters are really constant and which are variable fields. Our point of view (not shared by some authors [107]) is that the velocity of light $c = 1$ and the parameter $\ell$ should be really constant, because they determine the structure of $\mathcal{T}^+$, namely of the fundamental causal structure of the geometry.

In order to avoid serious problems with the quantization procedure, Planck's constant too has to be really constant. If one wants to avoid the proliferation of fundamental constants (an economy principle), an appealing choice is

$$m = 4, \qquad \ell = \nu \hbar^{1/2},$$ (6.54)

where $\nu$ is an adimensional factor, presumably of the order of one. We get in this way a classical theory ``prepared'' for quantization, and the quantization procedure should determine the possible values of $\nu$. As we have noted in Section 2.4, a similar situation also appears in classical theories with constant gravitational coupling $G$ and a fixed fundamental length $\ell$.

In ref. [5] the different choice $m = 1$ has been suggested, starting from considerations based on Mack's principle. One assumes that the forms $\omega^{[0r]}$ do not measure directly the acceleration of a frame, but the force per unit of gravitational mass acting on the object associated to the frame. The forms $\tilde \omega^{[0r]}$ introduced above measure the acceleration, namely the force per unit of inertial mass and the ratio $\phi^{-1}$ between the two forms gives the ratio between the inertial and the gravitational mass. According to Mack's principle, the inertia is determined by the matter present in the universe and is proportional to the field $\Phi$, that satisfies the field equation (6.49) in which the energy-momentum of the cosmic matter provides the source. In conclusion, we have, with suitable normalizations, $\Phi = \phi^{-1}$, namely $m = 1$.

Note that in the model we are considering, assuming that the coefficient $k_4$ vanishes or is proportional to $\phi^m$, the gravitational Lagrangian form is an homogeneous functions of degree $m$ of the structure coefficients. It follows that it has a simple behavior under the dilatations of the vector space $\mathcal{T}$, that we call total dilatations, since they imply dilatations of both the vertical and the horizontal subspaces. Note that the cone $\mathcal{T}^+$ is invariant under total dilatations, but not under separate vertical or horizontal dilatations. The infinitesimal total dilatations are given in terms of the infinitesimal parameter $\zeta$ by

$$\delta A_{\alpha} = \zeta A_{\alpha}, \qquad \delta \omega^{\alpha} ... ...ha}, \qquad \delta F_{\alpha \beta}^{\gamma} = \zeta F_{\alpha \beta}^{\gamma},$$

$$\delta \phi = \zeta \phi, \qquad \delta \chi_i = \zeta \chi_i,$$

$$\delta \lambda^H = (m - 4) \zeta \lambda^H, \qquad \delta \lambda^A = (m - 4) \zeta \lambda^A.$$ (6.55)

It is interesting to investigate the behavior of the other parts of the Lagrangian form under total dilatations. For the Dirac Lagrangian (5.36) we put

$$\delta \Psi = 2^{-1} (m-1) \zeta \Psi, \qquad \delta \omega^a = 0, \qquad \delta \Xi = \zeta \Xi,$$

$$\delta \lambda^D = (m - 4) \zeta \lambda^D.$$ (6.56)

Note that we have to assume that the basis vectors $A_a$ of the extended vector space $\mathcal{T}$ are not affected by the dilatations. We also have to assume a specific transformation property of the Higgs field $\Xi$. With the same assumptions we find that the Lagrangian (5.14) of the internal gauge fields is invariant and the Lagrangian (5.26) of the Higgs scalar field is also invariant if we disregard the potential term $- V(\Xi) \eta$.

In conclusion, if we put $m = 4$, the complete Lagrangian form of gravitation and elementary particles is invariant, apart from the terms containing the cosmological constant and the Higgs potential (one can redefine the Higgs potential in such a way that it includes the cosmological constant). If we take these terms into account, we have

$$\delta \lambda = - \delta (V(\Xi) \eta) - \delta(k_4 \eta) = \zeta (- 4 \lambda v^2 (\Xi^{\dagger} \Xi - v^2) + 4 k_4) \eta,$$ (6.57)

where we have adopted the expression (5.61) for the Higgs potential and we have assumed that $k_4$ is constant. This simple result is a further indication that $m = 4$ is an interesting choice.

Note that the potential term that is necessary in order to obtain the spontaneous symmetry breaking of the internal gauge symmetry is also responsible for the explicit symmetry breaking of the total dilatation symmetry. If one likes to preserve the total dilatation symmetry, one has to invent a different way to generate a nonvanishing vacuum expectation value of the Higgs field $\Xi$.

For instance, it could have a cosmological origin, as it happens for the Brans-Dicke field $\Phi$. However, a model of this kind would imply the existence of unobserved zero mass Higgs particles. Note that also the Brans-Dicke field should describe unobserved zero-mass particles. In any case, it is very difficult to modify the Standard Model of elementary particles without spoiling its very good agreement with the experimental observations. A serious discussion of these problems lies outside the scope of the present notes.

In analogy with eq. (4.41), one can define (using the real formalism) the 3-form

$$\theta_D = - \omega^i \wedge \sigma_i - 2^{-1} \omega^{[ik]} \wedge \sigma_{[ik]} + \Xi^T \pi^S.$$ (6.58)

Note that the internal gauge fields and the Dirac fields do not contribute. For $m = 4$ we have some relevant cancellations and we obtain (introducing the complex formalism for $\Xi$)

$$\theta_D = 6 \alpha \Phi \chi^i \eta_i - 2^{-1} A^i (\Xi^{\dagger} \Xi) \eta_i.$$ (6.59)

This form is conserved (namely it is closed) only if $\delta \lambda = 0$. Otherwise we have

$$d \theta_D = (- 4 \lambda v^2 (\Xi^{\dagger} \Xi - v^2) + 4 k_4) \eta.$$ (6.60)

Note that the right hand side vanishes if $k_4 = 0$ and $\Xi$ takes its vacuum expectation value.

By means of eq. (4.69) we obtain the more explicit formula

$$6 \alpha A_i (\Phi \chi^i) - 6 \alpha \Phi \chi_i \chi^i - 2^{-1} A^i A_i (\Xi^{\dagger} \Xi) + 2^{-1} \chi^i A_i (\Xi^{\dagger} \Xi)$$

$$= - 4 \lambda v^2 (\Xi^{\dagger} \Xi - v^2) + 4 k_4.$$ (6.61)

The same equation can also be obtained directly from the field equations, but the derivation given above helps us to understand their meaning and may suggest further developments.