Weak Scale From the Maximum Entropy Principle

The theory of multiverse and wormholes suggests that the parameters of the Standard Model are fixed in such a way that the radiation of the $S^{3}$ universe at the final stage $S_{rad}$ becomes maximum, which we call the maximum entropy principle. Although it is difficult to confirm this principle generally, for a few parameters of the Standard Model, we can check whether $S_{rad}$ actually becomes maximum at the observed values. In this paper, we regard $S_{rad}$ at the final stage as a function of the weak scale ( the Higgs expectation value ) $v_{h}$, and show that it becomes maximum around $v_{h}={\cal{O}}(300\text{GeV})$ when the dimensionless couplings in the Standard Model, that is, the Higgs self coupling, the gauge couplings, and the Yukawa couplings are fixed. Roughly speaking, we find that the weak scale is given by \begin{equation} v_{h}\sim\frac{T_{BBN}^{2}}{M_{pl}y_{e}^{5}},\nonumber\end{equation} where $y_{e}$ is the Yukawa coupling of electron, $T_{BBN}$ is the temperature where the Big Bang Nucleosynthesis starts and $M_{pl}$ is the Planck mass.


Introduction and Review of History of Universe
The theory of multiverse and wormholes [1,2,3,4,5] suggests that the parameters of our universe are fixed in such a way that the radiation of the universe S rad at the final stage becomes maximum, which we call the maximum entropy principle or the Big Fix. Here, "the final stage" means that the curvature term balances with the other contents of the universe, and the radius of the universe a is getting close to the critical value a * spending infinite time. See Fig.1 for example and see [2,3,4,5] for the details. This assertion can be checked in principle by changing the parameters of the Standard Model (SM) one by one, if we know how S rad is determined by those parameters. In general, this procedure is difficult to do because of the lack of our understanding of the history of the universe. However for a few couplings of the SM, we can estimate their effects on S rad under some assumptions, and we can actually check the principle. As a concrete example we consider the Higgs expectation value v h . Although there are many possibilities for the final stage of the universe, we assume the following scenario: Assumptions for the Final Stage of the Universe (I): The Dark Matter (DM) decays much earlier than the baryons. This guarantees that the radiation produced by the DM is negligible compared with that of baryons.
(II): The Cosmological Constant (CC) at the final stage of the universe is fixed to the critical value Λ cri so that the curvature term balances with the radiation produced by the baryon decay (see Fig.1). The maximum entropy principle predict that the dark energy should decrease from the present value to Λ cr in the future.
(III): Baryons decay with the life time τ B , and the radiation S rad is produced. In this paper, we assume that τ B = 10 36 year. (1) After that, the radiation and the CC balance with the curvature term while electrons, positrons and neutrinos annihilate into photons.
Based on these assumptions, we show an evidence of the Big Fix : When the Higgs self coupling, gauge couplings and Yukawa couplings are fixed, S rad becomes maximum around v h = O(300GeV) 1 .We first review how S rad is produced through the history of the universe. In the following argument, we denote the photon temperature by T .
• Stage 1: The baryon number N B is produced by the sphaleron process if we assume the standard leptogenesis scenario. Here we briefly summarize how N B is produced in the early universe. The number density of a particle species i is given by where g i is the degree of freedom, µ i is the chemical potential and the sign ± is − for boson and + for fermion. Then the difference between a particle and an anti-particle is given by where (4) We can eliminate the chemical potentials {µ i } by using the conservations of the total electromagnetic charge Q and the difference between the baryon number and the lepton number N B − N L . As a result, the baryon number N B is given by a function of m i T : See Appendix B in [5] for the explicit formula and detailed calculations. When T reaches the sphaleron decoupling temperature T sph , N B is fixed to N B ( m i T sph ). We can obtain T sph by using the recent numerical result [6]: Here, we have assumed that the Higgs self coupling and the gauge couplings are fixed. Thus, N B at T sph is given by Note that if we fix the gauge couplings and the Yukawa couplings, N B is just a constant because the quark masses m q and the gauge boson mass m W are given by • Stage 2: The ratio of neutrons to all nucleons X n is fixed by the following processes: 1)For T > 1MeV, protons and neutrons are in thermal equilibrium through the weak interactions, and X n at that time is given by where Q := m n − m p is the mass difference between a neutron and a proton.
2)The weak interactions are frozen out, and X n decreases through the beta decay until T reaches the temperature T BBN where the Big Bang Nucleosynthesis (BBN) starts.
Here, note that the life time of a neutron τ n depends strongly on v h , m e and Q − m e where m e is the mass of the electron. After T BBN , neutrons are rapidly converted to atomic nuclei. We discuss those processes in more detail in section3.2. See also [5].
• Stage 3: The radiation at the early universe becomes dilute, and the matter dominated era starts. After that, the dark energy becomes dominant. This is the era where we live. As discussed before, we assume that the dark energy becomes very small in the future.
• Stage 4: The DM decays sufficiently earlier than baryons.
• Stage 5: Baryons decay, and the radiation S rad is produced. S rad depends on X n , the masses of proton and helium nucleus m p , m m He and their life times τ p , τ He . Moreover, we must take into account the possibility that a pion produced by the decay of helium nucleus is scattered by the remaining nucleons, and loses its energy. We denote the energy loss through this process by a dimensionless parameter ǫ.
From the above argument, it is clear that we need to know the v h dependence of the following quantities in order to evaluate S rad as a function of v h : m p , m He , Q( or m n ) , X n , T BBN , τ p , τ He , ǫ.
In this paper, we use the phenomenological equations for m p , Q and τ p : The meanings of the parameters and the factor G are as follows.
• m u and m d are masses of an up quark and a down quark. Their typical values are (m u , m d ) = (2.3MeV, 4.8MeV) [9].
• Λ QCD is the scale where the QCD coupling becomes O(1), which we fix to 300MeV in this paper.
• M em is the electromagnetic energy of a neutron, and α, β, γ are numerical constants that should be determined by QCD. In principle, they can also be found from observations. The typical values we use in this paper are which explain the experimental results [9] well.
• G can be approximately calculated by using the effective interaction of the proton decay as Here, we neglect the electron mass m e in the formula for the proton life time (see Appendix B). In the limit m u,d ≪ m p , we expect that G depends linearly on m u,d mp : On the other hand, for m He , τ He and ǫ, it is difficult to determine their v h dependence because of the complicated effects of nuclear physics. However, by the numerical calculations (see Section2 and also [5]), we can show that these quantities effectively contribute to S rad through a single function c ǫ, τ He τp , m He mp whose typical value is of O(0.01). In this paper, we assume that c is a constant, namely, we neglect the v h dependence of m He , τ He and ǫ. The details are given in Section2 and Section5. Finally, we also assume that T BBN does not depend on v h in the main part of this paper, and fix it to 0.1MeV. The case where it depends on v h is discussed in Appendix A.
Summary : Assuming that are fixed to the phenomenologically reasonable values, we find that S rad has a global maximum around v h = O(300GeV) when the Higgs self coupling, the gauge couplings and the Yukawa couplings are fixed.
This paper is organized as follows. In Section2, we briefly review how the radiation of the universe is determined, and give a qualitative expression. In Section3, we discuss the v h dependence of S rad . In Section4, we consider the Big Fix. In Section5, we give summary and discussion.

Radiation of the Universe at the Final Stage
The radiation of the universe at the final stage can be obtained in principle by solving the following equations: dN Here N p (t) and N He (t) are the numbers of protons and helium nuclei, and ǫ represents the following effect: a pion produced by the nucleon decay of a helium nucleus is scattered by the remaining nucleons, loses its kinetic energy and produces less radiation after it decays [5].
The initial values of N p (t), N He (t) are given by where N B is the total baryon number and X n is the ratio of neutrons to all nucleons. By numerical calculations, we have obtained the qualitative expression of S rad [5]: Qualitative meaning of this equation is as follows: First, if there is no atomic nuclei, baryons are all protons having its life time τ p . If we simplify the situation so that these protons decay similitaneously, we can obtain the radiation S rad by the energy conservation from which we have In our scenario, because the universe is matter dominated until τ p , the Friedman equation indicates Thus, we obtain This expression is modified by the existence of the atomic nuclei. The 1 − c · X n term in Eq.(24) represents the such effects. If c is positive, the radiation decreases and vice versa. As discussed in Section1, N B and X n depend on the SM parameters as If the Higgs self coupling, the gauge couplings and the Yukawa couplings are fixed 2 , the v h dependence of S rad comes from m p , τ p and X n in Eq.(24): Here, note that because the baryon number N B depends on the SM parameters through m i v h , it is just a constant in this case. While the analytic expressions of m p , m π and τ p are given by Eq. (11) and (12), it is difficult to give the counterparts to m He , τ He and ǫ. However, because c varies by only a few percent for v h < 1TeV 3 , the v h dependence of c is not so important compared with that of X n . In this paper, we assume that c is a constant such as 1/50 or 1/100 [5]. We discuss this point in Appendix C.

v h Dependence of S rad
In this section, we discuss how S rad depends on v h . First, we consider m Here, for G we have used Eq.(14). We can understand Fig.2 qualitatively as follows. In the large v h region, because the current quark masses are larger than Λ QCD , G becomes a constant, and we have Therefore, m where η is given by (see Appendix B) As one can see from p decreases monotonically as a function of v h , which is crucial for obtaining the maximum of around v h = O(300GeV). This is because, as we will see in the next subsection,

Neutrons to Nucleons Ratio X n
In this subsection, we discuss the v h dependence of X n . At a temperature T ≫ 1MeV, neutrons and protons are in thermal equilibrium through the following six processes: and X n is given by .
The total reaction rate of the p → n process is given by [7] Γ(p → n) = 0.400 sec −1 × T 1MeV Here, and T ν is the neutrino temperature where S(x) := 1 + 45 2π 4 · ∞ 0 dy y 2 · x 2 + y 2 + y 2 3 x 2 + y 2 · 1 1 + exp As the universe expands, the above processes except for the beta decay decouple at T dec . We can obtain T dec by solving H = Γ(p → n).
Below T dec , X n decreases through the beta decay until T reaches T BBN where the BBN starts 4 . Thus, X n at T BBN is given by where τ n is the neutron life time, and the relation between T and t is given by the Friedman equation:  Here, N is the degrees of freedom and it is given by 43/4 when m e < T < m µ . By using the Fermi theory, we can calculate τ n as where F (x) := One can see that X n is a monotonically decreasing function of v h , and that X n increases for fixed v h when we increase M em . The latter behavior is easily understood; if we increase M em , the initial value of X n (which is given by Eq.(38)) and the life time of neutron τ n becomes large because Q becomes small.

Big Fix of v h
By using Eq.(31) and the results of the previous section, we can determine S rad as a function of v h . From Fig.2 and Fig.3, it is clear that S rad has a maximum around v h = O(300GeV). Fig.4   In the left (right) figure, we assume c = 1/50 (1/100). S rad has a maximum around v h = O(300GeV).
We can also check that the quantitative behavior of S rad does not change even if α, β and γ are varied by O(0.1). See Fig.5. It is interesting to see whether QCD predicts the such values of α, β and γ that S rad becomes maximum at v h = 246GeV precisely.
A few comments are in order. Originally, the c · X n term in S rad comes from the existence of helium nuclei, which is guaranteed by This is equivalent to where ∆ is the binding energy of a helium nucleus. However, when we change v h with the Yukawa couplings fixed at the observed values, ∆ and 2(Q − m e ) can become equal at some point v h = v 0 > 246GeV. This is because Q − m e is an increasing function of v h and ∆ is expected to be a decreasing function of v h 5 . Therefore, in the v h > v 0 region, atomic nuclei can not exist (see Fig.6). However, a simple analysis indicates that v 0 is greater than 1TeV, where X n is very close to zero (see Fig.3). Therefore, we can trust the so far analysis even in this region 6 .
5 The nucleon-nucleon interaction comes from the pion exchange, which becomes weak if the pion mass m π becomes large, thus ∆ is a decreasing function of v h because m π is an increasing function of v h (see Eq.(11)). 6 Although the v h dependence of ∆ is complicated, the upper bound of v 0 can be obtained by Q − m e = ∆(v h = 246GeV)/2 = 14MeV. For example, if we assume β = 1.4 and use (m u , m d ) = (2.3MeV, 4.8MeV) [9], M em becomes 2.2MeV, and v   On the other hands, in the v h < 246GeV region, there are three critical points where Q + m e , Q and Q − m e becomes zero, which we denote respectively by In the region v Q < v h < v β , the beta decay does not occur. But, this case is already included in the so far analysis. In the v < v Q region, proton becomes heavier than neutron. Even if such universe exists, there should also be atomic nuclei because of the isospin-symmetry. In the v h < v r region, protons becomes unstable, and again atomic nuclei cease to exist if v h is much smaller than v r . As a result, S rad increases as v h gets smaller. It is interesting to see whether the peak around v h = O(300GeV) is the global maximum or not.

Summary and Discussion
In this paper, we have shown that the radiation of the universe S rad at the final stage has a maximum around v h = O(300GeV) when the Higgs self coupling, the gauge couplings and the Yukawa couplings are fixed. The v h dependence of S rad is given by Eq.(24). We have seen that m p is a decreasing function for typical values of α, β and γ while 1 − cX n increases rather rapidly for smaller values of v h and becomes constant one around v h = 300GeV. Therefore, S rad becomes maximum at the scale where 1 − cX n becomes one. As we have seen in subsection 3.2, X n is the product of and Actually (53) determines where X n becomes zero (see Fig.7) and thus 1 − cX n becomes one. As a result, the maximum point of S rad is qualitatively given by This shows a surprising and mysterious relation between v h , T BBN , y e and M pl .
In conclusion, we mention the possibilities that we can make further predictions by combining the maximum entropy principle with the other principle such as the stability of the Higgs potential. For example, we can consider the U(1) gauge coupling g Y . As seen in Fig.4, S rad increases if M em is decreased. Because M em mainly depends on the g Y , the smaller values of g Y are favored by the maximum entropy principle. Thus, if there exists a lower bound of g Y , we can conclude that g Y is fixed to that value. From the recent analyses [10] based on the observed Higgs mass, it is possible that the Higgs potential is marginally stable up to the energy scale 10 17 ∼ 10 18 GeV. In other words, if g Y is smaller than the observed value, the Higgs potential is unstable. This means that the present value of g Y is at the lower bound, which is consistent with the above argument. It would be interesting to consider various use of the maximum entropy principle.  Figure 8: X n as a function of v h in the cases where the v h dependence of T BBN is included (Red), and where T BBN is fixed to that of v h = 246GeV (Blue). In the left and right panels, we assume M em = 1.8MeV and M em = 2.2MeV, respectively.
Here, we have put X n (1 − X n ) ≃ 0.1. This will be verified by seeing that X n has an O(0.1) value as a function of v h around v h = 300GeV (see Fig.3). By solving X d = 1 numerically, we can obtain Although determining B d as a function of v h is difficult, the qualitative behavior is easily understood. Because m 2 π ∝ v h , the strength of the nuclear force is roughly given by v −1 h . Therefore, B d should be a decreasing function of v h . Here, we use the recent result [8] 8 in which B d is calculated as a function of the current quark masses by using the effective chiral perturbation theory. The result in [8] agrees with the above heuristic argument. Now we consider X n when we take the v h dependence of T BBN into consideration. First we read off B d as a function of v h from [8], then by using Eq.(64), Eq.(45) and Eq.(44), we obtain X n as a function of v h . See Fig.8. 9 By using Eq.(24), we obtain S rad as a function of v h . See Fig.9 10 . We can see that the maximum of S rad slightly changes. 8 Although, in [8], B d is calculated in the 0.5 < m q /m (phy) q < 2 or 123GeV < v h < 492GeV region, where the chiral perturbation theory seems to be reliable, this region is enough to examine the maximum point of S rad . 9 One can see that the v h dependence of T BBN has an effect to decrease (increase) X n in the v h > (<)246GeV region. This can be understood intuitively; as discussed above, because B d is a decreasing function of v h , T BBN is also a decreasing function. Thus, if v h is large, T BBN becomes small, and the time when the BBN starts becomes large. As a result, the beta decay lasts for a longer time, and X n decreases. 10 Qualitatively, the radiation increases (decreases) in the 246GeV < v h (246GeV > v h ) region because X n decreases (increases) in this region.