_{1}

^{*}

A result from Kieffer, as outlined at the beginning of the article identifies two different candidates for initial time steps, delta t. We assert that this difference in time steps may be related to a specific early universe Lorentz Violation. The author asserts that the existence of early universe Lorentz violation in turn is assisting in a breakup of primordial black holes. And that also has a tie into Kieffer different time steps as outlined. And the wrap up is given in the final part of this document.

Our initial goal is to obtain, via a Kieffer Density function candidate minimum time steps which will be for the purpose of giving input into an uncertainty principle of the form [

Δ E Δ t ≈ 4 ℏ (1)

With a use of [

a ( t ) = a initial t ν ⇒ ϕ = ln ( 8 π G V 0 ν ⋅ ( 3 ν − 1 ) ⋅ t ) ν 16 π G

⇒ ϕ ˙ = ν 4 π G ⋅ t − 1 ⇒ H 2 ϕ ˙ ≈ 4 π G ν ⋅ t ⋅ T 4 ⋅ ( 1.66 ) 2 ⋅ g ∗ m P 2 ≈ 10 − 5 (2)

In doing all of this we are making full use of the following from [

ρ ( ϕ , ϕ ) ≈ { exp ( ± 3 M P 4 8 ⋅ V ( ϕ ) ) } ⋅ ϕ − Z ˜ − 2 (3)

Which after we isolate out Δ t makes use of Equation (1), which is derived as given in [

Δ E ≈ ℏ Δ t 2 γ t P 2 ⋅ 8 ℏ 2 γ t P 2 ( ℏ Δ t ) 2 ≡ 4 ℏ Δ t (4)

We will be applying Equation (4) to obtain Δ t , and then from this step applying Equation (1) to say foundational import issues of time flow in the beginning.

1a. Understanding the probability density functional as an Outgrowth of Kieffer’s derivation

Our assumption is that time, t, which becomes Δ t is extremely small. Hence without loss of generality we write, if as an example, Z ˜ ≈ 2 .

And we simplify time dependence by setting ν = 3 π 2 in Equations (2)-(4)

Then, without loss of generality, if we observe this, and set Θ as a probability density value of Equation (4), we then have [

( 1 4 ⋅ exp ( ± 3 8 ⋅ M P 4 V 0 ) ) ⋅ ( 1 − 4 ⋅ ( 8 π V 0 ( 3 π 2 ) ⋅ ( 9 π 2 − 1 ) ) ⋅ Δ t ) ≈ Θ (5)

If so, then we have a minimum time step of the form

Δ t ≈ 4 ( ( 3 π 2 ) ⋅ ( 9 π 2 − 1 ) 8 π V 0 ) ⋅ ( 1 − 4 Θ exp ( ± 3 8 ⋅ M P 4 V 0 ) ) (6)

Equations ((6), (9)) are keys to the formation of a Lorentz violation, for the reasons we will go into in the next section.

Now for the existence of an initial Lorentz violation, in part linked to Equation (6).

To do this we need to review the Lorentz violating energy-momentum relationship. In short we have that

E 2 = p 2 + m 2 − λ ˜ p 3 (7)

where the positive LV parameter λ ˜ is usually assumed of the order of Planck mass, λ ∼ 1/M (Planck mass). This Lorentz violating energy-momentum relationship leads to, according to [

d ρ = 8 π p c 3 h ⋅ 1 + m 2 p 2 − λ ˜ p exp ( [ c p k B T t e m p ] ⋅ 1 + m 2 p 2 − λ ˜ p ) − 1 ⋅ d p (8)

p ≈ 1 / λ ˜ is used if we integrate, Equation (6) and if we use the first order Romberg numerical integration scheme as given in [

ρ ≈ 8 π M P 3 c h ⋅ m 2 M P 2 exp ( [ c p k B T t e m p ] ⋅ m 2 M P 2 ) − 1 ≈ 4 M P 2 c ℏ ⋅ ( [ c m k B T t e m p ] − 1 2 ( c m k B T t e m p ) 2 ) → k B = c = ℏ = M P = 1 4 ⋅ ( [ m T t e m p ] − 1 2 ( m T t e m p ) 2 ) (9)

We will then in the next section interpret Equation (9) when we set Notice here that we have restrictions on particle manufacturing for the theory. This is in line with [

Δ E ≈ ℏ ⋅ ( 8 π V 0 ( 3 π 2 ) ⋅ ( 9 π 2 − 1 ) ) ⋅ ( 1 + 4 Θ exp ( ± 3 8 ⋅ M P 4 V 0 ) ) (10)

Futher more we can make the following identification m ≈ m g ⋅ N g .

We are then looking at [

ρ ≈ 4 ⋅ ( [ 10 − 65 N g ( T t e m p / T P ) ] − 1 2 ( 10 − 65 N g ( T t e m p / T P ) ) 2 ) ≈ 10 − 60 (11)

We then can up to a modeling round off make the following approximation This value of Equation (11) will lead to approximately if ( T t e m p / T P ) ~ 1 [

N g ≈ 10 57 × ( 1 ± ( 1 − 2 × 10 − 60 ) ) ≈ 2 × 10 57 − 2 × 10 − 3 (12)

We will be utilizing [

To whit the differences in Equations (9) and (10), will be such that we can now attend to the final piece of the puzzle which is primordial black holes being composed of Gravitons as Bose-Einstein condensates so the following is true, namely [

m ≈ M P N gravitons M B H ≈ N gravitons ⋅ M P R B H ≈ N gravitons ⋅ l P S B H ≈ k B ⋅ N gravitons T B H ≈ T P N gravitons (13)

If, so then we will have applying special relativity, and a further elaboration of Equation (13a) can be done as follows, namely [

m ≈ m g 1 − ( v g c ) 2 ≈ M P N gravitons ≈ 10 − 10 grams (14)

Given us, let us compare terms, and tell ourselves what we can expect if we have the variance in Equation (14) for BEC treatment of Gravitons.

We will first of all refer to an early universe treatment of the uncertainty principle is, in the startup of inflationary cosmology [

The value of time t will be set as t ~ (10^{−}^{32} s/t(Planck)) whereas we can utilize the ideas of having Planck time set ~ 5 × 10^{−}^{44} seconds, hence, t ~ 10^{12}, in Planck Units, whereas ℏ = G = l P = m P = k B = 1 , so then we will have, as reset as [

ρ ≈ γ ⋅ ( 10 24 ) 8 π + V 0 { 8 π ⋅ V 0 ⋅ ( 10 12 ) γ ⋅ ( 3 γ − 1 ) } 1 2 ⋅ ( γ π − 4 ⋅ π γ ) ≈ E effective ( 1000 km ) 3 ≈ E effective ( 6.25 × 10 40 ( l P = 1 ) ) 3 (15)

The interesting thing, is that the factor of roughly 10^-120 shows up in this situation so as to imply that there may be some linkage between setting the effective energy as roughly some proportional power value of Planck Mass [

10 2 m P ( gravitons ) ⇒ 10 67 gravitons per 10 5 m P ( blackhole ) (16)

Roughly put, one hydrogen atom is about 1.66 times 10^{−}^{24} grams. The weight of a Massive graviton is about 10^{−}^{65} grams [^{−}^{22} grams, or about 10^{44} gravitons, with each graviton about 6 × 10^{−32} eV/c^{2} after 10^{−}^{27} seconds, the following in the set of equations given below are Equivalent, and that these together will lead to a cosmological constant, Λ of the sort which we will be able to refer to later with 1 graviton ≈ 10 − 65 g . Then set to have a micro sized black hole of the size [

M P B H ( 10 − 43 ) ≈ 10 15 × ( 10 − 43 s 10 − 23 s ) g ≈ 10 − 5 g ≈ 1 M P (17)

Assuming that gravitons contribute to the Dark Energy value will lead to us using the Karen Freeze model, with gravitons being released in the early universe by the breakup of early universe black holes which have a maximum value of about 1 g, as opposed to the value of the Sun which has about 10^{33} grams, in first 10^{−32} seconds.

Let us first recall the Shalyt-Margolin and Tregubovich (2004, p.73) [

For sufficiently small γ . The above could be represented by [

Δ E ≈ ℏ Δ t 2 γ t P 2 ⋅ ( 1 ± ( 1 − 8 ℏ 2 γ t P 2 ( ℏ Δ t ) 2 ) ) ⇒ Δ E ≈ either ℏ Δ t 2 γ t P 2 ⋅ 8 ℏ 2 γ t P 2 ( ℏ Δ t ) 2 , or ℏ Δ t 2 γ t P 2 ⋅ ( 2 − 8 ℏ 2 γ t P 2 ( ℏ Δ t ) 2 ) (18)

Number of black holes, | Mass of black hole of size 1 Planck mass set aside for gravitons | Mass of black hole for 10^{8} gravitons | Radii of proto universe |
---|---|---|---|

10^{8} | 10^{−8} grams | 10^{−5} g = 1 Planck mass | 1000 Kilometers |

Volume of Universe is 10^{3} kilometers, cubed | Starting range for Mass of black hole for Gravitons | Assumed to be starting range of BH masses, at about 10^{−43} seconds | From less than a meter to 1000 Kilometers for constructing black holes which may be torn asunder by Karen Freeze’s criteria |

This would lead to a minimal relationship between change in E and change in time as represented by Equation (18), so that in the end we would write a limiting case as

Δ E Δ t ≈ 4 ℏ (19)

Having brought this up, let us then go to the Rosen [

T = ( ρ F / σ ) 1 / 4 ⋅ a ⌣ r 7 ( a ⌣ 4 + r 4 ) 2 (20)

With a ⌣ = 10 − 3 cm , ( ρ F / σ ) 1 / 4 = 1.574 × 10 32 K ( kelvin ) Then according to [

m ≈ 8 π R ( radius ) 3 ⋅ ρ 3 (21)

Here, the density function is given by Equation (15) and Equation (11), for our application and also we obtain for black holes a break up criteria for mass m Black holes if [

m ≈ − ( 4 π ρ 3 ) ⋅ ( 1 + 3 ⋅ ρ G p G ) ⋅ 8 m 3 M P 6 (22)

So we can have the start of breakup of black holes, if we have gravitons from 1/1000 of the mass of given black holes, and if black holes contribute DE according to when pressure is approximately equal to the negative value of the density which would lead to a Black hole contribution of Equation (23) to DE. As given below [

DE from black holes = 7 × 10 − 30 g / cm 3 = 7 × 10 − 6 g / ( 1000 km ) (23)

This rough value of DE, as given in Equation (23) will be directly compared to what we can expect as far as applying in Equation (23) as to comment directly on the Δ t time interval for the active generation of DE in the early cosmos.

Keep in mind that J. W. Moffat in [

This would be doable if the initial phases of creation of the universe follow [

E ( thermal energy ) = d ( dim ) ⋅ k B ⋅ T universe 2 (24)

The tie in, as can be stated is to make, after all the proximations, the following argument in [

Ψ ≈ Ψ initial ⋅ exp ( i m 2 c 2 ℏ 2 + R r r ⋅ r ) (25)

Whereas the approximation given is that we will be examining what if R r r is a constant. If so then, using the S.E. will be leading to an effective energy of

E e f f ≈ [ m c 2 2 + R r r ⋅ ℏ 2 2 m ] (26)

R r r ≈ 12 H initial 2 + 6 κ a min 2 → a min 2 ≈ 10 − 50 6 κ a min 2 , if κ ≠ 0 (27)

Where we are assuming there is a non zero flatness term.

This will be linkable to, if we are assuming temperature eventually appears,

E e f f ≈ [ k B d ( dim ) ⋅ T e f f 2 ] ≈ [ m c 2 2 + 3 ℏ 2 κ a min 2 m ] ≈ ℏ ⋅ ω effective ⇒ ω effective ≈ [ m c 2 2 ℏ + 3 ℏ κ a min 2 m ] ≈ [ k B d ( dim ) T e f f 2 ℏ ] (28)

The final step is to assume a non zero effective temperature we are in this situation directly using the idea of Lorenz invariance being broken in order to make sense of what Equation (28) is stating, we can also look at the following’ heuristic short hand, namely of Equations (13)-(14) being employed for the value of m above so as to have, say we are considering black holes of BEC configuration

m ≈ 2 ℏ ⋅ ω e f f c 2 ⋅ [ 1 ± 1 − 6 κ c 2 a min 2 ⋅ ω e f f 2 ] ≈ M P N g (29)

In such a situation, for gravitons being released from black holes due to a BEC condensate for black holes, as being destroyed by the Karen Freeze criteria, we must realize that a Lorentz violation would either tend to favor a very small release of gravitons being put into Equation (26) or a large value i.e., the Lorentz violation would then be favoring the term N g as being favored by the larger value as given in Equations (12)-(14). This leads to the following question to be raised. Assume we could release a breakup of primordial Black holes, do we have, say any conditions where.

After setting up the formulation of conditions for the creation of DE according to which we are looking at in the regime of space-time from less than a meter in radii to 1000 km in radii we have the formation of DE with the given frequency, for setting up DE, and then by default the Cosmological constant,

V ( ϕ ( t ) ) ≈ ω GravitongeneratedDE (30)

Is Equation (30) in any way commensurate with the Frequency stated as of Equation (28) and is this identification due to an early Universe Lorentz violation?

c ρ d S = ( R 00 − g 00 R ) 8 π | d S ⇔ ( − 3 a ¨ a − g 00 6 Λ ) 8 π | d S ≈ ( − 3 a ¨ a + g 00 6 ( a ¨ a + ( a ˙ a ) 2 + 2 κ a 2 ) ) 8 π | d S (31)

See this final take away as to what the cosmological constant is equivalent to, i.e. does the following make sense? If we multiply Equation (19) say by the cube of a Planck volume, and then can we work with the following derivation? One solution is as follows, [

Δ E Δ t ≈ ℏ ≡ ℏ ω Δ t ≈ ℏ ω ⋅ ( 2 3 a min ) 1 / γ ⇒ ω ≈ ℏ − 1 ⋅ ( 2 3 a min ) − 1 / γ (32)

The above is one such relationship. We need to reconcile it with Equation (32) in which we have the following provided that we have, at the point where κ = 0 ( flatness ) will lead to m not equal to

κ = 0 ( flatness ) ⇒ N g ≈ 10 50 / γ ( 3 / 2 ) 2 / γ (33)

m ≠ N g ⋅ M P if κ = 0 , but instead m ≈ M P N g (34)

What is set in Equation (19) is in line with also using

a ( t ) = a min t γ (35)

Leading to [

ϕ ≈ γ 4 π G ln { 8 π G γ ⋅ ( 3 γ − 1 ) ⋅ t } (36)

And what we will use later the “inflaton potential” we write as [

V = V 0 ⋅ { 8 π G V 0 γ ( 3 γ − 1 ) ⋅ t } γ 4 π G − 8 π G γ (37)

Furthermore, we have from applying a reference from the author, [

m ⋅ ∂ t ( a 2 − a 3 ) = 0 (38)

And then, a minimum time step we define via a minimum time step of

t = ( 2 3 a min ) 1 / γ (39)

Note that if the time as defined by Equation (39) is on the order of Planck time, i.e. 10^{−}^{44} seconds, we have then that γ ≈ 61 - 62 Given this value of 61 to 62 in gamma, it leads to having N g approximately of the value of 10 which is in line with, if m is in this case commensurate with Equation (14), i.e. an effective graviton mass. Furthermore gamma of the value of 61 or so would lead to an effective frequency 1.8549 × 10^{43} s^{−1} times 1/sqrt of 10. i.e. c/Planck Length times 1/sqrt of 10, which is normalized in Planck Units to be ~ 1/sqrt of 10.

In line with Equation (25), and normalized Planck units, this would correspond to 1.416784 (16) × 10^{32} K/sqrt 10, or about a breakup of Planck sized black holes at a temperature, due to Lorentz violation processes, early on of

T Plank BH Breakup ≈ T P / 10 ≡ 0.507 × 10 20 GeV (40)

Having this temperature for the energy is then equivalent, if we have a transition from curved to flat space to writing, if Gamma is ~61 and we have Planck time, after a Lorentz violating transformation from curved space to Flat space which we would write as

V = V 0 ⋅ { 8 π G V 0 γ ( 3 γ − 1 ) ⋅ t } γ 4 π G − 8 π G γ → l Planck ≡ ℏ ≡ G ≡ 1 V 0 ⋅ { 8 π V 0 γ ( 3 γ − 1 ) ⋅ t } γ 4 π − 8 π γ ≈ ω ∝ ( 2 3 a min ) − 1 / γ (41)

Which is in turn related to [

P G = γ ⋅ t 2 8 π G − V 0 ⋅ { 8 π G V 0 γ ⋅ ( 3 γ − 1 ) ⋅ t } γ 4 π G − 8 π G γ (42)

ρ B H = m p 4 32 π ⋅ ( m P m ) 4 ⋅ 1 | 1 + 3 P G ρ G | (43)

What we will be looking at would be having a glimpse of

P G ρ G ≈ γ ⋅ t 2 8 π G − V 0 ⋅ { 8 π G V 0 γ ⋅ ( 3 γ − 1 ) ⋅ t } γ 4 π G − 8 π G γ γ ⋅ t 2 8 π G + V 0 ⋅ { 8 π G V 0 γ ⋅ ( 3 γ − 1 ) ⋅ t } γ 4 π G − 8 π G γ (44)

Doing this would be at a minimum of having, a distance greater than or equal to

λ D E ≈ 10 30 l P (45)

Assume then we have, for the sake of argument

P G ρ G ≈ 1 − V 0 ⋅ { 8 π G 8 π G V 0 γ 2 ⋅ ( 3 γ − 1 ) } γ 4 π G − 8 π G γ t − 2 + γ 4 π G − 8 π G γ 1 + V 0 ⋅ { 8 π G 8 π G V 0 γ 2 ⋅ ( 3 γ − 1 ) } γ 4 π G − 8 π G γ t − 2 + γ 4 π G − 8 π G γ ≈ − 10 ζ (46)

The particulars of the coefficient of the right hand side of Equation (46), if ℏ = G = l P = t P = 1 , then if we set

γ 4 π − 2 γ 4 π − 1 = 0 (47)

So then we have if we wish to neutralize sensitivity to time itself at first approximation

γ = 4 π ⋅ ( 1 ± 2 ) 2 (48)

We can then look at the following

1 − V 0 ⋅ { 8 π 8 π V 0 16 π 2 ⋅ ( 1 ± 2 ) 4 ⋅ ( 12 π ⋅ ( 1 ± 2 ) 2 − 1 ) } ( 1 ± 2 ) − 2 ( 1 ± 2 ) 1 + V 0 ⋅ { 8 π 8 π V 0 16 π 2 ⋅ ( 1 ± 2 ) 4 ⋅ ( 12 π ⋅ ( 1 ± 2 ) 2 − 1 ) } ( 1 ± 2 ) − 2 ( 1 ± 2 ) ≈ − 10 ζ (49)

The above equation can be used, to locate appropriate values for V 0 in units where

ℏ = G = l P = t P = m P = c = 1

Given an approximate value for V 0 we will then proceed to come up with examining

ρ B H = m p 4 32 π ⋅ ( m P m ) 4 ⋅ 1 | 1 − 3 ⋅ 10 − ζ | M black hole primordial ~ 10 15 × ( t 10 − 23 s ) g ( grams ) (50)

The Lorentz symmetry breaking would be occurring at roughly a multiple of Planck time t, at a distance of say where we would be using we will first start off with the redone calculation as to the vacuum energy and how we rescale them to be in sync as to the observed value for vacuum energy which is of the present era. This methodology is consistent with the zero-point energy calculation, we start off with the following as given by [

1 2 ⋅ ∑ i ω i ≡ V ( volume ) ⋅ ∫ 0 λ ⌢ k 2 + m 2 k 2 d k 4 π 2 ≈ λ ⌢ 4 16 π 2 → λ ⌢ = M Planck ρ boson ≈ 2 × 10 71 GeV 4 ≈ 10 119 ⋅ ( ρ D E = Λ 8 π G ) (51)

In stating this we have to consider that ρ D E = Λ 8 π G ≈ ℏ ⋅ ( 2 π ) 4 λ D E 4 , so then that

the equation we have to consider is a wavelength λ D E ≈ 10 30 l Planck which is about 10^{30} times a Plank length radius of a space-time bubble. That would, we have after 10^{−}^{42} seconds

ρ D E = Λ 8 π G ≈ ℏ ⋅ ( 2 π ) 4 λ D E 4 (52)

We then have to consider how to reach the experimental conditions for when a nonsingular expansion point for Cosmology, will after 10^{−}^{42} seconds lead to Equation (4). That means a discussion of what Rosen and Israelit did in [^{30} expansion as to where we can at least measure the onset of DE.

Looking now at Rosen and Israelit, in terms of Thermodynamics of a non-singular universe [

A. We will be able to come up with an initial temperature of 10^{−}^{180} Kelvin, at a radius of about Planck length, in value. Almost absolute zero B. The temperature of space-time will be of the order of Planck Temperature after expansion of about 10^{30} times from the initial nonsingular configuration earliest phase, whereas we have the following Entropy value of [

S ~ 3 ⋅ [ 1.66 g ∗ ] 2 T 3 (53)

Here, the degrees of freedom term is defined in [

m inf 2 = d 2 V ( ϕ ) d ϕ 2 | ϕ = 0 , t = 3.9776 ⋅ t P = 16 π 3 π / 2 ⇒ m inf = 2.74635619187 ( Planckunits ) (54)

IMO the inflaton mass is 2.746356 times Planck Mass, and this is a starting value of inflaton mass att = 3.9776 Planck time. We should compare the inter relationship of this inflaton mass, in a volume of space with the results of Equation (50). The interrelationship of this inflaton mass, at a given time t ~ 3.997 Planck time, as compared Equation (50) for the breakup of black holes, and with the information given in Equations (8) and (9) of this document.

The author declares no conflicts of interest regarding the publication of this paper.

Beckwith, A. (2021) How Kieffer Density Matrix Formalism Aids Different Initial Time Steps, Leading to Lorentz Violations, and Breakup of Primordial Black Holes for GW Generation. Journal of High Energy Physics, Gravitation and Cosmology, 7, 1315-1327. https://doi.org/10.4236/jhepgc.2021.74081