Let us use the principle of conservation of momentum applied to the universe which says that a force F applied during a time Δt will move a mass m_u (the apparent mass of the universe) by increasing its speed by Δv. The luminous universe is expanding at the speed of light in vacuum c. Therefore, we can consider that in Equation (1), Δv = c.

F Δ t = m u Δ v → F = m u Δ v Δ t → F = m u c Δ t (1)

Let us suppose that the universe is expanding for a time Δt equal to the apparent age of the universe T_u. This value is given by the inverse of the Hubble constant which is about H 0 ≈ 72.1   km ⋅ s − 1 ⋅ MParsec − 1 [5] [9] . Let us note that 1 MParsec = 3.085677581 ´ 10²² m.

Δ t = T u = 1 H 0 ≈ 13.65 × 10 9   years (2)

If we look at the universe as a whole, and if we use Newton’s universal attraction equation to calculate the force F that the universe of apparent radius R_u applies on its own mass m_u, we get Equation (3). The constant G is the universal gravitational constant.

F = G m u 2 R u 2 (3)

In Equation (3), the value of R_u is the apparent radius of curvature of the universe.

R u = c T u = c H 0 ≈ 1.28 × 10 26 m (4)

Let us equate the forces F from Equation (1) and (3). Once we simplify and use Equation (2) and Equation (4), we get Equation (5).

m u = c 3 G H 0 ≈ 1.73 × 10 53 kg (5)

We get an equation which is the same as Carvalho [10] .

Let us use the principle of conservation of Energy to find the apparent mass m_u of the universe. Initially, at the Big Bang, there was no movement yet. All the energy contained in the universe was contained in the mass m_u of the universe. The total amount of energy E contained in the universe is given by Einstein formula.

E = m u c 2 (6)

Since the universe is expanding, the initial energy is converted to potential energy through a gravitational force F applied on a distance R_u (see Equation (4)) from the center of mass of the universe. Let us use Newton’s law.

F = G m 2 R u 2 → E = F R u = G m 2 R u (7)

By equating Equation (6) and Equation (7), and using Equation (4), we get Equation (8).

m u c 2 = G m u 2 R u where R u = c H 0 (8)

Isolating the apparent mass of the universe m_u from Equation (8), we get Equation (9).

m u = c 2 R u G = c 3 G H 0 ≈ 1.73 × 10 53 kg (9)

One will notice that this is the same equation as Equation (5).

Let us calculate the apparent mass of the universe m_u by using the Planck mass m_p. By definition, the Plank mass is defined by Equation (10) where h is the Planck constant, c is the speed of light, and G is the universal gravitational constant.

m p = h c 3 2 π G ≈ 2.18 × 10 − 8 kg (10)

Let us define m_ph as being the mass associated with the lowest energy photon in the universe [5] . When we look at the energy of a wavelength λ, the energy is at its lowest level when λ is the largest. The largest dimension of the universe is its apparent circumference. Therefore, we can associate a mass m_ph to a photon of wavelength λ = 2πR_u where R_u is the apparent radius of the universe.

m p h = h c λ c 2 = h c 2 π R u c 2 = h H 0 2 π c 2 ≈ 2.74 × 10 − 69 kg (11)

The Plank mass m_p is the geometric average between the smallest mass m_ph associated with the lowest energy photon and the biggest mass which is without any doubt the apparent mass of the universe m_u. From this fact, we get the following Equation.

m p = m p h m u ≈ 2.17 × 10 − 8 kg (12)

Equating Equations (12) and (10), and using Equation (11), we get m_u..

m u = c 3 G H 0 ≈ 1.73 × 10 53 kg (13)

Again, we get, without any surprise, the same equation as Equation (5).

We will calculate the apparent mass m_u of the universe by equating the gravitational energy of a photon with the mass-energy contained in a photon (being a corpuscle) [5] . Let us associate a mass m_ph (like in Equation (11)) with a photon of the lowest energy [5] that is at the periphery of the luminous universe (with a wavelength λ equal to the apparent circumference of the universe λ = 2πR_u). Then, if we place this photon at the periphery of the luminous universe, it will have an E_g gravitational energy.

E g = G m u m p h R u (14)

According to the special relativity, the mass-energy associated with this photon is E_m.

E m = m p h c 2 (15)

By equating Equations (14) and (15), replacing R_u with Equation (4), and isolating m_u, we get the same equation as Equation (5).

m u = c 3 G H 0 ≈ 1.73 × 10 53 kg (16)

Recently, with a new cosmological model, the precise values of the universal gravitational constant G and of the Hubble constant H₀ have been found [5] as a function of the classical electron radius r_e, the mass of the electron m_e, the fine-structure constant α, and β (see Equation (19)). Let us use Equations (17) and (18) to evaluate precisely m_u (for metrology purposes).

G = c 2 r e α 20 m e β ≈ 6.673229809 ( 86 ) × 10 − 11 m 3 ⋅ kg − 1 ⋅ s − 2 (17)

H 0 = c α 19 β r e ≈ 72.09548580 ( 32 )   km ⋅ s − 1 ⋅ MParsec − 1 (18)

In these two equations, β is defined as the ratio between the expansion speed of the material universe and the expansion speed of the luminous universe. According to our model, the material universe is embedded in a luminous universe, both being spherical and expanding with a speed proportional to their radius [5] .

β = 3 − 5 ≈ 0.76 (19)

with his special relativity theory, Einstein showed that if we accelerate any mass at rest m₀ to a speed v, its energy E would increase because of the Lorentz factor (the square root in Equation (20)) [11] . If v tends towards c, the energy E would tend towards infinity, which is impossible since it cannot get more energy than there is available in the universe. Therefore, v must be slower than c. It also implies that the expansion speed of the material universe is slower (explaining β in Equation (19)) than its luminous counterpart which is expanding at the speed of light (which is c for now). In Equation (20), we show what would happen if we expand Einstein’s formula in a series. The first term of the series is the energy at rest and the second one is the kinetic energy which is used in Newton’s classical mechanics [12] .

E = m c 2 = m 0 c 2 1 − v 2 c 2 ≈ m 0 c 2 + 1 2 m 0 v 2 + 3 8 m 0 v 4 c 2 + ⋯ with     v < c (20)

With Equation (17) and Equation (18), we modify Equation (5) to get Equation (21).

m u = m e β α 39 = 1.728098528 ( 26 ) × 10 53 kg (21)

This equation is one of the most precise of all since it relies on very well-known constants defined in the CODATA 2014 [8] . It also highlights the fact that there are very strong links between the universe’s dimensions and its constituents as such as the electron.

The electron charge q_e may be described as a function of the electron mass m_e, the electron radius r_e and the relative permeability of vacuum ε₀ [5] .

q e = − 4 π r e m e μ 0 ≈ − 1.60 × 10 − 19 C (22)

If we isolate the electron mass m_e from Equation (22) and if we put its value in Equation (21), we get an equation that gives the apparent mass of the universe as a function of the electron charge q_e. Its precision is equivalent to Equation (21).

m u = μ 0 q e 2 β 4 π r e α 39 = 1.728098528 ( 26 ) × 10 53 kg (23)

Let us use the definition of the Rydberg constant R_¥ as a function of the fine-structure constant α and the classical electron radius r_e to find the apparent mass of the universe.

R ∞ = α 3 4 π r e ≈ 1.10 × 10 7 m − 1 (24)

If we replace the classical electron radius r_e from Equation (23) with Equation (24), we get Equation (25) that gives the apparent mass of the universe m_u as a function of the Rydberg R_¥ constant and the electron charge q_e.

m u = μ 0 q e 2 R ∞ β α 42 = 1.728098528 ( 27 ) × 10 53 kg (25)

The classical way to calculate the apparent radius of curvature of the universe R_u consist in saying that the speed of light in vacuum c was constant and traveled a time-lapse equal to an age of the universe T_u which is a function of H₀ (see Equation (2)).

R u = c T u = c H 0 ≈ 1.28 × 10 26 m (26)

According to different sources, H₀ is between 67.8(9) km×s⁻¹×MParsec⁻¹ [15] and 77.6 − 4.3 + 4.8 km×s⁻¹×MParsec⁻¹ [16] . Most measurements of the Hubble constant H₀ rely on inaccurate methods such as the observation of stars. Uncertainties from different measurement results do not always overlap. For better accuracy, we will use H₀ from our Equation (18). It is compatible with Salvatelli, with H 0 ≈ 72.1 − 2.3 + 3.2   km ⋅ s − 1 ⋅ MParsec − 1 [9] .

Let us use Equation (18) in Equation (26) to evaluate precisely (for metrology purposes) the apparent radius of curvature of the universe R_u.

R u = r e α 19 β ≈ 1.2831078845 ( 57 ) × 10 26 m (27)

Let us define the apparent radius of curvature R_u as a function of the Rydberg constant R_¥. Let us isolate r_e from Equation (22) and use it in Equation (27). We get Equation (28) which gives the apparent radius of curvature of the universe R_u as a function of the electron charge q_e and of the electron mass m_e.

R u = q e 2 μ 0 4 π m e α 19 β ≈ 1.283107889 ( 13 ) × 10 26 m (28)

Like Equation (28) which uses well-known parameters from CODATA and β constant which can be calculated very accurately, we want to define the apparent radius of curvature R_u as a function of the Rydberg constant R_¥. Let us rewrite Equation (27) with Equation (24) to get the most precise way, up to now, to calculate the apparent radius of the universe R_u.

R u = 1 4 π R ∞ α 16 β ≈ 1.2831078902 ( 48 ) × 10 26 m (29)

In 1929, Edwin Powell Hubble found that galaxies distance themselves from one another at a speed proportional to the distance between them [4] . He deduced a law involving a constant called H₀. It represents the average recession speed v of galaxies per unit of distance Δr. Let us note that galaxies have their own freedom of movements. Some may move closer to each other and some will move away from each other. But overall, they will move away from each other because of the inflating movement of the universe.

v Δ r = 1 Δ t = H 0 = 72.1   km ⋅ MParsec − 1 ⋅ s − 1 (30)

As seen in Equation (18), H 0 ≈ 72.1   km ⋅ MParsec − 1 ⋅ s − 1 . Let us isolate Δt.

Δ t = 1 H 0 ≈ 13.65 × 10 9 years (31)

NASA is currently evaluating the age of the universe with Equation (31) [17] . This way of calculating the age assumes that the expansion rate of the universe is constant.

According to the model of Friedmann-Robertson-Walker [22] , the true age of the universe would be around 2/3 of Equation (31).

Δ t = 2 3 ⋅ H 0 ≈ 9.04 × 10 9 years (32)

The speed of the light in vacuum and the expansion speed of the material universe may have not been constant over time [5] . Our goal here is to calculate the age of the universe (of complex type) by simulating a return to the past by doing the integral of the inverse of the expansion speed v_m(r) of the material universe.

We will use some results of the work cited in [5] to solve the integral of the inverse of the speed v_m(r) of expansion of the material universe as a function of the radius of curvature of the universe. In this way, we will calculate the age of the universe Δt_u.

As mentioned before, the general theory of relativity predicts that the presence of a massive body changes near space-time and increases the index of refraction n(r) (which changes as a function of the distance r from the center of mass of the massive body) of the vacuum around a mass [25] . By moving away from that mass, the gravitational influence is being reduced and the speed of light tends toward c.

We apply the same principle to the universe which is certainly the biggest existing mass. Since the universe is expanding [4] , we move away from a certain center of mass and the density of the universe becomes lower over time. Like before, this causes the refractive index to diminish over time and lets the speed of light slowly increase over time tending toward an asymptotic speed that we named k [5] . Of course, the current speed of light in vacuum is c. To keep the total energy of the universe constant meanwhile the speed of light increases, the mass of the universe must slowly decrease over time. The asymptotic speed of light k (when r ® ¥) is given by Equation (33).

k = c 2 + 5 ≈ 2 c ≈ 6.17 × 10 8 m ⋅ s − 1 (33)

As soon as we try to calculate the speed of light for the time in the past or in the future, we must take into account that the speed of light v_L(r) changes as a function of the apparent radius of curvature r of the universe [5] . The value of Q is the gravitational potential for the universe and n(r) is the refractive index of the universe as a function of r.

v L ( r ) = k n ( r ) where     n ( r ) = 1 − 2 Θ / k 2 1 + 2 Θ / k 2       and       Θ = − G m u r ≤ 0 (34)

In this equation, there is a radius of curvature r = r_h for which the speed of light v_L(r_h) = 0. This position r_h is called the horizon of the universe. This is the position for which the denominator of the square root of Equation (34) becomes zero. In a similar way, in a black hole, the radius of curvature of the horizon r_h is obtained by the Schwarzschild’s radius where we replace c by k:

r h = 2 G m u k 2 ≈ 6.06 × 10 25 m (35)

It is the same principle as that of a black hole. In fact, the universe is the biggest existing black hole since it has the biggest mass.

Let us note that for a conventional black hole, its entire mass is located in its center of mass. However, for the universe, a big part of the black hole mass lays outside the boundaries of the horizon. The center of mass of the universe coincides with the center of mass of the black hole.

The expansion speed of the universe is currently the speed of light c [5] . Based on the principles of relativity, the matter cannot move at the speed of light without having infinite energy. Consequently, the previous assertion about the expansion of the universe can be true only for light (which we call the luminous universe). The material universe (containing the galaxies, intergalactic dust clouds, etc.) is expanding at a slower speed equal to βc. The factor β, must by necessity be less than 1 since we cannot surpass the speed of light which represents a speed boundary. According to our Equation (19), its value is about 0.76. The apparent radius of curvature r_u of the material universe is, therefore, a portion β of the apparent radius of curvature R_u of the luminous universe [5] .

r u = β R u ≈ 9.80 × 10 25 m (36)

The expansion speed v_m(r) of the material universe is β times the speed of light v_L(r) because matter must travel slower than light [11] .

v m ( r ) = β v L ( r ) = β k n ( r ) where     n ( r ) = 1 − 2 Θ / k 2 1 + 2 Θ / k 2       and       Θ = − G m u r ≤ 0 (37)

If we take the derivative of the expansion speed v_m(r) of the material universe with respect to the distance r, we get the Hubble constant H₀ [5] .

d v m ( r ) d r | r = r u = k β y r u ( 1 ( 1 + y ) 1 − y 2 ) = H 0 where       y = 2 G m u k 2 r u (38)

It is important to realize that we do not use the derivative of the expansion speed of the luminous universe to get H₀ since astronomers cannot observe that limit. Through their telescopes, they only see objects like stars and galaxies. Consequently, when Hubble defined its constant H₀, it was based on the derivative of the expansion speed v_m(r) of the material universe.

At the periphery of the luminous universe (at a distance r = R_u from the center of mass of the universe), light accelerates at an a_L(R_u) = cH₀ rhythm.

a L ( R u ) = d v L ( r ) d t | r = R u = ( d r d t ⋅ d v L d r ) | r = R u = c ( d v L d r ) | r = R u = c H 0 (39)

However, locally, at a distance r = r_u from the center of mass of the universe, light slowly accelerates at an a_L(r_u) = cH₀/β rhythm.

a L ( r u ) = d v L ( r ) d t | r = r u = ( d r d t ⋅ d v L d r ) | r = r u = c ( d v L d r ) | r = r u = c H 0 β (40)

Locally, at a distance r = r_u from the center of mass of the universe, matter from the material universe slowly accelerates at an a_m(r_u) = cH₀ rhythm.

a m ( r u ) = β d v L ( r ) d t | r = r u = β ( d r d t ⋅ d v L d r ) | r = r u = β c ( d v L d r ) | r = r u = β c H 0 β = c H 0 (41)

Figure 1 shows the module of the expansion speed v_m(r) of the material universe calculated from Equation (37). It also puts in evidence the Hubble constant H₀ as the slope evaluated precisely at the position r = r_u from the center of mass of the universe (at r = 0). Between r = 0 (at the Big Bang) and r = r_h (horizon of the universe), the dotted part of the curve shows that the expansion speed of the material universe would normally be negative and of imaginary type. Then, from r = r_h and infinite, the expansion speed of the material universe becomes of real type.

The negative and imaginary value of the speed v_m(r) we get with r < r_h is a mathematical way to show that in a physical world, matter inside the horizon evolves independently than outside the horizon. Here, we cannot see what is happening inside the horizon and vice versa. We could compare it to the cylindrical envelope of an optical fiber. Light can travel inside a fiber from one extremity to the other end, but we cannot see through the envelope (if the fiber is used in its curving limits). It is therefore impossible to know what is happening from outside the fiber and vice versa. Of course, the principles in cause in fiber are not the same. We only bring this example to do an image of the situation happening inside the horizon of a black hole.

Performing the integral of the inverse of v_m(r) with respect to the radius of curvature r, it is possible to calculate precisely the age of the universe more precisely (in its entirety with the real part and its imaginary part) than by using a single tangential projection. In Figure 1, the slope of the tangential projection gives the Hubble constant H₀ which can be put in Equation (31) to give the apparent age of the universe.

Let’s find the age of the universe Δt_u(r) by performing the integral of Equation (42) between the center of mass of the universe (at r = 0) and the apparent radius of curvature of the material universe r_u. The resulting value Δt_u(r) is of a complex type.

Δ t u ( t ) = ∫ 0 r u 1 v m ( r ) d r = Δ t h u ( r ) + Δ t 0 h ( r ) (42)

In Equation (42), the value of Δt_hu(r) represents the time elapsed between the horizon and the actual age of the universe (see Equation (43)). The resulting value is of real type.

Δ t h u ( r ) = ∫ r h r u 1 v m ( r ) d r (43)

In Equation (42), the value of Δt_0h(r) represents the time elapsed between the Big Bang and the horizon (see Equation (44)). The resulting value is of an imaginary type.

Δ t 0 h ( r ) = ∫ 0 r h 1 v m ( r ) d r (44)

Performing the integral calculation of Equation (42), we get Equation (45).

∫ 1 v m ( r ) d r = ( z ( r ) + 2 G ⋅ m u ln ( 2 [ k 2 r + z ( r ) ] ) ) β k 3 where     z ( r ) = k 4 r 2 − 4 G 2 m u 2 (45)

Consequently, the value of Δt_u(r) becomes:

Δ t u ( r ) = Δ t h u ( r ) + Δ t 0 h ( r ) ≈ ( 9.50 + 10.47 i ) × 10 9 yearswhere     i = − 1 (46)

This result is of a complex type. In Equation (46), the first part of the integral (shown in Equation (43)) is of a real type (between r_h and r_u). However, the second part of this latter one (shown in Equation (44)) is of an imaginary type (between 0 and r_h).

If we look carefully to Equation (37), for a radius smaller than the horizon r_h, the speed of light v_L(r_h) becomes of an imaginary type. For this reason, the time T_0h become of an imaginary type as well. This mathematical situation just means that time inside the horizon evolves in a completely independent way compared to time outside the horizon, which is why no one can see what happens inside the limits of the horizon of a black hole. The best example we could give is within a fiber optic cable. How light evolves inside the fiber cannot be seen from outside the fiber, and vice versa. Mathematics are a nice language which has to be interpreted to find sense in the real world.

When we wish to consider the time elapsed between position 0 of the Big Bang and the radius of curvature of the horizon r_h, we must calculate the module of the time elapsed çΔt_uç. We define this value as the apparent age T_u of the universe because it does not necessarily represent the true age of the universe. This number represents only an apparent age in the likely event that the Big Bang existed. It could also be seen as the time that elapsed if the speed of light had been constant over time since the Big Bang.

T u = | Δ t u | = | Δ h u + Δ t 0 h | = Δ t h u 2 + Δ t 0 h 2 ≈ 14.14 × 10 9 years (47)

We see that the value is only 4.25% over the value of Equation (31).

As for the calculation of power in electrical motors (with the real power, the inductive power and the apparent power), the age of the universe may be seen as follows: the “real” part of the age of the universe, the “imaginary” part of the age of the universe and the “apparent” age of the universe. The module of the two components (real and imaginary) can be calculated using the Pythagorean Theorem by finding the square root of the sum of the squares of the real part and the imaginary part of the age of the universe.

The approximation of the age of the universe will be made in 3 parts: the approximation of the real part of the age of the universe, the approximation of the imaginary part of the age of the universe and the calculation of the module of the apparent age of the universe. In Figure 2, we show the parallelogram built from these values.

Let’s perform the approximation of the real part Δt_hu of the age of the universe Δt_u.

For a radius of curvature r_h of the horizon, the square root in Equation (48) equals zero.

z ( r h ) = k 4 r h 2 − 4 G 2 m u 2 = 0 (48)

So, according to Equation (42) and Equation (45), we get Equation (49).

Δ t h u = z ( r u ) + 2 ⋅ G ⋅ m u ⋅ ln ( 2 ⋅ [ k 2 ⋅ r u + z ( r u ) ] ) β ⋅ k 3 − 2 ⋅ G ⋅ m u ⋅ ln ( 2 ⋅ k 2 ⋅ r h ) β ⋅ k 3 where       z ( r u ) = k 4 ⋅ r u 2 − 4 ⋅ G 2 ⋅ m u 2 (49)

Using Equation (35) with a few simplifications, we get Equation (50).

Δ t h u = 1 β ⋅ k ( r u 2 − r h 2 + r h ln ( r u 2 − r h 2 + r u r h ) ) (50)

Without changing anything, this same equation could be rewritten as follows.

Δ t h u = r u 2 ⋅ β ⋅ c ⋅ ( 1 + β 2 ) ⋅ y ≈ r u 2 ⋅ β ⋅ c ⋅ ( 1 + β 2 ) where     y = [ 2 ⋅ c k ⋅ r u ⋅ ( 1 + β 2 ) ⋅ ( r u 2 − r h 2 + r h ln ( r u 2 − r h 2 + r u r h ) ) ] ≈ 1 (51)

As shown, the content of the bracket is approximately equal to 1. By doing this approximation and using Equation (36), Equation (51) becomes Equation (52).

Δ t h u ≈ 1 4 H 0 ( 2 + β ) (52)

Further, in this document, we will use Equation (52) to perform the calculation of the apparent age of the universe. For now, let us show that this equation can be approximated to get an equation commonly used by some astronomers to calculate the actual age of the universe.

According to Equation (19), the value of β ≈ 0.76. Let’s use this approximation to rewrite Equation (52). After a few simplifications, we get:

Δ t h u ≈ 1 4 H 0 ( 2 + 76 100 ) ≈ 1 H 0 ( 69 100 ) ≈ 2 3 H 0 ≈ 9.04 × 10 9 years (53)

This last ratio can be deduced from the model of Friedmann-Robertson-Walker [22] . Therefore, Equations (52) and (53) represent good approximations of the real part of the age of the universe.

Now, let’s find the approximate value of the imaginary part T_0h of the age of the universe T_u. From Equation (42), Equation (45) and Equation (48) we get:

Δ t 0 h = [ 2 ⋅ G ⋅ m u ⋅ ln ( 2 ⋅ k 2 ⋅ r u ) β ⋅ k 3 ] − [ − 4 ⋅ G 2 ⋅ m u 2 + 2 ⋅ G ⋅ m u ⋅ ln ( 2 ⋅ − 4 ⋅ G 2 ⋅ m u 2 ) β ⋅ k 3 ] (54)

Using Equation (35) and performing a few simplifications, we get Equation (55).

Δ t 0 h = r h β ⋅ k ⋅ [ ln ( − i ) − i ] where i = − 1 (55)

Using the relationship in Equation (56), we rewrite Equation (55), which is the result of a purely imaginary type, to get Equation (57).

ln ( − i ) = − π 2 ⋅ i (56)

Δ t 0 h = − r h β ⋅ k ⋅ [ 1 + π 2 ] ⋅ i (57)

We can rewrite Equation (57) in the following way without changing anything:

Δ t 0 h = − r u 2 ⋅ β ⋅ c ⋅ ( 1 − β 2 ) ⋅ ( 1 + π 2 ) ⋅ x ⋅ i where     x = [ 2 ⋅ c k ⋅ r h r u ⋅ ( 1 − β 2 ) ] ≈ 1 (58)

As shown, the value of x is approximately 1. Equation (58) then becomes:

Δ t 0 h ≈ − 1 2 ⋅ H 0 ⋅ ( 1 − β 2 ) ⋅ ( 1 + π 2 ) ⋅ i (59)

Let’s calculate the apparent age of the universe by using Equation (47) with Equation (52) and Equation (59). After a few simplifications, we get Equation (60).

| Δ t u | = | Δ t h u + Δ t 0 h | = Δ t h u 2 + Δ t 0 h 2 ≈ 1 H 0 χ ≈ 1 H 0 where       χ = ( 1 2 ⋅ ( 1 + β 2 ) 2 + ( ( 1 − β 2 ) ⋅ ( 1 + π 2 ) ) 2 ) ≈ 1 (60)

The value of χ is approximately equal to 1. Consequently, we have shown that the integral of Equation (42) can be approximated by Equation (31). According to us, based on the approximation calculation made at Equation (60), Equation (31) represents only an apparent age T_u of the universe. In fact, it comes from the calculation of the module of a complex sum of the real part and the imaginary part of the age of the universe. We define T_u as the apparent age of the universe.

T u = | Δ t u | ≈ 1 H 0 ≈ 13.65 × 10 9 years (61)