How will our universe eventually end?
In modern standard cosmology, scenarios such as the "Big Rip" (where dark energy infinitely expands and tears the universe apart), the "Big Crunch" (where the universe contracts and crushes itself), or the "Big Bounce" (a cycle of contraction and expansion) are widely debated.
However, the Intergeometric Theory derives a completely different, fourth doomsday scenario mathematically from pure geometry and the law of energy conservation.
It is the "Big Reset"—a scenario where the universe suddenly vanishes one day, and at that exact moment, a new universe begins.
Why will the universe vanish? Because the universe we live in is not a "firm reality," but merely a "Giant Virtual Particle" existing on energy borrowed in advance from the vacuum.
1. The Universe is Using the Vacuum's "Credit Line"
According to quantum mechanics, even in an empty vacuum, "virtual particles" can be created by borrowing energy for a very short time, due to "Heisenberg's Uncertainty Principle."
In conventional cosmology, it is believed that after the universe was born as a virtual particle from quantum fluctuations, it underwent a vacuum phase transition (falling from a high energy state to a lower one). The released potential energy transformed the universe into "real particles," granting it eternal life.
However, calculations in this theory show that the universe never materialized through a phase transition. It is still a virtual particle carrying debt.
The definitive proof lies in the observational data of the "Higgs vacuum expectation value (ν)," which gives mass to elementary particles.
Originally, the empty vacuum had a "credit line" allowing it to borrow energy equivalent to two Higgs masses ( \(2M_H\) ). However, current observations show that this value has decreased, as described by the following equation:
\[\nu=2M_H-\left(\frac{3}{81}\right)\chi_0M_H\]
The latter half of this equation indicates that "because someone has already borrowed energy from the vacuum, the remaining available credit line has decreased." The borrower is none other than our universe itself.
Here, \(chi_0\) represents the "remaining balance ratio" of the debt the universe holds (in the early universe, \(\chi_0=100\%\) ).
So, how much total energy is the entire universe borrowing?
The reference energy density of the universe (standard critical density \(\rho_s\) ) is expressed by the following equation:
\[\rho_s=\frac{3}{81}\frac{M_H}{\frac{4\pi}{3}\left(Wl_H\right)^3}\]
- "3": The total energy of the universe (Matter is 1, Dark Energy is 2. See [another article] for details).
- "1/81": The dilution of energy due to inflation (spatial hierarchization).
- Denominator: The spherical volume of space stretched by a factor of W due to inflation.
(Note: When this standard critical density \(rho_s\) is converted to its current value, it perfectly matches actual observational data. For details, see the [Cosmological Constant Problem article].)
Although the density of the universe is calculated "per volume," the total amount of energy generating this density perfectly matches the "decrease in vacuum expectation value" ( \(3/81M_H\) ) mentioned earlier.
Astonishingly, our universe did not materialize through a phase transition; even at this very moment, it remains a "giant virtual particle" continuously borrowing energy from the vacuum.
2. Calculating the Lifespan of the Universe (Standard Hubble Time \(t_s\))
As long as the universe is a virtual particle with debt, it cannot escape the absolute rule of the "Uncertainty Principle" ( \(\Delta E\Delta t\sim\hbar\)). The product of the total borrowed energy and the time it can exist (lifespan) cannot exceed the limit of the Planck constant.
So, how much lifespan does our universe have left?
By substituting the energy density of the universe (standard critical density (\rho_s\) ) into Einstein's Friedmann equations, we find the "Standard Hubble Constant (expansion rate) \(H_s\) ," the baseline at which the universe maintains its most beautiful balance.
\[H_s=\sqrt{\frac{8\pi G}{3}\rho_s}\]
The reciprocal of this value is the reference time scale of the universe, the "Standard Hubble Time \(t_s\) ."
\[t_s=1/H_s=14.437\text{ Gyr (approx. 14.4 billion years)}\]
(Note: This standard Hubble time \(t_s\) serves as the baseline value for solving various mysteries of the universe, such as the Hubble Tension.)
Why can we definitively call this 14.437 Gyr the "lifespan of the universe"?
In the microscopic world, the time \(t_H\) a virtual Higgs particle can exist is expressed by the uncertainty principle as:
\[t_H=\frac{\hbar}{M_H c^2}\]
On the other hand, when we geometrically expand the macroscopic standard Hubble time \(t_s\) we just calculated, astoundingly, it transforms into the following equation with the exact same structure:
\[t_s=\frac{\hbar}{(l_H/l_s)M_H c^2}\]
(Note: \(l_H\) is the Higgs length, and \(l_s\) is the standard Hubble length obtained by multiplying \(t_s\) by the speed of light.)
What does this mean?
The universe never transformed into "real particles." The rapid expansion of space called inflation stretched the scale (applying the coefficient \(l_H/l_s\) ), and this merely extended its lifespan as a virtual particle from a microscopic instant to 14.4 billion years.
