### Infinite Speed and No-dimension Zone

Closer to (a = 90^{o}), particle cannot have any speed value other than infinity. Therefore, for our original assumption to hold (i.e., for Vx to remain fixed all along the wavelength), we have to assume that the wave-particle leaves the space-time and enters a no-time zone. With infinite speed (the assumed actual speed), the wave particle cannot remain in space-time. It has to leave and enter a no-dimension entity.

How are we to describe this format of wavelength? Where do we find a no-dimension zone for the particle to enter?

### A zone with no real dimension

Lorentz transformation for mass in motion** **is obtained by,

m = m_{0}/√ (1 - v ^{2}/c ^{2})

If the speed surpasses the speed of light (1 - v^{2}/c^{2) }turns negative. Then we are dealing with square root of a negative number which is an imaginary number. Here mass has to turn to an imaginary mass.

Mind you, space-time in this model is discrete, and therefore the singularity is in the neighbourhood and readily available to deportees of space-time. Besides, according to our previous assumption, the singularity has no dimension. We have also previously supposed that the singularity cannot contain mass. Keeping in mind the law of conversion of mass and energy, we can deduce that mass is converted into energy while joining the singularity.

On the other hand, according to the standard model of particles, subatomic particles are mass-less. The consensus is that Higgs boson adds the property of mass to them while particles accelerate in space-time (see section “Mass and Gravity“). This is another reason that particles should lose their mass while leaving space-time.To build a more objective argument, let us study particles with a propagation speed of 1/20 c. For the particle to reach to speed of light, the magnitude of a can be calculated as follows:

V = V _{x} √ (1+tang ^{2} a).

C = 1/20 c √ (1+ tang ^{2} a), and,

√ (1+ tang.^{2} a) = c / c/20, therefore,

20 = √ (1+ tang.^{2} a), then

400 = (1 + tang.^{2} a)

Thus tang.^{2} a = 399, hence tang. a = 19.98, which approximately corresponds to tangent of 87^{o},where the velocity reaches light speed.

#### Invariance Zone in the Proximity of the X-Axis

For this speed, we can redraw our diagrams as,

The equation V = V_{0} x √ (1 + tang.^{2} a) requires that as angle a increases beyond 87^{o}, further speed increase is not allowed. A particle with speeds higher than that c must be out of space-time. At a = 90^{o}, the speed of the particle-wave reaches infinity (tangent of 90^{o} = infinity).

We previously assumed that at the speed of light, mass must convert to energy. The amount of mass itself calculated from m = c^{2}/E. In addition, the momentum (p) of the wave-particle at this point is obtained using the following calculation: P = (E/c^{2}) × V

As the wave-particle departs from 87^{o} and approaches 90^{o} (i.e., hits the X-axis), tang.a will be infinity, so the speed will also be infinity. If V= ∞, then

P= (E/c^{2}) × V = (E/c^{2}) × ∞ = ∞

Momentum equals infinity and

E _{total} = √ (p^{2} c^{2} + m_{0}^{2} c^{4}) = ∞

It is therefore apparent that at the 90^{o} zone, the amount of energy has to be infinite.

**Mass and Schwarzschild radius**

In this scenario, as mass of the particle increases, it reaches to its Schwarzschild radius ratio. At this point the space-time is wrapped and a black hole appears. One can say that at this point, the particle joins the singularity.

Hawking radiation suggest that as particle leaves the black hole, it carries the information along. So it this senario particle reappears as the same particle.

#### Space around the X-Axis

If momentum is infinite at 90^{o}, then wave length disappears around the X-axis:

λ = h / p = h / ∞ = 0

And the wave number k turns to

k = 2 p/λ = 2 p/0 = ∞.

Δk = k- k0, Δk = ∞- k0, Δk = ∞

The momentum location relationship in this senario is in line with Born Reciprocity. In addition, we can write the Heisenberg uncertainty equation as follows: Δk × Δx ≥ h/2π

Therefore, Δk has an inverse relation with Δx: Δx ≥ 1 /Δk

If Δk equals infinity, then Δx can shrink to zero. In this case, there is no movement in space along the wavelength. We can interpret and speculate that the wave-particle does not encounter space at a = 87^{o} to 90^{o}. Alternatively, at the infinite energy point, we might express the following:

Ψ (E) = (E max) e^{i (kr^n - ω t)} = (E max) e^{i (kr^n - ω t)}

E = ∞ = (E max), then e^{i (kr^n - ω t)} = 1,

therefore i (kr^n - ω t) = 0

Or, k r^n =ωt, and since.

2 π/λ* r^n = 2 π/ T*t, hence, r^n = t* λ/T = 0

We can claim that there is no space at the 90^{o} . Then we can conclude the following:

Assertion WP2: Space disappears intermittently during the course of the particle’s journey along its wavelength.**Time around the X-Axis**

We write the Heisenberg uncertainty principle as follows:

Δt × ΔE ≥ h/2π

If ΔE = ∞, then Δt can reduce to zero, this demonstrates that time can remain put for wave-particle at.

a = 87^{o} to a = 90^{o}. ^{[4]}^{}

Assertion WP3: Time may stop intermittently for the objects during the course of their wave motion.

^{↑}