### Wave Function in this Model

### Mass, Energy-Information Phase Transition

As the speed increases so does the particle’s momentum and energy. Quantum field theory predicts that when the energy of the particle is enough, a new set of particles is created. However, we need to re-normalize our theory and ignore the extra particles. Here I will leave the energy increase until the speed of the particle hits speed of light. There I will assume that the particle disappear from space-time. This way we do not need to re-normalize along the way. Let us see if this assumption has some merit.

We can make the following argument for an electron. Einstein’s special theory of relativity says that nothing can travel at or above the speed of light in our space-time universe. The reason behind this is the relation between mass and its speed,

m = m_{0}/√ (1-v^{2}/c^{2})

Where m is the relativistic mass (mass in motion), m_{0} is the mass of an object at rest, v is velocity of the particle relative to the observer, and c is the speed of light. Acceleration augments the particle’s mass. At the speed of light, v^{2}/c^{2} equals 1. In this case, m_{0} will be divided by 0.

So mass will increase to infinity.

m = m_{0}/ √ (1-v^{2}/c^{2})

m = m_{0}/ √ (1- 1)

m = m_{0}/ 0 = ∞

By definition, infinity does not exist in space-time universe; besides, we would need an infinite amount of energy to move this infinite mass, which would extend beyond physics as we know it. Therefore, it has been concluded that nothing can travel at or faster than the speed of light. Here again we are faced with the notion of infinity, a notion that most physicists detest. To me this is another road sign indicating that we need to strive for a deeper understanding of reality. We must not ignore these signs. John Earman writes, “In 1960, Einstein found that, in a condition of static equilibrium, as the radius of the cluster approaches Schwarzschild radius, the particle would eventually have to move faster than light.”^{[3]}

Pilot wave model suggests that during the course of its wavelength, while traveling along a spatial dimension (X-axis) with linear speed v, the particle oscillates in another direction (the Z direction) as well.

The above diagram shows a wave-particle, which propagates along X-axis. We take the propagation speed along the X-axis uniform or constant. To satisfy the uniformity of speed along the X-axis, the particle has to accelerate in its wavelike motion along the Z-axis. The actual velocity of particle varies according to its position along the curve.

In other words, the actual speed will exceed the propagation speed as the particle leaves the peak and travels down the slope. The equations can be written as follows:

Δz/ Δx = tang. a

Δz = Δx tang. a

Δz/ Δt = tang. a Δx/ Δt If

lim Δt = 0,

Δz/ Δt = tang. a Δx/ Δt

In this situation we may write Δt gets near 0, therefore

Vz = tang. a Vx

Please note that in the above diagram, the angles a and a’ are equal because their sides are perpendicular to each other. Therefore, the actual speed of a particle at any point can be obtained by the following calculation:

V^{2} = Vz^{2} + Vx^{2} = Vx^{2}+ Vx^{2} tang.^{2} a = Vx^{2} (1+tang.^{2} a), Then;

V = Vx √ (1+tang.^{2}a) … (1) ^{3}

Regardless of propagation speed, as the particle approaches the X-axis there is a moment when its speed equals and exceeds the speed of light. For example, at an 61^{o} angle the tangent value is 1.802 that doubles the speed and from there it speeds up much faster .Therefore, there will be a time for any particle to reach the light speed as it travels toward the X-axis following its pilot wave regardless of its linear speed.

3) .Gozin

Somewhere around X-axis (a = 90^{o}), particles are in infinite speed zone.

V = Vx √ (1+tang.^{2}a) (1)

V = Vx √ (1+ ∞)

V = ∞

Since infinity is not defined in space-time physics, we would normally say that the speed equals maximum. But because this model includes singularity, in which infinite energy is allowed, I have chosen to leave infinity in place.

^{↑}