Why \(\mathbf{E=mc^2}\)?

 Why \(\mathbf{E=mc^2}\)?

by

Mircea Bidian

October 26, 2024

An important detail often overlooked is the origin of \(c^2\) in the energy equation \(E=mc^2\). This factor is generally accepted as given, traditionally considered a postulate, without further inquiry into its origins, implying an effect without a cause, which is impossible. Knowing Planck's constant \(h\), we can determine the mass of a quantum as: \[m=\frac{h}{c^2}=7.37249732 \times 10^{-51} \, \text{[kg]}\] Using the equation below, we can find the Event Horizon radius on a quantum level, denoted \(L\), which has the following value:\[L = \frac{G \cdot M}{c^2} = \frac{6.67128190 \times 10^{-11} \cdot 7.37249732 \times 10^{-51}}{(299792458)^2} = 5.47245892 \times 10^{-78} \, \text{[m]}\] The time \(T\) required to travel a distance equal to the radius \(L\) at the speed of light is: \[T=\frac{L}{c}=\frac{5.47245892 \times 10^{-78}}{299792458}=1.82541581 \times 10^{-86} \, \text{[s]}\] where \(T\) is in seconds, \(L\) is the radius in meters, and \(c\) is the speed of light in meters per second. The speed of æther at the Event Horizon equals the speed of light, and it is reasonable to assume that the angle of incidence for this speed is 1 radian. This assumption leads to an equality between the centrifugal force and gravitational force. The centrifugal force equation is:\[{F} = {m\omega^2 r} = \frac{m v^2}{r}\] Equating the gravitational and centrifugal forces, we have: \[\frac{G M m}{r^2} = \frac{m v^2}{r}\] When \(v\) equals \(c\) at the Event Horizon, this yields: \[r=\frac{G M}{v^2}=\frac{G M}{c^2}\] If the Event Horizon circumference is the length \(C_{EH}\) of a circle with radius \(L\), the spin time \(T_s\) required for light to cover this circumference will be \(2\pi T\): \[T_s=\frac{2\pi L}{c}=\frac{2\pi L}{\frac{L}{T}}=2\pi T=1.146942579362 \times 10^{-85} \, \text{s}\] The angular velocity \(\omega\) of the quantum is: \[\omega = \frac{2\pi}{T_s} = \frac{2\pi}{2\pi T} = \frac{1}{T} = 5.478203896 \times 10^{85} \, \text{s}^{-1}\] Next, we compute the angular momentum \(S\) of the quantum: \[S = I \omega = M L^2 \frac{1}{T} = \frac{M L^2}{T}\] Angular momentum, denoted here by \(S\), is the product of an object’s moment of inertia \(I\) and angular velocity \(\omega\). This results in: \[S = \frac{M L^2}{T} = \frac{7.37249732 \times 10^{-51} \cdot (5.47245892 \times 10^{-78})^2}{1.82541581 \times 10^{-86}} = 1.20953332 \times 10^{-119} \, \text{[kg⋅m⋅s}^{-1}\text{]}\] This angular momentum is a constant, denoted here as \(\Omega\): \[\Omega = 1.20953332 \times 10^{-119} \, \text{[kg⋅m}^{2}\text{s}^{-1}\text{]}\] With the quantum period, we can also determine its frequency \(f_\Omega\): \[f_\Omega=\frac{1}{T_s}=\frac{1}{2\pi T}=\frac{1}{2 \cdot \pi \cdot 1.82541581 \times 10^{-86}}=8.718832 \times 10^{84} \, \text{[Hz]}\] Planck's energy \(h\) is expressed as: \[h = 2\pi f \cdot \omega \cdot M \cdot L^2 = \omega^2 \cdot M \cdot L^2 = \frac{M L^2}{T^2} = 6.62607015 \times 10^{-34} \, \text{[kg⋅m}^2 \text{s}^{-2}\text{]}\] The angular momentum \(S\) remains as constant \(\Omega\), defined as the ratio of the product of mass and the square of radius over time. Linear momentum \(p\) of the quantum, as the product of mass and speed, is given by: \[p = m \cdot v = m \cdot c = \frac{M L}{T}\] With numeric values: \[p = \frac{7.37249732 \times 10^{-51} \cdot 5.47245892 \times 10^{-78}}{1.82541581 \times 10^{-86}} = 2.210219094 \times 10^{-42} \, \text{[kg⋅m⋅s}^{-1}\text{]}\] This linear momentum \(p\) is a constant, comparable to De Broglie’s photon momentum. In the final energy equation: \[E = h = 2 \cdot \pi \cdot f \cdot \omega \cdot M \cdot L^2 = m \cdot c^2\] Since \(m\) is known, we examine \(c^2\): \[c^2 = 2 \cdot \pi \cdot f \cdot \omega \cdot L^2\] This gives us a clearer understanding, as follows: \[c^2 = \frac{L^2}{T^2} \, \text{[m}^2 \text{s}^{-2}\text{]}\] The origin of \(c^2\) is thus defined by the ratio of the quantum radius squared \(L^2\) to time squared \(T^2\), elucidating much of the mystery and relating to angular momentum conservation. Notably, \(L^2\) is composed of a segment of the Event Horizon's circumference and the quantum radius, and \(T^2\) accounts for the quantum’s frequency in radians per second, confirming the structure of energy in this framework.

The mystery being largely clarified, we can’t help but wonder what exactly \(\frac{L^2}{T^2}\) represents. To explore this, we return to the initial statement that the total sum of angular moments is the cause of momentum conservation. As we can observe from the angular momentum relation, \[S = M \cdot c \cdot L\] we have the quantum’s speed at the Event Horizon represented by \(c\), given by the ratio \(\frac{L}{T}\), and distance \(L\), which is the radius or distance from the axis of rotation. Since the distance on the Event Horizon equals the quantum radius, the angle is 1 radian. Now we understand that \(L^2\) comprises a segment of the Event Horizon’s circumference and the quantum radius. \[S = M \cdot L \cdot \frac{L}{T}\] We are left to discover the nature of \(T^2\). From Planck’s constant definition, we have: \[h = 2 \cdot \pi \cdot f \cdot S = S \cdot \frac{1}{T}\] where \(\frac{1}{T}\) represents the number of angular moments or radians, given as: \[n_S = \frac{1}{T} = 5.47820389 \times 10^{85} \, \text{[s}^{-1}\text{]}\]

Thus, we have identified the roles of \(L^2\) and \(T^2\) in the energy equation.