Aeromash

https://www.scienceforums.net/topic/125748-доказательство-гипотезы-больших-чисел-дирака-proof-of-the-hypothesis-of-large-dirac-numbers/?do=findComment&comment=1186586

[math]E \sim T^{\circ}[/math] Energy is proportional to temperature. Which means that the temperature has the same degree as the energy in the system L. [math]L^{-11}=T^{\circ}[/math]

A good question takes a long time to answer. I am preparing a general table of the transition of the one-dimensional laws of the system L to the SI system.

A car moves at different speeds relative to the Earth's surface, but its speed does not exceed the square root of the gravitational potential at the Earth's surface. In this matter, it is a gravitationally bound object, which is one with the Earth. Unlike Tesla's car, which is part of the solar system.

Yes! Using unit analysis gives you proportionalities. But using unit analysis gives me equalities in L and LT systems. For the transition to the SI system need the gravitational constant G or, in electrodynamics, epsilon . Newton knew nothing about the constant G. Mass in his era had dimensions: [math]\frac{m^{3}}{s^{2}}[/math] In general, this is terrible, but the SI system is a mixture of the LMT, LT and L systems. Therefore, thinking that we are working in SI, in fact, we use the LT or L system. When we work in SI, then coefficients of the G or epsilon type appear in the formulas right away. When you calculate the volume in SI by multiplying the area by the height, you are actually working in the L system. If you added mass to the LT system, that is, switched to the LMT system, then the constant G immediately appears in the formulas. The larger the dimension of the system, the more different coefficients(constants) you need! This is what we observe in modern physics. System L does not need coefficients! This is its beauty. Vector if the radius is a vector. [math]\vec{r}[/math]

[math]\frac{\varphi^{2}}{G} =\frac{G^{2}M^{2}}{Gr^{2}}=F[/math] G is the constant of the transition from the LT system to SI.

You are talking about the speed of the car. And not about the square of the speed or the gravitational potential of the car. You are confusing physical quantities.

Thanks! My mistake! [math]L^{-12}=L^{-5}L^{-7}=F=qE[/math] [math]L^{-5}[/math] charge. [math]L^{-7}[/math] electric field strength. [math]L^{-6}[/math] Рotential.

[math]F=\frac{c^{4}}{G}[/math] Force in the SI system. [math]\varphi=v^{2}=G\frac{m}{r}[/math] The square of speed is the gravitational potential. Because we originally defined the system L in such a way that L^1 - is radius or line, L^2 - square or r^2, L^3 - volume or r^3, L^4 - time or period. Then we calculate charge q in L sistem. Its dimension turned out to be L^{-5}, like a mass m. The capacitance of the capacitor is: [math]C=L^{1}=L^{2}L^{-1}=\frac{L^{2}}{L^{1}}=\frac{S}{R}[/math] Where S is the area of the capacitor plates, R is the distance between the plates.

How it works. Let's say I don't know physics. But I know the correspondence of physical quantities for the system of dimension L. But I want to get the law that connects force and charge. Then we write in the system L. [math]L^{-12}=L^{-6}L^{-5}[/math] Then we look at the corresponding dimensions of physical quantities for the system L. [math]L^{-6}=E[/math] Electric potential. [math]L^{-5}=q[/math] Electric charge q. Then we get an equation relating charge to force. [math]F=qE[/math] Let's say I don't know Coulomb's law. But I can get it elementarily. [math]L^{-12}=L^{-5}L^{-5}L^{-2}[/math] [math]F=\frac{q\cdot q}{r^{2}}[/math] Let's say I don't know electrical laws in physics. But I can get these laws instantly. [math]L^{-5}=L^{1}L^{-6}[/math] Convert to normal view. [math]q=C \varphi[/math]

Does it mean not to continue?

[math]m=L^{-5}=L^{3}L^{-8}=V \rho[/math] [math]m=L^{-5}=L^{2}L^{-7}=R^{2}a[/math] [math]m=L^{-5}=L^{1}L^{-6}=R \varphi[/math] [math]Q=L^{-5}=L^{4}L^{-9}=It[/math] [math]m=L^{-5}=L^{6}L^{-11}=\frac{E}{\varphi} = \frac{L^{-11}}{(L^{-3})^{2}}=\frac{E}{v^{2}} [/math] [math]m=L^{-5}=L^{7}L^{-12}=\frac{F}{a}[/math] [math]Q=L^{-5}=L^{1}L^{-6}=CU[/math] To be continued...

Reducing everything to the system of dimensions L, we got the opportunity to obtain analytically any known and unknown equations in physics! I forgot to add Newton's law of gravity: [math]L^{-12}=\frac{L^{-5}L^{-5}}{L^{2}}=\frac{mM}{r^2}[/math] Let's write down the equations of mass or charge in a one-dimensional system L.

Thus, we get a one-dimensional system A, in which all basic physical quantities are expressed. Length or capacity C in electric: [math]L^1[/math] Square S: [math]L^2[/math] Volume V or magnetic permeability or electrical resistance R: [math]L^3[/math] Time, period or resistivity: [math]L^4[/math] specific heat: [math]L^5[/math] Inductance L: [math]L^6[/math] Magnetic flux Ф: [math]L^{-2}[/math] to be continued... Linear velocity or concentration: [math]L^{-3}[/math] Frequency, magnetic induction, conductivity: [math]L^{-4}[/math] Mass m or electric charge Q: [math]L^{-5}[/math] Gravitational potential or electric potential: [math]L^{-6}[\math] Linear acceleration a, field strength of magnetic, electric, gravitational or Planck's constant: [math]L^{7}[/math] Density: [math]L^{-8}[/math] Electric current I: [math]L^{-9}[/math] Viscosity: [math]L^{-10}[/math] Work A, energy E, amount of heat Q, temperature T: [math]L^{-11}[/math] Force F: [math]L^{-12}[/math] Surface tension: [math]L^{-13}[/math] Pressure P: [math]L^{-14}[/math] Power: [math]L^{-15}[/math] Enough for now. Let's look at the equation of force: [math]F=ma=L^{-12}=L^{-5}L^{-7}[/math] We can write it in various ways: [math]F=L^{-12}=L^{-5}L^{-7}=L^{-1}L^{-11}=L^{{-6}^2}=L^{2}L^{-14}[/math] And if we make a reverse transition to the LMT system. Then we get: [math]F=ma=\frac{E}{r}=\varphi^{2}=PS[/math] I think you already understood what this is about?

Just a little bit left, please wait. And you will see everything for yourself. Acceleration in the L system has the dimension [math]a=\frac{L}{T^2}=L^{-7}[/math] The gravitational potential is [math]\varphi=L^{-6}[/math] Energy is [math]E=m\cdot \varphi= L^{-11}[/math] Thus, we can convert all known physical quantities into a one-dimensional system in which each physical quantity has its own axis. Force F in system L is [math]F=ma=L^{-5}L^{-7}=L^{-12}[/math] Thus, we get a one-dimensional system A, in which all basic physical quantities are expressed. [math]L^1 \qquad length \qquad \qquad L^-1[/math] [math]L^2 \qquad square \qquad \qquad L^-1 \qquad magnetic flux \Phi [/math]