Aeromash

35 minutes ago, studiot said:

What is n ?

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

17 hours ago, studiot said:

Note at the bottom of the table the comment about temperature.

I did ask and you have not said how your L system handles temperature.

[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]

4 hours ago, studiot said:

Perhaps showing one simple example developed from start to fininsh, but involving as many as possible of the seven basic quantities in the LMTAθIN system.

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.

17 hours ago, swansont said:

A car moving on the surface of the earth can have a wide range of speeds without changing its gravitational potential.

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.

14 hours ago, swansont said:

Using unit analysis gives you proportionalities, at best. Not equalities.

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.

16 hours ago, SergUpstart said:

The force is a vector, not a scalar

Vector if the radius is a vector.

[math]\vec{r}[/math]

49 minutes ago, swansont said:

If you square GM/r you get G²M²/r² which is not the force (GmM/r²)

[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.

45 minutes ago, swansont said:

A car moving on the surface of the earth can have a wide range of speeds without changing its gravitational potential.

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.

40 minutes ago, studiot said:

L-¹² does not equal L^-6L^-5

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.

10 minutes ago, swansont said:

Using only dimensional analysis leads you to draw incorrect conclusions, such as "the square of speed is the gravitational potential. And the square of the gravitational potential is the force."

[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.

19 minutes ago, swansont said:

How do you know that charge isn't 1/L^4 and the force drops off as 1/r^4?

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]

17 hours ago, swansont said:

Please, no.

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]

[math]v=\frac{L}{T}=\frac{L}{L^4}=L^{-3}[/math]

Let us calculate the mass in the one-dimensional system L.

In the LT system, the mass is determined by the dimension

[math]m=\frac{L^3}{T^2}[/math]

Then in the one-dimensional system, the mass is equal to.

[math]m=\frac{L^3}{T^2}=\frac{L^3}{{L^{4}}^2}=\frac{L^3}{L^8}=L^{-5}[/math]

1 hour ago, swansont said:

Neither of these are true.

And none of your post addresses the use of the classical electron radius.

Because the classical radius of an electron has nothing to do with the real Universe.

Let's continue. Thus, we have defined 4 physical quantities. Length, area, volume and time.

Let us calculate the speed in the one-dimensional system L.

1 hour ago, SergUpstart said:

Well, then it follows simply from the definition of the gravitational radius R=GM/c^2 that the ratio of the gravitational radius of the universe to the gravitational radius of an electron is equal to the ratio of their masses.

But the mass of the electron is about 10^-30 kg, and the mass of the Universe is about 10^56 kg. So if we use gravitational radii, their ratio will be different.

Please go to the first post on the link to my book. In it, the main task was to find the exact value of the mass and radius of the Universe through fundamental constants.

The mass and radius of the Universe are superconstants that determine the numerical values of all other constants. Therefore, if we know the values, for example, the speed of light and the Milgrom constant or the Hubble constant, then we can calculate the mass and radius of the Universe.

My values of the mass of the Universe

[math]M_{u}=9.985 \cdot 10^{53}kg[/math]

and its radius

[math]R_{u}=7.414 \cdot 10^{26}m[/math]

1 hour ago, SergUpstart said:

Not so, the field strength is the gradient from the potential with a minus sign. And the force is the field strength multiplied by the mass of the test body ( if we are talking about the gravitational force in the Newton paradigm) or by the test charge (if we are talking about the electrostatic force).

The minus in the field gradient is accepted conventionally. Based on the assumption that the field decreases at infinity. In our case, the field grows with distance from the test body. See rotation curves of galaxies.

The fourth power of speed divided by G is force.

[math]\frac{c^4}{G}=F_{u}[/math]

or

[math]\frac{v^4}{G}=F[/math]

So. Let's make the transition from the LT system to the L system. I managed to make the transition from the SI system to the LMT system and then to the LT system myself. I found the transition to the L system in Mikhail Sokolnikov.

Let be

[math] L^0 \Rightarrow point[/math]

[math]L^1 \Rightarrow line \qquad or \qquad radius[/math]

[math]L^2 \Rightarrow square \qquad or \qquad r^2[/math]

[math]L^3 \Rightarrow volume \qquad or \qquad r^3[/math]

[math]L^4 \Rightarrow time[/math]

To be continued...

Let's rewrite the Dirac proportion in the LT system.

[math]m\sim r^2 \sim v^4 \Rightarrow \frac{L^3}{T^2} \sim \frac{L^2}{T^0} \sim \frac{L^4}{T^4}[/math]

And let's swap the members in order of ascending L.

[math]\frac{L^3}{T^2} \sim \frac{L^2}{T^0} \sim \frac{L^4}{T^4}\Rightarrow \frac{L^2}{T^0} \sim \frac{L^3}{T^2} \sim  \frac{L^4}{T^4}[/math]

Having passed to the LT system, we see how the terms are formed in the Dirac proportion.

Any member of the Dirac proportion can be expressed in one expression!

[math]\frac{L^{n+2}}{T^{n^{2}}}[/math]

As you can see, we could not fully express the Dirac ratio in the SI system of dimensions. But this became possible with the transition to the LMT system. Thus, we see that in the transition to the LT system, the physical laws become more "transparent".

Thus, the transition in physics to the system of L.

In this case, the new physical property will be determined by the degree with dimension L.

To be continued...

On 9/10/2021 at 5:42 PM, swansont said:

I'm not sure how you draw a conclusion using a number that has no physical meaning, and is basically used only for convenience.

The number matters. For example. If we transform the expression.

[math]\frac{m}{M_{u}}=\frac{r^2}{R_{u}^2}=\frac{v^4}{c^4}\Rightarrow m=\frac{M_{u}}{c^4}v^4=const\cdot v^4[/math]

Gives us the dynamic Milgrom equation from MOND theory for the Tully-Fisher relation for spiral galaxies!

Good. We well understand what the physical meaning of speed has!

[math]v[/math]
And what is the physical meaning of the square of speed?

[math]v^2[/math]
And what is the physical meaning of the fourth degree of speed?

[math]v^4[/math]

What is hidden behind these values?

Speed to the power n is no longer speed, but an absolutely different physical quantity.

[math]v^n[/math]

Visitors and astronomers know that the square of speed is the gravitational potential. And the square of the gravitational potential is the force.

[math]v^2=\varphi[/math]

[math]v^4=F[/math]

Thus, the Dirac(Tulli-Fisher) ratio 

[math]m \sim v^4[/math] 

 is encrypted

[math]m \sim F[/math]

40 minutes ago, SergUpstart said:

I briefly read the text that is indicated by the link in the first post of this topic. What I would like to note is that there is no mention of the costant alpha=1/137 in the text, and this is the most important physical constant.

There are a huge number of constants in physics. Which ones are the main and which are the minor ones? This answer does not depend on the opinion of the person. The Universe must answer what constants are fundamental. Have you seen the law relating fundamental constants in my book? In this law there is a constant the speed of light, Milgrom's constant, Hubble's constant, charge constant. There are constants unknown to us. But in this law there is no alpha and there is no Planck's constant. Which suggests that these constants do not belong to the main series of fundamental constants, or they are secondary or derived from fundamental constants.

7 minutes ago, swansont said:

Did Dirac use the classical electron radius, which we know to be incorrect? Once QM had been developed a bit, it was recognized that the electron (as with all fundamental particles) is a point particle.

You are absolutely right. Dirac used the classic radius. But he did not pretend to be accurate. He only noticed that the ratio of the classical radius to the radius of the Universe gives a value of 10 ^ 40. The same value is given by the ratio of the roots of the masses of the electron and the Universe. r -radius must be gravitational.

I do not know what's happening. But it doesn't work again. I will try to find another way to insert formulas.

15 minutes ago, SergUpstart said:

What do you think is the smallest particle and what is its radius???

Dirac meant the electron. But as it turned out, Dirac's formula works on any scale.

Дирак чуть раньше чем я заметил, что отношение массы самой мелкой частицы во Вселенной к массе Вселенной дает такое же по порядку число, что и отношение квадратов их радиусов. Если быть точным, то корни отношений масс дают тот же порядок, что и отношение радиусов. Т.е.:

Dirac a little earlier than I noticed that the ratio of the mass of the smallest particle in the Universe to the mass of the Universe gives the same order of magnitude as the ratio of the squares of their radii. To be precise, the roots of the mass ratios give the same order as the ratio of the radii. Those.:

 [math]\frac {\sqrt{m}}{\sqrt{M_{u}}}=\frac{r}{R_{u}}[/math]

or so

[math]\frac {m}{M_{u}}=\frac{r^2}{R_{u}^2}[/math]

Я же, путем построения собственной вселенной получил выражение:

I, by building my own universe, received the expression:

[math]\frac {m}{M_{u}}=\frac{r^2}{R_{u}^2}=\frac{v^4}{c^4}[/math]

И уже после публикации книги:

And after the publication of the book:

[math]\frac{1}{D_{m}}=\frac {m}{M_{u}}=\frac{r^2}{R_{u}^2}=\frac{v^4}{c^4}=z^4[/math]

Где Dm - большое число Дирака, показывающее какое число объектов массы m или радиуса r может содержаться во вселенной. z -классическое космологическое красное смещение, Mu- масса Вселенной, Ru- радиус Вселенной, m - масса гравитационно связанного объекта, r - его гравитационный радиус, v - линейная скорость его движения, вращения.
Продолжение следует....

Where Dm is a large Dirac number, showing how many objects of mass m or radius r can be contained in the universe. z is the classical cosmological redshift, Mu is the mass of the Universe, Ru is the radius of the Universe, m is the mass of a gravitationally bound object, r is its gravitational radius, v is the linear velocity of its motion or rotation.
To be continued ......

31 minutes ago, studiot said:

[math]{E^2} = {\left( {m{c^2}} \right)^2} + p{c^2}[/math]

Thank you for your comment, but the formula you indicated belongs to Einstein's theory of gravity, which on the scale of galaxies gives an error in mass of about 25%, and on the scale of the Universe the error is 70%.

9 minutes ago, studiot said:

There should be no space between [math] and \frac

It did not help.

43 minutes ago, studiot said:

I await the derivation of this with interest.

You will be amazed at the simplicity of this formula. The one who created our universe was a genius.

40 minutes ago, SergUpstart said:

This formula does not describe everything.

36 minutes ago, studiot said:

[math]\frac{{\sqrt m }}{{\sqrt {{M_u}} }} = \frac{r}{{{R_u}}}[/math]

There should be no space between [math] and \frac

I removed the space, but it didn't help.

Dirac a little earlier than I noticed that the ratio of the mass of the smallest particle in the Universe to the mass of the Universe gives the same order of magnitude as the ratio of the squares of their radii. To be precise, the roots of the mass ratios give the same order as the ratio of the radii. Those.:

 [math]\frac {\sqrt{m}}{\sqrt{M_{u}}}=\frac{r}{R_{u}}[/math]