SergUpstart Posted September 10, 2021 Share Posted September 10, 2021 9 minutes ago, Aeromash said: Dirac meant the electron. But as it turned out, Dirac's formula works on any scale. The electron does not have a certain radius and certain coordinates. An electron is not a ball. This follows from the Heisenberg uncertainty principle. Maybe you meant the radius of the orbit of an electron in a hydrogen atom??? But the electron does not have a clear orbit in the atom, at present physicists are talking not about orbits, but about orbitals. Link to comment Share on other sites More sharing options...

swansont Posted September 10, 2021 Share Posted September 10, 2021 7 minutes ago, Aeromash said: I do not know what's happening. But it doesn't work again. I will try to find another way to insert formulas. Dirac meant the electron. But as it turned out, Dirac's formula works on any scale. 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. Link to comment Share on other sites More sharing options...

Aeromash Posted September 10, 2021 Author Share Posted September 10, 2021 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. Link to comment Share on other sites More sharing options...

swansont Posted September 10, 2021 Share Posted September 10, 2021 1 hour ago, Aeromash said: 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'm not sure how you draw a conclusion using a number that has no physical meaning, and is basically used only for convenience. Link to comment Share on other sites More sharing options...

Aeromash Posted September 13, 2021 Author Share Posted September 13, 2021 (edited) 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] Edited September 13, 2021 by Aeromash Link to comment Share on other sites More sharing options...

Aeromash Posted September 13, 2021 Author Share Posted September 13, 2021 (edited) 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... Edited September 13, 2021 by Aeromash Link to comment Share on other sites More sharing options...

SergUpstart Posted September 13, 2021 Share Posted September 13, 2021 (edited) On 9/10/2021 at 4:14 PM, Aeromash said: 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. 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. 1 hour ago, Aeromash said: Visitors and astronomers know that the square of speed is the gravitational potential. And the square of the gravitational potential is the force. 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). Edited September 13, 2021 by SergUpstart Link to comment Share on other sites More sharing options...

Aeromash Posted September 13, 2021 Author Share Posted September 13, 2021 (edited) 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... Edited September 13, 2021 by Aeromash Link to comment Share on other sites More sharing options...

swansont Posted September 13, 2021 Share Posted September 13, 2021 2 hours ago, Aeromash said: Visitors and astronomers know that the square of speed is the gravitational potential. And the square of the gravitational potential is the force. Neither of these are true. And none of your post addresses the use of the classical electron radius. Link to comment Share on other sites More sharing options...

Aeromash Posted September 13, 2021 Author Share Posted September 13, 2021 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. Link to comment Share on other sites More sharing options...

swansont Posted September 13, 2021 Share Posted September 13, 2021 3 minutes ago, Aeromash said: Because the classical radius of an electron has nothing to do with the real Universe. But you said: 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. If you make the comparison, you are using the classical electron radius Also the radius of the observable universe is not constant, so any comparison to constants that one deems meaningful has to just be accidental, since that ratio changes in time. Link to comment Share on other sites More sharing options...

Aeromash Posted September 13, 2021 Author Share Posted September 13, 2021 (edited) [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] Edited September 13, 2021 by Aeromash Link to comment Share on other sites More sharing options...

swansont Posted September 13, 2021 Share Posted September 13, 2021 9 minutes ago, Aeromash said: v=LT=LL4=L−3 Time does not have dimensions of L^4 You can't just make stuff up Link to comment Share on other sites More sharing options...

Aeromash Posted September 13, 2021 Author Share Posted September 13, 2021 (edited) 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] Edited September 13, 2021 by Aeromash Link to comment Share on other sites More sharing options...

Aeromash Posted September 13, 2021 Author Share Posted September 13, 2021 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? Link to comment Share on other sites More sharing options...

Aeromash Posted September 13, 2021 Author Share Posted September 13, 2021 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. Link to comment Share on other sites More sharing options...

Bufofrog Posted September 13, 2021 Share Posted September 13, 2021 32 minutes ago, Aeromash said: 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: L−12=L−5L−5L2=mMr2 Let's write down the equations of mass or charge in a one-dimensional system L. This is all just so much gibberish. Link to comment Share on other sites More sharing options...

Aeromash Posted September 13, 2021 Author Share Posted September 13, 2021 (edited) [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... Edited September 13, 2021 by Aeromash Link to comment Share on other sites More sharing options...

swansont Posted September 13, 2021 Share Posted September 13, 2021 1 hour ago, Aeromash said: To be continued... Please, no. I think everybody understands that you can make a self-consistent substitution of independent units. It's precisely because you can't validly substitute length for charge because they describe different things (they are, in a sense, orthogonal) that you can make this swap without a contradiction in unit consistency. Everybody can be named Bruce. So what? Where's the science? Where is your model that can be used to predict behavior? Link to comment Share on other sites More sharing options...

Aeromash Posted September 14, 2021 Author Share Posted September 14, 2021 17 hours ago, swansont said: Please, no. Does it mean not to continue? Link to comment Share on other sites More sharing options...

swansont Posted September 14, 2021 Share Posted September 14, 2021 21 minutes ago, Aeromash said: Does it mean not to continue? Don’t continue posting unit conversions. Link to comment Share on other sites More sharing options...

Aeromash Posted September 14, 2021 Author Share Posted September 14, 2021 (edited) 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] Edited September 14, 2021 by Aeromash Link to comment Share on other sites More sharing options...

swansont Posted September 14, 2021 Share Posted September 14, 2021 20 minutes ago, Aeromash said: Let's say I don't know Coulomb's law. But I can get it elementarily. L−12=L−5L−5L−2 F=q⋅qr2 How do you know that charge isn't 1/L^4 and the force drops off as 1/r^4? IOW if it's ad-hoc, then there's no science in it. 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." Link to comment Share on other sites More sharing options...

Aeromash Posted September 14, 2021 Author Share Posted September 14, 2021 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. Link to comment Share on other sites More sharing options...

studiot Posted September 14, 2021 Share Posted September 14, 2021 1 hour ago, Aeromash said: 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. L-^{12} does not equal L^{-6}L^{-5} Link to comment Share on other sites More sharing options...

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