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Аbout relativistic mass, Special theory of relativity, Physics

Запись от IGPIGP размещена 13.11.2019 в 00:03
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Аbout relativistic inertional mass
There is shown a way to get an equation of Lorenz transformation for inertial mass relatively of speed. The respective an equation for gravitational mass is shown there:
https://www.cyberforum.ru/blog... g6564.html
Math is a tool. When it takes a leading role, it easily turns everything it touches into mathematics. And mathematics don't really care about nature. Making assumptions, they are interested in what follows from this more than in how adequately the assumption itself describes the picture. Here the reverse logic works easily - "If nature does not fit into the model, we will tweak it."
Relativistic mass is a consequence of the Lorentz transformations resulting from the absolutization of the speed of light. Moreover, Lorentz unveiled his transformation as a kind of joke. He perfectly understood the problems that the absolutization of the speed of light hides and did not regard his transformations as a real theory. Rather, he tried to say what could follow from the absolutization of the speed of light. That is why priority is given to Einstein, who boldly declared: "This is not idiocy at all! This is the truth." People love this and everyone is immediately imbued. The ether, which could not be found, left the open stage and became the subject of scholastic research.
In Einstein's work, the expression of relativistic mass is obtained through the complex enough logic. A pair of photons and the Doppler effect veil the fact of complete equivalence of inertial and gravitational masses. However, Einstein did not deny such equivalence. Followers did it for him. However, electromagnetic phenomena like quantum ones (Planck's formula e = hW) are not needed to obtain a relativistic transformation with respect to mass. Worse than that. This transformation is easily obtained from the Lorentz transformation with respect to speed and raises the question - what to do with it? This is where tricks with the split of masses begin - gravimass and inert. And even to the negation of the relativistic mass as such it came. Otherwise, the foundation of physics is collapsing. Let's first get the conversion relative to mass. This transformation can be safely addressed to Lorentz. My efforts here are minimal. Although in truth, in this form you will not find them, most likely not where. I have not found. So. Suppose we have a pair of bodies with masses m = m moving towards each other in the c.m. system with speeds Vc = Vc directed to the cm. Let's move to the system of one of the bodies. Let name center of mass as cm.
Denote the speed cm by Vc. The speed of the incident body in the cm system is also Vc. But the speed of the incident body in the observer system is according to Lorentz:
https://www.cyberforum.ru/cgi-bin/latex.cgi? v=\frac{2\cdot v _{c}}{1+\frac{{v _{c}}^{2}}{C^{2}}}
The speed of the center of mass is found from a simple quadratic equation. One of the roots is always higher than the speed of light and it is considered meaningless from a physical point of view (and that is also strange as it is not complex number or even the negative one… but ok - let it be), and one that makes sense leads to:
https://www.cyberforum.ru/cgi-bin/latex.cgi?v_{c }=\frac{C^{2}}{v}\cdot (1-\sqrt{1-\frac{v^{2}}{C^{2}}})
or more compact:
https://www.cyberforum.ru/cgi-bin/latex.cgi?v_{c }=C\cdot \sqrt{\frac{\gamma _{v}-1}{\gamma _{v}+1}}
Now we write the law of conservation of momentum in the following form. The observer understands that since the classical addition of speeds does not work, then most likely the mass of the incident body should seem to him to be different from m (the mass of the incident body, measured in his own system, i.e., rest). He prudently denotes this mass through M - (the mass of the twin incident body, which is measured by the observer or at least seeming to him from his system) and tries to record the impulse of the incident body. It is equal to https://www.cyberforum.ru/cgi-bin/latex.cgi?M\cdot v . On the other hand, in terms of an incident system, a system of mass m + M flies at the observer (here he flies to himself :-) but this is the property of cm) at a speed that we recently received. After simple transformations we get:
https://www.cyberforum.ru/cgi-bin/latex.cgi?M =m\cdot \frac{1+\sqrt{\frac{\gamma _{v}-1}{\gamma _{v}+1}}}{1-\sqrt{\frac{\gamma _{v}-1}{\gamma _{v}+1}}}
a little cumbersome but easily leads to a formula from Einstein:
https://www.cyberforum.ru/cgi-bin/latex.cgi?M =\gamma \cdot m _{0}
obtained together with famous formula of mass-energy through a pair of photons, Planck equation for kvant energy, Doppler effect and using Energy and Momentum conservation laws. As can be seen, the Momentum conservation law and the Lorentz transformation for speed known as relativistic law of speed addition, are quite enough. This means that the result is true both for the inertial center of mass and for the gravitational one. If the former is indisputable by the method of obtaining the formula, then the latter is easily obtained from contradictions from the non-coincidence of one and the second (as well as non-coincidence i/ g masses). It is easy to draw a line perpendicular to the line of approach of the bodies through the cm and to place on it a heavy observer who interacts symmetrically with both bodies. He’ll not notice any asymmetry leading to a rotation of the system relative to his system. But the observers which are moving with bodies m = m do not think so . They should not see a difference in the forces of gravity. Because. The inertia (laziness) of motion in a gravitational field, equivalent to the curvature of space, is built precisely on the fact that the "inertial resistance" (total force) and gravitational force are proportional to the same mass.
This gives rise to an insoluble contradiction with nowaday physics wich has rejected of relativistic mass as real physical a value which can be measured and observed. Please help me to resolve that a situation. Those who is not ready argue in presented terms please do not argue at all. I by my own can say something like: "The special theory of relativity is a holy knowledge to encroach on which is inconceivable". Sentences like this will not be useful.

Regards

post scriptum
***
I decided to give a complete conclusion of the formula, since decent in level people would seem to be unable to do it on their own.
Here I repeat the formula expressing the speed of the center of mass in the left body system, through the speed of the right body, observed from the left body system:
https://www.cyberforum.ru/cgi-bin/latex.cgi?v_{c }=\frac{C^{2}}{v}\cdot (1-\sqrt{1-\frac{v^{2}}{C^{2}}})
the speed of the right body V is not equal to zero and dividing by it we get:
https://www.cyberforum.ru/cgi-bin/latex.cgi?\frac{v_{c }}{v}=\frac{C^{2}}{v^{2}}\cdot (1-\sqrt{1-\frac{v^{2}}{C^{2}}})  \quad     \quad   \quad  [A]
namely, we need the relation on the left next.
we write the coordinate of the center of mass in the left observer system, where the OX axis is chosen to coincide with the line of approach of the left and right bodies, and the origin is the left body.
https://www.cyberforum.ru/cgi-bin/latex.cgi?X_{c }=\frac{m _{0}\cdot X _{0}+M\cdot X}{m _{0}+M}
considering that Xo = 0, and the masses are independent of time - we differentiate by time:
https://www.cyberforum.ru/cgi-bin/latex.cgi?v_{c }=\frac{M\cdot v}{m _{0}+M}
Express the mass of the right M (measured in the system of the left) through the mass of the left one: mo :
https://www.cyberforum.ru/cgi-bin/latex.cgi?M =\frac{m _{0}\cdot v _{c}}{v-v _{c}}
or
https://www.cyberforum.ru/cgi-bin/latex.cgi?M =\frac{m _{0}\cdot \frac{v _{c}}{v}}{1-\frac{v _{c}}{v}}
simplify the radical Rv
https://www.cyberforum.ru/cgi-bin/latex.cgi?R_{v }=\frac{\frac{v _{c}}{v}}{1-\frac{v _{c}}{v}}
let we substitute the right side of the expression instead of the speed relation [A] :
https://www.cyberforum.ru/cgi-bin/latex.cgi?R_{v }=\frac{\frac{c^{2}}{v^{2}}\cdot (1-\sqrt{1-(\frac{v _{c}}{v})^{2}})}{1-(\frac{c^{2}}{v^{2}})\cdot (1-\sqrt{1-(\frac{v _{c}}{v})^{2}})}
reducing on with с2/v2 we get:
https://www.cyberforum.ru/cgi-bin/latex.cgi?R_{v }=\frac{1-\sqrt{1-(\frac{v _{c}}{v})^{2}}}{\frac{v^{2}}{c^{2}}-(1-\sqrt{1-(\frac{v _{c}}{v})^{2}})}
which boils down to:
https://www.cyberforum.ru/cgi-bin/latex.cgi?R_{v }=\frac{1}{\frac{v^{2}}{c^{2}\cdot (1-\sqrt{1-\frac{{v _{c}}^{2}}{v^{2}}})}-1}
Let's deal with the radical:
https://www.cyberforum.ru/cgi-bin/latex.cgi?r_{v }=\frac{\frac{v^{2}}{c^{2}}}{1-\sqrt{1-\frac{{v _{c}}^{2}}{v^{2}}}}
there is a lot of work for a person with time and imagination, but without possessing either one, I propose a substitution:
https://www.cyberforum.ru/cgi-bin/latex.cgi?\frac{v}{c}=\sin(\alpha )
https://www.cyberforum.ru/cgi-bin/latex.cgi?r_{v }=\frac{\sin(\alpha )^{2}}{1-\cos(\alpha )}
https://www.cyberforum.ru/cgi-bin/latex.cgi?r_{v }=\frac{1-\cos(\alpha )^{2}}{1-\cos(\alpha )}
https://www.cyberforum.ru/cgi-bin/latex.cgi?r_{v }=\frac{(1-\cos(\alpha ))\cdot(1+\cos(\alpha ))}{1-\cos(\alpha )}=1+\cos(\alpha )
Returning to the expression for mass, we get the cherished:
https://www.cyberforum.ru/cgi-bin/latex.cgi?M =\frac{m _{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}
Размещено в Theory of relativity
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