Alexander Masterov

I repeat: light speed always and everywhere is same. Light speed is constant. Light speed is absolute, and in a vacuum does depend on no nothing. Look at this animation once again: You should see that the transit time of red spots are different in different directions, despite the fact that the rate of red spots is equal(same) in both directions. This animation are illuminating for inequality: [math]T_1'\neq T_2'[/math] [math]A^{-1}_{00}=A_{11}/|A|[/math] We have differing result for [math]|A|>1[/math] and [math]|A|<1[/math] This math does not explain anything.

Electromagnetism have much properties. Absoluteness of light speed is one of them. This means that the speed of light does not depend on a speed of a light source and a speed of a observer.I believe in it with great difficulty, but the facts show that: it is true. Are you offer me a justification for this expression?: [math] x^2 - (ct)^2 =x'^2 - (ct')^2=0 [/math]This expression is not correct. ([math]T_1'\neq T_2'[/math]) [math] x^2 - (ct)^2 =(x' - (c+v)t_1')(x' - (c-v)t_2')=0[/math] - it's correct. [math]t_1'+t_2'=2t[/math] [math]t_2'-t_1'=\frac{x'}{c-v}-\frac{x'}{c+v}=\frac{2x'v}{c^2-v^2}[/math] [math]t_1'=t-\frac{x'v}{c^2-v^2}[/math]; [math]t_2'=t+\frac{x'v}{c^2-v^2}[/math] [math](x' - (c+v)(t-\frac{x'v}{c^2-v^2}))(x' - (c-v)(t+\frac{x'v}{c^2-v^2}))=0[/math] [math](x' - (c+v)t+\frac{x'v}{c-v})(x' - (c-v)t-\frac{x'v}{c+v})=0[/math] [math](x'(c-v) - (c^2-v^2)t+x'v)(x'(c+v) - (c^2-v^2)t-x'v)=0[/math] [math](x'c - (c^2-v^2)t)(x'c - (c^2-v^2)t)=0[/math] [math](x'c - (c^2-v^2)t)^2=0[/math] [math]\frac{x'}{1-v^2/c^2}-ct=0[/math] [math]x'=x(1-v^2/c^2)[/math] Matrixes may be any.

This means that the electromagnetism have such properties. These properties bear no relation to Newton's mechanics and to a transformation of the real coordinates. (Sound effects no bear relation to the coordinate transformations.) I didn't say it.

Light speed of and sound speed has no relation to coordinate transformations. Relativism is a visual effect. Yes, if you do not take into account the Doppler effect. (And if I understand you correctly (what I'm not sure).)

The transformation of real coordinates no related to light speed and sound speed. Real coordinates are changed in consequence of the presence of acceleration, which can be measured. Go back to Newton and to Galileo. Forget Einstein.

The transformation of real coordinates (Galileo): [math]\begin{pmatrix}t\\ x \\ y \\ z \end{pmatrix}=\begin{pmatrix}1&0&0&0\\-v & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}t'\\x' \\ y' \\ z' \end{pmatrix}[/math] [math]\begin{pmatrix}t'\\ x' \\ y' \\ z' \end{pmatrix}=\begin{pmatrix}1&0&0&0\\v & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}t\\x \\ y \\ z\end{pmatrix}[/math] ???

Linearity is a special case of nonlinearity. (Linearity does not exist in reality.)Inertia is a special case of acceleration. (External actions and acceleration are taking place always.) Master Theory solves the problem in the general (non-inertial) case. Visual (of relativism) and sound effects must be treated separately from physical processes, if the field (or environment) in the problem are not considered. Go back to Newton and to Galileo. Forget Einstein.

It's not true. The finiteness of the light speed (and of sound) and visual relativistic effects (due to the specifications of EMF) have no relationship to the real coordinates. Real coordinates - on their own. Relativism, any properties of light and sound - on their own.

Master Theory defines two types of coordinates: real coordinates and visual coordinates. Real coordinates obey Newton's physics and Galilean transformations. These coordinates can be calculated by double integration of acceleration, but acceleration can be measured (weight on a spring). Visual coordinates exist on an equal footing with coordinates, which has we determine by ear. Ie - the visual coordinates have no great physical sense. Therefore, there is no point to philosophize about the rules of transformation of visual coordinates. Master Theory returns we to Newton and to Galileo.

??? Einstein's problems has an infinite number of solutions. This infinite number of solutions came after I took away the absoluteness from a cross-scale, which Einstein presented to it (without reason). [math]L' = L (1 - v^2/c^2)^{0.5+\alpha}[/math] [math]H' = H(1 - v^2/c^2)^\alpha[/math] [math]T' = T/(1 - v^2/c^2)^{ 0.5-\alpha}[/math] [math]\alpha[/math] - free parameter. (Light speed absoluteness for any [math]0\le\alpha\le 0.5[/math]) [math]\alpha=0[/math] - Einsteinian theory (cross-scale absoluteness). [math]\alpha=0.5[/math] - Master Theory (time absoluteness). Master Theory is valid in all reference frames. Non inertial reference frame Real coordinates: Real speed: Visual coordinates: Visual time: [math] t' = t - \frac{|\vec r(t')|}{c} [/math] Visual speed: [math]\vec v(t)=\frac{2\vec V(t')}{(\sqrt{1+4V^2/c^2}+1)(1+\vec n\vec V/c)}[/math] [math] \vec n = \frac{\vec r}{|\vec r|} [/math]

1. I have shown that the Minkowski metric is non-unique consequence of the absoluteness of the light speed. 2. Minkowski metric implies the relativity of time, contrary to experience. (Time is absolute.) 3. I decided the one case (between the infinite number) in which time is absolute and called it - "Master Theory". A reference system is inertial, if a stationary object in it does not feel an acceleration. As usual. (How would you answered your question usually?)

In these two images shows a clock (one the clock). The first picture corresponds to a stationary observer. The second image corresponds to a moving observer. It is assumed that the velocity of the red dot (the speed of light) in both pictures (both ways) is identical. (The images inappropriately shows it. I intend to fix it.) The constancy of the speed of red dots (the independence of this velocity on the speed of the observer) make a demonstration of the absoluteness of the light speed. Expression A^{v^{-1}} is not similar to (A^v)^{-1}. Is'nt it? Expression A^{v^{-1}} is equivalency A^{1/v}. _________________________________________________ You want to show me: how you do know how to do reverse Lorentz transformation? I understand you correctly? [math]x=\frac{x'-vt'}{\sqrt{1-v^2/c^2}}[/math] [math]t=\frac{x'-vt'/c^2}{\sqrt{1-v^2/c^2}}[/math] TO: [math]x'=\frac{x+vt}{\sqrt{1-v^2/c^2}}[/math] [math]t'=\frac{x+vt/c^2}{\sqrt{1-v^2/c^2}}[/math] ______________________________________________________ [math]\begin{pmatrix} ct \\ x \\ y \\ z \end{pmatrix}=\begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{-v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ \frac{-v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ 0 & \quad & 0 & \quad & 1 & 0 \\ 0 & \quad & 0 & \quad & 0 & 1 \end{pmatrix}\begin{pmatrix} ct' \\ x' \\ y' \\ z' \end{pmatrix}[/math] TO: [math]\begin{pmatrix} ct' \\ x' \\ y' \\ z' \end{pmatrix}=\begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ \frac{v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ 0 & \quad & 0 & \quad & 1 & 0 \\ 0 & \quad & 0 & \quad & 0 & 1 \end{pmatrix}\begin{pmatrix} ct \\ x \\ y \\ z \end{pmatrix}[/math] What does that prove?

It's not true.I use absoluteness of lite speed for Master Theory and not use [math] x^2 - (ct)^2 =x'^2 - (ct')^2=0 [/math]. Formuls of Master Theory satisfy absoluteness of lite speed, but not satisfy [math] x^2 - (ct)^2 =x'^2 - (ct')^2=0 [/math]. [math] x^2 - (ct)^2 =x'^2 - (ct')^2=0 [/math] - Unfounded limitation (restriction). No. [math] A^{v^{-1}}=A^{\frac{1}{v}} [/math] - It can be understood? Formuls of Master Theory satisfy absoluteness of lite speed, but not satisfy [math] x^2 - (ct)^2 =x'^2 - (ct')^2=0 [/math]. the screen - singular. Projectors - pair (couple).

You just need to drop a hint.If you are going to manipulate the Lorentz transformations, then: Lorentz transformations and absoluteness cross-scale are the result of [math]x^2 - (ct)^2 =x'^2 - (ct')^2=0[/math] whose validity requires proof. Are you sure that the mathematical expression [math](A^{v^{-1}})_{0,0}[/math] should be clear to everyone? I have the education an physics and applied mathematics. I have not seen such expressions before. Them meaning do not understand. I'm sure: you do not have answer my question and (in order to confuse the debate) you juggle with obscure mathematical expressions. It's not true. I showed that from absoluteness of the speed of light follow infinite much of consequences and not follow absolute cross-scales. (See above, post 2) On the screen can be mounted photographic recorder, which responds to the sequence of frames. The result of registration can not be individualized for each observer.

I understand the matrix. I understand what it means to multiply a matrix by another matrix. I understand that means the inverse matrix. (AA-1=1)But I prefer to communicate in the forum, using the language by which writed physical handbooks. This language is understood by all and it does not require knowledge of matrices. And this language is sufficient to describe any idea. The introduction of little-known concepts will not give us anything but - confusion. My question: Einstein bestowed the absoluteness to the cross-scale (but not for time) - what basis? It's simple question. If you have answer, then it's simple. Why? Why must you see an acceleration of time? Please give a precise reason why. I could write that this is obvious. That would is sufficient. It really is obvious, since this is consistent with everyday experience. But we can conduct a thought experiment: both the observer to do a project of a movie on a screen. If one sees an acceleration, and second sees a deceleration, then they will see a frames by equal sequence. If both the slowdown seen, for each of them will have private sequence of frames. It's impossible in our reality.

I'm in the firm belief that: if you are unable to understandably communicate your idea to any student, then you poorly understand your idea. Mediocrity of Russian Academy of Sciences often use this method: they create pseudoscientific language, full of obscure concepts and notation that allows them to create the appearance of science value of their work. We will not follow their example.

I see no need to use a matrix. You are trying to confuse the debate by scarce (insufficiently plain) mathematical notation. It's not OK.

- This equation true except momentarily. My english is poor. I'm sorry. I'm physicist and had a good math basis (programming mathematician, specializing in developing very sophisticated algorithms), I understand the matrix, but rarely use them. Matrix is a shortened way of writing linear equations (no more than). I think: long bed-sheet, filled formulas, not to do clear, but it to do muddle for me. The answer to my question on the surface. If you knew him you would not have to write long formulas. Long formulas are needed in order to confuse and deceive. Do not need the expression to obtain the relativistic formulas for me. In our reality is really a rule: if my interlocutor detected slow down of my time, then: I must an acceleration of see time of my partner. Einstein's theory argues that in such a situation, you (both) can watch the slowing of time. This ambiguity is impossible.

Essence of Master Theory are reduced to a simple idea: the absoluteness of a cross-scales of SRT is not justified. Therefore: we obtain opportunity deprive the cross-scale of this quality, that is, cross-scale can be relative. Thus get a free parameter for each value that you can build a separate theory of relativity, each of which will have as much right exist, as special relativity Einstein's theory has. Then I prove rigorously that time must be absolute (otherwise violates the principle of causality). So (of the entire infinite set of alternatives SRT) remain in force only one a theory which I called Master Theory.

In Master Theory: [math]T'_1+T'_2=T_1+T_2[/math] [math] \frac{L'}{c+v}+\frac{L'}{c-v}=\frac{2L}{c}[/math] ______________________________________________ Einstein theory: [math](T'_1+T'_2)\sqrt{1-v^2/c^2} =T_1+T_2[/math] [math] (\frac{L'}{c+v}+\frac{L'}{c-v})\sqrt{1-v^2/c^2} =\frac{2L}{c}[/math] ______________________________________________ Both cases: [math]T'_1\neq T'_2[/math]

Yes, accurate within inside out. (I.e. - no: light speed id absolute always.) We have cases: light catches up with the mirror(1) and a light runs to the mirror(2). Which gives we two different flight times.

Yes, if the observer is stationary. then a flight time is independent of direction: [math] T_1=T_2=\frac{L}{c} [/math] If the observer moves: then the flight time in different directions will be different for all theories (Einstein's theory included). [math] T_1=\frac{L}{c+v} [/math] [math] T_2=\frac{L}{c-v} [/math] [math]x'^2 - (ct')^2 = 0[/math] - is not correct.

Yes! Everyone satisfied to the fullest of it. I recall (for example) simple pendulum. ([math]T=2\pi\sqrt{\frac{L}{g}}[/math]) You offer to take into account the longitudinal Doppler effect?But this effect is complementary to the slowing of time, but - not cause of it. Your math is based on the dogma of Einstein's theory and vote self-explanatory. The origin of [math]x^2 - (ct)^2 = 0[/math] is not obvious. At the beginning of this theme I have shown that: order to meet the absoluteness of light speed no needs for it.

No! [math]\frac{d|r|}{dt}=0[/math] [math]\frac{d\phi}{dt}=const[/math] It is self-explanatory. Need explain it. It violates the principle of causality. If one observer sees time dilation, other observer is required to see the time acceleration. No options.

It is not true. When moving across a view axis the distance does not change (d|r|/dt=0). Nevertheless: the time dilation takes place. (Einstein allege.) Einstein allege: time can slow down (by acceleration), but time can not accelerate (by acceleration). If time slow down, then - it can not accelerate, can not return back. Traveler will have its own (slow) time on Earth. How can this be?