# Master Theory

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Einstein's theory compelled you deceive yourself.

Ok Alexander, it is clear we cannot convince that Einsteinian relativity holds well in nature. I cannot see this thread developing.

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Look at the formula of Master Theory again. Make sure that the problems that are solved by Einstein be resolved in another.

My solution is true.

Ok Alexander, it is clear we cannot convince that Einsteinian relativity holds well in nature. I cannot see this thread developing.
Try to answer the question: why Einstein gave the absoluteness to cross-scale, and not give it to the time? What is the reason?

The answer to this question will explain all and tells the truth for us.

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Look at the formula of Master Theory again. Make sure that the problems that are solved by Einstein be resolved in another.

My solution is true.

Try to answer the question: why Einstein gave the absoluteness to cross-scale, and not give it to the time? What is the reason?

The answer to this question will explain all and tells the truth for us.

what is cross-scale? You mean the space-time interval? The constant speed of light follows from Maxwell's equations which are invariant with respect to the Lorentz transformations and not the Galilean transformations. The extension that all non-gravitational physics be Lorentz invariant then demands that the space-time interval be the invariant object rather than both the time and space intervals separately. Intuitively you see this because the Lorentz transformations mix space and time. You do not have a totally agreed up on way of decomposing into space and time.

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what is cross-scale? You mean the space-time interval? The constant speed of light follows from Maxwell's equations which are invariant with respect to the Lorentz transformations and not the Galilean transformations. The extension that all non-gravitational physics be Lorentz invariant then demands that the space-time interval be the invariant object rather than both the time and space intervals separately. Intuitively you see this because the Lorentz transformations mix space and time. You do not have a totally agreed up on way of decomposing into space and time.
It conditions can be met without a use of the time dilation. It is done Master Theory.

For each of the intermediate values () we can construct theories have.

For a result is Einstein's Theory of Relativity.

For a result is Master Theory.

Edited by Alexander Masterov
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OK, so I will take your word for that. Now can you show that Maxwell's equations are invariant with respect to the "Master group"?

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I'm not eliminate that my theory is wrong. But (more than two years) no one proves the opposite. Nobody give solid argumentation to give a absoluteness to a cross-scale (but not for the time). What if you do it?
I think there are some aspects of what you're saying that are right, but that you go too far with certain things and they undermine everything you say.

There is another problem. Time in Einstein's theory can slow down only. (Because of this exist a twin paradox). Time can not be accelerated. And it is not clear: how to a time of a traveller return back to normal, after it has been slow.
Go and look at A World Without Time: The Forgotten Legacy of Godel and Einstein. Time doesn't slow down, clocks clock up local motion, not "the flow of time". It doesn't matter whether it's a mechanical clock, a quartz clock, or an atomic clock, that's what clocks do. So when a clock slows down, that local motion is occuring at a reduced rate. That's all there is to it. If you travel out and back through space your rate of local motion is reduced because of your motion through space. It's pretty obvious in the parallel-mirror light clock example wherein the Lorentz factor is derived from Pythagoras' theorem. The hypotenuse is the light path of length 1 because we're using natural units, the base is your speed as a fraction of c, and the height is the Lorentz factor, wherein there's a reciprocal to distinguish time dilation from length contraction.

What you seem to be missing in your presentation is the significance of pair production. Matter, such as an electron is quite literally created from light, in a lab. And an electron exhibits spin angular momentum and magnetic dipole moment. The Einstein-de Haas effect then demonstrates that spin angular momentum is of the same nature as classical angular momentum. So there's something going round and round in there, and it ain't cheese. Now think about why an electron can't go faster than light.

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OK, so I will take your word for that. Now can you show that Maxwell's equations are invariant with respect to the "Master group"?
I was able to convert Maxwell's equations in differential form into the integral form (and back). The last time I did it more than a quarter century ago.

I'm not expert of relativism. (You probably have noticed it.)

My specialty: non-linear dynamics (auto-oscillation, auto-wave, attractors ...).

Master Theory is a byproduct of other (more general) theory. (This happened fifteen years ago, roughly).

I have developed methods for analyzing nonlinear integro-differential equations. I easy to study analytically the equation, which have previously been extremely difficult for study.

Among those equations, which I could understand, I found one that can claim to a vacancy of a generator of matter. But my attempts to satisfy Einstein's theory proved futile.

Then I began to study Einstein's Special Theory of Relativity. I have found that Einstein wrongly gave to absolute cross-scales.

So to come into being my Master Theory.

Master Theory was after I completed the career of a scientist. Therefore, I have no intentions to develop this theory, but I hope that it will make other people do.

Edited by Alexander Masterov
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And I thought so earlier as well. But ... alas. I'm mistaken.

Yes.

But I was wrong. I (for some reason) decided that your line is vertical, while it is horizontal.

The distance between the lines:

The lines will seeming farther from the observer than it in real.

I just want to get your answer absolutely clear. Am I correct in saying that English is not your first language?

So will the observer on the train think that they are closer or farther apart?

Currently, from what I can tell, you are saying the observer on the train will see them as being closer, as H' < H.

=Uncool-

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I just want to get your answer absolutely clear. Am I correct in saying that English is not your first language?
Yes.

I stady German, but not use it.

My English is poor.

So will the observer on the train think that they are closer or farther apart?

Currently, from what I can tell, you are saying the observer on the train will see them as being closer, as H' < H.

Farther. (A visual spatial scales will be diminish.)

Object (which across-moves to a observation beam) will seem farther away than in real.

If the object approach to (run away from) the observer, then the observer will see a acceleration (deceleration) time.

Edited by Alexander Masterov
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Yes.

I stady German, but not use it.

My English is poor.

In that case, I'd suggest getting someone who knows English to write here; it can make communication easier.

Far. (A visual spatial scales will be diminish.)

OK. So the train observer thinks that the lines on the wall are separated by a distance greater than H. Now assume that the observer on the train is holding two paintbrushes. He thinks that they are vertically separated by a distance H. Will the station observer think that they are farther than H apart, or closer than H apart?

=Uncool-

Edited by uncool
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In that case, I'd suggest getting someone who knows English to write here; it can make communication easier.
I do not have that opportunity.
OK. So the train observer thinks that the lines on the wall are separated by a distance greater than H. Now assume that the observer on the train is holding two paintbrushes. He thinks that they are vertically separated by a distance H. Will the station observer think that they are farther than H apart, or closer than H apart?
The observer will have a visual effect: the brush will anoint formerly.

Observer it would seem that he has not reach out to the wall.

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The observer will have a visual effect: the brush will anoint formerly.

Observer it would seem that he has not reach out to the wall.

I'm afraid that your answer does not make sense. I cannot figure out what you are saying.

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I'm afraid that your answer does not make sense. I cannot figure out what you are saying.
1. apparent distance up to the wall more than in reality

2. paint-brush anoint without touching the walls (a visual effect)

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1. apparent distance up to the wall more than in reality

Does this mean that the station observer will think that the two brushes are closer together than the train observer does?

2. paint-brush anoint without touching the walls (a visual effect)

I'm afraid I have no idea what this sentence means.

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Does this mean that the station observer will think that the two brushes are closer together than the train observer does?
No.

This means that the station observer will think that apparent distance up to the the train observer more than in reality.

I'm afraid I have no idea what this sentence means.
This means that the distance from the paint brush to the wall will have a visual distortion. Edited by Alexander Masterov
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Relativistic Maxwell

Since any differential operator is local (in close proximity), and (consequently) a finiteness of a propagation speed of the fields (in infinitely small distance) can not make adjustments to the equation. Therefore, Maxwell's equations in differential form in the context of the Master Theory would not be changed in comparison with the classics. Namely:

But integral Maxwell (because of a nonlocality of a integral operators) will have a fundamentally different (non-classical, but - relativistic) form. This is explained by the fact that Stokes' and Gauss' theorems:

and

valid only for stationary fields, or fields, the speed of which is infinite. Application of these theorems (for the output of the integral form of Maxwell's equations) can not be correct in the relativistic case.

Edited by Alexander Masterov
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Suppose we have a closed-loop surface S, which have a characteristic size L. Then the integral expressions of the classical Maxwell's Equations:

valid only in cases where a characteristic time of variation of the fields (and electric charge) in these equations are much smaller than L/c.

A similar can be said about the other pair of expressions.

Edited by Alexander Masterov
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So Alex, what exactly is the full transformation?

In relativity, the full transformation is:

$t' = \gamma (t - \frac{vx}{c^2})$

$x' = \gamma (x - vt)$

$y' = y$

$z' = z$

Under your hypothesis, if you're given t, x, y, and z, what are they after your transformation?

=Uncool-

Edited by uncool
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Non inertial reference frame

Real coordinates:

$\vec r(t) =\vec r_0+ \int_{0}^{t}{\vec V(\tau) d\tau}$

Real speed:

$\vec V(t) =\vec V_0+ \int_{0}^{t}{\vec a(\tau) d\tau}$

$\vec V = \frac{\vec v}{1-v^2/c^2}$

Visual speed:

$\vec v = \frac{2\vec V}{\sqrt{1+4V^2/c^2}+1}$

Visual coordinates:

$\vec r^,(t') =\vec r^,_0+ \int_{0}^{t'}{\vec v(\tau) d\tau}$

Visual time:

$t' = t - \frac{|r(t')|}{c}$

Inertial reference frame

Real coordinates:

$x(t) =x_0+ Vt$

Visual coordinates:

$x^,(t) =x^,_0+ vt'$

$t' = t - \frac{x(t')}{c}=t - \frac{x_0+Vt'}{c}=\frac{t-x_0/c}{1+V/c}$ - Dopler's effect.

$y' = y\sqrt{1-v^2/c^2}$

$z' = z\sqrt{1-v^2/c^2}$

Edited by Alexander Masterov
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Visual coordinates:

$t' = t - \frac{Vx}{c^2}$

$x' = (x - Vt)/\gamma^2$

$y' = y/\gamma$

$z' = z/\gamma$

Real speed:

$\vec V = \frac{2\vec v}{1-v^2/c^2}$

Visual speed:

$\vec v = \frac{2\vec V}{\sqrt{1+4V^2/c^2}+1}$

$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$

So in that case, let's say that we have two observers A and B, such that B is moving at velocity v relative to A. In the frame of A, A's path is $(t, 0, 0, 0)$. Under your transformation, that path becomes $(t, -\frac{vt}{\gamma_v^2}, 0, 0)$. In other words, you get that if A thinks B is moving at velocity $v$, then B thinks A is moving at velocity $-\frac{v}{\gamma_v^2}$.

Is this correct so far?

=Uncool-

Edited by uncool
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I mistakenly clicked the button "submit post". So you see an intermediate result.

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I mistakenly clicked the button "submit post". So you see an intermediate result.

Then can you tell me the correct version of what I posted?

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I am not sure that I understand you. I try to answer you:

If:

Real coordinates:

$x(t) =Vt$

Then:

Visual coordinates:

$x^,(t) =vt'=\frac{vt}{1+V/c}=\frac{2Vt}{(\sqrt{1+4V^2/c^2}+1)(1+V/c)}$

Next:

Visual speed (with Dopler's effect):

$V' =\frac{2V}{(\sqrt{1+4V^2/c^2}+1)(1+V/c)}$

$V' =\frac{v(1-v^2/c^2)}{1+v/c-v^2/c^2}=v-\frac{v^2/c}{1+v/c-v^2/c^2}$

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I'm asking you to tell me the following:

Say observer A sees a timeline that takes up the coordinates (t, 0, 0, 0). Where will observer B (who is moving at velocity v relative to A) think that timeline goes? Under special relativity, the timeline will go to $(\gamma t, -v \gamma t, 0, 0)$. Where does (t, 0, 0, 0) go under your transformation?

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My translation program (English language) do not.

Try to formulate your questions shortly.

Edited by Alexander Masterov

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