He must be using an automatic translator program from Russian to English, he doesn't know much English. As a mater of fact, he doesn't know much physics either.
Master Theory (edition 2)
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He must be using an automatic translator program from Russian to English, he doesn't know much English. As a mater of fact, he doesn't know much physics either.
I have a problem with translation into Russian. My dictionaries do not always cope with the scientific terminology. I can not adequately understand the English text is not always a case.
For example: mater === "brain-tunic" or "mother".
This is not true. The space-time isotropy setted limits for a types of matter-generator. This generator must be hamiltonian. This means that a solution to his equations will always give a paired results. These decisions will differ a time direction. (For AlphaNumeric also.)
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Masterov, this is and was my fault. You were being generous when you said my English was good. A complement my English teachers would have had some disagreement with, though it is my first language.
I used the word "mater" incorrectly and it does indeed translate to "mother". I should have been using the word "matter" all along.
Spelling was never a strong suit for me and with the advent of spell checkers it has only become worse.
My apologies, for the confusion. I have since obtained a dictionary of my own. Digital of course. Now the problem will be remembering to use it....
Don't mention apologize. We are not linguists. (Both.)
No he didn't. What he alleged was the same thing that Lorentz alleged: that an accelerated frame has a slowed time rate relative to a non-accelerated frame.Masterov said:
Which means a non-accelerated frame has a speeded up time rate relative to an accelerated frame.
By "time rate" is meant the time recorded by otherwise identical clocks. I read about this when I was 10, and I thought I understood it then (but perhaps not). At least I thought I understood what the twin paradox meant--if you move through space faster than someone else you age less, relative to the slower traveller, who appears to age more quickly. Both travellers see themselves aging at a "normal" rate, and their clocks record time at the same rate locally.
That's impossible!
Return home will require the acceleration, which again slows down time. Subsequent braking will slow down again.
That's why it was claimed you were repeating one of the mistake of Dingle, because you were focusing on the term $$\frac{\partial \Delta t'}{\partial \Delta t} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$ when the Lorentz transform is more than that, it's $$\Delta t' = \frac{\Delta t - \frac{v}{c^2}\Delta x}{\sqrt{1 - \frac{v^2}{c^2}}}$$.
This is the same mistake as claiming a rotation left by 30 degrees isn't canceled by a rotation right by 30 degrees because $$\cos \, 30^{\circ} = \cos \, -30^{\circ} = \frac{\sqrt{3}}{2}$$ when the relevant expression is $$\begin{pmatrix} \cos \, 30 ^{\circ} & \quad & - \sin \, 30^{\circ} \\ \sin \, 30^{\circ} & \quad & \cos \, 30^{\circ} \end{pmatrix} \begin{pmatrix} \cos \, -30 ^{\circ} & \quad & - \sin \, -30^{\circ} \\ \sin \, -30^{\circ} & \quad & \cos \, -30^{\circ} \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} & \quad & - \frac{1}{2} \\ \frac{1}{2} & \quad & \frac{\sqrt{3}}{2} \end{pmatrix} \begin{pmatrix} \frac{\sqrt{3}}{2} & \quad & \frac{1}{2} \\ -\frac{1}{2} & \quad & \frac{\sqrt{3}}{2} \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} \frac{\sqrt{3}}{2} + \frac{1}{2}\frac{1}{2} & \quad & \frac{\sqrt{3}}{2} \frac{1}{2} - \frac{\sqrt{3}}{2}\frac{1}{2} \\ \frac{\sqrt{3}}{2} \frac{1}{2} - \frac{\sqrt{3}}{2}\frac{1}{2} & \quad & \frac{\sqrt{3}}{2} \frac{\sqrt{3}}{2} + \frac{1}{2}\frac{1}{2} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$
So you criticize relativity because you don't understand relativity.
And you don't understand relativity because, at a minimum, you don't understand the mathematics of the Lorentz transform.
Specifically, you don't understand that a Lorentz transform of v is canceled by a Lorentz transform of -v (in the same direction).
Attempted translation: "But if the velocity and the x direction are perpendicular, the time-dilation only accumulates."
Attempted translation: "But if original change in x is zero, the time-dilation only accumulates."
Response: Untrue. Please consult my previous post.
Restrict: $$0 < |v| < c$$
Define: $$\Lambda_x (v) = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{-v}{c^2 \sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 \\ \frac{-v}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & 0 & 0 \\ 0 & \quad & 0 & 1 & 0 \\ 0 & \quad & 0 & 0 & 1\end{pmatrix} $$
Define: $$\Lambda_y (v) = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & \quad & \frac{-v}{c^2 \sqrt{1 - \frac{v^2}{c^2}}} & 0 \\ 0 & \quad & 1 & \quad & 0 & 0 \\ \frac{-v}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & 0 \\ 0 & \quad & 0 & \quad & 0 & 1 \end{pmatrix}$$
So $$\Lambda_x (-v) \Lambda_x (v) = \Lambda_x (v) \Lambda_x (-v) = I$$
Likewise $$\Lambda_y (-v) \Lambda_y (v) = \Lambda_y (v) \Lambda_y (-v) = I$$
Likewise for $$\Lambda_z$$.
Also: $$\Lambda_x (-v) \Lambda_y (-v) \Lambda_y (v) \Lambda_x (v) = \Lambda_x (-v) \Lambda_y (v) \Lambda_y (-v) \Lambda_x (v) = \Lambda_x (v) \Lambda_y (-v) \Lambda_y (v) \Lambda_x (-v) = \Lambda_x (v) \Lambda_y (v) \Lambda_y (-v) \Lambda_x (-v) = I$$
This means $$\begin{pmatrix} \Delta t'' \\ \Delta x'' \\ \Delta y'' \\ \Delta z'' \end{pmatrix} = \Lambda_x (-v) \begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix}$$ and $$\begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda_x (v) \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix}$$ together mean $$\begin{pmatrix} \Delta t'' \\ \Delta x'' \\ \Delta y'' \\ \Delta z'' \end{pmatrix} = \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix}$$
Proof: $$\begin{pmatrix} \Delta t'' \\ \Delta x'' \\ \Delta y'' \\ \Delta z'' \end{pmatrix} = \Lambda_x (-v) \begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda_x (-v) \left( \Lambda_x (v) \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} \right) = \left( \Lambda_x (-v) \Lambda_x (v) \right) \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = I \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix}$$. Even if $$\Delta x = 0$$ or $$\Delta x' = 0$$ we always have $$\Delta t'' = \Delta t$$.
Alternate proof: $$\begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda_x (v) \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \frac{\Delta t - \frac{v}{c^2} \Delta x}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{\Delta x - v \Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \Delta y \\ \Delta z \end{pmatrix}$$, so $$\begin{pmatrix} \Delta t'' \\ \Delta x'' \\ \Delta y'' \\ \Delta z'' \end{pmatrix} = \Lambda_x (-v) \begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \begin{pmatrix} \frac{\Delta t' - \frac{-v}{c^2} \Delta x'}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{\Delta x' - (-v) \Delta t'}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \Delta y' \\ \Delta z' \end{pmatrix} = \begin{pmatrix} \frac{\left( \Delta t - \frac{v}{c^2} \Delta x \right) - \frac{-v}{c^2} \left(\Delta x - v \Delta t \right)}{\sqrt{1 - \frac{v^2}{c^2}}\sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{\left(\Delta x - v \Delta t \right) - (-v) \left( \Delta t - \frac{v}{c^2} \Delta x \right)}{\sqrt{1 - \frac{v^2}{c^2}}\sqrt{1 - \frac{v^2}{c^2}}} \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \frac{\Delta t - \frac{v}{c^2} \Delta x + \frac{v}{c^2} \Delta x - \frac{v^2}{c^2} \Delta t}{1 - \frac{v^2}{c^2}} \\ \frac{\Delta x - v \Delta t + v \Delta t - \frac{v^2}{c^2} \Delta x}{1 - \frac{v^2}{c^2}} \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \frac{\Delta t - \frac{v^2}{c^2} \Delta t}{1 - \frac{v^2}{c^2}} \\ \frac{\Delta x - \frac{v^2}{c^2} \Delta x}{1 - \frac{v^2}{c^2}} \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix}$$
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That is not true. Such a transformation wouldn't leave the space-time interval unchanged. A transform $$(\Delta t,\Delta x) \to (\Delta t',\Delta x')$$ must be such that $$-(\Delta t)^{2} + (\Delta x)^{2} = -(\Delta t')^{2} + (\Delta x')^{2} $$, assuming $$\Delta y$$ and $$\Delta z$$ are unchanged. This is the central tenant of Lorentz transformations.
So, is your problem that you :
A. Do not understand Transverse Doppler Effect?
or
B. Do not understand the reciprocity of time dilation?
You still aren't being clear.
Did you mean:
$$\begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda_y(-v) \begin{pmatrix} \Delta t \\ 0 \\ 0 \\ \Delta z \end{pmatrix} = \begin{pmatrix} \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ 0 \\ \frac{v \Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \Delta z \end{pmatrix}$$ ?
Well, $$\Delta x = 0, \; \frac{\Delta y'}{\Delta t'} = v, \; \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}}$$ but that doesn't stop me from computing:
$$\begin{pmatrix} \Delta t'' \\ \Delta x'' \\ \Delta y'' \\ \Delta z'' \end{pmatrix} = \Lambda_y(v) \begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda_y(v) \begin{pmatrix} \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ 0 \\ \frac{v \Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \Delta z \end{pmatrix} = \begin{pmatrix} \frac{1 - \frac{v^2}{c^2}}{1 - \frac{v^2}{c^2}} \Delta t \\ 0 \\ \frac{v \Delta t - v \Delta t}{1 - \frac{v^2}{c^2}} \\ \Delta z \end{pmatrix} = \begin{pmatrix} \Delta t \\ 0 \\ 0 \\ \Delta z \end{pmatrix} $$
And so the Lorentz transform in the opposite direction undoes the coordinate change, including any time dilation effects.
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Or did you mean:
$$\begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda_x(v) \begin{pmatrix} \Delta t \\ 0 \\ \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}} \Delta t \\ \Delta z \end{pmatrix} = \begin{pmatrix} \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{- v \Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \frac{v \Delta t}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \Delta z \end{pmatrix} $$ ?
Well, $$\Delta x = 0, \; \frac{\Delta y'}{\Delta t'} = v, \; \Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}}$$ but that doesn't stop me from computing:
$$\begin{pmatrix} \Delta t'' \\ \Delta x'' \\ \Delta y'' \\ \Delta z'' \end{pmatrix} = \Lambda_x(-v) \begin{pmatrix} \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \Lambda_x(-v) \Lambda_x(v) \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} \Delta t \\ 0 \\ \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}} \Delta t \\ \Delta z \end{pmatrix} $$.
Once again, the Lorentz transform in the opposite direction undoes the coordinate change, including any time dilation effects.
It still appears you are repeating the mistake of Dingle and so your criticisms of Special Relativity reveal only your lack of familiarity with Special Relativity and therefore miss the target.
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But you don't understand the TDE, it is a direct confirmation of time dilation.
Compare both teories:
If the observer is stationary.
then a flight time is independent of direction: $$T_1=T_2=\frac{L}{c} $$
If the observer moves:
then the flight time in different directions will be different for all theories (Einstein's theory included).
$$T'_1=\frac{L'}{c+v}$$
$$T'_2=\frac{L'}{c-v}$$
For both theories: $$T'_1\neq T'_2$$
In Master Theory:
$$T'_1+T'_2=T_1+T_2$$
$$\frac{L'}{c+v}+\frac{L'}{c-v}=\frac{2L}{c}$$
______________________________________________
Einstein theory:
$$(T'_1+T'_2)\sqrt{1-v^2/c^2} =T_1+T_2$$
$$(\frac{L'}{c+v}+\frac{L'}{c-v})\sqrt{1-v^2/c^2} =\frac{2L}{c}$$
______________________________________________
In Master Theory:
$$\frac{T'}{T}=1$$
$$\frac{L'}{L}=1-\frac{v^2}{c^2}$$
Einstein theory:
$$\frac{T'}{T}=\frac{1}{\sqrt{1-v^2/c^2}}$$
$$\frac{L'}{L}=\sqrt{1-v^2/c^2}$$
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$$x'^2 - (ct')^2 = 0$$ - is not correct.
If the observer is stationary.
then a flight time is independent of direction: $$T_1=T_2=\frac{L}{c} $$
If the observer moves:
then the flight time in different directions will be different for all theories (Einstein's theory included).
$$T'_1=\frac{L'}{c+v}$$
$$T'_2=\frac{L'}{c-v}$$
For both theories: $$T'_1\neq T'_2$$
In Master Theory:
$$T'_1+T'_2=T_1+T_2$$
$$\frac{L'}{c+v}+\frac{L'}{c-v}=\frac{2L}{c}$$
______________________________________________
Einstein theory:
$$(T'_1+T'_2)\sqrt{1-v^2/c^2} =T_1+T_2$$
$$(\frac{L'}{c+v}+\frac{L'}{c-v})\sqrt{1-v^2/c^2} =\frac{2L}{c}$$
______________________________________________
In Master Theory:
$$\frac{T'}{T}=1$$
$$\frac{L'}{L}=1-\frac{v^2}{c^2}$$
Einstein theory:
$$\frac{T'}{T}=\frac{1}{\sqrt{1-v^2/c^2}}$$
$$\frac{L'}{L}=\sqrt{1-v^2/c^2}$$
______________________________________________
$$x'^2 - (ct')^2 = 0$$ - is not correct.
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Compare both teories:
If the observer is stationary.
then a flight time is independent of direction: [/tex]T_1=T_2=\frac{L}{c} [/tex]
If the observer moves:
then the flight time in different directions will be different for all theories (Einstein's theory included).
$$T'_1=\frac{L}{c+v}$$
$$T'_2=\frac{L}{c-v}$$
Both cases: $$T'_1\neq T'_2$$
In Master Theory:
$$T'_1+T'_2=T_1+T_2$$
This is because "Master" theory is so stupid that it has no notion of time dilation/length contraction. This, in turn, is a consequence of the fact that the "Master" theory is so stupid that it still uses the Galilei transforms (absolute time) instead of the Lorentz transforms.
The above idiocies mean, amongst other things that :
1. you cannot explain the null result of the Michelson-Morley experiment
2. your theory fails the invariance of Maxwell's equations (we have already seen ample proof that you don't understand this issue).
This doesn't stop me from pointing out the errors in your theory.
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