Master Theory

[Alexander Masterov]

Again, in that case, what precisely does it mean to say that the speed of light is absolute?

Light speed of and sound speed has no relation to coordinate transformations. Relativism is a visual effect.

So your objection is equivalent to the statement that if the (0, 0) element of A^v (that is, the upper left element (which for this transformation matrix is [math] \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/math]) is greater than 1, then the (0, 0) element of (A^v)^-1 must be less than 1. Do you agree?

Yes, if you do not take into account the Doppler effect. (And if I understand you correctly (what I'm not sure).) Edited September 22, 2011 by Alexander Masterov

[uncool]

Light speed of and sound speed has no relation to coordinate transformations. Relativism is a visual effect.

Then what does it mean to say that they are absolute? You still have not answered that question.

Yes, if you do not take into account the Doppler effect. (And if I understand you correctly (what I'm not sure).)

And am I correct that you think that that is true for any matrix - that is, for any matrix A, A_0,0 > 1 implies that A^-1_0,0 < 1?

=Uncool-

[Alexander Masterov]

Then what does it mean to say that they are absolute? You still have not answered that question.

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.)

And am I correct that you think that that is true for any matrix - that is, for any matrix A, A_0,0 > 1 implies that A^-1_0,0 < 1?

I didn't say it. Edited September 22, 2011 by Alexander Masterov

[uncool]

This means that the electromagnetism have such properties.

What 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.)

So again, please explain precisely what you think it means for the speed of light to be absolute. Do you think that there is one preferred frame, relative to which light will expand directly from a single point, and that in all other frames, the center of the light-sphere will move with time?

I didn't say it.

So do you think it is true for any matrix? If not, then why can't it be true for this transformation matrix?

=Uncool-

Edited September 22, 2011 by uncool

[Alexander Masterov]

What such properties?

Electromagnetism have much properties. Absoluteness of light speed is one of them.

So again, please explain precisely what you think it means for the speed of light to be absolute.

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.

Do you think that there is one preferred frame, relative to which light will expand directly from a single point, and that in all other frames, the center of the light-sphere will move with time?

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]

 

So do you think it is true for any matrix? If not, then why can't it be true for this transformation matrix?

Matrixes may be any. Edited September 22, 2011 by Alexander Masterov

[uncool]

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.

But you've already agreed that under your theory, the speed of light is observer-dependent. That is, you've already agreed that in different frames, the speed of light is different.

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])

I'm trying to figure out what exactly you think.

[math] x^2 - (ct)^2 =(x' - (c+v)t_1')(x' - (c-v)t_2')=0[/math] - it's correct.

So is this your definition of the speed of light being absolute?

Matrixes may be any.

That doesn't answer my question. Is it true for any matrix that if the upper-left element is greater than 1, then the upper-left element of the inverse must be less than 1?

=Uncool-

[Alexander Masterov]

But you've already agreed that under your theory, the speed of light is observer-dependent. That is, you've already agreed that in different frames, the speed of light is different.

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.

So is this your definition of the speed of light being absolute?

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]

That doesn't answer my question. Is it true for any matrix that if the upper-left element is greater than 1, then the upper-left element of the inverse must be less than 1?

[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.

Edited September 22, 2011 by Alexander Masterov

[uncool]

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.

But that simply isn't true under Galilean transformations. The speed of light changes under Galilean transformations. Which means that it is observer-dependent - it is not the same for different observers. Which makes it non-absolute.

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]

That result is there in Galilean relativity and special relativity already. It is a trivial statement. I'm asking you to make a specific statement about your theory.

[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.

I'm asking you to make a specific answer, yes or no.

 

If you have a matrix A - any matrix A - and you know that A_0,0 > 1, does that mean that

A^-1_0,0 must be less than 1?

=Uncool-

[Alexander Masterov]

But that simply isn't true under Galilean transformations. The speed of light changes under Galilean transformations. Which means that it is observer-dependent - it is not the same for different observers. Which makes it non-absolute.

That result is there in Galilean relativity and special relativity already. It is a trivial statement. I'm asking you to make a specific statement about your theory.

Galilean transformations for real coordinates only. (Real coordinates to real coordinates.)

Relativism is visual effect only.

Edited September 22, 2011 by Alexander Masterov

[Alexander Masterov]

I'm asking you to make a specific answer, yes or no.

 

If you have a matrix A - any matrix A - and you know that A_0,0 > 1, does that mean that

A^-1_0,0 must be less than 1?

Even if such a rule exists, then to me about him is not known.

 

I admit that this rule can applies to a particular class of matrices.

Edited September 23, 2011 by Alexander Masterov

[uncool]

Galilean transformations for real coordinates only. (Real coordinates to real coordinates.)

Relativism is visual effect only.

What is a "visual effect"? What does that mean physically?

 

Even if such a rule exists, then to me about him is not known.

 

I admit that this rule can applies to a particular class of matrices.

Then what is special about transformation matrices that it should be true for them?

=Uncool-

Edited September 23, 2011 by uncool

[Alexander Masterov]

BBC news

22 September 2011 Last updated at 17:28 GMT

Speed-of-light experiments give baffling result at Cern

 

hyperz, my thanks to you for this news.

 

 

What is a "visual effect"? What does that mean physically?

Visual effects and sound effects have approximately equal ratio to the real coordinates.

Then what is special about transformation matrices that it should be true for them?

The transformation matrices have meaning for real coordinates only.

[uncool]

BBC news

22 September 2011 Last updated at 17:28 GMT

Speed-of-light experiments give baffling result at Cern

 

hyperz, my thanks to you for this news.

 

 

Visual effects and sound effects have approximately equal ratio to the real coordinates.

This means nothing. What physically do you mean by "visual effects"?

 

The transformation matrices have meaning for real coordinates only.

That doesn't say anything about why transformation matrices must have the property mentioned.

=Uncool-

[Alexander Masterov]

Visual effects and sound effects have approximately equal ratio to the real coordinates.

This means nothing.

This means that the visual effects (arising due to the special properties of EMF) do not have direct coupling to a transformation of coordinates (and in general - to the coordinates). In any case: properties of light have coupling to a coordinates transformation to the same extent that the sound to coupling to it.

What physically do you mean by "visual effects"?

Because of the finite speed of light we see distant stars and galaxies as they were millions of years ago. Everyone knows that stars and galaxies are now in a different place and differently look.

This is an example of visual coordinates and visual effects.

That doesn't say anything about why transformation matrices must have the property mentioned.

the property mentioned??? Edited September 23, 2011 by Alexander Masterov

[Alexander Masterov]

The mathematical expression:

 

[math]x^2-(ct)^2=(x')^2-(ct')^2=0[/math]

 

is not a consequence of the universality of the speed of light.

 

So the result is the following expression:

 

[math]x^2-(ct)^2=(x'-vt')^2-(ct')^2=0[/math]

 

Look at the following animation to see this:

 

 

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. The first expression allege that these times are identical.

 

Let solve one a school's puzzle:

 

1. Two travelers do walking on the road with equal speed ([math]v[/math]) in one direction at a distance ([math]L[/math]) from each other.

 

2. Between them runs a dog (speed of it [math]c[/math]).

 

QUESTION: How much time a dog runs forward, and how many - back?

 

ANSWER: [math]T_1=L/(c+v)[/math] and [math]T_2=L/(c-v)[/math]

 

But Einstein, Lorentz, Minkowski and others (who entrust to them) say that time will be the same. So is was alleged by the first expression.

==============================================================

 

 

Let talk about an electron-neutrino and an electron-antineutrino.

 

The latter occurs in a pair with a positron. (A first - in a pair with an electron.)

 

But those neutrinos (and antineutrinos) are not some separate particles.

 

When paired with an electron positron born. And nothing else. What is called the electron neutrino is a positron, whose rate is greater than the speed of light. Well, the electronic antineutrino is - an electron whose velocity is greater than the speed of light.

 

The mass of both are 0.5 MeV (0.28 eV obtained by SRT-formulas, when the mass of the speed dependent). But mass no depend on a speed today.

 

These neutrinos and antineutrinos are produced as a result of common (cascading) decay, when an particle decays into a pair of electron + positron. In some cases, the electron flies in the same direction, which gave birth to it. In this case, the electron (speed of which > c) is called the electron antineutrino.

 

If the positron flew forward (the electron - back) then the positron called by electron neutrino.

 

In those cases where the electron and positron fly away (to broadside direction), and result of their speed does not exceed the speed of light, we are witnessing the birth of an pair of electron+positron.

 

It would be strange to call the bus, which moves in some other way (not, as we call the bus, which stoped). But physics have it. We called neutrino all elementary particles (whose velocity is greater than the speed of light).

Edited September 29, 2011 by Alexander Masterov

[uncool]

Alex, the reason I asked you to express it in matrices is the following:

 

The only reasoning that leads to requiring that slowdowns for one person requires speedups for the other is that you think inverse matrices in general must have the property about the upper left element. You can't explain any reasoning as to why there should be anything special about the matrices that would require speedups. You are simply asserting, and it is false.

=Uncool-

[Alexander Masterov]

My programs can not interpret your message.

Your sentences are too long.

Too many words of one sentence, so translation is difficult.

Edited September 29, 2011 by Alexander Masterov

[uncool]

You have said that if one person sees the other person slow down, then the 2nd person must see the first speed up. You have not given a specific reason why.

=Uncool-

[Alexander Masterov]

If time of your companion will slow, then he will be see the acceleration of your time. I am sure that you argue with that you will not.

Alternatively: if your partner sees and slowing down your time - is impossible. Otherwise we would have the ambiguity.

Reasons time dilation does not matter.

Relativistic effect of delay could be a reason for it, but relativism can not slow down time for both.

Edited September 29, 2011 by Alexander Masterov

[uncool]

If time of your companion will slow, then he will be see the acceleration of your time. I am sure that you argue with that you will not.[/QUOTe]

I will, actually. That was the whole point of the matrices - to show that it does make sense to have time dilation in both directions.

Alternatively: if your partner sees and slowing down your time - is impossible. Otherwise we would have the ambiguity.

And why is ambiguity a problem? As long as cause always precedes effect - which is true in relativity - there is no problem with ambiguity. That is the reason it is called "The relativity of simultaneity".

=Uncool-

[Alexander Masterov]

And why is ambiguity a problem? As long as cause always precedes effect - which is true in relativity - there is no problem with ambiguity.

Slowing down time of your buddy is the reason that he sees the acceleration of your time.

The time delay of your buddy can not is the cause of the acceleration.

If the first case does not violate the principle of causality, the second must to do it.

That is the reason it is called "The relativity of simultaneity".

The lie can be called by different names, but this name can not convert it to truth. Edited September 30, 2011 by Alexander Masterov

[uncool]

Slowing down time your buddy is the reason that he sees the acceleration of your time.

You have not explained why this is necessary.

The time delay of your buddy can not is the cause of the acceleration.

If the first case does not violate the principle of causality, the second must to do it.

Again, relativity does not violate causality. Relativity states that if event A causes event B, then event B must be in the future light-cone of event A. If that is true, then no change of frame can cause event B to happen before event A, so causality is not violated.

=Uncool-

[Alexander Masterov]

Slowing down time of your buddy is the reason that he sees the acceleration of your time.

You have not explained why this is necessary.

I demonstrated to you the inadmissibility of the simultaneous slowing of time.

 

If one sees the acceleration, while the second sees a slowdown, and this is logical, then other option (two slowing )can not be logical.

The time delay of your buddy can not is the cause of the acceleration.

If the first case does not violate the principle of causality, the second must to do it.

Again, relativity does not violate causality. Relativity states that if event A causes event B, then event B must be in the future light-cone of event A. If that is true, then no change of frame can cause event B to happen before event A, so causality is not violated.

 

This x²-(ct)²=(x')²-(ct')²=0 (light-cone) is not correct.

 

 

This x²-(ct)²=(x'-vt')²-(ct')²=0 is correct.

 

 

Edited September 30, 2011 by Alexander Masterov

[uncool]

I demonstrated to you the inadmissibility of the simultaneous slowing of time.

No, you never have.

If one sees the acceleration, while the second sees a slowdown, and this is logical, then other option (two slowing )can not be logical.

First, under special relativity we don't see any two frames where one accelerates time relative to the other and the other slows down. Second, why does one exclude the other?

This x²-(ct)²=(x')²-(ct')²=0 (light-cone) is not correct.

You have yet to demonstrate this.

=Uncool-

[Alexander Masterov]

No, you never have.

???

First, under special relativity we don't see any two frames where one accelerates time relative to the other and the other slows down.

Yes. But reality has it and reality has not double time dilation.

Second, why does one exclude the other?

Similar ambiguity is impossible.

You have yet to demonstrate this.

#140 Edited September 30, 2011 by Alexander Masterov