Master Theory (edition 2)

[Masterov]

This doesn't stop me from pointing out the errors in your theory.

You do not indicate an error. You (in rough form) are allege about my error, but not able to substantiate his claim.

So behave religious fanatics. You are not a scientist. You - a religious fanatic who professes "scientific" theories such as Einstein's theory.

Scientist not only indicates an error opponent, but proves it.

The fanatic knows the correct answer, but can not prove its correctness.

Scientists theorize that knows the correct answer, and in the controversy is trying to prove it.

You did not interest me.

 

[Tach]

You do not indicate an error.

You have some very strong delusions.

You (in rough form) are allege about my error, but not able to substantiate his claim.

Really?Then how come you are unable to explain the null result of Michelson-Morley with your "theory"? How come that you can't show the invariance of the Maxwell equations? How come you don't understand the transverse Doppler effect? How come you can't explain why the muons formed high in the atmosphere reach the Earth? Eh?

So behave religious fanatics. You are not a scientist. You - a religious fanatic who professes "scientific" theories such as Einstein's theory.

LOL.

Scientist not only indicates an error opponent, but proves it.

I did, and so did rpenner, you are just too stupid to understand the proofs.

 

[Masterov]

You have some very strong delusions.

1. You are not proved it.

2. You importuned me (as a gypsy on market-place).

 

[Tach]

1. You are not proved it.

Of course not, this is why you are a crackpot, insensitive to all disproofs

 

[Masterov]

Of course not, this is why you are a crackpot, insensitive to all disproofs

From this point on, I'm going to ignore you.

 

[Tach]

From this point on, I'm going to ignore you.

That's ok, the mainstream physicists ignore your crackpot theories.

 

[rpenner]

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).

Moving across a view axis (the distance does not change):

$$\Delta t' = \frac{\Delta t}{\sqrt{1 - \frac{v^2}{c^2}}}$$

$$\Delta x = 0$$

I was referring to the case:

$$d|\vec r|/dt = 0$$

$$d\phi/dt = const$$

Only transverse Doppler effect.
Longitudinal Doppler effect is absent completely.

How is that a rebuttal of my disproof of your claim that "Return home will require the acceleration, which again slows down time. Subsequent braking will slow down again" ? (emphasis added)

You have either lost track of what you were claiming or dishonestly changed the subject.

 

[Masterov]

Acceleration is the acceleration, even if you call it braking. So: any inclusion of jet engines will result in a time dilation.

You have either lost track of what you were claiming or dishonestly changed the subject.

I did not change the subject, but tried to return to it.

 

[Masterov]

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.

 

[Tach]

.

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.

...which is a piece of garbage no one gives a sh^t about

 

[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.

 

[rpenner]

Wrong thread. All discussion of the OPERA neutrino experiment and evaluations of it's strength belong here: http://sciforums.com/showthread.php?t=110051

 

[Masterov]

Wrong thread. All discussion of the OPERA neutrino experiment and evaluations of it's strength belong here: http://sciforums.com/showthread.php?t=110051

It's new for me too. Thank you.

Master Teory meet with experimental approval.

Einstein theory meet with experimental disproof.

It cause joy to me.

 

[Tach]

It's new for me too. Thank you.

Master Teory meet with experimental approval.

Einstein theory meet with experimental disproof.

It cause joy to me.

You have no proof that the experiment confirms your garbage.
Besides, what will you do when the experiment is proven wrong?

 

[Masterov]

Einstein bestowed the absoluteness to a cross-scale by $$ x^2 - (ct)^2 =x'^2 - (ct')^2=0 $$, but this expression is not correct.($$T_1'\neq T_2'$$)
$$ x^2 - (ct)^2 =0$$ - the observer stood motionless.
$$x_1=ct$$
$$x_2=-ct$$
$$ (x' - (c+v)t_1')(x' - (c-v)t_2')=0$$ - the observer moved.
$$x_1'=(c+v)t_1'$$
$$x_2'=(-c+v)t_2'$$
$$x_1'=-x_2'=x'$$
$$t_1'+t_2'=2t$$
$$t_2'-t_1'=\frac{x'}{c-v}-\frac{x'}{c+v}=\frac{2x'v}{c^2-v^2}$$
$$t_1'=t-\frac{x'v}{c^2-v^2}$$; $$t_2'=t+\frac{x'v}{c^2-v^2}$$
$$(x' - (c+v)(t-\frac{x'v}{c^2-v^2}))(x' - (c-v)(t+\frac{x'v}{c^2-v^2}))=0$$

$$(x' - (c+v)t+\frac{x'v}{c-v})(x' - (c-v)t-\frac{x'v}{c+v})=0$$
$$(x'(c-v) - (c^2-v^2)t+x'v)(x'(c+v) - (c^2-v^2)t-x'v)=0$$
$$(x'c - (c^2-v^2)t)(x'c - (c^2-v^2)t)=0$$
$$(x'c - (c^2-v^2)t)^2=0$$
$$\frac{x'}{1-v^2/c^2}-ct=0$$
$$x'=x(1-v^2/c^2)$$

For SRT:
$$t_1'+t_2'=2t/\sqrt{1-v^2/c^2}$$
$$x'=x\sqrt{1-v^2/c^2}$$
$$x'^2 - (ct')^2\neq 0 $$
$$(x' - (c-v)t_1')(x' - (c+v)t_2')=0$$
Einstein was wrong.
______________________________________

Tach, I do not read your messages, because I have no translation of it. But your cackling do weary people who can (and therefore - are forced to) read your crow. You are importunate to an obscene.

You must roll into a tube your mustachioed uncle and his wacky theories and shove it up your ass. Place of it there.

I hope that the moderator will not to do summons to me (to force me to do courtesy for this boor).

 

[Masterov]

Easy mathematical problem:
Two travelers going same speed (v)
L - distance one another.
A dog runs Between them (c - speed of it).
How long does dog to run forth and how long - back?

A childs capable of solve this problem, but Sirs: Einstein, Lorentz, Minkowski, and all those who professed SRT - no capable.

 

[Tach]

For SRT:
$$t_1'+t_2'=2t/\sqrt{1-v^2/c^2}$$
$$x'=x\sqrt{1-v^2/c^2}$$
$$x'^2 - (ct')^2\neq 0 $$
$$(x' - (c-v)t_1')(x' - (c+v)t_2')=0$$
Einstein was wrong.
.

You don't have a clue what SRT is saying.

 

[AlphaNumeric]

$$ x^2 - (ct)^2 =0$$ - the observer stood motionless.

No, that means the object is moving at light speed. If $$v = \frac{dx}{dt}$$ and $$x = ct$$ then v=c. Simple.

That invalidates everything afterwards because you put in v incorrectly.

Einstein was wrong.

No, you just can't do simple algebra.

You must roll into a tube your mustachioed uncle and his wacky theories and shove it up your ass. Place of it there.

Is that some non-English phrase which doesn't translate properly?

I hope that the moderator will not to do summons to me (to force me to do courtesy for this boor).

No, I think they will 'do summons to you' for telling someone to shove it up their ass.

 

[Masterov]

No, that means the object is moving at light speed. If $$v = \frac{dx}{dt}$$ and $$x = ct$$ then v=c. Simple.

That invalidates everything afterwards because you put in v incorrectly.

No, you just can't do simple algebra.

Give prove (in SRT context): $$T_1'=T_2'$$

______________________________________

Trailer record of previous my post was addressed to Tach (not to you).

 

[Tach]

Give prove (in SRT context): $$T_1'=T_2'$$

In the frame of the train, $$T_1=T_2$$. This was known since Galilei. Are you THAT ignorant?

In any other frame, moving with respect to the train, $$T'_1 \ne T'_2$$. This was also known since Galilei. Are you THAT ignorant?