Three Experiments Challenging SRT

**przyk** · 07-15-12, 03:08 AM

Originally Posted by Masterov

It is error.

That's just repetition, not a rebuttal. There's nothing wrong with $x'^{2} \,-\, (ct')^{2} \,=\, x^{2} \,-\, (ct)^{2}$ . Your application of it in the wrong context is your error, as I explained. When you misapply a theory, that's your fault. Not the theory's.

**AlphaNumeric** · 07-15-12, 06:01 AM

Originally Posted by Masterov

Without your formula and without your scientific argumentations?
Rational conversation impossible.

przyk has provided formulas but you haven't responded to them with any level of understanding. The problem here is you, not us. You just assert things without evidence as you just did when you claimed it's banned in Germany to disagree with relativity. You don't provide any evidence and you ignore the evidence I provide that show you're mistaken.

Originally Posted by Masterov

I have not argued that there is a conspiracy.
It's you I impose this idea.

You are, you're just not using that word. You're claiming scientists know relativity is wrong, the LHC was a massive con job and there's a decree of silence for anything disagreeing with relativity. That is a conspiracy, whether you want to call it so or not.

Originally Posted by Masterov

I have a physics education, but actually I am a programmer mathematician.
I have been programming since 1979.

I don't believe you.

Originally Posted by Masterov

In 1986 I created a method for analyzing nonlinear dynamical systems with distributed parameters (not discrete). It is my graduate work at the university.

You mean continuous parameters, not distributed. Parameters in non-linear dynamical systems regularly have continuous domains. The Lorenz attractor and the logistic map being examples known to anyone who has studied non-linear systems.

[QUOTE=Masterov;2958185]In 1996 I wrote a monograph, which outlines methods for the analysis of strongly nonlinear dynamical systems.

Originally Posted by Masterov

COMMENT: people think that explore the nonlinear dynamical system is much more complicated than it is possible for linear systems.
In his book, I show that investigate the models of nonlinear dynamical systems (even with hysteresis) is simpler than that for linear systems.

That's demonstrably false. Linear systems are special cases of non-linear systems so a method which makes non-linear systems simple will work on linear systems too, you just set then non-linearity to zero in your results.

Originally Posted by Masterov

I was not allowed to publish my work, which I worked for over ten years.
I was not allowed to publish a monograph as a preprint even.

Firstly, the amount of time you spend on a piece of work is irrelevant. See point 11. Secondly, if threads like this are anything to go by you have very poor mathematical skills so if you did write such a thing I'm not in the least bit surprised it got rejected. There are other people on this forum who have put in the better part of a decade into their 'pet theory' and gotten rejected because it was rubbish. You've shown you don't even know what the origin of the Lorentz transforms are, as przyk has highlighted. And Lorentz transforms are linear operators!

If you could grasp Lorentz transforms properly you'd see how particle accelerators do test the mass-energy-momentum relationship in quantum field theory. Unfortunately it is outside of your ability to understand and therefore you cling to just one thing, one thing you think you understand, and demand everyone explains things in terms of it (the $m' = \gamma m$ expression). Your inability to understand evidence doesn't mean it isn't there.

Originally Posted by Masterov

It is error.

Correct so:

$(ct')^{2} - x'^{2} = (ct)^{2} - (x-vt)^{2}$

Otherwise: speed of light is not an absolute value, and an anisotropy in space.

As przyk said, you're just repeating assertions. Furthermore that claim of yours is explicitly falsified by experiments. The momentum-energy-mass version of $(ct')^{2} - x'^{2} = (ct)^{2} - x^{2}$ is $-m^{2} = -E^{2} + |p|^{2} = -(E')^{2} + |p'|^{2}$ . If you were right then this relationship would be altered to $-m^{2} = -E^{2} + |p-mv|^{2} = -(E')^{2} + |p'|^{2}$ . The relationship $-m^{2} = -E^{2} + |p|^{2} = -(E')^{2} + |p'|^{2}$ is used by particle physicists to predict the scattering amplitudes of colliding beams in accelerator detectors. The simplest frame to do the calculations is the centre of mass frame but measurements are done in the detector rest frame. As a result the Lorentz transform $-m^{2} = -E^{2} + |p|^{2} = -(E')^{2} + |p'|^{2}$ is applied, the calculations done and then the relationship is inverted to be back in the detector's rest frame. If $-m^{2} = -E^{2} + |p|^{2} = -(E')^{2} + |p'|^{2}$ was incorrect and actually it was as you claim then predictions for the outcome of experiments would be wildly inaccurate. In fact they are very accurate. While this doesn't prove relativity (nothing could do that) it does disprove claims inconsistent with the result, including your $(ct')^{2} - x'^{2} = (ct)^{2} - (x-vt)^{2}$ . That is demonstrably not how time and space behave.

**Masterov** · 07-15-12, 10:50 AM

Originally Posted by przyk

It is error.

That's just repetition, not a rebuttal. There's nothing wrong with $x'^{2} \,-\, (ct')^{2} \,=\, x^{2} \,-\, (ct)^{2}$ . Your application of it in the wrong context is your error, as I explained. When you misapply a theory, that's your fault. Not the theory's.

Do you deny the obvious.

**Masterov** · 07-15-12, 10:53 AM

Originally Posted by AlphaNumeric

Without your formula and without your scientific argumentations?
Rational conversation impossible.

przyk has provided formulas...

Speak for yourself.

**Masterov** · 07-15-12, 10:56 AM

Originally Posted by AlphaNumeric

I have not argued that there is a conspiracy.
It's you I impose this idea.

You are, you're just not using that word. You're claiming scientists know relativity is wrong, the LHC was a massive con job and there's a decree of silence for anything disagreeing with relativity. That is a conspiracy, whether you want to call it so or not.

I try to call a spade a spade.

**Masterov** · 07-15-12, 10:59 AM

Originally Posted by AlphaNumeric

I have a physics education, but actually I am a programmer mathematician.
I have been programming since 1979.

I don't believe you.

I do not care.

**Masterov** · 07-15-12, 11:16 AM

Originally Posted by AlphaNumeric

In 1986 I created a method for analyzing nonlinear dynamical systems with distributed parameters (not discrete). It is my graduate work at the university.

You mean continuous parameters, not distributed. Parameters in non-linear dynamical systems regularly have continuous domains. The Lorenz attractor and the logistic map being examples known to anyone who has studied non-linear systems.

My book offers analytical methods of analysis (analytical and numerical analysis), the integro-differential essentially nonlinear (with hysteresis) equations in which there is no discrete parameters and space (which identified the problem) is continuous and non-limited.

Download monography (Russian language).

**Masterov** · 07-15-12, 11:30 AM

Originally Posted by AlphaNumeric

COMMENT: people think that explore the nonlinear dynamical system is much more complicated than it is possible for linear systems.
In his book, I show that investigate the models of nonlinear dynamical systems (even with hysteresis) is simpler than that for linear systems.

That's demonstrably false. Linear systems are special cases of non-linear systems so a method which makes non-linear systems simple will work on linear systems too, you just set then non-linearity to zero in your results.

No.
Linearity is a special case of nonlinearity.
A linear function is a special case of approximation of the nonlinearity.
Not for any approximation of nonlinearity the solution can be written as a finite number of functions, whose properties are studied.
One method allows you to choose an approximation of the nonlinearity (in the class of parametric functions) for which the solution can be written as a finite number of known functions.

**Masterov** · 07-15-12, 11:37 AM

Originally Posted by AlphaNumeric

As przyk said, you're just repeating assertions. Furthermore that claim of yours is explicitly falsified by experiments. The momentum-energy-mass version of $(ct')^{2} - x'^{2} = (ct)^{2} - x^{2}$ is $-m^{2} = -E^{2} + |p|^{2} = -(E')^{2} + |p'|^{2}$ . If you were right then this relationship would be altered to $-m^{2} = -E^{2} + |p-mv|^{2} = -(E')^{2} + |p'|^{2}$ . The relationship $-m^{2} = -E^{2} + |p|^{2} = -(E')^{2} + |p'|^{2}$ is used by particle physicists to predict the scattering amplitudes of colliding beams in accelerator detectors. The simplest frame to do the calculations is the centre of mass frame but measurements are done in the detector rest frame. As a result the Lorentz transform $-m^{2} = -E^{2} + |p|^{2} = -(E')^{2} + |p'|^{2}$ is applied, the calculations done and then the relationship is inverted to be back in the detector's rest frame. If $-m^{2} = -E^{2} + |p|^{2} = -(E')^{2} + |p'|^{2}$ was incorrect and actually it was as you claim then predictions for the outcome of experiments would be wildly inaccurate. In fact they are very accurate. While this doesn't prove relativity (nothing could do that) it does disprove claims inconsistent with the result, including your $(ct')^{2} - x'^{2} = (ct)^{2} - (x-vt)^{2}$ . That is demonstrably not how time and space behave.

$(ct')^{2} - x'^{2} = (ct)^{2} - x^{2}$ - it is error.
Look again:

If speed of light is constant then: "time left" $\neq$ "time right"

Formula $(ct')^{2} - x'^{2} = (ct)^{2} - x^{2}$ provide two identical roots.
It is error.

**Kittamaru** · 07-15-12, 12:25 PM

Masterov, please, for the love of science... stop and think about what is being said before replying... oh, and you do know you can reply to multiple people/quotes within a single post, right?

**przyk** · 07-15-12, 12:37 PM

Originally Posted by Masterov

Do you deny the obvious.

No. Do you? Your equation does not have an invariant c.

For a ray travelling at the speed of light in the positive x direction, $x \,=\, ct$ , which implies $x^{2} \,-\, (ct)^{2} \,=\, 0$ .

So according to your equation, in a boosted frame:

$ \begin{eqnarray} (x' - vt')^{2} \,-\, (ct')^{2} &=& 0 \\ x' \,-\, vt' &=& \pm\, ct' \\ x' &=& (v \,\pm\, c) t' \,. \end{eqnarray} $

So according to your equation, the velocity of light transforms according to $c' \,=\, v \,\pm\, c$ . Is that your idea of "invariant"?

**Masterov** · 07-15-12, 12:39 PM

Originally Posted by Kittamaru

Masterov, please, for the love of science... stop and think about what is being said before replying... oh, and you do know you can reply to multiple people/quotes within a single post, right?

My English is not good enough for to read it.

**Masterov** · 07-15-12, 12:45 PM

Originally Posted by przyk

No. Do you? Your equation does not have an invariant c.

For a ray travelling at the speed of light in the positive x direction, $x \,=\, ct$ , which implies $x^{2} \,-\, (ct)^{2} \,=\, 0$ .

So according to your equation, in a boosted frame:

$ \begin{eqnarray} (x' - vt')^{2} \,-\, (ct')^{2} &=& 0 \\ x' \,-\, vt' &=& \pm\, ct' \\ x' &=& (v \,\pm\, c) t' \,. \end{eqnarray} $

$t'_0=x'/(c+v)$

$t'_1=-x'/(c-v)$

**przyk** · 07-15-12, 12:52 PM

Originally Posted by Masterov

$t'_0=x'/(c+v)$

$t'_1=-x'/(c-v)$

$c' \,=\, v \,\pm\, c$ . Your equation predicts a varying c.

**AlphaNumeric** · 07-15-12, 01:07 PM

I provide equations, illustrate their application is supported by reality and Masterov fails to retort it while also complaining I don't provide any equations. Thanks Masterov, you make it so much easier when you're blatantly dishonest.

**przyk** · 07-15-12, 01:14 PM

Here's another proof that your equation,

$ (x' - v't)^{2} \,-\, (ct')^{2} \,=\, x^{2} \,-\, (ct)^{2} \,, $

is wrong. When I was born, I was 0 years old and born 0 km from where I was born, so $t \,=\, 0$ and $x \,=\, 0$ , and therefore:

$ x^{2} \,-\, (ct)^{2} \,=\, 0 \,. $

Now, I'm 25 years old and I live about 756 km away from where I was born, and I'm sitting still, so $t' \,=\, 25\,\mathrm{yr}$ , $x' = 756\,\mathrm{km}$ , and $v \,=\, 0$ . In the units I'm using, $c \,\approx\, 9.46\,\cdot\,10^{12}\,\mathrm{km}/\mathrm{yr}$ . So:

$ (x' - vt')^{2} \,-\, (ct')^{2} \,\approx\, - 5.60\,\cdot\,10^{28}\,\mathrm{km}^{2}/\mathrm{yr}^{2} \,. $

Obviously here,

$ (x' - v't)^{2} \,-\, (ct')^{2} \,\neq\, x^{2} \,-\, (ct)^{2} \,. $

So your equation is wrong.

**przyk** · 07-15-12, 01:30 PM

By the way, we still haven't seen a meaningful reply to this post (which you largely ignored, despite it providing and linking to mathematics you asked for) or this post (which explains the error you keep making with Minkowski's equation, and which you obviously ignored).

Something from one of your previous posts you might like to recall:

Originally Posted by Masterov

Your efforts to stem the tide of scientific publications that will refute the SRT, will not give any results. Previously, you could control the printing press, but you can not undo the publicity on the Internet.

Publicity on Internet will make obvious (for all) falsity of SRT.

Unfortunately for you, you're wielding a double edged sword. If you keep making the same stupid mistakes, even when they're repeatedly explained to you, everyone on the internet gets to see that too. Just something you might like to consider before posting yet another of your knee-jerk replies.

**Masterov** · 07-15-12, 02:18 PM

Your qualifications are not enough to solve the problem of school physics course.

Shame!

Look my post again: Lorentz's error.

**AlphaNumeric** · 07-15-12, 03:24 PM

Anyone who has access to a book on special relativity or the internet can check that your repeated assertion Lorentz transforms are based on $x^{2} - (ct)^{2} = (x')^{2} - (ct')^{2} = 0$ is false. You obviously haven't even bothered to find out what Lorentz transforms do and what impact they have. My posts illustrating their application to experimental data and their consistency with said data back up what przyk has been saying. You keep saying $(x')^{2} - (ct')^{2} = 0$ but that isn't a general result. The sign of $x^{2} - (ct)^{2}$ determines whether a location is time-like, space-like or null with respect to the origin. Setting it equal to zero is the third case. This is the causal structure of space-time in special relativity and a central concept. One you have failed to find out about and refuse to learn even when it's put in front of you.

Do you think if you lie enough time reality will bend to your desires? Do you think you're going to convince professional physicists, which przyk and I both are, you're worth listening to by lying about work we actually do?

**arfa brane** · 07-15-12, 10:00 PM

You can derive the Lorentz factor from first principles using mirrors, reflected beams of light, and two observers.
All you need is Pythagoras and the assumption that the speed of light is the same for inertial observers.

You have one observer at say, A, the other at B and the mirror at C. A is stationary wrt "the ground", and B plus the mirror which is a distance h from B are stationary wrt each other.
B plus the mirror are comoving relative to A at constant velocity, v.
B emits light towards the mirror at time t, and sees it arrive back at t' so for B t' - t = 2h/c = t_B. During the same interval t' - t (from B's perspective), A sees B plus mirror move a distance vt_A, say along A's x axis. But A also thinks the light travels a greater distance.

If you draw the diagram ('meh') and use Pythagoras, you get the factor 1/√(1 - v²/c²) between B's and A's measured times for the path length of a light pulse, where B is stationary wrt mirror and A isn't.

Thereby demonstrating how, for observers with relative velocities, time is also relative.