# Three Experiments Challenging SRT

Discussion in 'Pseudoscience Archive' started by Masterov, Jun 12, 2012.

1. ### MasterovRegistered Senior Member

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Three Experiments Challenging Einstein’s Relativistic Mechanics and Traditional Electromagnetic Acceleration Theory
Liangzao FAN
Senior Research Fellow, Chinese Academy of Science
fansixiong@yahoo.com.cn

Abstract
First Experiment: The speed of electrons accelerated by a Linac was measured in order to clarify whether the Linac’s effective accelerating force depends upon the speed of electrons or not. Second experiment: High-speed electrons from a Linac bombarded a lead target and the increase of the target’s temperature was measured. Third experiment: High-speed electrons from a Linac were injected perpendicularly into a homogeneous magnetic field and the radius of circular motion of the electrons under the action of the Lorentzian deflecting force was measured. Analyses of all the three experiments prove: (1) The accelerator’s efficiency decreases as the speed of electrons increases and the measured speed of electrons is far less than calculated according to the traditional electromagnetic acceleration theory. (2) Results of the experiments do not accord with Einstein’s formulas of moving mass and kinetic energy but conform with the formulas in the newly developed Galilean relativistic mechanics. (3) The third experiment proves that the effectiveness of the Lorentzian deflecting force also depends upon the speed of the deflected electrons.

More: http://ivanik3.narod.ru/TO/DiHUALiangzaoFAN/3LiangzaoFAN.doc

3. ### MasterovRegistered Senior Member

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Let Coulomb force depends on the speed as follows: $F=eU(1-v^2/c^2)/d$
Then: $m_e\ddot{x}=eU(1-\dot{x}^2/c^2)/d$
Integration:
$\int d(\dot{x}/c)/(1-\dot{x}^2/c^2)=eUt/(m_ed)=Wct/d$
$W=eU/m_ec^2$
$(1/2)ln((1+\dot{x}/c)(1-v_o/c)/(1-\dot{x}/c)(1+v_o/c))=Wct/d$
$\tau=Wct/d=eUt/(m_edc)$
$d\tau=Wc\ dt/d=eU\ dt/(m_edc)$
$dt=(d/Wc)d\tau$
$(1+\dot{x}/c)(1-v_o/c)/(1-\dot{x}/c)(1+v_o/c)=e^{2\tau}$
$(1+\dot{x}/c)/(1-\dot{x}/c)=e^{2\tau}(1+v_o/c)/(1-v_o/c)=e^{2\tau}a$
$a=(1+v_o/c)/(1-v_o/c)$
$1+\dot{x}/c=e^{2\tau}a-e^{2\tau}a\dot{x}/c$
$\dot{x}/c=(e^{2\tau}a-1)/(e^{2\tau}a+1)$
So I found a speed as a function of time.

I'm looking for an time of flight distance d
$dx=(d/W)(e^{2\tau}a-1)d\tau/(e^{2\tau}a+1)$
$dx=(d/W)(e^{2\tau}a-1)de^{2\tau}/2e^{2\tau}(e^{2\tau}a+1)$
$dx=(d/W)(\xi a-1)d\xi /2\xi(\xi a+1)$
$\xi=e^{2\tau}$
$dx=(d/W)(\xi a+1-2)d\xi /2\xi(\xi a+1)$
$dx=(d/W)(1/2\xi-1/\xi(\xi a+1))d\xi$
$dx=(d/W)(1/2\xi-a(\xi a+1-\xi a)/a\xi(\xi a+1))d\xi$
$dx=(d/W)(1/2\xi-a(1/a\xi-1/(\xi a+1)))d\xi$
$dx=(d/W)(1/2\xi-1/\xi+a/(a\xi+1)))d\xi$
$dx=(d/W)(a/(a\xi+1)-1/2\xi)d\xi$
$x=(d/W)(ln(a\xi+1)-ln(\xi)/2)+C$
$x=(d/W)\ ln(a\ e^{\tau}+e^{-\tau})+C$
$d=(d/W)\ ln((a\ e^{\tau}+e^{-\tau})/(a+1))$
$W=ln((a\ e^{\tau}+e^{-\tau})/(a+1))$
$a\ e^{\tau}+e^{-\tau}=(a+1)e^W=2b$
$b=(a+1)e^W/2=e^W/(1-v_o/c)$
$a\ e^{2\tau}-2be^{\tau}+1=0$
$e^{\tau}=(b+\sqrt{b^2-a})/a$
$\tau=ln((b+\sqrt{b^2-a})/a)$
$\tau=ln((e^W+\sqrt{e^{2W}-1+v^2/c^2})/(1+v_o/c))$

Preliminary test:
1. Let the initial velocity is absent (a = 1), and the accelerating voltage (W) is very large....
$\tau=W\ ln(2)$
$eUt/(m_edc)=eU\ ln(2)/m_ec^2$
$t=ln(2)d/c$ - may be true.

Let the initial velocity equals the speed of light (the Coulomb force is zero):
$t=d/c$
Excellent!
The check was the result.

Next, substitute this:
$e^{\tau}=(b+\sqrt{b^2-a})/a$
$e^\tau=(e^W+\sqrt{e^{2W}-1+v_o^2/c^2})/(1+v_o/c)=(1-v_o/c)/(e^W-\sqrt{e^{2W}-1+v_o^2/c^2})$
$e^{2\tau}=(e^W+\sqrt{e^{2W}-1+v_o^2/c^2})/a(e^W-\sqrt{e^{2W}-1+v_o^2/c^2})$
to here:
$\dot{x}/c=(e^{2\tau}a-1)/(e^{2\tau}a+1)$
$v/c=(e^{2\tau}(1+v_o/c)-(1-v_o/c))/(e^{2\tau}(1+v_o/c)+(1-v_o/c))$Ïîëó÷èì ñêîðîñòü íà âûõîäå èç óñêîðÿþùåãî ïîëÿ:
$v/c=((e^W+\sqrt{e^{2W}-1+v_o^2/c^2})/(e^W-\sqrt{e^{2W}-1+v_o^2/c^2})-1)/((e^W+\sqrt{e^{2W}-1+v_o^2/c^2})/(e^W-\sqrt{e^{2W}-1+v_o^2/c^2})+1)$
$v/c=\sqrt{e^{2W}-1+v_o^2/c^2}/e^W$
$v/c=\sqrt{1-(1-v_o^2/c^2)e^{-2W}}$
$\sqrt{1-v^2/c^2}=\sqrt{1-v_o^2/c^2}e^{-W}$
$\gamma=\gamma_oe^W$
$\gamma=\gamma_oe^{eU/m_ec^2}$ --- final result.

Last edited: Jun 12, 2012

5. ### AlphaNumericFully ionizedRegistered Senior Member

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6,698
The first and third of those are done by accelerators already, since the on/off pulsing of the electromagnets must be timed precisely with the position of the particle bunches. This is so the accleration is done properly, both in terms of speeding up the particles and also in terms of keeping them moving in a circle which follows the accelerator's ring. The second experiment, smashing particles into stationary targets is precisely what RHIC and the LHC do.

All of the experiments have been done, the behaviour of particles observed and all of them fall in line with quantum field theory, which includes relativity,

You're so ignorant you don't even know the basics of physics. Didn't you even bother to find out how accelerators work?

7. ### Sylwester KornowskiNeutrinos are nonrelativisticRegistered Senior Member

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703
Masterov and AlphaNumeric they both are right. Accelerated particles IN REALITY change their internal structure. The increase in relativistic mass is the real effect. This causes that energy density of emitted energy by accelerated particle increases when speed increases. It is true that acceleration is more and more difficult when speed increases but true is also the fact that the Einstein formula m(relativistic) = m(rest)/sqrt(1 – vv/cc) is correct.

8. ### MasterovRegistered Senior Member

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SRT will be justified when it is proved that a temperature of target: T ~ Ue.
FAN's experiment deprives us of hope for it.

9. ### AlphaNumericFully ionizedRegistered Senior Member

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6,698
As usual you ignore evidence which shows you're mistaken and ignorant. The experiments you demanded to be done have been done. The behaviour of the particles is precisely as the relativistic model of quantum field theory says. Your claims that relativity would fail in such areas are demonstrably false. Now all you're doing is trying to change the subject. Again.

10. ### Sylwester KornowskiNeutrinos are nonrelativisticRegistered Senior Member

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Masterov, electrons are accelerated but they simultaneously emit energy. The energy density of emitted energy depends on speed. We must take into account this phenomenon. This phenomenon does not concern the SR. Due to the emission of energy by accelerated particle, there is not a simple relation between the U and speed of particles applied in the SR.

11. ### MasterovRegistered Senior Member

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728
For speed 0.5c<v<0.8c this problems absent, but relativity - present.

12. ### MasterovRegistered Senior Member

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Energy:

$F=eU(1-v^2/c^2)/d$

$m_e\ddot{x}=eU(1-\dot{x}^2/c^2)/d$

$\Delta E=\int_o^dFdx=\int_o^d m_e\ddot{x}dx=\int_o^d(eU(1-\dot{x}^2/c^2)/d)dx=(eU/d)\int_o^d(1-\dot{x}^2/c^2)dx$

insert it: $(1-\dot{x}^2/c^2)=(1-v_o^2/c^2)e^{-2Wx/d}$ to up.

$\Delta E=eU(1-v_o^2/c^2)\int_o^1e^{-2W\xi}d\xi$

$\Delta E=eU(1-v_o^2/c^2)(1-e^{-2W})/2W=m_ec^2(1-v_o^2/c^2)(1-e^{-2W})/2$

$E\ <\ m_ec^2/2\ =\ 255KeV$ - for all tames!

Last edited: Jun 13, 2012
13. ### BelieveHappy mediumValued Senior Member

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1,194
You seem to have skipped some steps in there. Mainly how you get from the 3rd line to the forth.

14. ### MasterovRegistered Senior Member

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Impulse:

$\Delta p=m_e(v-v_o)$
$(1-v^2/c^2)=(1-v_o^2/c^2)e^{-2W}$
$\Delta p=m_e(v-v_o)=m_e(c\sqrt{1-(1-v_o^2/c^2)e^{-2W}}-v_o)$

If: $v_o=0$
$p=m_ev=m_ec\sqrt{1-e^{-2W}}$

Testing:

Let W-small and $v_o=0$ (no-relativist):

$m_ev=m_ec\sqrt{2W}$

$v=c\sqrt{2Ue/m_ec^2}$

$v^2=c^2(2Ue/m_ec^2)$

$m_ev^2/2=Ue$

It's TRUE!

Last edited: Jun 13, 2012
15. ### AlphaNumericFully ionizedRegistered Senior Member

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3+4=7

It's true!

$\Lambda^{\top} \cdot \eta \cdot \Lambda = \eta$

It's true!

$g^{ab}R_{ab} = R$

It's true!

Simply throwing out equations doesn't mean that is how reality works. If I just listed a bunch of relativity equations would you take that to mean relativity is right? Hardly. So don't expect you trying that tactic will convince anyone.

16. ### James RJust this guy, you know?Staff Member

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32,024
Masterov:

Please do not spam the same post across several threads. If you insist on doing that, you will be banned from sciforums.

17. ### MasterovRegistered Senior Member

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General case for a reciprocally-perpendicular fields and for speed which is directed along the electric field:

$\vec F=(\vec E+q[\vec v\times \vec B])(1-v^2/c^2)$

Electron trajectory radius in a magnetic field

$R=m_ev/eB(1-v^2/c^2)$​

Last edited: Jun 14, 2012
18. ### MasterovRegistered Senior Member

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It's not spam.

FAN's experiments deserve individual theme.
I doing theoretical substantiations for FAN's experiments from Master Theory.

Master Theory is individual thema.
FAN's experiments confirm Master Theory.

Therefore formulas coincide.

Last edited: Jun 14, 2012
19. ### MasterovRegistered Senior Member

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I agree with you: errors Einstein and Lorentz simple and obvious.
We can only wonder that these errors were not detected for so long (one hundred years).

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21. ### martilloRegistered Senior Member

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877
Experiments of electrons at relativistic velocities in uniform magnetic fields give a different result than what you say.
Experimentally has been verified that the radius of the circular path verifies:
R=mv/eB.sqrt(1-v2/c2)
(sorry but I don't know Latex)
It varies with the term sqrt(1-v2/c2) and not just (1-v2/c2) as you state.
http://en.wikipedia.org/wiki/Kaufmann%E2%80%93Bucherer%E2%80%93Neumann_experiments
If you try to debunk Relativity Theory you must do something better.
The first two experiments you mention in the first post seems to give something interesting but the conclusion is only that there would be some "(c-v) effect" that should be studied further...

Last edited: Jun 14, 2012
22. ### MasterovRegistered Senior Member

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Let: $F=eU\sqrt{1-v^2/c^2}/d$

$m_e\ddot{x}=eU\sqrt{1-\dot{x}^2/c^2}/d$

$\int (d\dot{x}/c)/\sqrt{1-\dot{x}^2/c^2}=eUt/(m_ed)=Wct/d=\tau$

$\tau=Wct/d$
$W=eU/m_ec^2$

$arcsin(\dot{x}/c)=\tau+C$

$arcsin(\dot{x}/c)-arcsin(v_o/c)=\tau$

$arcsin(\dot{x}/c)=\tau+arcsin(v_o/c)$

$\dot{x}/c=sin(\tau+arcsin(v_o/c))$

$\dot{x}/c=\sqrt{1-v_o^2/c^2}sin(\tau)+(v_o/c)cos(\tau)$

$\dot{x}/c=\sqrt{1-v_o^2/c^2}sin(Wct/d)+(v_o/c)cos(Wct/d)$

I find time:

$d\tau=Wc\ dt/d$
$dt=(d/Wc)d\tau$

$dx=Wd\ sin(\tau+arcsin(v_o/c))d\tau$

$d=Wd(cos(arcsin(v_o/c))-cos(\tau+arcsin(v_o/c))$

$1=W(\sqrt{1-v_o^2/c^2}-cos(\tau+arcsin(v_o/c))$

$cos(\tau+arcsin(v_o/c))=\sqrt{1-v_o^2/c^2}-1/W$

$v/c=\dot{x}/c=sin(\tau+arcsin(v_o/c))$

$\sqrt{1-v^2/c^2}=cos(\tau+arcsin(v_o/c))=\sqrt{1-v_o^2/c^2}-1/W$

$\sqrt{1-v_o^2/c^2}-\sqrt{1-v^2/c^2}=1/W$

it's invalid.

Last edited: Jun 14, 2012
23. ### martilloRegistered Senior Member

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877
Well, arriving at an equation near to Einstein's equation: W= E=(m-m0)c2 where here m represents the "relativistic mass" varying with 1/sqrt(1-v2/c2). What does it means to you?

Last edited: Jun 15, 2012