Thank you!
Master Theory (edition 3)
- Thread starter Masterov
- Start date
My donkey does not like me.Masterov,
Ignore what the bad man says. Just think about how someday, scientists will generate gravitational redshift as a means of acceleration field generator propulsion. Someday, we will have starships that routinely travel to the moon, to Mars, Venus, and the asteroid belt where we will mine for ore. We will replace millions of transistors on IC chips with light emitting diodes. These IC LED packages will be used to generate gravity fields. All you have to do is ignore the bad man.
I was not disappointed by it.
My donkey uphill drags well.
I love my donkey for it.
Masterov,
Ignore what the bad man says. Just think about how someday, scientists will generate gravitational redshift as a means of acceleration field generator propulsion. Someday, we will have starships that routinely travel to the moon, to Mars, Venus, and the asteroid belt where we will mine for ore. We will replace millions of transistors on IC chips with light emitting diodes. These IC LED packages will be used to generate gravity fields. All you have to do is ignore the bad man.
Your world view is irrelevant because it doesn't correspond with reality. If you had some training in physics or just a smidgeon of commonsense you would have quit spamming public forums with your brand of nonsense. I view this as further proof of your inability to understand the literature and what it has to say about theories which are empirically falsified. You call me a bad man because I think you're a pariah. Ignorance is a choice. You choose ignorance. So does the originator of Master Theory.
Starships which routinely travel to the Moon, Mars, Venus, and the Asteroid Belt. To much. Star-ships which routinely travel to objects in our solar system? Expensive ride.
Last edited:
If you were not so blind to the grand scheme of the universe, maybe you would become more than just a hideous and hateful creature.
Consider particle-anti-particle creation-annihilation. An electron -Q and its anti-particle, the positron +Q annihilate to yield energy, gamma rays, photons. But where does energy come from? How do you create light?
+E is the energy of the big bang; -E is the gravitational energy of the space-time continuum. They are created, simultaneously, from nothingness: 0 = -E + E.
Gotta go!
Consider particle-anti-particle creation-annihilation. An electron -Q and its anti-particle, the positron +Q annihilate to yield energy, gamma rays, photons. But where does energy come from? How do you create light?
+E is the energy of the big bang; -E is the gravitational energy of the space-time continuum. They are created, simultaneously, from nothingness: 0 = -E + E.
Gotta go!
I'm only hideous and hateful to scientific illiterates such as yourself. You haven't learned anything in the two years physforum members have had to put up with your illiterate comments. Now your spreading your brand of bs to this site. Lucky I'm not a moderator. For you and all the 'dunce stool candidates' who disrespect the scientific literature. A good thing is SciForum has this Alternative section for 'round filed' nonsense.
Last edited:
The moderator jettisoned Pincho for his huge disconnect with reality and disrespect for the scientific literature over an extended period of time. I'd like to see more of that. Not my business.
I'm referring to the Zero Energy Universe.
http://arxiv.org/abs/gr-qc/0605063
Disrespect for the scientific literature, huh? The moderator should kick you off this website for making nothing but hate-filled personal attacks on others. I admit that I might speculate and interpret the physics a little beyond what I can back up with letter and verse of the scientific literature. But it is the moderator's decision if he wants to kill all discussion, and let the hatemongers roam freely.
On a more positive and productive note, it is my interpretation that the energy of the big bang +E, and the negative energy of gravity (the space-time continuum), -E, have a "creation/annihilation" relationship. The space-time continuum is, not the anti-particle, but instead the anti-energy of the big bang. The space-time continuum is a single universally large object that length contracts and frequency shifts to compensate for gravitationally large objects AND interconnects all particles (coordinate systems) into a relativistic observant medium. At the moment of the big bang, the energy content and the time dilation/length contracting space-time continuum were created. See attachment.View attachment 4579
Did you try to understand the paper or are you just quote mining?
The velocity you refer to is apparent velocity, a visual effect caused by the alignment of a fast moving object and a distant observer. It doesn't mean something is moving faster than light but rather a set of relativistic effects conspire to give the 'mirage' of the object being faster than light. The actual velocity, in the observer's frame, can be computed from the visual data using standard methods, as outlined in the paper or here.
Your quote of the paper even says it, that the apparent motion of 10c equates to an actual motion of 0.9999c.
Rather than just seeing what you want to see or being too illiterate/innumerate to understand things you quote take the time to learn some physics. I absolutely guarantee you'll go nowhere with any of the things you claim if you stick to the course you're on.
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
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
Last edited:
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
$$dx=(m_ed/ceU)(e^{2\tau}a-1)d\tau/(e^{2\tau}a+1)$$
$$dx=k(e^{2\tau}a-1)de^{2\tau}/2e^{2\tau}(e^{2\tau}a+1)$$
$$k=m_edc^2/eU$$
$$dx=k(\xi a-1)d\xi /2\xi(\xi a+1)$$
$$\xi=e^{2\tau}$$
$$dx=k(\xi a+1-2)d\xi /2\xi(\xi a+1)$$
$$dx=k(1/2\xi-1/\xi(\xi a+1))d\xi$$
$$dx=k(1/2\xi-a(\xi a+1-\xi a)/a\xi(\xi a+1))d\xi$$
$$dx=k(1/2\xi-a(1/a\xi-1/(\xi a+1)))d\xi$$
$$dx=k(1/2\xi-1/\xi+a/(a\xi+1)))d\xi$$
$$dx=k(a/(a\xi+1)-1/2\xi)d\xi$$
$$x=k(ln(a\xi+1)-ln(\xi)/2)+C$$
$$x=k\ ln(a\ e^{\tau}+e^{-\tau})+C$$
$$d=k\ ln((a\ e^{\tau}+e^{-\tau})/(a+1))$$
$$W=ln((a\ e^{\tau}+e^{-\tau})/(a+1))$$
$$a\ e^{\tau}+e^{-\tau}=(a+1)e^W=2b$$
$$b=(a+1)e^W/2=e^W/(1-v_o/c)$$
$$a\ e^{2\tau}-2be^{\tau}+1=0$$
$$e^{\tau}=(b+\sqrt{b^2-a})/a$$
$$\tau=ln((b+\sqrt{b^2-a})/a)$$
$$\tau=ln((e^W+\sqrt{e^{2W}-1+v^2/c^2})/(1+v_o/c))$$
Preliminary test:
1. Let the initial velocity is absent (a = 1), and the accelerating voltage (W) is very large....
$$\tau=W\ ln(2)$$
$$eUt/(m_edc)=eU\ ln(2)/m_ec^2$$
$$t=ln(2)d/c$$Ýòî ìîæåò áûòü ïðàâäîé.
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.
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
$$dx=(m_ed/ceU)(e^{2\tau}a-1)d\tau/(e^{2\tau}a+1)$$
$$dx=k(e^{2\tau}a-1)de^{2\tau}/2e^{2\tau}(e^{2\tau}a+1)$$
$$k=m_edc^2/eU$$
$$dx=k(\xi a-1)d\xi /2\xi(\xi a+1)$$
$$\xi=e^{2\tau}$$
$$dx=k(\xi a+1-2)d\xi /2\xi(\xi a+1)$$
$$dx=k(1/2\xi-1/\xi(\xi a+1))d\xi$$
$$dx=k(1/2\xi-a(\xi a+1-\xi a)/a\xi(\xi a+1))d\xi$$
$$dx=k(1/2\xi-a(1/a\xi-1/(\xi a+1)))d\xi$$
$$dx=k(1/2\xi-1/\xi+a/(a\xi+1)))d\xi$$
$$dx=k(a/(a\xi+1)-1/2\xi)d\xi$$
$$x=k(ln(a\xi+1)-ln(\xi)/2)+C$$
$$x=k\ ln(a\ e^{\tau}+e^{-\tau})+C$$
$$d=k\ ln((a\ e^{\tau}+e^{-\tau})/(a+1))$$
$$W=ln((a\ e^{\tau}+e^{-\tau})/(a+1))$$
$$a\ e^{\tau}+e^{-\tau}=(a+1)e^W=2b$$
$$b=(a+1)e^W/2=e^W/(1-v_o/c)$$
$$a\ e^{2\tau}-2be^{\tau}+1=0$$
$$e^{\tau}=(b+\sqrt{b^2-a})/a$$
$$\tau=ln((b+\sqrt{b^2-a})/a)$$
$$\tau=ln((e^W+\sqrt{e^{2W}-1+v^2/c^2})/(1+v_o/c))$$
Preliminary test:
1. Let the initial velocity is absent (a = 1), and the accelerating voltage (W) is very large....
$$\tau=W\ ln(2)$$
$$eUt/(m_edc)=eU\ ln(2)/m_ec^2$$
$$t=ln(2)d/c$$Ýòî ìîæåò áûòü ïðàâäîé.
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:
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!
__________________________
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!
$$F=eU(1-v^2/c^2)/d$$
$$m_e\ddot{x}=eU(1-\dot{x}^2/c^2)/d$$
$$\Delta E=\int_o^dFdx=\int_o^d m_e\ddot{x}dx=\int_o^d(eU(1-\dot{x}^2/c^2)/d)dx=(eU/d)\int_o^d(1-\dot{x}^2/c^2)dx$$
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!
__________________________
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!
martillo
Registered Senior Member
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:
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:
$$d\tau=Wc\ dt/d$$
$$dt=(d/Wc)d\tau$$
$$dx=Wd\ sin(\tau+arcsin(v_o/c))d\tau$$
$$d=Wd(cos(arcsin(v_o/c))-cos(\tau+arcsin(v_o/c))$$
$$1=W(\sqrt{1-v_o^2/c^2}-cos(\tau+arcsin(v_o/c))$$
$$cos(\tau+arcsin(v_o/c))=\sqrt{1-v_o^2/c^2}-1/W$$
$$v/c=\dot{x}/c=sin(\tau+arcsin(v_o/c))$$
$$\sqrt{1-v^2/c^2}=cos(\tau+arcsin(v_o/c))=\sqrt{1-v_o^2/c^2}-1/W$$
$$\sqrt{1-v_o^2/c^2}-\sqrt{1-v^2/c^2}=1/W$$
delirium, ravings, gibberish...
$$m_e\ddot{x}=eU\sqrt{1-\dot{x}^2/c^2}/d$$
$$\int (d\dot{x}/c)/\sqrt{1-\dot{x}^2/c^2}=eUt/(m_ed)=Wct/d=\tau$$
$$\tau=Wct/d$$
$$W=eU/m_ec^2$$
$$arcsin(\dot{x}/c)=\tau+C$$
$$arcsin(\dot{x}/c)-arcsin(v_o/c)=\tau$$
$$arcsin(\dot{x}/c)=\tau+arcsin(v_o/c)$$
$$\dot{x}/c=sin(\tau+arcsin(v_o/c))$$
$$\dot{x}/c=\sqrt{1-v_o^2/c^2}sin(\tau)+(v_o/c)cos(\tau)$$
$$\dot{x}/c=\sqrt{1-v_o^2/c^2}sin(Wct/d)+(v_o/c)cos(Wct/d)$$
I find time:
$$d\tau=Wc\ dt/d$$
$$dt=(d/Wc)d\tau$$
$$dx=Wd\ sin(\tau+arcsin(v_o/c))d\tau$$
$$d=Wd(cos(arcsin(v_o/c))-cos(\tau+arcsin(v_o/c))$$
$$1=W(\sqrt{1-v_o^2/c^2}-cos(\tau+arcsin(v_o/c))$$
$$cos(\tau+arcsin(v_o/c))=\sqrt{1-v_o^2/c^2}-1/W$$
$$v/c=\dot{x}/c=sin(\tau+arcsin(v_o/c))$$
$$\sqrt{1-v^2/c^2}=cos(\tau+arcsin(v_o/c))=\sqrt{1-v_o^2/c^2}-1/W$$
$$\sqrt{1-v_o^2/c^2}-\sqrt{1-v^2/c^2}=1/W$$
delirium, ravings, gibberish...
Last edited:
martillo
Registered Senior Member
Seems it's getting better now but you know, with the exception of a constant (m0.c2), seems you are arriving near to Einstein's equation:
W = E = m0[(1/sqrt(1-v^2/c^2) - 1/sqrt(1-v0^2/c^2)]c^2 = (m-m0)c^2
That is: E = (m-m0)c^2 where here m represents the called "relativistic mass" (not recommended anymore by modern relativistics...).
The interesting point of this is to arrive near to Einstein's equation but assuming the "relativistic factor" 1/sqrt(1-v2/c2) in the definition of the Electric and Magnetic Fields in spite of a "space-time curvature".
I propose that approach in my theory (trying to debunk Relativity as you but in a different way). I think I have already mentioned it to you but just for the case here is the link: http://www.geocities.ws/anewlightinphysics/
What I don't understand is what your derivation actually means since in my theory the Kinetic Energy gained by an electron is just the classical one K=mv2/2 (something I think you also proposed if I don't remember wrongly)...
W = E = m0[(1/sqrt(1-v^2/c^2) - 1/sqrt(1-v0^2/c^2)]c^2 = (m-m0)c^2
That is: E = (m-m0)c^2 where here m represents the called "relativistic mass" (not recommended anymore by modern relativistics...).
The interesting point of this is to arrive near to Einstein's equation but assuming the "relativistic factor" 1/sqrt(1-v2/c2) in the definition of the Electric and Magnetic Fields in spite of a "space-time curvature".
I propose that approach in my theory (trying to debunk Relativity as you but in a different way). I think I have already mentioned it to you but just for the case here is the link: http://www.geocities.ws/anewlightinphysics/
What I don't understand is what your derivation actually means since in my theory the Kinetic Energy gained by an electron is just the classical one K=mv2/2 (something I think you also proposed if I don't remember wrongly)...
Last edited:
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.
$$m_e\ddot{x}=eU\sqrt{1-\dot{x}^2/c^2}/d$$
$$\int (d\dot{x}/c)/\sqrt{1-\dot{x}^2/c^2}=eUt/(m_ed)=Wct/d=\tau$$
$$\tau=Wct/d$$
$$W=eU/m_ec^2$$
$$arcsin(\dot{x}/c)=\tau+C$$
$$arcsin(\dot{x}/c)-arcsin(v_o/c)=\tau$$
$$arcsin(\dot{x}/c)=\tau+arcsin(v_o/c)$$
$$\dot{x}/c=sin(\tau+arcsin(v_o/c))$$
$$\dot{x}/c=\sqrt{1-v_o^2/c^2}sin(\tau)+(v_o/c)cos(\tau)$$
$$\dot{x}/c=\sqrt{1-v_o^2/c^2}sin(Wct/d)+(v_o/c)cos(Wct/d)$$
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:
martillo
Registered Senior Member
Near to Einstein's equation ... That could mean that from the energy point of view could be equivalent to consider a "space-time curvature" than a variation of the fields of forces with the relativistic factor 1/sqrt(1-v2/c2) but I'm not sure about your derivation and what your arrived equation means...
Last edited:
martillo
Registered Senior Member
After thinking more your derivation doesn't make sense to me. You start with an electric field with parameters e, U and d and they simply vanish at the end. Seems you are simply recalculating Einstein's equation someway.
In my theory the kinetic energy gained by an electron is just the classical one K=mv2/2 even if accelerated by an electric field varying with the factor sqrt(1-v2/c2) and I think you also proposed something like this time ago (saying that the equation E=(m-m0)c2 is a wrong one).
Seems time to rethink things...
What I keep is that may be some of your steps could be applied to calculate something about an electron accelerated by an electric field varying with the factor sqrt(1-v2/c2)...
In my theory the kinetic energy gained by an electron is just the classical one K=mv2/2 even if accelerated by an electric field varying with the factor sqrt(1-v2/c2) and I think you also proposed something like this time ago (saying that the equation E=(m-m0)c2 is a wrong one).
Seems time to rethink things...
What I keep is that may be some of your steps could be applied to calculate something about an electron accelerated by an electric field varying with the factor sqrt(1-v2/c2)...
Last edited: