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Speed of Light = 0, ―So “Proven” by Special Relativity, Indisputably!

Author:

Rebigsol, Cameron

Category:

Research Papers

Sub-Category:

Relativity Theory

Language:

English

Date Published:

December 20, 2019

Downloads:

70

Keywords:

Lorentz factor, rest length, moving length

Abstract:

Mathematical verification according to the guidelines emphasized in the original paper of the theory of special relativity (the TSR) published in 1905 shows that the TSR forces the appearance of c=0 for speed of light. Such an outcome can only suggest that the TSR rejects its own second postulate, the absolute lifeline of the TSR. Rejecting this lifeline, the TSR must end up as being self-refuted. The TSR fundamentally relies on the following equation set for its calculation development: Advocating its second postulate, the TSR conceives the following two equations indifferently representing the same single and only light sphere in space for its entire derivation if the sphere is created at where the origin of the two frames meets, i.e. x=x’=0 at time t=t’=0: x^2+y^2+z^2=(ct)^2 and x'^2+y'^2+z'^2=(ct')^2. As such, these two equations must further require that the observer on each of the frame necessarily sees the center of the light sphere permanently coincide with the origin of his own frame. Now, a question inevitably surfaces up: What enables the coinciding seen by each observer to continue for all time t >0 and t’ >0 if the two origins must move away from each other at a nonzero speed v?

Comments

Lev Verkhovsky(Moscow, Russian Federation):
SRT is completely erroneous since it is based on the wrong kind of transformations: they have lost the scale factor characterizing the Doppler effect. First, Lorentz considered a more general form of transformations (with a scale factor), but then he, and also Poincare and Einstein equated it 1 without proper grounds. Their form was artificially narrowed, the formulas became incorrect. This led to a logical contradiction of the theory, to unsolvable paradoxes.
Accordingly, GRT is also incorrect.
For more details, see my brochure "Memoir on the Theory of Relativity and Unified Field Theory" (2000):
http://vixra.org/abs/1802.0136

Posted: May 08, 2020 @ 2:03:55 am
1 Replies

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Jan Zwarts(Netherlands):
Indeed. The Lorentz transformation includes no Doppler effect.
The transformations are direct, no light signals are used.
The result: pure relativistic time dilation and length contraction.

Using light signals for the observation creates Doppler effect.
This has a big impact. Total time dilation factor becomes gam + gam.v/c = sqr((c+v)/(c-v)), so
increases for observers moving apart (v>0) and decreases when they approach each other (v<0).
Approaching even means time acceleration !

Posted: May 20, 2020 @ 6:53:42 am
Jan Zwarts(Netherlands):
You call 5a-d equations.
But 5a and 5b are actually transformations.
They transform (x,t) to (x',t') for all x and t.
They can not be compared with a set equations like:
x + 2y= 5
x - y = 2
where x and y can be solved by substitution for instance.
You say in your paper:
"what the TSR has done is actually forcing the formation of an over-conditioned equation set that reads 5a-d."
That is great nonsense. You must understand that. If you don't, find yourself a good math book.
To solve the 4 a's you need exactly 4 conditions which have to be met. I called them boundary conditions.
I mentioned these conditions already several times.

You said:
Of course “r'+ and r'- are not in the X observer's frame” and that is why the x observer can detect moving length.
You should know that rAB in Eq.Re-A and Eq.Re-B is measured in the stationary frame where also tA and tB are defined.
For you is that frame X with time t.

Posted: January 24, 2020 @ 5:28:35 pm
1 Replies

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Cameron Wong(Vancouver, United States):
Sigh! Enough is enough!

Posted: January 25, 2020 @ 12:45:15 pm
Jan Zwarts(Netherlands):
You want to show a violation you have put in yourself ? That works very well of course.
You put x=0 and x'=ct' in equation 5a: x'= a11(x-vt), where you have not even solved a11.
This violates postulate 2 because for x'=ct' you must have x=ct in 5a.

Your equations (8) and (9) are also wrong. In two ways:
You assume that the t's are the same which is not the case according to eq.RE-A and eq.RE-B.
And r'+ and r'- are not in the X observer's frame.

And in (11) you use the Lorentz factor you have not even derived yet in your paper.

Posted: January 23, 2020 @ 4:42:04 pm
1 Replies

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Cameron Wong(Vancouver, United States):
You are not careful in reading before you refute other’s paper.

>>You put x=0 and x'=ct' in equation 5a: x'= a11(x-vt), where you have not even solved a11.

Immediately before the appearance of Eq. 5a-d, did you ever read in my paper the following sentence “Then, relativity obtains its solution set, which includes ????11 = ????44 = 1 √1 − ( ????⁄????) ⁄ 2 , the so called Lorentz factor ” ? Now tell me where you find x=0 in Eq 5a, which in my paper shows only 4 equations without x=0? When you find me using x=0 for Eq. 12, we are at the stage of verifying the validity of equation set of Eq. 5a-d, which is piled up by relativity, not me. I just simplify the displaying of its piling up with the elimination of y, y’, z, z’. If relativity is valid, it of course needs to show that it is good for x=0 as what Eq-12 is going to verify. At this stage, if you like, you can put the Lorentz factor in place of a11 in Eq. 12. I did not replace it just because I assume the reader what is going on, and I feel no need to take off the pant to fart.

Now, I notice one more miss from you: you seem either not knowing how to verify equations with answer which the equations lead you to have or at least think such verification is not necessary. Indeed, if you do not want to use x=0 to verify, you can use any other constant in place of x=0. How about x=1 (length unit)? Eq. 12 would lead you to have c=[(1/t’)^2+v^2]^0.5. With this, set of Eq. 5a-d is non-sense at t’=0 i.e, this equation set does not even have mathematical legitimacy to be established.

>> You assume that the t's are the same which is not the case according to eq.RE-A and eq.RE-B.
And r'+ and r'- are not in the X observer's frame.

You even have trouble understanding relativity. Of course “r'+ and r'- are not in the X observer's frame” and that is why the x observer can detect moving length. If they are in the x frame, it means the X frame and the X’ frame are the same. Then, where else can he find moving length if not from the X’ frame but only X frame and X’ frame are given to him? r'+ and r'- are just two different line segments for him to have found on the same frame but during the same time interval recorded by a clock next to him.

>>And in (11) you use the Lorentz factor you have not even derived yet in your paper.
Why should it be my responsible to derive it if I am the one to verify its validity after someone else has derived it?

Posted: January 24, 2020 @ 3:17:59 pm
Jan Zwarts(Netherlands):
I did not replace 5c and 5d by x=0 but by the condition [5c <==> 5d] (= postulate 2).
In eq. 13 you have the combination x=0 and x'=ct'. This is a violation of postulate 2.

To solve a11, a14, a41 and a44 we need 4 (boundary) conditions:
1) for x'= 0 we must have x=vt ; solves a14 as function of a11.
2) for x= 0 we must have x'=-vt' ; solves a44 as function of a11.
3) postulate 2: x=ct <==> x'=ct' ; solves a41 as function of a11.
4) postulate 1: LT = LT^-1 ; solves a11.

Posted: January 22, 2020 @ 5:09:08 pm
1 Replies

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Cameron Wong(Vancouver, United States):
“In eq. 13 you have the combination x=0 and x'=ct'. This is a violation of postulate 2” said you.


Violation of postulate 2? Thank you very much! That is exactly what I want to show. However, leading to this equation is not any calculation created by me, but is inevitably forced to come to by relativity’s mathematics.

Posted: January 23, 2020 @ 3:26:52 pm

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