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


Rebigsol, Cameron


Research Papers


Relativity Theory



Date Published:

December 20, 2019




Lorentz factor, rest length, moving length


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?


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
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

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
Cameron Wong(Vancouver, United States):
To refute my paper, you must present the following:

(1) To arrive at the solution a11, a41, a44, relativity did not rely on all the equations listed in (Eq. 5 a-d), but instead,
(2) Relativity uses the equation set you put together and presented to me: (I) my eq. 2a, (II) my equation 2b, (III) x=0.
(3) (Eg. Re-A) and (Eq. Re-B) will not lead to v=0, which is the speed between the X frame and X’ frame
(4) Relativity must fail when applied to situation that is examined with a ray that is parallel to both X axis, and X’ axis
(5) Each observer on the X axis, and X’ axis must consider they see the same and only sphere, and the origin of his own frame coincides with the center of the light sphere FOREVER, even if they would have been, for example, 10 light-years apart.

As to (1) above, you obviously thought it was me who put the equations together, and that is why you suggest to take the 5c and 5d away and replace them with your x=0 shown in (2) above.

As to (4) above, it is what your note (January 20, 2020 @ 7:03:21 pm) inevitably leads to. x^2+y^2+z^2= (ct)^2 is an equation for a sphere formed with light rays, what is wrong to see the ray starting from the origin and parallel to both x and x’ to bear the equation x^2=(ct)^?

As to (5) above, your x^2+ y^2+ z^2- (ct)^2 = x'^2+ y'^2+ z'^2- (ct')^2 is a necessary and sufficient condition for what (5) can be concluded.

To answer your “I wonder how you found a14, starting with x'= a11.x+ a14.t”: I have no objection of the way you find a14. What I object is that you suggest removing Eq. 5c and 5d and replace them with x=0 in solving Eq. set 5 a-d. This is not what is found with relativity’s paper. When you think your way leads to the Lorentz factor for the solution, you have repeated relativity's mathematical mistake: grouping equations in different clusters and feel whatever the final solution from the last cluster being commonly applicable to all groups.

Posted: January 22, 2020 @ 12:47:50 pm
Jan Zwarts(Netherlands):
Indeed, where I said 4a and 4b it should be 2a and 2b or 5a and 5b.
I said remove 5c and 5d because they are wrong in the way you present and use them.
You said : an over-conditioned equation set. So you counted 5c and 5d as two separate equations.
They must be replaced by the condition (second postulate) I mentioned several times.

You don't understand at all what my derivation of a44 is about.
I wonder how you found a14, starting with x'= a11.x+ a14.t
My way (similar as for a44): The point x'= 0 relative to the unprimed frame is x= vt.
Substitute: 0= a11.vt+ a14.t so a14= -a11.v

See here:
select site: › ~yakovenk › teaching › Lorentz
and open this pdf: Derivation of the Lorentz Transformation - University of Maryland

The first two steps are identical to mine finding a14 and a44.
The next step is more complicated. Mine is much simpler. I use postulate 2 direct where
they check it at the end.
There last step is similar to mine.

Again: the Lorentz transformation is defined for all x, y, z and t ,
where x^2+ y^2+ z^2- (ct)^2 = x'^2+ y'^2+ z'^2- (ct')^2 (the Lorentz invariance).

Posted: January 21, 2020 @ 5:06:40 pm

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