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**Author:**

Rebigsol, Cameron

**Category:**

Research Papers

**Sub-Category:**

Relativity Theory

**Language:**

English

**Date Published:**

December 20, 2019

**Downloads:**

182

**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?

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.

(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.

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: www2.physics.umd.edu › ~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).

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: www2.physics.umd.edu › ~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).

4a and 4b in your paragraph 1) should actually be Eq. 2a and Eq. 2b in my paper. For your operation to be realized, you have actually formed the equation set that contains the following equations (I) my eq. 2a, (II) my equation 2b, (II) x=0. What you do is actually to have replaced Eq. 5c and 5d in my paper with x=0 in set 5a-d but thought you solve the problem with what the second postulated stipulates. What school of mathematics grands you such freedom? At least, x=0 is not even what relativity proposes for its derivation.

Your paragraph 2) must then emphasize that x=ct and x'=ct' represent the same light sphere commonly shared by the observers in both frame. Thanks! That is what I said and that is what is expressed wit Eq 5c and d. But your previous reply tells me to remove both equations in order to solve the set. If you take away Eq 5c and d, you no longer have the right to use them for your derivation but still think you have this right like what you do now with paragraph 2). You are so contradicting to yourself!

Because your invalid operation of paragraph 1) and 2), you consequentially make your operation in paragraph 3) enjoy no credit.

Your note paragraph only suggests that at least on the x and x’ axes relativity is inapplicable. Your note makes me suspect that you may not even know how to write a general equation to express a straight line in a 3D coordinate system containing (x, y, z) or (x’, y’, z’).

My conclusion from the debate is that a self-refuted theory needs self-refuted warriors to defend

Your paragraph 2) must then emphasize that x=ct and x'=ct' represent the same light sphere commonly shared by the observers in both frame. Thanks! That is what I said and that is what is expressed wit Eq 5c and d. But your previous reply tells me to remove both equations in order to solve the set. If you take away Eq 5c and d, you no longer have the right to use them for your derivation but still think you have this right like what you do now with paragraph 2). You are so contradicting to yourself!

Because your invalid operation of paragraph 1) and 2), you consequentially make your operation in paragraph 3) enjoy no credit.

Your note paragraph only suggests that at least on the x and x’ axes relativity is inapplicable. Your note makes me suspect that you may not even know how to write a general equation to express a straight line in a 3D coordinate system containing (x, y, z) or (x’, y’, z’).

My conclusion from the debate is that a self-refuted theory needs self-refuted warriors to defend

Ok.

1) Determine a44:

For the primed frame the point x=0 is x'=-vt'= (with 4a) -a11.vt. So t'= a11.t

From 4b with x=0 we have t'= a44.t; conclusion a44= a11.

So the equations become 4a: x'= a11(x-vt) and 4b: t'= a41.x + a11.t

2) Apply postulate 2 : for a light ray with x=ct we must have x'=ct'.

Substitute this in 4a: ct'= a11.ct- a11.vt; Substitute in here 4b:

c.a41.x+ c.a11.t= a11.ct- a11.vt ==> c.a41.x= -a11.vt and with x=ct we find:

c^2.a41.t= -a11.vt so a41= -a11.v/c^2.

So the equations become 4a: x'= a11(x-vt) and 4b: t'= a11(t-x.v/c*2)

3) apply postulate 1 : we must have the reverse with v in opposite direction:

4ar: x= a11(x'+ vt') and 4br: t= a11(t'+ x'.v/c*2)

Substitute 4a and 4b in 4ar and it follows that a11= 1/(1-v^2/c^2)^0.5 , the Lorentz factor.

Note: You must not consider x^2+y^2+z^2= (ct)^2 as permanent restriction of x, y, z and t.

It's only the equation of a light ray.

We want the transformation to be valid for all x, y, z and t.

1) Determine a44:

For the primed frame the point x=0 is x'=-vt'= (with 4a) -a11.vt. So t'= a11.t

From 4b with x=0 we have t'= a44.t; conclusion a44= a11.

So the equations become 4a: x'= a11(x-vt) and 4b: t'= a41.x + a11.t

2) Apply postulate 2 : for a light ray with x=ct we must have x'=ct'.

Substitute this in 4a: ct'= a11.ct- a11.vt; Substitute in here 4b:

c.a41.x+ c.a11.t= a11.ct- a11.vt ==> c.a41.x= -a11.vt and with x=ct we find:

c^2.a41.t= -a11.vt so a41= -a11.v/c^2.

So the equations become 4a: x'= a11(x-vt) and 4b: t'= a11(t-x.v/c*2)

3) apply postulate 1 : we must have the reverse with v in opposite direction:

4ar: x= a11(x'+ vt') and 4br: t= a11(t'+ x'.v/c*2)

Substitute 4a and 4b in 4ar and it follows that a11= 1/(1-v^2/c^2)^0.5 , the Lorentz factor.

Note: You must not consider x^2+y^2+z^2= (ct)^2 as permanent restriction of x, y, z and t.

It's only the equation of a light ray.

We want the transformation to be valid for all x, y, z and t.

1 Replies

Your paragraph 2) must then emphasize that x=ct and x'=ct' represent the same light sphere commonly shared by the observers in both frame. Thanks! That is what I said and that is what is expressed wit Eq 5c and d. But your previous reply tells me to remove both equations in order to solve the set. If you take away Eq 5c and d, you no longer have the right to use them for your derivation but still think you have this right like what you do now with paragraph 2). You are so contradicting to yourself!

Because your invalid operation of paragraph 1) and 2), you consequentially make your operation in paragraph 3) enjoy no credit.

Your note paragraph only suggests that at least on the x and x’ axes relativity is inapplicable. Your note makes me suspect that you may not even know how to write a general equation to express a straight line in a 3D coordinate system containing (x, y, z) or (x’, y’, z’).

My conclusion from the debate is that a self-refuted theory needs self-refuted warriors to defend

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