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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?

Eq. 5c and 5d must both be removed.

x^2 is not equal to c^2.t^2 in general; x and t are independent variables which can have any value.

The same for x'^2 and c^2.t'^2.

What you do have with these relations is the CONDITION I mentioned twice before:

If x=ct then, and only then x'=ct' , the second postulate (constant light speed).

Meaning that if x#ct then x'#ct'.

After simply solving a44 (=a11), two unknowns remain, a11 and a41.

Which can be found with the postulates as I indicated already.

You need to do some substitutions. If you don't succeed yourself I can show you if you like.

x^2 is not equal to c^2.t^2 in general; x and t are independent variables which can have any value.

The same for x'^2 and c^2.t'^2.

What you do have with these relations is the CONDITION I mentioned twice before:

If x=ct then, and only then x'=ct' , the second postulate (constant light speed).

Meaning that if x#ct then x'#ct'.

After simply solving a44 (=a11), two unknowns remain, a11 and a41.

Which can be found with the postulates as I indicated already.

You need to do some substitutions. If you don't succeed yourself I can show you if you like.

1 Replies

If Eq. 5c and 5d are both removed, 5a and 5b cannot be solved. Then you make relativity meaningless. These two equations are the key for the original paper of relativity being able to continue its calculation. Without them, the second postulate has no mathematical expression.

“ x^2 is not equal to c^2.t^2 in general” should be a new discovery that you need to teach Einstein before he published his 1905 paper. In my opinion, if a ray is moving along the X- axis with 0 component on Y and Z, you must have x^2=(ct)^2.

“x and t are independent variables which can have any value”, Yes, you are right, but if you accept X^2+y^2+z^2=(ct)^2, with y=0, z=0, and c=constant, after you choose any value for x, you can no longer choose any value for t, but only one, which is t=x/c

According to your “What you do have with these relations is the CONDITION I mentioned twice before: If x=ct then, and only then x'=ct' , the second postulate (constant light speed)”, you should find no disagreement on my equation set Eq. 5-d when you see both equations are placed with Eq. 5a and Eq. 5b together to form one set.

“Meaning that if x#ct then x'#ct'.” Meaning is not good enough. We are talking about mathematical expression here. Please show work to prove x=/=ct and x’=/=ct’ with signs like =, >, or <

“After simply solving a44 (=a11), two unknowns remain, a11 and a41.” After you remove both Eq. 5c and 5d, how is your “simply solving a44 (=a11)” realized?”

“Which can be found with the postulates as I indicated already.” Unfortunately, if you remove Eq. 5c and 5d, you would have trashed the 2nd postulate

“ x^2 is not equal to c^2.t^2 in general” should be a new discovery that you need to teach Einstein before he published his 1905 paper. In my opinion, if a ray is moving along the X- axis with 0 component on Y and Z, you must have x^2=(ct)^2.

“x and t are independent variables which can have any value”, Yes, you are right, but if you accept X^2+y^2+z^2=(ct)^2, with y=0, z=0, and c=constant, after you choose any value for x, you can no longer choose any value for t, but only one, which is t=x/c

According to your “What you do have with these relations is the CONDITION I mentioned twice before: If x=ct then, and only then x'=ct' , the second postulate (constant light speed)”, you should find no disagreement on my equation set Eq. 5-d when you see both equations are placed with Eq. 5a and Eq. 5b together to form one set.

“Meaning that if x#ct then x'#ct'.” Meaning is not good enough. We are talking about mathematical expression here. Please show work to prove x=/=ct and x’=/=ct’ with signs like =, >, or <

“After simply solving a44 (=a11), two unknowns remain, a11 and a41.” After you remove both Eq. 5c and 5d, how is your “simply solving a44 (=a11)” realized?”

“Which can be found with the postulates as I indicated already.” Unfortunately, if you remove Eq. 5c and 5d, you would have trashed the 2nd postulate

With a44= a11 there are two unknowns left: a11 and a41.

And we have two conditions (the postulates) to solve them.

So the set is not over-conditioned but can exactly be solved.

2e postulate: if x=ct then, and only then x'=ct'. This results in a41= -a11.v/c^2.

1e postulate: reversed we must have: x= a11(x'+ vt') and t= a11(t'+ x'.v/c^2)

The solution of a11 is the Lorentz factor 1/(1-v^2/c^2)^0.5

And we have two conditions (the postulates) to solve them.

So the set is not over-conditioned but can exactly be solved.

2e postulate: if x=ct then, and only then x'=ct'. This results in a41= -a11.v/c^2.

1e postulate: reversed we must have: x= a11(x'+ vt') and t= a11(t'+ x'.v/c^2)

The solution of a11 is the Lorentz factor 1/(1-v^2/c^2)^0.5

4 Replies

Definition of over-conditioned equation set in algebra: number of equations in the set is more than number of unknowns. Before relativity arrives at the solution of 3 unknowns, i.e., a11, a44 and a41, relativity needs all four equations listed in (Eq. 5a-d). If it is not an over conditioned set, please tell me which one equation from the set can be taken away for relativity to arrive at the solution.

If you think 2e postulate is another necessary and sufficient condition for a11, a44 and a41 to be arrived, you must put it in the above set, and so is 1e postulate. If you solve equation set by grouping different equations one group at a time so that number of unknowns and equations match equally in each group, then use the solution set as a common solution set for all groups, you break the rule commonly accepted in mathematical practice. Indeed, if you plug the solutions from one group into another to replace the unknowns there, your solution may be rejected by the second group.

If you think 2e postulate is another necessary and sufficient condition for a11, a44 and a41 to be arrived, you must put it in the above set, and so is 1e postulate. If you solve equation set by grouping different equations one group at a time so that number of unknowns and equations match equally in each group, then use the solution set as a common solution set for all groups, you break the rule commonly accepted in mathematical practice. Indeed, if you plug the solutions from one group into another to replace the unknowns there, your solution may be rejected by the second group.

Definition of over-conditioned equation set: number of equations in the set is more than number of unknowns. Before relativity arrives the solution of 3 unknowns, i.e., a11, a44 and a41, relativity needs all four equations listed in (Eq. 5a-d). If it is not an over conditioned set, please tell me which one equation from the set can be taken away for relativity to arrive at the same solution.

If you think 2e postulate is another necessary and sufficient condition for a11, a44 and a41 to be arrived at, you must put it in the above set, and so is 1e postulate. If you solve equation set by grouping different equations one group at a time so that you can match the equations and solution numbers equally, and use the solution set as a common solution set for all groups, you break the rule commonly accepted in mathematical practice. Indeed, if you plug the solutions from one group into another to replace the unknowns there, your solution may be rejected by the second group.

If you think 2e postulate is another necessary and sufficient condition for a11, a44 and a41 to be arrived at, you must put it in the above set, and so is 1e postulate. If you solve equation set by grouping different equations one group at a time so that you can match the equations and solution numbers equally, and use the solution set as a common solution set for all groups, you break the rule commonly accepted in mathematical practice. Indeed, if you plug the solutions from one group into another to replace the unknowns there, your solution may be rejected by the second group.

If you think 2e postulate is another necessary and sufficient condition for a11, a44 and a41 to be arrived at, you must put it in the above set, and so is 1e postulate. If you solve equation set by grouping different equations one group at a time so that you can match the equations and solution numbers equally, and use the solution set as a common solution set for all groups, you break the rule commonly accepted in mathematical practice. Indeed, if you plug the solutions from one group into another to replace the unknowns there, your solution may be rejected by the second group.

Sorry, Zan! I did not repeat my same message 2 or 3 times. It is because I am not familiar with the operation of the message recording provided by this website and error happened. I'll watch out next time. My apology here. .

You found x'= a11(x-vt) and t'= a41.x + a44.t

It's also easy to find that a44= a11 (x=0 is for O': x'=-vt' = -a11.v.t).

Then eq. 4a,b must be considered as a condition: x=ct <==> x'=ct' (second postulate !).

This results in the relation a41= -a11.v/c^2.

To solve a11 the relativity principle (first postulate) must be used.

It's also easy to find that a44= a11 (x=0 is for O': x'=-vt' = -a11.v.t).

Then eq. 4a,b must be considered as a condition: x=ct <==> x'=ct' (second postulate !).

This results in the relation a41= -a11.v/c^2.

To solve a11 the relativity principle (first postulate) must be used.

2 Replies

Jan, Thank you very much for your comment, but may I try to understand your comment a little more in detail before I make a more confident reply? You said in the 5th line: “To solve a11…”, but in your second line of comment you already said, “It's also easy to find that a44= a11”. Indeed, as mentioned in my paper, a solution for a44 and a11 is already found when (Eq. 5a-d) is listed. Why should we now still pursue after “to solve a11”? Do you mean that a11 should have another value other than what relativity has reached?

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