I Principle of relativity in the proof of invariance of interval

[Sagittarius A-Star]

I found answer here:

https://en.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations#Rigorous_Statement_and_Proof_of_Proportionality_of_ds2_and_ds′2


But I can't understand how "violate the postulated homogeneity of spacetime"?

From homogeneity of spacetime and isotropy of space follows that all coefficients in a transformation between primed and unprimed coordinates must be constants. Therefore, if you transform the coordinates of the spacetime interval formula i.e. from unprimed to primed coordinates, then you must get a polynomial of second order. A theorem of algebra states, that two irreducible polynomials of the same degree (i.e. quadratic), which share all their zeros (invariance of the speed of light!), can differ by at most a constant factor $K$.

 

[Mike_bb]

Hello, Sagittarius A-Star!

1.
View attachment 356460
2.
View attachment 356458

See post

I decided to add more details to equations. Now equation contain coefficients a1 and a2 that depend on coordinates and time:

$$c^2\Delta t^2-\Delta x^2 = a1(v\Delta t , L, \Delta t)c^2\Delta t'^2-\Delta x'^2$$
$$c^2\Delta t'^2-\Delta x'^2 = a2(v\Delta t , L, \Delta t)c^2\Delta t^2-\Delta x^2$$

If $$a1(v\Delta t , L, \Delta t) = a2(v\Delta t , L, \Delta t)$$ then $$a1(v\Delta t , L, \Delta t) = a2(v\Delta t , L, \Delta t) = K$$

So:

$$c^2\Delta t^2-\Delta x^2 = K(c^2\Delta t'^2-\Delta x'^2)$$
$$c^2\Delta t'^2-\Delta x'^2 = K(c^2\Delta t^2-\Delta x^2)$$

Is it right?

Thanks!

 

[Mike_bb]

Sagittarius A-Star

In previous post I meant that coordinates and time are the same for two intervals in different coordinate systems (by Principle of Relativity). Am I right?

 

[bquuune]

Greetings Mike_bb,

I just saw the reply you made at https://math.stackexchange.com/ques...-infinitesimal-be-non-integer/5069096#5069096

I don't understand the proof you wrote there.
If you can prove
(1) $c^2\Delta t^2-\Delta x^2 = K(c^2\Delta t'^2-\Delta x'^2)$ and
(2) $c^2\Delta t'^2-\Delta x'^2 = K(c^2\Delta t^2-\Delta x^2)$

Then you can plug the second equation into the first and get
$c^2\Delta t^2-\Delta x^2 = K K(c^2\Delta t^2-\Delta x^2)$
So $K=1$ or $-1$.