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Relativistic geometry

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Date: Sun, 12 Jun 22 21:56:25 +0000
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From: r.hac...@tiscali.fr (Richard Hachel)

by: Richard Hachel - Sun, 12 Jun 2022 21:56 UTC

Well, I was thinking earlier about the relationship that binds proper time
to observable time (improper time) in accelerated relativistic frames of
reference.

I gave the equations to find the right solutions.

Proper time:
Tr=sqrt(2x/a)

Observable time:
To=(x/c).sqrt(1+2c²/ax)

If one wants to make a geometrical representation, one sees that geometry
is not at all (but then not at all) that which is taught to us by
physicists.

This is how the good doctor Richard Hachel (that's me) draws his geometry.

On the axis of the abscissas it is necessary to put the proper time (real
time). Let Tr.

On the y-axis, you have to put the distances covered. Let x.

Now, we must place the observable time To in the observing frame.

Where to place To in my schema?

It is very simple. To constantly is the hypotenuse of the values Tr
and x.

We therefore see that the progression of To is particular and not constant
as is Tr.

I'm crossposting here on the math forum.

There will be someone to consider that what I say is true, and to
understand its geometric meaning.

R.H.

"Gort, klaatu nikto barada." -- The Day the Earth Stood Still

tech / sci.physics.relativity / Relativistic geometry

Subject	Author
Relativistic geometry	Richard Hachel