I Why Are Equally Spaced Events Preserved in Transformation?

TonyEsposito
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Hi guys, I'm reading a book where the author, only form the invariance of the speed c draws conclusions about the transoformation from a system to another in inertial motion. The author shows a spacetime diagram (x,t) and the two dimensional light cone, he marks two events on the light cone (x1,t) and (-x1,t) and another event (0,t). He says since the three events are equally spaced in the first system and the transformation is linear they must be equally spaced also in the new system, why he assumes this? (the book still did not explained the invariance of the Minkwosky Norm), my brain is freezed now. Summing up the question: why he assumes that equally spaced events in a system must be equally spaced in another? thanks!
 
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TonyEsposito said:
i'm reading a book

What book? Please give specific references.
 
The Geometry of Spacetime by Callahan. Page 32 "the graphical solution".
 
TonyEsposito said:
He says since the three events are equally spaced in the first system and the transformation is linear they must be equally spaced also in the new system, why he assumes this?

It's not an assumption, it's a consequence of linearity. A linear transformation means that the new coordinates ##x', t'## are linear functions of the old coordinates ##x, t##. In other words, we must have

$$
x' = ax + bt + c
$$
$$
t' = dx + et + f
$$

where ##a, b, c, d, e, f## are all constants (and some might be zero, we don't know at this point). You should be able to show that if we have three points equally spaced in ##x, t##, such transformations must give three points equally spaced in ##x', t'##.
 
I'm afraid I'm not able to find a proof of this :( can you help me?
 
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