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Re: A Geometrical Proof of the Non-invariance of the Spacetime Interval



David Rutherford writes:

I think I can show that the spacetime interval as described in special
relativity (SR) is not invariant by using a geometrical argument.

That would be an astonishing feat!!

Here is a simple summary of the scenario you proposed:

Event Coordinates in F Coordinates in F'
E1 (0,0,0,0) (0,0,0,0)
E2 (vt,0,0,t) (0,0,0,t/gamma)

Now use the *definition* of the squared (spacelike) spacetime interval, i.e.

squared interval = (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2 - (c*t2-c*t1)^2

to obtain

In F (vt-0)^2+(0-0)^2+(0-0)^2-(ct-0)^2 = (v^2-c^2)t^2
In F' (0-0)^2+(0-0)^2+(0-0)^2-(c*t/gamma-0)^2 = (v^2-c^2)t^2

Clearly these are identical, so what is the perceived problem?

John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm