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

[David Rutherford <drutherford@SOFTCOM.NET>]

John Mallinckrodt wrote:
> 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?

I'm not claiming that the magnitude of the spacetime interval is not
invariant. I'm claiming that, since the assigned positions of E1 in F
and F' don't coincide in space (meaning they are not independent of any
reference frame), at times later than t = 0, lines drawn between E1 and
E2 representing the spacetime interval between E1 and E2 in F and F'
don't coincide in spacetime (meaning they are not independent of any
reference frame). In other words, they aren't the same line.

--
Dave Rutherford
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