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

[Michael Burns-Kaurin <mburns-k@SPELMAN.EDU>]

As far as I can tell, the definition of E1 is changed from occurring at the
zero of position and time to occurring at the origin of F', for all times,
at least in F'.  Then he states that E1 is no longer the same in both
frames, since he still keeps E1 at time zero in F.

Michael Burns-Kaurin
Spelman College





                      John Mallinckrodt
                      <ajm@CSUPOMONA.ED        To:       PHYS-L@lists.nau.edu
                      U>                       cc:
                      Sent by: Forum           Subject:  Re: A Geometrical Proof of the Non-invariance of
                      for Physics               the Spacetime Interval
                      Educators
                      <PHYS-L@lists.nau
                      .edu>


                      01/06/2003 03:20
                      PM
                      Please respond to
                      Forum for Physics
                      Educators






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