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



I think I can show that the spacetime interval as described in special
relativity (SR) is not invariant by using a geometrical argument. The
reason that it is not invariant is that the endpoints of the interval
are _not_ independent of inertial reference frames. To show that this is
the case, I'll give a simple example.

Imagine the standard relativistic configuration consisting of a
reference frame F' in uniform motion with speed v > 0 in the x-direction
relative to a reference frame F. Suppose an event E1 occurs at the
origins of F and F' when the origins coincide. The coordinates of E1 in
F and F' are then x = x' = 0, y = y' = 0, z = z' = 0 and t = t' = 0. Say
the coordinates of a second event E2 in F are x = vt, y = 0, z = 0 and
t = t. Since E1 was at the origin of F, these are also the components
of the spacetime interval from E1 to E2 in F. The Lorentz
transformations for this situation are

x' = gamma(x - vt)
y' = y
z' = z
t' = gamma(t - vx/c^2)

where gamma = 1/sqrt(1 - (v/c)^2). Plugging in the coordinates of E2 in
F, we get

x' = gamma(vt - vt)
y' = 0
z' = 0
t' = gamma(t - v(vt)/c^2)

or

x' = 0
y' = 0
z' = 0
t' = t/gamma

These are the coordinates of E2 in F'. Again, since E1 was at the origin
of F', the coordinates x', y', z' and t' are also the components of the
spacetime interval from E1 to E2 in F'. Notice that, even though the
coordinates of E2 are different in F and F', both sets of coordinates
describe the _same_ event in spacetime. Not so for E1! The location of
E1, in F and F', in this case, is at the origins of each frame, at all
times. At t = t, the origin of F is at x = 0, so that's where the
location of E1 is in F. But the origin of F' at t = t is at x = vt, so
that's the location of E1 in F' at t = t. The coordinates of E1 in F
and F' describe _different_ events in spacetime at t = t, because the
origins of F and F' are at _different_ locations at t = t. The locations
of the origins of F and F' at t = t are what's used as the _spatial_
coordinates of E1 for the endpoints of the interval, _not_ the spatial
coordinates of E1 at t = 0. Otherwise, x' would not be zero at t = t.
Unfortunately, you've got the endpoint E1 of the spacetime interval
located at two different places at t = t.

That means that the event E1 at the endpoint of the interval in question
is _not_ independent of any reference frame, thus the displacement
4-vector between the two events is likewise _not_ independent of any
reference frame.

I would like to hear your comments about the geometrical validity of
this argument. Thanks.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555/newtransform.pdf

Applications:
"4/3 Problem Resolution"
http://www.softcom.net/users/der555/elecmass.pdf
"Action-reaction Paradox Resolution"
http://www.softcom.net/users/der555/actreact.pdf
"Energy Density Correction"
http://www.softcom.net/users/der555/enerdens.pdf
"Proposed Quantum Mechanical Connection"
http://www.softcom.net/users/der555/quantum.pdf