,
Copyright © 1999 David E. Rutherford
All Rights Reserved
The transformation matrix,
,
replaces the Lorentz transformation matrix, and the transformation equations
replace the Lorentz transformation equations, where
Just as in special relativity, we find a contraction of lengths in the direction
of motion. As an example, we place a ruler along the x-axis of an
unprimed frame of reference with coordinates,
. A primed
reference frame with coordinates,
, is in uniform
motion with four-velocity
relative
to the unprimed frame. The spacetime vector between the endpoints of the
ruler at rest in the unprimed frame is
. To find
the components of the spacetime vector,
, in the
primed frame, we use Eqs. (2), where the coordinates are replaced by the
components of the spacetime vectors
or
The length measurement is made by noting the instantaneous positions
of the two endpoints of the ruler, therefore the time component of the spacetime
vector in the unprimed frame is zero. Since
the spacetime
vector in the unprimed frame is
. Substituting
the actual components into Eqs. (5), we get
and, therefore, the spacetime vector in the primed frame is
.
But, because we are only interested in the
component,
the length of the ruler in the primed frame is
Remembering now that
and since 
we have
therefore, we can write Eq. (7) as
where the component
, in this
case, is the proper length of the ruler.
We can use similar methods to find the time measurements between events in
two reference frames in uniform relative motion with respect to each other.
The spacetime vector between two events at the same place, but at different
times, in the unprimed frame is
. We wish
to find the components of the spacetime vector
between
these events as measured by an observer at rest in the primed frame. Inserting
the four-velocity
of
the primed frame relative to the unprimed frame and the components of the
spacetime vector
into Eqs.
(5), we get
or 
. But
since we are only comparing the time components, we have
or
The component,
, in this
case, is the proper time,
.
Whenever the time component of the spacetime vector is the only nonzero component, it is the proper time. Similarly, if the time component is zero, but the spatial components are nonzero, the spatial components are the components of the proper length. In the primed frame, in this example, neither the time nor the spatial components are zero, so neither is proper. The proper time and proper length are always the maximum values for the time or length, respectively.
In each of the examples above, the magnitude of the spacetime vector in both
frames of reference is the same. Due to the orthogonality of
, the magnitude
of the spacetime interval,
, in the
general case, is
or in the two cases above, the magnitudes of the spacetime vectors are
where 
is
the four-dimensional Kronecker delta. The equality in Eq. (15) holds true
not only for the spacetime vector, but for any arbitrary four-vector.
Copyright © 1999 David E. Rutherford
All Rights Reserved
E-mail: drutherford@softcom.net