The new spacetime geometry
REFERENCE: None
Recall the experiment shown below.
In Fig. 1, the view from the frame (S), there was no motion in the x, y, and z directions of the coordinate system so that
Dx = Dy = Dz = 0.
In Fig. 2, the view from the frame (S'), there was no motion in the y' and z' directions of the coordinate system so that
Dy' = Dz' = 0.
Recall that one of the forms of the time dilation equation was given by
Dt = Dt'[1 - v2/c2]1/2.
Squaring both sides of Eq. 3 and cleaning up will yield
-(cDt')2 + (vDt')2 = -(cDt)2.
But Dx' = vDt' so that Eq. 4 becomes
-(cDt')2 + (Dx')2 = -(cDt)2.
Observing Eq. 1 and Eq. 2 we can safely add the missing spatial increments to both sides of Eq. 5 to obtain
-(cDt')2 + (Dx')2 + (Dy')2 + (Dz')2
= -(cDt)2 + (Dx)2 + (Dy)2 + (Dz)2.
Note that the LHS has coordinates in (S), and the RHS has coordinates in (S'), and that the two sides have the same form, and are equal. We now define the invariant spacetime distance Ds to be
(Ds)2 = -(cDt)2 + (Dx)2 + (Dy)2 + (Dz)2.
Note that except for the -(cDt)2 term, (Ds)2 looks just like the Euclidean distance between two points in space. Eq. 7 is sometimes written in terms of differentials (which are just infinitesimally small deltas) like this:
ds2 = -(cdt)2 + dx2 + dy2 + dz2.
Just as the distance DS (with an upper-case S) in Euclidean space is defined to be
(DS)2 = (Dx)2 + (Dy)2 + (Dz)2,
Eq. 7 defines a new distance between two points: a spacetime distance. Just as Euclidean distance DS is invariant in Euclidean geometry, spacetime distance Ds is invariant in spacetime geometry. We give a name to the spacetime geometry described by Eq. 7 (and Eq. 8). We call it Minkowski space. Eq. 9 then describes Euclidean space--the one with which we are familiar.
The key to obtaining Eq. 7 was the time dilation relation of Eq. 3: Dt = Dt'[1 - v2/c2]1/2. But Eq. 3, you may recall, was obtained because of the absolute quality of the speed of light (c' = c). Thus the changing of the Euclidean definition of distance (Eq. 9) to the spacetime definition of distance (Eq. 7) is simply necessary, in order to preserve the constancy of the speed of light.
Just as there were three coordinates (x ,y ,z) in Euclidean space, there are now four coordinates (t, x, y, z) in Minkowski space. How do you graph Minkowski space? The problem, of course, is that we now have a fourth coordinate axis corresponding to t. Mathematics has no difficulty in adding a fourth orthogonal dimension to the traditional three-dimensional coordinate system. However, being creatures of three spatial dimensions, we have difficulty visualizing a fourth dimension. That problem is easily solved. The solution is to only graph one or two spatial dimensions with the time axis, as shown in Fig. 3. Note that the time axis is labeled ct instead of t and that it is the vertical, instead of the traditional horizontal axis. We use ct instead of t so that all of the axes have the same units: those of length.
As an example of the difference between a spacetime diagram and a traditional space diagram, contrast the plots of a particle in uniform circular motion in the x-y plane shown in Fig. 4: In the traditional diagram note that the point repeats its coordinates regularly, whereas in the spacetime diagram the point never repeats its coordinates. It will repeat spatial coordinates as regularly as the point in the traditional diagram, but it will never repeat its time coordinate. This is more in keeping with how things really are.
As a final note, we could also draw spacetime diagrams in the Newtonian view using Euclidean space. However, the geometry would still be Euclidean, not Minkowski, since the distance formula would not have the minus sign in front of the (cDt)2 term as it does in Eq. 7. It is precisely this minus sign that makes Minkowski space different from ordinary Euclidean space.