Shall we experiment a little?

Start with the well-known time dilation formula from SR:

t' = t / SQRT( 1 - v / c )             (1.1)

t = time in frame 0, t' = time in frame 1, t'' = time in frame 2, and so on. All frames have a constant rectilinear velocity relative to standing still (defined as standing still) frame 0. Frame 0 sees frame 1 to have velocity v, frame 1 sees frame 0 to have velocity -v. Displacement s = vt, s' = -vt'. 1ijk-space = 4-dim quaternion space, xyzt-spacetime = our well-known spacetime. Take in mind two events A and B and set the Origin in A.

t' = t / SQRT( 1 - v / c )

( 1 - v / c ) t' = t

t' - v t' / c = t' - s' / c = t

(c t') - s' = (c t) - 0 = ...             (1.2)

The 0 means s = 0, we just set s at zero.

SR states the interval (c t) - s is a constant for all frames:

(c t) - s = (c t') - s' = (c t'') - s'' = (c t''') - s''' = ... = constant             (1.3)

When you take the Lorentz transformations t' = (formula expressed in t and x) and s' = (formula expressed in t and x) and fill them in in (c t') - s' then you find (c t) - s.

In fact this is all there is to SR. It is called the invariance of the interval. When there was no SR then in an x-t-diagram would have hold the theorem of Pythagoras, (c t') + s' = constant for all frames. The t-axis and the axes x, y and z would be perpendicular to each other in all frames. Lengths along the t axis would have been treated just like lengths along x, y or z. So only this minus sign is the difference between Newton spacetime and Minkowski spacetime. Can quaternions provide this minus sign?

Usual is to take the space coordinates as real numbers and the time coordinate as an imaginary number (t provided with an i, the i from complex numbers) . Now instead take time t to be real and take displacement s to be the length of quaternion q = fi + gj + hk.

A rectangle with sides a meter and b meter has surface ab m (m = meter). A length of 1 meter squared is 1 m, the unit of surface. Likewise squaring length fi yields surface (fi) = f i = -f (because i = -1). Then the distance s is:

s = f i + g j + h k

= -f -g -h             (1.4)

Distances are defined in ijk space likewise as in xyz spacetime, except for that the distances count as negative.

The distance in 1ijk space between the origin and e + fi + gj + hk then is

= e -f -g -h             (1.5)

(We used real numbers e, f, g, h instead of a, b, c, d in order to avoid the confusing use of c as real number and as the speed of light.)

The distance in 1ijk space resembles a lot the invariant spacetime interval as it is in xyzt spacetime. What about the spacetime interval of xyzt space, but then expressed in 1ijk space? Real number e then corresponds with imaginary number t, and e' corresponds with t'.

(c t') - s' = (c t)             (1.6)

(c e') + f + g + h = (c e)             (1.7)

As you see, the so called invariance of the interval is in 1ijk space just the well-known theorem of Pythagoras.

I call this the real-imaginary swap. See more about it in paragraph The real-imaginary swap at page 4 of QQD.

1ijk is a richer space. In 1ijk space length a times width b is height c, where if a * b = c then b * a = -c. When changing from 1ijk to xyzt spacetime, the height dimension property is lost, leaving only a real number there. The two possibilities ab = c and ba = -c are reduced to ab = ba = c. Compare the transition from wavefunction to probability division by by taking the length (the modulus) and squaring it.

Something like this minus-sign one also meets in (1.5) at page 3 of QQD:

(-1 + i + j + k) (-1 + i + j + k) = -2 -2i -2j -2k = -2 ( -1 + i + j + k)

from the excel document baryoncollision.xls

A previous attempt

                        

Triangles 1 up to 6 are congruent.
1 means the surface of triangle 1, and so on.

a + 1 + 2 + 3 + 4 - b

= e + 3 + 4 + 5 + 6

a - b = e

a + (b * i) = e

Generalize this to

a + (b * i) + (c * j) + (d * k) = e

a - b - c - d = e

This resembles the invariance of the interval,

t - x - y - z = a