What's the Time?

Rant#21:

There are two types of quantities we are interested in: scalars which have a magnitude (think point), and vectors which have a magnitude and a direction (think arrow). There are two types of multiplications (products)of vectors (say A and B): the dot(scalar product: A . B =|A||B|cos(angle between the vectors)) and cross(vector product: A x B =|A||B|sin(angle between the vectors)) products. For the perpendicular axes we will consider the angles between the vectors as 90 degrees with the sin(90)=0 and the cos(90)=1. So all scalar products will zero and the vector products will equal 1|A||B| where |A| is the length but not the direction of A only, etc.. The vectors will be unit vectors with a length equal 1, so their only difference will be their direction. These directions are alternately labeled X or i, Y or j, Z or k. There is a good dscussion of the vector product on the web at SolitaryRoad.com (who invites email contact) and once again I got summarily ignored when I emailed him/her/it with what I thought was an interesting question (see below). Now the new thing I am ranting about is combining three Argand planes as described in Rant#6. Each of these planes are X vs. vX, Y vs. vY, and Z vs. vZ. each aligned to one of the three axes: X, Y, and Z. This somehow makes for six perdendicular axes in 3-dimensional space (not something you can visualize). The figure shows the Argand planes (recall that i has two usages with some confusion: i is the label of the unit X vector and also the imaginary quantity of the square root of minus one. This is why I chose to use X as the unit vector along the X axes and vX as the imaginary axis. The vis for virtual (and looks like the square root surd) and is the same as imaginary but less confusing than i and is perpendicular to X. The direction of the vector product is perpendicular to the plane of the two vectors involved in a right-handed sense. This is shown in the figure, it would be writen X x Y = Z but if the order of multiplying were reversed the result would be negative as Y x X = -Z.
Using these rules and noting the order and sign given by the circle in the figure, i x j=k and i x k =-j, we can construct a multiplication table in the order ROW VARIABLExCOLUMN VARIABLE. There are a couple of observations we can make about the table. First off, any axis is not perpendicular to itself, it's parallel with an angle of zero, so the sin(0)=0 and the cross product is zero. This accounts for the long (main) diagonal in the table. Second, and this is the question to SolitaryRoad.com, we have no knowledge of the cross product of a real axis with its imaginary (as in the Argand plane), I postulate that it is TIME as shown in the figure and in the Table in the diagonals in the upper left and lower right quadrants. The rest of the terms follow the |A||B|sin(90) rule.

Variable	X	Y	Z	vX	vY	vZ
X	0	Z	-Y	t	vZ	-vY
Y	-Z	0	X	-vZ	t	vX
Z	Y	-X	0	vY	-vX	t
vX	-t	vZ	- vY	0	vZ	-vY
vY	-vZ	-t	vX	-vZ	0	vX
vZ	vY	-vX	-t	vY	-vX	0

This table can be simplified by treating each row as a equation where Row Variable = Sum of (Table Entry x Column Variable). If we consider the terms in the Sum as scalar products where A.B = B.A and a lot of terms will cancel in each row.

• X = 0 X + Z Y - Y Z + t vX + vZ vY - vY vZ = t vX
  
• Y = -Z X + 0 Y + X Z - vZ vX + t vY + vX vZ = t vY
  
• Z = Y X - X Y - 0 Z + vY vX - vX vY + t vZ = t vZ
  
• vX = -t X + vZ Y - vY Z + 0 vx + vZ vY - vY vZ = -t X
  
• vY = -vZ X - t Y + vX Z - vZ vX + 0 vY + vX vZ = -t Y
  
• vZ = vY X - vX Y - t Z + vY vX - vX vY - 0 vZ = -t Z
  

Variable	X	Y	Z	vX	vY	vZ
X	0	0	0	t	0	0
Y	0	0	0	0	t	0
Z	0	0	0	0	0	t
vX	-t	0	0	0	0	0
vY	0	-t	0	0	0	0
vZ	0	0	-t	0	0	0

So now there are three equations in the real coordinates (X,Y,Z) and three in the corresponding virtual coordinates all equal to our postulated variable t, time, If each of the three virtual equations is divided by the square root of -1 = v, for example:

vX = -t X becomes vX/v = -t X/v but 1/v = -v then X = t vX. In other words there are only 3 equations. Alternatively we could have started with the three equations in the real variables (X,Y,Z) and obtained the 3 equations in the form vX = -tX which is a bit more satisfying because this says that the three space (Cartesian) coordinates themselves are related to their imaginary counterparts and the negative of time. This in a sense is what Einstein established about the concept of "spacetime." That events can be separated by a Pythagorean like theorem in which the distance is given by the square root of (X^2+Y^2+Z^2-t^2) where the space coordinates are summed but the time coordinate is subtracted.