sci.physics.relativity

> : >
> : > R^2 = (ct)^2 + (ix)^2 + (iy)^2 + (iz)^2 = (ct)^2 + (ir)^2
> : >
>
> The distance from the centre to the surface of a hypersphere.
>

Ok. That makes sense, mathematically anyway. So then the point
(x,y,z,t) is at a distance R from some origin where that hypersphere
is, where the values 'x', 'y' and 'z' are mapped to the axis X, Y and
Z respectively of the Space dimensions, and the value 't' is mapped
onto the Time axis. Cool.

I understand why Einstein (or whoever else) came up with that formula
because mathematically, it follows the same form as the one for the
Space dimensions:
x^2 + y^2 + z^2 = R^2
*except* that in that formula, R is a distance in the same Space
dimensions as x, y and z whereas in the formula
R^2 = (ct)^2 + (ix)^2 + (iy)^2 + (iz)^2
R is the *time* value T (as you confirmed below). It does not reflect
anything about the Space dimension where (x,y,z) is. It's like that
the distances in Space are irrelevant. One would think that the R
should be some value where both time and distance units are involved,
no? This formula confuses me.

> : The
> : (vt) is distance on a Space axis, say x, so it comes to: T^2 + x^2 =
> : (ct)^2, or T^2 = (ct)^2 - x^2 = (ct)^2 + (ix)^2.
> :
> : So does the R mean the same as the T? So then R is 'distance' in the
> : Time dimension... is that correct?
>
> Supposed to be, except Einstein and Minkowski were jokers and idiots,
> not mathematicians.
>
> : Another thing. The fact that the imaginary number 'i' appears in the
> : Space dimension should not be taken to mean that the Space dimensions
> : are imaginary in the sense that they do not exist. They do exist:
> : there are three such dimensions. That is clear. What is imaginary are
> : the Space axes. They are imaginary in the same way that a grid on a
> : city map are imaginary. Having said that, in the formula, the time 't'
> : is not imaginary, so then the Time axis is real, although we cannot
> : see or feel it. Am I understand this correctly?
>
> I can go back to where I was. That's imaginary, because I travel
> a negative distance. I seem to be having difficulty going back to
> when I was, but that was real enough.
>

I see what you mean, and I was thinking the same too, except that the
minus sign in the formula
R^2 = (ct)^2 - x^2 - y^2 - z^2
does not mean a negative direction of movement in space. When I go
back where I was, it is the 'x', 'y' and 'z' values that become
negative, right?

In any case, that formula does not feel to have a physical reality. No
wonder that the 'i' number is involved! ;-)

Jean