Hello again, my friend; well here it is:
http://img41.imageshack.us/img41/5448/specialrelativitydiagra.jpg
A Special Relativity diagram to demonstrate the relationship between two Inertial Frames of Reference moving with a constant relative velocity.
A and B represent a single axis (time or length) of Minkowski spacetime for each of the two IFoRs. They are drawn against a common background representing Proper Units
An observer at rest within each IFoR will be experiencing proper units (length and time) within that system; as shewn by the horizontal lines, labelled A and B.
But from each IFoR, the other frame's axis -- rotated according to their relative velocity -- will be reckoned in co-ordinate units, as shewn by the perpendicular projections from the coloured diagonals onto the observer's own axis.
The diagram is drawn to scale to represent two IFoRs with a constant relative velocity = 0.6c, giving \gamma = 1.25 and \frac{1}{\gamma} = 0.8
From this we can see exactly what Einstein was saying in http://www.bartleby.com/173/12.html" , when he writes:
But the metre-rod is moving with the velocity v relative to K. It therefore follows that the length of a rigid metre-rod moving in the direction of its length with a velocity v is
of a metre.
For the metre rod moving with the velocity 0.6c relative to B would be represented in the diagram by the number 1 on the red diagonal and we can see the projection onto B's x-axis (as it would be in this scenario) where it would be the green 1 co-ordinate unit.
And this agrees with {x^'} = \frac {x}{\gamma}
i.e. {x^'} = 0.8x
Similarly, in the second part of that same chapter, Einstein writes:
**Let us now consider a seconds-clock which is permanently situated at the origin (x' = 0) of K'. t' = 0 and t' = 1 are two successive ticks of this clock. The first and fourth equations of the Lorentz transformation give for these two ticks:
t = 0
and
As judged from K, the clock is moving with the velocity v; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but
seconds, i.e. a somewhat larger time.
And again we can see just how this works, for this time he is converting the time from the observer's frame, t (proper time units) into the observed time {t^'} (co-ordinate time units) which would be to take the blue, 1 proper unit and project it upwards onto the red diagonal line or, indeed, one could read it off the green co-ordinate scale on B's axis.
Not surprisingly this agrees with Einstein's own equation:
t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} or
t = \gamma({t^'}) = 1.25 co-ordinate units
I am fairly certain that this diagram meets all the conditions, references and relationships between the elements that compose SR as described in this thread.
For instance, the co-ordinate units are greater in number but reduced in size.
The principal problem that is brought to light here is that there are far more elements than at first appear.
Consider if you will:
1) We start with a solitary IFoR where the measurements are all, by definition, in proper units.
2) We add a 2nd IFoR moving at a constant velocity with respect to the first: both are measured in proper units and their times are identical and synchronous.
3) They then observe one another and upon doing so we find that the observed frames are rotated with respect to their observers, as shewn, but their units are still proper units.
4) When observing the rotated frames, their proper units are projected vertically onto the observer's frame of reference, being there-by transformed into co-ordinate units.
5) So we have the axis of each IFoR and its rotation, both measured in Proper units and its projection onto the other's axis measured in co-ordinate units.
6) And as all this is matched reciprocally by the other IFoR we have this duplicated giving 4 measures in proper units and two in co-ordinate units.
Is it any wonder that we become confused when trying to deal with this using only primed and unprimed symbols?

And this is with it all reduced to two dimensions...
This exercise has certainly helped me to understand how it all fits together – I only hope that I have understood it correctly?
Many thanks, once again to Rasalhague, and Matheinste whose patience and explanations have been a great help.
