Dirk Van de moortel wrote in message

wrote in message

I'm not sure I know exactly how to use the Minkowski diagram. For
example, let's say you have a stationary observer and a moving
observer, and an event somewhere, for which the coordinates are x and
t in the stationary frame and x' and t' in the moving frame. How do
you get t', for example, from the Minkowski diagram?

Ram.

Yes!
See http://users.telenet.be/vdmoortel/di...es/Lorentz.htm

Measuse slope of x'-axis w.r.t. x-axis. That gives you v.
Measure Ox' and Ot' (for instance in cm), multiply with
the calculated value
sqrt(1-v^2)/sqrt(1+v^2)
and you have the transformed coordinates x' and t'.
Verofy with the transformed values of the measured x and t.

See http://users.telenet.be/vdmoortel/di...es/Lorentz.png
and a somewhat interactive version
http://users.telenet.be/vdmoortel/di...es/Lorentz.htm
You can drag the point H around a bit, and also the 'top points'
of the t-axis and the t'axis.

Enjoy.
Dirk

Found my notes.

Take the standard drawing of the Minkovski diagram with perp.
x-axis and t-axis, and do some kid's analytic geometry....

The t-axis is represented by t = 1/v x.
The x-axis is represented by t = v x.

Take a point H with coordinates (X,T).
Take line through H, parallel with t'-axis:
t - T = 1/v ( x - X )

Intersection of this line with the x'-axis gives:
x = (T-X/v) / (v-1/v)
t = T + 1/v ( (T-X/v)/(v-1/v) - X )
The distance between (0,0) and this intersection is given by
D = sqrt( x^2 + t^2 ) .
Simplify this to
D = sqrt( (1+v^2)/(1-v^2) ) 1/sqrt(1-v^2) ( X - v T )
= sqrt(1+v^2)/sqrt(1-v^2) X'
where
X' = 1/sqrt(1-v^2) ( X - v T ) Lorentz!

So, to find X', just multiply this distance D with the scale factor
sqrt(1-v^2)/sqrt(1+v^2)

Likewise for the T'.

Dirk Vdm