On Mar 4, 3:51*pm, Eric Gisse wrote:
On Mar 4, 4:47 am, "
wrote:



On Mar 4, 2:17 pm, "Artful" wrote:

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.

Seehttp://en.wikipedia.org/wiki/Minkowski_diagram
Look at the diagram in the section entitled "Minkowski diagram in special
relativity" with the caption "In the theory of relativity both observers
assign the event at A to different times."

Artful:

I considered it, but it seems to contradict the equations for Lorentz
transformation. I mean, when I tried to get x' and t' through the
diagram and through Lorentz transformation, I got different things. I
expressed x' and t' using trigonometry from the diagrams, and I got
some kind of ugly mess. Can you point out my mistake? Or maybe there
is an analysis of how Minkowski diagrams work somewhere on the web?

Thanks,
Ram.

Lorentz transformations have nothing to do with spacetime diagrams.
Lorentz transformations are a specific type of transformation between
inertial reference frames, and the Minkowski/space-time diagram is a
characterization of the geometry of a manifold [they generalize to
conformal diagrams] by using null paths [the paths light travel along]
which is unrelated to frame transformations.

Your answer is somewhat confusing to me. Can Minkowski diagrams tell
you what x' and t' are? (When I say x' and t', I mean the coordinates
of an event from the frame of the moving observer)
This is from the Wikipedia page on Minkowski diagrams:
"Its main purpose is to allow for the space and time coordinates x and
t used by one observer to read off immediately the corresponding x'
and t' used by the other and vice versa."
Is this not correct?

And as I understand, the Lorentz transformation is supposed to tell
you x' and t' as well.

Shouldn't the two answers be the same, or am I missing something?


Thanks,

Ram.