"Dirk Van de moortel" wrote
in message ...

"Jeckyl" wrote in message
...
"Dirk Van de moortel"
wrote in message ...

"Jeckyl" wrote in message
...
"snapdragon31" wrote in message
...

[snip]

4. x' = x * sqrt(1 - v^2/c^2)
5. t' = t * sqrt(1 - v^2/c^2)

Lets just double check that .. LT tells us that (assuming we have that
at t=t'=0 x=x'=0, the corresponding event for (x,t) for S is (x',t')
for M, where
x' = gamma(x - vt)
t' = gamma(t - xv/c^2)
where gamma = 1/sqrt(1-v^2/c^2)
So when M (as seen from S) has gone a distance x = vt, M will be at
(x1',t1') in its own frame of reference
x1' = gamma(vt - vt)
x1' = 0 (as expected .. M has not moved relative to itself)
t1' = gamma(t - vtv/c^2)
t1' = gamma(t(1 - v^2/c^2))
t1' = gamma(t(1 - v^2/c^2))
t1' = t/gamma
and at that time, M will see S as being at (x2',t2')
x2' = -vt1'
x2' = -vt/gamma
x2' = -x/gamma
So that's all fine (other than getting the sign correct there)

All fine?
Yes all fine, provided x' = t' = x = t = 0

No .. all fine for all values where x = vt (ie for all spacetime events
along the path of M). Lorentz transforms do work for values other than 0
you know !!

You wrote:
| "...assuming we have that at | t=t'=0 x=x'=0, the
corresponding event for (x,t) for S is (x',t') for M,

Sorry .. what I mean was that we have when t=0, t'=0 and when x=0 x'=0 .. my
apologies for not expressing that correct, and thanks for spotting my
mistake in expressing that

But the rest of the analysis should be correct.