"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,
| where
| x' = gamma(x - v t)
| t' = gamma(t - x v / c^2)"

This is equivalent with
x = gamma(x' + v t')
t = gamma(t' + x v'/c^2)
so, the minute you write
x' = x * sqrt(1 - v^2/c^2)
in this context, you also write
t' = 0
and the minute you write
t' = t * sqrt(1 - v^2/c^2)
in this context, you also write
x' = 0 .

So the combined equations
x' = x * sqrt(1 - v^2/c^2)
t' = t * sqrt(1 - v^2/c^2)
are only valid if
x' = t' = x = t = 0,
in other words for only one event.

Dirk Vdm