Jeckyl skrev:
"Dirk Van de moortel" wrote
in message ...
"Jeckyl" wrote in message
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"snapdragon31" wrote in message
...
[snip]

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

So what do these equations mean?

x' = x * sqrt(1 - v^2/c^2)
---------------------------
x = gamma*(x' + vt')
when t'=0, x = gamma*x' (which is your equation)
That is, it tells us that the event with coordinates (x',0)
in the primed frame has the coordinates (gamma*x',?) in the unprimed frame.
To find the temporal coordinate, you have to use:
t = gamma(t'+vx'/c^2) = gamma*(vx'/c^2)

So: primed frame (x',0) = unprimed frame (gamma*x',gamma*(vx'/c^2))
================================================== =================

t' = t * sqrt(1 - v^2/c^2)
--------------------------
t = gamma*(t' + x'v/c^2)
when x' = 0, t = gamma*t' (which is your equation)
That is, it tells us that the event with primed frame coordinates (0,t')
has the unprimed frame coordinates (?,gamma*t')
To find the spatial coordinate, we must use:
x = gamma*(x' + vt') = gamma*vt'

So: primed frame (0,t') = unprimed frame (gamma*vt',gamma*t')
================================================== =============

Conclusion:
the (x',t'), and the (x,t) in the equations:
x' = x * sqrt(1 - v^2/c^2)
t' = t * sqrt(1 - v^2/c^2)
cannot be the coordinates of an event in the respective frames
unless the event has coordinates (0,0) in both frames.

What did you want to find?
Wasn't it the primed frame coordinates of the event with
unprimed frame coordinates (x,t), where x = vt?

x' = gamma(x - vt) = gamma(vt - vt) = 0
t' = gamma(t - vx/c^2) = gamma(t(1-v^2/c^2)) = t/gamma

So: unprimed frame (vt,t) = primed frame (0,t/gamma)
================================================== =============


Can't you see that this is incompatible with the equations:
x' = x * sqrt(1 - v^2/c^2)
t' = t * sqrt(1 - v^2/c^2)
which yields:
unprimed frame (vt,t) = primed frame (vt/gamma,t/gamma)
which is wrong for all t but zero. And then x' = x = t = 0 as well.


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

Right.
So your "double check" showed that all was not fine,
quite the contrary, the equations you checked proved to be wrong.

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 !!

But the equations you double checked do not work for values other than 0.

Did you possibly miss what Dirk said was not fine?

--
Paul

http://home.c2i.net/pb_andersen/