Re: t' = (t-uxcc)/sqrt(1 -uu/cc)
From: The Ghost In The Machine (ewill_at_sirius.athghost7038suus.net)
Date: 03/19/05
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Date: Sat, 19 Mar 2005 20:00:03 GMT
In sci.physics, Eugeniusz
<lala@magma-net.pl>
wrote
on Sat, 19 Mar 2005 19:27:48 +0100
<d1hqni$jhs$1@opal.futuro.pl>:
>
> Użytkownik "The Ghost In The Machine" <ewill@sirius.athghost7038suus.net>
> napisał w wiadomości news:veqsg2-768.ln1@sirius.athghost7038suus.net...
>> In sci.physics, Eugeniusz
>> <lala@magma-net.pl>
>> wrote
>> on Fri, 18 Mar 2005 16:33:18 +0100
>> <d1es4b$gmn$1@opal.futuro.pl>:
>> > The time t' > t always ,if velocity u>0
>> > It means ,that man become older in ( x'y'z't') system ,than man in
> (xyzt)
>> > system
>> >
>>
>> That's tau = (t - vx/c^2) / sqrt(1-v^2/c^2) [*], but you're correct
>> in that all processes in the moving system will appear to slow
>> down (including one's broadcast frequency!) relative to you.
>>
>> [*] the traditional symbol for velocity is v;
>> tau is a greek letter, not a product;
>> c^2 = c * c (or, in your notation, cc).
>>
>> --
>> #191, ewill3@earthlink.net
>> It's still legal to go .sigle
> *******************************************************
> Hi
> The hands of clock goes more quickly in (x'y'z't') system,than in (xyzt)
> system..
> Sincerelly yours E.W.
>
Depends on who's observing it. As far as observer A goes, B's
clock is running slow. B can say the same regarding A's clock.
The transformation is as follows.
xi = (x - vt) / sqrt(1-v^2/c^2)
tau = (t - vx/c^2) / sqrt(1-v^2/c^2)
If one computes the following, where g = 1/sqrt(1-v^2/c^2) for brevity:
xi + v * tau = (x - v*t)*g + v*(t - vx/c^2)*g
= x*g - v*t*g + v*t*g - v^2*x*g/c^2
= x*g*(1 - v^2/c^2)
= x/g
tau + v * xi / c^2 = (t - v*x/c^2)*g + v*g*(x-v*t)/c^2
= t*g - v*x*g/c^2 + v*g*x/c^2 - v^2*g*t/c^2
= t*g*(1-v^2/c^2)
= t/g
then one immediately sees that
x = (xi + v*tau) / sqrt(1-v^2/c^2)
t = (tau + v*xi/c^2) / sqrt(1-v^2/c^2)
which is exactly what one should get.
So both of them would say "Hey, your clock is the one that's slow!" :-)
-- #191, ewill3@earthlink.net It's still legal to go .sigless.
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