Re: lorenz transformation
- From: The Ghost In The Machine <ewill@xxxxxxxxxxxxxxxxxxxxxxx>
- Date: Fri, 21 Sep 2007 21:05:40 -0700
In sci.physics, Chrisd5000rouge
<christian.duez@xxxxxxxxxxx>
wrote
on Wed, 19 Sep 2007 13:19:28 -0700
<1190233168.853738.213500@xxxxxxxxxxxxxxxxxxxxxxxxxxx>:
Hello,
Can somebody clarify this for me?
if two spacecrafts approach at the speed of v=0.8c and w=0.8c, the
lorenz transformation gives their relative approach speed by u= (v+w)/
(1+vw/c²) = about 0.975c. ok fine, I understand it can not be more
than 1c.
Now let's suppose that they run on opposite directions at those same
speeds, is this relation still valid? If I use the same relation, I
would state v=0.8 and w= -0.8c, but then the result is u=0 ???. How
should it be calculated?
Your coordinate system is slightly confused. One might
use instead the subtraction formula
u = (v-w)/(1-vw/c^2)
for the sake of consistency. That way, spacecraft A could
have velocity 0.8c, and spacecraft B velocity -0.8c --
and they would approach in that case, yielding
u = 1.6 / 1.64 = 40/41 = 0.97561 c.
If both spacecraft have velocity 0.8c, then one gets
(0-0)/(1 - .64) = 0
which is a useful check to at least ensure the signs are close to
correct. ;-)
(Note that it is possible to derive either formula from the Lorentz.)
--
#191, ewill3@xxxxxxxxxxxxx
"Woman? What woman?"
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- References:
- lorenz transformation
- From: Chrisd5000rouge
- lorenz transformation
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