Natural Science Forum / Physics / General Physics / October 2007

View velocity from the center of mass of the system or from one of the
objects.  Sign only tells direction.

On Sep 19, 4:19 pm, Chrisd5000rouge <christian.d...@fulladsl.be>
wrote:

Go back to the first situation. A "transformation" is between two
points of view (formally, frames of reference). So first you have
to define what those two points of view are.

The first one might be some point in space from which it
appears that spaceship 1 is moving at 0.8c to the right,
and spaceship 2 is moving at 0.8c to the left.

Now we can take spaceship 1 as the other FoR and ask
how the velocity of spaceship 2 transforms to that frame.

I'll use the transformation of this form:

   u' = (u - v)/[1 - uv/c^2]

where u = velocity of object (spaceship 2) in first FoR,
u' = velocity of same object in second FoR, and
v = velocity of second FoR with respect to first FoR.

So u = -0.8c and v = +0.8c

u' = (-0.8c - 0.8c)/[1 + 0.64] = 0.976c

OK, now let's consider your second situation:

The relation is still valid, but what do you mean by "run
on opposite directions"? They already were moving in
opposite directions in the first situation. One was moving
right, the other moving left.

Are you asking how to use the transform if they are receding
instead of approaching? Nothing is changed. If "to the right"
is positive, then I'd still use u = -0.8c and v = +0.8c and I'd
get the same answer.

Which makes sense if you think about it. All we used was
the velocity of the spaceships, not their relative location. The
transformed velocity is the same whether the second spaceship
is ahead moving toward the first, behind moving away, or
somewhere off to the side.

                - Randy

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?

If two cars approach at the speed of v= 80 mph and w=80 mph, the
cuckoo malformation gives their relative approach speed by
u= (80+80)/ (1+1600/c²) = about 160 mph.

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=80 mph and w= -80 mph, but then the result is u=0.

--

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In sci.physics, Chrisd5000rouge
<christian.duez@fulladsl.be>
wrote
on Wed, 19 Sep 2007 13:19:28 -0700
<1190233168.853738.213500@g4g2000hsf.googlegroups.com>:

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.)