Re: Rel. Speed


"razyrel" <razyrel@xxxxxxxxxxxxxxxxxx> wrote in message news:f5vglc$4g5$1@xxxxxxxxxxx
"N:dlzc D:aol T:com (dlzc)" <dlzc@xxxxxxx> wrote in message news:eoHgi.472804$115.245105@xxxxxxxxxxxxxxx
Dear rayzel:

"razyrel" <razyrel@xxxxxxxxxxxxxxxxxx> wrote in message
news:f5vdmj$quq$1@xxxxxxxxxxx
...
Ok, then
how is the situation in the following extension of the
above experiment:

the ship shoots a torpedo that has an instant speed
of 0.4c, into the front direction of the ship, and a
second one of the same kind to the rear direction;
both shot exactly at the same time (ie. at the same
location) as in the scenario above. How far away,
relative to the stationary observer, will each of these
torpedos be after 10 seconds (in c units)?
Is it 0.4*10=4c for both torpedos, or will they differ?

You say 0.4c... is that as measured by the ship travelling at 0.8c?

Let's say it is the propery of these rockets,
ie. they are that fast if fired from rest (ie. if the ship's v=0).

http://hermes.physics.adelaide.edu.au/~dkoks/Faq/Relativity/SR/rocket.html

You say 10 seconds... is that measured by the stationary observer?

Yes

They could differ, depending on your answers.

Would be interessting to analyse the differences.

This is simple geometry.

The ship has equation of motion
x = v t
where
v = 0.8 c

The forward torpedo has equation
x' = u t'
The backward torpedo has equation
x' = - u t'
where
u = 0.4 c

After a time t', the distance between the torpedos, according
to the ship is
D'(t') = u t' - ( - u t' ) = 2 u t'

According to the stationary observer, using the Lorentz
transformation, the equation of motion of the forward
torpedo
x' = u t'
is transformed to
g ( x - v t ) = u g ( t - v x /c^2 )
giving
x = ( v + u ) / ( 1 + v u /c^2 ) t
in which you might recognize the relativistic velocity
composition formula.

Likewise, the backward torpedo equation
x' = - u t'
is transformed to
g ( x - v t ) = - u g ( t - v x / c^2 )
giving
x = ( v - u ) / ( 1 - v u / c^2 ) t

So after a time t the distance between the torpedos,
according to the stationary observer is
D(t) = ( v + u ) / ( 1 + v u / c^2 ) t' - ( v - u ) / ( 1 - v u / c^2 ) t'

Al you have to do now, is to fill in the values for u and v
and for c and t.

I get 2.4 10^9 meters after 10 seconds measured by the
ship, and 9.6 10^8 meters after 10 seconds measured by
the stationary observer.

Dirk Vdm


.

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