daniel350 said:
I need help with this as well; I'm trying to get my head around it and its so much... fun and frustration at the same time, fun because its so interesting, frustrating because I don't understand it as much as I wish too.
Here is my similar problem:
A spacecraft is moving at 80% the speed of light (0.8c) from planet X, a radio station sends a light beam parallel to the spacecraft .
What would the observer on planet X and the spacecraft pilot see in regards to 'watching' this light beam move towards the spacecraft .
In both situations light would obviously move at 1c, but how does this work for the spacecraft and observer... damn Counter-Intuitive addition of velocities... where does Time dilation and the Length Contraction come into effect, and how.
Thanks for any help you can provide,
regards
In the planet X rest frame:
Suppose as the spacecraft is leaving planet X, they synchronize clocks, so that the planet X clock reads t=0 and the spacecraft clock reads t'=0 at the moment it leaves planet X. Now suppose that planet X sends the signal 75 years later at t=75 on the planet X clock, and that at the same moment in this frame, the spacecraft is passing another planet, planet Y, which is at rest relative to planet X and 60 light years away in this frame (since the spacecraft was traveling at 0.8c in this frame, naturally 75 years after it left planet X it would be at a distance of 75*0.8=60 light-years from planet X). There is a clock at planet Y which is synchronized with the on at planet X in this frame, so it also reads t=75 when the spacecraft passes it. In this frame the spacecraft clock is slowed down by a factor of sqrt(1 - 0.8^2) = 0.6 due to time dilation, so it only reads t' = 75*0.6 = 45 when it passes planet Y.
The spacecraft starts out 60 light years away from planet X when the signal is sent at t=75, so after 300 years have passed in this frame the spacecraft will have traveled an additional 300*0.8c = 240 light years away, so 300 years later at t=375 years, it'll be 60 + 240 = 300 light-years away from planet X. And naturally if we assume the signal travels at c = 1 light year/year in this frame, the signal will also be 300 light-years away from planet X after 300 years have passed from the time it was sent, so t=375 years must be the time in the planet X frame that the signal reaches the craft. However, because of the time dilation factor, at t=375 the spacecraft clock only reads 375*0.6 = 225 years, so t'=225 years must be the time on the spacecraft 's clock when the signal reaches it.
In the spacecraft frame:
All frames agree on local events, so it must be true in this frame too that the planet Y clock reads t=75 at the moment the spacecraft is passing it, at which time the spacecraft clock reads t'=45, as we figured out before. However, in this frame the planet X clock does
not read t=75 at this moment, due to the
relativity of simultaneity. The way the relativity of simultaneity works is that if two clocks are synchronized and a distance L apart in their own rest frame, then in a frame where they are moving at speed v parallel to the axis between them, the clock in the rear will show a time that's ahead of the clock at the front by vL/c^2. So, in the spacecraft 's frame at the moment it's passing planet Y, the planet X clock must be behind by (0.8c)*(60 light-years)/c^2 = 48 years, so the planet X clock reads t=75-48=27 years at the moment the spacecraft is passing planet Y in the spacecraft frame. Since the planet X clock is slowed down by a factor of 0.6 in this frame, we know the planet X clock will take an additional 48/0.6 = 80 years to tick forward to t=75, i.e. the planet X clock will read 75 at t'=45+80=125 in the spacecraft frame. And again, all frames agree on local events, so it must be true in this frame as well that the signal was sent when the planet X clock read t=75, at t'=125 in the spacecraft frame.
Meanwhile, because of length contraction, if the distance between planet X and planet Y was 60 light years in their own rest frame, then in the spacecraft rest frame the distance is shrunk by a factor of sqrt(1 - 0.8^2) = 0.6, so the distance between planet X and planet Y is only 60*0.6 = 36 light-years in the spacecraft rest frame. So at t'=45 when the spacecraft was next to planet Y, planet X was 36 light-years away; and since planet X is moving at 0.8c away from the spacecarft in this frame, that means 80 years later at t'=125 when planet X was sending the signal in this frame, planet X must have been an additional 80*0.8=64 light-years away, for a total of 64+36=100 light-years away from the spacecraft at the moment the signal was sent at t'=125. So, under the assumption that light travels at c in this frame too, the signal will reach the spacecraft 100 years later at t'=225.
By the way, another way to approach this problem would be to use the Lorentz transformation, which says that if an arbitrary event has coordinates x,t in one frame, then in a second frame moving at v along the first frame's x-axis (and with the two frames having a common origin, so x=0,t=0 in the first frame coincides with x'=0,t'=0 in the second), this same event will have coordinates:
x'=gamma*(x - vt)
t' =gamma*(t - vx/c^2)
with gamma = 1/sqrt(1 - v^2/c^2)
In this problem v=0.8c so gamma=1/0.6=1.666..., which means for example that if the event of the signal being sent had coordinates (x=0, t=75) in the planet X frame, in the spacecraft frame the event of the signal being sent has coordinates:
x'=1.666...*(0 - 0.8*75) = -100
t' =1.666...*(75 - 0.8*0) = 125
And if the event of the signal reaching the spacecraft happened at (x=300, t=375) in the planet X frame, in the spacecraft frame this event has coordinates:
x'=1.666...*(300 - 0.8*375) = 0
t'=1.666...*(375 - 0.8*300) = 225
So, you can see that this agrees with the previous calculations, and it's a little simpler to figure out although if you use the Lorentz transform you don't get to think about the different contributions of time dilation, length contraction and the relativity of simultaneity.
