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-Vitaly-
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Homework Statement
Two observers A and B , in inertial frames, have identical clocks. B passes close to A
with a relative velocity 0.5c. t seconds after B has passed, A flashes him a light signal.
2t seconds after receiving this signal (as measured on his own clock), B acknowledges by
a return light signal to A. When does A receive it?
Homework Equations
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html"
The Attempt at a Solution
I'm new to relativity, so this question may seem trivial to some of you. Nevertherless here is what I tried to do.
-Consider A to be stationary. With coordinate x=0. Then B is moving with V=0.5c relative to x. So B's coordinate wrt to A is x'=0.5cT/y (T time, of A reference frame). where y=(1-(0.5c)^2/c^2)^0.5=(sqrt3)/2
-At time (A reference frame) T=t, A flashes a light signal to B, which travels at c. So it's coordinate is c*(T-t), in A reference frame, because the light beam is kinda delayed by t (i.e. if you put T=t, the coordinate of the light beam is 0).
-Now, I need to calculate how long it takes in ref frame A for the light to get to B. So,
x'=c(T-t) => cT/sqrt(3)=c(T-t) => T=t*sqrt(3)/(sqrt(3)-1)
-So the light beam reached B, now B delays 2t time interval in B's frame (T'=2t). So I need to calculate how long it is in A frame.I start by writing down what T' is in T terms.
T'=(T-Vx/c^2)/(sqrt(3)/2)=(T-0.5x/c)*2/sqrt(3)=(2Tc-x)/(c*sqrt(3)), but x=0 (?not sure?,I'll try yes)
And this is equal to 2t. => 2t=2T/sqrt(3) => T=t*sqrt(3)
-So now I know that the total delay until B flashes back from the start of the motion is (A reference frame) t+t*sqrt(3)/(sqrt(3)-1)+t*sqrt(3)=t(sqrt(3)-2)/(sqrt(3)-1)
-Now I need to find how long it takes for the light to get back to A from B.
This probably involves knowing the coordinate of B and the distance between the points. I don't know what to do next. And I have 16 of these questions to do till tomorrow (this is 3 and they get harder and harder
![Cry :cry: :cry:](/styles/physicsforums/xenforo/smilies/cry.png)
Thank you
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