# Rocket to a space station

1. Sep 19, 2006

### michael879

ok this is a rly simple problem I just cant figure it out. There is a rocket traveling from earth to a space station (which are in the same reference frame). Earth sends light signals at some interval to the rocket, which then sends the signals to the space station. When I work this out I get that the rocket sees the signals coming from earth as dt*gamma apart where dt is earth's interval. The rocket then sends the signals at dt*gamma which should be received by the space station at dt*gamma*gamma intervals. I know for a fact this is wrong (and I suspect the first one might be wrong too) since when I draw a space-time diagram of this situation the space station receives the signals at the same speed that earth sends them and the rocket sends and receives at a higher interval. What am I doing wrong?

2. Sep 19, 2006

### Staff: Mentor

Are we to assume that the rocket transmits its signals as soon as it receives the signals from Earth?
According to the rocket frame, the Earth transmits signals dt*gamma apart. Now figure out the time interval (according to the rocket) between the arrival of those signals at the rocket.

3. Sep 19, 2006

### michael879

damn, right, thanks a lot man.

4. Sep 22, 2006

### lightarrow

I'm sorry, maybe I have to study more, but there is something I don't understand: wouldn't this situation be the same as doppler red-shift of light emitted from earth? So the rocket should see the signals arriving to it with a lower frequency, because it is moving away from earth.

When it re-trasmit them immediately to the space-station, the station see them doppler blue-shifted because the rocket is approaching it; so, at the end, signals arrive to the station with exactly the same frequency they are trasmitted from earth (in the earth ref. frame):

f(rocket) = f(earth)*SQRT[(c-v)/(c+v)]

f(space-station) = f(rocket)*SQRT[(c+v)/(c-v)] = f(earth)*SQRT[(c-v)/(c+v)]*SQRT[(c+v)/(c-v)] =

= f(earth).

f=frequency
v=rocket's speed

Last edited: Sep 22, 2006
5. Sep 22, 2006

### Staff: Mentor

Sure it's the same. The observed frequency of incoming signals compared with the source frequency is the Doppler effect. (In solving this problem one essentially derives the relativistic Doppler formula.)