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Observing from Earth the clock in a spaceship

  • #1

Homework Statement


A spaceship leaves Earth at time ##t=0## with constant speed ##u##. Its clock is synchronized with the terrestrial one. At time T an earthling reads with an optical telescope the clock inside the spaceship. What value does he read?

Homework Equations


Lorentz equations.

The Attempt at a Solution


The solution says:
##c(T-T_e)=u T_e##
##T_e=\frac{T}{1+u/c}## and ##X_e=\frac{uT}{1+u/c}##
Apply a Lorentz transformation:
##t=\sqrt{\frac{1-u/c}{1+u/c}}T## at ##x=0##
And ##t## is the answer.

However when I tried to solve it I considered also the time needed for light to reach the Earth from the spaceship. As a consequence my solution would be ##t'## such that:
##t'=t+T-T_e##
Is it wrong?

Thank you very much.
 

Answers and Replies

  • #2
vela
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Can you explain in words what's being done in the solution? It appears you don't fully understand what those equations mean.
 
  • #3
Ok, I think I have taken a few steps forward.

I misunderstood which points was the Lorentz equation applied to. I thought there was a time dilatation in ##(T_e,X_e)##, but it is wrong, isn't it? In the Earth reference frame, when the spaceship emits the beam of light the clock time is ##T_e##, without any correction. If that is true, I applied a useless Lorentz transformation, and by chance I found the correct answer.

So:
solving ##x=ut## (spaceship) and ##x=-c(t-T)## (beam of light), I find ##(T_e,X_e)##: that is when the beam of light is emitted. Then I need to go back to ##(T,0)##, where the observer is. But why do I need a Lorentz transformation with boost ##u##? (I have to use that, haven't I?)

Thank you.
 
  • #4
vela
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Science Advisor
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Ok, I think I have taken a few steps forward.

I misunderstood which points was the Lorentz equation applied to. I thought there was a time dilatation in ##(T_e,X_e)##, but it is wrong, isn't it? In the Earth reference frame, when the spaceship emits the beam of light the clock time is ##T_e##, without any correction. If that is true, I applied a useless Lorentz transformation, and by chance I found the correct answer.

So:
solving ##x=ut## (spaceship) and ##x=-c(t-T)## (beam of light), I find ##(T_e,X_e)##: that is when the beam of light is emitted.
That's when and where the light is emitted as measured by the Earthbound observer.

Then I need to go back to ##(T,0)##, where the observer is. But why do I need a Lorentz transformation with boost ##u##? (I have to use that, haven't I?)
As long as light is emitted at ##(T_e, X_e)##, it will reach the observer at (T, 0). The question now is what is the image that was emitted?
 

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