Observing from Earth the clock in a spaceship

[giuliopascal]

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.

 

[vela]

Can you explain in words what's being done in the solution? It appears you don't fully understand what those equations mean.

 

[giuliopascal]

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.

 

[vela]

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?