Rocket moving away from the Earth

[Nugatory]

All the more so if I want to start a PhD course in theoretical physics at the end of the year.

(off-topic, so followups should go in a new thread)
You may want to read https://www.physicsforums.com/threads/so-you-want-to-be-a-physicist-a-22-part-guide.240792/

 

[PeterDonis]

- I know the proper length $l_0$.
- I know that the time interval between the return of the first and second signal is $\Delta T$.
- I know that the first signal comes back to Earth after a time of $T$.
- I know that the second signal comes back to Earth after a time of $T + \Delta T$.
- While the reflection time interval is equal to $\frac{\Delta T}2$
- The light signal obviously moves at speed $1$.

For the contraction of the lengths we have that the Earth measures:

$l = \frac{l_0}{\gamma}=l_0 \sqrt{1-v^2}$

All of this is fine. And the information above is already enough for you to answer the first part of the question--how far the rocket (more precisely, the rear end of the rocket, where the mirror that reflects the first signal is) is from Earth when the first signal arrives (at the mirror on the rear end of the rocket). There's not even any calculation involved; one of the specific quantities you list in the above is the answer to the question (given the units you have chosen, in which $c = 1$, which are the units I would choose as well).

I have to find $l$.

No, you don't, you just gave a formula for it above.

What you need to find is a formula for $\Delta T$ in terms of other quantities, including $v$. That will allow you to answer the second part of the question, which is, what is $v$.

The interval between the two signals reflections is $\frac{\Delta T} 2$.

Yes, this is correct (assuming that by "interval" you mean "time interval in the Earth frame"). What does this tell you about the distance between the two reflection events, in the Earth frame?

Is it correct to use the distance traveled by the rocket in this time interval as a length?

A length of what? If you mean a length of the rocket, no, of course not.

Or is it just a measure of how far the rocket has moved and has nothing to do with the length of the rocket according to earth?

Yes. However, that does not mean the length of the rocket has nothing to do with the problem. You are forgetting to take into account that the first signal reflects off a mirror at the rear of the rocket, while the second signal reflects off a mirror at the front of the rocket. That means the total distance, in the Earth frame, between the two reflection events is made up of two parts, one involving the motion of the rocket and one involving the length of the rocket. Can you see what the two parts are?

 

[PeroK]

It occurs to me that for the first part they might be asking where the rocket is when it receives the first signal. In which case it's simply $(c)\frac{T}{2}$ and not $(c+v)\frac{T}{2}$.

Then the second part calculates its speed at that time. And you have the rocket's position and speed at time $T/2$.

 

[PeterDonis]

It occurs to me that for the first part they might be asking where the rocket is when it receives the first signal.

That's how I was reading it. How else would you interpret "determine the rocket's distance from Earth at the moment of receiving the first signal"?

Then the second part calculates its speed at that time.

The rocket's speed is stated to be constant, so any calculation of its speed tells us its speed at any time we like. Knowledge of $\Delta T$ enables us to calculate the average speed between the first and second signal reflection, which is sufficient to answer the second part.

 

[PeroK]

That's how I was reading it. How else would you interpret "determine the rocket's distance from Earth at the moment of receiving the first signal"?

When the signal is received back on Earth. Originally, I didn't think light bouncing off a mirror was a signal being received by the ship. In any case, estimating the speed is the key part of the question.

 

[PeterDonis]

When the signal is received back on Earth. Originally, I didn't think light bouncing off a mirror was a signal being received by the ship.

Ah, I see. Well, one can always calculate both to be covered either way.

 

[Steve4Physics]

The OP seems to have put effort in. Much of the explanation has been given, but is 'scattered'. So I’m taking a bit of a liberty and posting this, hoping it will help and that it's OK with @PeroK and @PeterDonis.

Q1.

It takes time $T$ for the light to return from the rear mirror: $\frac T 2$ traveling outwards and $\frac T 2$ returning.
If the rear mirror is a distance $d$ from the source at the moment of reflection, this means: $d = speed * time = c\frac T 2$

This is the same calculation as used in elementary (sound) echo problems. It has nothing to do with relativity.

Q2.

Let P be the position of the rear mirror at the moment it reflects.

$\Delta T$ is the time for the light to travel from P to the front mirror and back to P: $\frac {\Delta T} 2$ traveling outwards and $\frac {\Delta T} 2$ returning.

Distance moved by front mirror in time $\frac {\Delta T} 2$ is $v \frac {\Delta T} 2$.

If $l$ is the (contracted) length (distance between mirrors as observed from earth) then in time $\frac {\Delta T} 2$ the light covers a total distance $l + v\frac {\Delta T} 2$

Also in time $\frac {\Delta T} 2$ we know light covers a distance $c\frac {\Delta T} 2$, giving the equation:

$c\frac {\Delta T} 2 = l + v\frac {\Delta T} 2$

No relativity needed up to this point! If we replace $l$ with $l_0 \sqrt{1-{(\frac v c})^2}$ we now have an equation from which v can be found.

Edit - unclear wording improved.

 

[Vanadium 50]

Comment 1: you didn't put in equations in the "relevant equations" part of the template. That was a mistake.

Comment 2: you don't seem to have a plan. You seem to be calculating random things hoping one will be the answer. This strategy will not get you successfully through a PhD program.

I suggest you start over. Write down the plan on how you will solve this. Write down the equations you need to execute this plan. Then execute the plan.

 

[Frostman]

I understand the problem, I find it hard to frame the situation.
For example, I didn't really get to the fact that yes, $c \frac{\Delta T} {2}$ is the distance traveled by the light, but it's not the length of the rocket, it's the length of the rocket plus the distance traveled by the rocket in that time interval.
The exercise itself is as simple as it is banal. In the end it was a question of remembering the equations of uniform motion and then understanding what those times actually represented.
I apologize to all of you, especially @PeroK for wasting your time.
In the next few times I will try to focus more on the knowledge I have available and to try, as far as I can, to understand the text better.