Rocket moving away from the Earth

[Frostman]

Homework Statement

A rocket of rest length $l_0$ moves away from the Earth with constant speed $v$. A light signal sent from Earth is reflected by mirrors placed one on the tail and one on the head of the rocket. The first reflected signal returns to the transmitting station after a terrestrial time interval $T$, while the second reflected signal arrives with a delay of $\Delta T$ compared to the first. Based on this information, determine the rocket's distance from Earth at the moment of receiving the first signal and its speed of departure.

Relevant Equations

Time dilation and length contraction

I first began to identify the various events in the problem. I call the rocket $S'$ and the Earth $S$.

1. Sending signal
2. Tail signal reflection
3. Head signal reflection
4. Tail signal return
5. Head signal return

For the Earth I know that:
$t_4 = T$
$t_5 = T + \Delta T$

Since the two events occur at the same spatial point for $S$, $S'$ will observe a dilation of the time interval.
$\Delta T_{54} = \Delta T$
$\Delta T_{54}' = \gamma \Delta T_{54}=\gamma \Delta T$

Furthermore, $S$ sees the rocket contracted by a gamma factor.
$l=\frac {l_0} \gamma$

By identifying with $d$ the distance traveled by the rocket we can write that ($c=1$):

$t_2=\frac d {1-v}$
$t_3=\frac {d+\frac {l_0} \gamma}{1-v}$

But I can't go beyond these considerations to find the two requests ...

 

[PeroK]

Why not focus on the $\Delta T$ first? What does $\Delta T$ depend on?

 

[PeroK]

PS finding the distance should be easy, so I'm not sure why there's a problem there.

 

[Frostman]

Why not focus on the $\Delta T$ first? What does $\Delta T$ depend on?

It depends on the time that the signal from the tail arrives on Earth ($t_4$) and on the time that the signal from the head arrives on Earth ($t_5$).
So if the tail-to-ground distance at instant $t_4$ is $x$, the distance that the head-to-ground signal must travel is $x + \frac {l_0} \gamma$. Right?
So:
$\Delta T = \frac {\frac {l_0} \gamma}1$

 

[Frostman]

PS finding the distance should be easy, so I'm not sure why there's a problem there.

If $t_4=T$, then the distance should be: $d=(1-v)T$, where $v$ is the speed of the rocket, right?

 

[PeroK]

If $t_4=T$, then the distance should be: $d=(1-v)T$, where $v$ is the speed of the rocket, right?

This makes no sense.

I assume they are asking for the distance to the rocket when the signal is received back on Earth. The questions are:

1) How far is the rocket when the first signal reaches the rocket?

2) How far is the rocket when the first signal is received back on Earth?

This is elementary kinematics.

 

[PeroK]

It depends on the time that the signal from the tail arrives on Earth ($t_4$) and on the time that the signal from the head arrives on Earth ($t_5$).
So if the tail-to-ground distance at instant $t_4$ is $x$, the distance that the head-to-ground signal must travel is $x + \frac {l_0} \gamma$. Right?
So:
$\Delta T = \frac {\frac {l_0} \gamma}1$

You're missing the obvious here: $\Delta T$ is the time it takes the second signal to go from the first mirror to the second mirror, then back to where the first mirror was.

That's the extra journey that the second signal takes.

 

[Frostman]

1) How far is the rocket when the first signal reaches the rocket?

$x_1 = (1-v)t_2$
Because as the signal reaches the mirror, the rocket moves.

2) How far is the rocket when the first signal is receiced back on Earth?

$x_2 = 1t_4=T$

But I need to fine $t_2$ so

 

[PeroK]

$x_1 = (1-v)t_2$
Because as the signal reaches the mirror, the rocket moves.

No. $x_1 = ct_2$

And $t_2 = T/2$, right? Is that not clear?

 

[Frostman]

No. $x_1 = ct_2$

Because $t_2$ already take into account the extra time the rocket has moved?

And $t_2 = T/2$, right? Is that not clear?

Because it takes the same time to come and go?

 

[PeroK]

Because $t_2$ already take into account the extra time the rocket has moved?

Because it takes the same time to come and go?

$T$ does not directly depend on the speed of the rocket. It simply tells you how far the rocket is at a point in time. You're not told how long the rocket has already been gone.

And, yes, the light is just bouncing back off a mirror. It must be the same time there and back.

 

[Frostman]

We have:

$t_1=0$
$t_2=\frac T 2$
$t_3=\frac {T+\Delta T} 2$
$t_4=T$
$t_5=T+\Delta T$

So the rocket's distance from Earth at the moment of receiving the first signal is:

$d=\frac T 2 + T = \frac 32 T$

Now I need to find the speed of the rocket. I suppose to extract this information by $\gamma$ in some time dilation or length contraction. I have those times for $S$. I have the proper length $l_0$.
If I measure the distance between the two mirror for $S$, can I use it as a measure of the length of the rocket? Or probably will be more long than it should be, because the mirror in the head is going on?

 

[PeroK]

We have:

$t_1=0$
$t_2=\frac T 2$
$t_3=\frac {T+\Delta T} 2$
$t_4=T$
$t_5=T+\Delta T$

So the rocket's distance from Earth at the moment of receiving the first signal is:

$d=\frac T 2 + T = \frac 32 T$

No. You're not thinking clearly about this problem.

It's hard to help if you make so many basic errors.

Perhaps you should work through the problem with a given $v = \frac 1 2 c$, say, and some initial distance from Earth: $10$ light-seconds, say. Then, work through the problem, calculating $T$ and the distance of the ship from Earth when the first signal arrives. Do the first part first. Leave the second mirror until you have formed a correct idea about what is happening.

Note: the first part is simple kinematics - you don't need to worry about SR.

 

[Frostman]

I try to describe the situation:
The rocket leaves the Earth at a time $t_1 = t_1' = 0$.
At a certain moment that we do not know the Earth sends a signal.
This signal arrives on the mirror present on the tail of the rocket at a time $t_2$.
When this signal returns to earth, it returns to a time $t_4=T$.
Since it is reflected by a mirror, the outward time $t_2$ is half of the total time spent in the return, so $t_2 = \frac T 2$.
Can we agree on this first part?

 

[PeroK]

The round trip time for a light signal to the rocket is $T$.
Since it is reflected by a mirror, the outward time and return times are both $T/2$.
Can we agree on this first part?

I'd agree with that.

You need to try with the example data I gave. You need to find some way to understand this and working an example is the obvious way.

 

[Frostman]

I'd agree with that.

You need to try with the example data I gave. You need to find some way to understand this and working an example is the obvious way.

When the first signal starts, the rocket is at a distance of 10 light-seconds, to which we add the distance to get from the Earth to the mirror and from the mirror to the earth, i.e.:

$d = 10 \text{ light-seconds} + c\frac T2 + c\frac T2$

 

[PeroK]

That's wrong. And, you haven't even tried to calculate $T$.

This is hard work!

Let me show you the answer to this. But, after that you have to start thinking for yourself.

We let $t = 0$ when the signal is sent:

The signal is at position $ct$ and the ship at position $10cs + vt$.

We then equate the two: $ct = 10cs + vt$ gives $t = \frac{10cs}{c - v} = 20s$.

The signal reaches the ship after $20s$ and so $T = 40s$.

Where is the ship at $t = 40s$?

We have $d = 10 cs + \frac 1 2 c (40s) = 30 cs$.

The ship is $30$ light seconds away when the signal returns to Earth.

Finally, how does that relate to $T$ and $v$?

When the signal reaches the ship, it was a distance $d_1 = \frac{cT}{2}$ from Earth.

And, when the signal arrived back on Earth, the ship was $d = d_1 + \frac{vT}{2} = \frac{cT}{2} + \frac{vT}{2}$.

That gives us a formula: $$d = \frac{(c+v)T}{2}$$ And, we can check that works in this example: $d = \frac{3cT}{4} = 30 cs$. Which is correct.

But, now it's up to you.

 

[PeroK]

When the first signal starts, the rocket is at a distance of 10 light-seconds, to which we add the distance to get from the Earth to the mirror and from the mirror to the earth, i.e.:

$d = 10 \text{ light-seconds} + c\frac T2 + c\frac T2$

In other words, that should have been: $$d = 10 \text{ light-seconds} + c\frac T2 + v\frac T2$$

 

[Frostman]

That's wrong. And, you haven't even tried to calculate $T$.

This is hard work!

Let me show you the answer to this. But, after that you have to start thinking for yourself.

We let $t = 0$ when the signal is sent:

The signal is at position $ct$ and the ship at position $10cs + vt$.

We then equate the two: $ct = 10cs + vt$ gives $t = \frac{10cs}{c - v} = 20s$.

The signal reaches the ship after $20s$ and so $T = 40s$.

Where is the ship at $t = 40s$?

We have $d = 10 cs + \frac 1 2 c (40s) = 30 cs$.

The ship is $30$ light seconds away when the signal returns to Earth.

Finally, how does that relate to $T$ and $v$?

When the signal reaches the ship, it was a distance $d_1 = \frac{cT}{2}$ from Earth.

And, when the signal arrived back on Earth, the ship was $d = d_1 + \frac{vT}{2} = \frac{cT}{2} + \frac{vT}{2}$.

That gives us a formula: $$d = \frac{(c+v)T}{2}$$ And, we can check that works in this example: $d = \frac{3cT}{4} = 30 cs$. Which is correct.

But, now it's up to you.

Okay, what I did is really stupid. I calculated the path of the light signal, not the distance covered by the rocket.
The equations of motion are:

$x_L = ct$
$x_R = x_{0R} + vt$

So when the light signal reaches the rocket we are at a distance

$d_1 = x_L = x_R$
$d_1 = \frac{cT}2$

So when the light signal returns to Earth the rocket has traveled a distance

$d_2 = x_R(T)-x_R(\frac T2) = \frac{vT}2$

Finally we have:

$d = d_1 + d_2 = \frac{(c+v)T}{2}$

I know that I have repeated what you have written, but I have tried to elaborate considering precisely the equations of motion and times that are considered.
The point is that I considered the distance traveled by the light.

 

[Frostman]

Now I need to find the speed of this rocket $v$.
If I want to use length contraction by knowing the proper length $l_0$ I should find what $S$ sees. What we have is some times.
- The rocket takes a time equal to $\Delta T' = \frac {l_0} v$ to cover the tail-head section.
- While the light takes a time equal to $\Delta T = c \frac {l_0} {\gamma}$ to cover that stretch.
Could this be correct reasoning? I am not convinced of the second point, but perhaps it is due to the same problem as before.

 

[PeroK]

Now I need to find the speed of this rocket $v$.
If I want to use length contraction by knowing the proper length $l_0$ I should find what $S$ sees. What we have is some times.
- The rocket takes a time equal to $\Delta T' = \frac {l_0} v$ to cover the tail-head section.
- While the light takes a time equal to $\Delta T = c \frac {l_0} {\gamma}$ to cover that stretch.
Could this be correct reasoning? I am not convinced of the second point, but perhaps it is due to the same problem as before.

First, everything is in the Earth frame, so there is no need to involve the rocket frame. The only relevance of SR is length contraction of the rocket.

Second, this is a similar kinematic problem to before: light reflecting off a receding mirror.

 

[Frostman]

Okay, then if I don't consider primed quantities. I can write that the difference in time between the tailed signal and the head signal is given by this reason: the head signal travels farther than the tailed one.

For $c=1$

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

$v= \frac{\sqrt(l_0^2-{\frac {\Delta T} 2}^2)}{l_0}$

 

[PeroK]

That's not right at all.

 

[Frostman]

Oh, again

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

$v=\frac{l_0}{\sqrt({\frac {\Delta T} 2}^2 + {l_0}^2)}$

I mixing every time. I should take the time of signal and value it for the rocket.

 

[PeroK]

Not even close. Try $v \approx 0$.

 

[Frostman]

Not even close. Try $v \approx 0$.

For $v \approx 0$ I find:

$v\frac {\Delta T} 2 = \frac {l_0} {\gamma} = \sqrt{1-v^2}{l_0} = (1-\frac 12 v^2 + o(v^2))l_0 \approx l_0$

 

[PeroK]

For $v \approx 0$ I find:

$v\frac {\Delta T} 2 = \frac {l_0} {\gamma} = \sqrt{1-v^2}{l_0} = (1-\frac 12 v^2 + o(v^2))l_0 \approx l_0$

You were supposed to see for yourself that that is a nonsensical answer.

I must be honest, this doesn't seem like a productive use of time. We have 26 posts and you have not posted anything worthwhile. You need to think about how you want to use this forum.

 

[Frostman]

I want to try to understand these exercises where events appear, I have a lot of difficulty in carrying out this type of exercises, once the events have been identified I have problems managing this information.
I am looking for help in this forum to chew these exercises better and find a correct approach to follow.
I'm not only interested in passing the exam, but taking home solid knowledge and skills.
All the more so if I want to start a PhD course in theoretical physics at the end of the year.
Honestly, I don't even like that it's a passive thing, in fact it's something that bothers me a lot.
You asked me to evaluate the case where the velocity is close to zero, what I conclude is that the length seen by $S$ and $S'$ coincides, as it should be.

 

[PeroK]

I want to try to understand these exercises where events appear, I have a lot of difficulty in carrying out this type of exercises, once the events have been identified I have problems managing this information.
I am looking for help in this forum to chew these exercises better and find a correct approach to follow.
I'm not only interested in passing the exam, but taking home solid knowledge and skills.
All the more so if I want to start a PhD course in theoretical physics at the end of the year.
Honestly, I don't even like that it's a passive thing, in fact it's something that bothers me a lot.
You asked me to evaluate the case where the velocity is close to zero, what I conclude is that the length seen by $S$ and $S'$ coincides, as it should be.

Well, okay, but you are misunderstanding basic kinematics to the point where the only way you get the answer is for me to do all the calculations with no contribution from you at all.

I'll put a note on the advisor's forum to see whether someone else can help, as I'm out of ideas.

 

[Frostman]

I try to follow the approach to the problem a bit, assuming that obviously I don't want the solution.
You told me not to consider the rocket's primate quantities. So all $\Delta T'$, $T'$, ...
What I'm trying to find is the speed of the rocket.
What do I have available?

- 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:

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

I have to find $l$. The interval between the two signals reflections is $\frac{\Delta T} 2$.

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

 

[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.