Spaceship is approaching the Earth with an unknown speed

[Frostman]

Homework Statement

A spaceship that is approaching the earth with an unknown speed, sends messages by communicating the time (of the spaceship) that is missing upon arrival, starting from the moment the message is sent.
In the first message, the missing time is $T_1'$, in the second message it is $T_2'$, and between the arrival of the two messages, a time interval $\Delta T$ elapses on earth.
1. What is the speed of the spaceship?
2. How long after (terrestrial), after the arrival of the second signal, will the starship arrive?

Relevant Equations

Lorentz Transformations

I started by finding the main events:

1. Sending the first message
2. Receipt the first message
3. Sending the second message
4. Receipt the second message

Now, what we know is the time by $S'$ (comoving frame with the spaceship) $T_1'$ and $T_2'$ remaining to arrive to the Earth measured at different times.
$\Delta T$ is the difference between $T_2$ and $T_1$, measured on the earth.
We can do this table

Events​	$S'$​	$S$​
A: Sending 1	$t_A'=0$ $x_A'=(c-v)T_1'$	$t_A=\gamma(t_A'+vx_A')=\gamma v(c-v)T_1'$ $x_A=\gamma(x_A'+vT_A')=\gamma (c-v)T_1'$
B: Receipt 1		$t_B = t_D-\Delta T$
C: Sending 2	$t_C'=T_2'-T_1'$ $x_C'=(c-v)T_2'$	$t_C=\gamma(t_C'+vx_C')=\gamma(T_2'-T_1'+v(c-v)T_2')$ $x_C=\gamma(x_C'+vt_C')=\gamma((c-v)T_2'+v(T_2'-T_1'))$
D: Receipt 2		$t_D = \Delta T + t_B$

I don't know if is it correct this scheme, and it could help me to find the speed and how soon will the spaceship arrive on earth.

 

[anuttarasammyak]

I cannot take meaning of "the time (of the spaceship) that is missing upon arrival, starting from the moment the message is sent. ". May I take $T'$s as reading of spaceship clock when messages are sent, i.e. time of event A or C in the spaceship FR?

If I may take so
$$v=\frac{X_2-X_1}{T_2-T_1}$$
$$T_2-T_1=\gamma(T_2'-T_1')$$
$$\triangle T + \frac{X_2-X_1}{c}=T_2-T_1$$
These equation give
$$\frac{v}{c}=\frac{1-Q}{1+Q}$$ where
$$Q=(\frac{\triangle T}{T_2'-T_1'})^2$$.

If values of $(T_1,X_1)$ or $(T_2,X_2)$ are given we get the arrival time.

I should appreciate it if you check these with your results.

 

[PeroK]

Homework Statement:: A spaceship that is approaching the Earth with an unknown speed, sends messages by communicating the time (of the spaceship) that is missing upon arrival, starting from the moment the message is sent.

By missing time, do you mean the estimated time until arrival at Earth? The spaceship must know how far away the Earth is and how fast it is moving towards Earth. So, the spaceship can calculate its ETA (estimated time of arrival) on Earth. I assume that $T_1$ and $T_2$ are the ship's estimated times of arrival at Earth.

 

[Frostman]

By missing time, do you mean the estimated time until arrival at Earth?

Yes. It is the time to arrive on earth.

The spaceship must know how far away the Earth is.

I suppose he does, he knows the time that is missing, so also how far away he is.

How fast it is moving towards Earth.

It is the first request.

So, the spaceship can calculate its ETA (estimated time of arrival) on Earth. I assume that T1$T_1$ and $T_2$ T2 are the ship's estimated times of arrival at Earth.

No, $T_1'$ and $T_2'$, they are ship's estimated times, they are sent to Earth these values.

 

[PeroK]

Yes. It is the time to arrive on earth.

I suppose he does, he knows the time that is missing, so also how far away he is.

It is the first request.No, $T_1'$ and $T_2'$, they are ship's estimated times, they are sent to Earth these values.

There is a language problem here. Let's try an example to make sure we agree.

The ship calculates it is 10 days from Earth ($T'_1 = 10$ days). It sends this message to Earth.

The next day the ship sends a message $T'_2 = 9$ days.

These messages are received on Earth at some times $t_3, t_4$, say. Then $\Delta T = t_4 - t_3$ is the time difference on Earth between receiving the messages..

We then have to calculate $v$ based on $\Delta T' = T_1 - T_2$ and $\Delta T = t_4 - t_3$.

This looks quite tricky.

One idea is to run through an example problem to see what happens. Let $v = 0.6c$, for example. Just to see when these messages arrive on Earth. This will also give you a test for the general answer.

That's my suggestion: do the problem with $v = 0.6c$, $T_1 = 8$ units, $T_2 = 6$ units.

Hint: use the ship frame and have events E_1 and E_2 be the sending of the messages, Events E_3 and E_4 be the messages received on Earth. And E_5 the ship arriving on Earth.

 

[Frostman]

We then have to calculate $v$ based on $\Delta T' = T_1 - T_2$ and $\Delta T = t_4 - t_3$.

Are $T_1$ and $T_2$ prime in $\Delta T'$? If we consider spaceship as $S'$ frame.
.

One idea is to run through an example problem to see what happens. Let $v = 0.6c$, for example. Just to see when these messages arrive on Earth. This will also give you a test for the general answer.

That's my suggestion: do the problem with $v = 0.6c$, $T_1 = 8$ units, $T_2 = 6$ units.

Hint: use the ship frame and have events E_1 and E_2 be the sending of the messages, Events E_3 and E_4 be the messages received on Earth. And E_5 the ship arriving on Earth.

Supposing that $T_1 = 8$ units, $T_2 = 6$ are $T_1' = 8$ units, $T_2' = 6$ (spaceship), we have:

$Event\ 1:$
$T_1' = 8$
$L_1' = vT_1'=4.8$ (distance from Earth for spaceship)

$Event\ 2:$
$T_2' = 6$
$L_2' = vT_2'=3.6$ (distance from Earth for spaceship)

$Event\ 3:$
$L_3' = vt_3'$

$Event\ 4:$
$L_4' = vt_4'$

Making the difference we get:
$\Delta L_{43} = v\Delta T$

Now if I make a boost for the $Events\ 1-2$ in $S$ (earth) we get:
$T_1=\gamma (T_1'+vL_1')=(\gamma+v^2)T_1'$
$T_2=\gamma (T_2'+vL_2')=(\gamma+v^2)T_2'$

These are the times that Earth ($S$) saw when spaceship ($S'$) sent their messages.

But I'm getting confusing now. I don't know how to get $T_3$ and $T_4$, I only know that their difference is $\Delta T$. How can I get this, my datas are $T_1'$, $T_2'$ and $\Delta T$.

 

[anuttarasammyak]

By clarification in post #4 I would revise post #2

Please find attached space time diagrams of Spaceship FR and Earth FR.

 

[PeroK]

You need to think. $T'_1$ and $T'_2$ are not times of events in the spaceship frame. If the time of event $E_1$ is $t'_1 = 0$, then the time of event $E_2$ is $t'_2 = T'_1 - T'_2$.

Note: personally I would not have put $'$ on $T'_1$ as this is data, not coordinates.

Second, if you can analyse a problem in one frame, why change frames? There's no reason to Transform to the Earth frame. You can use the ship frame and treat the Earth as a moving clock in that frame.

Let me help you get started with the specific problem. You really need to do this first, or you will get hopelessly lost with the general problem. I'll take $c = 1$:

$v = 0.6$, $\gamma = \frac 5 4$, $T'_1 = 8$, $T'_2 = 6$, $L'_1 = 4.8$, $L'_2 = 3.6$.

Where $L'_1$ is the distance to the Earth when the first message is send etc.

$t'_1 = 0$, $t'_2 = 2$

These are the times that the messages are sent in the spaceship frame.

$t'_3 = \frac{d'_1}{1+v} = \frac{4.8}{1.6} = 3$

$t'_4 = t'_2 + \frac{d'_2}{1+v} = 2 + \frac{3.6}{1.6} = \frac{17}{4}$

$\Delta t' = t'_4 - t'_3 = \frac 5 4$

That's all in the spaceship frame. And, note that we haven't actually used any SR yet - except that the speed of light is $c = 1$. That would be the same calculation in classical mechanics!

Now, treating the Earth as a moving clock, we have simply:
$$\Delta \tau = \tau_4 - \tau_3 = \frac{\Delta t'}{\gamma} = 1$$
Can you do the last bit and calculate the remaining time on the Earth clock ($\tau_R$) until the Earth reaches the spaceship?

 

[anuttarasammyak]

You shouldn't post the full answer!

Sorry, I deleted the lines. If it is not enough delete all the message please.

 

[Frostman]

You need to think. $T'_1$ and $T'_2$ are not times of events in the spaceship frame. If the time of event $E_1$ is $t'_1 = 0$, then the time of event $E_2$ is $t'_2 = T'_1 - T'_2$.

Note: personally I would not have put $'$ on $T'_1$ as this is data, not coordinates.

So it was what I'm saying in the first post.
$E_1$ is $t'_1 = 0$
$E_2$ is $t'_2 = T'_1 - T'_2$

They are datas, not coordinates, ok.

Second, if you can analyse a problem in one frame, why change frames? There's no reason to Transform to the Earth frame. You can use the ship frame and treat the Earth as a moving clock in that frame.

Let me help you get started with the specific problem. You really need to do this first, or you will get hopelessly lost with the general problem. I'll take $c = 1$:

$v = 0.6$, $\gamma = \frac 5 4$, $T'_1 = 8$, $T'_2 = 6$, $L'_1 = 4.8$, $L'_2 = 3.6$.

Where $L'_1$ is the distance to the Earth when the first message is send etc.

$t'_1 = 0$, $t'_2 = 2$

These are the times that the messages are sent in the spaceship frame.

I agree.

$t'_3 = \frac{d'_1}{1+v} = \frac{4.8}{1.6} = 3$

$t'_4 = t'_2 + \frac{d'_2}{1+v} = 2 + \frac{3.6}{1.6} = \frac{17}{4}$
$\Delta t' = t'_4 - t'_3 = \frac 5 4$

That's all in the spaceship frame. And, note that we haven't actually used any SR yet - except that the speed of light is $c = 1$. That would be the same calculation in classical mechanics!

Are these the times, always in spaceship, when Earth receipt the two messages?
Doesn't the speed you used in $t'_3$ and $t'_4$ violate the postulate that the limiting speed is $c$ ($c=1$)?

Now, treating the Earth as a moving clock, we have simply:
$$\Delta \tau = \tau_4 - \tau_3 = \frac{\Delta t'}{\gamma} = 1$$

Did you call $\tau$ because the receiving events is happening in the same place? If yes, so it is a proper time.
What spaceship see is a time dilation.

Can you do the last bit and calculate the remaining time on the Earth clock ($\tau_R$) until the Earth reaches the spaceship?

$\tau_R' = T_2'-t_4'=1.75$​

$T_2'$ as we know is the remaining time on the spaceship clock until the Earth reaches the spacehip.
$t_4'$ is the time on the spaceship clock when Earth received the second messages.
This is the remaining time on the spaceship clock ($\tau_R'$) until the Earth reaches the spacehip when Earth received the second message.
By applying the time dilation we get:

$\tau_R = \frac{T_2'-t_4'}{\gamma}=1.4$
​

I hope this reasoning is correct.

 

[Frostman]

Sorry, I deleted the lines. If it is not enough delete all the message please.

Thank you, but I need to understand deeply.

 

[PeroK]

Are these the times, always in spaceship, when Earth receipt the two messages?
Doesn't the speed you used in $t'_3$ and $t'_4$ violate the postulate that the limiting speed is $c$ ($c=1$)?
Did you call $\tau$ because the receiving events is happening in the same place? If yes, so it is a proper time.
What spaceship see is a time dilation.

Yes, everything can be done in the spaceship frame. Stay in that frame!

It's simple kinematics that if light is moving at $c$ in one direction and the Earth is moving at $v$ in the opposite direction, then the distance between them reduces at a rate of $c + v$. This is in the frame in which the Earth is moving at $v$. Nothing is moving faster than $c$ for that to happen.

The maximum separation speed in SR is, therefore, $2c$.

I used $\tau$ to emphasise that we are staying in the ship frame and treating the Earth as a moving clock. You could use $t$ is you like, but $\tau$ is better as it is the Earth's proper time.

Note: we can do this because events E_3, E_4, E_5 are all colocated on Earth. So, we can use the Earth's proper time at those events.

$\tau_R' = T_2'-t_4'=1.75$

$T_2'$ as we know is the remaining time on the spaceship clock until the Earth reaches the spacehip.
$t_4'$ is the time on the spaceship clock when Earth received the second messages.
This is the remaining time on the spaceship clock ($\tau_R'$) until the Earth reaches the spacehip when Earth received the second message.

You cannot mix up proper time, data and coordinate time in one equation like that! Instead, you want:
$$\tau_R = \tau_5 - \tau_4 = \frac{t'_5 -t'_4}{\gamma}$$

 

[Frostman]

Yes, everything can be done in the spaceship frame. Stay in that frame!

It's simple kinematics that if light is moving at c in one direction and the Earth is moving at v in the opposite direction, then the distance between them reduces at a rate of c+v. This is in the frame in which the Earth is moving at v. Nothing is moving faster than c for that to happen.

The maximum separation speed in SR is, therefore, 2c.

I used τ to emphasise that we are staying in the ship frame and treating the Earth as a moving clock. You could use t is you like, but τ is better as it is the Earth's proper time.

Note: we can do this because events E_3, E_4, E_5 are all colocated on Earth. So, we can use the Earth's proper time at those events.
You cannot mix up proper time, data and coordinate time in one equation like that! Instead, you want:
τR=τ5−τ4=t5′−t4′γ

Ok, so what I have to find is the $t'_5$, the time where spaceship arrives on Earth and it is simply $t'_5=\frac{d_1}{v}=8$

$\tau_R= \tau_5−\tau_4=\frac{t'_5−t'_4}{\gamma}=3$
​

Correct?

 

[PeroK]

Ok, so what I have to find is the $t'_5$, the time where spaceship arrives on Earth and it is simply t5′=d1v=8

$\tau_R= \tau_5−\tau_4=\frac{t'_5−t'_4}{\gamma}=3$
​

Correct?

Yes. Now you have a series of calculations for a given $v = 0.6c$. You must now tackle the problem for any $v$. This time, however, you are aiming to express $v$ in terms of $\Delta T' = T'_2 - T'_1$ and $\Delta T = \Delta \tau = \tau_4 - \tau_3$.

What you've done so far should be very useful. And, when you get an answer you can check it for the specific problem you have done.

 

[Frostman]

Yes. Now you have a series of calculations for a given $v = 0.6c$. You must now tackle the problem for any $v$. This time, however, you are aiming to express $v$ in terms of $\Delta T' = T'_2 - T'_1$ and $\Delta T = \Delta \tau = \tau_4 - \tau_3$.

What you've done so far should be very useful. And, when you get an answer you can check it for the specific problem you have done.

Ok, I can rewrite all in terms of $T'_2$, $T'_1$ and $\Delta T$:

$t_1'=0$
$t_2'=T_1'-T_2'$
$t_3'=\frac{vT_1'}{1+v}$
$t_4'=T_1'-T_2'+\frac{vT_2'}{1+v}$

$\Delta t' = t_4'-t_3' = T_1'-T_2'+\frac{vT_2'}{1+v} - \frac{vT_1'}{1+v}=T_1'-T_2'+\frac{v}{1+v}(T_2'-T_1')=(\frac{v}{1+v}-1)(T_2'-T_1')$

Now remembering that

$\Delta T = \frac{\Delta t'}{\gamma}=(\frac{v}{1+v}-1)(T_2'-T_1')\sqrt{1-v^2}$

$\Delta T = (\frac{v}{1+v}-1)(T_2'-T_1')\sqrt{1-v^2}$

$\frac{\Delta T}{(T_2'-T_1')}=(\frac{v}{1+v}-1)\sqrt{1-v^2}$

$(\frac{\Delta T}{(T_2'-T_1')})^2=\frac{1-v^2}{(1+v)^2}$

Naming:

$Q=(\frac{\Delta T}{(T_2'-T_1')})^2$

We have:

$\frac{1-v^2}{1+v^2+2v} = Q$

$1-v^2=Q+Qv^2+2Qv$

$v_{1,2}=\frac{-Q\pm\sqrt{Q^2-(1+Q)(Q-1)}}{1+Q}$

$v_{1,2}=\frac{-Q\pm 1}{1+Q}$

$v_1 = \frac{-Q - 1}{1+Q}=-1$
$v_2 = \frac{-Q + 1}{1+Q}=\frac{1-Q}{1+Q}=\frac{(T_2'-T_1')^2-\Delta T^2}{(T_2'-T_1')^2+\Delta T^2}$

Why do I have two speeds? Is it all correct?

 

[PeroK]

That all looks fine. $v = -1$ is clearly a spurious solution from solving a quadratic. We have $v > 0$.

What about the remaining time until the ship reaches Earth?

 

[anuttarasammyak]

Why do I have two speeds? Is it all correct?

1+v is canceled out both from denominator and numerator of the equation so
$$\frac{1-v}{1+v}=Q$$

 

[Frostman]

That all looks fine. $v =1$ is clearly a spurious solution from solving a quadratic. We have $v > 0$.

What about the remaining time until the ship reaches Earth?

$\tau_R=\frac{t_5'-t_4'}{\gamma}={(T_1'-T_1'+T_2'-\frac{vT_2'}{1+v})}{\sqrt{1-v^2}}$
$\tau_R=(T_2'-\frac{vT_2'}{1+v}){\sqrt{1-v^2}}$
$\tau_R=T_2'\frac{\sqrt{1-v}}{\sqrt{1+v}}$
$\tau_R=T_2'\frac{\Delta T^2}{(T_2'-T_1')^2}$

Is it ok?

 

[PeroK]

$\tau_R=\frac{t_5'-t_4'}{\gamma}={(T_1'-T_1'+T_2'-\frac{vT_2'}{1+v})}{\sqrt{1-v^2}}$
$\tau_R=(T_2'-\frac{vT_2'}{1+v}){\sqrt{1-v^2}}$
$\tau_R=T_2'\frac{\sqrt{1-v}}{\sqrt{1+v}}$
$\tau_R=T_2'\frac{\Delta T^2}{(T_2'-T_1')^2}$

Is it ok?

You have one mistake there. The last formula shouldn't have squares. Also, note that $T'_1 > T'_2$.

And what happens if you try $T'_1 - T'_2 = 2$ and $\Delta T = 1$? Do you get $v = 0.6$ and $\tau_R = 3$?

 

[Frostman]

1+v is canceled out both from denominator and numerator of the equation so
1−v1+v=Q

Yes, thank you.

$Q=\frac{\Delta T^2}{(T_2'-T_1')^2}$

$\frac{\Delta T^2}{(T_2'-T_1')^2}=\frac{1-v}{1+v}$

$v=\frac{(T_2'-T_1')^2-\Delta T^2}{(T_2'-T_1')^2+\Delta T^2}$

Can you confirm?

 

[anuttarasammyak]

Can you confirm?

That is fine.

 

[Frostman]

You have one mistake there. The last formula shouldn't have squares. Also, note that $T'_1 > T'_2$.

And what happens if you try $T'_1 - T'_2 = 2$ and $\Delta T = 1$? Do you get $v = 0.6$ and $\tau_R = 3$?

I get $\tau_R = -3$, because in denominator I have $-2$ and numerator is $6$. Have I forgotten any signs?

 

[PeroK]

I get $\tau_R = -3$, because in denominator I have $-2$ and numerator is $6$. Have I forgotten any signs?

You used $T'_2 - T'_1$, which (as I pointed out) is negative. And this became a problem when you took the square root.

You need to be more careful. Note that also, you used $\frac{v}{1+v} - 1$. This simplies to: $-\frac{1}{1+v}$, which is also negative.

You need to start noticing these things at this level of physics.

 

[PeroK]

$\Delta t' = t_4'-t_3' = T_1'-T_2'+\frac{vT_2'}{1+v} - \frac{vT_1'}{1+v}=T_1'-T_2'+\frac{v}{1+v}(T_2'-T_1')=(\frac{v}{1+v}-1)(T_2'-T_1')$

For example. That is not wrong. But, you should notice that:
$$(\frac{v}{1+v}-1)(T_2'-T_1') = \frac{T'_1 - T'_2}{1+v}$$
Which is better, in my opinion.

 

[Frostman]

Okay, I understand. I'll make all the counts on my book and I'll repost the result, now with too many posts I lose the my thread.

 

[Frostman]

It's just a sign as you pointed out.
The results are:

$v=\frac{(T_1'-T_2')^2-\Delta T^2}{(T_1'-T_2')^2+\Delta T^2}$

$\Delta \tau_R = T_2'\frac{\Delta T}{T_1' - T_2'}$
​

Thank you PeroK and anuttarasammyak. I always make a mistake in setting the problem and I take absurd paths when the easiest way is not there. I think I have to do more and more practice.