Special relativity and Lorentz Transformation Exercise

AI Thread Summary
The discussion focuses on a special relativity problem involving a rocket's journey as analyzed through Lorentz transformations. Participants clarify the definitions of the stationary frame (S) and the moving frame (S'), emphasizing the importance of correctly identifying the rest frame of the rocket. They suggest using non-relativistic mechanics initially to build intuition before applying the Lorentz transformations for accurate calculations. The conversation highlights the complexity of the problem, with users sharing strategies for organizing information and deriving equations to find the speeds and return time of the rocket. Overall, the thread provides insights into tackling relativistic motion problems while encouraging a methodical approach to problem-solving.
R3ap3r42
Messages
30
Reaction score
3
Misplaced Homework Thread -- Moved from the Technical Forums
Summary:: Special relativity and Lorentz Transformations - I got this problem from a first-semester course at university. I have been struggling for a few days and decided to get some help.

A rocket sets out from x = x' = 0 at t = t' = 0 and moves with speed u in the negative x'-direction, as measured in S'.
After 1 year has elapsed in S', the rocket turns around.
According to observers in S, the rocket turns around at x = −1/5 light-year and it returns to the origin in S' at x = 4 light-years.

i) Calculate v/c.

ii) Calculate u/c.

iii) At what time in S does the rocket return to the origin in S'?
 
Physics news on Phys.org
Please show your working. We can help you with your homework but we're not going to do it for you. You also need to state what are S and v, since you haven't defined those.
 
Hi there,
This is not exactly homework, it is just a hobby for me.
I am a software engineer who enjoys doing physics for fun...

I got stuck pretty much at the start. I tried a few things but they all seem to get to a dead-end.

1642692812909.png

S is the stationary frame and S' is the one moving at v in relation to S.
 
Last edited:
R3ap3r42 said:
I got stuck pretty much at the start. I tried a few things but they all seem to get to a dead-end.
Welcome to PF. All schoolwork-type questions go in the Homework Help forums, per the PF rules.

I tried to enhance your picture, but no luck so far. Could you please take a better picture or scan your work and try again? Thank you.
 
Done. Let me know if it is still not clear.
 
R3ap3r42 said:
Summary:: Special relativity and Lorentz Transformations - I got this problem from a first-semester course at university. I have been struggling for a few days and decided to get some help.

A rocket sets out from x = x' = 0 at t = t' = 0 and moves with speed u in the negative x'-direction, as measured in S'.
After 1 year has elapsed in S', the rocket turns around.
According to observers in S, the rocket turns around at x = −1/5 light-year and it returns to the origin in S' at x = 4 light-years.

i) Calculate v/c.

ii) Calculate u/c.

iii) At what time in S does the rocket return to the origin in S'?
I suggest doing this problem using non-relativistic mechanics first. Then, if you are comfortable doing that, try the relativistic version.
 
Are you sure about those S and S primes? Which one is the rest frame of the rocket?
 
PeroK said:
I suggest doing this problem using non-relativistic mechanics first. Then, if you are comfortable doing that, try the relativistic version.
PS I had a look at this. It's quite tricky. I did it non-relativistically first to get a grip on what was happening kinematically. That's important, I think, to really understand what's going on before you start.

The main solution used the Lorentz Transformations. The trick is choosing which transformation to use first. If you write them all down, then one is simpler than the others.

I also calculated the relativistic speeds of the rocket in the S frame and double-checked the answer that way.
 
Chestermiller said:
Are you sure about those S and S primes? Which way me is the rest frame of the rocket?
##u## in the problem is the speed of the rocket (both ways) in the S' frame. Perhaps it should have been ##u'##. The speed of the rocket in the S frame is different before and after the turnaround.
 
  • #10
Chestermiller said:
Are you sure about those S and S primes? Which one is the rest frame of the rocket?
Yes, this is an exact extract from Manchester University First Semester Physics for the 2019 exam.
 
  • #11
R3ap3r42 said:
Yes, this is an exact extract from Manchester University First Semester Physics for the 2019 exam.
As above, try writing down all four Lorentz Transformations. I.e. we have two events: ##E_1## is the turnaround and ##E_2## is when the rocket gets back to the ##S'## spatial origin. We have the LT and inverse LT of both of these. I wrote them all down first.

Not the easiest problem I've ever seen.
 
  • #12
I'm still struggling to read that.

There's a fairly simple recipe for solving problems of this nature.
  1. Do not use length contraction or time dilation. They are special case formulae and easy to misapply where they are inappropriate.
  2. Write down the coordinates of any events of interest in either frame - typically either ##(x,t)## or ##(x',t')##, but actually any two of the four will do.
  3. Apply the Lorentz transforms to get the coordinates in the frame you don't have them.
  4. Decide which frame you want to be working into get the answer requested. Use coordinate differences in that frame
Looking at the problem statement, it looks like you don't actually have all of the coordinates you'll need - a lot of them are implied by things like the fact that ##x=4## when the rocket reaches ##x'=0##. So the first step, I think, is to think about writing down the coordinates of events you do know and seeing how you can go about filling in the blanks. A table like this might help:$$\begin{array}{|l|c|c|c|c|}
\hline
\mathrm{Event}&x / \mathrm{ly}&t/ \mathrm{years}&x'/ \mathrm{ly}&t' / \mathrm{years}\\
\hline
\mathrm{Rocket\ leaves}&0&0&0&0\\
\mathrm{Rocket\ turns\ around}&-1/5&&&1\\
\mathrm{Rocket\ returns}&4&&0&\\
\hline
\end{array}$$
 
Last edited:
  • Informative
Likes berkeman
  • #13
PeroK said:
##u## in the problem is the speed of the rocket (both ways) in the S' frame. Perhaps it should have been ##u'##. The speed of the rocket in the S frame is different before and after the turnaround.

The rocket moves at -u and then u in the S' frame we need to the expression for the speed of the rocket in the S frame is trivial w1 = (v-u)/(1-vu/c^2) and w2 = (v+u)/(1+vu/c^2)
I think we need a system of equation to get rid of u first so we can get v/c
I am getting stuck on which equations to use. My attempts resulted in the wrong answer multiple times.For the S frame, the rocket traveled at speed w1 on the first leg of the journey for a time t. So the first equation I assumed was:

w1*t = 1/5*c

but t = γ*t' and t'=1 so

w1*γ = 1/5*c

This gives me the wrong result for v/c.
 
  • #14
R3ap3r42 said:
The rocket moves at -u and then u in the S' frame we need to the expression for the speed of the rocket in the S frame is trivial w1 = (v-u)/(1-vu/c^2) and w2 = (v+u)/(1+vu/c^2)
I think we need a system of equation to get rid of u first so we can get v/c
I am getting stuck on which equations to use. My attempts resulted in the wrong answer multiple times.
It should be possible this way, but it looked messy, so I used the LT's instead.
R3ap3r42 said:
For the S frame, the rocket traveled at speed w1 on the first leg of the journey for a time t. So the first equation I assumed was:

w1*t = 1/5*c

but t = γ*t' and t'=1 so

w1*γ = 1/5*c

This gives me the wrong result for v/c.
This is wrong. Time dilation is a special case where events are colocated in the relevant frame. You can't apply time dilation across events that have different spatial coordinates in the relevant frame, You must use the full LT.
 
  • #15
Can you give me a hint?

The correct answers are:
a) 0.894c
b) 0.984c
c) 4.47 years
 
  • #16
Here's a hint and something that doesn't seem to be in textbooks. If you generate the quantity ##\gamma v##, which happens quite often. Note that:
$$\gamma v = w \ \Rightarrow \ \frac{v^2}{1 - v^2} = w^2 \ \Rightarrow \ v^2 =\frac{w^2}{1 + w^2}$$$$\Rightarrow \ v = \frac{w}{\sqrt{1 + w^2}} \ \text{and} \ \gamma = \sqrt{1 + w^2}$$That's how you extract ##v## and ##\gamma## from ##\gamma v##. Note that I've used ##c=1##.
 
  • #17
R3ap3r42 said:
Can you give me a hint?

The correct answers are:
a) 0.894c
b) 0.984c
c) 4.47 years
That's what I get. The hint is to write down the inverse Lorentz transformations:
$$E_1 = (t_1, x_1) = \dots, \ E_2 = (t_2, x_2) = \dots$$
And, you'll need the two events in the S' frame:
$$E'_1 = (t'_1, x'_1) = \dots, \ E'_2 = (t'_2, x'_2) = \dots$$Where these can be specified as much as possible.
 
  • #18
I may be missing something... that is what I see from the post...
1642703236186.png


Do I need any extension on my browser?
 
  • #19
Nevermind I was using Brave it works on Chrome.
 
  • Like
Likes PeroK
  • #20
R3ap3r42 said:
Nevermind I was using Brave it works on Chrome.
MathJax is a Javascript package, and quite a computationally hefty one, so it's lazily loaded. Sometimes it's a bit too lazy and you need to refresh the page to get it to realize it needs to do something.
 
  • #21
Ibix said:
I'm still struggling to read that.

There's a fairly simple recipe for solving problems of this nature.
  1. Do not use length contraction or time dilation. They are special case formulae and easy to misapply where they are inappropriate.
  2. Write down the coordinates of any events of interest in either frame - typically either ##(x,t)## or ##(x',t')##, but actually any two of the four will do.
  3. Apply the Lorentz transforms to get the coordinates in the frame you don't have them.
  4. Decide which frame you want to be working into get the answer requested. Use coordinate differences in that frame
Looking at the problem statement, it looks like you don't actually have all of the coordinates you'll need - a lot of them are implied by things like the fact that ##x=4## when the rocket reaches ##x'=0##. So the first step, I think, is to think about writing down the coordinates of events you do know and seeing how you can go about filling in the blanks. A table like this might help:$$\begin{array}{|l|c|c|c|c|}
\hline
\mathrm{Event}&x / \mathrm{ly}&t/ \mathrm{years}&x'/ \mathrm{ly}&t' / \mathrm{years}\\
\hline
\mathrm{Rocket\ leaves}&0&0&0&0\\
\mathrm{Rocket\ turns\ around}&-1/5&&&1\\
\mathrm{Rocket\ returns}&4&&0&\\
\hline
\end{array}$$
That is my understanding but I fail to generate the needed equations from that.
 
  • #22
R3ap3r42 said:
That is my understanding but I fail to generate the needed equations from that.
You need to use the Lorentz Transformations.
 
  • #23
R3ap3r42 said:
That is my understanding but I fail to generate the needed equations from that.
You should be able to complete the table in terms of ##v##. You'll then have to think about how you could get some expressions for ##u## (hint: think about what observers using frame S say about the rocket's motion).
 
  • #24
Chestermiller said:
Are you sure about those S and S primes? Which one is the rest frame of the rocket?
Neither S nor S' is the rest frame of the rocket.
 
  • #25
I am still struggling so I will post what I have so far...

1642707967694.png


Are these correct?
 
  • #26
vela said:
Neither S nor S' is the rest frame of the rocket.
The question is as posted in the exam. Word for word.
 
  • #27
R3ap3r42 said:
I am still struggling so I will post what I have so far...

View attachment 295780

Are these correct?
You should be able to write down ##t'_2##.

I don't think you'll get a solution very easily using velocity transformations. Lorentz transformations are the way to go. I might have said that already.
 
  • #28
Ibix said:
A table like this might help:$$\begin{array}{|l|c|c|c|c|}
\hline
\mathrm{Event}&x / \mathrm{ly}&t/ \mathrm{years}&x'/ \mathrm{ly}&t' / \mathrm{years}\\
\hline
\mathrm{Rocket\ leaves}&0&0&0&0\\
\mathrm{Rocket\ turns\ around}&-1/5&&&1\\
\mathrm{Rocket\ returns}&4&&0&\\
\hline
\end{array}$$
I find spacetime diagrams really helpful as well to keep things straight.

sr.png


event O = rocket leaves
event A = rocket turns around
event B = rocket returns

As @PeroK noted, it should be pretty clear what ##t'_B## (PeroK's ##t'_2##) is equal to from the diagram on the left.
 
  • Like
Likes BvU and R3ap3r42
  • #29
Spacetime diagrams are an excellent tool.

@R3ap3r42 - the point with the Lorentz transforms is that you can use them to complete each row of the table in terms of ##v## because in each case you know two things, ##x## and either ##x'## or ##t'##. The Lorentz transforms are two equations, and there are two things you want to know..

Only at that point do you bring in the fact you've already observed, that ##x'## at the turn around is ##u\times 1##. You'll also need the value of ##t'## at the return, which you should be able to deduce). Then you can solve for ##v## then ##u##.
 
  • Like
Likes R3ap3r42
  • #30
OK, this helps a little more.
I was happy for a minute there until I got again to something wrong...
Where did I mess it up this time?
1642753511118.png
 
  • #31
Did you get that ##t'_B =2## years? Note that that is also true in classical mechanics.

I did suggest writing down all of the LTs and finding the simplest one. That was:$$x_B = \gamma(0 +vt'_B)$$That gives you a value for the quantity ##\gamma v##. I explained in a previous post how to deal with that.
 
  • Like
Likes R3ap3r42
  • #32
I finally got it.
a) and c) were straightforward and match the expected result.
For b) I got 2 different values depending on if I use O to A or A to B to get u.
Is this correct? If so the question should have specified which direction right?

1642759326626.png


My sincere thanks to @PeroK, @Ibix, @vela etc. I really appreciate the tough-love. I can honestly say I have learned loads just with this one question.
 
  • Like
Likes PeroK and BvU
  • #33
Here's my minimialist solution (natural units):
$$x_B = \gamma(0 + vt'_B) = 2\gamma v \ \Rightarrow \ \gamma = \sqrt 5, v = \frac 2 5 \sqrt 5$$$$t_B = \gamma(t'_B) = 2\gamma = 2\sqrt 5$$$$x_A = \gamma(x'_A + vt'_A) \ \Rightarrow x'_A = -\frac{11}{25}\sqrt 5$$$$x'_A = -ut'_A \ \Rightarrow \ u = \frac{11}{25}\sqrt 5$$To double check, we calculate the speeds of the rocket in the S frame:
$$u_A = \frac{-u + v}{1-uv} = -\frac{1}{3}\sqrt 5 \ \Rightarrow \ t_A = \frac{x_A}{u_A} = \frac{3}{25}\sqrt 5$$$$u_B = \frac{u + v}{1+uv} = \frac{21}{47}\sqrt 5$$And we can confirm that:$$x_B = u_At_A + u_B(t_B-t_A) = 4$$
 
Last edited:
  • Like
Likes Ibix and R3ap3r42
  • #34
R3ap3r42 said:
For b) I got 2 different values depending on if I use O to A or A to B to get u.
Is this correct? If so the question should have specified which direction right?
You should have gotten the same speed for both directions. Remember that ##u## is the speed of the rocket as seen by an observer at rest in ##S'##. The speed you calculated using OA, for instance, is the speed of the rocket an observer at rest in ##S## would see.
 
  • Like
Likes R3ap3r42
  • #35
Yes, I understand that now. I got the values described by @PeroK, which are the velocities S sees the rocket coming and going.
The question asked u in the S' frame so only one value for both legs of the trip.
Thanks a lot for all the help.
 
  • #36
I'd just want to echo something @vela said upthread: spacetime diagrams are incredibly useful for visualising these problems. They are essentially displacement-time graphs (if you've come across those) with the time axis up the page. You can draw one for each frame and mark events and worldlines on them and they are a great tool for seeing roughly what the solution must be before you start plugging in numbers.
 
Back
Top