Special relativity and Lorentz Transformation Exercise

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SUMMARY

The discussion focuses on solving a problem related to special relativity and Lorentz transformations, specifically involving a rocket moving in the S' frame. The rocket starts at the origin and travels with speed u, turning around after 1 year in S', with observers in S noting its turnaround at x = -1/5 light-year and return at x = 4 light-years. Key calculations include determining v/c and u/c, with the correct answers being 0.894c and 0.984c respectively, and the time taken for the rocket to return being 4.47 years.

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  • Understanding of Lorentz transformations in special relativity
  • Familiarity with the concepts of time dilation and length contraction
  • Basic knowledge of kinematics and reference frames
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  • #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.
 
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  • #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?

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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.
 
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  • #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:$$x_B = u_At_A + u_B(t_B-t_A) = 4$$
 
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  • #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.
 
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  • #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.
 

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