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I had some doubt solving the following problem regarding :

There is a rocket travelling away in space, and an asteroid approaching it with a speed 0.24c relative to the rocket. The distance is 14.4 light minutes as measured from the rocket. How much time does the rocket have to escape the trajectory ?

This part is extremely easy, and the answer is 60 min which i got correct.

The next part of the question is, how much time passed on a clock on the asteroid.

Case 1 : Considering the Rocket in rest S and asteroid S': At first I considered the rocket to be at rest and asteroid moving right to left. So velocity was negative and so was delta x ( because of change in direction ).Thus I got the time on asteroid to be 58.25 min. Case 1

Case 2 : Considering the Asteroid in rest S and Rocket S' : This is what I did. Assuming, here the asteroid was at rest, and rocket moving from right to left. In this case the time was known i.e. 58.25 min. I wanted to get back the time on the rocket. I got 56 min instead of my original 60 min. Case 2 Correction: in the list line I wrote speed instead of time.

The time spent in the rocket is not matching in case 1 and case 2, and I'm inclined to believe, this is because of the relativity of simultaneity. But shouldn't lorentz transformation, automatically take care of that ? Based on this problem, I think there is a large fallacy in my understanding of lorentz transformations. Some people have pointed out that case 2 is nothing but the inverse lorentz transformation of case 1. However, I'm failing to understand the fallacy here. Why can I not do Case 1 and Case 2 separately and get the same result independently ?

Thanking you.

EDIT : Is it my definition of Events and their locations in Case 2, that is causing the problem ? If I put del x = 0, then the problem gets resolved again.

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  • $\begingroup$ I may be confused here, but since we are only given a closing speed wouldn't symmetry give the same time to close for both rocket FOR and asteroid FOR? $\endgroup$
    – Adrian Howard
    Commented May 22, 2021 at 12:01
  • $\begingroup$ If you show the your lorentz calculations in detail, then people will be able to point out, where you made any mistakes $\endgroup$
    – silverrahul
    Commented May 22, 2021 at 12:47
  • $\begingroup$ I've added my lorentz transformations, and you please check it ? Thank you. $\endgroup$
    – Nakshatra Gangopadhay
    Commented May 22, 2021 at 13:29
  • $\begingroup$ I think in the case of with respect to asteroid frame, the time interval $58.25min$ is proper time interval. If we are sitting on particle, then everything moves but particle remains at rest. So I think $\Delta x=0$. Thus , we get $\Delta t'=\gamma\Delta t=60min\; (approx)$. Then we get consistent result. $\endgroup$
    – Iti
    Commented May 22, 2021 at 14:26
  • $\begingroup$ I think I found the problem. In the first case, I considered rocket to be S and asteroid to be S'. If I wanted to make the asteroid at rest and rocket to be moving, I could just take the inverse lorentz transformation, which you've said. $\endgroup$
    – Nakshatra Gangopadhay
    Commented May 22, 2021 at 15:01

1 Answer 1

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Yes, your problem is that you have not properly taken into account the relativity of simultaneity. The rocket calculates the elapsed time for the asteroid by assuming that the time t=0 both on the horizon and at the rocket- in other words, t=0 is the baseline time for the calculation everywhere in the S frame of reference. However, in the S' frame, the plane of simultaneity is tilted compared with the S frame. So if it is t'=0 at the horizon where the rocket is located at t=0, it is not t'=0 at the asteroid. Use the transformations to determine what time it is at the asteroid in S' when the rocket is at t=0, t'=0, and you will see that you need to take a different start time for your calculation.

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  • $\begingroup$ Thank You so much. I did the calculation, and got the starting time. Added that back, and got correct answer. I basically ruined Case 2 with overthinking, instead of applying inverse lorentz transformation, I made up a new case, and accounted only for dilation, instead of simultaneity. In Case 2, 58.25 min should have been proper time instead. Using that I could have easily solved the problem, thank you so much $\endgroup$
    – Nakshatra Gangopadhay
    Commented May 22, 2021 at 15:26
  • $\begingroup$ My pleasure. Almost every time someone appears on this website to be confused about special relativity it is because they have not taken simultaneity properly into account, so I always consider it first when encountering a new problem. $\endgroup$
    – Professor Sushing
    Commented May 22, 2021 at 15:52

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