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Time dilation

  1. Nov 25, 2004 #1
    suppose there are 2 starships, A and B, moving towards each other wuth a relative velocity of 0.999c. they pass each other at time t=t0. after moving a certian distance (relative), they stop(relative to each other), turn around and again statr moving towards each other with the same velocity, of 0.999c. this time they pass each other at time, t=t1.
    (neglecting the time taken for them to stop, turn around, and accelerate to the same rel. velosity.)

    if the time interval, t1-t0, for this entire event as measured by an observer on starship A is equal to 20 units of time, then what will be the time measured by an observer on starship B.
  2. jcsd
  3. Nov 25, 2004 #2


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    20 units of time.
  4. Nov 27, 2004 #3

    with that i think you have dissmised the whole concept of time dilation, and that would surely give einstein a huge setback.
  5. Nov 27, 2004 #4
    well actually this is a pretty serious question, as my physics teacher was also not able to give me a satisfactory answer, so i was hoping if some, in fact most, of you smart people uot there could help me get a satisfactory answer to this.
  6. Nov 27, 2004 #5


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    The answer to your question depends on a piece of information you failed to give. When you say the starships turn around after traveling a certain distance, you never say which frame this distance is measured from. Starship A, Starship B, or some other frame.
  7. Nov 27, 2004 #6
    on a different note, is it true that time actually slows down when you start running as opposed to just sitting down. that is, when looking at a clock while running it appears slower than if you are just sitting down looking at? i read this somewhere, then i tried it on my digital watch, and i think it worked.
  8. Nov 29, 2004 #7
    Dave's answer makes the most sense to me, assuming they're both observing from the same relative point on the oposite starships.

    I know that time slows down, but it would slow down on both the ships at the same rate. Thus, the time measured on one ship would equal the time measured on the other.
  9. Nov 29, 2004 #8
    the distance is also relative. and according to my physics proffessor(although i was not satisfied by his answer) we dont need to define a third frame of reference.

    as per your question, i had made some calculations, and if we take the time on A as proper time, then the time on B will be around 448 units of ttime, but if we consider the time on B as proper time, then the time on B would come out as (i dont remember the exact value) 0.82 or something of that order.

    if we consider a third frame of reference, which is stationary with respect to the point in space where they pass each other, both the times, then the time measured by both the shi9ps would come out to be same.

    but time dilation depends on the relative velocity of the two bodies, so if we are given the relative velocity, then do we actually need to define , which frame of reference is stationary, or if there is a third frame of reference, with respect to which we are measuring time dilation.
  10. Dec 6, 2004 #9


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    I... I... :yuck:

    Alright, problems with your experiment:
    You are in the same frame as your watch and would notice no 'watch 'time dilation
    You're travelling at a very low fraction of c, the time dilation is almost unobservable at this speed (ESPECIALLY to the human eye)
    You running is still an inertial frame and is not any different than you not running as far as your watch is concerned.
  11. Dec 6, 2004 #10
    Of course you do – It’s not that you need to give Janus a distance
    in a frame of reference. Or even a Third frame
    You need to see that you already have four frames of reference:
    A, A-reverse, B and B-reverse. And only A to B and A-r to B-r are defined.

    Demonstration by adding a 5th frame of reference C
    With point C for t=t0 at first passing same point C t=t1 (t based on spaceship A)
    And that after 10 units of time (let say at least 10 seconds) Space ship A has only gone 10 feet!! And returns to C in the same 10 units of time then for our relative light speed comparisons we can consider all three frames C A and A-reverse to be the same.
    As for Spaceship B using an instantaneous change of speed and reversal is fine.(Avoids any calculations for acceleration). The returning speed gives us the same time dilation as the outbound. And Spaceship B time change would only be a fraction of the 20 units of time.

    However as your problem also allows that it could be Ship B that never really leaves the passing point C. Then for the traveling ship A to have 20 units of time pass, Ship B and location C will need massive amounts of time to pass.

    Finally if after 10 units of time the distance from C to both A and B is the same very far distance for both. When they return they will both show 20 units of time, yet the clock at C will show much more. C won't care what direction they are going.

    Don’t get me wrong C is not “Stationary” it can be a spaceship that is falling behind A or B (Same as running ahead of A-reverse or B-reverse) but it must remain in a constant frames of reference to work your calculations from. Otherwise danger Will Robinson – your location is Lost In Space and you won’t know where the two passing points at t0 and t1 really are.

    Randall B
  12. Dec 7, 2004 #11
    my actual problem is not concerned with my location in space.

    all i wanna know that if in the future it becomes possible to attain velocities of the order of c, and if then this is performed practically, then irrespective of the location in space, or the frames of references, what will be the value of time interval measured by the observer on starship B.

    there have been many approaches, for solving this, but nobody has given me a definite value (for answer), and i wanna know, whether the theory of relativity can give a definite answer to this question.
  13. Dec 7, 2004 #12
    It seems to me that both ships pass the same distance at the same velocity, and so, to an outside viewer the same amount of time. I can not see why it should not appear to be a passage of 20 units of time in both ships.

    Edit: on a more intresting note for those who do not know, although the ships are moving .999c past each other, their combined speed relative to each other is still less than c itself.
    Last edited: Dec 7, 2004
  14. Dec 7, 2004 #13
    Yes, it can! Proper time [itex]\tau[/itex] is defined as the total time elapsed by an observer's wristwatch as they move between two events in spacetime. The events of coincidence of both observers at times [itex]t_0[/itex] and [itex]t_1[/itex] will be denoted by [itex]E_0[/itex] and [itex]E_1[/itex] respectively. Hence, the time interval as measured by either observer is found by evaluating the integral [itex]\tau =\int_{E_0}^{E_1}d\tau[/itex]. Although SR allows us to calculate in the non-inertial frames of A or B, the components of the metric tensor look particularly nice in inertial coordinates. Establish an inertial frame C, at rest relative to the points of coincidence. In this frame, [itex]\tau =\int_{E_0}^{E_1}\sqrt{dt^2-dx^2}[/itex]. The coordinate differences for both observers are identical apart from the reversal of the sign of the [itex]dx[/itex] which has no effect due to the power of 2. Therefore, both observers will measure the same time interval [itex]\tau = t_1 - t_0[/itex].
  15. Dec 7, 2004 #14


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    From the way you worded the problem, it sounds like it is symmetrical: ie, a person sitting at the crossing point watching the two spacecraft would see each doing exactly the same thing but in opposite directions. In that case, each ship will agree with the other every time they pass. That gives you Dave's answer: 20.
  16. Dec 7, 2004 #15
    i agree. :approve:
    moreover, imho the problem can't have different answer: the two starships are identical - the problem is symmetric so, due to the isotrophy of space and the principle of relativity, the total proper time mesaured in both starships is same.
    Last edited by a moderator: Dec 7, 2004
  17. Dec 7, 2004 #16
    Russ - JD
    Your answer of 20 is fine but only in the special case where:
    speed of Spaceship A-reverse as measured from Spaceship A is .999c
    ( release probe from A before the "turnaround" to send measurements)

    The problem as stated in post 1 can allow this measurement to be ANYTHING!
    Including Zero!
    That is why Vikas is coming up with so many different answers in post #8.

    If you do your calculations adding the above requirement. Then and only then you can use any one of the four reference frames as "stationary" (ie.spaceship A stands still for 10 units of time before going after B at .999c) Then SR will give you only one answer - the same as Dave's. You won't be able to come up with the other answers you had in post #8.

    As long as you are definitive about this last element of the problem SR will give you one and only one definite answer.

    Randall B
  18. Dec 9, 2004 #17
    Did this give you and your physics teacher a satisfactory answer.
    ... do the ships turn around simultaneously?? Which one turns first?
    .. how about if you look from the other ship which one turns first?
    .From where and when do they turn simultaneously?

    Did you try holding the Delta v for ship A at Zero?
    What is the Delta v for ship B?
    Even if you do the measurements from ship B (be careful and complete) it will still show the time for ship B to be much less than the 20 Units on A.

    Hold the Delta v for Ship B to zero
    Now even measuring from Ship B (Still need to be careful and complete) it will show the time on ship B to much much more than the 20 Units of elapsed time on Ship A.

    When you "get" all these and understand "simultaneous" - Then you will start to know SR.

    Randall B
    PS: Being carefull and complete can be tricky - stick with it.
    ..Be sure you know what time it is and where you are when you change referance frames.
  19. Dec 11, 2004 #18
    this was exactly the point i wanted get clarified. in the above two conditions we get two different answers depending on which condition do we choose, but what i am asking is that when this experiment is actually performed, we dont have to tell the WATCH, on starship B, which condition have we chosen, and in that case, the time measured by the watch will be ........

    kindly fill the blank with appropriate value of time.
  20. Dec 11, 2004 #19
    Alas! you still have missed the point.
    It's not two conditions - they are infinite.
    What can we do if the person wearing the watch from direct observation will not tell us:


    This is most easily known by the guy with the watch and untill he sets it at one number we can only guess any number as the speed change for B. Then what do you expect the Time for B with A=20 units to be! It can be anything as well.

    Now once you do define this simple piece of info you will be locked into only one answer for all kinds of things. Max seperation from any ref Frame. What frame gives a common time on both A and B and what that time is. And where is Space that observation is. And of course the time on B with A going for 10 Units plus 10 Units.

    Otherwise unless you can show how to get one exact answer for x in x= 5+4+y with no additional info about y. Then no one will be able to help you with this one either.

    Try this - set a number for Delta B or Delta A (they will be the same) then try to come up with more than one answer to your question. You will see that this Delta is the most important "relative speed" in your problem.

  21. Dec 13, 2004 #20
    i know that.
    i guess my question(doubt) was not a very correct one, and that is why all this confusion was created.
  22. Dec 13, 2004 #21
    It's not that question is incorrect - Just not complete in all required details.
    The question can also be "What are the various times for Delta B time given each of several different Delta v for Shipa.

    Note: My statement that setting:
    " a number for Delta B or Delta A (they will be the same) "
    Was way incorrect! They are only the same in the case that Delta t for B is the same 20 Units as Clock A had. You will see that quick as you work a couple, like the extreme case where you hold the Delta v for Ship to 0 (zero) and your problem turns into the classic "Twins" problem. And your set up helps show why it is just a problem and not a 'Paradox'.

    Have you resolved your (DOUBTs) about SR.??
  23. Dec 17, 2004 #22
    well, i wont say that my doubts have been cleared, but i think i will have to study it in a bit more detail before i can contue my side of arguments on it.
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