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Lorentz Contraction/ Simultan Problem

  1. Jan 9, 2012 #1
    Hi all, this is not a HW problem but one I made up myself, (so you won't find it in any txtbooks or solutions etc.). It also is much more advanced than for someone on the physics HW site, as I can see, so I thought it best to put here.

    I'm having trouble working out simultaneity and Lorentz Contraction for the following case:


    PROBLEM: Earth sees a rocket moving at β=+0.99999c (in the direction of the positive x-axis). At t=0, x=0, the coordinates of the rocket are t'=0, x'=0 (origins coincide). In the Earth frame, there is an obstacle at 2000 m at t=0. The obstacle does not move from this coordinate the whole time, as viewed from Earth. At t=t'=0, a light from the obstacle travels back towards the rocket at -c (the negative denotes velocity and c is light speed). Now, the rocket travels towards the obstacle, but the rocket won't know the obstacle is there until the rocket recieves the light pulse from the obstacle.

    !!!!!!!!!!!
    MY QUESTION when the rocket recieves the pulse from the obstacle, how much time will the rocket have to avoid the obstacle and how far will the obstacle be from the rocket?
    !!!!!!!!!!

    If you skip down to SUMMARY I will have the main discussion there, also, my questions and problems with my own analyses. Below is how I got my numbers:

    Parameters

    β for rocket is = 0.99999 thus γ=223.6073568 (decimals matter, as I panfully learned)


    Origin

    Earth says rocket is at: x=0, t=0
    Rocket says Rocket is at: x'=0, t'=0

    Earth says obstacle is at: x=2000 m, t = 0

    Use Lorentz on the above to get to the coords

    x'=γ(x-βct); t'=γ(t-βx/c)]

    to get the coords for the rocket below:

    Rocket says obstacle is at: x' = 4.47214(10^5) m, t' = -.0014907 secs



    Rocket Recieves Light from obstacle at

    Earth says: [condition is, in Earth's frame]: βct = 2000-ct

    x_rocket_rcvs_light = βct = 999.995 m
    t_rocket_rcvs_light = 2000/[c(1+β)] = 3.33335(10^-6) secs.



    Use Lorentz on the above to get to the coords

    x'=γ(x-βct); t'=γ(t-βx/c)]

    to get the coords for the rocket below:

    rocket says

    x'_rocket_rcvs_light ≈ 0
    t'_rocket_rcvs_light = 1.4907(10^-8) secs


    At the time the rocket recieves the light from the obstacle, the obstacle is at:

    Earth says obstacle is at (obstacle does not move in Earth's frame):

    x_obstacle = 2000 m
    t_obstacle = 3.33335(10^-6) secs


    Use Lorentz on the above to get to the coords

    x'=γ(x-βct); t'=γ(t-βx/c)]

    to get the coords for the rocket below:


    Rocket says obstacle is at:


    x'_obstacle = 223608 m
    t'_obstacle = -7.453(10^-4) secs


    If the rocket were to continue and collide with the obstacle, this would happen at,

    Earth says:

    x_rocket_no_swerve = 2000m
    t_rocket_no_swerve = 2000/(0.99999c) = 6.6667(10^-6) secs



    Use Lorentz on the above to get to the coords

    x'=γ(x-βct); t'=γ(t-βx/c)]

    to get the coords for the rocket below:


    x'_rocket_no_swerve = 0
    t'_rocket_no_swerve = 2.236(10^-8) secs


    Notice that, all the x's are all zero, as the rocket will always feel itself at the origin of its own frame. In the rocket's frame, the rocket will recieve the light from the obstacle at the rocket's own origin , and if the rocket hits the obstacle, the rocket will do so at the origin of the rocket's frame. So this is a good check I have used the Lorentx Equations correctly, at least up to there.


    SUMMARY:


    Earth:
    The Earth says the rocket starts at x=0,t=0, then the rocket recieves the light from the obstacle at x=999.995 m, t=3.33335(10^-6)secs. The obstacle is at x=2000, t= 3.33335(10^-6)secs at this point. If the rocket plows into the obstacle, the Earth will say that this will happen at x=2000 m, t = 6.6667(10^-6) secs.

    Rocket:

    The rocket says the rocket starts at x'=0,t'=0, then the rocket recieves the light from the obstacle at x'= 0 , t'= 1.49(10^-8) secs. The obstacle is at x = 223608 m, t= -7.453(10^-4) secs at this point. If the rocket plows into the obstacle, the rocket will say that this will happen at x'=0 m, t = 2.236(10^-8) secs.


    QUESTIONS

    1. I'm having some trouble understanding this. I know that what is simultan in one frame is not in another. Here, when the rocket recieves the light pulse, the Earth says the rocket is at 999.995 m and the obstacle is at 2000m, and the time these two things happen occurs, for the Earth at the same time of 3.33335(10^-6) secs. But, in the rocket frame, the rocket will recieve the light signal first, and then some time later, the obstacle will be at the pos, time Earth sees for the obstacle at 2000 m, 3.33335(10^-6) secs. But, when the rocket recives the light signal, he will know the obstacle is there. Notice that for the second coordinate set, the rocket recieves the signal at 745 μsecs before the origins even line up (that's from the negative from the time transform) In the rocket's frame, once the rocket recieves the light signal, far does the rocket say the obstacle is and how much time does the rocket think the rocket needs to get to the obstacle? How do I use the Lorentz Transforms correctly for this? .

    2. Now, the distance from the Earth to the obstacle is 2000m, but shouldn't this length be shorter according to the rocket's frame? This 2000m is a proper length, since it does not move in the Earth's frame. The rocket should see the 2000 m shrink. But, if you notice, when the rocket recieves the light pulse from the obstacle, the rocket thinks the obstacle is at 223608m !

    I tried to use the famous Lorentz Length Contraction Formula of L=Lo/γ (which assumes the origins line up at x=x'=0, t=t'=0) for 2000m, and the Length Contraction Formula gives me a length of 8 m or so for the rocket's frame. If I do an inverse lorentz transformation, then I get the 2000 m coordinate of the obstacle contracted down to 8 m as viewed by the rocket, in agreement with the Lorentz Contraction Formula. But why should I use an inverse lorentz transformation? Why didn't the standard lorentz transformation give a length contraction on the coordinate? Why did the coordinate of the obstacle appear farther away in the frame of the rocket using the standard Lorentz Transformaion?


    3. Also, the rocket says it recieved the light pulse 7.453(10^-4) secs BEFORE the origins were ever aligned. Haven't I eliminated this problem by trying to force the origins to be x'=0, t'=0?
     
  2. jcsd
  3. Jan 9, 2012 #2

    PAllen

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    I haven't read all you've written, but approximate answer is obvious. Assuming γ for .99999c, and giving answer to your first bold question from point of view of rocket:

    rocket will get signal from obstacle tiny bit after (1000m/c)/γ seconds after passing earth. If it does nothing to avoid obstacle, it will hit it slightly after (2000m/c)/γ seconds. The obstacle will appear (using conventional interpretations) slightly more than 1000m/λ away when signal is received.
     
  4. Jan 10, 2012 #3

    ghwellsjr

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    Just as you have to calculate the progress of the rocket in the Earth frame to identify the event of the rocket arriving at the obstacle, you have to do a similar thing in the rocket frame to identify events of interest. So for example, once you have calculated the event in the Earth frame of the rocket arriving at the obstacle, you can transform that event into the rocket's frame which will tell you how much time transpired from the start in the rocket's frame. Then you use the same rocket speed (relative speeds are the same in both frames) to calculate the distance the obstacle has to travel in the rocket's frame and this will identify the event for the obstacle at the start. You can use the negative of that same distance at the end of the scenario with the final time in the rocket's frame to identify the event for the Earth at the end. Once you get the hang of it, I'm sure you'll be able to figure out the other events of interest.
     
  5. Jan 10, 2012 #4
    Hi all,

    Here's what I did. I took the advice of ghwellsjr and PAllen.

    The Earth says the rocket will recive the signal at 999.995 m, 3.33335(10^-6) secs. The collision would occur at 2000 m, 6.6667(10^-6) secs.

    Rocket Frame

    Length Contraction Method

    Rocket recieves signal at: 999.995/γ = 4.47210 m
    Collision at: 2000/γ = 8.9442495 m.
    Thus the distance from signal reception to collision is 8.9442495 - 4.47210 = 4.472147 m.
    The time it would take the rocket to reach the obstacle over this distance would be L'/v=t' or 4.472147/0.99999c = 1.4907(10^-8) secs.

    Time Dilation Method

    Rocket recieves signal at: 3.33335(10^-6)/γ = 1.4907(10^-8) secs
    Collision at: 6.6667(10^-6)/γ = 2.981431(10^-8) secs.
    Thus the time to go from signal reception to collision is 2.981431(10^-8) - 1.4907(10^-8) = 1.490115 secs. The distance the rocket needs to cover over this interval is thus L'=vt' or 1.490115 *0.99999c = 4.4721 m.

    So the times and distances from reception to collision match in both methods.

    But why didn't the Lorentz Transforms give the same answers as above? Below is a table where the rocket's coords are derived from pure Lorentz Transforms. For example, why does a pure Lorentz Transform say that, in the rocket's frame, when the rocket recieves the signal, that the obstacle is at 223608 m,-7.453(10^-4) secs (bold below)? The coordinate of the obstacle is much farther away than the Earth's rest frame coordinate. Why, also, the negative time?

    I really appreciate everyone's help on this. It took a lot of effort.





    Earth says

    _________________________x________t

    Rocket starts at:___________0,_______0
    Obstacle is at:_____________2000,____0
    Rocket receives signal at____999.995,__3.35(10^-6) secs.
    Obstacle is at______________2000,____3.35(10^-6) secs
    Rocket/obstacle collision_____2000,____6.67(10^-6) secs

    Rocket says

    _________________________x'____________t'

    Rocket starts at:___________0,__________0
    Obstacle is at:__________447214,__-.0014907 secs
    Rocket receives signal at____0,___ 1.4907(10^-8) secs
    Obstacle is at:_______ 223608 m,__-7.453(10^-4) secs
    Rocket/Obstacle Collision____0, 2.236(10^-8) secs
     
  6. Jan 10, 2012 #5

    ghwellsjr

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    If you transform the events for determining times into the rest frame of the rocket, you should get the x coordinates all to be zero, otherwise, the rocket is not at rest. Also, any event in one frame that you transform into a second frame will get you back the same coordinates when you change the sign of the velocity so it shouldn't surprise you that the same event in one frame does not correspond to the equivalent concept in the other frame. For example, a pair of events can show time dilation when transforming in one direction but not when going back in the other direction. You need different events (where the locations are zero) in order to show reciprocal time dilation in the other direction.
     
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