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Muon length contraction confusion

  1. Jun 2, 2015 #1
    Muon created by fast cosmic ray has speed 2.99X10^8 m/s and life time 2.2 micro sec. According to this numbers muon should travel only 0.66Km with respect to earth in his life time. however it travels more than 10 km (10.4km) due to time dilation. in muon frame of reference, earth travels only 0.66km. now if we consider muon created 20.8km above earth, so in muon frame of reference earth seems to be at distance 1.32km and if we consider one more muon created on earth exactly the first one decade. will the observer in first muon frame of reference see that earth cross him and then second muon decade and the observer on earth see that the observer in first muon frame still not crossed even though the second muon is decade ?
     
  2. jcsd
  3. Jun 2, 2015 #2

    PAllen

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    I can see English is not your first language. As a result, I am not sure I fully understand your scenario. However, you should know I was specifically confused, then amused at your use of decade. Obviously, you meant decayed. However, decade means 10 years, which was particularly confusing given your discussion of times and distances. Hopefully, someone else can untangle your language.
     
  4. Jun 2, 2015 #3
    m sorry for that, its decayed and not decade.
     
  5. Jun 2, 2015 #4
    hopefully now you can understand it.
     
  6. Jun 3, 2015 #5
    the situation given below may not actual happens.Just want to know the results in of scenarios.

    Muon created by fast cosmic ray has speed 2.99X10^8 m/s and life time 2.2 micro sec. According to this numbers muon should travel only 0.66Km with respect to earth in his life time. however it travels more than 10 km (10.4km) due to time dilation. In muon frame of reference, earth travels only 0.66km. now if we consider muon created 20.8km above earth, so in muon frame of reference earth seems to be at distance 1.32km and if we consider one more muon created on earth exactly when the first one decayed. will the observer in first muon frame of reference see that earth crossed him when the second muon (muon on earth) decayed. The observer on earth see that the observer in first muon frame of reference still not crossed when the second muon is decayed ?
     
  7. Jun 3, 2015 #6

    Ibix

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    I'm afraid it is not at all clear what you are asking.

    I understand that you are imagining a muon created in the upper atmosphere heading down towards earth. You are also imagining a second muon, which I think is created in the upper atmosphere at the same time that the first one decays, and follows the first one down towards the earth. You seem to be assuming that both muons decay in the middle of the atmosphere, half way to earth.

    You seem to be asking if, when the second muon decays, whether an observer moving beside the first muon would have struck the earth's surface.

    Is this correct?

    If so, you need to tell us whether you mean "at the same time" according to the muon frame, or according to the earth frame. These are not the same thing.
     
  8. Jun 3, 2015 #7
    Actually I am assuming the second muon on earth (stationary to earth). My confusion is, in the muon frame of reference, observer observes the distance between muon and earth is 0.66km since muon is stationary and from earth the distance in 10.4km since muon moving with speed 2.99X10^8 m/s. now I am just considering that muon creation at altitude 20.8 km, double the distance we considered in previous case. so the observer in the muon frame of reference will also observer the earth is at 1.32km (double the previous distance). If yes, the we are considering second muon to be created on earth when first one is decayed so the condition will get revered since muon is now created on earth.
     
  9. Jun 3, 2015 #8

    Ibix

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    I have not checked your numbers, but this sounds reasonable. I'm assuming by "muon frame of reference" you mean the frame of the first muon.

    OK.

    I have no idea what you are trying to say here, I am afraid. What "condition" are you talking about? Also, as I said in my last post, "at the same time" is a different thing in different frames. In which frame are the first muon's decay and the second muon's creation "at the same time"?
     
  10. Jun 3, 2015 #9
    In the earth frame of reference.
     
  11. Jun 3, 2015 #10
    Your approach is good. We will go step by step.

    Consider two frame for reference 1 and 2 (will not consider muon decay here). Frame-1 is moving with speed 2.99X10^8 m/s with respective Frame-2. Frame-1 is at 20.8KM apart from Frame-2 (distance measured in Frame-2). Will the Frame-1 also measure same distance of Frame-2 i.e 20.8KM?
    If answer for above is yes, then will imagine that when they are 20.8 KM apart, muon is created in Frame-1. It will have life time of 2.2 micro sec in Frame-1 and 34.30 micro second in Frame-2 (due to time dilation). So in muon decay duration, frame-1 will travel 0.66km towards Frame-2 and Frame-2 will travel 10.4KM towards Frame-1. Is this correct? if yes, what will be the distance between this two frames when measured from frame-1 and frame-2?
     
  12. Jun 3, 2015 #11

    ghwellsjr

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    I think some spacetime diagrams might help to understand your issue. But first, I would like to change the parameters slightly to make them fit better with my software that draws the diagrams. Instead of muons that decay in 2.2 microseconds, I'm going to use muons that decay in 3.3 microseconds which happens to be 1 light-kilometer. And instead of the normal muons being created 10.4 kilometers above the surface of the earth, they will be created at 16 kilometers. Also, I'm going to draw my diagrams with the time axis horizontal and the distance axis vertical which I think makes more sense when we are considering muons that are traveling vertically down toward the earth.

    OK, here's the first diagram for the earth rest frame with a red muon being created 16 kilometers above the blue earth and traveling downward at 99.8%c:

    MuonLC1.PNG

    The dots mark off 1 light-kilometer increments of time (3.3 microseconds). Notice that the muon is Time Dilated so it can reach the surface of the earth. Gamma at 99.8%c is 15.8 which enables the muon to survive the trip. Also notice that time on the earth is not dilated in the earth rest frame.

    Now let's transform to the muon rest frame:

    MuonLC2.PNG

    If you consider just the earth time increment before the collision, you will see a similar mirror image (reverse) pattern to the first frame.

    But now you want to see a similar reverse pattern if you have the muon start at double the distance compared to another stationary muon created on the earth when the first one decays:

    MuonLC3.PNG

    Now we transform to the muon rest frame (note the scale difference):

    MuonLC4.PNG

    If we had drawn in a continual pattern of time increments for the earth, I think you would see that there could be the reverse pattern relationship that you are asking about, but it can't be a muon that is created when the first one decays according to the earth rest frame, it would have to be during the time increment that occurred just before the collision in the second diagram (the one that is shown partially in the lower left corner).

    Does this make sense to you? Any questions?
     
  13. Jun 3, 2015 #12
    Sorry, I am not able to understand diagrams. could you please explain in detail ?
     
  14. Jun 3, 2015 #13
    Also please answer my two questions from latest post.
     
  15. Jun 3, 2015 #14

    Nugatory

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    You're overlooking the relativity of simultaneity here. If the second muon was created on earth at the same time that the first muon decayed according to an observer moving with the first muon, then the second muon was not created at the same time that the first muon decayed according to an observer on earth. Thus, the bolded text above is ambiguous.
     
  16. Jun 3, 2015 #15

    ghwellsjr

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    Have you ever been on a treadmill? It will draw a chart of your progress as a function of time. That's all these diagrams are except the time scale is very short. Each grid mark and each dot represents a time interval of 1 light-kilometer which is 3.3 microseconds which is the lifetime of a muon according to my modified parameters.

    So in the first diagram, the surface of the earth (blue) is at sea level or 0 kilometers and its clock ticks at the same rate as the Coordinate Time. The red line is the muon created at an altitude of 16 kilometers and it travels down to sea level in 16 light-kilometers which is about 53 microseconds but because its speed is 99.8%c, its time is dilated by a factor of 15.8 so instead of decaying in 3.3 microseconds, it survives for over 52 microseconds.

    Do you understand the first diagram?
     
  17. Jun 3, 2015 #16

    ghwellsjr

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    Your terminology doesn't make sense to me. Instead of talking about things happening to the frames, think of a frame as a coordinate system (an x-y chart showing distance along one axis and time along the other axis). Talk about what the objects are doing in the frame, where they are (distance coordinates) at different coordinate times. You need to define your scenario according to one frame, say the earth rest frame, and then use the Lorentz Transformation process to see what it looks like in a second frame moving with respect to the first frame. You can't guess, you have to do the calculations. That's what my software does for you automatically but you can check to make sure that my second diagram is correctly transformed from my first diagram. Same for my fourth and third diagrams.

    Can you rephrase your questions in this way?
     
  18. Jun 3, 2015 #17
    Is the half-life of muon (2.2 microseconds) in the frame of the muon or of the earth-bound observer?
     
  19. Jun 3, 2015 #18

    PAllen

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    Half lives are always given in particle rest frame.
     
  20. Jun 3, 2015 #19

    ghwellsjr

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    I'm treating the muons' lifetimes to be 3.3 microseconds (1 light-kilometer) just so that my drawing program can show the time intervals as either coordinate times or Proper Times for all muons (moving or at rest) as indicated by the interval between a pair of dots on a worldline.
     
  21. Jun 3, 2015 #20
    So, if a muon is moving at 0.99c with respect to the earth bound observer, then according to the earth bound observer, the half-life of muon is 2.2 * 1/sqrt(1-0.99^2) microseconds, which is about 15.6 microseconds. Muons moving at 0.99c for 15.6 microseconds will travel about 4.6 km. Again, 15.6 microseconds is a half-life; some muons will live longer and some will not live as long. The ones that live longer will travel more than 4.6 km. Hence, it is expected then that a significant percentage of the muons would reach the surface of the earth. If we assume that we are talking about that part of the atmosphere that is 9.2 km above earth, then we would expect about 25% of the muons to reach the surface of the earth.

    Or, let's consider the frame of reference of the muon. The half-life is 2.2 microseconds, and the earth is traveling at the muon at a speed of about 0.99c. The distance between the earth's surface and the muon from the muon's frame of reference is about 1/7th of the distance observed by an earth-bound observer - about 1300 m. So, from the average muon's frame of reference, the earth is speeding towards it at 0.99c. In 2.2 microseconds, the earth has traveled about 653 m. In 4.4 microseconds (by which time about 25% of the muons are still alive), the earth would have traveled about 1306 m - and crashing into the muons that survived that long.
     
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