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Particle Decay After a Certain Distance

  1. Oct 13, 2016 #1
    1. The problem statement, all variables and given/known data
    A certain elementary particle lives only a time [T][/0] = 5 sec (proper time) before disintegrating. What velocity must the particle have if it is to reach the Earth from the Sun before disintegrating? Distance between the Earth and Sun is 1.5x10^11 meters.



    2. Relevant equations
    I dont necessarily know which equations would be most relevant.

    3. The attempt at a solution
    I have been working on this problem for several hours now and cannot seem to get an answer, I have tried every which way but no luck so far.
     
  2. jcsd
  3. Oct 13, 2016 #2

    kuruman

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    Hi PatrickStar and welcome to PF.

    Surely you can do better than this. Maybe you are new, but we do expect to see some initial effort on your part.
    What topics are you covering in class now or have covered recently? Make a list of these equations, eliminate the ones you think are irrelevant and post the rest.
    Show all the ways you have tried. We will steer you away from blind alleys and point out what looks promising.
     
  4. Oct 13, 2016 #3
    Sorry, excuse my lack of effort, I have just been working on this question for nearly 7 hours and have no sight of solving it and am pretty defeated.
    I have tried using several equations, such as v=d/t where I subbed out t for the [t][/0]/(sqrt(1-v^2/c^2) to account for the time dilation that would occur when the particle approaches the speed of light but that didn't get me anywhere a long with several other strategies. A push in the right direction would be very much appreciated.
     
  5. Oct 13, 2016 #4

    kuruman

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    Using d/t is the correct equation. Look at the question from the particle's point of view. The particle lives for 5 s in its own frame, so that's the t in the denominator. What is the distance in the particle's own frame that the particle covers in that amount of time?
     
  6. Oct 13, 2016 #5
    The distance would be the total length, i.e. 1.5x10^11 wouldn't it? Or would it be 5 seconds * its Velocity?
     
  7. Oct 13, 2016 #6

    kuruman

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    Nope. That's the distance s measured by someone at rest with respect to the Sun-Earth system. Remember, the particle "sees" that distance moving relative to it.
     
  8. Oct 13, 2016 #7
    So the particle will only see the distance of how fast its traveling V * how long its traveling for t?
     
  9. Oct 13, 2016 #8

    kuruman

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    That's the left hand side of the equation, v*t. What is the right hand side? In other words, what is an expression for the Earth-Sun distance when viewed from a frame moving at velocity v relative to the Earth-Sun system?
     
  10. Oct 13, 2016 #9
    I'm not sure I'm following.
     
  11. Oct 13, 2016 #10

    kuruman

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    Can you answer the following question?
    You have a meter stick that is moving relative to you with velocity 0.8c. You measure the length of the stick while it's moving. What do you get for an answer?
     
  12. Oct 13, 2016 #11
    Ohh so are you saying that we have to use the length contraction of L`= L/gamma ?
     
  13. Oct 13, 2016 #12
    Never mind, I just tried using that and didn't really see anything promising so I'm assuming that's not what you were alluding to?
     
  14. Oct 13, 2016 #13

    kuruman

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    Precisely. The Earth-Sun distance is shortened by gamma in the particle's reference frame.
     
  15. Oct 13, 2016 #14

    kuruman

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    Look again. One side of the equation is v*t. The other side is L/γ and γ has v in it. So ...
     
  16. Oct 13, 2016 #15
    I tried solving for v in that case and I got stuck. My process is as follows:
    vt = L/gamma
    vt = L * sqrt(1-v^2/c^2)
    vt/L = sqrt(1-v^2/c^2)
    (v^2 * t^2)/L^2 = 1-v^2/c^2
    then to v^2/c^2 = 1 - (v^2 * t^2)/L^2
    but after that I just feel like im going no where
     
  17. Oct 13, 2016 #16

    kuruman

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    Not so. Multiply things out and you'll get a fourth order polynomial in v that looks like Av4+Bv2+C = 0. Replace v2 with y and you have a quadratic equation in y that you can solve. Then v is the appropriate square root of y.
     
  18. Oct 13, 2016 #17

    kuruman

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    L/gamma is not L * sqrt(1-v^2/c^2). gamma = sqrt(1-v^2/c^2)

    On edit: It actually is. Sorry for confusing you. gamma = 1/sqrt(1-v^2/c^2). I must have had a brain lapse.
     
    Last edited: Oct 13, 2016
  19. Oct 13, 2016 #18
    So it would be vt = L/(1/sqrt(1-v^2/c^2)) or is it just L/(sqrt(1-v^2/c^2)) ?
     
  20. Oct 13, 2016 #19
    As of now, I have worked it out to y^2/c^2 - y + L^2/t^2 = 0 after subbing y = v^2. Does that look right so far?
    I went with vt = L / (1/sqrt(1-v^2/c^2))
     
  21. Oct 13, 2016 #20

    kuruman

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    An easy way to remember which way it goes is that γ = sqrt(1-v^2/c^2) is greater than 1. So if you want a shorter length you need to put it in the denominator.
     
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