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Homework Help: A problem on time taken by relativistic particles

  1. Apr 4, 2012 #1
    1. The problem statement, all variables and given/known data
    a) What is the speed of proton that has relativistic γ factor of 10^10? Write as (1-ε)c

    b) A proton with γ=10^10 is chased by a photon across a galaxy. Observer A at one end of the galaxy (xA=0) sees the proton at t1=0 and the photon at t2 = 1.33 x 10^-8 s. According to A, how far apart are the two particles at t1?

    c) Another observer B is at another end of the galaxy (xB=10^21 m). The two observers are at rest relative to each other. At what times will B see the particles go past?

    d) How many years does it take the proton to cross the galaxy according to A and B?

    2. Relevant equations

    L = Lo [itex]\sqrt{1 - v^2/c^2}[/itex]
    Δt = Δto/[itex]\sqrt{1 - v^2/c^2}[/itex]

    3. The attempt at a solution
    a) Y = 1/[itex]\sqrt{1 - v^2/c^2}[/itex]
    1 - v^2/c^2 = 1/Y^2
    v^2/c^2 = 1 - 1/Y^2
    v^2/c^2 = 1 - 1/(10^10)^2
    v^2/c^2 = 1- 1/10^20
    v/c = sqrt(1 - 1/10^20)
    v/c = (1 - 5 * 10^-21)
    v = (1 - 5*10^-21) c
    Is this correct?

    b) The photon covers the distance in t2 - t1 = 1.33 x 10^-8 s
    The distance = c t^2
    = 3 x 10^8 x 1.33 x 10^-8
    = 3.99 m
    Is this correct?

    c) Can we say that the observer B will see the proton pass by at t3 = t1 + xB/v and the photon pass by at t4 = t2 + xB/c ?
    If not, then what is the correct method?

    d) Can we simply divide distance (which is 10^21 m) by speed and convert into years? If not, then what is the correct method?

    For c and d, I am not sure if we need to use Lorenz transformation equations. The observers are at rest relative to each other. So I think we do not need.
     
  2. jcsd
  3. Apr 4, 2012 #2
    For part c, is it possible to express the answers as combination of a large number and a small number? I want to express t3 and t4 in such a way that it is easy to see the difference between them.
     
    Last edited: Apr 4, 2012
  4. Apr 4, 2012 #3
    Please see if this is correct
    c) t3 = time at which observer B sees the proton fly by
    t4 = time at which observer B sees the photon fly by

    t3 = t1 + 10^21/(1-ϵ)c
    t3 = 3.33 * 10^12/(1-ϵ)
    t3 = 3.33 * 10^12[1 + 1/(1-ϵ) - 1]
    t3 = 3.33 * 10^12[1 + ϵ/(1-ϵ)]
    Since ϵ << 1, therefore using the approximation 1-ϵ = 1, we get
    t3 = 3.33 * 10^12(1 + ϵ)
    t3 = 3.33 * 10^12(1 + 5*10^-21)
    t3 = 3.33 * 10^12 + 1.67 * 10^-8 s
    t4 = t2 + 10^21/c
    t4 = 1.33 * 10^-8 + 3.33 * 10^12
    t4 = 3.33 * 10^12 + 1.33 * 10^-8 s

    d) For both observers, the time taken by the cosmic ray is approximately 3.33 * 10^12 s
    = 10^5 years
     
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