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Homework Help: Light Problem

  1. Oct 9, 2006 #1
    A flash of light is emitted at point O and is later reabsorbed at point P. In frame S, the line OP has a length l and makes an angle theta with the x axis. In a frame S' moving relative to S with a constant velocity v along the x axis:
    How much time tau' elapses between emission and absorption of the light?

    The answer is (1-betacos(theta))gamma*l/c, where beta=v/c

    How does one get this answer?
     
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  3. Oct 9, 2006 #2

    OlderDan

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    It took some effort, but I managed to do this. I used length contraction in the x direction to figure out the coordinates of P in S' when the pulse is emitet. Since S' is moving relative to points O and P in S, the pulse receiver is moving through S' toward the y' axis a distance v*tau' as the light pulse is moving a distance c*tau'. It's just vector addition to find the distance the pulse travels to the receiver. You get a quadratic equation for tau' that is most easily handled in terms of the combination c*tau'/gamma. Solve the quadratic and simplify and you've got it.
     
  4. Oct 9, 2006 #3
    Could you give some more detail on "length contraction in the x direction to figure out the coordinates of P in S' when the pulse is" emitted?

    Code (Text):
    [tex]l= l_p/gamma[/tex]
    Also,
    (c*tau')^2=(v*tau')^2+(x)^2 ?...
     
    Last edited: Oct 9, 2006
  5. Oct 9, 2006 #4

    OlderDan

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    If O is the origin, then the coordinates of P in S are
    x = l*cos(theta)
    y = l*sin(theta)

    Let the origins of S and S' coincide when the pulse is emitted. Then the coordinates of P in S' are found by contracting in the direction of motion

    x' = x/gamma
    y' = y

    You could use Pythagoras to find the distance from O to P in S', but the light pulse will not go that far because P is moving through S' with speed v. You need to figure out how far it is from the origin to where the light pulse intercepts P. That will be at
    (x' - v*tau', y') = (l*cos(theta)/gamma - v*tau', l*sin(theta))
    Use Pythagoras to find the distance from O this point and set it equal to c*tau'
     
  6. Oct 16, 2006 #5
    [(l*cos(theta)/gamma - v*tau')^2 + (l*sin(theta))^2 = d^2 = (c*tau')^2
    [(l*cos(theta)/gamma - v*tau')^2 + (l*sin(theta))^2]^(1/2)/c=tau'

    "most easily handled in terms of the combination c*tau'/gamma"
     
    Last edited: Oct 16, 2006
  7. Oct 16, 2006 #6

    OlderDan

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    The blue is all you need, and then some algebra skills and one well known trig identity.
     
  8. Oct 16, 2006 #7

    OlderDan

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    The sin^2 = 1-cos^2 is the first step after squaring terms. Combine the cos^2 terms and you will have a factor of 1-1/gamma^2 which is beta^2. Replace any v with beta*c and you should have something in the form of a quadratic
    0 = x^2 + bx + c where b = 2*l*beta*cos(theta), c = l^2[1-beta^2cos(theta)^2] and x = c*tau'/gamma.
    Use the quadratic formula. The beta*cos(theta) contributions to the discriminant cancel leaving you with l*sqrt(1). Only the + root is kept. Solve for tau' and your done.
     
    Last edited: Oct 16, 2006
  9. Oct 16, 2006 #8

    OlderDan

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    Yes. FOIL it
     
  10. Oct 16, 2006 #9

    OlderDan

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    square the whole thing to get rid of the ^(1/2) and combine the v^2tau'^2with the c^2tau'^2 into a term with a gamma^2
    use the sin^2 = 1-cos^2 and combine the cos^2 terms
     
  11. Oct 16, 2006 #10

    OlderDan

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    Keep going :!!)
    [l^2cos^2(theta)/gamma^2 - 2v*tau'*l*cos(theta)/gamma + l^2sin^2(theta)]=c^2tau'^2 - v^2tau'^2 = c^2tau'^2/gamma^2
     
  12. Oct 16, 2006 #11

    OlderDan

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    c^2tau'^2 - v^2tau'^2 = (c^2 - v^2)tau'^2 = c^2(1 - v^2/c^2)tau'^2= c^2tau'^2/gamma^2.

    Replace that v in the linear term with beta*c
     
  13. Oct 16, 2006 #12
    l^2cos^2(theta) - 2beta*c*tau'*l*cos(theta)gamma + [l^2{1-cos^2(theta)}]gamma= c^2tau'^2

    l^2cos^2(theta) - 2beta*c*tau'*l*cos(theta)gamma + [{l^2-l^2cos^2(theta)}]gamma= c^2tau'^2

    What do I need to do to get the below?
    At some point, I need to combine the cos^2 terms.

    "quadratic
    0 = x^2 + bx + c where b = 2*l*beta*cos(theta), c = l^2[1-beta^2cos(theta)^2] and x = c*tau'/gamma. (-> 0= (c*tau'/gamma)^2 + 2*l*beta*cos(theta)(c*tau'/gamma) + l^2[1-beta^2cos(theta)^2] )

    a=1?

    Use the quadratic formula."

    *I still need to get the above but I'm going to insert things into the quadratic formula now.

    http://mathworld.wolfram.com/images/equations/QuadraticEquation/equation5.gif [Broken]

    c*tau'/gamma= (-2*l*beta*cos(theta)) +/- [(2*l*beta*cos(theta))^2-4(1){l^2[1-beta^2cos(theta)^2]}]^(1/2)) / (2(1))

    You said, "The beta*cos(theta) contributions to the discriminant cancel leaving you with l*sqrt(1). Only the + root is kept. Solve for tau' and your done." ...?

    c*tau'/gamma = (-2*l*beta*cos(theta)) + l ?

    Again, the answer is tau' = (1-betacos(theta))gamma*l/c
     
    Last edited by a moderator: May 2, 2017
  14. Oct 16, 2006 #13

    OlderDan

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    When the file appears, you can see it all. Check my work!!!

    I can't get this LaTeX stuff to appear either
     

    Attached Files:

    Last edited: Oct 16, 2006
  15. Oct 16, 2006 #14

    OlderDan

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    Yes you do. They are both part of the "c" term
     
  16. Oct 17, 2006 #15
    Last edited: Oct 17, 2006
  17. Oct 17, 2006 #16

    OlderDan

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    I started by saying the distance from the origin to the position where the pulse is received is c*tau'. The unlabeled diagram (in S') shows this as the red arrow. The black arrow is v*tau'. The black slanted line is the separation of O and P when the pulse leaves O. Its components are l*cos(theta) and l*sin(theta). The radical is the sum of the squares of the components of c*tau'
     
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