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Finding Relative Velocity

  1. Sep 15, 2016 #1
    1. The problem statement, all variables and given/known data
    An event occurs in S at x = 6 x 10^8 meters and S' at x' = 6 x 10^8 meters. Find the relative velocity of the systems. Assume that S refers to an inertial system (x,y,z,t), and S' refers to an inertial system (x',y',z',t') moving along +x axis with speed v relative to S. The origins coincide at t = t' = 0.

    2. Relevant equations
    c = 3 x 10^8
    Lorentz Factor = (1-(v/c)^2)^(-1/2)
    x' = 1/LF ( x - vt)
    t' = 1/LF ( t - vx / c^2)

    3. The attempt at a solution
    I am having trouble because x' - x = 0 and t = 0 so the equations aren't working out.
    x' = 1/LF ( x - vt )
    1 = (1-(v/c)^2)^(-1/2) (1 - v(0))
    1 = (1-(v/ 3x10^8)^2)^(-1/2)
    (1-(v/ 3x10^8)^2)^1/2 = 1
    1-(v/ 3x10^8)^2 = 1
    -(v/ 3x10^8)^2 = 1 - 1 = 0
     
  2. jcsd
  3. Sep 16, 2016 #2

    Orodruin

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    Nobody says the event occurs at t = 0 or t' = 0.

    Edit: Of course, there is no unique solution to this problem. There exists a solution for every possible inertial frame as ##x = x_0## and ##x' = x_0## defines two non-parallel lines in Minkowski space regardless of the relative velocity ##v## (unless ##v = 0## for which the statement is just trivial). In order to fix the relative velocity, you need to know the time of the event in at least one of the frames.
     
    Last edited: Sep 16, 2016
  4. Sep 16, 2016 #3
    Hi Orodruin. Thanks for the reply.
    The time for t' is 1 seconds. And then it asks for when t' is 4 seconds.

    Could you help me set this up?
     
  5. Sep 16, 2016 #4

    Orodruin

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    What does the inverse of the Lorentz transformation tell you?
     
  6. Sep 16, 2016 #5
    The inverse (1-(v/c)^2) ^ 1/2 and is always less than 1 so there is time dilation? The clock should run slower for t' than for t.
     
  7. Sep 16, 2016 #6

    Orodruin

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    That is not the inverse Lorentz transformation. That is the Lorentz factor.
     
  8. Sep 16, 2016 #7
    The inverse of the Lorentz transformation tells me that the relative velocity of the systems are equivalent?
     
  9. Sep 16, 2016 #8

    Orodruin

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    No, I am asking you to use the inverse transform to relate the coordinates of the events.
     
  10. Sep 16, 2016 #9
    x = (x' + vt') / (1-(v/c)^2)^1/2

    6x10^8 = (6x10^8 + v(1)) / (1-(v/c)^2)^1/2

    (6x10^8)(1-v/(3x10^8))^1/2 - 6x10^8 = v

    How do I determine the t for the event occurring in S?
     
  11. Sep 16, 2016 #10
    t = (t' + vx'/c^2) / (1-(v/c)^2)^1/2

    t = (1 + (6.10^8)v / c^2) / (1-(v/(3x10^8))^1/2
     
  12. Sep 16, 2016 #11

    Orodruin

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    You dont nees to. You have one equation and one unknown.

    Also, do not forget the units of all the quantities. Units are important.
     
  13. Sep 16, 2016 #12
    Thanks a lot for your time.

    If I solve for v, is that the answer then?
    v = (6x10^8)(1-v/(3x10^8))^1/2 - 6x10^8
    so I have to get the v on the right to the left.
     
  14. Sep 16, 2016 #13
    You have one algebraic equation in one unknown, v. So solve the equation. This is where the math comes in.
     
  15. Sep 16, 2016 #14
    I see. Haven't done math in a while too.
    I will give it a try. Thanks.
     
  16. Sep 16, 2016 #15
    v = (6x10^8) (1-(v/c)^2)^1/2 - 6x10^8
    I can't solve for v because squaring both sides cancel them out.
    Where did I go wrong?
     
    Last edited: Sep 16, 2016
  17. Sep 16, 2016 #16
    My advice to you is to solve the equation algebraically first, and then plug in the numbers:
    $$x=\frac{(x'+\frac{v}{c}(ct'))}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$$
    Solve for v/c.
     
  18. Sep 16, 2016 #17
    Ah okay!
     
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