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Finding Lorentz boost speed

  1. Feb 16, 2015 #1
    1. The problem statement, all variables and given/known data

    In the inertial frame of observer A two events occur at the same position a time 10 ns apart. In the frame of the observer B moving with respect to RA, one event occurs 1m away from the other. What is the difference in time between the two events in B's frame.
    Solve by finding the Lorentz boost speed v that connects the two frames.

    2. Relevant equations

    Lorentz transformations:

    T = γT'
    L = L' / γ

    [t] = [vγ γ][t']
    [x] [γ vγ][x'] <--- supposed to be the Lorentz transform matrix

    3. The attempt at a solution

    I've attempted this a couple of different ways:

    First, in B's frame:

    T' = 10^-8 , X' = 0m
    T = ? , X = 1m

    If I attempt to use the length contraction formula to find γ, I find γ = 0 as X' = 0
    So then I don't have enough info to find T. So i figured this can't be right!

    Next, I tried using the transformation matrix

    [t] = [vγ γ][t']
    [x] [γ vγ][x']

    Plug values above, and I find I get stuck with an equation with loads of v's that i can't simplify.

    Any help??

    Thanks
    [
     
  2. jcsd
  3. Feb 16, 2015 #2

    TSny

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    Note that the time interval between the two events is 10-8 s in frame A, not frame B. For frame B, the time between the two events, ΔT', is initially unknown.

    The two events occur at the same place in A.

    According to frame B, how far does frame A move during the time ΔT'?
     
  4. Feb 16, 2015 #3

    TSny

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    I think I misinterpreted which frame you are calling the primed frame. From the quote above, I first thought you were taking B to be the primed frame.

    However, what you wrote makes more sense if you are taking A to be the primed frame. It's always a good idea to be very explicit about which frame is the primed frame.

    Anyway, there are two events. So, I'm not sure which event you are referring to when you write T' = 10^-8, X' = 0. Or are you referring to intervals of time and distance here?
     
  5. Feb 16, 2015 #4
    Ok, fair enough, I'm taking A to be the primed frame.

    If we call the two events P and Q, we have the following information given to us:

    tp = tp' = 0
    xp = xp' = 0

    So I placed event P at the origin of my spacetime diagram.

    tq is unknown, tq' = 10-8 secs
    xq = 1m , xq' = 0m

    Frame A would move distance x = v*ΔT in observer B's reference
     
  6. Feb 16, 2015 #5

    TSny

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    OK, that clears it up nicely.

    This might be better written as Δx = v*ΔT. Do you know the value of Δx? Can you relate ΔT to ΔT' using γ?
     
  7. Feb 16, 2015 #6
    Yes,

    v = Δx / ΔT = 1 / ΔT'γ

    ΔT' = 10-8

    10-8v = 1/γ

    10-8v = (1 - v2)1/2

    10-16v2 = 1 - v2

    10-16v2 + v2 = 1

    v(1 + 10-8) = 1

    v = 1 / (1 + 10-8) = 0.9999999...

    Doesn't seem right to me
     
  8. Feb 16, 2015 #7
    I worked it out earlier using another method (invariance of spacetime distance), but I'd like to get the correct answer with the one above too.

    ds'2 = cdt'2 - dx'2

    ds'2 = c(10-8)2 - 0

    ds'2 = 3*108 * (10-8)2 = 3*10-8

    ds'2 = ds2

    ds2 = cdt2 - dx2 = cdt2 - 1

    3*10-8 + 1 = cdt2

    dt = [(3*10-8 + 1) / 3*108]1/2

    dt = 5.77*10-5 secs

    Does that look right as well? I'm not getting the answers the match up which is frustrating
     
  9. Feb 16, 2015 #8

    TSny

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    You left out the c2 in γ.
     
  10. Feb 16, 2015 #9

    TSny

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    Should the c be squared on the right side?
     
  11. Feb 16, 2015 #10
    Getting somewhere, I've been trying to use c=1 units but seems like they are confusing everything.

    10-16v2 = 1 - v2/c2

    v2(10-16 - 1/c2) = 1

    v = 1.06*108 = 0.353c

    Now seems like my other method is wrong.

    That's the equation I have down in my notes, what should it be? :S
     
  12. Feb 16, 2015 #11
    Ah typo,

    should be v2 = 10-16 + 1/c2

    Which gives v = 17320 ms-1

    That agrees with the value I got from the invariable spacetime distance method.
     
  13. Feb 16, 2015 #12

    TSny

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    I don't think this is what you really meant to type. You had
    You are correct that the minus sign on the left should be a plus sign in the second equation above..

    You should get a much larger value for v.
     
    Last edited: Feb 16, 2015
  14. Feb 16, 2015 #13
    Yeah, I'm being dopey, should be 1/(10-16+1/c2)

    Gives me v = 0.316c
     
  15. Feb 16, 2015 #14

    TSny

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    Looks good.
     
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