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Homework Help: Lorentz Transformations problem

  1. Jun 5, 2014 #1
    Hello all,

    I have an exam on Monday and am having trouble with this problem, any help would be greatly appreciated!

    Q: A straight stick of length L' is at rest in the moving S' frame. The stick appears to have length L in the S frame. The S' frame is moving at a velocity √(2/3) c. Calculate the following quantities:

    gamma = ______
    Δx' = ______L
    Δy' = ______L

    And calculate the length of the stick L' as observed in the S' frame.

    ΔL' = ______L

    And at what angle (with respect to the x'-axis) is the stick L' observed to be in the S' frame?

    angle' = _____degrees

    Relevant equations:

    1/Sqrt[1 - (β])^2]
    Δx' = gamma(Δx - vΔt)

    My attempt:

    the gamma factor is = 1/√[1 - (√[2/3])^2] which is 1.73

    then I use Δx' = gamma * x since Δt = 0

    so Δx' would be gamma/2 in units of L? - (divided by 2 due to the nature of a 30 60 90 triangle?) so Δx' = .865 L?

    Then Δy' is equal to √(3)/2 L because there is no contraction of length in the y direction because the motion is along the x-axis, so √(3)/2 is the only factor applied due to the nature of a 30 60 90 triangle.

    as for the next two im sort of lost.

    Thanks for any help.

    Attached Files:

    Last edited: Jun 5, 2014
  2. jcsd
  3. Jun 5, 2014 #2


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    Homework Helper
    Gold Member

    Hello Meekay,

    Welcome to Physics Forums! :smile:

    I'm a little confused about the delta on L'. Should that just be L'?

    Ignoring minor rounding differences, that looks reasonable to me. :approve:

    That also looks correct. :approve:

    The Pythagorean theorem should come in useful. Follow that up with your favorite inverse trigonometric function (arctan, for example). :smile:

    [Edit: By the way, if you're clever and keep all answers in terms of fractions and roots, the last answer -- the angle -- should be obvious even without a calculator.]
    Last edited: Jun 5, 2014
  4. Jun 5, 2014 #3
    Thank you for your your reassurance and help. And yes it is L' and not a delta L'. I got a bit carried away with deltas I suppose. And wow that last part was very simple, I feel pretty dumb now, I was trying to use a relativistic type equation.

    So i got L' = 1.22 L . If im right, that does make sense because the stick is at rest in the S' frame and its length is observed to be contracted in the S frame.

    And for the angle I got ~ 45 degrees using sin^-1(delta y'/L')

    Thanks again for the help!
  5. Jun 6, 2014 #4


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    Gold Member

    'Looks good to me! :smile:

    [Edit: btw, if you keep things in terms of fractions and roots, you'll find that the angle is not just approximately 45 deg, it's exactly 45 deg.]
  6. Jun 6, 2014 #5
    Awesome, thanks. And okay I gotcha, I will next time. I need to brush up on my trig.
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