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Relativistic Spaceship moving past Earth

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data
    A relativistic spaceship is moving "horizontally" past Earth at 99% speed of light (c), and the water in the ship's swimming pool rises at 5 m/s. What is the tilt of the water surface with respect to the horizontal, AS OBSERVED FROM EARTH?


    2. Relevant equations



    3. The attempt at a solution
    I tried using simple trigonometry to find the angle between the velocity of the ship and the velocity of the water, but cannot translate this to find an angle between the acceleration of the earth vector and the velocity of the water.
     
    Last edited by a moderator: Feb 2, 2012
  2. jcsd
  3. Feb 2, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi PermaTrashed! Welcome to PF! :smile:

    I'm not sure that you're seeing the issue here.

    The astronauts are pouring water into the pool, and in their frame the water level is a horizontal line moving vertically at 5 m/s.

    So write the equation for the water surface in x,y,z,t then convert that into x',y',z',t' and put t' = 0 to find what the water looks like at any fixed time on Earth. :wink:
     
  4. Feb 2, 2012 #3
    Wow I do have it wrong then, I totally thought the water rising was due to the gravitational field of Earth, it doesn't say anything about astronauts pouring water into it, thanks a lot!
     
  5. Feb 2, 2012 #4
    if the direction of the velocity of water is parallel to the ships velocity, you will need relativistic velocity addition, which is by no means easy to derive:
    http://en.wikipedia.org/wiki/Velocity-addition_formula
    otherwise if the water moves perpendicular to the spaceship, the answer is simply 5 m/s
     
  6. Feb 2, 2012 #5
    But it's not asking for the velocity, it's asking for the tilt of the water WRT the horizontal axis in earth's frame
     
  7. Feb 2, 2012 #6

    tiny-tim

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    yup! :smile: use the method i suggested …
     
  8. Feb 2, 2012 #7
    What I don't understand is that from reference of Earth, the tilt of the surface of water won't change, it still will be horizontal, just moving in the diagonal direction. Is this angle (the angle at which it travels diagonally) what I'm looking for???
     
  9. Feb 2, 2012 #8

    tiny-tim

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    No.

    y' = y, but t' ≠ t.

    What is the equation (in x y z and t) for the water surface in the spaceship frame?
     
  10. Feb 2, 2012 #9
    in the spaceship frame, y'=5.0t where t=ζ(t'+vx'/c^2)?? Then in the earth's frame x=ζ(x'+vt), then its just the angle between x and y?
     
  11. Feb 2, 2012 #10

    tiny-tim

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    Nooo, you're already confusing me :confused:

    in any frame, either use y and t, or use y' and t'.

    Start again.​
     
  12. Feb 2, 2012 #11
    Right, so y'=5.0m/s(t') in the spaceship frame, and then x=ζ(x'+vt)?
     
  13. Feb 2, 2012 #12

    tiny-tim

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    just write it y'=5t' …

    now translate y' and 5t' into x y and t :smile:
     
  14. Feb 2, 2012 #13
    Word, that makes more sense, so t'=ζ(t-vx/c2) and then I find the angle between y=5t' and x=ζ(x'+vt) then plug in t' and x'=0?
     
  15. Feb 3, 2012 #14
    before you can use the Lorentz transformations to find out the contraction of the length, you will need the relative velocity of the two objects, and for that you need the relativistic velocity addition i just wrote, although moving at 99% of the speed of light, you probably get something very similar as a result.
     
  16. Feb 3, 2012 #15

    tiny-tim

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    (just got up :zzz: …)
    (why are you using zeta (ζ ) instead of the usual gamma (γ)? :confused:)

    that's very confusing :redface:

    just write y' = 5t' as y = 5γ(t-vx/c2),

    then put t = constant (eg 0) :smile:
     
  17. Feb 3, 2012 #16
    Sorry I couldn't find the gamma key, so that all seems correct then? Thanks for all the help!
     
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