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Special Relativity Question

  1. Jan 14, 2006 #1
    Hello.
    I am doing a special relativity question.
    I got a particle traviling at some speed v. It travels 20000 light years from reference of galaxy in 30 years from the reference of the particle.
    I need to find the speed of the particle.
    Well, i tried to use time dialation and length contraction formulas but i end up with 3 unknown and 2 equations.
    So after thinking a little more i made up another equation:
    30 years=Dialated length/velocity.
    Of course the units were correct years into seconds and so on.
    Then i bassicle get 2 v's, one inside the gamme and the other one on the bottom of the fraction, but the answer i get is C. So i made a mistake somewhere...any ideas?
    Thx
     
  2. jcsd
  3. Jan 14, 2006 #2

    robphy

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    Can you show some details of your work?
    If you can think trigonometrically (using rapidities), the answer and its method of solution should become obvious.
     
  4. Jan 14, 2006 #3

    Doc Al

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    Nothing wrong with this approach. Can't tell where your mistake was unless you show more details. (I assume by "dilated length" you actually mean contracted length: the galactic distance traveled as measured in the particle frame.)
     
  5. Jan 14, 2006 #4
    well here is what i got:
    Let say that time taken to go the distance in the frame of reference of a particle is T. and it is in seconds.
    The contracted distance is not known to us but we do know that proper distance is D in meters.
    Then we have
    T=(D/sqrt(1-v^2/c^2))/v
    then:
    vT=D/sqrt(1-v^2/c^2)
    so we square it all
    we got
    v^2*T^2=d^2 - d^2*c^2/v^2
    now...
    we need to solve for v, but to do that i will need to use power of 4....and i think that that is a good indicator that i am wrong....if anyone seeas a mistake in above logic plz tell me other i will solve it out...but that seems about right that i do have a mistake in there.
     
  6. Jan 14, 2006 #5
    scratch my last response i made a mistake in there, but i am stil getting c....
    here is how
    assume t being time, and d being proper distance.
    t=d*gamma/v
    t^2v^2=d^2*(1-v^2/c^2)
    t^2v^2+d^2v^2/c^2=d^2
    v^2(t^2+d^2/c^2)=d^2
    now since t^2 is so much smaller then d^2/c^2 it is insignificant
    then when u solve u get c for the speed.
    I think i might have messed something up
    note that t is the time from the perpective of the partice, d is the distance from the inertial reference,and i need to know velocity of the particle.
    Any ideas where i made a mistake?
    note that proper distance is 23 000 light years and relative time is 30 years, that is why it is insignificant and does nothing
     
  7. Jan 14, 2006 #6

    lightgrav

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    well, Tv = d/gamma would lead to T^2 v^2 = d^2 ( 1 - v^2 / c^2) ...

    you're deleting the only epsilon that distinguishes its speed from c.
    (yea, it is really fast, apparently, but not quite that fast!)
    just keep a few more digits.
     
  8. Jan 14, 2006 #7
    well sig figs will not affect it at all, i mean the diffrence between the 2 values is something like 10 ^10....there is no way to keep sig figs there....u sure that that is correct?
    Basicly here is the initial problem, maybe i interpreted it incorrecly?
    a particle travels 23 000 light years proper distance in 30 years from the particles frame of reference.
    i need to find it;s speed, when i did as i said above it gives me practically almost c, without any use for sig figs since they are at a 10 ^10 off the mark.....
     
  9. Jan 14, 2006 #8

    lightgrav

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    closer to 10^6 . write v/c = d(1-e)
     
  10. Jan 14, 2006 #9
    how did u get that v/c equation? and what is e?
    sry for my stupid question:(
     
  11. Jan 14, 2006 #10

    lightgrav

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    that was a suggestion for you to try ... e , epsilon? is 0 << e << 1 ,
    so you can expand sqrt(1-e) = 1 - e/2 , that sort of thing . yours is tc/d .
     
  12. Jan 14, 2006 #11
    how did u get there exectly?, cuz i don;t see how u can isolate v/c by itself......
    i mean the sqrt(1-e) = 1 - e/2 that is an estimate right?
    and why do u say e=tc/d should it not be v^2/c^2?
     
    Last edited: Jan 14, 2006
  13. Jan 15, 2006 #12

    Doc Al

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    Here's how I'd do it:
    [tex]L_0 / \gamma = v t[/tex]

    Rewrite in terms of [itex]\beta = v/c[/itex]:
    [tex](L_0/c) \sqrt{(1 - \beta^2)} = \beta t[/tex]
    or:
    [tex](L_0/(ct)) \sqrt{(1 - \beta^2)} = (\beta)[/tex]

    Square:
    [tex](L_0/(ct))^2 (1 - \beta^2) = \beta^2[/tex]

    Now just solve for [itex]\beta[/itex], saving your approximations for the last step.
     
  14. Jan 15, 2006 #13

    robphy

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    Here's my trig/rapidity solution:
    cT=c(proper time)="hypotenuse of a [Minkowski]-right triangle"
    L=(distance between earth-at-rest and galaxy-at-rest in earth frame)="opposite leg"=cT*sinh(rapidity)=[itex]cT*\sinh(\theta) [/itex]
    [itex]\beta=\tanh(\theta) [/itex]
    So,
    [tex]\beta=\tanh\left(\sinh^{-1}\left(\displaystyle\frac{L}{cT}\right)\right) [/tex]
    [add]
    Spacetime Trigonometry:
    [itex]\beta=\tanh(\theta) [/itex]
    [itex]\gamma=\cosh(\theta) [/itex]
    [itex]\beta\gamma=\sinh(\theta) [/itex]
    Quiz: what is the interpretation of [itex]\exp(\theta)[/itex]?
    [/add]
     
    Last edited: Jan 15, 2006
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