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Reletivity question

  1. Sep 25, 2008 #1
    1. The problem statement, all variables and given/known data
    An event has spacetime coordinates (x, t) = (1200 m, 2.0 μs) in reference frame S. What are
    the event’s spacetime coordinates
    (i) in reference frame S′ that moves in the positive x-direction at 0.8c and
    (ii) in reference frame S′′ that moves in the negative x-direction at 0.8c?


    2. Relevant equations
    x'=[tex]\gamma[/tex]x-vt
    t'=[tex]\gamma[/tex]t-vx/c^2
    gamma=1/root(1-beta^2)
    beta=v/c

    3. The attempt at a solution
    ok this is wat i did since v=0 i did.
    x'=[tex]\gamma[/tex]x=(1200/(1-.8^2))=2000m
    t'=[tex]\gamma[/tex]t=(2x10^-6/(1-.8^2)=3.33[tex]\mu[/tex]s

    x"=x-[tex]\gamma[/tex]x=1200-x'=-800m
    t"=t'=3.33[tex]\mu[/tex]s

    im not sure about wat ive done just having doubts, could someone plz confirm, or help me fix my mistake... thx
     
    Last edited: Sep 25, 2008
  2. jcsd
  3. Sep 25, 2008 #2
    anyone have any ideaS?
     
  4. Sep 25, 2008 #3
    anyone?
     
  5. Sep 25, 2008 #4
    please be patient.
    most of it looks right, but I think you may have made a mistake when you were solving for x".

    You also have your basic equations wrong. Check here:
    https://www.physicsforums.com/showthread.php?t=110015
    Your lucky that it didn't affect your answers though because v=0.

    for example it should be
    x'=(gamma) (x-vt)
    NOT
    x'=(gamma)x - vt
     
  6. Sep 25, 2008 #5

    gabbagabbahey

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    Gold Member

    For part (i), v=+0.8c while for part (ii) v=-0.8c ...why did you have v=0 before?
     
  7. Sep 25, 2008 #6
    ya i know i stuffed that part up... well becasue in my mind v=0 coz the event isnt moving?? and beta=0.8c???
     
  8. Sep 25, 2008 #7
    coz if i change it to x'=gamma(x-vt)=(1200-((.8*3x10^8)*2x10^-6))/0.6=1200m????
    if thats wat ur ur suggesting gabba.. beat=v/c=>v=beta*c=2.4x10^8m/s?
     
  9. Sep 25, 2008 #8
    and t' would equal -2us?
     
  10. Sep 25, 2008 #9
    na v has to equal 0... so x'=1200/0.6=2000m
    t'=2x10^-6/.6=3.33us
    im not 100% sure with x" and t"
    but it makes sense to me if x"=x-x'=-800m
    and because the direction doesnt effect the time then the time in both reference fromes should be equal.. thus t"=t'=3.33us??

    can anymore ppl confirm this??
     
  11. Sep 25, 2008 #10

    Doc Al

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    Staff: Mentor

    These parts look OK to me.

    Take the same approach for S''. What's the only thing that changes in the Lorentz transformations?
     
  12. Sep 25, 2008 #11
    oh ok then ummm the only thing that changes is v??
     
  13. Sep 25, 2008 #12
    so x"=x-x'=1200-1200=0m
    but does t" still equal t'=2us?
     
  14. Sep 25, 2008 #13
    can i ask doc al, y do u think that v= something when the event isnt moving?? coz its the complete opposite of what me and perillux think..
     
  15. Sep 25, 2008 #14

    Doc Al

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    Staff: Mentor

    Yes. If for S', v = +.8c, what will v equal in the formulas for S''?

    I have no idea where you got these values.
     
  16. Sep 25, 2008 #15

    Doc Al

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    Staff: Mentor

    Events don't move. v is the speed of the frame, not the event.

    You need to read up on the Lorentz transformations and what they mean.
     
  17. Sep 25, 2008 #16
    sorry i kinda just thought wat would make sense to me... so for x" and t" instead of beta=1/root(1-.8^2) its -.8? but then doesnt that negative cancel with the square??

    so x"=(1/root(1--.8^2))(1200-(0.8*3x10^8)(2x10^-6))=562m?? i dotn think thats rite..
     
  18. Sep 25, 2008 #17

    Doc Al

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    Staff: Mentor

    The value of gamma (not beta!) doesn't change, since the v is squared. But that's not the only place that v appears.
     
  19. Sep 25, 2008 #18
    ok i found another eqn in book under lorentz... im sorry my book is really bad, and just formulas and not words... ct'=x'=gamma*t(c-v)=2x10^-6(3x10^8-(0.8*3x10^8))/0.6=200m??? y am i getting different answers for the same things??
     
  20. Sep 25, 2008 #19
    ok yes so if x' and t' are correct then all x" and t" is....
    x"=(1200-((-.8*3x10^8)*2x10^-6))/0.6=2800m
    t"=(2x10^-6-((-.8*3x10^8)*1200)/(3x10^8)^2))/0.6)=8.66us?
     
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