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Relativity q

  1. May 2, 2006 #1
    A spacecraft (Lo = 80 m) travels past a space
    station at speed 0.7c. Its radio receiver is on
    the tip of its nose. The space station sends a
    radio signal the instant the tail of the spacecraft
    passes the space station.

    (a) What is the length of the spacecraft in the
    reference frame of the space station?
    L=80(1-.7^2)^.5 = 57m (fine)

    (b) What is the time taken for the radio signal
    to reach the nose of the spacecraft,
    according to those on the space station?

    t= s/v = 57/c = 1.9x10^-7 s (They get 6.3x10^-7 s (consequential 'errors' result) which can be obtained by t= 57 / 0.3c - but this logic seems to contradict einstein's 2nd postulate since the signal must move relative to the spacecraft at c)



    (c) How far from the space station is the nose
    of the spacecraft when it receives the radio
    signal from the reference frame of the
    space station?

    S(t) = s(moved) + s(length of ship)

    s= 0.7c *(1.9x10^-7) + 57 = 97m
    their answer: 190m

    (d) What is the time taken for the radio signal
    to reach the nose, according to those on
    the spacecraft?

    t = s / v = 97 / c = 3.2x10^-7s

    their answer: 2.67x10^-7s

    What am i doing wrong?

    Thanks.
     
    Last edited: May 2, 2006
  2. jcsd
  3. May 2, 2006 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Correct.

    But in the time it takes the light signal to move 57 m. the rocket ship has moved.

    You have to use the Lorentz transformation here to determine the time in the space station frame:

    [tex]t = \frac{t' + \frac{vx'}{c^2}}{\sqrt{1 - v^2/c^2}}[/tex]

    where [itex]t' = 80/c[/itex] and [itex]x' = 80[/itex]

    [tex]t = \frac{80/c + \frac{.7c*80}{c^2}}{\sqrt{1 - .7^2}}[/tex] = 190/c sec.


    Again, use the Lorentz transformation:

    [tex]x = \frac{x' + vt'}{\sqrt{1-v^2/c^2}}[/tex]

    [tex]x = \frac{80+.7*80/c}{\sqrt{1-.7^2}}[/tex]

    [tex]x = 190 m.[/tex]


    Again, use the Lorentz transformation to relate space time coordinates in one frame to the other.

    [tex]t' = \frac{t - \frac{vx}{c^2}}{\sqrt{1 - v^2/c^2}}[/tex]

    where [itex]t = 190/c[/itex] and [itex]x = 190[/itex]


    [tex]t' = \frac{190/c - \frac{.7c*190}{c^2}}{\sqrt{1 - .7^2}}[/tex] = 80/c sec. = 2.67e-7 sec.

    AM
     
    Last edited: May 2, 2006
  4. May 2, 2006 #3

    Curious3141

    User Avatar
    Homework Helper

    No problems there.

    You're considering everything from the space station's perspective. Let's say the pulse hits the nose of the craft at time t. In that time t, the light pulse would've travelled a distance ct from the space station. The tail of the craft would've travelled 0.7ct from the station. Since the craft is 57 meters long from the perspective of the station,

    ct = 0.7ct + 57

    t = 57/(0.3c) = 6.3*10^(-7)s, as required.

    Look at what the answer for b) above. The required distance is simply ct, which is :

    d = ct = (c)(57/0.3c) = 57/0.3 = 190 meters.

    The simplest way to work this out is to consider that from the perspective of the craft's occupants, the light pulse is travelling at c towards the nose. The distance to be covered is L0 = 80 meters.

    Hence, t = 80/c = 2.67*10^(-7)s

    A more complicated way to do it would be to transform the time found in b) with the Lorentz transform, but this is needlessly complicated. And if you use the intermediate results, there's a significant round off error.
     
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