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Question on both Time Dilation and Lorentz Contraction

  1. May 15, 2008 #1
    Hi guys,

    There's a couple questions that I've been trying to figure out that I have not been able to make much progress on.

    1. Time Dilation

    the equation states that T' = Lorentz Factor * T0
    So the Lorentz factor will always be > 1 because v < c

    So then, if, let's say, there's 2 events, A and B. There are 2 reference frames, a guy on earth and a guy moving in a spaceship at a very high speed.

    Event A: a stopwatch on earth displays t = 0 s
    Event B: a stopwatch on earth displays t = 30 s

    So, on earth, T0 = 30
    Let's assume the lorentz factor = 2

    Doesn't this mean, then, according to the equation, that the guy on the spaceship measures 60 seconds between events A and B? Doesn't this mean that in the duration of 30 seconds passing on earth, 60 seconds has passed for the guy in the spaceship, meaning that the guy on the spaceship is aging faster than the guy on earth, completely counter to the established theory?

    2. Lorentz Contraction Derivation

    I understand that to derive the Lorentz contraction, you use

    l0 = x2-x1 reference frame of earth
    l = x2' - x1' reference frame of spaceship

    so then, since x1 = L (x1' + vt')
    x2 = L (x2' + vt')

    you can write x2-x1 = L (x2'-x1')
    So, l = l0/L, the Lorentz contraction formula.

    My question is, why couldn't you have also used the reverse set of equations

    x1' = L (x1-vt)
    x2' = L (x2-vt)

    to get x1'-x2' = L (x1-x2)
    and then, in this case, l = l0 * L, which is the exact opposite result?

    I think I'm probably thinking about this the wrong way, so if someone could give me some help that would be really appreciated.
  2. jcsd
  3. May 15, 2008 #2
    1] you plugged in for the wrong T's; the guy on the spaceship experiences 15 seconds.

    2] you can use (1) to derive 2. If the guy on the spaceship experiences half the time, and they both agree on his velocity (relative to light as they must), then the distance must be half as along.
  4. May 15, 2008 #3

    Doc Al

    User Avatar

    Staff: Mentor

    That's right.
    Realize that the effect is completely symmetric: Earth observers will measure 60 earth seconds for a spaceship clock to measure 30 seconds.

    Don't confuse this with the "twin paradox", where the symmetry is broken.

    Again, length contraction is completely symmetric. In the first case, your "stick" is in the earth frame and is being measured by the spaceship observers (who measure the position of the ends at the same time)--the stick is measured to have contracted.

    In the second case you have the stick in the spaceship frame, so it is earth observers who see it contracted.
  5. May 15, 2008 #4
    At the same time the observer on Earth sees the stopwatch of the spaceship record 30 seconds when his own watch records 60 seconds. Each sees the other's clock as running slower. This is the origin of the "twin's paradox" which is not really a paradox and can be resolved. There are many many threads on the twin's paradox here in this forum.

    Lo = x2 - x1 Length of rod on Earth as measured by Earth observer
    L' = x2' - x1' Length of rod on Earth as measured by Spaceship observer

    Gamma factor: y = sqrt(1-v^2/c^2) = 2

    Lorentz tranformation equation: x' = (x-vt)*y as per ref http://en.wikipedia.org/wiki/Lorentz_transformation

    L' = (x2' - x1') = ((x2-vt2) - (x1-vt1))*y -->

    L' = (x2' - x1') = ((x2-x1) - vt1 + vt2)*y -->

    In the Earth Frame say x2 = 8 and x1 =2 and t2=t1=3 then

    L' = (x2' - x1') = (8-2 -3+3)*y = 6*y = 12

    which is an incorrect result! What went wrong? The ends of the rod were measured simultaneously in the frame of the Earth observer at t1=t2=3 but those two events were not simultaneous in the frame of the spaceship observer. The relativity of simultaneity tells us that clocks in the other frame are out of sync according to to their displacement from the origin proportional to t+x*v/c^2 so we can rewrite the length contraction formula as:

    L' = (x2' - x1') = ((x2-v(t2+x2*v/c^2)) - (x1-v(t1+x1*v/c^2)))*y -->

    L' = (x2' - x1') = ((x2(1-v^2/c^2)-vt2) - (x1(1-v^2/c^2)-vt1))*y -->

    L' = (x2' - x1') = ((x2/y -vt2) - (x1/y -vt1)) -->

    L' = (x2' - x1') = ((x2-x1)/y -vt2 +vt1)) -->

    L' = (x2' - x1') = (x2-x1)/y = (8-2)/2 = 3

    which is now the correct result. The reason the inverse equation cannot be applied directly is that two events that are simultaneous in one frame are not simultaneous in other frame, so we cannot simply assume vt1=vt2 and cancel them out when looking from another frame.
  6. May 15, 2008 #5
    wow, thanks for the responses. they really cleared it up.
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