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Finding the contraction formula using proper time

  1. Apr 30, 2005 #1
    If a passing space ship sends a beam from the back to the front in their time of L/c andthen immediately the beam is reflected from front to back, can an stationary observer using proper time determine the length contraction formula?

    I don't see how to set this up. Thanks for any help.
     
  2. jcsd
  3. Apr 30, 2005 #2
    Let S' be the frame in which the ship is stationary, and let S be the frame where the ship's speed is v.

    Define events:
    E1, the light flashes
    E2, the light is reflected

    So the coordinates for E1 are:
    x' = x = 0 and t'= t = 0

    and the coordintates for E2 are:
    x' = L and t' = L/c
    x = Y(L +BL) = YL(1+B) and t = Y(L/c + BL/c) = YL/c(1+B)

    (Y = gamma, B = beta)

    So in S (the unprimed frame):

    The distance between the two events is YL(1+B). But the ship moved between the time of the events a distance of vt. So the distance between the events is the length of the ship + vt. So:

    the length of the ship (as measured in S) = x - vt
    = YL(1+B) - YLB(1+B)
    = YL*(1-B)*(1+B)
    = L/Y
     
  4. May 1, 2005 #3
    That is all well and good, but the problem was in the book prior to defining the Lorenz Transformation. The idea is that knowing about time dialation, can you derive the formula for length contraction? The space ship involves three events. The light beam is sent from the back, it reaches the front end of the ship and is reflected, it returns to the back of the ship.
     
  5. May 1, 2005 #4

    robphy

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    Here's a proof using a moving simplified Michelson-Morley apparatus (two light clocks).

    Along the transverse direction, you will derive the time-dilation effect.
    Along the longitudinal direction, you argue like this:
    in order for the round-trip time to be the same in both directions (as required by the principle of relativity when compared to a similar apparatus at rest), the length along the longitudinal direction must be contracted.
     
  6. May 1, 2005 #5
    robphy,

    But he says you can't use the Lorentz Transforms. And the "principle of relativity" is necessary and sufficient for the LTs being true. So you can't use the principle of relativity.
     
  7. May 1, 2005 #6
    robert Ihnot,

    Your two posts don't seem to be asking for the same thing. In the first one you don't say anything about knowing the time dilation formula. And you never say whether we can assume the speed of light is constant or what (if anything) the stationary observer measures or finds out about the space ship experiment.

    You need to give us the EXACT wording of the problem in your book and any assumptions that we can or can not make.
     
    Last edited: May 1, 2005
  8. May 1, 2005 #7

    robphy

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    "Special Relativity" implicity involves Lorentz Transformations.

    Using the LTs explicitly is one way to obtain results in SR.
    However, at this point, the student does has not seen the LTs explicitly.
    So, using a thought experiment (as I've described), one can obtain the length-contraction effect using algebra and geometry [which is consistent with using the LTs, when it is introduced].
     
  9. May 2, 2005 #8
    jdavel: robert Ihnot,

    Your two posts don't seem to be asking for the same thing. In the first one you don't say anything about knowing the time dilation formula. And you never say whether we can assume the speed of light is constant or what (if anything) the stationary observer measures or finds out about the space ship experiment.

    You need to give us the EXACT wording of the problem in your book and any assumptions that we can or can not make.


    The author, Leo Sartori has already gone into the use of clocks and that the speed of light is observed as a constant by all observers. He makes the point that knowing time dilation, one can deduce length contraction.

    I might add it seems then that robphy: Along the transverse direction, you will derive the time-dilation effect. Along the longitudinal direction, you argue like this:in order for the round-trip time to be the same in both directions (as required by the principle of relativity when compared to a similar apparatus at rest), the length along the longitudinal direction must be contracted. This sounds like the right idea, but I would like more detail.

    Here is the problem: "A spaceship whose length in its own rest frame is L(0) moves at velocity V relative to the earth. Let L be the length of the spaceship as measured in the earth's rest frame, S.
    (A) A light pulse emitted at the rear of the spaceship (event E(1)) arrives at the front (event E(2)). In the spaceship frame, S', the time interval between E(1) and E(2) is t'-t'(1) = L(0)/c. Find the time interval between the same two events in frame S, in terms of L, V, and c. (Note that this time interval is not proper* in either frame. Picture** is mentioned.)

    (B) The light pulse is reflected and arrives at the rear of the spaceship (event E(3)). Find the time interval between E(2) and E(3) in frame S.

    (C) Applying a proper* time argument to the interval between E(1) and E(3), show that L and L(0) are related by the lenght contraction formula L= L(0)/Y" (where Y represents gamma as used in jdavel's example.)

    *"An interval between events that happen at the same place is called a proper time interval. Any other interval is called improper.

    ** The picture refers to three diagrams describing the rays in Michelson's experiment. One horizontal, one vertical, and one along a diagonal.
     
    Last edited: May 2, 2005
  10. May 2, 2005 #9
    robertIhnot,

    Ok, I see what he wants.

    You can assume that the speed of light (in either direction) as measured in either S or S' is always the same, namely it's c.

    So, for part a) The time interval betwen E1 and E2 as measured from the earth frame will just be the distance the beam travels divided by the speed of the beam. So, the denominator of the fraction is, of course, c. Now you get to do the hard part. What's the distance (measure from earth) that the light beam travels to get from the back to the front of the ship?

    Hint: First it has to get to where the front of the ship was when it started traveling forward from the back. How long will that take? Then it has to go a little further because the front of the ship moved during that time.
     
  11. May 2, 2005 #10
    jdavel: Now you get to do the hard part. What's the distance (measure from earth) that the light beam travels to get from the back to the front of the ship?

    The beam would have to travel L (as measured in the Earth Frame + Vt, where t is the time it takes the beam to get there. So, starting from 0 in the stationary system, we have ct = L+Vt, which gives us
    t =L/(c-v).
     
    Last edited: May 2, 2005
  12. May 2, 2005 #11
    robertIhnot,

    That would be hard if I didn't have your answer!

    But as the light goes back, the back of the ship is coming forward to meet it at just the speed it was going away from the beam in your calculation. You found the time between E1 and E2 (measured from the earth) is:

    t(1,2) = L/(c-v)

    so

    t(2,3) = L/(c+v)

    Now it's your turn again.

    What's t(1,3) as measured in the earth frame?
     
  13. May 3, 2005 #12
    The notation is a little confusing, but the time of the total trip is L/(c-v) + L/(c+v)=
    [tex]\frac{2Lc}{c^2-v^2}[/tex]
     
  14. May 3, 2005 #13
    robertIhnot,

    Sorry about the notation; t(n,m) is the time (measured in S, the earth rest frame) between events En and Em.

    So you got:

    t(1,3) = 2Lc/(c2 - v2) (clumsy notation again, but you know what I mean!)

    Dividing the top and bottom by c2 first will make the next step easier:

    t(1,3) = (2L/c)/(1 - v2/c2)

    Your turn. Divide this equation by the equation for t'(1,3) (the proper time between E1 and E3, measured on the space ship) and look what you get!
     
  15. May 3, 2005 #14
    The time measured on the ship is 2L(0)/c, so division yields [tex]\frac{L(1-(v/c)^2)}{L_0}[/tex]
     
    Last edited: May 3, 2005
  16. May 3, 2005 #15
    robertIhnot,

    So that's equal to the ratio of the two time meausrements. But remember, the problem said that "....knowing time dilation, one can deduce length contraction". So rewrite the ratio of the times using the time dilation equation and set that ratio equal to what you just calculated.

    Edit: I think you have an error in your ratio. Either the L and L(0) need to be reversed or the other factor needs to be in the denominator. Right?
     
    Last edited: May 3, 2005
  17. May 3, 2005 #16
    Yes, thank you. I did have that inverted some how:[tex] \frac{2L/c)}{1-(v/c)^2} *\frac{1}{2L_0/c} =\frac{L/L_0}{1-(v/c)^2}[/tex]

    Since the ratio T/T' = [tex]1/\sqrt{1-(v/c)^2}[/tex] We can set the two terms equal giving [tex]L=L_0\sqrt{1-(v/c)^2}[/tex]

    Thus going back to the original problem: "A spaceship whose length in its own rest frame is L(0) moves at velocity V relative to the earth. Let L be the length of the spaceship as measured in the earth's rest frame, S."

    Thus, the interpretation is that we measure this same spaceship as shorter than measured in its own rest frame.

    jdavel, thanks very much for your help!
     
    Last edited: May 3, 2005
  18. May 4, 2005 #17
    robert,

    You're welcome.

    One last thing. The calculations you did required that the speed of light you used to find the time interval between E1 and E3 in the earth frame was the same as the one used to find the time interval in the spaceship frame. In both cases it was c. Be sure you see how important that was in getting the result; just assuming length contraction wouldn't have been enough.

    jdl
     
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