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Derive length contraction formula given a two system experiment.

  1. Feb 17, 2014 #1
    1. The problem statement, all variables and given/known data

    Show that the experiment depicted in Figure 2.11 and
    discussed in the text leads directly to the derivation of
    length contraction.

    Figure 2.11:
    HQc2xgs.png

    2. Relevant equations

    d=v*t

    Requested result: L=L0[itex]\sqrt{1-\frac{v^2}{c^2}}[/itex]

    3. The attempt at a solution

    In K the distance the light pulse travels is 2t1v+L and the total time for the pulse to return to its origin is that distance over c.

    In K' the distance the light pulse travels is 2L0 and the total time for the pulse to return to its origin is that distance over c.

    I can also say that t1=[itex]\frac{vt_1+L}{c}[/itex] then solve for t1 to substitute that into total time in K frame. At this point I'm not sure where to go or even if what I've done is useful. I am tempted to set the total times equal to one another and solve for L but I when I've tried this I can get it to look like it is supposed to although there are familiar pieces.

    Any suggestions? Thank you!
     
  2. jcsd
  3. Feb 18, 2014 #2

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    Since nobody has responded yet, I'll give it a helping try.

    If I may offer some guidance, there are a couple of things to keep in mind.

    First, one of your goals is to eventually find t2 as a function of L, c and v (without having t1 or another t2 in that same expression).

    You've already found the expression [itex] t_1 = \frac{v t_1 + L}{C} [/itex] which is good! :approve: But as you can see, there are two t1s in that equation, one on the left hand side and other on the right. Now, using algebra, solve that equation for t1. Eventually, and similarly, work your way to an equation that has t2 on the left hand side, and a function of L, c and v on the right hand side.

    One possible approach to doing this is to first find Δt1 and Δt2 (where Δt1 is equal to t1 -- you practically already have this). Then find

    t2 = Δt1 + Δt2

    That's one possible approach, but it's not the only way. Again, one way or another, find an expression for t2 as a function of L, c and v, without having t1 or another t2 in the same expression.

    And second: Don't forget that the time intervals between events as measured by Mary's clocks are not the same as the time intervals measured by Frank's clocks.

    The round trip time interval that Mary measures is 2L0/c. What is this time interval as measured by Frank's clocks?
     
  4. Feb 18, 2014 #3
    Ah great, thank you. I did end up actually solving for t1=[itex]\frac{L}{c-v}[/itex]

    t=2t1=[itex]\frac{2L}{c-v}[/itex]

    Also, yes, they measure the time between events differently. For some reason in my physics stupor last night I had the bright idea that setting them equal to one another would allow me to find that difference and somehow then relate that to the dilation of the distance. That's obviously not going to work.

    The closest I come doing that is L=L0(1-[itex]\frac{v}{c}[/itex]). Close but no cigar.
     
  5. Feb 18, 2014 #4

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    Very nice! Excellent. :smile:

    Wait, back up a little bit.

    The time, Δt1, that it takes for the light to reach the mirror, as measured by Frank, is greater than Δt2, the time it takes for the light to go from the mirror back to the front of the stick. (It's not as simple as multiplying Δt1 by 2 to get the total time interval. You must treat each direction separately.)

    In the forward case, the stick/mirror and the light are moving in the same direction, and that's why it takes longer. However, on the reverse path, the stick is moving in the opposite direction as the light. So it doesn't take very long for the light to travel back to the front of the stick.

    Yes, before the final step, you will have to invoke time dilation in Mary's frame. Mary is measuring t2' as her overall time interval, not t2. You'll have to express t2' in terms of t2 before all things are said and done. Once you have that expression you should be able to put everything together to form a relationship between L and L0.

    Yes, that's not quite right. Keep trying though, you're getting closer! :wink:
     
  6. Feb 18, 2014 #5
    Ohh! I was thinking we were interested in the time it took for the light to leave the source, hit the mirror, and return to the source. Silly me!

    So, the distance that light travels between t1 and t2 is L-vt2-vt1. I say this because t2 seems to be the total time from the pulse until it returns to the stick and not just the difference between t1 and t2. The time being that distance divided by c. The distance the light travels over the whole of t2 is L-vt2.

    After some algebra:

    t2=[itex]\frac{L}{c+v}[/itex]

    edit: made a mistake; I think this should be right.
     
    Last edited: Feb 18, 2014
  7. Feb 18, 2014 #6

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    I think you mean (if you are using the approach I suggested in a previous post), [itex] \Delta t_2 = \frac{L}{c+v} [/itex]

    where t2 = Δt1 + Δt2

    Δt2 is the time it takes for the light to go from the mirror back to the front of the stick.

    t2 (not to be confused with Δt2) is defined in the attached diagram. Its definition is part of the original problem statement. t2 is the round trip interval (front of stick to mirror to front of stick) as measured by Frank.

    So now, find the expression for t2, which is the round trip time interval of the light from the front of the stick (starting at time t = 0) to the mirror, then back to the front of the stick, as measured by Frank.
     
  8. Feb 18, 2014 #7
    I see what you mean by the difference between Δt2 and t2.

    t=[itex]\frac{2L}{c^2-v^2}[/itex]

    You say I will have to invoke time dilation in Mary's frame. I'm not sure how to go about doing that without simply multiply t by [itex]\sqrt{1-\frac{v^2}{c^2}}[/itex] to get t'. Or is that what you meant?
     
  9. Feb 18, 2014 #8

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    Beautifully close! :thumbs: But you're missing a 'c' in there somewhere. It's almost perfect, except for that 'c' that accidentally dropped out. (And that's t2, just to be clear.)

    Yes, more or less.

    You know that [itex] t'_2 = \frac{2L_0}{c} [/itex], in Mary's frame of reference. So, transform that to be in terms of t2 (Frank's frame of reference).

    (In other words, you already know the "primed" time: it's 2L0/c. Now you're looking for the "unprimed" time term.)
     
  10. Feb 18, 2014 #9
    Not sure where a c may have dropped out.

    t2=Δt1+Δt2=[itex]\frac{L}{c-v}[/itex]+[itex]\frac{L}{c+v}[/itex]=[itex]\frac{2L}{c^2-v^2}[/itex]

    Right?

    I did notice when trying to use time dilation to equate these two that I ended up with an extra c that would likely have canceled out with whatever c you are talking about. I'm just not sure where to get it :P
     
  11. Feb 18, 2014 #10

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    So far, so good! :smile:

    ..And that's where things take a turn for the worse.

    Let's step back and look at:
    [tex] t_2 = \frac{L}{c-v} + \frac{L}{c+v}. [/tex]
    Find a common denominator for the terms on the right hand side so they can be added together. Before adding the terms together though, express them as separate terms with a common denominator. Only then add them together and see what happens. :smile:
     
    Last edited: Feb 18, 2014
  12. Feb 18, 2014 #11
    HAH! Wow. It's always the simplest things...

    And it worked! Multiplying t' by the lorentz factor γ and setting that equal to t2 I can solve for L=L0γ-1.

    Sound about right?

    If so, thank you very much!
     
  13. Feb 18, 2014 #12

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    Perfect! :approve:

    Good job! :smile:
     
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