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Doubt regarding Length Contraction's derivation

  1. Dec 29, 2008 #1
    I have a doubt regarding the proof of Length Contraction.

    The proof of length contraction formula goes as follows:
    You choose a frame S (as shown in figure 1), suppose w.r.t. S, AB is moving with velocity v. Length of AB is L in AB frame and L1 in S frame. At t=0, a light pulse (shown with red color in figure) is left at A, it strikes B and reflects back to A.
    Figure 1: http://img166.imageshack.us/my.php?image=figure1fz0.png

    Time taken for first reflection t1 = L1/(c-v) and similarly for return t2 = L1/(c+v).
    So total time taken equals t = t1 + t2 = 2cL1/(c2-v2).
    Suppose time taken for these reflections in frame AB is t’, so t = t’γ (from Time Dilation).
    And in AB frame, t’ = 2L/c. Now eliminate t and t’ from these 3 equations to get the formula for Length Contraction:
    t = 2cL1/(c2-v2) and t = t’γ give t’γ = 2cL1/(c2-v2).
    Put t’ from t’ = 2L/c into this equation:
    2Lγ/c = 2cL1/(c2-v2)
    L1 = L/γ, this is the formula for length contraction.

    Now, my doubt starts, consider figure 2, and suppose that we are not looking the second reflection.
    Figure 2: http://img525.imageshack.us/my.php?image=figure2ia2.png
    Then t = t1 = L1/(c-v) and t’ = L/c. And t = t’γ.
    Now eliminate t and t’ from these 3 equations:
    t’γ = L1/(c-v), put t’ = L/c to get Lγ/c = L1/(c-v)
    this gives: L1 = Lγ(1 - v/c) = L [(c-v)/(c+v)]1/2 and not L1 = L/γ.
    Is there any mistake if we choose to not to consider the second reflection?
    Further suppose that we only consider the second reflection, in that case t = t2 and hence finally we will get:
    L1 = L [(c+v)/(c-v)]1/2.
     
    Last edited: Dec 30, 2008
  2. jcsd
  3. Dec 29, 2008 #2

    Janus

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    The problem with not using the whole round trip of the light pulse is that you have to consider the Relativity of Simultaneity. Now you can't directly compare t and t'. Clocks placed at A and B would show the same time according to A and B, but would show different times according to the observer in relative motion to A and B.
     
  4. Dec 30, 2008 #3
    Thanks a lot for replying. I didn't understand why we cannot compare t and t', do you mean that t = t’γ is wrong, why is that so? AB is moving with speed v w.r.t. S and event is occuring in AB, so the time measurement of S will be dilated by a factor of γ and hence t = t’γ. Can some please more clarify where I'm going wrong?
     
  5. Dec 31, 2008 #4
    OK, since I'm not getting any response I'll reframe my problem:

    The Problem:
    Why we cannot consider only one reflection of light in the derivation of Length Contraction formula as shown above?

    Attempt to the problem:
    As shown above the derivation in both cases results in different formulas and I've no idea where I'm mistaking.
     
  6. Dec 31, 2008 #5

    Janus

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    I'll try to write out a more complete answer later when I have more time, but for now, as I said before, the answer lay in the Relativity oF Simultaniety.
     
  7. Dec 31, 2008 #6

    Doc Al

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    The "time dilation" formula (Δt = γΔt’) only applies to time measurements made on a single clock in the prime frame.

    When you use the round trip light pulse, Δt’ is measured on the clock at A. So the time dilation formula applies.

    But when you only consider the one-way trip, Δt’ = L/c is measured using two prime frame clocks, one at A (t'1 = 0) and the other at B (t'2 = L/c). So the time dilation formula cannot be applied in a simple manner. As Janus stated, the key is the relativity of simultaneity.

    According to the unprimed S frame, the clocks at A and B are not in synch. Clock A is ahead of clock B by an amount equal to Lv/c^2. So, when clock A reads 0, clock B is reading -Lv/c^2. Thus when clock B reads L/c, the actual amount of time that elapsed on clock B during the one-way trip (according to frame S) is L/c + Lv/c^2. Since that's a time elapsed on a single clock (clock B), you can apply the time dilation formula. Crank it out, and you'll get the usual formula for length contraction.

    To properly analyze the one-way trip involves understanding simultaneity and clock synchronization. That's why we stick to the round trip thought experiment!
     
  8. Jan 1, 2009 #7
    Thanks Janus and DocAl for clearing the doubt, I've now realized what Janus wanted to tell earlier after DocAl's explanation.

    Yes, it did work and I got the Length Contraction's formula.

    Based on what I concluded from this I want to tell something, please check whether that's correct or not.

    There are four different relativistic effects of Special Theory of Relativity which occur "independently" in any system, they are:

    1. Time Dilation
    2. Length Contraction
    3. Relativity of Simultaneity
    4. Mass Defect

    In this case it happens as follows:

    1. Clock A is dilated by γ from clock in S frame.
    2. Length AB is Contracted by a factor of 1/γ in S frame.
    3. Clock at A is ahead of clock at B by Lvγ/c2 in frame S.
    4. In frame S, mass of AB is larger by a factor of γ.

    Please tell whether my conclusions are correct or not. And thanks again for clearing the doubt.
     
    Last edited: Jan 2, 2009
  9. Jan 1, 2009 #8

    Doc Al

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    Here's my take on your list. The first three are perfectly correct: Those are the basic "effects" of special relativity. Taken together, they are equivalent to the Lorentz transformations.

    As for the fourth item on your list, you are thinking of so-called "relativistic mass", not mass defect. While the concept does have some limited use, you'd be better off sticking with plain old mass (also called the invariant or rest mass) and relativistic momentum to understand the kinematics and dynamics of fast moving particles. In modern treatments of relativity the term mass (with no qualifications) almost always refers to the invariant mass.

    (Mass defect is the difference between the mass of a nucleus and the sum of the masses of its individual protons and neutrons.)
     
  10. Jan 2, 2009 #9
    OK, since I have just begun to learn Special Relativity, I'm not at all familiar with mass defect, so I'll get into that later. And one last thing the length L was in AB frame, so B will lag by Lvγ/c2 and not by Lv/c2.

    Thanks again for helping me.
     
  11. Jan 2, 2009 #10

    Doc Al

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    According to frame S observers, at any given instant the time shown on clock A will be Lv/c^2 ahead of the time shown on clock B.
     
  12. Jan 5, 2009 #11
    Relativity of Simultaneity:

    |S| A|-----------**-----------|B ->v t=0
    |S| |-------A|*--------------------*|B t=t1

    Here light is left from center of AB, which reaches to A and B, Striking of light to A and B are 2 events. Length of AB is L in it's frame, i.e. L is the rest length and L1 in S frame. AB is moving with v w.r.t. S. Now analyze the condition from:

    Frame AB:
    Both the lights from the center travel the distance L/2 and hence time taken by them to reach A and B is same, equals to L/2c. And hence w.r.t. AB, the two events (light striking A and B) are simultaneous.

    Frame S:
    The light striking A takes time L1/2(c+v) and the light striking B takes L1/2(c-v). We had to take L1 instead of L because of the length contraction of AB w.r.t. S. The time difference in the two events is:
    L1/2(c-v) – L1/2(c+v) = L1v/(c2-v2) and not Lv/(c2-v2). Note that there is no effect of time dilation here.

    Well, for some reason this is not exactly working. It works when we take t = L/c – Lvγ/c2 and not L/c + Lvγ/c2. But logically you seem to be right or may you are mistaking something. Please someone help me out, if needed I will post the complete mathematical work that why it is working either way and not in the suggested way.
     
    Last edited: Jan 5, 2009
  13. Jan 5, 2009 #12

    Doc Al

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    OK.

    Since B is moving away from the light, the time should be L1/2(c-v).
    Of course.
    (Be sure to correct your error.) There's no need to apply time dilation, because you are calculating things completely from frame S.

    Don't confuse:
    (1) The time it takes for light to travel from A to B according to S.
    (2) The difference in clock readings on clocks at A and B according to S.
    In (2), the clock readings are compared at the same instant according to S.

    It works for me. If you post exactly what you did, we can find your error. (I suspect you are making the same error as above.)

    In outline, it should go:
    Δt' = L/c + Lvγ/c2 [note typo; correction below]
    Δt = L1/(c-v) = γΔt'

    Edit: There's a typo in the second to last equation; It should have read:
    Δt' = L/c + Lv/c2
     
    Last edited: Jan 7, 2009
  14. Jan 7, 2009 #13
    Please check whether I'm right or not:

    If Event E1 happens at A and Event E2 happens at B at time t=0 in frame AB, simultaneously. Then in frame S, if E1 happens at t=0 then E2 will happen at Lvγ/c2, i.e. E2 happens later than E1 in frame S.

    I've corrected that error, that was only just typing mistake.

    Well this is where I was confused, thanks for correcting me. Just for confirmation, (1) is Δt and (2) is Δt', right?

    Wonderfully, some mathemagic is going on here. I get the formula for Length Contraction when:

    1. I consider equations given by you as above, but Δt' = L/c + Lv/c2 and not Δt' = L/c + Lvγ/c2.
    2. When, Δt = L1/(c-v) - Lvγ/c2 and Δt' = L/c and Δt = γΔt'.

    Both are WRONG and both conditions give the correct formula for Length Contraction. In 1, you were already telling that I must add Lv/c2 and not Lvγ/c2; but you changed after I gave the complete derivation of Relativity of Simultaneity. Please help me out.
     
    Last edited: Jan 7, 2009
  15. Jan 7, 2009 #14

    Doc Al

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    Right. Since, according to frame S, clock B is behind clock A by Lv/c^2.

    Yes, in the sense that (2) is a difference in prime frame clock readings.


    In my last post I had a typo. (See my original statements in post #6)
    That first equation should have read:
    Δt' = L/c + Lv/c2

    (Sorry about that.)
     
  16. Jan 9, 2009 #15
    Got it finally, Lvγ/c2 is the time added to the dilated time. And hence Δt' = L/c + Lv/c2. Anyway I've a question regarding myself as a physicist:

    I want to go for higher studies in theoretical physics like field theories, string theory, etc. Which vision or image of Special Relativity will you recommend through which I must look at the world:
    (a) Time Dilation, Length Contraction and Relativity of Simultaneity, OR
    (b) Lorentz Transformation only.
    Two are same but doubts and confusions may arise in different cases, one may help while other may give me a typo, so please answer this question.

    And thanks again for helping me out. The doubt is officially solved now, but suggestions are still welcome.
     
  17. Jan 9, 2009 #16

    Doc Al

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    Don't limit yourself to one approach.

    In my opinion, there are three ways to understand basic special relativity and all three are important:

    (1) Using the Lorentz transformations. For certain problems, this is the best approach, but often it leaves intuition behind.

    (2) Using the relativistic behavior of clocks and measuring rods. This can be extremely helpful in developing intuition. The three relativistic behaviors are: Length contraction, time dilation, and clock desynchronization (or the relativity of simultaneity). Some problems can be solved much quicker if you have a solid understanding of these relativistic effects.

    (3) Using space-time diagrams. This is where you begin to develop a real understanding of relativity.

    You might find this summary helpful: Basic Equations of Special Relativity
     
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