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Length Contraction, Time Dilation and The Principle of Relativity

  1. Dec 31, 2006 #1
    Time dilation and length contraction are expressed by the following equations:


    However, according to the principle of relativity, you can't tell which is the fixed frame of reference and which the moving frame, so you can swap L and L' and T and T' around in the equations, am I right so far?

    Well, if that is the case, then time is dilated in system K' as seen from K, but also in K as seen from K' (I'm dropping length contraction here, as the time dilation is the interesting part). Now consider the Twin Paradox. Twin One stays on earth, Twin Two flies off, comes back, and turns out to be younger. However, according to the principle of relativity, one could also say that Twin One flew off with earth while Twin Two stayed put in the spaceship, and thus Twin One should now be younger.

    I must be missing something, or the Twin Paradox wouldn't be so widely used. I'd greatly appreciate it if someone could tell me what exactly that is.
  2. jcsd
  3. Dec 31, 2006 #2

    Doc Al

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    The time dilation and length contraction formulas describe observations made between two uniformly moving inertial frames. As such, they are completely symmetric: Alice judges Bob's clocks to run slow, while Bob judges Alice's clocks to run slow. These conclusions are completely consistent. (To fully understand how this behavior of clocks and measuring sticks can be consistent, one must be sure to consider not just time dilation and length contraction, but the relativity of simultaneity as well.)

    One thing you're missing in viewing the "twin paradox" situation is that the traveling twin does not remain in a single, uniformly moving inertial frame: The travel twin must accelerate, which breaks the symmetry.

    There is plenty more to be said. Do a search on "twin paradox" here on PF (use Google) and you'll find many threads. You can also read the excellent sci.physics FAQ: The Twin Paradox
    Last edited: Dec 31, 2006
  4. Dec 31, 2006 #3
    For one of the, IMHO, best explanations of the false paradox using special relativity I refer to the Wikipedia Twin Paradox article.

    The article explains the distinction between what an observer sees and calculates. It first explains the situation from what an observer actually sees, a so called "Doppler View". From here the article explains how we can construct from this data a plane of simultaneity by taking the distances into account. And here, hopefully, the confusion about simultaneity at a distance becomes evident when the traveler changes direction.
    Last edited: Dec 31, 2006
  5. Dec 31, 2006 #4
    You mean this link.
  6. Dec 31, 2006 #5
    Yes, my mistake, I linked to the talk page instead. I corrected to the proper link.
  7. Dec 31, 2006 #6


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    Careful here, In order to grasp the Twin paradox you have to take all Relativistic effects into account; Time dilation, length contraction and Relativity of Simultaneity.

    For example. if the distance between Earth and the turn-around point is 10ly as measured by Twin 1, it would be measured as 5ly as measured by Twin 2 if the relative velocity between the two twins was 86.6% of c

    Thus, Twin 1 expects Twin 2's clock to record 11.55 yrs for the round trip because Twin 2's clock ran half as fast, and Twin 2 expects to record 11.55 yrs for the round trip, becuase the distance between the two points has been 1/2th.

    And, as already has been pointed out, the rules change for the twin undergoing accleration while he is accelerating. You can't use the simple time dialtion formula in this case.
  8. Jan 10, 2007 #7
    I apologize for the late reply. I've been pondering this question a bit, and I gather the problem's to do with acceleration. I haven't been able to find any information on acceleration in Special Relativity that I could properly understand. I'd much appreciate it if anyone could point me in the right direction.

    I read the "Resolution of the Paradox in Special Relativity" on the Wikipedia page, but I don't fully understand. I've been looking at the diagram in that section, but I'm not grasping it. What exactly are these simultaneity planes? What exactly are those dots? Is it just simple time dilation, that the distance between the traveling twin's dots are larger than the distance between the stationary twin's dots?

    I assume it's more than just the change of velocity. Because if Twin One is accelerating relative to Twin Two, then Twin Two also looks to be accelerating from Twin One's perspective. The only thing I can come up with is the cause of the acceleration. The Twin in the spaceship has force applied to him, the Twin staying behind doesn't. But I'm not sure if that can be applied here, as I'm conjuring that from classical mechanics (F=m*a).

    Well, I guess my post is a bit of a mess, but I hope you can understand my confused ramblings. Any help in getting my thoughts a bit straightened out would be much appreciated.
  9. Jan 10, 2007 #8
    Look here for your answer
  10. Jan 15, 2007 #9
    Upon researching this a little more, I stumbled upon this post by jtbell:


    It's basically what I was looking for, a detailed explanation of the matter, but I came across a problem.
    The post states the following:

    [tex]t^\prime = \gamma (t - vx/c^2)[/tex]

    Sure, just our Lorentz transformation. But then, filling it in:

    [tex]t^\prime = 1.6667 [5 - (0.8)(4)] = 3[/tex]

    If you read the post, you'll know that in this case, v = .8c and x = 4. That being said, this equation says that v/c^2=.8 and thus .8c/c^2=.8, which doesn't sound right to me. I am aware that in the post c is set to be equal to 1, but if you set it to be any other number, this equation is false, isn't it?
  11. Jan 15, 2007 #10


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    I'm not sure what you think the difficulty is.

    A dimensional analysis shows that

    [tex]\frac{vx}{c^2} = \frac{\frac{m}{s} \, m} {\left(\frac{m}{s}\right)^2} = \frac{\frac{m^2}{s}}{\frac{m^2}{s^2}} = s[/tex]

    so the units are correct for the added term, and gamma is dimensionless as it should be.
  12. Jan 15, 2007 #11
    It's not the equation in variables that seems to be incorrect, but the values entered. It states that .8c/c^2=.8 and thus that c/c^2=1, which is correct in this case, since c is set to be 1, but in general, this statement seems false to me.
  13. Jan 15, 2007 #12


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    Actually what is done is that the units are left out completely.

    In this case t is measured in years, x in lightyears and velocity in lightyears/year (thus c= 1ly/yr)

    Thus the equation reads.

    [tex]t' = 1.667(5y- \frac{(4ly)(0.8ly/y)}{(1ly/y)^2}[/tex]

    So the post does not say that 0.8c/c^2 = 0.8, but that it equals 0.8/c and that this multiplied by the distance x is a distance divided by velocity, which is measured in time.
    Thus you get:
    [tex]t' = 1.667(5y-4(0.8)y)[/tex]= 3y

    It will work out no matter what units you use as along as all the units are consistant.

    The original post simply left out the units in the equation because they had already been defined. It was a shortcut to writing it out fully.
  14. Jan 16, 2007 #13
    Yes, I suppose I hadn't looked at it that way. Thanks. So, if you were to express things in seconds, meters and m/s instead, you should get the same result? I'm actually getting different, ridiculous results, but they might just be the effect of misplaced brackets, etc.
  15. Jan 16, 2007 #14


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    If you would like to post what your doing, perhaps we can take a look see for you...
  16. Jan 16, 2007 #15
    Well I would have been content just knowing whether I was doing it wrong, but if you're willing to take the time, here goes.

    t' = (5/3) * ((5 y * 365.25 days/y * 24 h/day * 3600 s/h) - ((4 ly * 9.4605284e15 m/ly) * (.8 * 2.9979e8 m/s)) / (2.9979e8 m/s)) = -5.0456e16 s

    I'm aware that this is extremely messy, but since this is how I have to enter calculations into my calculator, I posted it like this.
  17. Jan 16, 2007 #16


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    I think your problem is here;
    You didn't square the denominator.
  18. Jan 16, 2007 #17
    Ah, quite right. Then it comes to be 94 675 014.95 seconds = 3.00013427 years. Thanks for pointing that out.
  19. Jan 16, 2007 #18
    I've once more stumbled upon something I don't quite grasp in that explanation. The following is said:

    Which I can follow. But then the next step is this:

    In which the calculated t' from the previous calculations is used. However, why use that t'? Isn't that the time in the travelings twin's reference frame as seen by the stationary twin and not as seen by the traveling twin himself? Furthermore, isn't the time dilation supposed to be symmetrical, since v is constant? Thus I would expect that in frame S, t=5 and t'=3 and in frame S', t'=5 and t=3. Have I understood things so poorly?
  20. Jan 22, 2007 #19
    Well, that last problem has been corrected, my physics teacher helped me realize where the flaw was in my chain of thought. I am now ready to continue, but I have another question. Where the generalized transformations are presented, using [tex]x_0[/tex] and [tex]t_0[/tex], etc., what exactly is calculated using these equations? I think I understand that these are simply two Lorentz transformations, subtracted. One represents the event we're interested in and one represents an event we happen to know the co-ordinates of, correct? Why are we subtracting, however? How does that remove the problem of the frames not being synchronized?
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