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On the Notations of Length Contraction and Time Dilation

  1. Oct 22, 2007 #1
    Now, if I'm not mistaken, from the Lorentz transformations one derives the following equations for time dilation and length contraction, respectively:

    T'=gT
    L'=gL

    Where g is the gamma factor, forgive my sloppy representations, but I'm not well versed in Latex.

    However, the equation for length contraction is always written as such:

    L=L'/g

    Which at first was very confusing to me because I thought that the equation was L'=L/g (the notations L and L' are also frequently swapped for others). Perhaps I am the exception to the rule, but this, to me, was quite a pitfall when trying to learn about Special Relativity. My question really is this: why this difference in notation? Why write these two equations, which are essentially very similar, in different ways?
     
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  3. Oct 22, 2007 #2

    robphy

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    I don't know the answer to why the equations are presented that way.... It's possibly historical... but I think I have an answer below. I'll admit that, in that notation, I tend to be confused myself. That's why I avoid thinking in terms of "formulas" first. Rather, I draw a spacetime-diagram, then interpret trigonometrically. This leads to the correct formulas.. backed up by some geometrical intuition which I can fall back on.

    From that intuition, I would say that the quantities on the left are in terms of "what you measure in your reference frame" in terms of an invariant quantity (e.g. the elapsed time along an inertial worldline or the rest-length of between a pair of spacetime-parallel worldlines).
     
  4. Oct 22, 2007 #3

    Dale

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    To expand on robphy's comments. Once you have a good geometric grasp then I think it naturally makes more sense to view the equations as a single transformation on spacetime rather than two seperate formulas for space and time.

    (t',x') = A.(t,x) where A is the Lorentz transform matrix and (t,x) and (t',x') are event coordinates in spacetime.

    This approach incorporates the relativity of simultaneity (which is missing from your above formulation), makes sure that you don't get confused about the different forms, and works for events that do not occur at the same location or time in a particular frame.
     
    Last edited: Oct 22, 2007
  5. Oct 22, 2007 #4

    robphy

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    One should be careful not to regard the time-dilation problem as somehow analogous or complementary to the length-contraction problem... despite the apparent similarities of the formulae.

    From a drawing of their spacetime diagrams [which I encourage you to do],
    - in the time-dilation problem, we measure the adjacent leg (representing the apparent elapsed time between events along the moving observer's worldline) obtained from projecting a timelike hypotenuse (representing elapsed time between events of the moving observer)
    - in the length-contraction problem, we measure the spacelike hypotenuse (representing the apparent separation between spacetime-parallel worldlines) which projects onto the moving observer's space [a spacelike line of simutaneity] (representing the "rest-length", the proper-separation between those spacetime-parallel worldlines)

    Although these triangles are geometrically similar, they are not physically similar because in one case we measure the adjacent-side and in the other case the hypotenuse. Furthermore, there are more than two events in question... i.e. we're not somehow measuring different components of the same spacetime displacement. So, these effects are actually two separate problems... which could be solved [if desired] using a single Lorentz transformation on two sets of spacetime events (one timelike separated and the other spacelike separated), which each lead to some algebraic simplifications... resulting in the apparent similarity of the "formulae".
     
    Last edited: Oct 22, 2007
  6. Oct 22, 2007 #5

    JesseM

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    T and L represent time and length in the object's own rest frame, while T' and L' represent the corresponding time and length in the frame where the object is moving with velocity v, correct? If so your second equation is backwards, it should be L=gL' (or equivalently L'=L/g)...after all, gamma is always larger than 1, but a ruler is shorter in a frame where it's moving, not longer.
     
  7. Oct 22, 2007 #6

    Dale

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    Perfectly demonstrating the confusion!!!
     
  8. Oct 22, 2007 #7
    Actually I meant L (T) and L' (T') the other way around from what you described. If one were to describe things as you do, T and T' flip around as well, I would think.


    DaleSpam, robphy, thanks for your replies. You're taking my question to way deeper levels than I had intended, though, to the point where I don't follow anymore. My experience in dealing with Special Relativity in geometrical terms is limited to the most basic Minkowski diagrams. I've had a look or two at the geometrical ways of describing Special Relativity, but so far I've not got very far in understanding them. My SR teacher gave away copies of a book in which the approach is purely geometrical, but at some point or other, it loses me. Thus, so far, my understanding of Special Relativity is purely mathematical. A lot of what you are saying, therefore, is rather lost on me.

    But I am digressing. Back to my initial question. Do you see any benefit in writing the equations for time dilation and length contraction as they are written?
     
  9. Oct 22, 2007 #8

    JesseM

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    OK, but either way, if you write both equations with the primed term on the left, then one will feature the unprimed term multiplied by gamma and the other will feature the unprimed term divided by gamma...if you're saying L' and T' represent the length and time in the object's own rest frame, the equations would be:

    T' = T/g
    L' = Lg

    Although normally they're written to give you the length and time in your frame in terms of the length and time in the object's own rest frame, like this:

    T = T'g
    L = L'/g

    In words, what this is saying is that in your frame the tick of a moving clock will take longer than the tick of the same clock in its rest frame, while in your frame the distance between marks on a ruler will be shorter than the distance between those marks in the ruler's rest frame.

    I would say the benefit is that in both equations you're consistent about which frame's length/time is on the right side and which frame's length/time is on the right side. The alternative would be something like this:

    T = T'g
    L' = Lg

    ...which would be more confusing, I think.
     
    Last edited: Oct 22, 2007
  10. Oct 22, 2007 #9
    Then my initial assumption was wrong and I am back at my pitfall, because if I use the Lorentz transformations, I arrive at the equations I stated in my first post. How do you find these equations, then?
     
  11. Oct 22, 2007 #10

    JesseM

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    Deriving the length contraction equation is a bit trickier than the time dilation equation...with time dilation you can just take two events on the worldline of a clock and compare the time between them in the clocks' rest frame with the time between them when you transform the events to a different frame. But with length contraction, you have to keep in mind that length depends on your definition of simultaneity, it's the distance between the front and back of the object "at a single moment" in whatever frame you're using. So one way to do it would be to take the worldline of the front of your ruler, and the worldline of the back, pick an event at the back of the ruler (you can make this the origin for simplicity), find the event on the worldline of the front of the ruler that's simultaneous with this event in the ruler's rest frame, then find the different event on the worldline of the front of the ruler that's simultaneous with this event in the frame where the ruler is moving at velocity v, and in each case look at the coordinate distance between the event at the back and the event at the front in the appropriate frame. So in this case you're dealing with three different events, not just two as you were when you derived the time dilation equation.
     
  12. Oct 22, 2007 #11

    robphy

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    Here's the spacetime diagram, with time running upwards.
    Our worldline is [tex]OT[/tex].
    The moving rod has worldlines [tex]OU_1[/tex] and [tex]OU_2[/tex].

    Sorry for all of the words... but I am trying to be precise.



    [tex]
    \]
    \begin{picture}(200,200)(0,0)
    \unitlength 2mm
    {
    \put(1,58){$\perp$}
    \put(1,5){$\theta$}
    \put(5,1){$\theta$}
    \qbezier(0,50)(0,25)(0,0)\put(0,0){O}
    \qbezier(0,0)(0,60)(0,60)\put(0,60){T}
    \qbezier(0,0)(30,0)(60,0)\put(30,0){S}
    \qbezier(0,60)(20,60)(40,60)

    \qbezier(0,0)(20,30)(40,60)\put(40,60){U1}
    \qbezier(30,0)(50,30)(70,60)\put(70,60){U2}
    \qbezier(0,0)(30,20)(60,40)\put(55,35){S1}
    \put(48,30){$\perp$}
    }

    {
    \qbezier[100](0,0)(30,30)(60,60)
    \put(60,60){L}
    }
    \end{picture}
    \[
    [/tex]

    Time-dilation:
    Given the proper-time [tex]OU_1[/tex] along the moving worldline,
    we wish to determine the apparent-time [tex]OT[/tex],
    where we regard [tex]T[/tex] as the event simultaneous with [tex]U_1[/tex].
    In doing so, [tex]TU_1[/tex] is a spacelike vector Minkowski-perpendicular to the timelike vector [tex]OT[/tex].
    In particular, the Minkowski-angle at [tex]T[/tex] is a Minkowski-right-angle, and [tex]OTU_1[/tex] is a Minkowski-right-triangle.
    Using "spacetime trigonometry", we have
    adjacent=hypotenuse*cosine [where the hypotenuse is the side opposite the right-angle]
    [tex](OT)=(OU_1)\cosh\theta[/tex],
    the time-dilation formula.
    (In ordinary notation, substituting [tex]\cosh\theta=\gamma[/tex] and [tex]\tanh\theta=(v/c)[/tex], we have
    [tex](OT)=(OU_1)\gamma[/tex].)


    Length-contraction:
    Given the proper-length [tex]OS_1[/tex] separating the spacetime-parallel moving worldlines,
    [so [tex]OS_1[/tex] is a spacelike-vector Minkowski-perpendicular to [tex]OU_1[/tex] and [tex]S_1U_2[/tex] and the Minkowski-angle at [tex]S_1[/tex] is a Minkowski-right-angle. ],
    we wish to determine the apparent-length [tex]OS[/tex],
    where we regard [tex]S[/tex] as the event on [tex]S_1U_2[/tex] that is simultaneous with [tex]O[/tex].
    In doing so, [tex]OS[/tex] is a spacelike vector Minkowski-perpendicular to the timelike vector [tex]OT[/tex]... and it can be shown that the Minkowski-angle between the spacelike vectors [tex]OS[/tex] and [tex]OS_1[/tex] is numerically equal to the Minkowski-angle between the timelike-vectors [tex]OT[/tex] and [tex]OU_1[/tex].
    So, [tex]OS_1S[/tex] is a [similar] Minkowski right-triangle.
    Using "spacetime trigonometry", we have
    adjacent=hypotenuse*cosine [where the hypotenuse is the side opposite the right-angle]
    [tex](OS_1)=(OS)\cosh\theta[/tex],
    the length-contraction formula. (In ordinary notation, substituting [tex]\cosh\theta=\gamma[/tex] and [tex]\tanh\theta=(v/c)[/tex], we have
    [tex](OS_1)=(OS)\gamma[/tex]
    or
    [tex](OS)=\frac{(OS_1)}{\gamma}[/tex].)
     
    Last edited: Oct 22, 2007
  13. Oct 22, 2007 #12
    Remember that the lenght contracts and time slows down. Also remember that g>1. Then you can write down the equations immediately.
     
  14. Oct 22, 2007 #13

    Dale

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    Sorry about that. I know that Wikipedia is often denigrated, but you might want to start with http://en.wikipedia.org/wiki/Lorentz_transform

    I particularly like the first formula in the "Matrix Transform" section. Once you get the idea that it is simply a linear transform you can just use standard linear algebra. The form of the matrix (for motion along the x-x' axis) is also particularly simple and symmetric. But the point is that using this form you won't get lost, you will always correctly get time dilation and length contraction, and you will correctly include the relativity of simultaneity.
     
  15. Oct 24, 2007 #14
    Thanks for your replies, guys. I think I understand that my derivation of the length contraction is wrong because you need to incorporate the added problem of making the measurements of the front and back of your ruler simultaneous.

    For the rest, I'm going to have to take your word for it. I cannot follow what you are saying, robphy. Too many things are unfamiliar to me - spacelike and timelike, spacetime trigonometry, Minkowski-right. Basically the same goes for that Wikipedia page, DaleSpam. I'm not familiar with matrices or linear algebra or the hyperbolic functions. I'm sure it's beautifully simple to those who know what it all means, to me it's gibberish, for now at least.

    My initial misunderstanding is solved, however, and for that I thank you.
     
  16. Oct 27, 2007 #15

    A.T.

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    An alternative to Minkowski diagrams are Epstein diagrams. They show the geometrical relationship between length contraction and time dilatation pretty well.
     
  17. Oct 27, 2007 #16

    robphy

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    For the novices in relativity,
    here are some earlier discussions of the Epstein diagrams and other alternative diagrams for spacetime:
    https://www.physicsforums.com/showthread.php?t=180176
    https://www.physicsforums.com/showthread.php?t=151075
    ..the bottom line is that while they may useful for visualizing certain features in special relativity for beginners, they are of limited use if these beginners who wish to proceed to more advanced topics.
     
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