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Educational value of spacetime diagrams

  1. Dec 23, 2007 #1

    Dale

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    I was thinking about how one would teach special relativity using a spacetime-geometric approach. I think you would first show spacetime diagrams in a single frame and describe velocity as the slope of the worldline. Then when you moved onto other frames it would be natural to leave the slope of a light cone at a 45º angle.

    However, I believe that the natural tendency of students would be to try to preserve absolute simultaneity by drawing the spactime diagram as attached. Such a diagram would preserve c as a frame invariant, would show length contraction, no time dilation, and absolute simultaneity. How could you get to the Lorentz transform from here?
     

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  2. jcsd
  3. Dec 23, 2007 #2
    diagram

    I've done this before. First note that your diagram was flawed when you say
    t = t'. No! What results is a squashed "egg crate" made of rhombi. The light's slope is preserved. I've shaded the transformed original square. Note that you can't have length contraction without time dilation.
     

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    Last edited: Dec 23, 2007
  4. Dec 24, 2007 #3

    robphy

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    People learning relativity have to be guided to properly draw spacetime diagrams.

    One method is to use radar experiments to operationally define the notions of elapsed time and length. (See, e.g., Bondi's k-calculus.) When drawn on a spacetime diagram, it leads to pictures like the one shown by Helios.

    One method I have been working on is using spacetime diagrams of light clocks.
     
  5. Dec 24, 2007 #4

    Dale

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    Thanks for the input, I am sorry that I was not clear. Yes, I know that the diagram is flawed, and what the correct diagram looks like. But I think that this is the kind of flawed diagram that a student would produce when first exposed to relativity. I am looking for geometrical reasons why it is flawed. I think that might help introduce the relativity of simultaneity.
     
  6. Dec 24, 2007 #5

    JesseM

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    Your diagram wouldn't make c invariant in both directions--if you draw a diagonal through the origin in the opposite direction it intersects with x'=-1 well before t'=1.
     
  7. Dec 24, 2007 #6

    Dale

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    That seems interesting. Of course, since the spacetime diagrams are limited to a single spatial direction you would be limited to a light clock whose axis is oriented parallel to the direction of travel. Such a clock would experience both time dilation and length contraction, so it may be difficult to tease apart the two effects.

    If you have such a derivation that works I would be very interested.
     
  8. Dec 24, 2007 #7

    Dale

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    Ah, of course. You are right, the way I have it drawn the light pulse travels at 3c in the negative direction. So the simultaneity lines can be drawn from the intersection of equidistant position lines and light lines in each direction. I guess you would have to introduce Einstein synchronization earlier for that to make sense to a student.
     
  9. Dec 24, 2007 #8

    Fredrik

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    Everyone can see that the t' axis (the line where x'=0) is the line through the origin with slope 1/v:

    [tex](-\infty,\infty)\ni\tau\mapsto\tau\begin{pmatrix}1 \\ v\end{pmatrix}[/tex]

    What the students need to understand somehow is that the x' axis (the line where t'=0) is the line through the origin with slope v. I think it's very easy to guess that this is the case, once you've opened your mind to the possibility that the x' axis may not coincide with the x axis. After all, the world line of a "ray of light" has slope 1 in the unprimed coordinate system, and if it's to have slope 1 in the primed system as well, the natural choice is a line with slope v. It's a "natural" guess, because it puts the world line of that ray of light exactly half-way between the t' and x' axis (in the diagram). If the x' axis is anywhere else, then how can the slope of the ray of light in the primed system be 1?

    However, I think that if you want to prove this rigorously, you're going to have to resort to algebra. First note that [itex]\Lambda[/itex] must be linear. (Books usually don't bother to prove this, so maybe you shouldn't either). Any linear operator can be represented by a square matrix, so let's write

    [tex]\Lambda=\begin{pmatrix}a & b\\c & d\end{pmatrix}[/tex]

    The components are constrained by the fact that [itex]\Lambda[/itex] must map the light cone onto itself. If we exclude parity and time-reversal for the moment, there must exist real numbers [itex]\alpha[/itex] and [itex]\beta[/itex] such that

    [tex]\Lambda \begin{pmatrix}1 \\1\end{pmatrix} = \alpha\begin{pmatrix}1 \\1\end{pmatrix}[/tex]

    and

    [tex]\Lambda \begin{pmatrix}1 \\-1\end{pmatrix} = \beta\begin{pmatrix}1 \\-1\end{pmatrix}[/tex]

    These equations imply that

    [tex]a+b=c+d=\alpha[/tex]

    [tex]a-b=c-d=\beta[/tex]

    [tex]2a=\alpha+\beta=2c[/tex]

    [tex]2b=\alpha+\beta=2d[/tex]

    [tex]\Lambda=\begin{pmatrix}a & b\\b & a\end{pmatrix}[/tex]

    The form of [itex]\Lambda[/itex] is further constrained by Einstein's first postulate. The t' axis in the unprimed system is a line through the origin with slope v, so the t axis in the primed system must be a line through the origin with slope -v. So there must exist a real number [itex]\gamma[/itex] such that

    [tex]\Lambda\begin{pmatrix}1 \\0\end{pmatrix}=\gamma\begin{pmatrix}1 \\-v\end{pmatrix}[/tex]

    and from this we get

    [tex]a=\gamma[/tex]

    [tex]b=-\gamma v[/tex]

    and finally

    [tex]\Lambda=\gamma\begin{pmatrix}1 & -v\\-v & 1\end{pmatrix}[/tex]

    [itex]\gamma[/itex] can be dermined from the condition

    [tex]\Lambda(v)^{-1}=\Lambda(-v)[/tex]

    which is also implied by the first postulate. The result is

    [tex]\gamma=\frac{1}{\sqrt{1-v^2}}[/tex]

    so

    [tex]\Lambda=\frac{1}{\sqrt{1-v^2}}\begin{pmatrix}1 & -v\\-v & 1\end{pmatrix}[/tex]

    Now that we have found the explicit form of the (homogeneous) Lorentz transformation in 1+1 dimensions (excluding parity and time-reversal), we can find the x' axis by calculating

    [tex]\Lambda^{-1}\begin{pmatrix}0\\1\end{pmatrix}[/tex]
     
    Last edited: Dec 24, 2007
  10. Dec 24, 2007 #9

    robphy

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    physics.syr.edu/courses/modules/LIGHTCONE/LightClock/ has the animated spacetime diagrams of light clocks and http://arxiv.org/abs/physics/0505134 has the paper which was written before the animations were created. The paper features the spacetime diagram of a "longitudinal light clock" [drawn on a planar spacetime diagram], which was the seed idea of the whole presentation.... eventually expanded to the "circular (spherical) light clocks".

    "The Mechanical Universe" video series drew similar diagrams... however, since they only draw spacetime diagrams of the usual "transverse light clocks", they missed the geometric interpretation of the invariance of a particular type of "area" seen in the longitudinal and circular light clocks. Similar results were obtained by Brill-and-Jacobson and Mermin... see https://www.physicsforums.com/showthread.php?p=747083#post747083 .
     
    Last edited: Dec 24, 2007
  11. Dec 24, 2007 #10
    no length contraction without time dilation

    I consider that the statement "Note that you can't have length contraction without time dilation is to strong holding only in the case of clocks synchronized following Einstein's clock synchronization procedure.
    As a counterexample I could quote
    C.Leubner, Elementary relativity with "everyday" clock synchronization, Eur.J.Phys. 13, 170 (1992) largely quoted in the literature of the subject.
    which leads to absolute simultaneity and length dilation without violating the the postulates on which special relativity is built.
    Once the discusion started I could offer other counterexamples.
    I start that thread because I have no "next door" physicists to discuss with them the problem.
    Thanks in advance for your contribution.
     
  12. Dec 25, 2007 #11

    Dale

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    Thanks for this, I found it very useful. Luckily, linear algebra is also very geometrical, so I think that we can follow this approach geometrically even for students who may not have been exposed to linear algebra yet.

    So, geometrically, this matrix describes the standard rhombus pattern shown by Helios above. It has two parameters that can be adjusted to change the "shear" and the scale of the rhombuses. The points of the rhombuses follow the light cone.

    Geometrically this equation constrains the "shear" of the rhombus to be related to the velocity. I think that we can introduce this constraint even before we introduce the light-cone constraint since it is simply the definition of velocity and would apply even for Galilean relativity. This constraint should be the easiest to understand.

    This, for me, is the key and is very useful. This is the final constraint which determines the scale of the rhombuses. I couldn't figure out how the scale of the rhombuses could be infered from the second postulate, and this shows that it is not the second postulate, but the first, which fixes the scale.
     
  13. Dec 25, 2007 #12

    JesseM

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    How could different observers all agree that light moved at c in both directions (the second postulate) and yet not disagree about simultaneity? For example, suppose I am on a train moving relative to you, and I set off a flash at the midpoint of the train. In my frame if the light is assumed to travel at the same speed in both directions, I must conclude the light from the flash hit each end of the train simultaneously, but if you also assume light travels at the same speed in both directions in your own frame, then since one end of the train is moving towards the point where the flash was set off while the other end is moving away from it, the light must hit the two ends non-simultaneously. Are you saying that the Leubner paper provide a way to deal with this situation such that both observers say that the second postulate is satisfied in their own frame, yet they both agree on whether the light hit the two ends simultaneously or non-simultaneously? If so, how does it work?
     
  14. Dec 25, 2007 #13
    absolute simultaneity?

    Before to answer please tel me if you have seen Leubner's paper?
     
  15. Dec 25, 2007 #14

    JesseM

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    No, is the paper available online?
     
  16. Dec 25, 2007 #15
    everyday clock synchronization

    The clock synchronization proposed by Leubner goes as follows:
    "the radio is open and you fix your clock according to "at the sound of the last tone it will be 12 o'clock" neglecting the time during which the radio signal propagates from the radio station to the point where your wrist watch is located.
    Regards
     
  17. Feb 14, 2008 #16

    Fredrik

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    I'm bumping this thread just to say that I was wrong. [itex]\gamma[/itex] is the 00-component of [itex]\Lambda[/itex], so we can get it from the 00-component of the condition

    [tex]\Lambda^T\eta\Lambda=\eta[/tex]

    There's no need to use the first postulate.

    In 3+1 dimensions we get

    [tex]\gamma=\frac{1}{\sqrt{1-\vec{v}^2}}[/tex]

    where the components of [itex]\vec{v}[/itex] are defined by saying that [itex]\Lambda[/itex] takes the time axis to the line

    [tex]\tau\mapsto\tau\begin{pmatrix}1\\ -v^1\\ -v^2\\ -v^3\end{pmatrix}[/tex]
     
  18. Feb 16, 2008 #17

    Fredrik

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    D'oh. I'm such a moron. Now I have to bump this again just to say that I was wrong when I thought I was wrong.

    The problem is that we need to use something like [itex]\Lambda(\vec{v})^{-1}=\Lambda(-\vec{v})[/itex] to get the condition [itex]\Lambda^T\eta\Lambda=\eta[/itex] (from preservation of the light-cone at the origin) instead of [itex]\Lambda^T\eta\Lambda=k\eta[/itex], where k is an arbitrary real number.
     
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