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Is the Loedel Diagram a Special Case of the Minkowski Spacetime Diagram?

  1. Dec 9, 2007 #1

    Jorrie

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    The main attraction of the Loedel spacetime diagram is that it treats the reference frame and the first moving frame symmetrically and hence they have identical scales in geometric units. The Minkowski diagram uses different scales for the orthogonal and the non-orthogonal axes, where the time axis of the second frame is defined by an angle so that [itex]v = \tan(\phi)[/itex] (clockwise convention), where [itex]\phi[/itex] can range from [itex]-\pi/4[/itex] to [itex]\pi/4[/itex] (the light-cone).

    The Loedel diagram uses an angle defined by [itex]v = \sin(\varphi)[/itex], where [itex]\varphi[/itex] can range from 0 to [itex]\pi/2[/itex], spanning symmetrically around the vertical through the origin, as shown in attached figure 1.** This gives a light-cone identical to the Minkowski diagram.

    One can view the Loedel diagram as a Minkowski diagram with three inertial frames: the orthogonal axes X,T plus two frames X0,T0 and X1,T1 moving at [itex]v_1[/itex] and [itex]v_0[/itex] respectively, where [itex]v_1 = -v_0 = \tan(\varphi/2)[/itex] relative to X,T. The calibration of these two axis systems is identical and relate to the orthogonal axis as

    [tex]
    \frac{\mbox{scale(X0,T0)}}{\mbox{scale(X,T)}} = \sqrt{\frac{1+v_0^2}{1-v_0^2}},
    [/tex]

    where `scale' refers to linearly plotted length per geometric unit along the relevant axis.

    It is now possible to add another coordinate system, say X2,T2, moving at a Minkowski velocity [itex]v_2[/itex] relative to X,T and plotted at an angle defined by [itex]v_2=\tan(\theta)[/itex] (see attached figure 2**).

    The equivalent `Loedel velocity' ([itex]v_{2L}[/itex]) is measured from X0,T0 and is related to [itex]v_2[/itex] through the relativistic subtraction:

    [tex]
    v_{2L} = \left(\frac{v_2-v_0}{1-v_2v_0}\right).
    [/tex]

    The scale of the X2,T2 axes is not the same as the scale of the first two, but the scale relative to the X0,T0 axes can be easily found as:

    [tex]
    \frac{\mbox{scale(X2,T2)}}{\mbox{scale(X,T)}}\times \frac{\mbox{scale(X,T)}}{\mbox{scale(X0,T0)}} = \sqrt{\frac{1+v_{2}^2}{1-v_{2}^2}} \times \sqrt{\frac{1-v_{0}^2}{1+v_{0}^2}}.
    [/tex]

    The orthogonal axes X,T are now essentially 'eliminated' and we are left with a Loedel diagram with three inertial frames on it.

    If [itex]v=\sin(\varphi)=0[/itex], then [itex]v_0 = v_1 = 0[/itex] and [itex]v_{2L}[/itex] reduces to [itex]v_2[/itex], the 'Minkowski velocity'. Likewise, the relative scale of the X2,T2 axes then reduces to that of the Minkowski spacetime diagram. Hence the "title question'': Is the Loedel diagram a special case of the Minkowski spacetime diagram? Or is it perhaps the other way round?

    It is clear that the Loedel diagram wins in the simplicity stakes when there are only two reference frames involved, but the Minkowski diagram is simpler to use when additional reference frames are required. After all, additional Loedel frames need to be converted to 'Minkowskian' before being added. This probably explains why the Loedel diagram has never really caught on, except perhaps in beginners teaching.

    ** Attachments successfully uploaded.
     

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    Last edited: Dec 10, 2007
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  3. Dec 9, 2007 #2

    Chris Hillman

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    OT comments about why wise users avoid downloading PDFs

    Hi, Jorrie,

    Be aware that opening pdfs can be dangerous. Be aware also that at least one website put up by a PF user has been flagged by Google as containing malware. So I for one feel that there are good reasons to avoid pdfs made available at a website which is not known to follow good security practices.

    I hasten to add that the website you cited is authored by Burt Jordaan, an aerospace engineer in Praetoria, has not been flagged by Google, and seems fairly mainstream at a glance, so chances are that everything is fine there. Nonetheless, I for one am reluctant to download any of the pdfs available there.

    Suggest you make your own gif figure and upload it as an attachment to your post; as I understand it, the mentors will scan this for malware before making it "live". This should provide some protection to PF users.
     
  4. Dec 9, 2007 #3

    robphy

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    I need a little time to respond to the OP in depth.
    For now... there's more to a "Minkowski spacetime diagram" than just being simply a "diagram" , a "graphical representation", or something akin to a nomograph. There is a geometry that is invariant under the Lorentz Transformation.

    Concerning PDFs... one might be able to use a web-based PDF viewer
    http://www.google.com/search?q=pdf+viewer+online
    which handles the PDF and gives the browser graphic images of the pages.
     
  5. Dec 9, 2007 #4

    Jorrie

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    Loedel attachments still an issue

    Hi Chris.

    I understand your concerns and will upload the two figures as soon as I can establish a workable connection with PF again. Still not sure whether the problem is on my side or if PF has a server problem. (All other sites appear to work normally.)

    To ease concerns, the hosting company of the website where I posted the pdf is well-known and uses a database to where everything is uploaded. Some processing is done before things go live, much the same as on PF. But I agree that one should rather err on the safe side...

    Jorrie
     
  6. Dec 9, 2007 #5

    Jorrie

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    Loedel and Lorentz invariance

    The Loedel diagram sports the same invariant features than the Minkowski diagram, at least for the first two inertial frames. I think it does the same for additional frames that are added correctly, but I will appreciate your input on that.

    Jorrie
     
  7. Dec 9, 2007 #6

    robphy

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    Last edited: Dec 9, 2007
  8. Dec 9, 2007 #7

    Jorrie

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    Loedel or Brehme?

    Rob, yes I'm sure it's Loedel. I first noticed the difference in Lindner's paper: "A Bridge to Modern Physics", http://www.wooster.edu/Physics/Lindner/Ph205Fall2006/ModernPhysics.pdf, page 35. (Hope it's a safe site!:wink:)

    The Brehme drops lines orthogonal onto the closest axis and has some peculiar characteristics (Shadowitz). The Loedel retains the parallel to axis projection, just like Minkowski. Both Brehme and Loedel maintains the invariance of the speed of light and of the spacetime interval, but the Loedel is preferred above the Brehme by most I've seen.

    Lastly, nope, it's not connected to a "center of mass frame" - just straight Loedel, with my "extension" to add additional inertial frames.

    Jorrie
     
  9. Dec 10, 2007 #8

    robphy

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    Some more thoughts...

    It seems that the Loedel and Brehme diagrams cling to Euclidean-orthogonality and Euclidean angles...which are foreign to the Lorentz-group (except in the spatial-rotation subgroup).

    In fact, your manuscript describes the Minkowski diagram by
    "The Minkowski diagram uses different scales for the orthogonal and the non-orthogonal axes".
    Of course, from a Minkowskian-spacetime viewpoint an observer's time axis is Minkowski-orthogonal to his space-axis"... and the claim of different scales is due to improperly imposing a Euclidean-metric on the diagram, as opposed to the Minkowski-metric with its "circles" (the hyperbolas of constant interval). By the way, a similar claim of different scales should be made with a Galilean spacetime diagram.... since any two Galilean-spacetime vectors with the same temporal-component have the same norm (i.e. tick off the same amount of Galilean proper time].

    In Fig.1, using the color-codes you employ,
    the red observer has axes X1 and T1, and blue has X0 and T0.
    In your diagram,
    X1 is drawn Euclidean-orthogonal to T0,
    and
    X0 is drawn Euclidean-orthogonal to T1.
    This is what one would expect from drawing the spacetime diagram of the two observers from an observer in the center of mass frame (assuming they all had the same rest mass)... where the two observers are symmetrically related to this frame.
    However, note that this Euclidean-orthogonal relationship doesn't hold in general... as you can see indirectly in your Fig.2 where a third observer's axes are drawn, although none of his axes are Euclidean-orthogonal to any of the axes already drawn. More directly, upon applying a boost to Fig. 1, the above axes are no longer drawn Euclidean-orthogonal to each other.

    In addition, what do the non-relativistic limits of these diagrams look like? I suspect that the Euclidean-orthogonality relationships above are lost altogether.

    At this point I don't see anything wrong with such diagrams... but I don't see their long-term benefit. They may be useful for a particular type of problem... but I suspect that the handholding needed to explain the diagram is possibly comparable to that needed to explain the Minkowski diagram. As you know, the Minkowski approach (inspite of its claimed difficulties*) is more versatile and (in hindsight) naturally generalizable to general relativity. So, the effort is probably better spent studying the Minkowski diagram.

    *I claim that some of the claimed difficulties of the Minkowski diagram can be overcome by paying a little more attention to the [noneuclidean] geometry.... which can be motivated more physically by "operational" definitions... although this hasn't been done much in the literature.
     
  10. Dec 10, 2007 #9

    Jorrie

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    I think the same would hold for any two observers that move symmetrically in opposite directions, irrespective of rest masses.

    Would it not be basically the same as the non-relativistic limit (c -> infinity) for the Minkowski diagram?
    [Edit] BTW, what does the non-relativistic limits look like on a Minkowski diagram?[/Edit]

    I agree that the Minkowski spacetime diagram is more general and hence the preferred one for tuition.
     
    Last edited: Dec 10, 2007
  11. Dec 10, 2007 #10

    robphy

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    Yes... I used "center of mass" or "center of momentum" because they are more familiar "centers". Maybe "center of velocity" is an appropriate term.

    Yes... but my point is that the apparent desire for Euclidean-orthogonal axes is not longer satisfied. Minkowski- (or in the nonrelativistic case, Galilean-) orthogonality is what is important. For both diagrams, in the nonrelativisitc limit, the light cone effectively flattens out, resulting in the spatial axes coinciding with the light cone.


    Yes.
    ...although I feel that it is still worthy to clearly and fully lay out these alternative diagrams [e.g. Loedel, Brehme, Epstein] ( as described in an earlier thread of ours https://www.physicsforums.com/showthread.php?p=1211719#post1211719 ) and, in my viewpoint, dispel the claims of difficulty attributed to the Minkowski spacetime diagram [which, by the way, is about to turn 100 years old! http://www.spacetimesociety.org/conferences/2008/ ]

    If the point of the Loedel diagram is to analyze a certain kind of problem in a convenient frame of reference, then I would think that it would be helpful if it used Minkowskian-geometry and trigonometry... i.e. rapidity (the Minkowskian-angle) and the hyperbolic trig functions rather than the unnatural-for-spacetime Euclidean ones.
     
    Last edited: Dec 10, 2007
  12. Dec 10, 2007 #11

    Jorrie

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    Minkowski-year project

    I can understand the last sentence, but for c -> infinity, the basis of the time axes (cT) is problematic, since there can be no temporal 'movement' for anything. Does that not destroy the Minkowski geometry as well? (Just curious, not in defense of the Loedel approach!)

    Maybe we can consider a joint project on this for the Minkowski-year(?).
     
    Last edited: Dec 11, 2007
  13. Dec 11, 2007 #12

    robphy

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    Some assorted comments:

    The line x=vt represents the worldline of a particle traveling with speed v.
    So, if v=cmax, the maximum signal speed, then as cmax tends to infinity the line becomes more horizontal (on a Minkowski spacetime diagram with the usual orientation where loosely speaking "time is upwards").

    In the designation of "cT" as a coordinate,
    the "c" is first any nonzero velocity so that "cT" has units of length. Here "c" is merely conversion-constant.

    Then, second, for convenience in SR, this conversion-constant "c" is chosen to equal the maximum-signal-speed cmax (a physical quantity) so that light(which travels as the maximum signal speed of SR) travels along "45-degree" lines in the diagram.

    In taking the nonrelativistic limit, one should decouple the two c's so that only cmax tends to infinity.

    There is certainly "temporal movement" in the sense that proper-time advances.... it's just that in the nonrelativistic case the elapsed proper-time between two events doesn't depend on the spacetime-path taken.

    Realize that the nonrelativistic limit of a (Minkowski) spacetime diagram is simply the familiar distance-vs-time graph (a Galilean spacetime diagram) from PHY101.

    A collaboration on this would be fun and fruitful.
     
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