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Determination of Lorentz transform from euclid geometry

  1. Aug 9, 2014 #1
    A first stage of the determination.

    We have a body of length L = AB, which moves along the x-axis with velocity v,
    say that coming to us.

    A -------- B v <---
    | h - vertical distance
    | ./ - a light converges to us with the speed eq. c
    | /
    O ---------> x

    At some time t = 0, we see the point A at the x = 0;
    which means, of course, that the point A is actually in a different place,
    because the time of flight of light is: dt = h/c, that is, the body moved to the x = -vdt - it's obvious?

    But we are interested how we see at time t = 0 the whole body,
    which is where we see the other end - the point B?
    x_B (t = 0) = ?
    Last edited: Aug 9, 2014
  2. jcsd
  3. Aug 9, 2014 #2

    Doc Al

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    Realize that the Lorentz transformations relate calculated positions and times after the travel time of light is accounted for, not raw observations of what some particular observer literally "sees". (Usually one assumes observers everywhere in a frame, so that light travel time can be neglected.)
  4. Aug 9, 2014 #3
    You can assume such method of time sync., ie.:
    t = 0 is a time of the emission moment of a light (at a point),
    instead of the reception moment by the observer.

    I intend only to recognize the true nature and validity of the Lorentz transformation, using only elementary geometry; and this should work good... after all we can see the moving bodies from a distance,
    under some conditions, which are not so difficult to recognise.
    Last edited: Aug 9, 2014
  5. Aug 9, 2014 #4

    Doc Al

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    You'll need more than just geometry--you'll need some physics.
  6. Aug 9, 2014 #5


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    Lorentz transformations are rotations in hyperbolic 4-S of Minkowski.
  7. Aug 9, 2014 #6
    What we are missing... what we still need to determine the observed position of the point B, for the given conditions?
  8. Aug 9, 2014 #7
    And what of that?
    You might as well say that the complex numbers lie in the complex plane... the real numbers on a line, vectors in the vectors space, ect.
  9. Aug 9, 2014 #8


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    I don't see how, even in principle, you can hope to derive ##ds^2=-c^2 dt^2+dx^2+dy^2+dz^2## from ##ds^2=dx^2+dy^2+dz^2##
  10. Aug 10, 2014 #9


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    What of that?
    Minkowski space is more than a mathematical object cf.- "real number", "vector". Minkowski space has the
    physicality of S.R. Note the confusion of "Dale Spam" above. In principle you cannot derive the interval equation from an elliptical space its source is hyperbolic thus the minus sign.

    With all due respect,

    Last edited: Aug 10, 2014
  11. Aug 10, 2014 #10


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    I agree with your determination of where point A is when it is viewed at t=0 by O but unless you know that the length of the body is contracted and by how much, how can you determine where point B is at t=0?

    I believe this is the physics that Doc Al said you would need in post #4.
  12. Aug 10, 2014 #11


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    I'm not sure what you're trying to do, the remark about hypervolic rotations was terse, but got to the essence of the issue - at least one staement about what you were trying to do.

    To go into a bit more detail on this approach:

    Consider rotations in a 2d Euclidean space, with cartesian coordinates (x,y). You can write:

    x' = cos(theta) x - sin(theta) y
    y' = -sin(theta) x + cos(theta) y

    Rotations preserve distances i.e x^2 + y^2 = x'2 + y'2

    Now consider a lorentz boost in a 2d Minkowski space, with coordinates (t,x) You can write

    x' = cosh(rapidity) x - sinh(rapidity) c t
    c t' = -sinh(rapidity) x + cosh(rapidity) c t

    See the wiki article on rapidity for more info on rapidity, http://en.wikipedia.org/wiki/Rapidity

    Note that tanh(rapidity) = v/c

    The lorentz boost transformation above preserves the Lorentz interval, i.e x^2 - (ct)^2 = x'^2 - (c' t)^2, just as the rotation preserved distance. It's easy to verify both statements with a bit of algebra.

    Notice that for a light beam either x = ct or x = -ct. The lorentz interval is just (x-ct)(x+ct), so a zero Lorentz interval between two events is a mathematical characteristic of them being connected by a light beam.

    WHen you preserve the Lorentz interval, you also preserve the nature of lightlike separation, with light moving at a constant velocity c. This happens because the Lorentz interval between two events is zero if they are connected by a lightbeam, and if two events are connected by a lightbea, the Lorentz interval is zero, both conditions are equivalent, as either one implies the other.

    Because the Lorentz transform preserves the Lorentz interval, it insures that any two events that are connected by a lightbeam moving at a velocity c in (t,x) coordinates are connected by a lightbeam moving at velocity c in (t', x') coordinates.

    Thus the self consistency of Lorentz transformations should be obvious at this point. As others have remarked, starting with this mathematical insight, you also might want to do some physics, The mathematical existence and motivation of the transform is hopefully is clearer now (?), but the work to apply them to physics still needs to be done.
  13. Aug 10, 2014 #12
    Distance from A to B is equal L: |AB| = L;
    It can be rather easily measured, thus we have no problem with this.

    In this stage we don't know yet whether the length L is a contracted version of some: L' = k*L, where k is unknown; just the L is given.

    This is a further stage... now we want just to predict the observed position of the point B.

    At this stage, we are interested in: how will look the moving object? And that's all;
    ie. we have to solve a problem of visualization the moving object.
    Last edited: Aug 10, 2014
  14. Aug 10, 2014 #13
    OK. I solved something already and discovered an equation - the condition of the visibility a moving point x at the same time of reception:

    ## h^2 + (x - vt)^2 = (h - ct)^2 ##

    for x = 0 we get: t = 0, this is the point A at x = 0, thus we observe the body at a time moment:
    ## t_o = h/c = const ##

    The t parameter is an emisson time moment at a point x of the body: x = [0, L].

    for any other x <> 0 we get a negative t, this just means we see the image from the past - the more distance to the point of emission, the more time needed for a light to reach us at the same time moment of observation: t_o.

    The equation is a hyperbola wrt the parameters (x,t), thus we probably have some good symptom of the Lorentz transform.
    Last edited: Aug 10, 2014
  15. Aug 10, 2014 #14

    Doc Al

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    Exactly. (Or the equivalent.)
  16. Aug 10, 2014 #15


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    I think you are trying to analyze what is called the Terrill effect which indicates that an object like a rod will appear elongated as it approaches and contracted as it recedes. But I don't think your equations or your idea has anything to do with the Lorentz Transform or with the actual visualizations. It's a very complex subject and threw a lot of people off for a long time.
  17. Aug 10, 2014 #16


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    Euclidean geometry and Lorentz transformations

    Hello Atto:

    The point of my post on Lorentz transformations was that in the end one
    needs the curved space time of Lobochevsky to give the transformations physical
    substance. Perhaps the cleverest derivation of the transformations is that of
    Sir Arthur Stanley Eddington in his "Space, Time and Gravitation" BTW still must
    reading for relativity fans.
    Eddington used as a model for derivation nothing more than swimmers travelling up and
    down a stream and across a stream. Initially the stream has velocity 0 relative to the
    shore then he considers the swimmers (longitudinal and transverse making complete
    circuits). The Lorentz transformations fall out when you calculate the RATIO of the
    end net velocities! It's a fun thing just remember that the observer role is the stream not
    either of the swimmers.
  18. Aug 10, 2014 #17
    For what reason do you think so?

    The condition of the 'visibility' of a body is only one:
    the light from the body parts must reach us at the same moment of time.

    And this is just the equation of the hyperbola i wrote.
  19. Aug 10, 2014 #18
    By the way: someone probably is looking for the power of the devil in this transformation.
    Sorry, but there is nothing in it beyond a simple geometry.
  20. Aug 10, 2014 #19


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    Doesn't the hyperbola indicate that the perceived length of the object is symmetrical as it approaches and as it recedes?
  21. Aug 11, 2014 #20
    No. The observed image of a moving object isn't symmetric.

    You must solve the equation, ie. write it in an explicit form: t = t(x),
    and then you just compute the times for two symmetric points, ie: x+ = a, x- = -a.

    And the observed position of a moving point x, is just a position of the point at a time moment of emission:

    x_o = x - vt; where t is the computed time of emission for a point x.
  22. Aug 11, 2014 #21


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    Then why don't you just do it and show us how it looks as the object approaches, passes and recedes?
  23. Aug 11, 2014 #22
    OK. Here you are computed the two points:

    For simplicity, let: c = 1; h = 1;

    and: v = 0.8, gamma = 1.66666666...

    The solutions:

    approaching: x = 1: t = -1.201
    receding : x = -1: t = -0.270

    Thus the observed positions are just that:
    +1' = 1 + 0.8*1.2 = 1.96
    -1' = -1 + 0.8*0.27 = -0.784
  24. Aug 11, 2014 #23


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    Can you please show all the details of the derivations? Pretend like I'm really ignorant.
  25. Aug 11, 2014 #24
    I forgot a version compatible with a 'contracted' body.

    When the body is physically shortened then the point x is actually in the position: x/gamma = x*0.6, for v = 0.8;

    therefore we must compute for x = +/-0.6, instead of +/-1:
    x = 0.6; t = -0.312
    x =-0.6; t = -0.12

    and now we observe:
    +1' = 0.6 + 0.8*0.312 = 0.85
    -1' = -0.6 + 0.8*0.12 = -0.5

    The equation:
    [tex]y^2 + (x-vt)^2 = (y-ct)^2[/tex]
    and we simply solve for the variable t:

    ## y^2 - 2yct + (ct)^2 - y^2 - x^2 +2xvt - (vt)^2 = 0 ##
    ## t^2(c^2-v^2) - 2t(yc - xv) - x^2 = 0 ##

    ## t = \frac{(yc-xv) \pm \sqrt{x^2(c^2-v^2) + (yc-xv)^2}}{c^2-v^2} ##

    for y = 1, c = 1 this is:
    ## t = \frac{(1-xv) - \sqrt{x^2(1-v^2) + (1-xv)^2}}{1-v^2}=\frac{(1-xv) - \sqrt{x^2-2xv+1}}{1-v^2}##
    Last edited: Aug 11, 2014
  26. Aug 12, 2014 #25


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    You've lost me. I thought you were trying to determine the x-location of point B when the image of point B and the image of point A both arrived simultaneously at point O at time t=0. We already agreed that point A is at some positive x-value under this circumstance. It would help if you would describe your variable in those terms. I can't even tell if this is supposed to be for point B or point A.

    Did you use your equation below to determine the values of t? Why are there two of them? Does one of them apply to point B and the other to point A?

    What are the primes for? And what do these apply to?

    Please provide a detailed explanation of each step along the way and also an overview of what you are doing.
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