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Derivation of Hyperbolic Representation from Lorentz/Minkowski equations in SR

  1. Sep 10, 2010 #1
    This is a carryover from a previous thread:

    https://www.physicsforums.com/showpost.php?p=2875138&postcount=68

    Sports Fans:

    I am familiar with the Minkowski equations and the Lorentz transformations in one or two dimensions:

    A) In algebraic form
    (1) t2 - x2 = t'2 - x'2

    (2) t' = [itex]\gamma[/itex](t - xv/c2)
    (3) x' = [itex]\gamma[/itex](x - vt)

    Now, by using equations (2) and (3) and doing the right substituions, one gets equation (1) and by starhaus:

    Thus [itex]\tau^2[/itex] is invariant which is saying the same thing, no matter where on the hyperbola your x and t are, wherever you go on that same hyperbolic curve, [itex]t^2 - x^2 = \tau^2[/itex]

    Notice I am using the timelike format.

    Now, the equivalent Lorentz equations in hyperbolic format are:

    B) In hyperbolic form.
    (4) t' = -xsinh [itex]\vartheta[/itex] + tcosh [itex]\vartheta[/itex]
    (5) x' = xcosh [itex]\vartheta[/itex] - tsinh [itex]\vartheta[/itex]

    where

    (6) sinh [itex]\vartheta[/itex] = [itex](v/c)\gamma [/itex]
    (7) cosh [itex]\vartheta[/itex] = [itex]\gamma[/itex]

    and where

    (8) [itex]\vartheta[/itex] = tanh-1(v/c)

    Now, I have tried, tried and tried and I cannot get from A) to B)

    In the figure posted we have the t-axis vertical and the x-axis horizontal and this is a time-like representation:

    In this equation that starthaus presented:

    c2t2 - x2 = [itex]\tau^2[/itex]
    [itex]\tau [/itex] should be the t intercept at x = 0

    any point on the hyperbola should be represented by the parametric equations of

    x = [itex]\tau[/itex]sinh [itex]\phi[/itex] and
    t = [itex]\tau[/itex]cosh [itex]\phi[/itex]

    where tanh-1 [itex]\phi[/itex] = x/t for any point (x, t) on the hyperbola

    How do I make the connection?
     

    Attached Files:

  2. jcsd
  3. Sep 10, 2010 #2

    Fredrik

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    Hey, didn't I tell you that once already? :biggrin:

    https://www.physicsforums.com/showthread.php?p=2678572

    (Start at "This is what I wrote in my notes:". After the first sentence, you can skip the stuff about adding velocities, and continue at "The definition of rapidity...".)
     
  4. Sep 11, 2010 #3
    You have explicitly stated c in some equations, but not in others. Here are the same equations with all the c's explicitly stated:

    Now starting from the Lorentz transformation for time:

    [tex]t' = \gamma(t-xv/c) = t(\gamma) - (x/c)(v/c)\gamma = t\, cosh\, \vartheta - (x/c)\, sinh\, \vartheta [/tex]

    which is (2) --> (4)

    and starting from the Lorentz transformation for distance:

    [tex]x' = \gamma(x-vt) = x(\gamma) - (vt)\gamma = x(\gamma) - ct(v/c)\gamma = x\, cosh\, \vartheta - ct\, sinh\, \vartheta [/tex]

    which is (3) --> (5).

    Now I am sure you are looking for something much deeper than that, but I thought we might as well get the algebraic relationship between the hyperbolic and regular expressions established in a (hopefully, barring typos) consistent way first.
     
    Last edited: Sep 11, 2010
  5. Sep 11, 2010 #4
    I think that should be:

    (not tanh-1)
     
  6. Sep 11, 2010 #5
    Fredrik -

    That post: https://www.physicsforums.com/showpost.php?p=2678572&postcount=36 is way beyond me. This may disqualify me from going further but that is the case.

    I haven't a clue what you wrote. Might as well be speaking Martian or some other alien tongue.

    Doc
    stevmg
     
  7. Sep 11, 2010 #6
     
    Last edited: Sep 11, 2010
  8. Sep 11, 2010 #7
    Continuation from post #6 above,

    I still can't get from A) to B) so my error isn't withing the typos but within the brain!

    Doc
    (by the way, when you address me in a post you can and should call me "Doc" as that's what everyone in the Air Force used to call me (better than some of the other phrases that I can't repeat here on PF.)
    stevmg
     
  9. Sep 11, 2010 #8

    Fredrik

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    I didn't expect you to understand all of it, but I'm surprised that you that you claim to understand nothing of it. This is the essential part:

    What part of that is impossible for you to understand? The third and fifth lines are the results that take you from A) to B).

    Explanation of the equalities:

    First line: 1) Definition of rapidity ([itex]\tanh\phi=v[/itex]). 2) Definition of tanh. 3) The identity [itex]\cosh^2x-\sinh^2x=1[/itex], which is easy to derive from the definitions of sinh and cosh. 4) For all real numbers A,B with A non-zero, we have (A+B)/A=1+B/A.

    Second line: 1) The result from the first line. 2) Definition of [itex]\gamma[/itex].

    Third line: 1) The square root of the result in the second line.

    Fourth line: 1) The identity [itex]\cosh^2x-\sinh^2x=1[/itex] again. ([itex]\cosh^2[/itex] should of course be [itex]\cosh^2\phi[/itex]. I left out the variable by accident). 2) The result from the third line. 3) Definition of [itex]\gamma[/itex] and the fact that 1=A/A for all non-zero real numbers A. 4) For all real numbers A,B,C with C non-zero, we have A/C+B/C=(A+B)/C. 5) Definition of [itex]\gamma[/itex].

    Fifth line: 1) The square root of the result in the fourth line.
     
    Last edited: Sep 11, 2010
  10. Sep 11, 2010 #9
    I'll download it to paper and with my calculus book that goes into hyperbolic functions and their derivatives I will pour over it until it sticks in my brain.

    Below, I have posted a part of a lecture that sort of proves it, too but this author just made a wild a-- guess which turns out to be true (equation 32) or [itex]\beta = v/c[/itex] which is just a WAG. Nothing like knowing the answer before you take the test. How do you think we doctors got through medical school? Certainly not on out of genius.

    Lorentz-hyperbolictransformations.jpg
     
  11. Sep 11, 2010 #10

    Fredrik

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    I wouldn't describe (32) as a guess. It's the definition of rapidity. (The same one I'm using, except that I'm calling it [itex]\phi[/itex] instead of [itex]\beta[/itex]. I don't like calling it [itex]\beta[/itex], because [itex]\beta[/itex] is often defined as v/c). Of course, when it's presented this way, it looks like we were just lucky that the definition turned out to be useful.

    I actually made the edit that added the quote above before I saw this below: :smile:
     
  12. Sep 11, 2010 #11
    LOL. Pesky c's. Missed a couple myself. Things are still not quite right.

    (2) ct' = (ct - xv/c2)

    should be (2) ct' = [itex] \gamma [/itex](ct - xv/c)

    (3) x' = (x - vct)

    should be (3) x' = [itex] \gamma [/itex](x - vt)

    (4) t' = -xsinh [itex]\vartheta[/itex] + ctcosh [itex]\vartheta[/itex]

    should be (4) ct' = -x sinh [itex]\vartheta[/itex] + ct cosh [itex]\vartheta[/itex]

    For some reason you dropped some gammas too, that I have put back in. You can probably go from there now.

    (1),(5),(6),(7) and (8) are OK.

    I have edited post #3 for the "c" I missed in equation (1) so all the equations in #3 should be exactly correct now, but they are presented slightly differently to the equations presented here.

    The complete and hopefully finally correct set of equations should be:

    (1) c2t2 - x2 = c2t'2 - x'2

    (2) ct' = [itex] \gamma [/itex](ct - xv/c)
    (3) x' = [itex] \gamma [/itex](x - vt)

    (4) ct' = -x sinh [itex]\vartheta[/itex] + ct cosh [itex]\vartheta[/itex]
    (5) x' = xcosh [itex]\vartheta[/itex] - ctsinh [itex]\vartheta[/itex]

    (6) sinh [itex]\vartheta[/itex] = [itex](v/c)\gamma [/itex]
    (7) cosh [itex]\vartheta[/itex] = [itex]\gamma[/itex]

    (8) tanh [itex]\vartheta[/itex] = (v/c)

    (9) x = c[itex] \tau[/itex] sinh [itex]\phi[/itex]
    (10) ct = c[itex] \tau[/itex] cosh [itex]\phi[/itex]

    (If they are not correct, PM me. We can't spend the whole thread correcting typos :tongue: )
     
    Last edited: Sep 11, 2010
  13. Sep 11, 2010 #12
    To yiuop and Fredrik -

    Using the equations [itex](ct)^2 - x^2 = \tau^2[/itex] I KNOW there has to be a more direct and intuitive way to prove the hyperbolic assertions.

    Using the timelike diagram where the ct axis is up and down and the x-axis ls the left and right, [itex] sinh \vartheta = x/\tau [/itex] and [itex] cosh \vartheta = ct/\tau [/itex] there has got to be a way of working [itex] v [/itex] and [itex] c [/itex] into this.

    [itex] x = \tau sinh \vartheta [/itex] and [itex] t = \tau cosh \vartheta[/itex]

    [itex] tanh \vartheta = x/t [/itex]

    [itex] \vartheta = arctan x/t [/itex]

    Now, let's get from [itex] x/t[/itex] to [itex] v/c [/itex]
     
    Last edited: Sep 11, 2010
  14. Sep 11, 2010 #13
    Everyone, go to post 12 above. All else is superfluous. I could not edit, correct and re-edit in time.

    Doc
     
  15. Sep 11, 2010 #14

    Fredrik

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    This is the equation of a specific hyperbola in the diagram, so it's not a good starting point if you want to find a result that's valid for all events in the diagram.

    I would say that about post 8. I think you should start by writing down the definitions of sinh, cosh and tanh, and using them to prove that cosh2x-sinh2x=1 for all x. Then study every equality in post 8 carefully. I don't think there is a way to do it that is simpler than the way I did it.
     
  16. Sep 11, 2010 #15
    Perhaps we should start with two images:
    268761_f520.jpg
    271083_f520.jpg

    Source: http://hubpages.com/hub/Hyperbolic-Functions
     
  17. Sep 11, 2010 #16
    Maybe this is a place to start (using c=1 as recommended by Fredrick :wink: ).

    The hyperbolic identity (similar to the trigonometric identity) is defined as:

    [tex]cosh^2(\phi) - sinh^2(\phi) = 1[/tex]

    From this we can create:

    [tex]\tau^2 cosh^2(\phi) - \tau^2sinh^2(\phi) = \tau^2[/tex]

    We also know the Minkowski metric is:

    [tex]t^2 -x^2 = \tau^2[/tex]

    So we can say:

    [tex]\tau^2 \, cosh^2(\phi) - \tau^2 \, sinh^2(\phi) = t^2 -x^2[/tex]

    From the above equation it is easy to make the natural associations:

    [tex]\tau \, cosh (\phi) = t [/tex]

    [tex]\tau \, sinh(\phi) = x[/tex]

    and:

    [tex]v = \frac{x}{t} = \frac{\tau \, sinh(\phi)}{\tau \, cosh(\phi)} = tanh(\phi)[/tex]

    Now we already know that gamma is equal to [itex]t/\tau[/itex] so from [itex]\tau \, cosh(\phi) = t [/itex] we can get [itex]cosh (\phi) = (t/ \tau) \rightarrow \, cosh(\phi) = \gamma [/itex] and from [itex]\tau \, sinh(\phi) = x[/itex] we get [itex]sinh(\phi) = (x/\tau) = \gamma(x/t) = \gamma v[/itex] which is the celerity or proper velocity (w). The mysterious quantity [itex]\phi[/itex] is defined as the hyperbolic velocity angle or rapidity and is obtained from the preceding equation as [itex]\phi = asinh(w)[/itex]. The rapidity can also be obtained in various other forms from any of the earlier equation that include [itex]\phi[/itex].
     
    Last edited: Sep 12, 2010
  18. Sep 12, 2010 #17
    Passionflower - I have seen those diagrams at hubpages before

    How does that get one from
    [itex] x = \tau sinh \vartheta [/itex] and [itex] t = \tau cosh \vartheta[/itex]

    [itex] tanh \vartheta = x/t [/itex]

    [itex] \vartheta = arctan x/t [/itex]

    Now, let's get from [itex] x/t[/itex] to [itex] v/c [/itex]?

    Make the connection for me. Right now all I see is "true - true - not related."

    Doc
     
  19. Sep 12, 2010 #18
    yiuop, I again remove the quotes so that we have the text in "real time" and we can work with it, is that OK? You liked that as I used it before and also I tried your "quotes within quotes" and it didn't work for me, as nothing ever does.

    The hyperbolic identity (similar to the trigonometric identity) is defined as:

    [tex]cosh^2(\phi) - sinh^2(\phi) = 1[/tex] - OK, even though I am not being sarcastic, that's rocket science!

    From this we can create:

    [tex]\tau^2 cosh^2(\phi) - \tau^2sinh^2(\phi) = \tau^2[/tex] beautiful!

    We also know the Minkowski metric is:

    [tex]t^2 -x^2 = \tau^2[/tex] Thank God for starthaus!

    So we can say:

    [tex]\tau^2 \, cosh^2(\phi) - \tau^2 \, sinh^2(\phi) = t^2 -x^2[/tex] by algebraic substitution

    From the above equation it is easy to make the natural associations:

    Now, where does this come from?
    [tex]\tau \, cosh (\phi) = t [/tex]

    [tex]\tau \, sinh(\phi) = x[/tex]

    Oh yes! I think I postulated that in my prior post but that was based on the geometry of [itex](ct)^2 - x^2 = \tau^2[/itex]. That's the basic geometric definitions of hyperbolic cosines and sines.

    and:

    [tex]v = \frac{x}{t} = \frac{\tau \, sinh(\phi)}{\tau \, cosh(\phi)} = tanh(\phi)[/tex]
    By algebraic division and equality. Good, so far we are on the exact same page. Ge that, folks? [itex]v = x/t[/itex]

    Now we already know that gamma is equal to [itex]t/\tau[/itex] How did we get this?

    is that because [itex]\gamma = 1/\sqrt{1 - v^2)}[/itex] We are assuming c = 1 for ease of calculations
    = [itex]t/\sqrt{t^2 - x^2)}[/itex]
    = [itex]t/\tau[/itex]

    THERE! I was right! It is s-o-o-o nice to be such a genius!


    so from [itex]\tau \, cosh(\phi) = t [/itex] we can get [itex]cosh (\phi) = (t/ \tau) \rightarrow \, cosh(\phi) = \gamma [/itex] and from [itex]\tau \, sinh(\phi) = x[/itex] we get [itex]sinh(\phi) = (x/\tau) = \gamma(x/t) = \gamma v[/itex] which is the celerity or proper velocity (w). The mysterious quantity [itex]\phi[/itex] is defined as the hyperbolic velocity angle or rapidity and I already knew this. starthaus got me going on this - a way of adding up velocities in hyperbolic functions directly is obtained from the preceding equation as [itex]\phi = asinh(w)[/itex]. The rapidity can also be obtained in various other forms from any of the earlier equation that include [itex]\phi[/itex].

    Now, Passionflower, starthaus, DrGreg, et al... This is what I have been looking for!

    You have to approach me like I'm four years old and assume I know less than nothing, because that's true. What is obvious to you and unspoken is totally oblivious to me. But guess what now? It isn't any more because I was brought from A [tex]\rightarrow[/tex] B and could understand every algebraic step on the way and it was intuitive. I AM teachable. Lot of work, but it can be done!!

    You know something, if I ever got to learn this stuff I would be the best damn teacher in the world because I would know every nook and cranny and blind alley the students would go through and would know how to get them out of it. When I taught medicine, I was the same way. I read every pertinent text and made every mistake you could make so when my students and residents made mistakes I was already there, knew what they did wrong, why they did it wrong and got them going on the right track.

    Such is the price of a learning disability, when you get it right, you really get it right.

    yiuop, DrGreg, Passionflower, starthaus - thanks for your help. Couldn't have done it without you.

    Doc
     
  20. Sep 12, 2010 #19

    Fredrik

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    Start doing what? It's not clear what you're trying to do in this post. Are you trying to explain how someone who hasn't yet done the calculations I did in #8 might have guessed that [itex]\tanh\phi=v[/itex] would be a useful definition?

    This is just the equation of a hyperbola. If you're going to use it, you should explain why it's relevant. The Minkowski metric has the property that such hyperbolas are preserved by linear transformations that preserve the metric (i.e. Lorentz transformations), but is that relevant for what you're trying to do?
     
    Last edited: Sep 12, 2010
  21. Sep 12, 2010 #20

    Fredrik

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    I don't see how those diagrams could be helpful. Doc, what you want to prove is a simple algebraic fact. You just want to show that the definition [itex]\tanh\phi=v[/itex] implies [itex]\cosh\phi=\gamma[/itex] and [itex]\sinh\phi=\gamma v[/itex]. The easiest way to do that is to do exactly what I did in #8. I explained every equality in that post, so there's no reason for you to ignore it.
     
  22. Sep 12, 2010 #21
    Fredrik - not ignoring it. Have downloaded it onto paper and will pour over it.

    Do not avoid your input into my future threads or postings. Trust me - all is appreciated and so is yours (most highly, I say.)

    And thanks for calling me "Doc" - reminds me of the good old days when the troops called me that as I was their "father protector."

    Doc
     
  23. Sep 12, 2010 #22
    Fredrik -

    I have not ignored what you posted and have included it below in all germane posts regarding the subject from you. I will place the quotes in this post and then, in the following post (so that I can get PF to "swallow it" without kicking me out as too long) I will discuss it.

    Go to post #23, https://www.physicsforums.com/showpost.php?p=2877552&postcount=23

    Doc
     
  24. Sep 12, 2010 #23
    Continued from post #22 above, https://www.physicsforums.com/showpost.php?p=2877463&postcount=22

    Fredrik-

    I tried numerous times to answer you but PF would not let me continue with symbols, even with small posts as it would lock up and go no further. Very frustrating.

    You have the posts I quoted above. I will write out what I was trying to say but "in prose" as the symbols do not work.

    Basically we have the equation of a hyperbola tau^2 = t^2 - x^2 (for below, tau = SQRT(t^2 - x^2) with y intercept of +- tau at x = 0. There are no x-intercepts.

    sinh phi = x/tau
    cosh phi = t/tau, tanh phi = x/t

    gamma = 1/SQRT(1 - v^2)

    If v = x/t, tanh phi = x/t = v (this is "proven" and not assumed)

    Then gamma = 1/SQRT(1 - x^2/t^2) = t/tau

    Substituting back into the Lorentz transformations:

    x' = xt/tau - v(t/tau)t = xcosh phi -tsinh phi
    t' = (t/tau)t - (t/tau)xv = tcos phi -xsinh phi or the more familiar -xsinh phi + tcos phi which can be represented in matrix format which I cannot do as yet on PF.

    This makes intuitive sense. For a given tau we have a hyperbola which represents all x's and t's which are different coordinates in different frames of reference but which "go back" to the same "proper time" or "tau." If one were totally inertial and not moving, x would = 0. If one were moving, for the initial FR where x does = 0, then any x and t on the hyperbola would represent the coordinates which would calculate back to that initial inertial frame where x = 0. It would take a speed of x/t = v to get to that point from x = 0 and t = tau.

    I hope this makes sense and there are no assumptions other than what Minkowski himself projected that tau^2 = t^2 - x^2.

    This is for timelike phenomena. Spacelike should be similar but the t's and x's would be reversed and the hyperboa would be "left and right" and a proper distance would be
    distance^2 = x^s - t^2 and x = tau cosh phi while t = tau sinh phi.

    I hope I am right.

    Doc
     
    Last edited: Sep 12, 2010
  25. Sep 12, 2010 #24
    That was a freakin' bear!

    By the way, what is this "[tex]\Lambda[/tex]" from the above post with the matrices? I've seen it before but haven't got a clue what it represents!

    stevmg
    Doc
     
  26. Sep 12, 2010 #25

    jtbell

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    [itex]\Lambda[/itex] is a common symbol for the matrix form of the Lorentz transformation.
     
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