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Derivation of Gamma

  1. Aug 4, 2008 #1
    [tex] \gamma = csc(cos^{-1}(v/c)) [/tex]

    [tex] \gamma^{-1} = sin(cos^{-1}(v/c)) [/tex]

    Is there something wrong with this derivation? If not, why isn't this the more formal way of writing it, it seems a lot more simple?
  2. jcsd
  3. Aug 4, 2008 #2
    I'm unclear what you mean by derivation, but I think you want the following hyperbolic trig relation:

    \gamma = \cosh(\tanh^{-1}(v/c)) = \frac{1}{\sqrt{1-(v/c)^2}}
  4. Aug 4, 2008 #3
    By derivation I just mean a derived form of.

    Yes, or the normal trig relation:
    [tex]\gamma = csc(cos^{-1}(v/c))[/tex]

    In any case, why is this not used, as it's much more simple?
  5. Aug 4, 2008 #4
    I don't think that "simpler" version is correct. You need something that expresses the hyperbolic "distance" of spacetime. For example for a given event that occurs at time t and distance x, the length of a straight path to that event is:

    s^2 = (ct)^2 - x^2

    It's this arc length that is invariant (constant) given other observer velocities, and therefore defines a hyperbola in coordinates ct, and x. With sines and cosines you have sine^2 + cosine^2 = 1 ... there's no minus sign there to express this hyperbolic relationship. You can do it with complex number angles but in that case you are really just saying "I need hyperbolic sine and cosine".
    Last edited: Aug 4, 2008
  6. Aug 4, 2008 #5
    I'm not so sure that it matters, I've tried it with a couple different examples.

    Consider a space ship traveling from Earth to the nearest star system outside of our solar system: a distance d = 4.45 light years away, at a speed v = 0.866c (i.e., 86.6 percent of the speed of light, relative to the Earth). The Earth-based mission control reasons about the journey this way (for convenience in this thought experiment the ship is assumed to immediately attain its full speed upon departure): the round trip will take t = 2d / v = 10.28 years in Earth time (i.e. everybody on earth will be 10.28 years older when the ship returns). The flow of time on the ship and aging of the travelers during their trip will be slowed by the factor \epsilon = \sqrt{1 - v^2/c^2}, the reciprocal of the Lorentz factor. In this case \epsilon = 0.500 \, and the travelers will have aged only 0.500×10.28 = 5.14 years when they return.

    [tex]t' = t/(sin(cos^{-1}(v/c)))[/tex]

    Plugging in t=5.14, and v=.866c, yields the same as the normal time dilation equation.
  7. Aug 4, 2008 #6
    Let u=v/c and see how many trig identities you can come up with. :smile:


  8. Aug 4, 2008 #7
    I assumed without checking that the hyperbolic functions were required, but if I do the trig, you are right and the csc(cos^{-1}()) works too. I'd guess they aren't used since the hyperbolic functions are just more convienent than your identities (which makes logical sense to me since the metric is hyperbolic).
  9. Aug 5, 2008 #8


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    As suggested by others, the hyperbolic functions are more natural for special relativity.

    Given two future-pointing unit-timelike 4-vectors (tangent to the observer worldlines) [itex]\hat t[/itex] and [itex]\hat u[/itex],
    [tex]\gamma = \hat t \cdot \hat u = \cosh(\theta)[/tex]...
    where [tex]\tanh\theta=v/c[/tex] (as seen in Peeter's post).

    Note that, using hyperbolic-trig functions, some of your Euclidean-intuition (via analogies) carries over into special-relativity. Observe that the "dot-product" is associated with the "cosine" function of the "angle-between" two vectors... and that the "slope" is a associated with a "tangent" function. The analogues don't stop there.

    The Euclidean functions (with real-valued angles) don't enjoy such geometric interpretations in special-relativity... unless you are working solely in a hyperplane spanned by spacelike vectors.
    (What would (say) the composition of velocities formula look like with with OP's formulation?)
  10. Aug 5, 2008 #9
    If [tex] cos \phi = \frac {v} {c} [/tex]

    then [tex] sin \phi = \sqrt {1 - cos^2 \phi} = \sqrt {1 - \frac {v^2} {c^2} } = \frac {\sqrt {c^2 -v^2}} {c} = \frac {c'} {c}[/tex]

    and [tex] \gamma = \frac {1} {sin \phi} = \frac {1} {\sqrt {1 - \frac {v^2} {c^2} } } = \frac {c} {\sqrt {c^2 -v^2}} = \frac {c} {c'}[/tex] .
  11. Aug 5, 2008 #10

    I don't understand. Are you agreeing or disagreeing?
  12. Aug 5, 2008 #11
    I agree of course, but the consequence of this geometrical consideration is - the speed of light is in the moving frame [tex]$ c' = \sqrt {c^2 - v^2} $[/tex] with respect to rest frame.

    Einstein wrote in http://www.fourmilab.ch/etexts/einstein/specrel/www/" [Broken]

    Last edited by a moderator: May 3, 2017
  13. Aug 5, 2008 #12


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    You misunderstand Einstein here, he was talking about a coordinate system which was created by doing a Galilei transformation on an ordinary inertial frame--this is not a genuine inertial frame in SR. A while ago on this thread I provided my own summary of what Einstein was doing in that paper, I'll repost it here and put the parts related to the section you quote in bold:
    Last edited by a moderator: May 3, 2017
  14. Aug 5, 2008 #13
    1) This derivation does not work because x' and t are dependent variables x'=(V-v)t and

    2) The assumption 1/2(tau0 + tau2) = tau1 can not be true

    because x'/(V-v)+x'/(V+v) can not be equal to x'/(V-v)+x'/(V-v) unless v=0.
  15. Aug 5, 2008 #14
    It seems to me that the word "sign" would have been more appropriate than "direction" here.

    Also, I'm not quite sure I follow why the length of a vertical rod travelling horizontally might depend upon the magnitude of the velocity (although you imply later that it doesn't without actually mentioning so [Phi(v)=Phi(-v)=1]).


  16. Aug 5, 2008 #15


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    That equation doesn't appear in what I wrote--are you substituting V for c or something? Please quote the section of what I wrote that contains the equation you're talking about so it'll be clear.
    Again, where do the terms x'/(V-v)+x'(V+v) and x'/(V-v)+x'/(V-v) [which unless you've written it wrong is just equal to 2x'/(V-v)] appear in what I wrote?
  17. Dec 5, 2008 #16
    I used original notation V instead of c.
    You write for example following:"...its t' coordinate as a function of x' is t'(x') = x'/(c-v)".

    And as You can see, it means: dx'=(c-v)dt'.

    It is just here, if You put Sx'=St'*(1/(c-v)).

    "With a little algebra, this reduces to:

    (1/2)*St'*(1/(c-v) + 1/(c+v)) = Sx' + St'*(1/(c-v))"
  18. Dec 5, 2008 #17


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    So what did you mean when you said "This derivation does not work because x' and t are dependent variables"? What specific step in the derivation do you think doesn't work because t' is a function of x'?
    Why would you do that? Sx' and St' refer to the slopes of a 2D plane (slope along the x' axis and slope along the t' axis) in the 3D space defined by variables (x', t', tau), with the 2D plane being oriented in such a way as to include three points in this space corresponding to the x', t', and tau coordinates of three events: the event of the light being sent from the origin of the k frame, the event of the light being reflected by a mirror at rest in the k frame, and the event of the light returning to the origin of k. Why do you think these slopes would obey the relation Sx' = St'*(1/(c-v))?
    Last edited: Dec 5, 2008
  19. Dec 6, 2008 #18
    You do not need 2D-plane because x and t are dependent. That means, You need only one independent variable for this problem.

    And further if You compare two formulas for xi, You can see that Einstein changed Beta^2 to Beta in the second set of formulas without any mathematical (or physical) reason.

    He writes Beta instead of c^2/(c^2 - v^2) and this is Beta^2.

    Last edited: Dec 6, 2008
  20. Dec 6, 2008 #19


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    You're being very vague. Maybe you think I don't "need" it, but certainly there's nothing inherently wrong with graphing two dependent variables on different axes, so what specific step in my proof do you think is incorrect? Also, you're talking vaguely about x' and t' without specifying whether you're solely talking about pairs of x',t' coordinates that happen to lie on the path of the original light beam from the origin (it is only coordinates on this line that were meant to obey the relation t'(x') = x'/(c-v)), or whether you're talking about the x', t' coordinates of arbitrary events (including the event with tau-coordinate tau2 of the light that had been reflected by the mirror returning to the origin, which does not lie on the line of the original light beam and whose x' and t' coordinates would not obey the relation t'(x') = x'/(c-v), yet by construction this event must be part of the plane).
    Please be specific. What original formula for xi contains c^2/(c^2 - v^2)? What is the new formula for xi in the "second set of formulas"?
  21. Dec 6, 2008 #20
    I'm talking about differential equation: 1/2(1/(c-v) + 1/(c+v))*(dtau/dt')
    = dtau/dx' + (1/c-v)*(dtau/dt').
    Here is x' a function of t' and in this case one must use chain rule for derivation i.e. d/dx'*dx'/dt'. So the equation changes to: 1/2(1/(c-v) + 1/(c+v))*(dtau/dt')
    = (1/c-v)*(dtau/dt') + (1/c-v)*(dtau/dt')=(2/(c-v))*(dtau/dt').

    These are the original formulas in Einstein paper and in one of your previous posts (I quoted it).

  22. Dec 6, 2008 #21


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    That's the equation he derives from the earlier equation 1/2[tau(0, 0, 0, t0') + tau(0, 0, 0, t0' + xm'/(c-v) + xm'/(c+v))] =
    tau(xm', 0, 0, t0' + xm'/(c-v)), under the assumption that x is "infinitesimally small". Are you claiming there is an error in the derivation, or do you agree that that equation follows from the earlier one?
    The chain rule for derivation of what from what? By the way, note that although I avoided the use of calculus in my derivation by making use of the fact that if you zoom in on an infinitesimally small region of a smooth function it'll look like a flat plane, you can derive the equation above using calculus--see this thread as well as DrGreg's post #6 on this thread.
    You just said something about the chain rule, but here you don't appear to have made use of the chain rule or any other calculus, you just did some algebra (and you forgot to carry over the 1/2 from the first equation, the final one should be (1/(c-v))*(dtau/dt'), not (2/(c-v))*(dtau/dt')).
    Sorry, when I went to respond to your post it didn't include your own quote from my post in the text box, so I didn't notice you had quoted the context of the formulas already, my mistake. Anyway, notice that the first formula contains the unknown function a(v) and the second formula contains the unknown function Phi(v). I defined a(v) in this paragraph:
    And I defined Phi(v) in this one:
    Naturally if Phi(v) = ac/squareroot(c^2 - v^2) and Beta is defined as c/squareroot(c^2 - v^2), this means Phi(v) = a*Beta. So, the second equation you quoted:

    (2) xi = Phi(v) * Beta * (x - vt)

    ...becomes xi = (a*Beta)* Beta * (x - vt) = a * Beta^2 * (x - vt). And Beta^2 = c^2/(c^2 - v^2), so this gives:

    xi = a * (c^2/(c^2 - v^2)) * (x - vt)

    And from the coordinate transformation between K and Kg we have x' = (x - vt), so this gives the first equation you quoted:

    (1) xi = a*(c^2/(c^2 - v^2))*x'

    Is this OK, or do you still have an objection?
    Last edited: Dec 6, 2008
  23. Dec 6, 2008 #22


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    A note on the circle trig vs. hyperbolic trig relations.

    The key relation is of course the invariant proper time:
    [tex] d\tau^2 = dt^2 - dx^2[/tex]
    (in [tex] c = 1[/tex] units.)

    You can thus parametrize the time and space components via:
    [tex] dx = d\tau \sinh(\vartheta)[/tex] and [tex] dt = d\tau \cosh(\vartheta)[/tex].

    The hyperbolic pseudo-angle [tex]\vartheta[/tex] is a frame parameter which we can relate to the velocity relative to the initial frame (where [tex]\vartheta= 0[/tex]) by:
    [tex] \frac{v}{c}=\beta = \tanh(\vartheta)[/tex]

    Note this parametrization works because [tex]\cosh^2(\vartheta) - \sinh^2(\vartheta) = 1[/tex]. We could however make use of circle trig functions via the identity:
    [tex] 1 = \sec^2(\theta) - \tan^2(\theta)[/tex]
    and thus parametrize frames via an angle:

    [tex] dx = d\tau \tan(\theta)[/tex] and [tex]dt = d\tau \sec(\theta)[/tex]

    Note then that the boost velocity is: [tex] \beta = sin(\theta)[/tex]

    You can view this another way by rewriting the proper time metric in the form:
    [tex] dt^2 = d\tau^2 + dx^2[/tex]
    whence this circle-trig. angle denotes the mixing of proper time and space:
    [tex] d\tau =dt \cos(\theta)[/tex] and [tex]dx = dt \sin(\theta)[/tex].

    The problem with this regular trig approach is that the angle [tex]\theta[/tex] does not linearly parametrize the one dimensional group of frame transformations. In short you don't get the correct velocity addition relationship by adding angles:
    [tex]\beta_{1+2} \ne \sin(\theta_1 + \theta_2)[/tex].
    (Note also you must artificially restrict the angle to [tex]-\pi/4 < \theta < \pi/4[/tex])
    This is because the (differential) frame time [tex]dt[/tex] is not the invariant quantity under frame transformations. (Mathematically it is because the secant and tangent functions are not components of the exponential transformation matrix.)

    You do get the correct velocity addition by adding pseudo-angles in the hyperbolic case:
    [tex] \beta_{1+2} = \tanh(\vartheta_1 + \vartheta_2)[/tex]
    where [tex]\beta_1 = \tanh(\vartheta_1)[/tex] and [tex]\beta_2 = \tanh(\vartheta_2)[/tex]

    I find that this helps clarify the distinction between Galilean and Special Relativites. Velocity is the linear parameter for Galilean boosts, this [tex]\vartheta[/tex] psuedo-angle is the linear parameter for Lorentz boosts. And of course note that the small angle approximation also occurs in hyperbolic trig. giving us the small velocity non-relativistic limit:
    [itex] \frac{V}{c} = \tanh(\vartheta) \approx \vartheta[/itex] when [itex] |\vartheta|<< 1[/itex].
  24. Dec 6, 2008 #23
    If one variable is a function of one another variable the chain rule must be used. Consequently, there is an error in Einstein derivation of SR.

    No, it's all totally wrong! Einstein said a=Phi(v) and you try to define another "value" for this constant. Einstein himself found a=Phi(v)=1.
  25. Dec 6, 2008 #24


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    Only if you are taking the derivative of the first variable with respect to the second variable. Where is Einstein doing that? Where am I doing that in my analysis of his proof?
    Einstein never says "a = Phi(v)", he just says "a is a function Phi(v) at present unknown", but it's clear from the context he was just speaking sloppily and actually meant that the two functions are related by a constant. If this isn't obvious to you, note that he writes:
    and then just a few lines later he writes:
    So from the context, it's very clear he intends that [tex]\phi (v) = a \frac{c}{\sqrt{c^2 - v^2}}[/tex].
  26. Dec 6, 2008 #25
    Everywhere. Try to replace dx' with (c-v)dt' and you can see what the consequence is.

    Einstein says "a is a function Phi(v)" and not "a is a function of Phi(v)". That means clearly a=Phi(v) .

    What he later writes is not consequence of mathematics but only an arbitrary construction to match his assumption. In addition he requires that Tau=0 if t=0 and in this case, a is an arbitrary constant without any relation to v and c.
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