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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:

    [tex]
    \gamma = \cosh(\tanh^{-1}(v/c)) = \frac{1}{\sqrt{1-(v/c)^2}}
    [/tex]
     
  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:

    [tex]
    s^2 = (ct)^2 - x^2
    [/tex]

    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:

    Regards,

    Bill
     
  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

    robphy

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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 "ON THE ELECTRODYNAMICS OF MOVING BODIES" (1905)

     
  13. Aug 5, 2008 #12

    JesseM

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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: Aug 5, 2008
  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]).

    Regards,

    Bill
     
  16. Aug 5, 2008 #15

    JesseM

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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

    JesseM

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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

    JesseM

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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).

     
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