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Hyperbolic Triangles

  1. May 16, 2014 #1
    Hello, I'm going through Landau and Lifshitz "The Classical Theory of Fields" this summer with a friend and in section 4 I've come to a bit of a math problem.

    Assume you have an inertial frame [itex]K'[/itex] moving at speed [itex]V[/itex] relative to an inertial frame [itex]K[/itex] in the [itex]x[/itex]-direction. In order for invariant intervals we require:

    [itex]
    (ct)^2 - x^2 = (ct')^2 - (x')^2
    [/itex]

    For this to be true:

    [itex]
    x = x'\cosh{\Psi}+ct'\sinh{\Psi}
    [/itex]
    [itex]
    ct=x'\sinh{\Psi}+ct'\cosh{\Psi}
    [/itex]

    Where [itex]\Psi[/itex] is the angle of rotation in the [itex]xt[/itex] plane. Which makes sense. Now if we just look at the origin of the [itex]K'[/itex] frame moving these can be reduced to:

    [itex]
    x=ct'\sinh{\Psi}
    [/itex]
    [itex]
    ct=ct'\cosh{\Psi}
    [/itex]

    and dividing the equations yields:

    [itex]
    \tanh{\Psi}=\frac{x}{ct}
    [/itex]

    But the speed [itex]V[/itex] is given by [itex]V=\frac{x}{t}[/itex] so:

    [itex]
    \tanh{\Psi}=\frac{V}{c}
    [/itex]

    The next part is what is a bit confusing. If this were regular trigonometry to find both [itex]\sin{\Psi}[/itex] and [itex]\cos{\Psi}[/itex] would just require the construction of a triangle. In this case however, they end up with very, very similar results except where the hypoteneuse would be they end up with something like this:

    [itex]
    \sinh{\Psi}=\frac{\frac{V}{c}}{\sqrt{1-({\frac{V}{c}})^2}}
    [/itex]
    [itex]
    \cosh{\Psi}=\frac{1}{\sqrt{1-({\frac{V}{c}})^2}}
    [/itex]

    If anyone can point me in the right direction of deriving these relationships either algebraically or using geometry that would be extremely helpful!
     
    Last edited: May 16, 2014
  2. jcsd
  3. May 16, 2014 #2

    PeterDonis

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    Staff: Mentor

  4. May 16, 2014 #3

    xox

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    [tex]\sinh{\Psi}=\frac{\tanh{\Psi}}{\sqrt{1-\tanh^2{\Psi}}}[/tex]
    [tex]\cosh{\Psi}=\frac{1}{\sqrt{1-\tanh^2{\Psi}}}[/tex]
     
  5. May 16, 2014 #4
    Okay hmm, assuming the Wikipedia section on "Comparison with circular functions" is correct then this is what I make of it all:

    http://imgur.com/4eL9du1

    Sorry for the mess, I've been working all day and am too tired to be perfect.
     
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