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Physics LibreTexts

3.6: Some kinematic identities

  • Page ID
    13034
  • [ "article:topic", "authorname:crowellb" ]

    Skills to Develop

    • List of various kinematics equations and identities

    In addition to the relations

    \[D(v) = \sqrt{\frac{1+v}{1-v}}\]

    and

    \[v_c = \frac{v_1 + v_2}{1 + v_1 v_2}\]

    the following identities can be handy. If stranded on a desert island you should be able to rederive them from scratch. Don’t memorize them.

    \[v = \frac{D^2 - 1}{D^2 + 1}\]

    \[\gamma = \frac{D^{-1} + D}{2}\]

    \[v\gamma = \frac{D - D^{-1}}{2}\]

    \[D(v)D(-v) = 1\]

    \[\eta = \ln D\]

    \[v = \tanh \eta\]

    \[\gamma = \cosh \eta\]

    \[v\gamma = \sinh \eta\]

    \[\tanh (x+y) = \frac{\tanh x + \tanh y}{1 + \tanh x \tanh y}\]

    \[D_c = D_1 D_2\]

    \[\eta _c = \eta _1 + \eta _2\]

    \[v_C \gamma _c = (v_1 + v_2)\gamma _1 \gamma _2\]

    The hyperbolic trig functions are defined as follows:

    \[\sinh x = \frac{e^x - e^{-x}}{2}\]

    \[\cosh x = \frac{e^x + e^{-x}}{2}\]

    \[\tanh x = \frac{\sinh x}{\cosh x}\]

    Their inverses are built in to some calculators and computer software, but they can also be calculated using the following relations:

    \[\sinh^{-1}x = \ln \left ( x + \sqrt{x^2 + 1} \right )\]

    \[\cosh^{-1}x = \ln \left ( x + \sqrt{x^2 - 1} \right )\]

    \[\tanh^{-1}x = \frac{1}{2}\ln \left ( \frac{1 + x}{1 - x} \right )\]

    Their derivatives are, respectively, \(\left ( x^2 + 1 \right )^{-1/2}\), \(\left ( x^2 - 1 \right )^{-1/2}\) and \(\left ( 1 - x^2 \right )^{-1}\).

    Contributor