$$\require{cancel}$$

# 3.6: Some kinematic identities

Skills to Develop

• List of various kinematics equations and identities

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

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 $$\left ( 1 - x^2 \right )^{-1}$$.