Answer to Question #100586 in Mechanics | Relativity for Kishwer

Question

How about some practice working with hyperbolic trig function ? Recall that
SinhQ= e^Q- e^Q / 2
CoshQ= e^Q+e^/2
TanhQ sinhQ/CoshQ
a) verify that cosh^2- sinh^2= 1
B) tanhQ1+tanhQ/1+tanhQ2= tanh(Q1+Q2)
So that the law of composition of velocities from last wrrk just mrans that boost parameters add.

Answer 1

a) For simplicity, $Q=x.$ Verify:

$\text{cosh}^2x-\text{sinh}^2x=\bigg[\frac{e^x+e^{-x}}{2}\bigg]^2-\bigg[\frac{e^x-e^{-x}}{2}\bigg]^2=\\ \space\\ =\frac{1}{4}[e^{2x}+e^{-2x}+2e^xe^{-2x}]-[e^{2x}+e^{-2x}-2e^xe^{-2x}]=\\ \space\\ =\frac{4}{4}=1.$

b) For simplicity, $Q_1=x, Q_2=y, \text{tanh}=\text{th}$.

On the one hand:

$\frac{\text{th}x+\text{th}y}{1+\text{th}x\cdot\text{th}y}=\frac{\frac{e^x-e^{-x}}{e^x+e^{-x}}+\frac{e^y-e^{-y}}{e^y+e^{-y}}}{1+\frac{e^x-e^{-x}}{e^x+e^{-x}}\cdot\frac{e^y-e^{-y}}{e^y+e^{-y}}}=\\ \space\\ =\frac{\frac{e^{x+y}-e^{-x+y}+e^{-y+x}-e^{-x-y}+e^{x+y}+e^{-x+y}-e^{-y+x}-e^{-x-y}}{(e^x+e^{-x})(e^y+e^{-y})}}{\frac{[e^{x+y}+e^{-x+y}+e^{x-y}+e^{-x-y}]+[e^{x+y}-e^{x-y}-e^{-x+y}+e^{-x-y}]}{(e^x+e^{-x})(e^y+e^{-y})}}=\\ \space\\ =\frac{2e^{x+y}-2e^{-x-y}}{2e^{x+y}+2e^{-x-y}}=\frac{e^{(x+y)}-e^{-(x+y)}}{e^{(x+y)}+e^{-(x+y)}}.$

On the other hand, what we need to get is

$\text{th}(x+y)=\frac{e^{(x+y)}-e^{-(x+y)}}{e^{(x+y)}+e^{-(x+y)}}.$

We see that the right part of the previous step is the same as the last one. Verified!