Answer to Question #99743 in Mechanics | Relativity for AbdulRehman

Question

How about some practice working with hyperbolic trig functions? Recall
that
sinh θ =
e
θ − e
−θ
2
,
cosh θ =
e
θ + e
−θ
2
,
and
tanh θ =
sinh θ
cosh θ
.
(a) Verify that cosh2
θ − sinh2
θ = 1.
(b) Verify that
tanh θ1 + tanh θ2
1 + tanh θ1 tanh θ2
= tanh(θ1 + θ2),
so that the law of composition of velocities from last week just means
that “boost parameters add.”

Answer 1

a)

$\cosh^2{\theta}=(0.5(e^\theta+e^{-\theta}))^2=0.25(e^{2\theta}+2+e^{-2\theta})$

$\sinh^2{\theta}=(0.5(e^\theta-e^{-\theta}))^2=0.25(e^{2\theta}-2+e^{-2\theta})$

$\cosh^2{\theta}-\sinh^2{\theta}=0.25(2-(-2))=1$

b)

$\tanh{θ_1} + \tanh{θ_2}=\frac{e^{θ_1}-e^{-θ_1}}{e^{θ_1}+e^{-θ_1}}+\frac{e^{θ_2}-e^{-θ_2}}{e^{θ_2}+e^{-θ_2}}$

$1+\tanh{θ_1} \tanh{θ_2}=1+\left(\frac{e^{θ_1}-e^{-θ_1}}{e^{θ_1}+e^{-θ_1}}\right)\left(\frac{e^{θ_2}-e^{-θ_2}}{e^{θ_2}+e^{-θ_2}}\right)$

$\frac{}{}$

$\frac{\tanh{θ_1} + \tanh{θ_2}}{1+\tanh{θ_1} \tanh{θ_2}}=\frac{\frac{e^{θ_1}-e^{-θ_1}}{e^{θ_1}+e^{-θ_1}}+\frac{e^{θ_2}-e^{-θ_2}}{e^{θ_2}+e^{-θ_2}} }{1+\left(\frac{e^{θ_1}-e^{-θ_1}}{e^{θ_1}+e^{-θ_1}}\right)\left(\frac{e^{θ_2}-e^{-θ_2}}{e^{θ_2}+e^{-θ_2}}\right)}=\frac{e^{θ_1+θ_2}-e^{-θ_1-θ_2}}{e^{θ_1+θ_2}+e^{-θ_1-θ_2}}$

$\frac{\tanh{θ_1} + \tanh{θ_2}}{1+\tanh{θ_1} \tanh{θ_2}}=\tanh({θ_1+θ_2})$