Consider a particle in hyperbolic motion,

$\mathbf{x}\left(t\right)\mathbf{=}\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{+}{\left(\mathrm{ct}\right)}^{\mathbf{2}}}\mathbf{,}\mathbf{y}\mathbf{=}\mathbf{z}\mathbf{=}\mathbf{0}$

(a) Find the proper time role="math" localid="1654682576730" $\mathit{\tau }$as a function of $\mathit{\tau }$, assuming the clocks are set so that$\mathit{\tau }\mathbf{=}\mathbf{0}$ when$\mathit{\tau }\mathbf{=}\mathbf{0}$ . [Hint: Integrate Eq. 12.37.]

(b) Find x and v (ordinary velocity) as functions of$\mathit{\tau }$ .

(c) Find${\mathit{\eta }}^{\mathbf{\mu }}$ (proper velocity) as a function of $\mathit{\tau }$.

Write the expression for the Lorentz contraction.

$\mathit{Y}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}}$ …… (1)

Here, v is the velocity, and c is the speed of light.

(a)

Write the expression for the velocity of a particle.

$v=\frac{dx\left(t\right)}{dt}$

Substitute $\sqrt{b^2+c^2{t}^2}$for $x\left(t\right)$in the above expression.

localid="1654683753767" $$\begin{gathered} v=\frac{d}{dt}\left(\sqrt{b^2+c^2{t}^2}\right) \\ v=\frac{1}{2\sqrt{b^2+c^2{t}^2}}\frac{d}{dt}\left(b^2+c^2{t}^2\right) \\ v=\frac{1}{2\sqrt{b^2+c^2{t}^2}}\left(2c^{2}t\right) \\ v=\frac{c^{2}t}{\sqrt{b^2+c^2{t}^2}} \end{gathered}$$

Squaring on both sides in equation (1).

$$\begin{gathered} {\gamma }^2={\left[\frac{\begin{array}{c}\\ 1\end{array}}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}\right]}^2 \\ {\gamma }^2=\frac{1}{1-\frac{v^2}{c^2}} \end{gathered}$$

Substitute $\frac{c^{2}t}{\sqrt{b^2+c^2{t}^2}}$ for v in the above expression.

localid="1654687319733" $$\begin{gathered} {\gamma }^2=\frac{1}{1-{\displaystyle \frac{{\left(\begin{array}{c}{\displaystyle \frac{c^{2}t}{\sqrt{b^2+c^2{t}^2}}}\end{array}\right)}^2}{c^2}}} \\ {\gamma }^2=\frac{1}{\left[1-{\displaystyle \frac{c^4{t}^2}{\left(b^2+c^2{t}^2\right)}}\times {\displaystyle \frac{1}{c^2}}\right]} \\ {\gamma }^2=\frac{1}{1-{\displaystyle \frac{c^2{t}^2}{b^2+c^2{t}^2}}} \\ {\gamma }^2=\frac{b^2+c^2{t}^2}{b^2} \end{gathered}$$

Write the relation between proper time and the time interval.

$d\tau =\sqrt{1-\frac{v^2}{c^2}dt}$

Integrate the proper time $d\tau$.

localid="1654685358421" $$\begin{gathered} \int d\tau =\tau =\int \sqrt{1-\frac{v^2}{c^2}dt} \\ \tau =\int \sqrt{1-\frac{v^2}{c^2}dt} \end{gathered}$$

Substitute $1-\frac{v^2}{c^2}=\frac{1}{{\gamma }^2}$and ${\gamma }^2=\frac{b^2+c^2{t}^2}{b^2}$in the above expression.

localid="1654685793217" $$\begin{gathered} \tau =\int \sqrt{\frac{1}{{\gamma }^2}dt} \\ =\int \sqrt{\frac{{\displaystyle \frac{1}{b^2+c^2{t}^2}}}{b^2}}dt \\ =\int \sqrt{\frac{b^2}{b^2+c^2{t}^2}}dt \\ =\int \frac{b}{\sqrt{b^2+c^2{t}^2}}dt \end{gathered}$$

To further solve the integration,

$\tau =\frac{b}{c}\mathrm{In}\left(ct+\sqrt{b^2+c^2{t}^2}\right)+k$ …… (2)

It is given that:

$\tau =0$then $t=0$

Solve as,

$$\begin{gathered} \tau =\frac{b}{c}\mathrm{In}\left(\mathrm{c}\left(0\right)+\sqrt{{\mathrm{b}}^2+{\mathrm{c}}^2{\left(0\right)}^2}\right)+k \\ 0=\frac{b}{c}\mathrm{In}\left(\mathrm{b}\right)+k \\ k=-\frac{b}{c}\mathrm{In}\left(\mathrm{b}\right) \end{gathered}$$

Substitute $-\frac{b}{c}\mathrm{In}\left(\mathrm{b}\right)$ fork in equation (2).

$$\begin{gathered} \tau =\frac{b}{c}\mathrm{In}\left(\mathrm{ct}+\sqrt{{\mathrm{b}}^2+{\mathrm{c}}^2{\mathrm{t}}^2}\right)-\frac{\mathrm{b}}{\mathrm{c}}\mathrm{In}\left(b\right) \\ \mathrm{\tau }=\frac{\mathrm{b}}{\mathrm{c}}\left[\mathrm{In}\left(\frac{\mathrm{c}+\sqrt{{\mathrm{b}}^2+{\mathrm{c}}^2{\mathrm{t}}^2}}{\mathrm{b}}\right)\right] \end{gathered}$$

Therefore, the proper time as a function of tis $\tau =\frac{b}{c}\left[\mathrm{In}\left(\frac{\mathrm{c}+\sqrt{{\mathrm{b}}^2+{\mathrm{c}}^2{\mathrm{t}}^2}}{\mathrm{b}}\right)\right]$.

(b)

Write the equation for the x ordinary velocity as a function of $\tau$.

$$\begin{gathered} \sqrt{x^2-b^2}+x=b{e}^{\frac{c\tau }{b}} \\ \sqrt{x^2-b^2}=b{e}^{\frac{c\tau }{b}}-x \end{gathered}$$

Squaring on both sides in the above expression.

$$\begin{gathered} {\left(\sqrt{x^2-b^2}\right)}^2={\left(b{e}^{\frac{c\tau }{b}}-x\right)}^2 \\ {x}^2-b^2=b^2{e}^{\frac{2c\tau }{b}}+x^2-2xb{e}^{\frac{2c\tau }{b}} \\ 2xb{e}^{\frac{2c\tau }{b}}=b^2\left[1+e^{\frac{2c\tau }{b}}\right] \\ x=\frac{b^2}{2}\left[\frac{1+e^{\frac{2c\tau }{b}}}{b{e}^{\frac{2c\tau }{b}}}\right] \end{gathered}$$

On further solving,

$$\begin{gathered} x=\frac{b}{2}\left[\frac{1+e^{\frac{2c\tau }{b}}}{e^{\frac{2c\tau }{b}}}\right] \\ x=b\left[\frac{e^{-\frac{c\tau }{b}}+e^{\frac{c\tau }{b}}}{2}\right] \\ x=b\cosh \left(\frac{c\tau }{b}\right) \end{gathered}$$

As it is kown that:

$$\begin{gathered} v=\frac{c^{2}t}{\sqrt{b^2+c^2{t}^2}} \\ v=\frac{c}{x}\sqrt{x^2-b^2} \end{gathered}$$

Substitute $b\cosh \left(\frac{c\tau }{b}\right)$for x in the above expression.

$$\begin{gathered} v=\frac{c}{b\cosh \left({\displaystyle \frac{c\tau }{b}}\right)}\sqrt{{\left(b\cosh \left(\frac{c\tau }{b}\right)\right)}^2-b^2} \\ =c\frac{\sqrt{{\cosh }^2\left({\displaystyle \frac{c\tau }{b}}\right)-1}}{\cosh \left(\frac{c\tau }{b}\right)} \\ =c\frac{\sinh \left({\displaystyle \frac{\mathit{c}\mathit{\tau }}{\mathit{b}}}\right)}{\cosh \left({\displaystyle \frac{c\tau }{b}}\right)} \\ =c\tanh \left(\frac{c\tau }{b}\right) \end{gathered}$$

Therefore, the x and v (ordinary velocity) as a function of $\tau$is $x=b\cosh \left(\frac{c\tau }{b}\right)$and $v=c\tanh \left(\frac{c\tau }{b}\right)$repectively.

(c)

It is known that:

${\eta }^{\mu }=\gamma \left(c,v,0,0\right)$

Substitute $\frac{\sqrt{b^2+c^2{t}^2}}{b}$for $\gamma$and $c\tanh \left(\frac{c\tau }{b}\right)$for v in the above equation.

$$\begin{gathered} {\eta }^{\mu }=\left(\frac{\sqrt{b^2+c^2{t}^2}}{b}\right)\left(c,c\tanh \left(\frac{c\tau }{b}\right),0,0\right) \\ {\eta }^{\mu }=\frac{x}{b}\left(c,c\tanh \left(\frac{c\tau }{b}\right),0,0\right) \\ {\eta }^{\mu }=\frac{b\cosh \left({\displaystyle \frac{c\tau }{b}}\right)}{b}\left(c,c\tanh \left(\frac{c\tau }{b}\right),0,0\right) \\ {\eta }^{\mu }=c\left[\cosh \left(\frac{c\tau }{b}\right),\sin \left(\frac{c\tau }{b}\right),0,0\right] \end{gathered}$$

Therefore, the proper velocity ${\eta }^{\mu }$as a function of $\tau$is

${\eta }^{\mu }=c\left[\cosh \left(\frac{c\tau }{b}\right),\sin \left(\frac{c\tau }{b}\right),0,0\right]$.