201903030504 Problem 2.12

From $\displaystyle{\frac{\mathrm{d}V}{1-V^2/c^2}}=f(\tau)\,\mathrm{d}\tau$ ,

upon integration it becomes

$V(\tau)=c\tanh \phi(\tau)$

where $c\,\phi (\tau)=\displaystyle{\int_0^\tau f(\tau)\,\mathrm{d}\tau}$ .

$\begin{aligned} \displaystyle{\frac{\mathrm{d}V}{1-\frac{V^2}{c^2}}} & = f(\tau)\,\mathrm{d}\tau \\ \displaystyle{\int\frac{1}{1-x^2}}(c\,\mathrm{d}x) & = \displaystyle{\int_0^\tau}f(\tau)\,\mathrm{d}\tau \\ c\,\mathrm{argtanh} x & =\displaystyle{ \int_0^\tau f(\tau)\,\mathrm{d}\tau} \\ \mathrm{argtanh} x & = \displaystyle{\frac{1}{c}\int_0^\tau f(\tau)\,\mathrm{d}\tau} \\ \tanh \bigg(\displaystyle{\frac{1}{c}\int_0^\tau f(\tau)\,\mathrm{d}\tau} \bigg) & = x = \displaystyle{\frac{V}{c}} \\ V(\tau) & = c\tanh \phi(\tau) \\ \end{aligned}$

Similarly one can obtain

$\dot{t}\equiv \displaystyle{\frac{\mathrm{d}t}{\mathrm{d}\tau}}=\displaystyle{\frac{1}{\sqrt{1-V^2/c^2}}}=\cosh \phi (\tau)$ ,

$\dot{x}\equiv \displaystyle{\frac{\mathrm{d}x}{\mathrm{d}\tau}}=\displaystyle{\frac{V}{\sqrt{1-V^2/c^2}}}=c\sinh \phi(\tau)$ .

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