# Four-velocity: another example

Required math: calculus

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 3; Problem P3.2.

Another example of four-velocity in special relativity. We start with an object whose velocity (3-d velocity, that is) in an inertial frame is

$\displaystyle v_{x}\left(t\right)=\sqrt{1-\frac{1}{\left(gt+1\right)^{2}}} \ \ \ \ \ (1)$

where ${g}$ is a constant with relativistic units of ${\mbox{m}^{-1}}$, and ${t\ge0}$ is the time measured in the inertial frame (so it’s not the proper time of the object). The suffix ${x}$ indicates the motion is along the ${x}$ axis, as usual.

The relation between a proper time interval ${d\tau}$ and the time interval ${dt}$ measured in a frame moving at speed ${v_{x}}$ along the ${x}$ axis with respect to the object is given by the time dilation formula

$\displaystyle d\tau=dt\sqrt{1-v_{x}^{2}} \ \ \ \ \ (2)$

This gives us the time component of the four-velocity:

 $\displaystyle u^{t}$ $\displaystyle =$ $\displaystyle \frac{dt}{d\tau}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{1-v_{x}^{2}}}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1+gt \ \ \ \ \ (5)$

We can integrate this to get ${\tau}$ in terms of ${t}$:

 $\displaystyle \frac{dt}{1+gt}$ $\displaystyle =$ $\displaystyle d\tau\ \ \ \ \ (6)$ $\displaystyle \frac{1}{g}\ln\left(1+gt\right)$ $\displaystyle =$ $\displaystyle \tau+\tau_{0} \ \ \ \ \ (7)$

where ${\tau_{0}}$ is the constant of integration. If we require ${\tau=0}$ when ${t=0}$, then ${\tau_{0}=0}$ and we get

$\displaystyle g\tau=\ln\left(1+gt\right) \ \ \ \ \ (8)$

From this we get

$\displaystyle u^{t}=e^{g\tau} \ \ \ \ \ (9)$

From the definition of four-velocity, we have

 $\displaystyle u^{x}$ $\displaystyle \equiv$ $\displaystyle \frac{dx}{d\tau}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{dx}{dt\sqrt{1-v_{x}^{2}}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{v_{x}}{\sqrt{1-v_{x}^{2}}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1+gt\right)\sqrt{1-\frac{1}{\left(gt+1\right)^{2}}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\left(1+gt\right)^{2}-1}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{e^{2g\tau}-1} \ \ \ \ \ (15)$

We can now find ${x}$ and ${t}$ as functions of ${\tau}$:

 $\displaystyle t\left(\tau\right)$ $\displaystyle =$ $\displaystyle \frac{1}{g}\left(e^{g\tau}-1\right)\ \ \ \ \ (16)$ $\displaystyle x\left(\tau\right)$ $\displaystyle =$ $\displaystyle \int_{0}^{\tau}\sqrt{e^{2g\tau^{\prime}}-1}d\tau^{\prime}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{g}\left[\sqrt{e^{2g\tau}-1}-\arctan\left(\sqrt{e^{2g\tau}-1}\right)\right] \ \ \ \ \ (18)$

using software to do the integral.