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.

3 thoughts on “Four-velocity: another example

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