Four-velocity: an example

Required math: calculus

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 3; Problem P3.1.

We’ve already looked at the four-velocity in special relativity, but it’s worth a second look from a different angle. We can instead define the four-velocity in terms of two events separated by an infinitesimal spacetime interval ${d\mathbf{s}}$. The four-velocity is defined as the derivative of ${d\mathbf{s}}$ with respect to the proper time ${\tau}$, so that

$\displaystyle u^{i}\equiv\frac{ds^{i}}{d\tau} \ \ \ \ \ (1)$

Since the components ${ds^{i}}$ transform using the Lorentz transformation, then so do the components ${u^{i}}$ of the four-velocity.

Since the spacetime interval is invariant (it has the same value in all inertial frames) the relation

$\displaystyle ds^{2}=-dt^{2}+dx^{2}+dy^{2}+dz^{2} \ \ \ \ \ (2)$

holds in all inertial frames. In particular, it holds in the observer’s rest frame, in which ${dt=d\tau}$, so we have

$\displaystyle ds^{2}=-d\tau^{2} \ \ \ \ \ (3)$

In this rest frame, then, we get

 $\displaystyle u^{i}$ $\displaystyle =$ $\displaystyle \left(\frac{d\tau}{d\tau},0,0,0\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1,0,0,0\right) \ \ \ \ \ (5)$

The square of ${\mathbf{u}}$ is then

 $\displaystyle \mathbf{u}\cdot\mathbf{u}$ $\displaystyle =$ $\displaystyle \eta_{ij}u^{i}u^{j}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -1 \ \ \ \ \ (7)$

where ${\eta_{ij}}$ is the metric used in special relativity:

$\displaystyle \eta_{ij}=\begin{cases} -1 & i=j=t\\ 1 & i=j=x,y,z\\ 0 & \mbox{otherwise} \end{cases} \ \ \ \ \ (8)$

Since this is true in the rest frame and the square of a four-vector is an invariant, it is true in all frames.

As an example, suppose we have an object that moves along a worldline given by (in some inertial frame)

$\displaystyle x\left(\tau\right)=\frac{1}{g}\left[\cosh\left(g\tau\right)-1\right] \ \ \ \ \ (9)$

(${y=z=0}$ and ${g}$ is a constant). The ${x}$ component of the four-velocity is then

 $\displaystyle u^{x}$ $\displaystyle =$ $\displaystyle \frac{dx}{d\tau}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sinh\left(g\tau\right) \ \ \ \ \ (11)$

Using ${\mathbf{u}\cdot\mathbf{u}=-1}$ we can find the ${t}$ component:

 $\displaystyle \left(u^{x}\right)^{2}-\left(u^{t}\right)^{2}$ $\displaystyle =$ $\displaystyle -1\ \ \ \ \ (12)$ $\displaystyle \sinh^{2}\left(g\tau\right)+1$ $\displaystyle =$ $\displaystyle \left(u^{t}\right)^{2}\ \ \ \ \ (13)$ $\displaystyle u^{t}$ $\displaystyle =$ $\displaystyle \cosh\left(g\tau\right) \ \ \ \ \ (14)$

using the identity ${\cosh^{2}x-\sinh^{2}x=1}$. From this we can get the time in the inertial frame:

 $\displaystyle u^{t}$ $\displaystyle =$ $\displaystyle \frac{dt}{d\tau}\ \ \ \ \ (15)$ $\displaystyle t\left(\tau\right)$ $\displaystyle =$ $\displaystyle \frac{1}{g}\sinh\left(g\tau\right) \ \ \ \ \ (16)$

The velocity of the object as seen in the inertial frame is

 $\displaystyle v_{x}$ $\displaystyle =$ $\displaystyle \frac{dx}{dt}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{u^{x}}{u^{t}}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \tanh\left(g\tau\right) \ \ \ \ \ (19)$

Since tanh is bounded by ${\pm1}$, the velocity never exceeds 1, so never exceeds the speed of light.

We can invert the relation between proper time ${\tau}$ and inertial time ${t}$ to get

$\displaystyle g\tau=\sinh^{-1}\left(gt\right) \ \ \ \ \ (20)$

Using the relations (derived from ${\cosh^{2}x-\sinh^{2}x=1}$)

 $\displaystyle \sinh\left(\sinh^{-1}x\right)$ $\displaystyle =$ $\displaystyle x\ \ \ \ \ (21)$ $\displaystyle \cosh\left(\sinh^{-1}x\right)$ $\displaystyle =$ $\displaystyle \sqrt{1+x^{2}}\ \ \ \ \ (22)$ $\displaystyle \tanh\left(\sinh^{-1}x\right)$ $\displaystyle =$ $\displaystyle \frac{x}{\sqrt{1+x^{2}}} \ \ \ \ \ (23)$

we get

 $\displaystyle u^{x}\left(t\right)$ $\displaystyle =$ $\displaystyle gt\ \ \ \ \ (24)$ $\displaystyle u^{t}\left(t\right)$ $\displaystyle =$ $\displaystyle \sqrt{1+\left(gt\right)^{2}}\ \ \ \ \ (25)$ $\displaystyle v\left(t\right)$ $\displaystyle =$ $\displaystyle \frac{gt}{\sqrt{1+\left(gt\right)^{2}}} \ \ \ \ \ (26)$

Again, note that ${\mathbf{u}\cdot\mathbf{u}=-1}$ and also that as ${t\rightarrow\infty}$, ${v\rightarrow1}$ so again the velocity remains less than ${c}$.

15 thoughts on “Four-velocity: an example”

1. Pingback: Four-velocity again | Physics pages

1. gwrowe Post author

As requested, I deleted your question about the physics, but it’s worth pointing out that, yes, Latex does work in comments. See “Instructions for commenters” in the Welcome menu at the top of the page for instructions.

2. BogdanT

very nice problem. This problem really tests your knowledge on the 4 vectors. I did it using the components given in the original problem. You actually started with the premise that you know the wordline function . Is there a way to post my own solution ?

1. gwrowe Post author

Sorry, no. I don’t accept posts from anyone else.
However, it’s easy enough (and free) to set up your own blog at wordpress.com.