Four-velocity of a particle moving in a circle

Question

If we have a particle moving in a circle then we will have $x$ and $y$ values of $$x = r\cos(\varphi)$$ $$y = r\sin(\varphi)$$As $ds = tv$ and $ds = \varphi r$ we can say that $$x = r\cos\bigg(\frac{tv}{r}\bigg)$$ $$y = r\sin\bigg(\frac{tv}{r}\bigg)$$ I am aware that the equation for four-velocity is $$\mathbf U = \frac{d\mathbf X}{d\tau}$$ however I am confused as to how to get a worldline for the particle to differentiate it. Can anyone help me out at this point?

Answer 1

So the particle moves on the curve $t \mapsto x^\mu (t)$, where $t$ is the time coordinate of the observer and

$$ x^\mu (t) = \begin{pmatrix} ct \\ r \cos (\frac{vt}{r})\\ r \sin (\frac{vt}{r}) \end{pmatrix} \,. $$ Now, you should calculate the proper time $\tau$ which is given by

$$ \tau = \int_{0}^t \frac{1}{c} \sqrt{-\eta(\dot{x},\dot{x})} dt' = \int_{0}^t \frac{dt'}{\gamma(t')} ,\quad \text{with} \quad \dot{x} \equiv \frac{d x}{dt'}\,. $$ In this case $\gamma(t)=\gamma=1/\sqrt{1-v^2/c^2}$. So, the proper time is the usual one, namely

$$ \tau = \frac{1}{\gamma} t\,. $$ All right! Now, you can calculate the four velocity:

$$ U^\mu = \frac{d x^\mu}{d\tau} = \gamma \frac{d x^\mu}{d t} = \gamma \begin{pmatrix} c\\ -\beta \cos(\frac{vt}{r})\\ \beta \sin(\frac{vt}{r}) \end{pmatrix}\,, $$ where $\beta := \frac{v}{c}$.