I have a conceptual problem relating to non-Lorentzian changes of coordinates in flat spacetime. I would be grateful is someone could point out my error.

Let's change to Rindler coordinates in $1+1$-dimensional special relativity, \begin{equation} x^\mu = \begin{pmatrix} t \\ x \end{pmatrix} = \begin{pmatrix} X\sinh[\alpha \tau] \\ X \cosh[\alpha \tau]\end{pmatrix}. \end{equation} Now, let's consider the four velocity, \begin{equation} v^\mu = \frac{dx^{\mu}}{d\tau} = \begin{pmatrix} \frac{dt}{d\tau} \\ \frac{dx}{d\tau} \end{pmatrix} = \begin{pmatrix} \alpha X\cosh[\alpha \tau] \\ \alpha X\sinh[\alpha \tau] \end{pmatrix}. \end{equation} Now let's calculate the magnitude of the four-velocity using the Minkowski metric, \begin{equation} |v| = g_{\mu\nu}v^\mu v^\nu = g_{\mu\nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} = -\left(\frac{dt}{d\tau}\right)^2 + \left(\frac{dx}{d\tau}\right)^2 = -(\alpha X)^2. \end{equation} I've always been taught to set \begin{equation} \frac{dt}{d\tau} = \gamma, \quad \frac{dx}{d\tau} = \gamma \mathbf{v}, \end{equation} where $\mathbf{v}$ is the three-velocity and $\gamma = (1-\mathbf{v})^{-1/2}$. This makes the magnitude of the four-velocity always equal to $-1$ in the original coordinates. This is a Lorentz scalar in that it is invariant under Lorentz coordinate transformation. However, now I am left with the situation that \begin{equation} -1 = -(\alpha X)^2. \end{equation}

Question: By changing coordinates I have gone from a constant function on spacetime to one that varies in coordinates. To me, this seems wrong. A constant function in one set of coordinates should be constant in all coordinates. Was I wrong in the inclusion of $\gamma$? Or am I wrong in this final assertion? Should I set $X = \pm 1/\alpha$? But surely I can change $X$ as I like? Should this not be considered a function on spacetime?

  • 1
    $\begingroup$ If you are on a specific trajectory, $X$ is a function of $\tau$ and you need to differentiate it. $\endgroup$
    – Javier
    Commented Mar 1, 2018 at 17:06
  • $\begingroup$ Thanks. The answer below gives me the situation when $X$ is constant, which is the constantly accelerated trajectory. However, your answer tells me that I don't have to make this choice, but in that case $X$ is a function of $\tau$ in general. Hopefully I've understood. Thanks to all. $\endgroup$
    – Matta
    Commented Mar 1, 2018 at 17:49
  • $\begingroup$ Wait, that's right, $X$ is constant for this trajectory. What you just found out is that Rindler coordinates with parameter $\alpha$ are adapted to the trajectory with $X=1/\alpha$. $\endgroup$
    – Javier
    Commented Mar 1, 2018 at 17:57

1 Answer 1


In Minkowski spacetime coordinates $(t, x)$ the worldline of a body in hyperbolic motion, with a constant proper acceleration $\alpha$ in the $+x$ direction, as function of proper time $\tau$ is described by
$t =\ \frac{1}{\alpha} \ \sinh (\alpha \tau)$
$x = \ \frac{1}{\alpha} \ \cosh (\alpha \tau)$
The path in Minkowski is an hyperbola of equation
$x^2 -t^2 = 1 / \alpha^2$

The velocity is
$v = (\cosh (\alpha \tau), \sinh (\alpha \tau))$
The acceleration is
$a = \alpha (\sinh (\alpha \tau), \cosh (\alpha \tau))$

With $\eta_{\mu \nu} = diag(-1, 1)$ metric tensor in 1+1-dimensional Minkowski, we have
$v^2 = -1$ time-like vector
$a^2 = \alpha^2$ space-like vector
$a \cdot v = 0$ velocity and acceleration are orthogonal

The worldline refers to a body with a specific acceleration, however a Rindler reference frame is made up of a continuous set of worldlines each with a constant acceleration, so that the Rindler coordinates are $(\tau, \alpha)$ with $-\infty \lt \tau \lt \infty$ and $\alpha \gt 0$. A constant $\alpha$ in Rindler coordinates picks up a stationary observer in that reference frame.
The $X$ coordinate in the question is equal to $1 / \alpha$ as per definition of Rindler coordinates, so there is no contradiction in your calculation of the squared velocity.


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