We have the following relation between the spatial part of the coordinate acceleration $A$ in an inertial frame and the proper acceleration $a$ in the comoving frame (supposing $1+1$ dimensions for simplicity): $$ a\left(\tau\right) = \gamma^3 A\left(t\right) \,, $$ where $\gamma$ is the Lorentz factor. When $a\left(\tau\right) = a$ is constant, i.e. uniform acceleration, we can integrate this twice to obtain the trajectory $x$ in function of $t$ in spacetime (these will ultimately yield the well-known Rindler coordinates).

I want to explore non-uniform acceleration but am struggling how to tackle this problem. More precisely, suppose that we want to study a linear acceleration; is the time variation in the proper acceleration the coordinate time ($t$) or the proper time ($\tau$)? In other words, do we consider $a\left(t\right) = a_{0}t$ or $a\left(\tau\right) = a_{0}\tau$? For the former, we can proceed as before by integrating twice to obtain the trajectory in spacetime but in the latter, this is not straightforward. In the latter case, we can integrate once to obtain: $$ \frac{\text{d}x}{\text{d}t} = \frac{a_{0}\int\tau\, \text{d}t}{\sqrt{1+\left(a_{0}\int\tau\, \text{d}t\right)^2}}\,.$$ Note that we cannot solve the integral since we do not know the proper time as a function of the time coordinate. Moreover, it holds that $$\frac{\text{d}\tau}{\text{d}t} = \frac{1}{\gamma} = \sqrt{1-\left(\frac{\text{d}x}{\text{d}t}\right)^2} = \frac{1}{\sqrt{1+\left(a_{0}\int\tau\, \text{d}t\right)^2}}\,.$$ From the last expression, I wanted to derive a differential equation for $\tau\left(t\right)$ such that I can solve the proper time as a function of time but I can't get to that differential equation.

To summarize, I have two questions:

  1. If we suppose a non-uniform acceleration, do we specify the proper acceleration as a function of the time coordinate or as a function of the proper time?
  2. If as a function of proper time, is there a way to obtain a differential equation for $\tau\left(t\right)$?

Thanks in advance.

  • $\begingroup$ "We have the following relation between the acceleration A in an arbitrary (non-inertial) frame and the proper acceleration a in the comoving frame" This relation does not hold in general, if ever. Consider a Rindler observer. In the non-inertial Rindler frame the Rindler observer's coordinate acceleration is $A=0$, but their proper acceleration $a\ne 0 = \gamma^3 A$ $\endgroup$
    – Dale
    Commented Dec 6, 2023 at 16:34
  • $\begingroup$ @Dale, how would this possible? Scalar are invariant under coordinate transformations. If in one frame $a \neq 0$, then in the other, it cannot be zero as far as I know. $\endgroup$
    – Kabouter9
    Commented Dec 6, 2023 at 16:37
  • $\begingroup$ The proper acceleration $a$ is a scalar, but the coordinate acceleration $A$ is not. The coordinate acceleration $A$ is a coordinate-dependent quantity, it is not a tensor of any type. $\endgroup$
    – Dale
    Commented Dec 6, 2023 at 16:39
  • $\begingroup$ Maybe I phrased it incorrectly but $A$ is spatial part of the coordinate acceleration. $\endgroup$
    – Kabouter9
    Commented Dec 6, 2023 at 16:40
  • $\begingroup$ Yes, that is what I understood. See my comments above $\endgroup$
    – Dale
    Commented Dec 6, 2023 at 16:41

1 Answer 1


The natural way is to specify proper acceleration with respect to proper time. This uniquely specifies your trajectory, given that you give its initial spacetime event and corresponding normalised 4-velocity.

Geometrically, this is the Minkowski analogue of reconstructing a curve in the plane, knowing its curvature as a function of its arc length. This uniquely specifies the curve up to a congruence, which is why you need to specify the initial point and corresponding tangent vector.

Concretely, in an inertial frame with coordinates $(t,x)$, and writing the proper time derivatives as $\dot{}$ and the coordinate time derivatives as $'$, you get the proper velocity and acceleration: $$ \begin{align} U &= (\dot t,\dot x) & A &= (\ddot t, \ddot x) \end{align} $$ so the definition of proper acceleration gives the ODE: $$ \begin{align} \ddot x &= a\dot t & \ddot t &= a\dot x \end{align} $$ This is a second order linear non-autonomous ODE, and the initial condition specifies $x(0),y(0),\dot x(0), \dot y(0)$ with the constraint that: $$ \dot t^2-\dot x^2 = 1 $$ which if obeyed for the initial condition is conserved by the equations of motion. Up to a Poincaré transformation, you can assume that: $$ \begin{align} t(0) &= 0 & x(0) &= 0 & \dot t(0) &= 1 & \dot x(0) = 0 \end{align} $$

You can use the relation to decouple the equations: $$ \begin{align} \ddot x &= a\sqrt{1+\dot x^2} & \ddot t &= a\sqrt{\dot t^2-1} \end{align} $$ You can therefore reconstruct your trajectory with no ambiguity by separating the variables and antiderivating. For coordinate time, this gives: $$ \int^\dot t \frac{du}{\sqrt{u^2-1}} = \cosh^{-1}(\dot t) \\ \cosh^{-1}(\dot t) = \int ad\tau \\ t(\tau) = \int_0^\tau \cosh\left(\int_0^{\tau_1} a(\tau_2)d\tau_2\right)d\tau_1 $$ For completeness, the same argument holds for the spatial coordinate: $$ x(\tau) = \int_0^\tau \sinh\left(\int_0^{\tau_1} a(\tau_2)d\tau_2\right)d\tau_1 $$ Notice that for constant proper acceleration, you recover hyperbolic motion.

As you can see the natural parameter is proper time. If you want to express everything in terms of coordinate time, the best way is to solve things with proper time and then invert $t(\tau)$ to get your desired dependence.

If you really want to parametrize with coordinate time, then $$ \frac{d}{dt} = \frac{1}{\dot t}\frac{d}{d\tau} $$ You get: $$ \begin{align} x' &= \frac{\dot x}{\dot t} \\ x'' &= \frac{\ddot x \dot t-\dot x\ddot t}{\dot t^3} \end{align} $$ Plugging in the previous equations of motion gives you your equation: $$ x'' = \frac{a}{\gamma^3} \\ \gamma = \frac{1}{\sqrt{1-x'^2}} $$ The same previous initial conditions give: $$ \begin{align} x(0) &= 0 & x'(0) &= 0 \end{align} $$ Following the same method: $$ \int^{x'}\frac{du}{\sqrt{1-u^2}^3} = \frac{x'}{\sqrt{1-x'^2}} \\ \frac{x'}{\sqrt{1-x'^2}} = \int a dt \\ x(t) = \int_0^t\frac{dt_1}{\sqrt{\left(\int_0^{t_1} a(t_2) dt_2\right)^{-2}+1}} $$ If you want proper time, you just use with initial condition $\tau(0) = 0$: $$ \frac{d\tau}{dt} = \sqrt{1-x'^2} \\ \tau(t) = \int_0^t \frac{dt_1}{\sqrt{1+\left(\int_0^{t_1} a(t_2)dt_2\right)^2}} $$ Once again, you can check the consistency with hyperbolic motion.

Hope this helps.

  • $\begingroup$ Thank you for your answer. Helped a lot! Only thing I don't get is how the definition of the proper acceleration implies that system of differential equations. $\endgroup$
    – Kabouter9
    Commented Dec 7, 2023 at 12:36
  • 1
    $\begingroup$ Glad it helped. You just use the the fact that $n = (\dot x,\dot t)$ is the unit normal vector to $U$, so $A = an$ with $a$ the proper acceleration by definition. It's the Minkowski analogue of the curvature for the 2D Frenet basis. $\endgroup$
    – LPZ
    Commented Dec 7, 2023 at 12:58
  • $\begingroup$ Sorry for the belated reply but I have two more related questions. First, to be sure, if you write $\int_{0}^{t_{1}} a(t_{2}) dt_{2}$, you really mean $a(t_{2})$ and not $a(\tau(t_{2}))$ right? Secondly, suppose that I want to solve the inverse problem; given the velocity $dx/dt$, from your equation $x(t)$, I can find an equation for $a(t)$ but how can I write this in function of proper time? Thanks in advance. $\endgroup$
    – Kabouter9
    Commented Feb 19 at 13:15

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