Let's have following scenario:

A spaceship departs from the Earth and constantly accelerates in a straight line for 60 000 000 seconds at 1g (10 m/s^2), then immediately starts decelerating at 1g for another 60 000 000 seconds, coming to a halt relative to the Earth (edit: time measured by the traveler).

When I do a "classical" calculation, the ship accelerates to a top speed of 600 000 000 m/s (or 2c), the average speed of the entire trip is 300 000 000 m/s, duration 120 000 000 seconds and distance travelled 36 x 10^15 meters.

How is the duration, distance, average and top speed of the trip calculated using relativity from the point of view of the traveler, and the Earth? (if possible, I'd appreciate a step-by-step description of how you calculate the values in the different frames of reference, I'd like to understand what's going on instead of just being given a number).

Is there any reference frame that would give similar results to the "classical" values (well, except the top speed being 2c...)?

  • $\begingroup$ One of the first things you are going to have to do is clarify what you mean by 'for 60 000 000 seconds'. By whose watch? Don't forget that elapsed times can be frame dependent. 60megaseconds in the Earth frame will not be the same as 60megaseconds on the spaceship. $\endgroup$ Commented Feb 16 at 13:04
  • $\begingroup$ Apologies for that omission - I meant as measured by the traveler. $\endgroup$
    – Tondo PX
    Commented Feb 16 at 14:22
  • $\begingroup$ You may have a look at en.wikipedia.org/wiki/Hyperbolic_motion_(relativity). There are all formulas you need (proper acceleration=1g in your case). $\endgroup$
    – Batiatus
    Commented Feb 16 at 17:36

1 Answer 1


With objects having a max speed of $c$ it is not really possible to constantly accelerate something, at least as measured by an inertial observer. The best you can do is have an object move with a constant proper acceleration, meaning in their frame they feel a constant acceleration (e.g. like us on earth feeling a constant proper acceleration of $g$) but according to an inertial observer the acceleration of the object slow down as it approaches $c$.

I think the best way to think of this is that all objects have a 4-velocity $\mathbf U = (U_t, U_x, U_y, U_z)$ with a constant magnitude, $\mathbf U\cdot \mathbf U = c^2$, and forces/gravity can only act by rotating the velocity, and have no way to change the magnitude. (For simplicity I'll stick to 1 spatial dimension, $\mathbf U = (U_t, U_x)$). A constant proper acceleration corresponds to a constant rate of rotation. When an object accelerates what is actually going on is that some of the "time" component of the 4-velocity $U_t$ is being rotated into to the "space" component $U_x$ so we see the object move slower through time (time dilation) and faster through space.

A very important detail I left ambiguous is that the constant magnitude of $\mathbf U$ is with respect to the hyperbolic Minkowski dot product $\mathbf U\cdot\mathbf U = U_t^2 - U_x^2$ rather than the circular Euclidean dot product $\mathbf r\cdot\mathbf r = r_x^2 + r_y^2$, and so this magnitude preserving "rotation" is not circular, but hyperbolic. Although this difference leads to alot of different behaviour, the math works out very similarly. A vector with constant Euclidean magnitude $\mathbf r\cdot \mathbf r = r_x^2 + r_y^2 = r^2$ can be parameterized in circular polar coordinates: $$\mathbf r = r(\cos\theta, \sin\theta)$$ enter image description here

A vector with a constant Minkowski magnitude such as the 4-velocity $\mathbf U\cdot\mathbf U = U_t^2 - U_x^2 = c^2$ can be parameterized in hyperbolic polar coordinates, where the hyperbolic angle $\xi$ is called the rapidity. $$\mathbf U = c(\cosh\xi, \sinh\xi)$$ enter image description here

The key difference between the two is that circular rotations move the vector around a circle and therefore regularly bring it back to the same point with rotations of $\theta = 2\pi$, but hyperbolic rotations move the vector along the hyperbola which extends off to infinity: $\cosh\xi\to\infty, \sinh\xi\to\infty$ as $\xi\to\infty$. Although this means $U_x = c\sinh\xi$ approaches $\infty$ as the object is accelerated (rotated), this is the spatial velocity with respect to their proper time $\tau$ (since their time slows down they travel large distances while barely moving through time). The spatial velocity with respect to the inertial observer's time $t$ is given by $$v = c\frac{U_x}{U_t} = c\tanh\xi$$ Since $\tanh\xi\to 1$ as $\xi\to\infty$, the object's speed $v$ will approach $c$ as it is accelerated. You can see this in the above picture as the slope of the hyperbola approaching $45^\circ$ as it extends outwards.

Like I mentioned earlier a constant proper acceleration corresponds to a constant rate of rotation of the 4-velocity, so the rapidity is: $$\xi = \frac{\alpha\tau}{c} + \xi_0$$ Where $\alpha$ is the proper acceleration, $c$ is the speed of light, $\tau$ is the proper time of the accelerated observer, and the initial rapidity $\xi_0$ corresponds to the initial motion of the object. Therefore the 4-velocity of an accelerated observer is: $$\mathbf U(\tau) = c(\cosh(\alpha\tau/c + \xi_0), \sinh(\alpha\tau/c + \xi_0))$$ I won't get into derivation which shows why the proper acceleration is divided by $c$, but you can think of it as a way to make the rapidity dimensionless.

Now for your scenario:

  • First the ship accelerates with a constant proper acceleration of $\alpha = g$ for proper time interval $\Delta\tau$. Assuming the ship begins with a velocity of $v = 0$ at proper time $\tau = 0$ the initial rapidity is: $$c\tanh(\alpha\cdot0/c + \xi_0) = 0 \implies \xi_0 = 0$$ Therefore the rapidity as a function of $\tau$ is: $$\xi_1(\tau) = \frac{g\tau}{c}$$
  • At the end of the first part of the flight the ship has reached a top rapidity of: $$\xi_1(\Delta\tau) = \frac{g\Delta\tau}{c}$$
  • Next the ship accelerates in the opposite direction, having proper acceleration $\alpha = -g$ for the same length of time $\Delta\tau$. The initial rapidity of this part is the final rapidity of the last part, $\xi_0 = \frac{g\Delta\tau}{c}$, so that the rapidity as a function of $\tau$ (measured so that $\tau=0$ is the end of the first half) is: $$\xi_2(\tau) = \frac{-g\tau}{c} + \frac{g\Delta\tau}{c}$$

To calculate the top speed, the final rapidity of the first half of the flight is plugged into the expression for speed: $$v_\mathrm{top} = c\tanh(g\Delta\tau/c)$$ Using the values $c = 3\times10^8m/s$, $g = 10 m/s^2$, $\Delta\tau = 6\times10^7 s$, this gives a value of $$v_\mathrm{top} \approx 0.964c = 2.89\times 10^8 m/s$$

To calculate the time and distance travelled during the flight according to the inertial observer the 4-velocity is integrated over both parts: $$\begin{align*} \Delta \mathbf X &= \int_{0}^{\Delta\tau}\mathbf U(\tau)\mathrm d\tau \\ &= \int_{0}^{\Delta\tau}c(\cosh(\alpha\tau/c + \xi_0), \sinh(\alpha\tau/c + \xi_0))\mathrm d\tau \\ &= \left[\frac{c^2}{\alpha}(\sinh(\alpha\tau/c + \xi_0), \cosh(\alpha\tau/c + \xi_0))\right]_0^{\Delta\tau} \\ &= \frac{c^2}{\alpha}\bigg(\sinh(\alpha\Delta\tau/c + \xi_0) - \sinh(\xi_0), \; \cosh(\alpha\Delta\tau/c + \xi_0) - \cosh(\xi_0)\bigg) \end{align*}$$

Plugging in the values for the 2 parts, $$\Delta \mathbf X_1 = \frac{c^2}{g}\bigg(\sinh(g\Delta\tau/c), \; \cosh(g\Delta\tau/c) - 1\bigg)$$ $$\Delta \mathbf X_2 = \frac{c^2}{-g}\bigg(\sinh(-g\Delta\tau/c + g\Delta\tau/c) - \sinh(g\Delta\tau/c), \; \cosh(-g\Delta\tau/c + g\Delta\tau/c) - \cosh(g\Delta\tau/c)\bigg) = \frac{c^2}{g}\bigg(\sinh(g\Delta\tau/c), \; \cosh(g\Delta\tau/c)\bigg)$$

The total time is found by combining the time components of the 4-displacements (divided by $c$): $$\Delta t_\mathrm{total} = 2\frac{c}{g}\sinh(g\Delta\tau/c) \approx 2.18\times10^8 s$$

The total distance is found by combining the absolute value of the spatial components (but since they're both positive they can just be added as is): $$\Delta x_\mathrm{total} = \frac{c^2}{g}(2\cosh(g\Delta t/c) - 1) \approx 5.87\times10^{16} m$$

The average speed is just the total distance divided by the total time: $$\bar v = \frac{\Delta x_\mathrm{total}}{\Delta t_\mathrm{total}} \approx 0.899c = 2.70\times10^8m/s$$

If you would like to learn more about this you can check out

  • $\begingroup$ Thank you for a thorough answer. It is a bit above my level but I'm determined to get to the bottom of it even if it may take some time. $\endgroup$
    – Tondo PX
    Commented Feb 27 at 10:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.