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An astronaut is on an odd mission. Each day, the spacecraft is accelerated by 10% of the speed of light, and at night the astronaut sleeps and will loose all memories of the previous day. Each day when the astronaut awakes, they are unaware of the ship's current velocity. Moreover, there's no way to measure the absolute velocity (because there is no such thing) and observation of the outside universe is not allowed. Nonetheless, the mission is carried out each day.

Why, after two weeks, is the ship not traveling at 140% the speed of light?

Or, phrased as a direct question: If there is no such thing as a zero-velocity frame of reference, then how can declaring that another frame of reference has a superluminal velocity even be meaningful?

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    $\begingroup$ Each day, the spacecraft is accelerated by 10% of the speed of light That assumption is where you go wrong. A spaceship experiencing constant proper acceleration undergoes “hyperbolic motion” and never exceeds the speed of light in any reference frame. $\endgroup$
    – G. Smith
    Dec 11, 2020 at 21:26
  • $\begingroup$ Any speed has to be relative to something, and the astronaut cannot tell acceleration from uniform gravity. $\endgroup$
    – Adrian Howard
    Dec 12, 2020 at 0:44
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    $\begingroup$ While you leave me downvotes please leave a comment on how the question can be improved. Thanks. $\endgroup$
    – Brent
    Dec 12, 2020 at 18:17

2 Answers 2

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The answer to almost any thought experiment of this type is to go through and add "relative to what?" any time a speed is mentioned.

When you say the spaceship is accelerated by 10% of the speed of light you are saying the velocity changes by 30,000 m/s. But 30,000 m/s relative to what other object?

If you answer "relative to the Earth" then the answer is: as the spaceship approaches the speed of light relative to Earth it takes more and more energy to increase the speed. Ultimately no amount of energy will allow the spaceship to reach the speed of light according to a given observer because $E_{\text{relative to Earth}} = mc^2/\sqrt{1-v_{\text{rel. Earth}}^2/c^2}$

If you answer "relative to the speed the spaceship was going at the start of the day" then there is no problem. If you then ask "what is the speed of the spaceship at the end of the day relative to its speed at the beginning of the day?" then the answer is obviously 30,000 m/s each day. But if you further ask "what is the speed of the spaceship at the end of the day relative to Earth?" you have to use relativistic velocity addition formula.

Say the spaceship's speed relative to Earth at the beginning of the day is $v_{\text{start, rel Earth}} = 0.9 c$, and at the end of the day the velocity of the spaceship relative to the start of the day is $v_{\text{end, rel start}} = 0.1c$, then the speed at the end of the day relative to Earth is $$v_{end, rel Earth} = \frac{v_{\text{start, rel Earth}} + v_{\text{end,rel start}}}{1+v_{\text{start,rel Earth}}\times v_{\text{end, rel start}}/c^2} $$ $$ = \frac{0.9c+0.1c}{1+0.9\times 0.1} = 0.917c$$ which is still less than $c$.

The relativistic velocity addition formula guarantees that as long as one inertial observer sees an object going less than the speed of light every other inertial observer will also see it going less than the speed of light.

Remember: velocity is only defined between two objects. Every time you think about the speed of an object ask yourself "relative to what?"

Extra reading https://en.wikipedia.org/wiki/Space_travel_using_constant_acceleration

The insight that a spaceship can just keep its rocket on and keep accelerating forever is a good one. In fact, as you speed up relative to your destination planet the distance from the planet to you is contracted. This makes it possible in principle to travel between stars in a matter of years for the person on the spaceship. However, the time dilation makes the journey take hundreds of years for someone watching from the planet.

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  • $\begingroup$ "When you say the spaceship is accelerated by 10% of the speed of light you are saying the velocity changes by 30,000 m/s. But 30,000 m/s relative to what other object?" I'm confused by this. Velocity is impossible to determine without outside observation, but acceleration isn't. An accelerometer measures acceleration without a second frame of reference. $\endgroup$
    – Brent
    Dec 11, 2020 at 22:21
  • $\begingroup$ Good. The concept you are looking for is "proper acceleration." One way of thinking about it is in every bit of time (measured by the person on the spaceship) you are adding a little bit of velocity $a \,\text{d}\tau$ relative to the spaceship an instant ago. Basically you turn the velocity addition formula into a differential equation: $v_{\text{rel Earth}}(\tau+d\tau) = \frac{v_{\text{rel Earth}}(\tau)+a d\tau}{1+v_{\text{rel Earth}}(\tau)\times a d\tau/c^2}$. And so it still is the case that the speed relative to any specific observer will always be less than $c$. $\endgroup$
    – Luke Pritchett
    Dec 11, 2020 at 22:27
  • $\begingroup$ Another way: Suppose you run your rockets so your proper acceleration is $1.04 \text{ m/s}^2$ (measured with your own accelerometer. You let your rockets run for 8 hours (measured with your own clock) then go to sleep. What is your final speed relative to your starting speed? It is not $v_f = a\times t = 30,000 \text{ m/s}=0.1 c$. Using the formula for proper acceleration it is actually $v_{f\text{ rel. start}} = c\times \tanh(at/c) =0.09967 c$. If you repeated this for $n$ days then your speed relative to the very beginning would be $c\times\tanh(nat/c)$ which is always less than $c$ $\endgroup$
    – Luke Pritchett
    Dec 11, 2020 at 22:38
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    $\begingroup$ I see. So even though you can experience 1.4c acceleration (as some constant force over some arbitrarily large time), due to relativistic effects, your relative velocity will still be less than c. I guess you would need to experience infinite acceleration to reach c, which checks out. $\endgroup$
    – Brent
    Dec 11, 2020 at 22:40
  • $\begingroup$ Yes, though I'd caution you to be careful of thinking of it as experiencing 1.4c of acceleration. The 1.4c number only comes up if you naively use $\Delta v = at$ to get the change in velocity. But that formula is a Newtonian approximation that isn't correct at relativistic speeds. $\endgroup$
    – Luke Pritchett
    Dec 11, 2020 at 22:46
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This elaborates on the comment by @GSmith and the answer by LukePritchett.

https://www.desmos.com/calculator/tjngj63cat
describes a piecewise-inertial approximation to a uniformly accelerated observer.

robphy-accelerated-worldline

  • After each tick of the accelerated-observer's clock ($\Delta\tau$),
    the velocity changes by a fixed amount [from the current velocity in the accelerated-observer's frame] $\Delta v$.
  • This velocity change (from the previous velocity) is associated with a fixed increment of rapidity (the Minkowski-angle) $\Delta\theta$.
  • This fixed amount corresponds to the fixed area of each thin hyperbolic-sector,
    where the sides are the [unit] 4-velocities between successive ticks.
  • The worldline is constructed by advancing along the current velocity for one tick.
  • This is an approximation to the hyperbolic worldline of a uniformly accelerated observer.

Observe that the 4-velocity asymptotically approaches the speed of light.
Zoom out to explore.

(The hyperbolic-trajectory of the worldline is on a spacetime diagram.
The unit-hyperbola with the 4-velocities [unit tangent vectors] really should be drawn at the tangent space of each event along the trajectory. It's drawn on the diagram at the origin for convenience. At each tick-event, I have highlighted the tangent vectors [along the 4-velocity], scaled down for visual convenience.)

robphy-accelerated-worldline-limiting

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