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This keeps me awake at night, and i can't seem to find the solution.

Introduction: Suppose an observer is currently at rest, and a space ship accelerates close to the speed of light (current velocity of spaceship from observers point of view = 0.99c). The observer can now see that the spaceship travels close to the speed of light, however, the spaceship still see light passing the spaceship with the speed of light.

Now the question is: Due to the fact that the spaceship still see light passing at "c", does that mean that the spaceship would be able to increase its velocity close to what the spaceship now perceive to be the speed of light? And would the observer only see that as a small increase in velocity (current velocity of spaceship from observers point of view = 0.9999c) of the spaceship?

Lorsayden
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2 Answers2

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Yes, basically. The spaceship can keep on increasing its velocity, though only in increasingly smaller increments. When dealing with relativistic speeds, there are two variables that are arguably better for picturing motion.

One is the gamma factor $$ \gamma = \frac{1}{\sqrt{1-v^2/c^2}} $$ which approaches infinity as $v$ approaches the speed of light. The spacecraft can always increase its gamma factor; particles in particle accelerators have been going basically the speed of light for a while now, but the newer, bigger accelerators accelerate them to a higher gamma factor.

The other is rapidity, defined as $$ \varphi = \text{cosh}^{-1}(\gamma) $$ This has two very nice properties:

  • For small, non-relativistic speeds, $\varphi \approx v/c$ and is basically the same thing as speed
  • Rapidities add the together the way people usually think velocities do. If the spaceship is traveling at $\varphi_1$ in the frame of some planet and shoots a missile traveling at $\varphi_2$ in the frame of the spaceship, the missile is traveling at $\varphi_1 + \varphi_2$ in the frame of the planet. In its own frame of reference, the spaceship will increase its rapidity at a constant rate if it applies constant thrust.
jwimberley
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Basically, yes.

You can always speed up in your own reference frame, you'll just experience more and more time dilation to keep $c$ constant.

You could use the Lorentz transformations to calculated the perceived change in speed.

$u$ - spaceship velocity
$v$ - reference frame velocity

After Acceleration:
observer reference frame $S$: $u = 0.9999c$.
reference frame at spaceship's initial speed $S'$: $v = 0.99c,u' = ?$.

$$ u' = \frac{u - v}{1 - \frac{uv}{c^2}} $$ $$ = \frac{0.9999c - 0.99c}{1 - \frac{0.99c\times0.9999c}{c^2}} $$ $$ = \frac{0.0099c}{1 - 0.989901} $$ $$ = 0.0099c \div 0.989901 $$ $$ = 0.98029507872c $$

So from the observer's perspective, the spaceship's speed increased by only $0.0099c$ whereas from the spaceship's perspective their speed increased by ~ $0.98c$.

j__
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