Is there a formula that gives the position of an object depending on the time, but which doesn't allow the object to surpass the speed of light?

Question

I have found these two formulas:

$v = at + v_0$

$x = \frac{1}{2}at^2 + v_0t + x_0$

• a is the acceleration
• v is the velocity
• x is the position
• t is the time
• $v_0$ is the initial velocity
• $x_0$ is the initial position

The problem is that with an acceleration of $10$ m.s$^{-2}$ (for example) the object would surpass the speed of light in $3 \times 10^7$ s = $11$ months and $11$ days, which should not be possible.

Is there some other formula that gives the position of an object depending on its acceleration and on the time, but which works and does not allow the speed of the object to surpass the speed of light?

Answer 1

Have a look at the article by Phil Gibbs on the relativistic rocket. This describes the motion of a rocket that is accelerating with a constant acceleration. In this context constant acceleration means the crew of the rocket feel a constant acceleration. Technically the rocket has a constant four-acceleration.

Anyhow, the velocity of the rocket as observed by a non-accelerating observer is given by:

$$ v = \frac{at}{\sqrt{1 + (at/c)^2}} $$

where $a$ is the acceleration measured by the occupants of the rocket and $t$ is time as measured by the non-accelerating observers.

At long times, when $(at/c)^2 \gg 1$ the velocity is approximately:

$$ v \approx \frac{at}{\sqrt{(at/c)^2}} = c $$

So at long times the velocity approaches $c$, though it never reaches it.

Answer 2

Is there some other formula ... which ... does not allow the speed ... to surpass the speed of light?

That would be the equations of special relativity mentioned by sahin in a comment.

^{Image from Loodog?}

Another factor you have to take into account with classical mechanics is to work out how a constant force can be applied to your object over 11 months and 11 days without affecting it's mass (therefore no fuel onboard) and without the additional speed giving rise to large opposing forces such as friction.

Answer 3

Look at sparknotes.com/physics/specialrelativity/dynamics/…, you can see $dE/dx=F$ - if your force is constant, it is the energy that increases constantly. $E=\gamma(v)m_0c^2$, you can deduce the $v$. Beacause of laziness I used mathomatic, and it gives me something like this:

\begin{equation} v=c\sqrt{1-\left(\dfrac{m_0 c^2}{F\cdot x + m_0 c^2}\right)^{2}} \end{equation}

If you check it for x=0 and x=inf, you get reasonable results.

Answer 4

which should not be possible.

Indeed, uniform coordinate acceleration $a$ is inconsistent with special relativity however, uniform proper acceleration $\alpha$ is consistent.

The proper acceleration is the acceleration of the object according to an attached accelerometer.

For 1D motion, the relationship between $\alpha$ and $a$ is given by

$$\alpha = \gamma^3 a = \frac{a}{\left( 1 - \frac{v^2}{c^2} \right)^{\frac{3}{2}} }$$

Since the Lorentz factor goes to infinity as $v \rightarrow c$, $a$ must go to zero if $\alpha$ is to remain finite.

If, from an inertial reference frame, an object were observed to have uniform coordinate acceleration $a$, the object's proper acceleration would be arbitrarily large as the object's speed approached $c$ in this frame.

Is there some other formula that gives the position of an object depending on its acceleration and on the time, but which works and does not allow the speed of the object to surpass the speed of light?

In the context of special relativity, one must be careful to distinguish between the proper and coordinate accelerations since, as described in the link above, they are not the same thing.

Assuming by, "its acceleration", you mean its (constant) proper acceleration, then, with zero initial conditions, the formula is

$$x(t) = \frac{c^2}{\alpha}\left[\sqrt{1 + \frac{\alpha^2 t^2}{c^2}}-1 \right]\;,\; t \ge 0$$

See that, as $t$ gets very large, $x(t)$ asymptotically approaches $ct$, i.e., the speed asymptotically approaches $c$.