Will a light beam ever reach a slowly accelerating observer in this scenario?

Question

I had a little thought experiment. Suppose an observer is trapped in an elevator. Initially, the elevator is at rest in outer space. A light source is located $1 \ \text{m}$ below the elevator. It releases a light beam to the elevator the moment it takes off with a constant acceleration of $1 \ \mathrm{mm/s^2}$. Will the light beam ever reach the observer in the elevator?

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I think the answer depends on the definition "acceleration" here. If it refers to coordinate acceleration, then the answer is yes — since the observer is very close to the light source and the acceleration is small.

But if it refers to proper acceleration, then I believe the answer is no — not when one inspects the Rindler coordinate chart.

I am not sure how to make sense of this result intuitively. What is the magnitude of coordinate acceleration in this case? I think coordinate acceleration should be bigger than proper acceleration. But since they are related by the formula $a=\alpha/\gamma^3$, where $a$ is coordinate acceleration and $\alpha$ is proper acceleration, coordinate acceleration is actually smaller than proper acceleration.

Answer 1

Suppose we have an observer with constant proper acceleration $a$ then it is straightforward to calculate the position of that observer in the "Earth frame". It is convenient to set the position of the observer to $x = c^2/a$ at time $t = 0$ as then the position is simply given by:

$$ x = \frac{c^2}{a}\sqrt{1 + \left(\frac{at}{c}\right)^2} $$

You can find this and similar equations in Phil Gibbs's article on the relativistic rocket. He does not give a derivation but it's not hard to do. We'll modify this by taking a factor of $at/c$ outside the square root to get:

$$ \begin{align} x &= \frac{c^2}{a}\frac{at}{c}\sqrt{\left(\frac{c}{at}\right)^2 + 1} \\ &= ct\sqrt{1 + \left(\frac{c}{at}\right)^2} \end{align} $$

Anyhow, suppose we start a light ray from the origin at $t = 0$ so its position in the Earth frame is:

$$ x_\ell = ct $$

And we get your lead on the light by subtracting the position of the light from your position:

$$ \begin{align} \Delta x &= ct\sqrt{1 + \left(\frac{c}{at}\right)^2} - ct \\ &= ct \left( \sqrt{1 + \left(\frac{c}{at}\right)^2} - 1 \right) \end{align} $$

And it should be obvious that $\Delta x$ is always greater than zero because:

$$ \sqrt{1 + \left(\frac{c}{at}\right)^2} > 1 $$

so the light beam will never catch you.

In the calculation above we gave the accelerating observer a head start of $c^2/a$ and the light beam could not catch them. That means in your example the light beam will never reach the person trapped in the elevator if, and only if, that person has a head start of $c^2/a$. However for $a = 1~\textrm{mm}\ \textrm{s}^{-2}$ the lead needed is about ten thousand light years so that's one hell of an elevator shaft.

What you've discovered is that the Rindler geometry contains a coordinate horizon at $x = -c^2/a$ and light rays emitted from any position at or behind the horizon can never reach the origin. In fact this nicely shows how coordinate horizons are not real singularities since the Rindler geometry is just flat spacetime written in accelerating coordinates, and flat spacetime obviously contains no singularities.

Answer 2

In Rindler coordinates, a light beam from the origin never catches up with any of the accelerating observers. The proper acceleration of the accelerating observers is equal to $a_p = c^2/x$, where $x$ is the spatial distance from the origin, when the coordinate time is zero. The light ray will catch up with any observer that has a head start of less than $c^2/a_p$ from the origin.

A light source is located $1 \ \text{m}$ below the elevator. It releases a light beam to the elevator the moment it takes off with constant acceleration of $1 \ \mathrm{mm/s^2}$. Will the light beam ever reach the observer in the elevator?

An observer with a constant proper acceleration of $a_p = 1 \ \mathrm{mm/s^2}$ ($= 0.001 \ \mathrm{m/s^2}$) will need a head start of $$x = c^2/a_p = 299792458^2/0.001 = 299792458000 \ \text{light seconds},$$

to stay permanently ahead of the light signal. Since the OP has specified a head start of just $1 \ \text{m}$, the light beam will catch up with the observer in that particular instance.

So far I have only considered constant proper acceleration, but the OP also asked about constant coordinate acceleration. In the latter case we can do the analysis simply using the regular Newtonian equations of motion (if we limit ourselves to velocities less than $c$). For a particle that is initially at rest, the relationship between position and constant coordinate acceleration ($a_c$) is given by:

$$x = x_0 + \frac{1}{2} \ a_c t^2,$$

where $x_0$ is the original position relative to the origin or the 'head start'. If a light signal starts out from the origin at time $t=0$, it will catch with the accelerating particle when: $$ct = x_0 + \frac{1}{2} \ a_c t^2.$$

Solving for $t$ gives us the catch up time:

$$t = \frac{c - \sqrt{c^2 - 4 a_c x_0}}{2 a}.$$

Clearly, the light signal cannot catch up with a particle that has constant coordinate acceleration $a_c$, if the particle's head start is greater than $\dfrac{c^2}{4 a_c}$, because the term in the square root becomes imaginary in that situation. This is exactly $1/4$ the head start required by a particle with constant proper acceleration. This makes sense, because constant coordinate acceleration means the proper acceleration is continually increasing.

Answer 3

The confusion arises because acceleration is not Lorentz-invariant.

Frst of all, you should have provided a link to Rindler coordinates.

At first look, this appears to be the hare and the tortoise explanation of why a rabbit could never catch Achilles, or Zeno's o-so-sophist explanation why Achilles can never reach a stationary object because he always has to cross one half of the remaining distance.

However, Rindler noticed something subtler than that. Because the speed of light has a finite value, continuous constant acceleration will get you there in a finite time.

Continuous 1G acceleration will reach c in 16 months. The trick is, Rindler is not talking about proper acceleration, because proper acceleration cannot be maintained forever, as seen from an unaccelerated inertial frame.

The hard part for the ship is maintaining 1G acceleration for 16 months, as seen from Earth. In fact, it's impossible because of time dilation. As the ship becomes relativistic, months stretch out (as seen through the ship's window from a telescope on Earth). Maintaining 1G acceleration from the point of view of Earth becomes increasingly difficult, even though proper acceleration is constant.

Mission Control would tell them, "You have to speed up! Your acceleration is no longer 1G," even though those on board think it is.

Nevertheless, if it WAS posible, as the link above says:

a light signal emitted at or after a certain critical time will not catch up with the accelerating particle;

So NO, this wouldn't happen to a flash bulb one meter below an elevator.

It would happen if the elevator had a head start and was almost moving at c. As @KDP correctly notes, light will never catch up if the head start was greater than 89,875,517,873,681,764,000 km.