# Thought Experiment: An object falling indefinitely under gravity?

So the other day, I devised this thought experiment:

Consider an infinite tunnel, and you drop a coin of mass $m$ into it. Considering the effect of gravity to be applicable and neglecting air drag and other viscous forces, it seems that from Newton's second law the coin would accelerate indefinitely since force ($mg$) is constantly being applied. But on the other hand I also know nothing can go faster than speed of light, as stated by Einstein's theory of relativity so how can I describe the above experiment where it seems like the coin would approach the speed of light?

I read somewhere else that when a object made to accelerate indefinitely then as per Mass-Energy Equivalence the mass of object begins to increase exponentially and hence greater and greater force is required to increase object's speed.

But somehow the above explanation doesn't seem to explain much in my thought experiment.

I would like to know the actual reason why the above scenario can't happen.

• – Phoenix87 Jan 28 '15 at 16:41
• @Phoenix87 Thanks! It seems to make some sense, I'll try to understand it. – ritvik1512 Jan 28 '15 at 16:45
• You will never find an infinite tunnel with a constant force of gravity through it. You've proposed an impossible situation and asked why the result is also impossible – Jim Jan 28 '15 at 16:53
• Please note that the MathJax formatting with the $<...>$ is for math, not for emphasis on words. – Kyle Kanos Jan 28 '15 at 17:45

Suppose you are an observer stationary in this infinite gravitational field, so you feel a gravitational force $g$ just as you do standing on the surface of the Earth. The only difference is that this acceleration $g$ is constant and doesn't change as you go higher or lower.

In that case your spacetime is described by the Rindler metric (as Phoenix87 mentions in a comment). You'll see the metric written in a variety of coordinates. The most physically intuitive, if not the easiest to work with is:

$$ds^2 = -\left(1 + \frac{g}{c^2}z \right)^2 c^2 dt^2 + dz^2$$

where I've left in the factors of $c$ that we traditionally set to unity. The variable $z$ is the distance relative to your stationary position i.e. at your position $z = 0$. There is a coordinate singularity at $z = - \tfrac{c^2}{g}$, i.e. where $1 + \tfrac{g}{c^2}z$ becomes zero, which is akin to a black hole event horizon.

If you drop a coin you will observe it accelerate downwards away from you towards the coordinate singularity. However as the coin approaches the horizon it will slow asymptotically and never reach the horizon even if you wait an infinite time. This is closely analogous to what happens if you dropped your coin into a black hole, where you would observe the coin to slow and freeze as it approached the horizon.

There is another singularity at $z = + \tfrac{c^2}{g}$ that is akin to a white hole. If you throw your coin upwards (to positive $z$) then no matter how fast you throw it, i.e. no matter how closely its velocity approaches the speed of light, the coin will never reach $z = + \tfrac{c^2}{g}$. So in effect your entire universe consists of the range $- \tfrac{c^2}{g} \le z \le \tfrac{c^2}{g}$

• Well written! This was the crucial step I was missing! In fact, @drone scientist's answer and your answer seem to suggest the same thing, did I comprehend your answer correctly? Thanks, again. – ritvik1512 Jan 28 '15 at 17:17
• No. Drone's answer is purely Newtonian and does not take general relativity into account. What he's saying is that there is no way to physically construct an infinite gravitational field, which is of course quite true but that shouldn't stop us doing the thought experiment. – John Rennie Jan 28 '15 at 17:22
• Ah, I see. So the most plausible explanation is that the coin will slow down asymptotically. – ritvik1512 Jan 28 '15 at 17:24
• Wait, can you again explain the analogy with the black hole event horizon? – ritvik1512 Jan 28 '15 at 17:26
• @RitvikChoudhary: Assuming you have an infinite gravitational field the coin will initially accelerate away from you, so it speeds up as you'd exect. Howver as the coin approaches $z = -\tfrac{c^2}{g}$ it wil decelerate and it will approach $z = -\tfrac{c^2}{g}$ asymptotically. – John Rennie Jan 28 '15 at 17:26