Gravitational Collapse of a Spherical Shell?

A while ago at high school, I was once walking down the hallway when I noticed my beloved physics teacher heading towards me wearing a big smile on his face. I knew what was coming, another one of his classic “try to solve this problem, if you can” moments. However, the level of his excitement was even higher than per usual, for he didn’t have a solution that time. The question was rather simple: “Imagine you have a ring of uniform mass density and of some initial radius somewhere in free space (i.e. without any external forces) and without any repulsive interactions within the mass. How long would it take the ring to collapse into singularity? (Not assuming General Relativity)”

My initial response was somewhere on the line of “cool, I’ll just use a bit of calculus and be done, right?”, but as it turned out, the problem was actually quite subtle. The math kept giving me infinite accelerations and instant times of collapse. It was a nightmare. I kept asking myself how could something this simple be so daunting at the same time? Later I came to realise that I was basically trying to calculate something rather unphysical, there is simply no such thing in nature and never could be.

The main difficulty in this problem lay in the fact that the mass of the ring was proportional to its radius, while using Newton’s inverse square law resulted in infinite force acting on every point of the ring. I was a bit disheartened, yet during the process of trying to solve the original problem, when working on some other symmetrical models, I actually managed to solve something else of similar interest. I had simply switched the ring for a spherical shell of potentially zero thickness. This time, on the other hand, I got back a fascinating result that I want to share with you, my dear reader.

There are at least two ways of solving this problem; one using a bit of vector calculus with the aid of some physical intuition, and the other using brute force vector calculus from start to finish. I will take the path of least action, i.e. the first route.


Potential Energy of the Shell

Let me introduce the problem to you one more time, just in case you are actually smart and have skipped the boring introduction. Imagine a spherical shell of zero thickness, of uniform mass density, of mass M, and of initial radius R. We are only considering the (Newton’s) gravitational force here, by which each point of the shell acts on all the others and vice versa. The shell is located in an empty universe with no present external forces, or internal forces other than gravity. The question is, given M and R, how long will it take the shell to collapse into a single point?

Let’s start by familiarising ourselves with one of the famous results in physics called The Shell Theorem, dating back to Sir Isaac Newton himself, which will help us tremendously. He proved, assuming his inverse square law, that an infinitesimally thin spherical shell produces no gravitational field inside the shell, while on the outside it acts as if the sphere were a point mass. Even as one of the founding fathers of calculus, he actually took a geometrical approach to the problem and solved it without the need of integration. Yet this approach is rather lengthy, hence the use of calculus is helpful. One can prove the given theorem by applying brute force multivariable calculus, considering how every infinitesimal surface element interacts with one another, but it might be more beneficial to us to prove this theorem with the use of something much more general, by Gauss’s law.

In our case, Gauss’s law for gravity takes the form of

(1)   \begin{equation*} \oiint_{S} \mathbf{g} \cdot d \mathbf{S} = - 4 \pi GM, \end{equation*}

where on LHS, \mathbf{g} is the gravitational field, S denotes the closed surface (of arbitrary shape) enclosing \mathbf{g}, and on RHS, G is the gravitational constant, and M the mass enclosed by S.

For the symmetrical case of a spherical surface (enclosing the mass M) the calculation is quite trivial, since LHS of Eq.(1) becomes just 4\pi r^2 |\mathbf{g}|, where r is the distance from the centre of the surface. The absence of enclosed mass inside the hallow sphere implies \mathbf{g} = \mathbf{0}, and on the outside of the shell the field becomes

(2)   \begin{equation*} \mathbf{g} = - \frac{GM}{r^2}\, \widehat{\mathbf{r}}, \end{equation*}

with the radial unit vector \widehat{\mathbf{r}}.

It has therefore the same form as the gravitational field of a point mass, so the spherical shell acts outside as if all its mass were concentrated at its centre. From this it is easy to calculate a gravitational potential at its surface, or the energy per unit mass needed to bring an object from infinite distance to distance r away from its centre, as

(3)   \begin{equation*} \phi = - \int_{\infty}^{r} \mathbf{g} \cdot d \mathbf{r} = - \frac{GM}{r}. \end{equation*}

The negative sign ensures that gravity binds objects together. Furthermore, the potential energy (or self-energy) of the shell structure can be calculated by integrating the potential over the mass of the sphere

(4)   \begin{equation*} U = \int_M \phi \,dM = - \frac{GM^2}{2r}. \end{equation*}

This step probably looks a bit tricky, however, it is just a gravitational analogy of the expression for energy stored in a charged capacitor or a conducting coil in electromagnetism. Instead of capacitor charging, we are accumulating mass.


Equation of Motion

To find the equation of motion for the shell collapsing on itself one can utilise Newton’s laws of motion, nonetheless, it is easier to derive it via the law of energy conservation. That is why we have calculated the gravitational potential energy of the spherical shell in the first place, for during our process this energy converts into kinetic energy of the infinitesimally small elements constituting the shell. The kinetic energy of the system is easily obtained

(5)   \begin{equation*} T = \frac{1}{2} \int_{M} \dot r^2 dM = \frac{1}{2} M \dot r^2, \end{equation*}

where we have adopted the dot notation for time derivative, so that \dot r \equiv \frac{d}{dt}r is the velocity by which the shell collapses.

The conservation of energy is simply

(6)   \begin{equation*} T + U = const. \end{equation*}

The constant term depends on the initial condition, i.e. that the shell starts collapsing at time t = 0. Hence, by rearranging, we obtain the following

(7)   \begin{equation*} \dot r^2 = \frac{GM}{r} - \frac{GM}{R}. \end{equation*}

Finally, by taking the square root of both side of the equation, and by remembering the fact that the velocity must be negative, for r decreases during the collapse, we get the final form of the equation of motion

(8)   \begin{equation*} \dot r = - \sqrt{ \frac{GM}{r} - \frac{GM}{R}}. \end{equation*}

This is a first order non-linear differential equation which looks difficult to solve, but as it turns out, it is manageable.


Solving for the Equation of Motion

In this section, we are going to channel our inner mathematicians, and turn back to physics once the differential equation is solved. To start off, we are going to make G=M=R=1 just to make mathematical manipulations easier. This transforms our equation into

(9)   \begin{equation*} \dot r = - \sqrt{ \frac{1}{r} - 1}. \end{equation*}

Since this is a separable differential equation, we can further transform it into its integral form as

(10)   \begin{equation*} - \int \frac{1}{\sqrt{ \frac{1}{r }- 1}} dr= \int dt. \end{equation*}

We are going to focus on LHS and the next step is going to be the following u substitution

(11)   \begin{equation*} u = \sqrt{ \frac{1}{r} - 1} \rightarrow dr = - 2 r^{2} \sqrt{ \frac{1}{r} - 1}\, du, \end{equation*}

which yields LHS to be

(12)   \begin{equation*} 2 \int \frac{1}{(u^2 + 1)^2} du. \end{equation*}

By taking a closer look at this expression we can use the fact that \textrm{tan}^2(w)+1 = \textrm{sec}^2(w) by making another w substitution

(13)   \begin{equation*} u = \textrm{tan}(w) \rightarrow du = \textrm{sec}^2 (w) dw, \end{equation*}

resulting in the following standard integral

(14)   \begin{equation*}  \begin{split} 2 \int \frac{ \textrm{sec}^2 (w)}{(\textrm{tan}^2(w)+1)^2} dw &= 2 \int \textrm{cos}^2(w) dw\\ &= w + \textrm{sin}(w) \textrm{cos}(w) + C. \end{split} \end{equation*}

Now as the difficult part is behind us we can plug back in the substitutions, starting with w then u;

(15)   \begin{equation*} w = \textrm{arctan}(u) = \textrm{arctan}\Big(\sqrt{ \frac{1}{r} - 1}\Big). \end{equation*}

After plugging in, the Eq.(14) can be further simplified by drawing out the right-angled triangle corresponding to angle represented by w (as seen in the figure below). With this we arrive to

(16)   \begin{equation*} \begin{split} \textrm{sin}(w) &= \sqrt{r} \sqrt{ \frac{1}{r} - 1}\\ \textrm{cos}(w) &= \sqrt{r}. \end{split} \end{equation*}


The right-angled triangle relating our w and u substitution to r.


In the end, we obtain following relationship for t;

(17)   \begin{equation*} t = \textrm{arctan} \Big(\sqrt{ \frac{1}{r} - 1}\Big) + r\sqrt{ \frac{1}{r} - 1} + C, \end{equation*}

where C vanishes because of our initial condition r(t = 0) = 1 [r(t = 0) = R].


Duration of the Collapse

Back to physics, after making Eq.(17) dimensionally correct, we finally have the expression for time t it has taken the process to shrink from the initial radius R to r \in (0,R]

(18)   \begin{equation*} t = \sqrt{\frac{R^3}{GM}} \, \Big[ \textrm{arctan} \Big(\sqrt{ \frac{R}{r} - 1}\Big) + \sqrt{ \frac{r}{R} - \frac{r^2}{R^2}} \, \Big], \end{equation*}

and by letting r \rightarrow 0 we find the time of the gravitational collapse simply as

(19)   \begin{equation*} t_{collision} = \frac{\pi}{2} \sqrt{\frac{R^3}{GM}}. \end{equation*}


The Final Result

It is wonderful to see such an elegant result in the end, yet it is even more interesting when you realise that it somehow resembles an orbital period described by Kepler’s laws. In fact, after the collapse, if the singularity started expanding symmetrically back to form a mirror image of the initial shell, then stopped, collapsed again, and expanded back to the same state as in t=0, it would take time

(20)   \begin{equation*} T = 2\pi \, \sqrt{\frac{R^3}{GM}}, \end{equation*}

which is exactly the mathematical expression of Kepler’s third law.

Hence, if the process oscillated like this (collapse-mirror expansion-mirror collapse-expansion), it would take the exact same time as it would some small object to make a single orbit around the shell with an axis of length R!

However, my favourite part about all of this is that because of the Shell Theorem, the gravitational force at distance R from the centre of the shell does not change during the whole process, meaning if a small object was orbiting this shell before time t=0 at distance R, it would still be orbiting it after it had started collapsing, and both processes would have the same period T. The orbiting object would also always “touch” only its starting point on the sphere, after being fully expanded every whole or half period.

Ultimately, I don’t think there is a deeper connection between the final result and Kepler’s laws. For example, if we had a solid sphere of uniform density, mass M, and radius R, and dug a tunnel through its centre, it would again take the same time period of T to fall to the other side of the tunnel and then fall back to the starting position. This is not hard to prove, hence I’m leaving it as an exercise to the reader.



I would like to thank my dear, former teacher of physics, Martin Kapoun, of the Johannes Kepler Grammar School in Prague, for showing me how much pain and enjoyment can a little physics problem bring. I also want to tell him that I am sorry I could not solve the original problem he had given me. Maybe in a few years time I’ll try again.


The featured image was borrowed from this address:

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