Is there a singularity theorem in higher-dimensional Newtonian gravity?

Question

In classical Newtonian gravity with 3 spatial dimensions, it's hard to get two particles to exactly collide, since at short distance the centrifugal force (~1/$r^3$) beats the gravitational attraction (~$1/r^2$). As a consequence, two particles can collide only if the angular momentum is exactly zero, which is measure zero.

In Newtonian gravity with $d$ spatial dimensions, the centrifugal force still goes like ~$1/r^3$ but the gravitational attraction goes like ~$1/r^{d-1}$. This means that for $d>4$, if two particles have negative total energy, they will collide regardless of their angular momentum. For two particles, having negative total energy is a sufficient condition to guarantee a collision after finite time.

My question is whether this is still true with more than two particles?

Conjecture: in $d>4$-dimensional Newtonian gravity, a collection of any number of point particles with negative total energy will experience a "singularity" (a distance=0 collision of at least two particles) within a finite time.

Answer 1

This should follow from the following Virial-type computation.

For convenience we assume all particles have the same mass; this is not essential.

Let $x_i$ denote the position vector of the $i$th particle, then Newton's law of universal gravitation, suitably normalized, gives

$$ \ddot{x_i} = (d-2) \sum_{j \neq i} \frac{ x_j - x_i}{|x_j - x_i|^d } $$

This gives

$$ x_i \cdot \ddot{x_i} = (d-2) \sum_{j \neq i} \frac{ x_j \cdot x_i - |x_i|^2 }{|x_j - x_i|^d} $$

summing over all particles we get

$$ \sum_{i} x_i \cdot \ddot{x_i} = (d-2) \sum_{i,j, i\neq j} \frac{ x_j \cdot x_i - |x_i|^2 }{|x_j - x_i|^d} = -\frac{d-2}{2} \sum_{i,j, i\neq j} |x_i - x_j|^{2-d} \tag{1}$$

Conservation of energy, on the other hand, states that

$$ - \sum_{i,j, i\neq j} |x_i - x_j|^{2-d} + \sum_i |\dot{x_i}|^2 = C < 0 \tag{2}$$

(by negative energy assumption). Summing (1) and (2) and using that $d \geq 4$ so that $(d-2)/2 \geq 1$ we get

$$ \frac{d^2}{dt^2} \sum_{i} |x_i|^2 < 0 $$

Which shows that $\sum |x_i|^2$ must impact $0$ at some finite time (either future or past).

(Note: by "dropping particles at infinity" and using time-symmetric property of Newtonian mechanics we see that there exists trajectory where there is only one time of impact in the past, and none in the future.)

Answer 2

Following the lines Willie followed, but allowing for unequal masses $m_i$, set $I(x) = \langle x, x \rangle $ where $\langle v, w \rangle = \Sigma m_i v_i \cdot w_i$ is the so-called mass metric built so that the kinetic energy is $K =\frac{1}{2}\langle v, v \rangle$ with $v = \dot x$. Here $x$ and $v$ lie in N copies of a d-dimensional Euclidean space. Then Newton's equations read $\ddot x = - \nabla V (x)$ where the gradient $\nabla$ is with respect to the mass metric. Suppose now that the potential $V$ is homogeneous of degree $-\alpha$. Using Euler's identity $-\alpha V (x) = \langle x, \nabla V (x) \rangle$ a fun elementary computation yields the virial identity $\ddot I = 4H +(2 \alpha - 4) V$, where $H = K +V$ is the total energy, which is conserved. (This identity is called by mathematical celestial mechanicians the Lagrange-Jacobi identity.) Now for $d$-dimensional ``Newtonian' gravity we have $V < 0$ and $\alpha = d-2$ so that Lagrange-Jacobi reads $\ddot I = 4H +(8-4d)U$, with $U = -V > 0$. Conclusion: for $d > 4$ and $H \le 0$ we have $\ddot I < 0$ and thus all such solutions either begin and end in total collision $I =0$, (or have singularities before these total collision times), as you conjectured. (Asides. If $d< 4$ then the virial identity implies that $\ddot I > 0$ for $H \ge 0$ and consequently bounded solutions have $H < 0$. The case $d =4$ is a wonderful bounding case between these extremes. At the critical dimension $4$ solutions must lie on spheres $I = const.$ and have energy $H =0$ in order to be bounded.)