Skip to main content
fixed typo; thanks @Adam B
Source Link
Willie Wong
  • 39k
  • 4
  • 94
  • 176

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

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

Conservation of energy, on the other hand, states that

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

(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.)

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

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

Conservation of energy, on the other hand, states that

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

(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.)

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.)

Source Link
Willie Wong
  • 39k
  • 4
  • 94
  • 176

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

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

Conservation of energy, on the other hand, states that

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

(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.)