Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Given a collection of point-particles, interacting through an attractive force $\sim \frac{1}{r^2}$.

Knowing only $m_1a=\sum_i \frac{Gm_1m_i}{r^2}$ and initial conditions we can deduce the motion of the system.

Consequently we can observe that three quantities remains constant A) center of mass of the system B) total energy C) angular momentum

How can we derive these 3 facts directly from $m_1a=\frac{Gm_1m_2}{r^2}$ ?

Are these quantities conserved for any attractive force $\sim\frac{1}{r^n}$ ?

Given any monotonically decreasing force for $r$ in $(0,\infty)$, which are the conserved quantities?

share|improve this question
I think for B you mean the net momentum of the system. Unless the initial center of mass is stationary, the center of mass will move at a constant velocity. –  Omega Centauri Feb 2 '11 at 19:26
By the center of mass, do you mean $\sum_i \left[ m_i \vec{x}_i - m_i \dot{\vec{x}}_i t \right] / \sum_i m_i$? –  QGR Feb 2 '11 at 19:37
Seems like homework. Is this homework? –  Rafael S. Calsaverini Feb 2 '11 at 23:00
@all Is there an analytical solution for the n-body problem with forces other then 1/n^2 ? –  user1708 Feb 3 '11 at 9:35
add comment

5 Answers

up vote 7 down vote accepted

Actually, this is the basic stuff every mechanics textbook should have.

  1. Center of mass is the most basic it needs just Newton laws:
    Second: $m_i\ddot{\vec{r}_i}= \vec{F}_i$
    And third: $\sum_i\vec{F}_i = 0$
    Summing over i one obtains:
    $\frac{d^2}{dt^2}\left(\sum_i m_ir_i\right) = \sum_i\vec{F}_i = 0$

  2. For the total energy you need those forces to be potential and independent on time:
    $F_i = -\frac{d}{d\vec{r}_i}U(\vec{r}_1,\vec{r}_2,...)$
    Then just take the total energy : $E = \sum_i \frac{m|\vec{r}_i|^2}{2}+U$
    And differentiate with respect to time:
    $\frac{dE}{dt} = \sum_i m_i(\ddot{\vec{r}}_i\dot{\vec{r}}_i)+\sum_{i} \frac{dU}{d\vec{ r}_i}\dot{\vec{r}}_i = 0$

  3. Finally, for the angular momentum $U$ must be rotational invariant:
    $U(R\vec{r}_1,R\vec{r}_2,R\vec{r}_3,...) = U(\vec{r}_1,\vec{r}_2,\vec{r}_3,...)$
    where R is the rotation matrix. Consider now very small (infinitesimal) rotation:
    $R\vec{r}_i = \vec{r}_i+\left[\delta\vec\phi\times\dot{\vec{r}}_i\right]$
    Substituting and expanding, one can get:
    Which works for every angle $\delta\phi$, so the following must hold $\sum_i\left[\frac{dU}{d\vec{r}_i}\times\vec{r}_i\right]=0$  
    Finally take the angular momentum:
    $\vec{M} = \sum_i m_i \left[\vec{r}_i\times\dot{\vec{r}}_i\right]$
    And differentiate with respect to time:
    $\frac{d\vec{M}}{dt} = \sum_i \left[\vec{r}_i\times m_i\ddot{\vec{r}}_i\right]+ \sum_i m_i \left[\dot{\vec{r}}_i\times \dot{\vec{r}}_i\right] = \sum_i\left[\frac{dU}{d\vec{r}_i}\times\vec{r}_i\right]=0$

Phew... You can see that I never used the "pairwise" interactions. And I only needed the basic assumptions about the potential energy.

Concerning "extra" things that you cans say for your systems -- check the Virial theorem

share|improve this answer
add comment

These quantities will be conserved by any system whose laws respect translational and rotational symmetry and are static in time. Respecting translational symmetry means that interactions don't depend on the positions of individual particles, just on differences of those positions. Respecting rotational invariance in turn means that interactions depend only on the magnitude of those differences. Being static means that the interactions don't depend explicitly on time or velocities. So this means that the two-point interaction must look like $V({\mathbf x}, {\mathbf y}) = f(|{\mathbf x - \mathbf y}|)$ (with $f$ suitably smooth). So yes, in particular your $r^{-n}$ potential will work.

Note that in general you could also consider more general interactions e.g. three- or more-point interactions. But to conserve the said quantities the interactions must still be singlet states of the respective groups.

share|improve this answer
add comment

The total momentum is also conserved.

Noether's theorem relates conserved quantities to continuous symmetries of the system.

For the inverse square law, there's also a less known symmetry $x \to e^\lambda x$, $t \to e^{2\lambda} t$.

But we shouldn't expect any more because of chaos; this isn't an integrable system.

share|improve this answer
add comment

All codes of the N-body simulation make use of instantaneous propagation of the speed of gravity. The other answers follow the same criterion that is good enough for local simulations.

As a consequence of the finite speed of propagation of gravity and the motion of the masses there is a dissipative component.

As the bodies are never static, the masses continuously lose energy to space because there is no known mechanism that reverses the direction of flow.

(An equivalent effect applies to the radiation of charged particles by virtue of its continued accelerated motion.)

from Jorgen Kalckar and Ole Ulfbeck experiment (a) (energy of two masses and a spring): ...

The difference between these two energies is lost by each mass: it is taken away by space-time, in other words, it is radiated away as gravitational radiation

gravity can behave like a wave: gravity can radiate ... All this follows from the expression of universal gravity when applied to moving observers, with the requirement that neither observers nor energy can move faster than c.

(a) motionmountain free online book, ch 18, pag 526/527: Motion in General Relativity, Gravitational Waves.

share|improve this answer
The usual theory is not aplicable in the case of space expansion , i.e. those quantities are not conserved. I offered 2 distinct ways that show that 'Energy Conservation Laws', in the context of n-body system, have limits of applicability. Please, be kind and leave a notice on the why(s) of your possible disagreement. –  Helder Velez Mar 4 '11 at 18:59
add comment

There are many more conserved quantities in a potential N-body problem but not all of them are additive on particles. This fact is not related to the form of potentials at all.

share|improve this answer
Can you name one ? –  user1708 Feb 2 '11 at 15:54
N ordinary differential equations of the second order have as many integrals of motion as the initial conditions or so. Non additive integrals of motion do not have special names, contrary to the additive ones. –  Vladimir Kalitvianski Feb 2 '11 at 16:04
Locally, they do have many integrals of motion, but globally, this isn't true in general because of chaos. –  QGR Feb 2 '11 at 19:15
See a simple example in my question post. –  Vladimir Kalitvianski Feb 11 '11 at 21:21
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.