Advanced Classical Mechanics/Many-Particle Systems

Here we will look a group of particles with forces that exert forces among themselves and have external forces exerted upon them as well. If we focus on particle $$i$$, we have

$$ \dot{\vec {p}}_i = \sum_j \vec{F}_{ji} + F_i^{(e)} $$

where the first term is the force that particle $$j$$ exerts on particle $$i$$ summed over the other particles. The second term denotes the external force on particle $$i$$.

Weak law of action and reaction -- total momentum
Let's assume the weak law of action and reaction which states that $$\vec{F}_{ij}=-\vec{F}_{ji}$$ and sum the change in the momentum of all the particles

$$ \sum_i \dot{\vec {p}}_i = \frac{d^2}{dt^2} \sum_i m_i \vec{r}_i = \sum_i F_i^{(e)} + \sum_{i,j} F_{ji} $$

The second term vanishes because each force between each pair of particles cancels out. If we define the position of the center of mass of the system as

$$ \vec{R} = \frac{\sum_i m_i \vec{r}_i}{\sum_i m_i} = \frac{1}{M} \sum_i m_i \vec{r}_i $$

we have

$$ M \frac{d^2 \vec{R}}{dt^2} = \sum_i \vec{F}_i^{(e)} = \vec{F}^{(e)}. $$

The total momentum is

$$ \vec{P} = \sum_i m_i \frac{d \vec{r}_i}{dt} = M \frac{d \vec{R}}{dt}, $$

so if $$\vec{F}^{(e)}=0$$ then $$\vec{P}$$ is conserved.

Strong law of action and reaction -- total angular momentum
The total angular momentum of the system is given by

$$ \vec{L} = \sum_i \vec{r}_i \times \vec{p}_i $$

and its rate of change is

$$ \dot{\vec{L}} = \sum_i \left (\dot{\vec{r}}_i \times \vec{p}_i + \vec{r}_i \times \dot{\vec{p}}_i \right ) = \sum_i \vec{r}_i \times \dot{\vec{p}}_i $$

where the first term vanishes because the linear momentum and velocity point in the same direction.

Substituting in the forces yields

$$ \dot{\vec{L}} = \sum_i \vec{r}_i \times \vec{F}_i^{(e)} + \sum_{i,j} \vec{r}_i \times \vec{F}_{ji}. $$

If we look at the final term more closely we have

$$ \sum_{i,j} \vec{r}_i \times \vec{F}_{ji} = \sum_{i<j} \vec{r}_i \times \vec{F}_{ji} + \vec{r}_j \times \vec{F}_{ij} = \sum_{i<j} \left ( \vec{r}_i - \vec{r}_j \right ) \times \vec{F}_{ji}, $$

so if we assume that the force between the particles points in the same direction as the displacement between the particles (strong law of action and reaction), the forces between the particles do not affect the total angular momentum of the system. This type of force is called a central force. In this case we have $$\sum_{i,j} \vec{r}_i \times \vec{F}_{ji} =0$$ so

$$ \dot{\vec{L}} = \sum_i \vec{r}_i \times \vec{F}_i^{(e)} = \vec{N}^{(e)}. $$

A central force satisfies the strong law of action and reaction (not necessarily the weak law). Gravity and electric forces are central and also satisfy the weak law of action and reaction but magnetic forces generally satisfy neither.

Let's take a closer look at the total angular momentum of the system of particles by defining the displacement of a particle relative to the center of mass.

$$ \vec{r}_i = \vec{s}_i + \vec{R} ~\rm{and}~ \vec{v}_i = \vec{u}_i + \vec{V} ~\rm{where}~ \vec{V} = \frac{d \vec{R}}{dt}. $$

The total angular momentum is given by

$$ \vec{L} = \sum_i m_i \vec{s_i} \times \vec{u}_i + \sum_i m_i \vec{s}_i \times \vec{V} + \sum_i m_i \vec{R} \times \vec{v}_i + \sum_i m_i \vec{R} \times \vec{V}. $$

The two middle terms vanish because

$$ \sum_i m_i \vec{s}_i = \sum_i m_i \left ( \vec{r}_i - \vec{R} \right ) = M \left (\vec{R} - \vec{R}\right ) = 0 $$

from the definition of the center of mass. We are left with

$$ \vec{L} = \sum_i m_i \vec{s_i} \times \vec{u}_i + M \vec{R} \times \vec{V}. $$

That is the angular momentum of a system is the angular momentum of the center of mass plus the angular momentum about the center of mass.

Uniform Gravitational Field
A familar possibility for the external forces and torques is a uniform gravitational field. In this case we have

$$ \vec N^{(e)} = \sum_i m_i \vec r_i \times \vec g = M \vec R \times \vec g. $$

Let's calculate the change in the velocity of the center of mass

$$ \dot \vec V = \frac{1}{M} \sum_i m_i \dot \vec v_i = \frac{1}{M} \sum_i m_i \vec g = \vec g $$

and the change in the angular momentum

$$ \dot \vec L = M \vec R \times \vec g = \frac{d}{dt}\left ( \sum_i m_i \vec{s_i} \times \vec{u}_i \right ) + M \left( \vec V \times \vec V + \vec R \times \dot \vec V \right ) = \frac{d}{dt}\left ( \sum_i m_i \vec{s_i} \times \vec{u}_i \right )  + M \vec R \times \vec g $$

so

$$ \frac{d}{dt}\left ( \sum_i m_i \vec{s_i} \times \vec{u}_i \right )  = 0. $$

In an uniform gravitational field, the internal angular momentum of a body is conserved. Think about footballs and frisbees!

Work and Kinetic Energy
Let's examine the total work performed on the system of particles. We have

$$ W_{12} = \sum_i \int_1^2 \vec{F}_i \cdot d \vec{s}_i = \sum_i \int_1^2 \vec{F}^{(e)}_i \cdot d \vec{s}_i + \sum_{i,j} \int_1^2 \vec{F}_{ji} \cdot d \vec{s}_i. $$

We also have

$$ W_{12} = \sum_i \int_1^2 \vec{F}_i \cdot d \vec{s}_i = \sum_i \int_1^2 m_i \dot{\vec{v}}_i \cdot \vec{v}_i dt = \sum_i \int_1^2 d \left ( \frac{1}{2} m_i v_i^2 \right ) $$

so

$$ W_{12} = T_2 - T_1 ~{\rm where}~ T=\frac{1}{2} \sum_i m_i v_i^2. $$

We can also look at the energy using the center of mass to get

$$ T = \frac{1}{2} \sum_i m_i \left ( \vec{u}_i + \vec{V} \right )^2 = \frac{1}{2} \sum m_i \vec{u}_i^2 + \sum m_i u_i \cdot \vec{V} + \frac{1}{2} M V^2. $$

The middle term vanishes for the reasons outlined earlier.

Potential Energy
First let's look at the external forces

$$ \sum_i \int_1^2 \vec{F}^{(e)}_i \cdot d \vec{s}_i = - \sum_i \int_1^2 \vec{\nabla}_i V_i \cdot d \vec{s}_i = - \sum_i \left .V_i \right |_1^2 $$

where $$ \nabla_i \equiv \frac{d}{d \vec{r}_i} $$ if the external forces are conservative. N.B.: in this section, $$V$$ denotes potential energy not the velocity of the center of mass.

To look at the interparticle forces, let's assume that they too are conservative and that the potential only depends on the distance between the particles so

$$ V_{ij} = V_{ij} \left ( \left | \vec{r}_i - \vec r_j \right | \right ). $$

Such a potential has two nice properties. First,

$$ \vec{F}_{ji} = -\vec{\nabla}_i V_{ij} = \vec{\nabla}_j V_{ij} = - \vec{F}_{ij} $$

so it satisfies the weak law of action-reaction. Second we have

$$ \vec{\nabla}_i V_{ij} \left ( \left | \vec r_i - \vec r_j \right | \right ) = \left ( \vec r_i - \vec r_j  \right ) f \left ( \left | \vec r_i - \vec r_j \right | \right ) $$

so it is central and satifies the strong law of action-reaction. If we use this particular type of force we have

$$ \sum_{i,j} \int_1^2 \vec F_{ji} \cdot d \vec r_i = - \sum_{i<j} \int_1^2 \left ( \vec \nabla_i V_{ij} \cdot d \vec r_i + \nabla_j V_{ij} \cdot d \vec r_j \right ) $$

Let's define $$\vec r_{ij} = \vec r_i - \vec r_j$$. We have

$$ \frac{dV}{d \vec r_{ij}} = \frac{d V}{d \vec r_i} = \nabla_i V = -\nabla_j V $$

and

$$d \vec r_{ij} = d \left ( \vec r_i - \vec r_j \right ) = d \vec r_i - d \vec r_j.$$

If we substitute these results into the expression for the work we have

$$ - \sum_{i<j} \int_1^2 \left ( \vec \nabla_i V_{ij} \cdot d \vec r_i + \nabla_j V_{ij} \cdot  d \vec r_j \right ) = -\sum_{i<j} \int_1^2 \left ( \vec \nabla_{ij} V_{ij} \cdot d r_{ij} \right ) = - \left. \sum_{i<j} V_{ij} \right |_1^2 = - \frac{1}{2} \left. \sum_{i\neq j} V_{ij} \right |_1^2 $$

so we can definte the total potential energy to be the sum of the potential energies of each particle due to the external force plus the sum of the potential energies of each pair of particles:

$$ V = \sum_i V_i + \frac{1}{2} \sum_{i\neq j} V_{ij}. $$

The Virial Theorem
Let's look at particular and common form of central force. In particular, let's assume that

$$ V_{ij}(\vec r_i, \vec r_j) = k \left | \vec r_{ij} \right |^n. $$

and that there is no external potential. This means than

$$ \nabla_i V_{ij} = n V_{ij} \frac{\vec r_{ij}}{\left | \vec r_{ij} \right|^2}. $$

We can define

$$ K = \sum_i \frac{1}{2} m_i \vec r_i^2 $$

as a measure of the size of the system. Let's calculate the second derivative of $$K$$ with respect to time

$$ \frac{d^2 K}{d t^2} = \sum_i \left ( m_i \dot \vec r_i^2 + m_i \vec r_i \cdot \ddot \vec r_i  \right ) = 2 T - \sum_i \vec r_i \cdot\sum_j \nabla_i V_{ij}. $$

Again let's group the particles in the final summation in pairs

$$ \frac{d^2 K}{d t^2} = 2 T - \sum_{i<j} \left ( \vec r_i - \vec r_j \right ) \nabla_i V_{ij} = 2 T - \sum_{i<j} \left ( \vec r_i - \vec r_j \right ) n V_{ij} \frac{\vec r_{i} - \vec r_j}{\left | \vec r_i - \vec r_j \right|^2} = 2 T - n V. $$

If we have a bunch of particles in equilibrium so they are not expanded or contracting on average than the left-hand side of the equation above vanishes. The most familar example of this is the force of gravity with $$n=-1$$, so for a self-gravitating system in equilibrium we have $$2 T + V=0$$. This relationship is used to weigh distant galaxies and clusters of galaxies.

If $$n<0$$ the potential energy of the system vanishes as the particles get further apart, so the total energy of the system $$E=T+V$$ is the negative of the amount of energy needed to break up the system and leave the particles stationary. Let's calculate the total energy of a system in equilibrium in terms of its kinetic energy. We have

$$ E = T + V = \left ( 1 + \frac{2}{n} \right ) T $$

We find that only if $$-2<n<0$$ is the energy of the system negative. For $$n=-2$$ the total energy of the equilibrium system is zero, so the equilibrium is marginally stable; even a tiny bit of additional energy is enough to break it apart. For $$n<-2$$ the equilibrium is unstable. The grouping of particles has more energy than if they separate.

Thankfully the force of gravity has $$n=-1$$. If $$n$$ were less than -2, there would be no self-gravitating equilibrium systems such as stars, galaxies etc. In fact theories of gravity in higher dimensions find that $$n=2-D$$ where $$D$$ is the number of spatial dimensions. Apparently $$D=3$$ is kind of special.