Mechanics of a System of Particles
- Classical Mechanics 1.1
Mechanics of a Particle
Let $\b{r}$ be the position vector of a particle, then the velocity $\b{v}$ and acceleration $\b{a}$ are given by
\[ \begin{align*} \b{v} &= \odv{\b{r}}{t} \nl \b{a} &= \odv{\b{v}}{t} = \odvn{2}{ \b{r} }{t} \end{align*} \]
The linear momentum $\b{p}$ of a particle with mass $m$ is defined as
\[ \b{p} = m\b{v} \]
The mechanics of the particle is contained in Newton’s second law of motion, which states that there exists frames of reference in which the motion of a particle is described by
\[ \b{F} = \odv{\b{p}}{t} \]
where $\b{F}$ is the net force acting on the particle. Such a frame of reference is called an inertial or Galilean system. If the mass $m$ is constant, we can also write this as
\[ \b{F} = m\odv{\b{v}}{t} = m\b{a} \]
Many of the important conclusions of mechanics can be expressed in the form of conservation theorems, and the first is the conservation of linear momentum of a particle:
If $\b{F} = 0$, then $\b{p}$ is conserved.
The angular momentum $\b{L}$ of a particle with respect to a point $\b{O}$ is defined as
\[ \b{L} = \b{r} \times \b{p} \]
where $\b{r}$ is the position vector of the particle with respect to the point $\b{O}$. The moment of force or torque $\b{N}$ with respect to the point $\b{O}$ is defined as
\[ \b{N} = \b{r} \times \b{F} \]
We can also express the angular momentum in terms of the torque:
\[ \begin{align*} \odv{\b{L}}{t} &= \odv{}{t} (\b{r} \times \b{p}) \nl &= \b{v} \times m\b{v} + \b{r} \times \b{F} \nl &= \b{N} \end{align*} \]
And we similarly have the conservation of angular momentum of a particle:
If $\b{N} = 0$, then $\b{L}$ is conserved.
Next, the work done by a force $\b{F}$ on a particle is defined as
\[ W_{12} = \int_1^2 \b{F} \cdot \dd{\b{s}} \]
If the mass $m$ is constant, we can also write this as
\[ \begin{align*} W_{12} &= \int_1^2 m\odv{\b{v}}{t} \cdot \odv{\b{r}}{t} \dd{t} \nl &= m\int_1^2 \odv{}{t} \left( \frac{v^2}{2} \right) \dd{t} \nl &= \frac{m}{2} \left( v_2^2 - v_1^2 \right) \end{align*} \]
Thus we define the kinetic energy $T$ of a particle as
\[ T = \frac{1}{2}mv^2 \]
and get the work-energy theorem:
\[ W = \Delta T \]
Let’s talk about the conservative forces. A force $\b{F}$ is called conservative if the work done by the force from a fixed point to another fixed point is independent of the path taken. This is equivalent with:
\[ \oint \b{F} \cdot \dd{\b{s}} = 0 \]
And by the Stokes theorem, this is equivalent with $\curl \b{F} = 0$. Thus, by the Poincaré lemma, we can write the force as the gradient of a scalar function $U$:
\[ \b{F} = -\grad U \]
The function $U(\b{r})$ is called the potential energy of the force, and can be calculated by setting the potential energy at a fixed point to zero:
\[ U(\b{r}) = -\int_{\b{r}_0}^{\b{r}} \b{F} \cdot \dd{\b{s}} \]
By the gradient theorem, we can write the work done by a conservative force as
\[ W = -\Delta U \]
Therefore, we can also write the work-energy theorem as
\[ \Delta (T + U) = 0 \]
This means that the sum of the kinetic energy and potential energy is conserved, and we call this the total mechanical energy. However, if a force is conservative but depends explicitly on time, then the total mechanical energy is not conserved; physically, displacement(path integral) is nonzero only when time flows.
Mechanics of a System of Particles
In generalizing the mechanics of a particle to a system of particles, we must distinguish between the external forces acting on the system and the internal forces acting between the particles of the system. The equation of motion for the $i$-th particle is given by:
\[ \dot{\b{p}}_i = \b{F}_i^\text{ext} + \sum_{j \neq i} \b{F}_{ij} \]
where $\b{F}_i^\text{ext}$ is the external force acting on the $i$-th particle, and $\b{F}_{ij}$ is the internal force acting on the $i$-th particle due to the $j$-th particle. Summing over all particles, we get the equation of motion for the system:
\[ \odvn{2}{}{t} \sum_i m_i \b{r}_i = \sum_i \b{F}_i^\text{ext} + \sum_{i\neq j} \b{F}_{ij} \]
By the Newton’s third law of motion, the internal forces satisfy $\b{F}_{ij} = -\b{F}_{ji}$, and thus the sum of the internal forces is zero. Writing the total mass of the system as $M = \sum_i m_i$ and the center of mass $\b{R}$ as
\[ \b{R} = \frac{1}{M} \sum_i m_i \b{r}_i \]
we can rewrite the equation of motion for the system as
\[ M \odvn{2}{}{t} \b{R} = \sum_i \b{F}_i^\text{ext} = \b{F}^\text{ext} \]
This means that the center of mass of the system behaves like a single particle with mass $M$ under the influence of the external forces. The total linear momentum of the system is given by
\[ \b{P} = \sum_i m_i \odv{\b{r}_i}{t} = M \b{V} \]
where $\b{V} = \odv{\b{R}}{t}$ is the velocity of the center of mass. And we have the conservation of linear momentum of a system of particles:
If $\b{F}^\text{ext} = 0$, then $\b{P}$ is conserved.
The total angular momentum of the system with respect to a point $\b{O}$ is given by
\[ \b{L} = \sum_i \b{r}_i \times \b{p}_i \]
Let’s define $\b{r}_{ij} = \b{r}_i - \b{r}_j$ as the position vector of the $i$-th particle with respect to the $j$-th particle.
\[ \begin{align*} \odv{\b{L}}{t} &= \sum_i \odv{}{t} (\b{r}_i \times m_i \odv{\b{r}_i}{t}) \nl &= \sum_i \b{r}_i \times \b{F}_i^\text{ext} + \sum_{i\neq j} \b{r}_i \times \b{F}_{ij} \nl &= \sum_i \b{r}_i \times \b{F}_i^\text{ext} + \sum_{i < j} \b{r}_{ij} \times \b{F}_{ij} \end{align*} \]
If the internal forces between the particles, in addition to being equal and opposite, also lie along the line joining the particles – a condition known as the strong form of Newton’s third law – then the second term vanishes. In this case, we can write the angular momentum of the system as
\[ \odv{\b{L}}{t} = \sum_i \b{r}_i \times \b{F}_i^\text{ext} = \b{N}^\text{ext} \]
And we have the conservation of angular momentum of a system of particles:
If $\b{N}^\text{ext} = 0$, then $\b{L}$ is conserved.
Let’s see the system from the point of view of the center of mass. Let’s define
\[ \b{r}^\prime_i = \b{r}_i - \b{R} \nl \b{v}^\prime_i = \b{v}_i - \b{V} \]
Then we have a good property:
\[ \sum_i m_i \b{r}^\prime_i = 0 \nl \sum_i m_i \b{v}^\prime_i = 0 \]
This means that the center of mass is at the origin in this new coordinate system. Using this, we can rewrite the angular momentum of the system as
\[ \begin{align*} \b{L} &= \sum_i \left( \b{r}^\prime_i + \b{R} \right) \times m_i \left( \b{v}^\prime_i + \b{V} \right) \nl &= \b{R} \times M\b{V} + \sum_i \b{r}^\prime_i \times m_i \b{v}^\prime_i \end{align*} \]
By writing $\b{p}^\prime_i = m_i \b{v}^\prime_i$,
\[ \b{L} = \b{R} \times \b{P} + \sum_i \b{r}^\prime_i \times \b{p}^\prime_i \]
This means that the angular momentum of the system can be decomposed into two parts:
- The angular momentum of the center of mass
- The angular momentum of the particles with respect to the center of mass
This emphasizes that $\b{L}$ depends on the choice of the point $\b{O}$, only through $\b{R}$. Only if $b{R}$ is rest with respect to $\b{O}$, $\b{L}$ will be independent of the point of reference.
The total kinetic energy of the system is given by
\[ \begin{align*} T &= \sum_i T_i = \sum_i \frac{1}{2} m_i \left\vert \b{v}_i^\prime + \b{V} \right\vert^2 \nl &= \sum_i \frac{1}{2} m_i \left( v_i^{\prime 2} + 2\b{v}_i^\prime \cdot \b{V} + V^2 \right) \nl &= \frac{1}{2} M V^2 + \sum_i \frac{1}{2} m_i v_i^{\prime 2} \end{align*} \]
This means that the total kinetic energy of the system can also be decomposed into two parts:
- The kinetic energy of the center of mass
- The kinetic energy of the particles with respect to the center of mass
Now let’s seek for the work done by the forces acting on the system. Assume that every force acting on the system is conservative, then
\[ \begin{align*} W_{12} &= \sum_i \int_1^2 \b{F}_i^\text{ext} \cdot \dd{\b{s}}_i + \sum_{i \neq j} \int_1^2 \b{F}_{ij} \cdot \dd{\b{s}}_i \nl &= -\sum_i \int_1^2 \grad_i U_i \cdot \dd{\b{s}}_i - \sum_{i \neq j} \int_1^2 \grad_i U_{ij} \cdot \dd{\b{s}}_i \end{align*} \]
where $U_i$ is the potential energy of the external force acting on the $i$-th particle, and $U_{ij}$ is the potential energy of the internal force acting between the $i$-th and $j$-th particles. $\grad_i$ means that we take the gradient with respect to the coordinates of the $i$-th particle. By the Newton’s third law, we have $\grad_i U_{ij} = -\grad_j U_{ji}$,
\[ \begin{align*} W_{12} &= -\sum_i \Delta U_i - \sum_{i < j} \int_1^2 \left( \grad_i U_{ij} \cdot \dd{\b{s}}_i + \grad_j U_{ji} \cdot \dd{\b{s}}_j \right) \nl &= -\sum_i \Delta U_i - \sum_{i < j} \int_1^2 \grad_i U_{ij} \cdot \left( \dd{\b{s}}_i - \dd{\b{s}}_j \right) \nl &= -\sum_i \Delta U_i - \sum_{i < j} \int_1^2 \grad_{ij} U_{ij} \cdot \dd{\b{s}}_{ij} \nl &= -\sum_i \Delta U_i - \sum_{i < j} \Delta U_{ij} \end{align*} \]
This means that we can write the whole potential energy of the system as
\[ U = \sum_i U_i + \sum_{i < j} U_{ij} \]
We often set $U_{ij}=U_{ji}$, so that the potential energy is symmetric with respect to the particles.
\[ U = \sum_i U_i + \frac{1}{2} \sum_{i \neq j} U_{ij} \]
Thus, we can write the work done by the forces acting on the system as
\[ W = -\Delta U \]
And also get the conservation of total mechanical energy:
\[ \Delta (T + U) = 0 \]
Let’s look for the internal potential energy more closely. If it depends only on the distance between the particles, i.e. $U_{ij} = U_{ij}(r_{ij})$, then we can write the force as
\[ \b{F}_{ij} = -\grad_i U_{ij}(r_{ij}) = -\pdv{U_{ij}}{r_{ij}} \grad_i r_{ij} = -\pdv{U_{ij}}{r_{ij}} \frac{\b{r}_{ij}}{r_{ij}} \]
This means that the internal forces are central forces, and they satisfy the strong form of Newton’s third law. Internal potential is generally not zero, and it may vary as the system changes with time. However, for rigid bodies, in other words for constant $r_{ij}$, $\dd{\b{s}}_{ij}$ can only be perpendicular to $\b{r}_{ij}$, and so for central forces, it is perpendicular to $\b{F}_{ij}$, which means that the work done by the internal forces is zero, resulting in the constant internal potential energy. Thus, we can completely disregard the internal potential energy of a rigid body system.
$U = \sum_i U_i$ for a rigid body system.
