Newton’s formulation

Conservative forces

We have $\vec{F} = m \vec{a}$, where we assume $m$ is constant.

It can be shown that, if a force, $\vec{F} = \vec{F}(x, y, z)$, on a particle is conservative, then there exists a function, $U = U(x, y, z)$, such that $$\vec{F} = - \nabla U$$

In component form, $$\begin{matrix}F_x(x,y,z) & = & -\frac{\partial U}{\partial x}\\F_y(x,y,z) & = & -\frac{\partial U}{\partial y}\\F_z(x,y,z) & = & -\frac{\partial U}{\partial z}\end{matrix}$$

The function $U$ is called the potential energy associated with the force $F$.

Conservation of energy

The kinetic energy, $T$, of the particle is related to the velocity $\vec{v} = [\dot{x}, \dot{y}, \dot{z}]$ in the normal way, $$T= \tfrac{1}{2} m v^2$$ where $v^2 = \dot{x}^2 + \dot{y}^2 + \dot{z}^2$.

The total mechanical energy of the system is then \begin{align*}E &= T + U\\ &= \tfrac{1}{2} m v^2 + U(x, y, z)\end{align*}

We can consider the rate of change of the total energy with respect to time, \begin{align*}\frac{\mathrm{d}E}{\mathrm{d}t} &= m \frac{\mathrm{d}}{\mathrm{d}t}(\vec{v} \cdot \vec{v}) + \frac{\mathrm{d}}{\mathrm{d}t} U(x, y, z)\\ &= m \vec{a} \cdot \vec{v} + \nabla U \cdot \vec{v}\\ &= \left( m \vec{a} + \nabla U \right) \cdot \vec{v}\\ &= \left( m \vec{a} - \vec{F} \right) \cdot \vec{v}\\ &= \left( 0 \right) \cdot \vec{v}\\ &= 0\end{align*}

Thus, the total mechanical energy is conserved.

Conservation of momentum

The definition of linear momentum of a particle, $\vec{p}$, is mass times velocity, so we can re-write Newton's Second Law as \begin{align*}\vec{F} &= m \vec{a}\\ &= m \frac{\mathrm{d} \vec{v}}{\mathrm{d}t}\\ &= \frac{\mathrm{d}}{\mathrm{d}t}(m \vec{v})\\ &= \frac{\mathrm{d} \vec{p}}{\mathrm{d}t}\end{align*}

Thus, it's clear that linear momentum of a single particle is conserved if and only if the net force on the particle is zero.

Systems of particles

The concepts of potential, kinetic and total mechamical energy can be extended to systems of many particles, using Newton's Third Law - forces between particles are equal in magnitude and opposite in direction.

Thus, if we have a system of $N$ particles, then the internal force on the $i$th particle, $\vec{F_i}$, is the sum of all the forces on that particle by all of the other particles in the system. That is, $$\vec{F_i} = \sum_{i = 1}^{N} \vec{F_{ij}}$$ where $\vec{F_{ij}}$ is the force on the $i$th particle, due to the $j$th particle.

We note that $$\left\{\begin{matrix} \vec{F_{ji}} &=& -\vec{F_{ij}} \\ \vec{F_{ii}} &=& 0 \end{matrix}\right.$$

If there are no external forces acting on the system, then the total force on the system is the sum of all of the internal forces acting on all of the particles \begin{align*} \vec{F} &= \sum_{i = 1}^{N} \vec{F_{i}}\\ &= \sum_{i = 1}^{N} \sum_{j = 1}^{N} \vec{F_{ij}}\\ &= \sum_{i < j} \vec{F_{ij}} + \sum_{i > j} \vec{F_{ij}} \quad \text{ using } \vec{F_{ii}} = 0\\ &= \sum_{i < j} \vec{F_{ij}} - \sum_{i > j} \vec{F_{ji}} \quad \text{ using } \vec{F_{ji}} = -\vec{F_{ij}}\\ &= \sum_{i < j} \vec{F_{ij}} - \sum_{j > i} \vec{F_{ij}} \quad \text{ re-labelling the second sum }\\ &= \sum_{i < j} \left(\vec{F_{ij}} - \vec{F_{ij}} \right)\\ &= 0\end{align*}

Since the force is just the rate of change of linear momentum, this just states that, where there is no external force, $$\frac{\mathrm{d} \vec{p}}{\mathrm{d}t} = \vec{F} = 0$$ meaning the momentum is conserved.

We'll postpone the generalisation of conservation of energy to many particles to later on in this course, where we derive it as a consequence of time symmetry.