To develop the Newton’s formulation of classical equation, first we need to discuss the essential description of a physical system. At an elementary level, one can think of a macro-system as a collection of $N$ point particles, the $j$th one located at $\vec{r}_j(t)$ at time $t$, with a velocity given by $$\vec{v}_j = \frac{d\vec{r}_j}{dt}.$$ The momentum of the $j$th particle is defined as $\vec{p}_j = m_j\vec{v}_j$.
First, we list the three laws of classical mechanics formulated by Newton:
- In absence of external forces, a body would either be at rest or execute motion in a straight line with a constant velocity $\vec{v}$.
- The action of an external force on a body is to induce a change in momentum, $\vec{p}$. More precisely, $\vec{F}_\text{ext} = \frac{\text{d}\vec{p}}{\text{d}t}$.
- The force exerted by body $B$ on body $A$ is equal in magnitude and opposite in direction to the force exerted by body $A$ on body $B$.
Note, that of the three Newtonian laws, only the second law has all the physics. The other two laws can be seen to be applications of this law.
Can you derive Newton’s first and third laws starting from the second law?
Before moving to a further discussion of multiple particles, let us discuss a few properties at the one particle level. These properties would also go through for multiple particles.
Work Done by External Force
Work done by an external force which moves a particle from point 1 to 2 along a path $\vec{s}$ is given by $$W = \int_1^2 \vec{F}\cdot d\vec{s}$$ $$=m\int_{t_1}^{t_2} \frac{d\vec{v}}{dt}\cdot \vec{v}\,dt$$ $$=m\int_{t_1}^{t_2} \frac{d}{dt}\left(\frac{\vec{v}\cdot\vec{v}}{2}\right)\,dt$$ $$=\frac{1}{2}m\vec{v}(t_2)\cdot\vec{v}(t_2) - \frac{1}{2}m\vec{v}(t_1)\cdot\vec{v}(t_1)$$
This quantity is called the kinetic energy, $T=\frac{1}{2}m\vec{v}\cdot\vec{v}$.
This also tells us that the work done along a closed path is 0. By elementary vector calculus, the force is the gradient of a scalar function. This scalar function is called the potential, $V\left(\vec{r}\right)$, and $\vec{F} = -\vec{\nabla} V\left(\vec{r}\right)$.
Conservation of Energy
Consider the quantity $H(\vec{r}, \vec{p}) = T(\vec{p}) + V(\vec{r})$, which we shall call the total energy. We derive the time-evolution of this quantity, $$\frac{dH}{dt} = \nabla_{\vec{p}} T(\vec{p})\cdot \frac{d\vec{p}_j}{dt}+ \nabla_{\vec{r}} V(\vec{r})\cdot\frac{d\vec{r}}{dt}$$ $$= \frac{\vec{p}}{m}\cdot \frac{d\vec{p}}{dt}+ \nabla_{\vec{r}} V(\vec{r})\cdot\frac{d\vec{r}}{dt}$$ $$= \vec{v}\cdot \vec{F} - \vec{F}\cdot\frac{d\vec{r}}{dt}$$ $$= 0$$
Thus, we see that the total energy is a constant of motion.