Newton's Laws of Motion

Tue, 25 Feb 2025 00:00:00 +0000

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.

Description of Multiple-Particle Systems

Tue, 25 Feb 2025 00:00:00 +0000

Till now the bulk of our discussion after the basic Newtonian laws was in terms of a single particle system. Those are the simplest and often the most boring. The real complexity and interesting phenomena relate to many-particle systems. How do we understand such systems?

If the dimensionality of the space is $d$ and the number of particles are $N$, then the multiparticle position vector can be thought of as a $dN$-dimensional vector. If $\vec{r}_j$ is the position of the $j$th particle (for $j=1,2,\ldots,N$), then the total configuration can be obtained by concatenating each of the position vectors together. The $c$th coordinate for the $j$th particle becomes the $d(j-1) + c + 1$th element of the configuration space vector: $$\vec{q}_{d(j-1)+c+1} = \vec{r}_{j, c}.$$

The velocity vector is the time-derivative of the configuration point, and the momentum vector is obtainedby multiplying by the diagonal mass matrix: $$\vec{p} = M\frac{d\vec{q}}{dt}$$

Definition of Phase-Space

When we come to the statistical mechanics part of the course, the molecular basis of thermodynamics will be developed in terms of energy as a function of the momenta and positions.

This multidimensional space obtained by a combination of the positions $\vec{q}$ and the momenta $\vec{p}$ is called the phase space. Every problem is, in addition to the Hamiltonian, specified by an initial position, $\vec{q}(0)$, and an initial momentum, $\vec{p}(0)$. Newton’s second law can be rewritten as follows: $$\frac{d\vec{q}(t)}{dt} = M^{-1}\vec{p}(t)$$ $$\frac{d\vec{p}(t)}{dt} = -\nabla V(\vec{q}(t)) = \vec{F}(\vec{q}(t))$$ It is well-known that a $o$th order differential equation solved with the initial values of the first $o$ derivatives is equivalent to $o$ coupled first-order differential equations. This is just the same decomposition of the second-order Newton’s law into two first-order differential equations.

The dynamical trajectories of a multiparticle system can be interpreted as a single trajectory in the phase-space. Because of conservation of energy, all points on this phase-space trajectory would have the same energy.

Classical Mechanics: The Basic Laws of the Universe | Bose Research Group

Newton's Laws of Motion

Work Done by External Force

Conservation of Energy

Description of Multiple-Particle Systems

Definition of Phase-Space