A short review of Classical Mechanics

Let's roll a ball down the slope. How to describe its motion? Classical Mechanics is a machinery that allows us to answer all these question. Let's reformulate the question from a marble to an electron. Now let's assume that the electron is rolling down some sort of an atomic hill.

What goes in that motion? What are the principles that allow us to tell how the electron would behave? Quantum mechanics is the foundation that allows us to answer all these questions.  To understand two model systems, a particle in a box and a harmonic oscillator model you should learn how to develop a Lagrangian formulation for classical mechanics and apply quantum mechanics to that.

Figure 1 – portrait of Sir Issac Newton

Motion of macroscopic objects is dictated by realm of classical mechanics which is governed by Newtonian mechanics since it’s based on Newton’s Laws of Motion. There are three main laws that govern Newtonian mechanics. The first law tells us that when viewed from an inertial reference frame, an object at rest tends to stay at rest, and an object in uniform motion tends to stay in uniform motion unless acted upon by a net external force. The Newton’s second law tells us that when a net force f is applied on an object it is equal the time rate of change of momentum leading to the famous equation.

$$F=ma$$

The third law tells us that every action corresponds to equal and opposite reaction.

One of mathematical formulations that is very useful is Lagrangian formulation. This formalism is based on the principle of stationary action. The Lagrange L of a particle is defined as the difference between its kinetic energy T and the potential energy V. Now, in generalized coordinates, the energy is given by.

$$L=T-V=\frac{m}{2}{{\dot{q}}^{2}}-V\left( q,t \right)$$

In the previous equation q stands for the generalized coordinate of space. Now, from this, we can define quantity called the action S. This quantity is defined as the integral of the Lagrangian between two instances of time that is t1 and t2.

$$S=\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{L\left( \dot{q},q,t \right)dt}$$

The principle of least action states that the classical part take by the system, between times t1 and t2, is the one for which the action is stationary. That is the change is zero. In other words we’re attempting to take the path of least action. Mathematically for the small change δq this principle states that differential change in action is zero.

Remember the end points are fixed at q1 and q2 therefore, the perturbation has the condition that δq1 and δq2 are both zero.

Figure 2 – Motion of a particle from q1 at time t1 to q2, at time t2 in the external potential V(x,t).

As seen in previous figure among the several possible paths in q and t that the particle can traverse, the one denoted in red is classical path that the particle chooses to traverse along, the other path curves are not taken by the particle.

$$\delta S=S\left[ \bar{q}+\delta q \right]-S\left[ {\bar{q}} \right]=0$$

If we expand the Lagrangian expression in Taylor series that leads to the emergence of two new terms.

$$\begin{align} & S\left[ \bar{q}+\delta q \right]=\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{L\left( \dot{q}+\delta \dot{q},q+\delta q,t \right)dt} \\ & =\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{\left( L\left( \dot{q},q,t \right)+\delta \dot{q}\frac{\partial L}{\partial \dot{q}}+\delta q\frac{\partial L}{\partial q} \right)}dt \\ & =S\left[ q \right]+\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{\left( \delta \dot{q}\frac{\partial L}{\partial \dot{q}}+\delta q\frac{\partial L}{\partial q} \right)dt} \\ & \frac{d}{dt}\frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q}=0 \\ \end{align}$$

Now we will use two examples to bring out the distinct difference between classical mechanics and quantum mechanics. The first example will be a free particle and the second one will be a particle that is in a harmonic potential well. Now let’s look at the first example of a free particle. Now let’s consider that the free particle is allowed to move in one dimension that is along the x axis. Now this expedience is no external potential. So we can write:

$$\begin{align} & V(x,t)=0 \\ & m\frac{{{d}^{2}}x}{d{{t}^{2}}}=0 \\ & \int{{{d}^{2}}x}={{C}_{1}} \\ & \dot{x}={{C}_{1}} \\ & x\left( t \right)={{C}_{1}}t+{{C}_{2}} \\ \end{align}$$

To determine the constants we need initial conditions. Some information about starting time of the experiment. Let’s say that at time t = 0 the particle is at reference position x = 0 and let’s assume that the particle is moving with velocity v (dx/dt = v).

$$\begin{align} & \dot{x}\left( t \right)={{C}_{1}} \\ & x\left( t \right)={{C}_{1}}t+{{C}_{2}} \\ & \text{At t = 0 }\Rightarrow \text{x=0}\Rightarrow \text{\dot{x}=v} \\ & {{C}_{1}}=v \\ & 0={{C}_{1}}\cdot 0+{{C}_{2}}\Rightarrow {{C}_{2}}=0 \\ & x\left( t \right)=vt \\ \end{align}$$

The resulting equation of motion tells us that we can determine the position of the body at any given time.

Now let’s assume the following example. Let’s assume there is a potential, and the potential of l is harmonic potential field. Now, a harmonic potential field is one where the external potential goes as the square of the displacement. Imagine a spring and try to stretch it the energy from its position is well described by a harmonic potential well. Once again, we can derive the equation of motion from the Lagrangian equation. Now here it turns out that the solution of the equation of motion consists of two sinusoids with a phase difference of pi over two.

$$\begin{align} & V\left( x,t \right)=\frac{k}{2}{{x}^{2}} \\ & m\frac{{{d}^{2}}x}{d{{t}^{2}}}+kx=0\to \frac{{{d}^{2}}x}{d{{t}^{2}}}+{{\nu }^{2}}x=0 \\ & x\left( t \right)=A\sin \left( \nu t \right)+B\cos \left( \nu t \right) \\ & t=0\Rightarrow x=0\Rightarrow \dot{x}={{v}_{0}} \\ & 0=B \\ & \dot{x}\left( t \right)=A\nu \cos \left( \nu t \right) \\ & {{v}_{0}}=A\nu \Rightarrow A=\frac{{{v}_{0}}}{\nu } \\ \end{align}$$

Hamiltonian formulation of classical mechanics describes the equation of motion, using a different quantity, H, called the Hamiltonian. Now the Hamiltonian is defined as the sum of the kinetic and the potential energy.

$$H=T+V$$

For the second example that we studied the Hamiltonian of the particle, it turns out that it would be independent of time.

$$\begin{align} & H=\frac{m}{2}{{{\dot{q}}}^{2}}+V\left( q,t \right) \\ & H=\frac{1}{2}mv_{0}^{2}{{\cos }^{2}}\left( vt \right)+\frac{v_{0}^{2}}{2{{v}^{2}}}{{\sin }^{2}}\left( vt \right)=\frac{1}{2}mv_{0}^{2} \\ \end{align}$$

From these two example, we can infer some key conclusions about classical mechanics:

• Classical mechanics predicts that the particle motion is completely deterministic.
• This is, the conditions of a particle at any given instant of time will chart out its future trajectory.
• The Lagrangian formulation teaches us that the particle traverses along a path such that its action S is an extremum that is minimum.
• A particle that is free from influence of any external potential will maintain a constant velocity, as proposed by Newton’s first law.
• Motion of a particle in time independent potential field will be governed by constraint of total energy.