Introduction to Classical Mechanics

Classical mechanics is the study of the motion of objects under the influence of forces. It deals with the regime where objects are much larger than atoms and move at speeds much smaller than the speed of light \(c\). Below is a table of the different regimes of physics.

	Small	Large
Slow	Quantum Mechanics	Classical Mechanics
Fast	Quantum Field Theory	Special Relativity

Some people treat special relativity as a part of classical mechanics, but we deal with it separately here.

Newton’s Laws of Motion

1. An object will remain at rest or in uniform motion in a straight line unless acted upon by a force.
2. The rate of change of momentum of an object is proportional to the force acting on it.
3. For every action, there is an equal and opposite reaction.

Newton’s first law defines what it means to be in an inertial frame of reference. An inertial frame is one in which Newton’s first law holds true, as opposed to a non-inertial (accelerating) frame. As an example, if a bus suddenly brakes, then in the frame of the bus, passengers are thrown forward even though no force is acting on them, thus violating Newton’s first law. Therefore the bus frame is non-inertial. From the perspective of the inertial frame of the road, it is the bus that slows down (from a braking force), and the passengers continue moving forward, obeying Newton’s first law.

Newton’s second law is given by

\[\mathbf{F} = \frac{\text{d}\mathbf{p}}{\text{d}t} \, ,\]

where the momentum is defined as mass times velocity \(\mathbf{p} = m \mathbf{v}\) (we ignore relativistic effects). If the mass is constant, we can then write \(\mathbf{F} = m \mathbf{a}\), where \(\mathbf{a} = \text{d}\mathbf{v}/\text{d}t\) is the acceleration. Obviously, this law is only useful if we know the forces acting on an object - these come from other laws of physics, such as those of electromagnetism or gravity.

Newton’s third law states that forces always come in pairs. If object A exerts a force on object B, then object B exerts an equal and opposite force on object A. As an example, when you push against a wall, the wall pushes back on you with an equal force. We shall see that this law forms the basis of conservation of momentum.

(Inertial) Frames of Reference

A frame of reference is a coordinate system used to describe the motion of an object. In classical mechanics, we often use inertial frames, which are non-accelerating. For example, consider Alice on a station platform and Bob on a train moving at constant velocity \(\mathbf{v}\).

Alice’s coordinate system \(\mathbf{r}_A = (x_A, y_A, z_A)\) is fixed to the station, while Bob’s coordinate system \(\mathbf{r}_B = (x_B, y_B, z_B)\) is fixed to the train. If the two orgins coincide at time \(t = 0\), then the transformation between the two coordinate systems is given by

\[\mathbf{r}_A = \mathbf{r}_B + \mathbf{v}t \, .\]

Hopefully this is clear from the diagram below.

An object's motion in time is described in Alice's frame of reference by $\mathbf{r}_A(t)$, and in Bob's frame of reference by $\mathbf{r}_B(t)$. If the origins of the two frames coincide at time $t = 0$ and Bob is moving at constant velocity $\mathbf{v}$ with respect to Alice, then what are the transformations for the velocities $\mathbf{v}_A = \dot{\mathbf{r}}_A$ and $\mathbf{v}_B = \dot{\mathbf{r}}_B$ and the accelerations $\mathbf{a}_A = \ddot{\mathbf{r}}_A$ and $\mathbf{a}_B = \ddot{\mathbf{r}}_B$?

We already know that $\mathbf{r}_A = \mathbf{r}_B + \mathbf{v}t$. Differentiating once with respect to time gives the transformation for the velocity $\dot{\mathbf{r}}_A = \dot{\mathbf{r}}_B + \mathbf{v}$. Differentiating again gives the transformation for the acceleration $\ddot{\mathbf{r}}_A = \ddot{\mathbf{r}}_B$. Therefore, Alice measures the same acceleration as Bob, but the observed velocity of the object is modified by the velocity of the train.

The simple addition of velocities should hopefully make sense. If Bob on the train throws a ball at speed \(u < v\) in the direction of the train, then Alice will see the ball moving at speed \(v + u\) in the direction of the train. If Bob throws the ball in the opposite direction, then Alice will see the ball moving at speed \(v - u\) in the direction of the train.

Newton’s first law states that an object has zero acceleration in if no forces act on it. We have shown that the acceleration of an object is the same in all frames moving at constant velocity with respect to each other. Therefore, if Newton’s first law holds in one inertial frame, it holds in all inertial frames, and can be treated as a definition of an inertial frame.

This is part of a more general principle known as the principle of relativity, which states that the laws of physics are the same in all inertial frames. All physics experiments conducted by Bob on the train will give the same results as those conducted by Alice on the platform.

Conservation Laws

Often we are given a force acting on an object and asked to find the motion of the object. This is done by solving Newton’s second law, which is a second-order differential equation. However, if we only care about the initial and final states of the object, we can use conservation laws to simplify the problem.

Conservation of Momentum

The conservation of momentum states that the total momentum of a system is conserved if no external forces act on it.

To see that this is a consequence of Newton’s third law, consider two objects A and B which only interact with each other (no external forces). If the force on A due to B is \(\mathbf{F}_{AB}\), then the force on B due to A is \(\mathbf{F}_{BA} = -\mathbf{F}_{AB}\) (equal and opposite).

We have from Newton’s second law that \(\text{d}\mathbf{p}_A/\text{d}t = \mathbf{F}_{AB}\) and \(\text{d}\mathbf{p}_B/\text{d}t = \mathbf{F}_{BA} = -\mathbf{F}_{AB}\). Adding these equations gives

\[\frac{\text{d} (\mathbf{p}_A + \mathbf{p}_B)}{\text{d}t} = \mathbf{F}_{AB} + \mathbf{F}_{BA} = 0 \, .\]

Therefore, the total momentum \(\mathbf{p}_A + \mathbf{p}_B\) is conserved.

Show that the total momentum of a system of multiple objects is conserved if no external forces act on it (i.e., a generalisation of the argument above for two objects).

The total momentum of the system is $\mathbf{P} = \sum_i \mathbf{p}_i$ (with the sum running over all objects). The force on object $i$ due to object $j$ is $\mathbf{F}_{ij}$, and the force on object $j$ due to object $i$ is $\mathbf{F}_{ji} = -\mathbf{F}_{ij}$. Therefore, the total force on object $i$ is $$ % \frac{\text{d}\mathbf{p}_i}{\text{d}t} = \sum_{j=1}^N \mathbf{F}_{ij} \, . but with i \neq j \frac{\text{d}\mathbf{p}_i}{\text{d}t} = \sum_{j\neq i} \mathbf{F}_{ij} \, . $$ Summing over all objects gives for the total momentum $\mathbf{P} = \sum_{i=1}^N \mathbf{p}_i$ $$ \frac{\text{d}\mathbf{P}}{\text{d}t} = \sum_i \frac{\text{d}\mathbf{p}_i}{\text{d}t} = \sum_i \sum_{j\neq i} \mathbf{F}_{ij} \, . $$ In this sum, for every force $\mathbf{F}_{ij}$ there is an equal and opposite force $\mathbf{F}_{ji} = -\mathbf{F}_{ij}$, so the sum is zero. Therefore, $\text{d}\mathbf{P}/\text{d}t = 0$ and the total momentum is conserved.

Conservation of Energy

The conservation of energy states that the total energy of an isolated system is conserved.

If we integrate Newton’s second law \(\mathbf{F} = \text{d}\mathbf{p}/\text{d}t\) over distance, we get

\[\begin{align*} \int \mathbf{F} \cdot \text{d}\mathbf{r} &= \int \frac{\text{d}\mathbf{p}}{\text{d}t} \cdot \text{d}\mathbf{r} \\ &= \int m \frac{\text{d}\mathbf{v}}{\text{d}t} \cdot \text{d}\mathbf{r} \\ &= \int m \frac{\text{d} \mathbf{r}}{\text{d}t} \cdot \text{d}\mathbf{v} \\ &= \int m \mathbf{v} \cdot \text{d}\mathbf{v} \, . \end{align*}\]

The rule for differentiating a dot product is \(\text{d}(\mathbf{a} \cdot \mathbf{b}) = \mathbf{a} \cdot \text{d}\mathbf{b} + \mathbf{b} \cdot \text{d}\mathbf{a}\) (which can be proven by writing out the dot product in terms of components and applying the product rule). From this, we have that \(\text{d}(\mathbf{v} \cdot \mathbf{v}) = \text{d} (v^2) = 2 \mathbf{v} \cdot \text{d}\mathbf{v}\), so

\[\int \mathbf{F} \cdot \text{d}\mathbf{r} = \int \text{d} \left(\frac{1}{2} m v^2 \right) \, .\]

If we integrate from some starting position \(\mathbf{r}_0\) and velocity \(\mathbf{v}_0\) to some final position \(\mathbf{r}\) and velocity \(\mathbf{v}\), we get

\[\int_{\mathbf{r}_0}^{\mathbf{r}} \mathbf{F} \cdot \text{d}\mathbf{r}^{\prime} = \frac{1}{2} m v^2 - \frac{1}{2} m v_0^2 \, .\]

We call the quantity \(\frac{1}{2} m v^2\) the kinetic energy \(E_K\) of the object. The term \(\int \mathbf{F} \cdot \text{d} \mathbf{r}\) is known as the work done \(W\) by the force on the object. This is equation is a statement of the work-energy theorem - the work done on an object is equal to the change in its kinetic energy.

For a special class of forces known as conservative forces, the work done is independent of the path taken. This means that the integral of the force over distance is a function of the initial and final positions only, and not the path taken. We can therefore define a potential energy \(U\) as

\[U(\mathbf{r}) = - \int_{\mathbf{r}_0}^{\mathbf{r}} \mathbf{F} \cdot \text{d}\mathbf{r}^{\prime} \, .\]

We can then rewrite the work-energy theorem as \(E_K + U = \text{constant}\), which is the conservation of energy. Note that the choice of the reference point \(\mathbf{r}_0\) is arbitrary, and only adds a constant to the potential energy. Since we only care about changes in energy, this constant is irrelevant.

We have defined \(U = - \int \mathbf{F} \cdot \text{d}\mathbf{r}\), or in differential form \(\text{d}U = - \mathbf{F} \cdot \text{d}\mathbf{r}\). We can also write \(\text{d} U = \mathbf{\nabla} U \cdot \text{d}\mathbf{r}\) (if you expand the dot product, you will see that this is just chain rule). Equating the two, we see that the force is the gradient of the potential energy \(\mathbf{F} = - \mathbf{\nabla} U\).

Prove that the following statements are equivalent: (i) the work done by a force is path-independent, (ii) the curl of the force is zero ($\mathbf{\nabla} \times \mathbf{F} = 0$), and (iii) the force can be written as the gradient of a scalar potential $\mathbf{F} = - \mathbf{\nabla} U$.

The equivalence of (i) and (ii) are a consequence of Stokes' theorem. The work done by a force $\mathbf{F}$ over a closed loop $C$ is $$ \oint_C \mathbf{F} \cdot \text{d}\mathbf{r} = \int_S (\mathbf{\nabla} \times \mathbf{F}) \cdot \text{d}\mathbf{S} \, , $$ where $S$ is a surface bounded by the path $C$. If the work done is path-independent, then the work done over a closed loop is zero - a closed loop represents going from point A to point B one way and then coming back another way, and the work done each way is equal and opposite. Since the equation above is true for any surface $S$, the integrand must be zero, i.e., $\mathbf{\nabla} \times \mathbf{F} = 0$.

The equivalence of (ii) and (iii) comes from the fact that the curl of a gradient is zero, i.e., $\mathbf{\nabla} \times \mathbf{\nabla} U = 0$. Therefore, if $\mathbf{F} = - \mathbf{\nabla} U$, then $\mathbf{\nabla} \times \mathbf{F} = 0$. The converse is also true - if $\mathbf{\nabla} \times \mathbf{F} = 0$, then $\mathbf{F}$ can be written as the gradient of a scalar potential.