A Comparison of the Newtonian, Lagrangian, and Hamiltonian Formalisms of Classical Mechanics

Abstract⌗

Physical systems can be modelled mathematically in many different ways, however three main formalisms have prevailed as the most commonly used within classical mechanics. Each of these have their own strengths and weaknesses in modelling different systems stemming from their underlying principles making them each more suited to certain situations than others. During this report, I utilise the simple pendulum to demonstrate how each formalism is used and where each of their strengths lie, as well as some insight into how they are each defined.

Introduction⌗

Classical Mechanics is the physical theory which describes the motion of objects, such as projectiles, over time. This is achieved by deriving equations which represent the motion of the system as functions of time (R.G. Lerner and George L. Trigg, 1991)¹. With these, we are able to mathematically describe the way in which a system evolves over time and make predictions about its state in the future.

While the equations of motion are often in terms of Cartesian coordinate vectors, it is sometimes simpler to use other coordinate systems depending on the situation. For example, a swinging pendulum often uses polar coordinates with the angle between the pendulum & the vertical and the length of the pendulum’s string.

Differential Equations⌗

While we can’t easily describe an object’s position at all points in time, we can often describe how it’s position changes.This means that the equations of motion of a system will almost always come from the solutions to some differential equation(s) that we are able to form. For example, Newton’s second law, $\vec{F} = m\vec{a}$ (Resnik, Robert and Halliday, David, 1966)², is actually the second-order differential equation $\vec{F} = m\ddot{\vec{s}}$, where $\vec{s}$ is the object’s displacement vector and each dot above it represents a derivative with respect to time. This is solved by finding some other expression for the force and setting the 2 expressions equal to each other to produce another differential equation which leads to the equations of motion.

Forces vs. Energy⌗

To derive a system’s equations of motion, there are multiple different routes that one can take. The first is to consider the forces acting within a system - commonly called Newtonian mechanics - and make use of Newton’s laws (the second law in particular) to form a differential equation which, when solved, will produce the equations of motion. The entire system is modelled as a collection of particles - point masses - which have corresponding vectors for the force acting on them, their velocity, and their acceleration. Knowing the force, velocity, and acceleration of a particle in this model is then enough to derive its equations of motion by solving a differential equation.

Comparatively, analytical mechanics considers the total energy of a system - most commonly consisting of its Kinetic and Potential energy. These scalars are then used to derive the system’s equations of motion by some property of their variation. The two dominant branches are Lagrangian mechanics and Hamiltonian mechanics. These branches differ by the choice of coordinates used to represent the system - with Lagrangian mechanics making use of generalised coordinates and corresponding generalised velocities, and Hamiltonian mechanics using coordinates and corresponding momenta.

The Scenario⌗

To demonstrate how each formalism works, I will use each of them to derive the equation of motion for a simple pendulum attached to a stationary support by an inextensible, always-taut string and swinging exclusively on one plane. The pendulum has a mass $m$kg and the string is of length $l$m, and experiences standard Earth gravity $9.81$ms$^{-2}$. Air resistance & friction will be considered negligible for the sake of simplicity and reintroducing these to the derivations will be left as an exercise to the reader.

Newtonian Mechanics⌗

As stated above, Newtonian mechanics is built around forces which act upon an object. Enumerating the relevant forces does require some kind of coordinate system though, so that will be the first step. Usually, Cartesian coordinates would be used and the equations of motion would become split into $x$ and $y$ parts which may be combined later but, in this situation, we have a better option. By recognising that the angle $\theta$ made by the pendulum string with the vertical is the only part of this problem that changes, we can instead use this as the basis of our singular coordinate. Because the pendulum will always be on the same arc, we can describe its position completely as a position along it, making our coordinate $x = l\theta$, where $\theta$ will be negative on one side of the vertical line and positive on the other. We can trivially differentiate this twice to find that $\ddot{x} = l\ddot\theta$ is our acceleration.

The only forces we must account for now to find a sufficient resultant force are those that act parallel to the pendulum’s motion, as those are the only ones which can influence its acceleration. This means that we only need to consider forces acting tangent to the pendulum’s arc as it swings. The only force within this system is the pendulum’s weight acting straight downwards. By forming a right-angle triangle and using basic trigonometry, the relevant component of this can be resolved as $-mg\sin\theta$. This force is negative as it acts opposite to the direction of the pendulum’s displacement - it is a restoring force pulling the system towards its equilibrium position at $\theta = 0$.

With this, our differential equation $F = m\ddot{x}$ can be written as $-mg\sin\theta = ml\ddot\theta$. Rearranging this for the acceleration $\ddot\theta$, we find our equation of motion to be $$\ddot\theta = -\frac{g}{l}\sin\theta$$

Lagrangian Mechanics⌗

As previously mentioned, Lagrangian mechanics models systems by considering their energy. It does this by manipulating one main quantity of the system: the Lagrangian, $\mathcal{L} = T - U$ (Torby 1984)³, where $T$ is the kinetic energy of the system and $U$ is its potential energy. The generalised coordinates are usually represented as $q$, with their corresponding generalised velocities being $\dot{q}$. Once the system’s Lagrangian is determined the equations of motion can be derived in different ways, but the most common of these is via the Euler-Lagrange equations (Hand & Finch 1998)⁴, namely

$$\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial{\mathcal{L}}}{\partial{\dot{q}_i}}\right) = \frac{\partial{\mathcal{L}}}{\partial{q_i}}$$

for the $i$th coordinate. Solving these differential equations will then lead to the equations of motion for the system.

The number of coordinates required to model a system is entirely dependent on the number of degrees of freedom within it. A degree of freedom is any condition which can be changed to influence the system in some way, and are usually not prohibitively difficult to find. One way of determining the number of degrees of freedom for a one-particle system is by expressing the particle’s position in terms of Cartesian coordinates and seeing which variables appear in each expression - in this case $x = l\sin\theta$ and $y = -l\cos\theta$, within which the only variable is $\theta$ (remember, $l$ is just some constant which cannot change during the system’s existence). Hence, for the simple pendulum there is only one degree of freedom and hence only one coordinate is necessary.

This generalised coordinate is $q = l\theta$. The kinetic energy of the system comes only from the pendulum, so can trivially be determined to be $T = \frac{1}{2}m\dot{q}^2 = \frac{1}{2}ml^2{\dot\theta}^2$. The potential energy also comes only from the pendulum, but we must know two things first: where to place the point of “zero” potential energy, and an expression for the pendulum’s vertical displacement from that point. I will choose to place $U = 0$ at the plane of the pendulum’s support, and some simple trigonometry leads to the pendulum’s displacement from it being $-l\cos\theta$, which is negative as the pendulum is always beneath the support. From this, our potential energy $U = -mgl\cos\theta$. Note that stating the potential energy in this way results in negative energy (i.e. $U < 0$), but this does not lead to an incorrect derivation of the equation of motion due to taking partial derivatives. A pedantic reader may instead use $U = mgl(1 - \cos\theta)$ and observe that it results in the same equation of motion as we find here (which implies that $U$ need only represent a change in energy from some initial condition, not the actual exact energy of the system at all times & points). This leads to our complete Lagrangian $$\mathcal{L} = \frac{1}{2}ml^2{\dot\theta}^2 + mgl\cos\theta$$

Now, we must determine the Euler-Lagrange equation that we need to solve for our equation of motion. $\frac{\partial{\mathcal{L}}}{\partial\dot\theta} = ml^2\dot\theta$, and the time derivative of this will equal $ml\ddot\theta$. By the same process, $\frac{\partial{\mathcal{L}}}{\partial\theta} = -mgl\sin\theta$, leading to our final differential equation $ml^2\ddot\theta = -mgl\sin\theta$. After rearranging this, we arrive at the same equation of motion as before:

$$\ddot\theta = -\frac{g}{l}\sin\theta$$

Hamiltonian Mechanics⌗

Like Lagrangian mechanics, Hamiltonian mechanics is an analytical formalism and therefore concerns itself with the energies in a system. By construction, Hamiltonian mechanics modifies the Lagrange equations into a simpler form which is much nicer to solve analytically. The key difference is that it replaces Lagrange’s generalised velocities $\dot{q}_i$ with generalised momenta (Hamilton, William Rowan, Sir, 1833)⁵ in an attempt to produce simpler differential equations to solve for the equations of motion.

For many systems, Lagrangian mechanics will result in second-order Ordinary Differential Equations (ODEs) to solve, however these can be transformed into systems of first-order ODEs by introducing an auxiliary variable. In general, $n$ second-order ODEs can be transformed into $2n$ first-order ODEs by this method. Given some Lagrangian $\mathcal{L}(q_i, \dot{q}_i, t)$, Hamiltonian mechanics defines generalised momenta as the auxiliary variables defined as

$$p_i = \frac{\partial{\mathcal{L}}}{\partial{\dot{q}_i}}$$

and are known as canonical momenta or conjugate momenta. By starting from some Lagrangian, one can define the central quantity of Hamiltonian mechanics, the Hamiltonian $\mathcal{H}$ via a Legendre Transform (an explanation of which is outside the scope of this paper) of $\mathcal{L}$ to turn the Lagrangian energy function $E_L = \sum\limits_{i=1}^{n}\left(\dot{q}_i\frac{\partial{\mathcal{L}}}{\partial{\dot{q}_i}}\right) - \mathcal{L}$

into the Hamiltonian, which is usually expressed as $$\mathcal{H} = \displaystyle\sum_{i=1}^{n}(p_i\dot{q}_i) - L$$

Combining these, the set of differential equations to solve known as Hamilton’s Equations can be derived (Arnol’d 1989, pg. 65-66)⁶ as

$$\frac{\partial{\mathcal{H}}}{\partial{\dot{p}_i}} = \dot{q}_i \space\text{and}\space -\frac{\partial{\mathcal{H}}}{\partial{q_i}} = \dot{p}_i$$

in the case of a time-independent $\mathcal{H}$ & $\mathcal{L}$, which can be used together to arrive at the equations of motion for a mechanical system.

To apply this to the pendulum system, recall the derived Lagrangian from before, $\mathcal{L} = \frac{1}{2}ml^2{\dot\theta}^2$. This leads to the conjugate momentum

$$p = ml^2\dot\theta$$

by differentiating partially with respect to $\theta$. From here, the Hamiltonian can be calculated as $(ml^2 \dot\theta)\dot\theta - \frac{1}{2}ml^2 {\dot\theta}^2 + mgl\cos\theta = ml^2{\dot\theta}^2 - \frac{1}{2}ml^2{\dot\theta}^2 + mgl\cos\theta = \frac{1}{2}ml^2{\dot\theta}^2 + mgl\cos\theta$, hence

$$\mathcal{H} = \frac{p^2}{2ml^2}-mgl\cos\theta$$

From here, the partial with respect to $p$ is $\frac{\partial{\mathcal{H}}}{\partial{p}} = \frac{p}{ml^2}$ and the partial with respect to $\theta$ is $\frac{\partial{\mathcal{H}}}{\partial{\theta}} = mgl\sin\theta$. This leads to our system of Hamilton’s equations being

$$\frac{p}{ml^2} = \dot\theta \space\text{and}\space -mgl\sin\theta = ml^2\ddot\theta$$

The first equation is simply the definition of our conjugate momentum seen earlier so is not useful to us, however the second one is useful to us. We can solve the second equation to arrive once more at the same equation of motion for the simple pendulum:

$$\ddot\theta = -\frac{g}{l}\sin\theta$$

Conclusion⌗

From these derivations, it should be clear which of the three formalisms is more suited to a simple situation such as this one. In a situation with very few forces & particles involved, Newtonian mechanics clearly wins. It has the least number of prerequisite steps before arriving at a differential equation to solve for the equations of motion, and calculating the resultant force on a particle often involves much simpler calculations than calculating energies. This simplicity is not a given, however - the more chaotic a system becomes, the harder it is to model it under Newtonian mechanics. For example, the double pendulum is much better modelled via the Lagrangian due to greater ease in expressing the system’s energies rather than its forces. The resulting equations are still quite ugly and have no known closed form solutions, but it is much to arrive at the Euler-Lagrange equations than to use Newton’s second law. The same is true of the inverted pendulum, where the respective derivations for each formalism vary wildly in their length.

For most human-scale systems, it is still excessive to use a Hamiltonian approach over a Lagrangian. Where Hamiltonian mechanics sees the most use is in modelling quantum systems, in particular through the Hamiltonian operator $\hat{H}$. As an operator, $\hat{H}$ corresponds to a system’s total energy $\hat{T} + \hat{V}$ where $\hat{T}$ is the kinetic energy operator and $\hat{V}$ is the potential energy operator. The Hamiltonian in this form allows quantum systems to be modelled much more easily due to energy being more easy to determine for most systems at this scale. It also fits much more nicely into the rest of quantum mechanics due to a general preference for operators when working with such systems.

Overall, most systems one encounters in everyday life can be and are modelled more than sufficiently with Newtonian mechanics. For most of the remainder, Lagrangian mechanics is the nicer analytical formalism to work with - Hamiltonian mechanics generally is used for the most complex of systems which are already the most difficult to model conceptually - to say nothing of their difficulty to be modelled mathematically. However, as the three are (almost) completely equivalent to each other, most physicists will simply use whichever one they are more comfortable with as a matter of preference - even if it means slightly longer derivations.