The Three Gables

In this instance our Three Gables refer to men of the mathematical world rather than a house in Harrow Weald; Newton, Hamilton and Lagrange.

Let us take the example of the simple pendulum and the three mathematicians' approaches to finding the equations of motion for this dynamical system. The pendulum weight has mass $m$, $\mathbf{x}$ denotes its position vector, $\dot{\mathbf{x}}$ its velocity vector (i.e. $\frac{d(\mathbf{x})}{dt}=\dot{\mathbf{x}}$) and $\mathbf{p}$ is the momentum vector of the pendulum mass.

Approaches
Newton: Relate a force $\mathbf{F}(\mathbf{x}, \dot{\mathbf{x}}, t)$ to the resulting motion using Newton's 2nd law: $F=ma$ or more appropriately $$\frac{d}{dt}m\dot{\mathbf{x}} = \mathbf{F}(\mathbf{x}, \dot{\mathbf{x}}, t).$$

Lagrange: Construct the Lagrangian function $\mathcal{L}(q, \dot{q},t)$ which can normally be given by $\mathcal{L} = T - V$ where $T$ and $V$ are the kinetic and potential energy of the system respectively. The equations of motion are given by the Euler-Lagrange equation: $$ \frac{d\mathcal{L}}{dq_i} - \frac{d}{dt} \left( \frac{d\mathcal{L}}{d\dot{q}_i} \right) = 0.$$

Hamilton: Construct the Hamiltonian function $H(\mathbf{x}, \mathbf{p},t)$, which is a function of a set of conjugate variable pairs. The Hamiltonian contains the dynamics of the system and Hamilton's equations provide the equations of motion: $$ \frac{d\mathbf{x}_i}{dt} = \frac{\partial H}{\partial \mathbf{x}_i}, $$ $$ \frac{d\mathbf{p}_i}{dt} = \frac{\partial H}{\partial \mathbf{p}_i}. $$

1: Newtonian Mechanics
This is the familiar method for most of us. I have dropped the bold font for vectors from here on.

The force on the pendulum mass is given by $F = -kx$ where $k$ is a constant given by $$k = \omega^2 m $$. Thus we can write $$F=-\omega^2 m.$$

Using Newton's 2nd law we get the equation of motion: $$ m \frac{d}{dt} \dot{x} = -m \omega^2 x, $$ which is equivalent to $$ \frac{d^2 x}{dt^2} = - \omega^2x.$$

The solution to this equation is the standard, two constant oscillatory equation: $$ x(t) = x_0 sin(\omega t + \phi_0)$$ where $x_0$ and $\phi_0$ are constants determined by the inital values of $x(t)$ and $\dot{x}(t)$.

2: Lagrangian Mechanics
The pendulum mass has a kinetic energy: $$T = \frac{1}{2} m \dot{x}^2 $$ and a potential energy given by $$V = \frac{1}{2}m \omega^2 x^2.$$

The Lagrangian in this case may be found using $\mathcal{L} = T-V$: $$\mathcal{L} = T-V = \frac{1}{2} m \dot{x}^2 - \frac{1}{2}m \omega^2 x^2. $$

Using the Euler-Lagrange equation: $$ \frac{d\mathcal{L}}{dx_i} - \frac{d}{dt}\left( \frac{d\mathcal{L}}{d\dot{x}_i}\right) = 0.$$
we can do this step by step:
$$ \frac{d\mathcal{L}}{dx_i} = -\omega^2 mx,$$
and:
$$ \frac{d\mathcal{L}}{d\dot{x}_i}=m\dot{x}.$$

Substituting into the Euler-Lagrange equation we have:
$$-\omega^2 m x - \frac{d}{dt}(m\dot{x}) = 0$$
which is equivalent to:
$$\frac{d^2 x}{dt^2} = -\omega^2 x.$$

Which unsurprisingly is equivalent to our Newtonian formulation and has the same standard oscillatory solution.

3: Hamiltonian
Here we need to use the conjugate pair of position $x_i$ and momentum $p_i$. Given the Lagrangian $\mathcal{L}$, the conjugate momentum and Hamiltonian can be derived as follows:
$$p_i = \frac{\partial \mathcal{L}}{\partial \dot{q}_i} = \frac{\partial \mathcal{L}}{\partial \dot{x}_i} = m\dot{x},$$
$$H = \sum_i \dot{q_i}{p_i} - \mathcal{L} = \sum_i \dot{x}_i p_i - \mathcal{L},$$
which leads to:
$$H = \dot{x}p_x - \mathcal{L} = m\dot{x}^2 - \mathcal{L},$$
and finally we have the Hamiltonian:
$$H = \frac{p^2_x}{2m} + \frac{1}{2}m \omega^2 x^2.$$

Using Hamilton's equations:
$$\frac{dx}{dt} = \frac{\partial H}{\partial p_x} = \frac{p_x}{m}, $$
$$\frac{d p_x}{dt}=-\frac{\partial H}{\partial x} = -m \omega^2 x. $$

Note that the last equation is equivalent to Newton's second law.

Remarks
Using Newtonian mechanics, given the function $\mathbf{F}$, we derive the equations of motion, which must be solved to give the explicit dependence of position $x$ (and thus velocity $\dot{x}$) on the independent parameter $t$.

The physics is constructing the form of $\mathbf{F}$ for a gvien system.

Using Lagrangian mechanics the variables $q_i$ can be any convenient set of parameters that describe the system state. We treat $q_i$ and $\dot{q}_i$ as independent variables, despite the fact that they are not. For $n$ degrees of freedom, the Euler-Lagrange equations give us $n$ second-order differential equations.

The physics is constructing the form of $\mathcal{L}$ for a given system.

Using Hamiltonian mechanics a set of conjugate variables that describe the system state must be carefully chosen. We treat the conjugate pair $x,p_x$ as independent variables despite the fact that they are related. For $n$ degrees of freedom, Hamilton's equations give us $2n$ first-order differential equations. Representing the dynamics using first-order differential equations has some advantages when considering linear methods.

The Hamiltonian is conserved if the independent variable does not appear explicitly in the Hamiltonian. This can be shown:

$$\frac{dH}{dt} = \frac{\partial H}{\partial x}\frac{dx}{dt} + \frac{\partial H}{\partial p_x}\frac{d p_x}{dt} + \frac{\partial H}{\partial t}$$

substituting in Hamilton's equations the first two terms cancel:
$$\frac{dH}{dt} = \frac{\partial H}{\partial x}\frac{\partial H}{\partial p_x} + \frac{\partial H}{\partial p_x}\frac{\partial H}{\partial x} + \frac{\partial H}{\partial t} =  \frac{\partial H}{\partial t}.$$

If the Hamiltonian does not depend explicitly on $t$, then the Hamiltonian is conserved:
$$\frac{dH}{dt} = \frac{\partial H}{\partial t} = 0.$$

For our system the Hamiltonian could be written: $$H = T + V, $$
which shows that the Hamiltonian appears to be the total energy of the system.


References
[1] Linear Dynamics, Lecture 1: Review of Hamiltonian Mechanics, Andy Wolski, 2006 [Cockcroft lecture series]
[2] Introduction to Beam Dynamics, Rob Appleby, 2011 [Cockcroft lecture series]

LaTeX in Blogger

I used a script after <head> in the html layout file to enable LaTeX for equations, details can be found here or on the MathJax website.

Here is a test:

$F=ma$ is our first equation
combining with
$a=\frac{v^2}{r}$
gives $$F=\frac{mv^2}{r}$$