The Classical State

Quote

It seems probable that most of the grand underlying principles have been firmly established and that further advances are to be sought chiefly in the rigorous application of these principles to all the phenomena which come under our notice.

— Albert Michelson, 1894

lol

Anyways, in the first quarter of the 20th century people realized that the laws of physics known up to that point didn’t do a great job explaining a wide range of phenomena involving electrons, atoms, and light. After a great deal of effort, a new theory (and a new law of motion) emerged in 1924:— quantum mechanics. It is now part of the basic framework for understanding atomic, nuclear, and subnuclear physics. As well as condensed matter (or “solid-state”) physics. Although, I admit to having no idea what that last one is. The laws of motion (Galileo, Newton, etc.) that came before it are called classical mechanics.

Classical mechanics are now understood as an approximation of quantum mechanics, but since much of the structure of quantum theory is inherited from classical theory, it is still absolutely worth going over!

The Principle of Least Action

Take a baseball and throw it straight up in the air. After a few seconds the baseball come back down. Because how high the baseball is depends on the time since it’s been thrown, we can denote it as a function of time: $x(t)$. This is the notation my textbook uses which kinda pisses me off because I’d like to use $y$. For height. Actually let’s just do that. Denote the height of the baseball, as a function of time, as $y(t)$; this is the trajectory of the baseball. If we plot $y$ as a function of $t$ (the usual caveat of ignoring air resistance and pretending we’re in a universal gravitational field blah blah blah applies here) then any trajectory has the form of a parabola and there are an infinite number of possible trajectories. Whichever one it follows depends on the $p$ (momentum) of the baseball as it leaves your hand.

Buuuut if we say that the baseball returns to your hand at exactly $\Delta t$ seconds after it leaves your hand, there is ONE trajectory the ball can follow. Now, while there is a simple way to figure out this trajectory for a baseball moving in a uniform gravitational field, we want a method which can be applied to a particle moving in any potential field $V(y)$. (Potential field means “field” of “potential energy” btw) Let us begin with Newton’s law $F=ma$, which is actually a second order differential equation

$$m\underset{acceleration}{\underbrace{\frac{d^{2}y}{d{t}^2}}}=\underset{Force}{\underbrace{-\frac{dV}{dy}}}\text{\hspace{1em}(1.1)}$$

Show Work

If position is $y(t)$, velocity ($v$) is how fast $y$ changes; $\frac{dy}{dt}$. Acceleration is the derivative of velocity so that’s how we get that. Force is a little more interesting. $V(y)$ is potential energy, that is how much stored energy the system has because of where it is. The reason $V(y)=mgy$ is because being higher up translates to more potential energy. (This is also why a stretched spring $V(x)=\frac{1}{2}k{x}^2$ where $x$ is large, has more energy.) $V$ is a function of $y$ because potential energy depends on where you are, not how fast you are moving. If we think of $V(y)$ as a hill with a ball sitting on it, we can intuit that the ball would roll downhill. The “steepness” of the hill is it’s derivative; $\frac{dV}{dY}$. Interestingly, force is defined as $-\frac{dV}{dy}$, here the minus sign ensures the force points downhill, i.e., toward lower potential energy.

Just a concrete example here if $V(y)=mgy$, then the slope is $\frac{dV}{dy}=mg$ and the force is $-mg$, and we can confirm this because we know the force of gravity is $-mg$. Just keep this in mind because thinking like this lets us ignore forces entirely and just derive everything from potential and kinetic energy via the Lagrangian $\mathcal{L}=T-V$. But that comes later… :)

Let’s reëxpress this second-order equation as a pair of first order equations, where $m$ is the mass and $p$ is the momentum of the baseball:

$$\begin{array}{rlrl}\frac{dy}{dt}& =\frac{p}{m}\text{,}& \frac{dp}{dt}& =-\frac{dV}{dy}\end{array}\text{\hspace{1em}(1.2)}$$

Show Work

If momentum is defined as $p$ = $mv$ where $v=\frac{dy}{dp}$, $p:=m\frac{dy}{dt}$.

So from the definition of $p$ we get $p=m\frac{dy}{dt}⟹\frac{dy}{dt}=\frac{p}{m}$. That’s our first equation. Next, if we start with $p=m\frac{dy}{dt}$ again, we can differentiate $p$ and take the derivative w.r.t time to get

$$\frac{dp}{dt}=m\frac{d^{2}y}{d{t}^2}$$

And then plug in newtons law to get

$$m\frac{d^{2}y}{d{t}^2}=-\frac{dV}{dy}$$

Yielding

$$\frac{dp}{dt}=-\frac{dV}{dy}$$

We want to find a solution to these equations such that $y(t_0)=Y_i$ and $y(t_0+\Delta t)=Y_f$, where $Y_i$ and $Y_f$ are the initial height of your hand when the baseball leaves it, and the final height of your hand when you catch the baseball, respectively.

The textbook talks about using a computer by splitting the problem into intervals but quite frankly, I couldn’t care less so I’m going to skip that. This is all to say there’s this thing called the Principle of Least Action. It says that the action $S$ is stationary at any trajectory $y_n$ (wait! what do the braces mean?) satisfying the conditions of motion $F=ma$, at every time $t_n$.

So to solve the trajectory of the baseball, we just program a computer to find the set of points $y_n$ (ohh its braces cos its a set of points 😌😌😌 so its like $y_0,y_1,\dots y_N$) which minimizes the quantity:

$$Q=\sum _k{\left(\frac{\Delta S}{\Delta y_k}\right)}^2\text{\hspace{1em}(1.3)}$$

$S$ is “action,” defined as $\int \mathcal{L},dt$ where $\mathcal{L}=T-V$ where $T$ is kinetic energy and $V$ is potential energy. But when we discretize it, our integral becomes a sum, $S\approx {\sum }_n{\mathcal{L}}_n\Delta t$. Dividing by $\Delta y_k$ is essentially asking “how much $S$ changes if I tweak $y_k$” and squaring it is just a mathematical trick because sometimes it could be negative and we want to avoid that and if we use $\text{abs}()$, its “sharper” near zero. An ML background might give insight in why we want to avoid a “sharp” gradient. The minimum is obtained at $Q=0$, where $S$ is stationary. $Q$ just measures how “badly” the whole path “violates” the principle of least action. Anyways, this set of points, joined by straight line segments, gives us the approximate trajectory of the baseball.

Problem: Dyre’s Dilemma

In discussing the motion of the baseball, we have been ignoring a lot of details about baseballs, such as the composition of the interior, the pattern of the stitching, and the brand-name printed on the surface. Instead, the baseball has been treated as though it were essentially a structureless point of mass $m$. It is necessary to make idealizations like this in physics; the real world is otherwise too complicated to describe. But sometimes an idealization misses something crucial. See if you can find what goes wrong in the following argument, which tries to prove that a rolling wheel (or, for that matter, a rolling baseball) can never come to rest through friction with the ground.

“Proof”: […] [T]he forward momentum of a wheel in the positive $x$-direction can only be eliminated by a force applied in the opposite direction. But the only place this force could be applied by friction is the point where the wheel touches the ground. And a force in the negative $x$-direction, applied at this point, will have the effect of making the wheel spin faster! Therefore, the wheel will never come to rest due to friction. QED.

Is this reasoning correct? Can you solve Dyre’s Dilemma?

Yea basically what Dyre’s dilemma does wrong is to treat friction as only affecting rotation via torque, while ignoring that ALSO reduces forward motion. Fundamentally, the idealized model assumes no energy dissipation, so there is no mechanism for the wheel to lose kinetic energy and come to rest. In reality, deformation and internal friction convert mechanical energy into heat, which is why objects actually stop.

Euler-Lagrange and Hamilton’s Equations

Note

Man, what would be really nice to have (for this section in particular, and perhaps future sections) is a plugin that turns math equations into manim animations that can be played. This would be both an Obsidian plugin, and have some thing that lets me view them if I were to export to HTML…

Basically, the Euler-Lagrange equations are the second-order form of the equations of motion, while Hamilton’s equations are the first-order form. In either form, the equations of motion are a consequence of the Principle of Least Action. We shall now re-write those equations in a very general way, which can be applied to any mechanical system, including those which are more complicated than a baseball.

I’m switching using $y$ to $x$ as the coordinate. I feel like $y$ was intuitive because we were talking about height specifically, but $x$ feels more general. It’s worth nothing that everything that follows generalizes further, like you can swap $x$ for $q_i$ and you can describe any mechanical system with any number of degrees of freedom.

We begin by writing

$$\underset{\text{function of whole path}}{\underbrace{S[x_i]}}=\sum _{n=0}^{N-1},\underset{\text{step size: }\Delta t}{\underbrace{ϵ}}\mathcal{L}(x_n,{\dot{x}}_n)\text{\hspace{1em}(1.4)}$$

Explainer

$x_i$ is the entire path, meaning the list of positions $x_0,x_1,\dots ,x_N$. Because indexing starts at 0 and ends at $N$, the total number of points is $\text{len}(x_i)=N+1$. However, the sum runs from $n=0$ to $N-1$ because each term corresponds to an interval between two consecutive points, using $x_n$ and $x_{n+1}$. There are only $N$ such intervals, even though there are $N+1$ points. The quantity $ϵ$ is the timestep $\Delta t$, and the velocity at each step is approximated by ${\dot{x}}_n\approx \frac{x_{n+1}-x_n}{ϵ}$. The action $S$ is therefore computed by summing the Lagrangian evaluated on each small segment of the path, which is a discrete approximation of the continuous integral $S=\int \mathcal{L},dt$.

Think of it like this… let $x$ be a list containing the path $[x_0,x_1,\dots ,x_N]$. Then $N=\text{len}(x)-1$ gives the number of intervals. Let $ϵ=dt$. Define velocity as $\text{velocity}(x_n,x_{n+1})=(x_{n+1}-x_n)/ϵ$. Define the Lagrangian as $\mathcal{L}(x_n,v_n)=T(v_n)-V(x_n)$. Then initialize $S=0$, and for each $n$ from $0$ to $N-1$, compute $v_n=\text{velocity}(x[n],x[n+1])$ and update $S\leftarrow S+ϵ\cdot \mathcal{L}(x[n],v_n)$. This loop runs exactly $N$ times, once per interval, which is why the sum stops at $N-1$.

where

$$\mathcal{L}(x_n,{\dot{x}}_n)=\frac{1}{2}m{\dot{x}}_n^2-V(x_n)\text{\hspace{1em}(1.5)}$$

and where

$${\dot{x}}_n\equiv \frac{x_{n+1}-x_n}{ϵ}\text{\hspace{1em}(1.6)}$$

$\mathcal{L}(x_n,{\dot{x}}_n)$ is known as the Lagrangian function. Then the principle of least action requires that, for each $k$, $1\le k\le N-1$

$$\begin{array}{rl}0& =\frac{d}{d{x}_k}S(x_i)=\sum _{n=0}^{N-1}ϵ\frac{d}{d{x}_k}\mathcal{L}(x_n,{\dot{x}}_n)\\ & =ϵ\frac{\partial }{\partial x_k}\mathcal{L}(x_k,{\dot{x}}_k)+\sum _{n=0}^{N-1}ϵ\frac{\partial \mathcal{L}(x_n,{\dot{x}}_n)}{\partial {\dot{x}}_n}\frac{d{\dot{x}}_n}{d{x}_k}\end{array}\text{\hspace{1em}(1.7)}$$

and, since

$$\frac{d{\dot{x}}_n}{d{x}_k}=\{\begin{array}{ll}\frac{1}{ϵ}& n=k-1\\ -\frac{1}{ϵ}& n=k\\ 0& \text{otherwise}\end{array}\text{\hspace{1em}(1.8)}$$

this becomes

$$\frac{\partial }{\partial x_k}\mathcal{L}(x_k,{\dot{x}}_k)-\frac{1}{ϵ}\{\frac{\partial }{\partial {\dot{x}}_k}\mathcal{L}(x_k,{\dot{x}}_k)-\frac{\partial }{\partial {\dot{x}}_{k-1}}\mathcal{L}(x_{k-1},{\dot{x}}_{k-1})\}=0\text{\hspace{1em}(1.9)}$$

Recalling that $x_n=x(t_n)$, this last equation can be written

$$(\frac{\partial \mathcal{L}(x,\dot{x})}{\partial x}{)}_{t=t_n}-\frac{1}{ϵ}\{{\left(\frac{\partial \mathcal{L}(x,\dot{x})}{\partial \dot{x}}\right)}_{t=t_n}{\left(\frac{\partial \mathcal{L}(x,\dot{x})}{\partial \dot{x}}\right)}_{t=t_{n-ϵ}}\}=0\text{\hspace{1em}(1.10)}$$

Anyways, this is the Euler-Lagrange equation for the baseball. It becomes simpler when we take the $ϵ\to 0$ limit (the “continuum” limit). In that limit, we have:

$$\begin{array}{rl}{\dot{x}}_n=\frac{x_{n+1}-x_n}{ϵ}& \to \dot{x}(t)=\frac{dx}{dt}\\ S=\sum _{n=1}^{N-1}ϵ\mathcal{L}(x_n,{\dot{x}}_n)& \to S={\int }_{t_0}^{t_0+\Delta t}dt\mathcal{L}(x(t),\dot{x}(t))\end{array}\text{\hspace{1em}(1.11)}$$

where the Lagrangian function for the baseball is

$$\mathcal{L}(x(t),\dot{x}(t))=\frac{1}{2}m{\dot{x}}^2(t)-V(x(t))\text{\hspace{1em}(1.12)}$$

and the Euler-Lagrange equation, in the continuum limit, becomes

$$\frac{\partial \mathcal{L}}{\partial x(t)}-\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{x}(t)}=0\text{\hspace{1em}(1.13)}$$

For the Lagrangian of the baseball, the relevant partial derivatives are

$$\begin{array}{rl}\frac{\partial \mathcal{L}}{\partial x(t)}& =\frac{dV(x(t))}{dx(t)}\\ \frac{\partial \mathcal{L}}{\partial \dot{x}(t)}& =m\dot{x}(t)\end{array}\text{\hspace{1em}(1.14)}$$

which, when substituted into (1.13), give

$$m\frac{{\partial }^{2}x}{\partial t^2}+\frac{dV}{dx}=0\text{\hspace{1em}(1.15)}$$

This is just the Newton’s second law in second-order form (1.1) again,

We now want to rewrite the Euler-Lagrange equation in first-order form. Of course, we already know the answer, which is (1.2), but let us “forget” this answer for a moment, in order to introduce a very general method. The reason the Euler-Lagrange equation is second-order in the time derivatives is that $\partial \mathcal{L}/\partial \dot{x}$ is first-order in the time derivative.

Why rewrite it as a pair of first-order equations?

First, a pair of first-order equations is easier to integrate numerically than one second-order equation. But also, quantum mechanics does not describe a system’s position and velocity; it uses its position and momentum. The state space of a classical system is $(x,p)$, not $(x,\dot{x})$, and this space (called phase space) turns out be the right framework for the quantum theory we’re leading to.

We are now going to declare war on $\dot{q}$s. We like the $p_i$s. — My Prof.

That is to say $\dot{x}$ is velocity, which is a derivative — it depends on where the particle is at two nearby moments in time. The momentum $p$ is a property of the state right now. Trading $\dot{x}$ for $p$ as the variable of the the point of the Hamiltonian.

So let us define the momentum corresponding to the coordinate $x$ to be

$$p=\frac{\partial \mathcal{L}}{\partial \dot{x}}\text{\hspace{1em}(1.16)}$$

This gives $p$ as a function of $x$ and $\dot{x}$, but, alternatively, we can solve for $\dot{x}$ as a function of $x$ and $p$, i.e.

$$\dot{x}=\dot{x}(x,p)\text{\hspace{1em}(1.17)}$$

Next, we introduce the Hamiltonian function

$$\mathcal{H}(p,x)=p\dot{x}(x,p)-\mathcal{L}(x,\dot{x}(x,p))\text{\hspace{1em}(1.18)}$$

Since $\dot{x}$ is a function of $x$ and $p$, $\mathcal{H}$ is also a function of $x$ and $p$.

The reason for introducing the Hamiltonian is that its first derivatives with respect to $x$ and $p$ have a remarkable property; namely, on a trajectory satisfying the Euler-Lagrange equations, the $x$ and $p$ derivatives of $\mathcal{H}$ are proportional to the time-derivatives of $p$ and $x$. To see this, first differentiate the Hamiltonian with respect to $p$,

$$\begin{array}{rl}\frac{\partial \mathcal{H}}{\partial p}& =\dot{x}+p\frac{\partial \dot{x}(x,p)}{\partial p}-\frac{\partial \mathcal{L}}{\partial \dot{x}}\frac{\partial \dot{x}(p,x)}{\partial p}\\ & =\dot{x}\end{array}\text{\hspace{1em}(1.19)}$$

where we have applied (1.16). Next, differentiating $\mathcal{H}$ with respect to $x$,

$$\begin{array}{rl}\frac{\partial \mathcal{H}}{\partial x}& =p\frac{\partial \dot{x}(x,p)}{\partial x}-\frac{\partial \mathcal{L}}{\partial x}-\frac{\partial \mathcal{L}}{\partial \dot{x}}\frac{\partial \dot{x}(p,x)}{\partial x}\\ & =-\frac{\partial \mathcal{L}}{\partial x}\end{array}\text{\hspace{1em}(1.20)}$$

Using (1.13) (and this is where the equations of motion enter), we find

$$\begin{array}{rl}\frac{\partial \mathcal{H}}{\partial x}& =-\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{x}}\\ & =-\frac{dp}{dt}\end{array}\text{\hspace{1em}(1.21)}$$

Thus, with the help of the Hamiltonian function, we have rewritten the single 2nd order Euler-Lagrange equation (1.13) as a pair of 1st order differential equations

$$\begin{array}{rl}\frac{dx}{dt}& =\frac{\partial \mathcal{H}}{\partial p}\\ \frac{dp}{dt}& =-\frac{\partial \mathcal{H}}{\partial x}\end{array}\text{\hspace{1em}(1.22)}$$

which are known as Hamilton’s Equations.

Note: When $\mathcal{L}$ has no explicit time dependence —that is to say the laws governing the system don’t change with time—it turns out that $\frac{d\mathcal{H}}{dt}=0$. That is to say the Hamiltonian is conserved. This is “Noether’s theorem” for time-translation symmetry. I mention this because the point is just that $\mathcal{H}$ being the total energy and $\mathcal{H}$ being conserved are the same fact stated two different ways.

For a baseball, the Lagrangian is given by (1.12), and therefore the momentum is

$$p=\frac{\partial \mathcal{L}}{\partial \dot{x}}=m\dot{x}\text{\hspace{1em}(1.23)}$$

This is inverted to give

$$\dot{x}=\dot{x}(p,x)=\frac{p}{m}\text{\hspace{1em}(1.24)}$$

and the Hamiltonian is

$$\begin{array}{rl}\mathcal{H}& =p\dot{x}(x,p)-\mathcal{L}(x,\dot{x}(x,p))\\ & =p\frac{p}{m}-\left[\frac{1}{2}m{\left(\frac{p}{m}\right)}^2-V(x)\right]\\ & =\frac{p^2}{2m}+V(x)\end{array}\text{\hspace{1em}(1.25)}$$

Note that the Hamiltonian for the baseball is simply the kinetic energy plus the potential energy; i.e. the Hamiltonian is an expression for the total energy of the baseball.

Why is $\mathcal{H}=T+V$ when $\mathcal{L}=T-V$?

I asked my Professor this and he straight up said “There are many ways the universe could be constructed that could be satisfying to you, a descendant of primates on earth […] But out here in the real world, where the rubber meets the road […] It’s like they tell kindergarteners; you get what you get and you don’t throw a fit.”

But it’s actually not arbitrary, rather it’s forced by the definition (1.18). If we substitute the baseball $\mathcal{L}$ directly:

$$\mathcal{H}=p\dot{x}-(T-V)=p\dot{x}-T+V$$

For a particle, $T=\frac{1}{2}m{\dot{x}}^2$ and $p=m\dot{x}$, so $p\dot{x}=m{\dot{x}}^2=2T$. Therefore:

$$\mathcal{H}=2T-T+V=T+V$$

The sign flip on $V$ comes from the definition; the $T$ works out because of the specific relation $p\dot{x}=2T$ for kinetic energy.

Substituting $\mathcal{H}$ into (1.22), one finds

$$\begin{array}{rl}\frac{dx}{dt}& =\frac{\partial }{\partial p}\left[\frac{p^2}{2m}+V(x)\right]=\frac{p}{m}\\ \frac{dp}{dt}& =-\frac{\partial }{\partial x}\left[\frac{p^2}{2m}+V(x)\right]=-\frac{dV}{dx}\end{array}\text{\hspace{1em}(1.26)}$$

which is simply the first-order form of Newton’s Law (1.2).

Classical Mechanics in a Nutshell

All the machinery of the Least Action Principle, the Lagrangian Function, and Hamilton’s equations, is overkill in the case of a baseball. In that case, we knew the equation of motion from the beginning. But for more involved dynamical systems, involving, say, wheels, springs, levers, and pendulums, all coupled together in some complicated way, the equations of motion are often far from obvious, and what is needed is some systematic way to derive them.

For any mechanical system, the generalized coordinates $\{q^i\}$ are a set of variables needed to describe the configuration of the system at a given time. These could be a set of cartesian coordinates of a number of different particles, or the angular displacement of a pendulum, or the displacement of a spring from equilibrium, or all of the above. The dynamics of the system, in terms of these coordinates, is given by a Lagrangian function $\mathcal{L}$, which depends on the generalized coordinates $\{q^i\}$ and their first time-derivatives $\{{\dot{q}}^i\}$. Normally, in non-relativistic mechanics, we first specify

The Lagrangian

$$\mathcal{L}(\{q^i\},\{{\dot{q}}^i\})=\text{Kinetic Energy}-\text{Potential Energy}\text{\hspace{1em}(1.27)}$$

One then defines

The Action

$$S=\int dt\mathcal{L}(\{q^i\},\{{\dot{q}}^i\})\text{\hspace{1em}(1.28)}$$

From the Least Action Principle, following a method similar to the one we used for the baseball, we derive

The Euler-Lagrange Equations

$$\frac{\partial \mathcal{L}}{\partial q^i}-\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial {\dot{q}}^i}=0\text{\hspace{1em}(1.29)}$$

These are the 2nd-order equations of motion. To go to 1st-order form, first define

The Generalized Momenta

$$p_i\equiv \frac{\partial \mathcal{L}}{\partial {\dot{q}}^i}\text{\hspace{1em}(1.30)}$$

which can be inverted to give the time-derivatives ${\dot{q}}^i$ of the generalized coordinates in terms of the generalized coordinates and momenta

$${\dot{q}}^i={\dot{q}}^i(\{q^n,p_n\})\text{\hspace{1em}(1.31)}$$

Viewing $\dot{q}$ as a function of $p$ and $q$, one then defines

The Hamiltonian

$$\mathcal{H}(\{q^i,p_i\})\equiv \sum _i{p}_i{\dot{q}}^i-\mathcal{L}(\{q^i,{\dot{q}}^i\})\text{\hspace{1em}(1.32)}$$

Usually the Hamiltonian has the form

$$\mathcal{H}(p,q)=\text{Kinetic Energy}+\text{Potential Energy}\text{\hspace{1em}(1.33)}$$

Finally, the equations of motion in 1st-order form are given by

Hamilton’s Equations

$$\begin{array}{rl}{\dot{q}}^i& =\frac{\partial \mathcal{H}}{\partial p_i}\\ {\dot{p}}_i& =-\frac{\partial \mathcal{H}}{\partial q^i}\end{array}\text{\hspace{1em}(1.34)}$$

Problem Set

Example: The Plane Pendulum

Our pendulum is a mass $m$ at the end of a weightless rigid rod of length $l$, which pivots in a plane around the point P. The “generalized coordinate”, which specifies the position of the pendulum at any given time, is the angle $\theta$.

We get our Lagrangian to be

$$\mathcal{L}=\frac{1}{2}m{l}^2{\dot{\theta }}^2-(V_0-mgl\cos (\theta ))$$

where $V_0$ is the gravitational potential at the height of point P, which the pendulum reaches at $\theta =\pi /2$. Since $V_0$ is arbitrary, we will just set it to $V_0=0$.

Next we get the Action to be $S={\int }_{t_0}^{t_1}dt\left[\frac{1}{2}m{l}^2{\dot{\theta }}^2+mgl\cos (\theta )\right]$. The next step is to derive the Euler-Lagrange Equations since we have $\frac{\partial \mathcal{L}}{\partial \theta }=-mgl\sin (\theta )$ and $\frac{\partial \mathcal{L}}{\partial \dot{\theta }}=m{l}^2\dot{\theta }$, therefore $m{l}^2\ddot{\theta }+mgl\sin (\theta )=0$ is the Euler-Lagrange form of the equations of motion.

Then we need to get The Generalized Momentum, which is $p=\frac{\partial \mathcal{L}}{\partial \dot{\theta }}=m{l}^2\dot{\theta }$. Then to get the Hamiltonian, we insert $\dot{\theta }=\frac{p}{m{l}^2}$ into $\mathcal{H}=p\dot{\theta }-\left[\frac{1}{2}m{l}^2{\dot{\theta }}^2+mgl\cos (\theta )\right]$ to get

$$\mathcal{H}=\frac{1}{2}\frac{p^2}{m{l}^2}-mgl\cos (\theta )$$

Thus, Hamilton’s Equations are

$$\begin{array}{rl}\dot{\theta }& =\frac{\partial \mathcal{H}}{\partial p}=\frac{p}{m{l}^2}\\ \dot{p}& =-\frac{\partial \mathcal{H}}{\partial \theta }=-mgl\sin (\theta )\end{array}$$

which are easily seen to be equivalent to the Euler-Lagrange equations.

Problem

Two pointlike particles moving in three dimensions have masses $m_1$ and $m_2$ respectively, and interact via a potential $V({\vec{x}}_1-{\vec{x}}_2)$. Find Hamilton’s equations of motion for the particles.

We know that $\mathcal{L}=T-V$. $T$ is $\frac{1}{2}{m}_i|{\dot{\vec{x}}}_i{|}^2$, and $V$ is $V({\vec{x}}_1-{\vec{x}}_2)$. Thus

$$\mathcal{L}=\frac{1}{2}{m}_1|{\dot{\vec{x}}}_1{|}^2+\frac{1}{2}{m}_2|{\dot{\vec{x}}}_2{|}^2-V({\vec{x}}_1-{\vec{x}}_2)$$

Now let’s derive our Euler-Lagrange Equations. Since $\frac{\partial \mathcal{L}}{\partial q^i}-\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial {\dot{q}}^i}=0$, for ${\vec{x}}_1$ and ${\vec{x}}_2$,

$$\begin{array}{rlr}\frac{\partial \mathcal{L}}{\partial {\vec{x}}_1}-\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial {\dot{\vec{x}}}_1}=0& & \frac{\partial \mathcal{L}}{\partial {\vec{x}}_2}-\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial {\dot{\vec{x}}}_2}=0\\ -{\nabla }_{{\vec{x}}_1}V({\vec{x}}_1-{\vec{x}}_2)-m_1{\ddot{\vec{x}}}_1=0& & -{\nabla }_{{\vec{x}}_2}V({\vec{x}}_1-{\vec{x}}_2)-m_2{\ddot{\vec{x}}}_2=0\\ m_1{\ddot{\vec{x}}}_1=-{\nabla }_{{\vec{x}}_1}V& & m_2{\ddot{\vec{x}}}_2=-{\nabla }_{{\vec{x}}_2}V\end{array}$$

Wait I’m a moron we totally didn’t need that. Anyways, since

$$\begin{array}{rl}{\dot{q}}^i& =\frac{\partial \mathcal{H}}{\partial p_i}\\ {\dot{p}}_i& =-\frac{\partial \mathcal{H}}{\partial q^i}\end{array}$$

We’ll, need to use

$$\mathcal{H}(\{q^i,p_i\})\equiv \sum _i{p}_i{\dot{q}}^i-\mathcal{L}(\{q^i,{\dot{q}}^i\})$$

Whereby

$$p_i\equiv \frac{\partial \mathcal{L}}{\partial {\dot{q}}^i}$$

So, let’s do this. The conjugate momenta are

$${\vec{p}}_i\equiv \frac{\partial \mathcal{L}}{\partial {\dot{\vec{x}}}_i}$$

So

$$\begin{array}{rlr}{\vec{p}}_1=m_1{\dot{\vec{x}}}_1& & {\vec{p}}_2=m_2{\dot{\vec{x}}}_2\end{array}$$

inverting, ${\dot{\vec{x}}}_i={\vec{p}}_i/m_i$. The Hamiltonian is

$$\begin{array}{rl}\mathcal{H}& ={\vec{p}}_1\cdot {\dot{\vec{x}}}_1+{\vec{p}}_2\cdot {\dot{\vec{x}}}_2-\mathcal{L}\\ & =\frac{|{\vec{p}}_1{|}^2}{m_1}+\frac{|{\vec{p}}_2{|}^2}{m_2}-\frac{|{\vec{p}}_1{|}^2}{2m_1}-\frac{|{\vec{p}}_2{|}^2}{2m_2}+V({\vec{x}}_1-{\vec{x}}_2)\\ & =\frac{|{\vec{p}}_1{|}^2}{2m_1}+\frac{|{\vec{p}}_2{|}^2}{2m_2}+V({\vec{x}}_1-{\vec{x}}_2)\end{array}$$

Hamilton’s equations then give

$$\begin{array}{rlr}{\dot{\vec{x}}}_1=\frac{\partial \mathcal{H}}{\partial {\vec{p}}_1}=\frac{{\vec{p}}_1}{m_1}& & {\dot{\vec{x}}}_2=\frac{\partial \mathcal{H}}{\partial {\vec{p}}_2}=\frac{{\vec{p}}_2}{m_2}\\ {\dot{\vec{p}}}_1=-\frac{\partial \mathcal{H}}{\partial {\vec{x}}_1}=-{\nabla }_{{\vec{x}}_1}V({\vec{x}}_1-{\vec{x}}_2)& & {\dot{\vec{p}}}_2=-\frac{\partial \mathcal{H}}{\partial {\vec{x}}_2}=-{\nabla }_{{\vec{x}}_2}V({\vec{x}}_1-{\vec{x}}_2)\end{array}$$

Note that ${\nabla }_{{\vec{x}}_2}V({\vec{x}}_1-{\vec{x}}_2)=-{\nabla }_{{\vec{x}}_1}V({\vec{x}}_1-{\vec{x}}_2)$, so ${\dot{\vec{p}}}_1=-{\dot{\vec{p}}}_2$. This is newton’s third law.

Problem

Suppose, instead of a rigid rod, the mass of the plane pendulum is connected to point P by a weightless spring. The potential energy of the spring is $\frac{1}{2}k(l-l_0{)}^2$, where $l$ is the length of the spring, and $l_0$ is its length when not displaced by an external force. Choosing $l$ and $\theta$ as the generalized coordinates, find Hamilton’s equations.

TODO: this problem....

The Classical State

Prediction is a rather important thing in physics because the only reliable test of a scientific theory is the ability to predict the future. Stated “rather abstractly,” the process of prediction works as follows: By a slight disturbance (viewing a thing implies that photons are bouncing off of that thing, after all this is what it means for an object to be “illuminated”) known as a measurement, an object is assigned a mathematical representation which we will call its physical state. The laws of motion are mathematical rules by which, given a physical state at a particular time, one can deduce the physical state of some object at a later time.

The later physical state is the prediction, which can be checked by a subsequent measurement of the object. From the discussion so far, its easy to see that what is meant in classical physics and by the “physical state” of a system is simply its set of generalized coordinates and the generalized momenta $\{q^a,p_a\}$. These are supposed to be obtained, at some time $t_0$, by the measurement process. Given the physical state at some time $t$, the state at $t+ϵ$ is obtained by the rule.

$$\begin{array}{rl}{q}^a(t+ϵ)& =q^a(t)+ϵ{\left(\frac{\partial H}{\partial p_a}\right)}_t\\ p_a(t+ϵ)& =p_a(t)-ϵ{\left(\frac{\partial H}{\partial q^a}\right)}_t\end{array}\text{\hspace{1em}(1.35)}$$

In this way, the physical state at any later time can be obtained (in principle) to an arbitrary degree of accuracy, by making the time-step $ϵ$ sufficiently small (or else, if possible, by solving the equations of motion exactly).

Note that the coordinates $\{q^a\}$ are alone are not enough to specify the physical state, because they are not sufficient to predict the future. Information about the momenta $\{p_a\}$ is also required. Blah blah blah the space of all possible $\{q^a,p_a\}$ is known as phase space. For a single particle moving in ${\mathbb{R}}^3$ (three-dimensions), there are three components of position and three components of momentum, so the “physical state” is specified by 6 numbers $(x,y,z,p_x,p_y,p_z)$, which can be viewed as a point in ${\mathbb{R}}^6$ phase space. Likewise, the physical state of a system of $N$ particles consists of $3$ coordinates for each particle ($3N$ coordinates in all), and $3$ components of momentum for each particle ($3N$ momentum components in all), so the state is given by a set of $6N$ numbers, which can be viewed as a single point in ${\mathbb{R}}^{6N}$, that is $6N$-dimensional space.

Classical mechanics fails to predict correctly the behavior of both light and matter at the atomic level and is replaced with quantum mechanics. But classical and quantum mechanics lowkey have a lot in common. They both assign physical states to objects, and these physical states evolve according to 1st-order diff-eqs. The difference between them lies in the contrast between a physical state as understood by classical mechanics, the “classical state”, and its quantum counterpart, the “quantum state”. This difference will be explored in the next new chapters…