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diff --git a/source/know/concept/hamiltonian-mechanics/index.md b/source/know/concept/hamiltonian-mechanics/index.md new file mode 100644 index 0000000..610f8bd --- /dev/null +++ b/source/know/concept/hamiltonian-mechanics/index.md @@ -0,0 +1,308 @@ +--- +title: "Hamiltonian mechanics" +date: 2021-07-03 +categories: +- Physics +- Classical mechanics +layout: "concept" +--- + +**Hamiltonian mechanics** is an alternative formulation of classical mechanics, +which equivalent to Newton's laws, +but often mathematically advantageous. +It is built on the shoulders of [Lagrangian mechanics](/know/concept/lagrangian-mechanics/), +which is in turn built on [variational calculus](/know/concept/calculus-of-variations/). + + +## Definitions + +In Lagrangian mechanics, use a Lagrangian $L$, +which depends on position $q(t)$ and velocity $\dot{q}(t)$, +to define the momentum $p(t)$ as a derived quantity. +Hamiltonian mechanics switches the roles of $\dot{q}$ and $p$: +the **Hamiltonian** $H$ is a function of $q$ and $p$, +and the velocity $\dot{q}$ is derived from it: + +$$\begin{aligned} + \pdv{L(q, \dot{q})}{\dot{q}} = p + \qquad \quad + \pdv{H(q, p)}{p} \equiv \dot{q} +\end{aligned}$$ + +Conveniently, this switch turns out to be +[Legendre transformation](/know/concept/legendre-transform/): +$H$ is the Legendre transform of $L$, +with $p = \partial L / \partial \dot{q}$ taken as +the coordinate to replace $\dot{q}$. +Therefore: + +$$\begin{aligned} + \boxed{ + H(q, p) \equiv \dot{q} \: p - L(q, \dot{q}) + } +\end{aligned}$$ + +This almost always works, +because $L$ is usually a second-order polynomial of $\dot{q}$, +and thus convex as required for Legendre transformation. +In the above expression, +$\dot{q}$ must be rewritten in terms of $p$ and $q$, +which is trivial, since $p$ is proportional to $\dot{q}$ by definition. + +The Hamiltonian $H$ also has a direct physical meaning: +for a mass $m$, and for $L = T - V$, +it is straightforward to show that $H$ represents the total energy $T + V$: + +$$\begin{aligned} + H + = \dot{q} \: p - L + = m \dot{q}^2 - L + = 2 T - (T - V) + = T + V +\end{aligned}$$ + +Just as Lagrangian mechanics, +Hamiltonian mechanics scales well for large systems. +Its definition is generalized as follows to $N$ objects, +where $p$ is shorthand for $p_1, ..., p_N$: + +$$\begin{aligned} + \boxed{ + H(q, p) + \equiv \bigg( \sum_{n = 1}^N \dot{q}_n \: p_n \bigg) - L(q, \dot{q}) + } +\end{aligned}$$ + +The positions and momenta $(q, p)$ form a phase space, +i.e. they fully describe the state. + +An extremely useful concept in Hamiltonian mechanics +is the **Poisson bracket** (PB), +which is a binary operation on two quantities $A(q, p)$ and $B(q, p)$, +denoted by $\{A, B\}$: + +$$\begin{aligned} + \boxed{ + \{ A, B \} + \equiv \sum_{n = 1}^N \Big( \pdv{A}{q_n} \pdv{B}{p_n} - \pdv{A}{p_n} \pdv{B}{q_n} \Big) + } +\end{aligned}$$ + + +## Canonical equations + +Lagrangian mechanics has a single Euler-Lagrange equation per object, +yielding $N$ second-order equations of motion in total. +In contrast, Hamiltonian mechanics has $2 N$ first-order equations of motion, +known as **Hamilton's canonical equations**: + +$$\begin{aligned} + \boxed{ + - \pdv{H}{q_n} = \dot{p}_n + \qquad + \pdv{H}{p_n} = \dot{q}_n + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-canoneq"/> +<label for="proof-canoneq">Proof</label> +<div class="hidden"> +<label for="proof-canoneq">Proof.</label> +For the first equation, +we differentiate $H$ with respect to $q_n$, +and use the chain rule: + +$$\begin{aligned} + \pdv{H}{q_n} + &= \pdv{}{q_n}\Big( \sum_{j} \dot{q}_j \: p_j - L \Big) + \\ + &= \sum_{j} \bigg( \Big( \dot{q}_j \pdv{p_j}{q_n} + p_j \pdv{\dot{q}_j}{q_n} \Big) + - \Big( \pdv{L}{q_n} + \pdv{L}{\dot{q}_j} \pdv{\dot{q}_j}{q_n} \Big) \bigg) + \\ + &= \sum_{j} \Big( p_j \pdv{\dot{q}_j}{q_n} - \pdv{L}{q_n} - p_j \pdv{\dot{q}_j}{q_n} \Big) + = - \pdv{L}{q_n} +\end{aligned}$$ + +We use the Euler-Lagrange equation here, +leading to the desired equation: + +$$\begin{aligned} + - \pdv{L}{q_n} = - \dv{}{t}\Big( \pdv{L}{\dot{q}_n} \Big) = - \dv{p_n}{t} = - \dot{p}_n +\end{aligned}$$ + +The second equation is somewhat trivial, +since $H$ is defined to satisfy it in the first place. +Nevertheless, we can prove it by brute force, +using the same approach as above: + +$$\begin{aligned} + \pdv{H}{p_n} + &= \pdv{}{p_n}\Big( \sum_{j} \dot{q}_j \: p_j - L \Big) + \\ + &= \sum_{j} \bigg( \Big( \dot{q}_j \pdv{p_j}{p_n} + p_j \pdv{\dot{q}_j}{p_n} \Big) + - \Big( \pdv{L}{q_j} \pdv{q_j}{p_n} + \pdv{L}{\dot{q}_j} \pdv{\dot{q}_j}{p_n} \Big) \bigg) + \\ + &= \dot{q}_n + \sum_{j} \Big( p_j \pdv{\dot{q}_j}{p_n} + - 0 \pdv{L}{q_j} - p_j \pdv{\dot{q}_j}{p_n} \Big) + = \dot{q}_n +\end{aligned}$$ +</div> +</div> + +Just like in Lagrangian mechanics, if $H$ does not explicitly contain $q_n$, +then $q_n$ is called a **cyclic coordinate**, and leads to the conservation of $p_n$: + +$$\begin{aligned} + \dot{p}_n = - \pdv{H}{q_n} = 0 + \quad \implies \quad + p_n = \mathrm{conserved} +\end{aligned}$$ + +Of course, there may be other conserved quantities. +Generally speaking, the $t$-derivative of an arbitrary quantity $A(q, p, t)$ is as follows, +where $\ipdv{}{t}$ is a "soft" derivative +(only affects explicit occurrences of $t$), +and $\idv{}{t}$ is a "hard" derivative +(also affects implicit $t$ inside $q$ and $p$): + +$$\begin{aligned} + \boxed{ + \dv{A}{t} + = \{ A, H \} + \pdv{A}{t} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-diff-t"/> +<label for="proof-diff-t">Proof</label> +<div class="hidden"> +<label for="proof-diff-t">Proof.</label> +We differentiate via the multivariate chain rule, +insert the canonical equations, +and eventually recognize the PB definition: + +$$\begin{aligned} + \dv{A}{t} + &= \sum_{n} \Big( \pdv{A}{q_n} \pdv{q_n}{t} + \pdv{A}{p_n} \pdv{p_n}{t} \Big) + \pdv{A}{t} + \\ + &= \sum_{n} \Big( \pdv{A}{q_n} \dot{q}_n + \pdv{A}{p_n} \dot{p}_n \Big) + \pdv{A}{t} + \\ + &= \sum_{n} \Big( \pdv{A}{q_n} \pdv{H}{p_n} - \pdv{A}{p_n} \pdv{H}{q_n} \Big) + \pdv{A}{t} +\end{aligned}$$ +</div> +</div> + +Assuming that $H$ does not explicitly depend on $t$, +the above property naturally leads us to an alternative +way of writing Hamilton's canonical equations: + +$$\begin{aligned} + \dot{q}_n = \{ q_n, H \} + \qquad \quad + \dot{p}_n = \{ p_n, H \} +\end{aligned}$$ + + + +## Canonical coordinates + +So far, we have assumed that the phase space coordinates $(q, p)$ +are the *positions* and *canonical momenta*, respectively, +and that led us to Hamilton's canonical equations. + +In theory, we could make a transformation of the following general form: + +$$\begin{aligned} + q \to Q(q, p) + \qquad \quad + p \to P(q, p) +\end{aligned}$$ + +However, most choices of $(Q, P)$ would not preserve Hamilton's equations. +Any $(Q, P)$ that do keep this form +are known as **canonical coordinates**, +and the corresponding transformation is a **canonical transformation**. +That is, any $(Q, P)$ that satisfy: + +$$\begin{aligned} + - \pdv{H}{Q_n} = \dot{P}_n + \qquad \quad + \pdv{H}{P_n} = \dot{Q}_n +\end{aligned}$$ + +Then we might as well write $H(q, p)$ as $H(Q, P)$. +So, which $(Q, P)$ fulfill this? +It turns out that the following must be satisfied for all $n, j$, +where $\delta_{nj}$ is the Kronecker delta: + +$$\begin{aligned} + \boxed{ + \{ Q_n, Q_j \} = \{ P_n, P_j \} = 0 + \qquad + \{ Q_n, P_j \} = \delta_{nj} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-cantrans"/> +<label for="proof-cantrans">Proof</label> +<div class="hidden" markdown="1"> +<label for="proof-cantrans">Proof.</label> +Assuming that $Q_n$, $P_n$ and $H$ do not explicitly depend on $t$, +we use our expression for the $t$-derivative of an arbitrary quantity, +and apply the multivariate chain rule to it: + +$$\begin{aligned} + \dot{Q}_n + &= \{Q_n, H\} + = \sum_{n} \bigg( \pdv{Q_n}{q_n} \pdv{H}{p_n} - \pdv{Q_n}{p_n} \pdv{H}{q_n} \bigg) + \\ + &= \sum_{n, j} \bigg( \pdv{Q_n}{q_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{p_n} + \pdv{H}{P_j} \pdv{P_j}{p_n} \Big) + - \pdv{Q_n}{p_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{q_n} + \pdv{H}{P_j} \pdv{P_j}{q_n} \Big) \bigg) + \\ + &= \sum_{n, j} \bigg( \pdv{H}{Q_j} \Big( \pdv{Q_n}{q_n} \pdv{Q_j}{p_n} - \pdv{Q_n}{p_n} \pdv{Q_j}{q_n} \Big) + + \pdv{H}{P_j} \Big( \pdv{Q_n}{q_n} \pdv{P_j}{p_n} - \pdv{Q_n}{p_n} \pdv{P_j}{q_n} \Big) \bigg) + \\ + &= \sum_{j} \bigg( \pdv{H}{Q_j} \{Q_n, Q_j\} + \pdv{H}{P_j} \{Q_n, P_j\} \bigg) +\end{aligned}$$ + +This is equivalent to Hamilton's equation $\dot{Q}_n = \ipdv{H}{P_n}$ +if and only if $\{Q_n, Q_j\} = 0$ for all $n$ and $j$, +and if $\{Q_n, P_j\} = \delta_{nj}$. + +Next, we do the exact same thing with $P_n$ instead of $Q_n$, +giving an analogous result: + +$$\begin{aligned} + \dot{P}_n + &= \{P_n, H\} + = \sum_{n} \bigg( \pdv{P_n}{q_n} \pdv{H}{p_n} - \pdv{P_n}{p_n} \pdv{H}{q_n} \bigg) + \\ + &= \sum_{n, j} \bigg( \pdv{P_n}{q_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{p_n} + \pdv{H}{P_j} \pdv{P_j}{p_n} \Big) + - \pdv{P_n}{p_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{q_n} + \pdv{H}{P_j} \pdv{P_j}{q_n} \Big) \bigg) + \\ + &= \sum_{n, j} \bigg( \pdv{H}{Q_j} \Big( \pdv{P_n}{q_n} \pdv{Q_j}{p_n} - \pdv{P_n}{p_n} \pdv{Q_j}{q_n} \Big) + + \pdv{H}{P_j} \Big( \pdv{P_n}{q_n} \pdv{P_j}{p_n} - \pdv{P_n}{p_n} \pdv{P_j}{q_n} \Big) \bigg) + \\ + &= \sum_{j} \bigg( \pdv{H}{Q_j} \{P_n, Q_j\} + \pdv{H}{P_j} \{P_n, P_j\} \bigg) +\end{aligned}$$ + +Which is equivalent to Hamilton's equation $\dot{P}_n = -\ipdv{H}{Q_n}$ +if and only if $\{P_n, P_j\} = 0$, +and $\{Q_n, P_j\} = - \delta_{nj}$. +The PB is anticommutative, +i.e. $\{A, B\} = - \{B, A\}$. +</div> +</div> + +If you have experience with quantum mechanics, +the latter equation should look suspiciously similar +to the *canonical commutation relation* $[\hat{Q}, \hat{P}] = i \hbar$. + + + +## References +1. R. Shankar, + *Principles of quantum mechanics*, 2nd edition, + Springer. |