Alphabeta Math

1 · Prerequisites

2 · Summary

The companion page to cm-math-canonical records explicit worked checks for its canonical-calculus and Hamilton-Jacobi results. Everything here is mathematics: the examples carry domain: mathematics, depend only on the mathematical items of the parent page and on mathematical suppliers, and make no physical or empirical claim.

The free particle: the complete integral $S = q α - t α^{2} / (2 m)$ , its characteristics and its canonical labels checks the free particle Hamiltonian $H = p^{2} / (2 m)$ : the family $S = q α - t α^{2} / (2 m)$ solves the Hamilton-Jacobi equation, has $S_{q α} = 1$ , produces the characteristics $q (t) = β + t α / m$ with momentum $p = α$ , and satisfies $d q \land d p = d β \land d α$ . The linear shear: the second-kind generating function $S (q, P) = q P + λ q^{2} / 2$ and the matrix test takes the second-kind $S (q, P) = q P + λ q^{2} / 2$ , generates the shear $(q, p) \mapsto (q, p - λ q)$ , and verifies the matrix test $D Ψ^{T} J D Ψ = J$ by direct multiplication. The unit cylinder shift: explicit local generating function, loop integral and global failure specialises the cylinder-shift proposition to $a = 1$ : it records the loop integral $1$ that makes $d θ_{C}$ non-exact, the generating function $S = θ P - θ$ on a chart, and the global failure of any single-valued generating function. The isotropic oscillator: the explicit flow, $Y^{T} J Y = J$ and $\det Y = 1$ computes the flow of $X_{H}$ for $H = (p^{2} + q^{2}) / 2$ , differentiates it to verify $Y^{'} = J H e s s H Y$ , and checks $Y^{T} J Y = J$ and $\det Y = 1$ explicitly.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

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The free particle: the complete integral $S = q α - t α^{2} / (2 m)$ , its characteristics and its canonical labels

Example

Let $m > 0$ and let $H : R \times R \times R \to R$ , $H (t, q, p) : = p^{2} / (2 m)$ , be the free-particle Hamiltonian (The Hamiltonian field of a $C^{2}$ function on a phase domain is $J \nabla H$ , and the Poisson bracket is $\sum_{i} (\partial_{q^{i}} f \partial_{p_{i}} g - \partial_{p_{i}} f \partial_{q^{i}} g)$ ), and let $S : R \times R \times R \to R$ , $S (t, q, α) : = q α - t α^{2} / (2 m)$ . Then:

$S$ solves the Hamilton-Jacobi equation $S_{t} + H (t, q, S_{q}) = 0$ on $R^{3}$ and is a complete integral at every point, with $S_{q α} = 1$ (The Hamilton-Jacobi equation and a complete integral of it);
the Jacobi constants are $β : = S_{α} = q - t α / m$ , so the characteristics are $q (t) = β + t α / m$ with momentum $p = α$ ; these satisfy $q^{'} = p / m$ , $p^{'} = 0$ , and $β$ is constant along them (A complete integral produces Hamilton characteristics, the Jacobi constants, a local converse and canonical $(β, α)$ labels);
on $R^{3}$ at fixed $t$ one has $d q \land d p = d β \land d α$ : the pair $(β, α)$ is canonically conjugate to $(q, p)$ .

Facts & Assumptions

Given: The mass $m > 0$ , the free-particle Hamiltonian $H = p^{2} / (2 m)$ and the function $S = q α - t α^{2} / (2 m)$ on $R^{3}$ .

[F1]

Partial derivatives (Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ): $S_{t} = - α^{2} / (2 m)$ , $S_{q} = α$ , $S_{α} = q - t α / m$ , $S_{q α} = 1$ .

[F2]

Complete integral and characteristics (The Hamilton-Jacobi equation and a complete integral of it, A complete integral produces Hamilton characteristics, the Jacobi constants, a local converse and canonical $(β, α)$ labels): a $C^{3}$ function with $S_{t} + H (t, q, S_{q}) = 0$ and $\det S_{q α} \neq 0$ is a complete integral, and the characteristic equations $q^{'} = H_{p} (t, q, S_{q})$ , $p = S_{q}$ then satisfy Hamilton's equations, with $S_{α}$ constant.

[F3]

Two-form calculus (A phase domain, its canonical one-form and its standard symplectic two-form, The pullback of a differential form, The wedge product of differential forms): on a phase domain with coordinates $(q, p)$ the standard form is $d q \land d p$ , and $d β \land d α$ is computed from the differentials $d q, d α$ by bilinearity and alternation.

Verification

technique · direct; compute the derivatives, check the Hamilton-Jacobi equation and the characteristics, then compare the two two-forms

1.1F1F2algebra

By [F1], $S_{t} + H (t, q, S_{q}) = - α^{2} / (2 m) + α^{2} / (2 m) = 0$ on $R^{3}$ ; since $S_{q α} = 1 \neq 0$ , $S$ is a complete integral at every point by [F2], proving clause 1.

1.2F1F2algebra

Clause 2: by [F1], $β = S_{α} = q - t α / m$ , so for fixed $α$ and $β$ the characteristic curve is $q (t) = β + t α / m$ with $p (t) = S_{q} = α$ ; then $q^{'} = α / m = H_{p} (t, q, p)$ and $p^{'} = 0 = - H_{q}$ , and $β$ is constant along the curve by construction.

2.1F3step 1.2algebra∎

Clause 3: at fixed $t$ , $d β = d q - (t / m) d α$ and $d p = d α$ ; hence $d β \land d α = (d q - (t / m) d α) \land d α = d q \land d α = d q \land d p$ , since $d α \land d α = 0$ .

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The linear shear: the second-kind generating function $S (q, P) = q P + λ q^{2} / 2$ and the matrix test

Example

Let $n = 1$ , let $λ \in R$ , and let $S : R^{2} \to R$ , $S (q, P) : = q P + λ q^{2} / 2$ . Then:

$S$ is a generating function of the second kind with $S_{q P} = 1 \neq 0$ , and the local map it generates is $Ψ_{λ} (q, p) = (Q, P) = (q, p - λ q)$ , defined on all of $R^{2}$ (Generating functions of the first and second kind for a local canonical map);
$Ψ_{λ}$ is canonical, both by the generating-function theorem (The local map generated by a first- or second-kind generating function is a canonical $C^{2}$ local diffeomorphism) and by direct verification of the matrix identity $D Ψ_{λ}^{T} J D Ψ_{λ} = J$ (A $C^{1}$ map between phase domains preserves the standard two-form iff $D C^{T} J D C = J$ , and a canonical diffeomorphism carries $X_{H \circ C}$ to $X_{H}$ ).

Facts & Assumptions

Given: The real number $λ$ and the function $S (q, P) = q P + λ q^{2} / 2$ .

[F1]

Derivatives (Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ): $S_{q} = P + λ q$ and $S_{P} = q$ , so $S_{q P} = 1$ and the second-kind relations $p = S_{q}$ , $Q = S_{P}$ read $p = P + λ q$ , $Q = q$ ; solving, $P = p - λ q$ and $Q = q$ , so the generated map is $Ψ_{λ} (q, p) = (q, p - λ q)$ on all of $R^{2}$ , a $C^{\infty}$ diffeomorphism (Generating functions of the first and second kind for a local canonical map, $C^{k}$ Euclidean maps and diffeomorphisms).

[F2]

Matrix test (A $C^{1}$ map between phase domains preserves the standard two-form iff $D C^{T} J D C = J$ , and a canonical diffeomorphism carries $X_{H \circ C}$ to $X_{H}$ , The standard two-form of a phase domain is closed and nondegenerate, with contraction $ι_{v} ω = \sum_{i} a_{i} d p_{i} - \sum_{i} b_{i} d q^{i}$ , Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes): a $C^{1}$ map $Ψ$ between phase domains is canonical iff $D Ψ^{T} J D Ψ = J$ ; here $J = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ .

Verification

technique · direct; read off the generated map from the defining relations and multiply the two-by-two matrices

1.1F1algebra

Clause 1 is [F1]: $S_{q P} = 1 \neq 0$ , and the relations $p = S_{q} = P + λ q$ , $Q = S_{P} = q$ give $Ψ_{λ} (q, p) = (q, p - λ q)$ , which is a linear isomorphism of $R^{2}$ and therefore a $C^{\infty}$ diffeomorphism.

2.1F1F2step 1.1algebra∎

Clause 2: the derivative is $D Ψ_{λ} = (\begin{array}{cc} 1 & 0 \\ - λ & 1 \end{array})$ , and $(\begin{array}{cc} 1 & - λ \\ 0 & 1 \end{array}) (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}) (\begin{array}{cc} 1 & 0 \\ - λ & 1 \end{array}) = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}) = J$ by direct block multiplication; hence $Ψ_{λ}$ is canonical by [F2], while the generating-function theorem gives the same conclusion because $d S = p d q + Q d P$ and $d^{2} S = 0$ force $d q \land d p = d Q \land d P$ .

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The unit cylinder shift: explicit local generating function, loop integral and global failure

Example

Take the cylinder $C = (R / Z) \times R$ with the shift $Φ (θ, I) = (θ, I + 1)$ , i.e. $a = 1$ in The symplectic shift of the cylinder changes $λ$ by the closed, non-exact form $a d θ$ and has no global generating function, and with the descended forms $λ_{C}, ω_{C}$ of that proposition. Then:

$Φ^{*} λ_{C} = λ_{C} + d θ_{C}$ and $Φ^{*} ω_{C} = ω_{C}$ ;
the loop $γ (t) = π (t, 0)$ , $t \in [0, 1]$ , satisfies $\int_{0}^{1} d θ_{C} (γ^{'} (t)) d t = 1 \neq 0$ , confirming that $d θ_{C}$ is not exact on $C$ ;
on the chart ${θ \in (0, 1)}$ the second-kind generating function $S (θ, P) = θ P - θ$ generates the shift; globally no such $C^{1}$ function exists.

Facts & Assumptions

Given: The cylinder $C$ , the shift $Φ$ , the descended forms and the loop $γ$ .

[F1]

The proposition The symplectic shift of the cylinder changes $λ$ by the closed, non-exact form $a d θ$ and has no global generating function establishes clauses 1 and 2 in general form for $a \neq 0$ ; for $a = 1$ the identities read $Φ^{*} λ_{C} - λ_{C} = d θ_{C}$ and $\int_{0}^{1} d θ_{C} (γ^{'} (t)) d t = 1$ . [F1]

[F2]

The second-kind relations and the local chart (Generating functions of the first and second kind for a local canonical map): for $θ$ in an open interval of length one, $S (θ, P) = θ P - θ$ satisfies $S_{θ} = P - 1$ , $S_{P} = θ$ and $S_{θ P} = 1$ , so it generates the map that sends $(θ, I)$ with $I = P - 1$ to $(Q, P) = (θ, I + 1)$ ; on the chart $S$ is a $C^{\infty}$ function.

Verification

technique · direct; specialise the general identities to $a=1$, compute the loop integral, and check the generating relations on a chart

1.1F1

Clause 1 is [F1] specialised to $a = 1$ : $Φ^{*} λ_{C} = λ_{C} + d θ_{C}$ and $Φ^{*} ω_{C} = ω_{C}$ ; clause 2 is the same proposition's non-exactness statement, whose proof integrates $d θ_{C}$ along $γ$ to $1$ .

2.1F1F2algebra∎

Clause 3: on the chart $θ \in (0, 1)$ the derivatives of [F2] satisfy the second-kind relations $p = S_{θ} = P - 1$ and $Q = S_{P} = θ$ with $S_{θ P} = 1 \neq 0$ ; the generated map is $(θ, I) \mapsto (θ, I + 1)$ , which is the shift, so $S$ generates $Φ$ on the chart. Globally, The symplectic shift of the cylinder changes $λ$ by the closed, non-exact form $a d θ$ and has no global generating function clause 3 (with $a = 1$ ) excludes any global $C^{1}$ generating function: a $1$ -periodic lift $\hat{S}$ would satisfy ${\hat{S}}_{P} = θ + m$ for a constant integer $m$ and ${\hat{S}}_{θ} = P - 1$ , hence $\hat{S} (θ, P) = θ P + m P + c (θ)$ with $c^{'} = - 1$ , so $c (1) - c (0) = - 1 \neq 0$ by the fundamental theorem of calculus, contradicting the periodicity $c (1) = c (0)$ .

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The isotropic oscillator: the explicit flow, $Y^{T} J Y = J$ and $\det Y = 1$

Example

Let $n = 1$ , $U = R^{2}$ with coordinates $(q, p)$ , and $H (q, p) = (p^{2} + q^{2}) / 2$ . Then:

$X_{H} = (p, - q)$ and its flow is $Φ_{t} (q, p) = (q \cos t + p \sin t, - q \sin t + p \cos t)$ , defined for all $t \in R$ ;
the derivative $Y (t) = D Φ_{t} (q, p) = (\begin{array}{cc} \cos t & \sin t \\ - \sin t & \cos t \end{array})$ satisfies $Y^{'} (t) = J H e s s H Y (t)$ , $Y (0) = I_{2}$ , $Y (t)^{T} J Y (t) = J$ and $\det Y (t) = 1$ for all $t$ , in accordance with The derivative of the Hamiltonian flow satisfies $Y^{T} J Y = J$ and $\det Y = 1$ .

Facts & Assumptions

Given: The Hamiltonian $H (q, p) = (p^{2} + q^{2}) / 2$ on $R^{2}$ and the matrix $J = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ .

[F1]

Hamiltonian field (The Hamiltonian field of a $C^{2}$ function on a phase domain is $J \nabla H$ , and the Poisson bracket is $\sum_{i} (\partial_{q^{i}} f \partial_{p_{i}} g - \partial_{p_{i}} f \partial_{q^{i}} g)$ ): $H_{q} = q$ , $H_{p} = p$ , so $X_{H} = (H_{p}, - H_{q}) = (p, - q)$ ; hence the flow solves $q^{'} = p$ , $p^{'} = - q$ .

[F2]

Trigonometric calculus (The derivatives of sine and cosine are cosine and minus sine, The addition formulas for sine and cosine, Sine and cosine defined by their real power series): $\sin^{'} = \cos$ , $\cos^{'} = - \sin$ , and $Φ_{t}$ as displayed is the solution of the linear system with $Φ_{0} = i d$ , since differentiating the displayed expressions gives $q^{'} = - q \sin t + p \cos t = p (t)$ and $p^{'} = - q \cos t - p \sin t = - q (t)$ .

[F3]

The matrix identity (The derivative of the Hamiltonian flow satisfies $Y^{T} J Y = J$ and $\det Y = 1$ , The standard two-form of a phase domain is closed and nondegenerate, with contraction $ι_{v} ω = \sum_{i} a_{i} d p_{i} - \sum_{i} b_{i} d q^{i}$ , The Hessian matrix and critical points of a scalar field): for a $C^{2}$ Hamiltonian the flow derivative satisfies $Y^{'} = J H e s s H Y$ , $Y^{T} J Y = J$ and $\det Y = 1$ , the Hessian being the matrix of second partial derivatives; here $H e s s H = I_{2}$ .

Verification

technique · direct; verify the displayed flow by differentiation and multiply the two-by-two matrices

1.1F1F2algebra

Clause 1: by [F1] the field is $X_{H} = (p, - q)$ and the flow solves the linear system; the displayed $Φ_{t}$ satisfies $Φ_{0} = i d$ and, by [F2], its derivatives are $q^{'} = p (t)$ and $p^{'} = - q (t)$ , so by uniqueness it is the flow, defined for all real $t$ .

1.2F3algebra

Clause 2: differentiating $Y (t)$ entrywise gives $Y^{'} (t) = (\begin{array}{cc} - \sin t & \cos t \\ - \cos t & - \sin t \end{array})$ , while $J H e s s H Y = J Y = (\begin{array}{cc} - \sin t & \cos t \\ - \cos t & - \sin t \end{array})$ ; thus $Y^{'} = J Y$ with $Y (0) = I_{2}$ .

2.1F3step 1.2algebra∎

The orthogonality of $Y (t)$ gives $Y^{T} J Y = J$ by direct multiplication, and $\det Y = \cos^{2} t + \sin^{2} t = 1$ ; hence both conclusions of [F3] hold explicitly for this flow.

Cm Math Canonical — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

The free particle: the complete integral $S = q α - t α^{2} / (2 m)$ , its characteristics and its canonical labels

Example

Facts & Assumptions

Verification

The linear shear: the second-kind generating function $S (q, P) = q P + λ q^{2} / 2$ and the matrix test

Example

Facts & Assumptions

Verification

The unit cylinder shift: explicit local generating function, loop integral and global failure

Example

Facts & Assumptions

Verification

The isotropic oscillator: the explicit flow, $Y^{T} J Y = J$ and $\det Y = 1$

Example

Facts & Assumptions

Verification

Sources

Cm Math Canonical — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

The free particle: the complete integral S=qα−tα2/(2m)S=qα−tα2/(2m), its characteristics and its canonical labels

Example

Facts & Assumptions

Verification

The linear shear: the second-kind generating function S(q,P)=qP+λq2/2S(q,P)=qP+λq2/2 and the matrix test

Example

Facts & Assumptions

Verification

The unit cylinder shift: explicit local generating function, loop integral and global failure

Example

Facts & Assumptions

Verification

The isotropic oscillator: the explicit flow, YTJY=JYTJY=J and det⁡Y=1detY=1

Example

Facts & Assumptions

Verification

Sources

The free particle: the complete integral $S = q α - t α^{2} / (2 m)$ , its characteristics and its canonical labels

The linear shear: the second-kind generating function $S (q, P) = q P + λ q^{2} / 2$ and the matrix test

The isotropic oscillator: the explicit flow, $Y^{T} J Y = J$ and $\det Y = 1$