Difference between revisions of "User:Nik"

Revision as of 10:05, 18 January 2015

Foreword

I use \(Q\:/\:P\) for the transformed system instead of \(\widetilde{q}\:/\:\widetilde{p}\) because it's easier to write in Latex.

Problem

Let \( \Phi \in C^\infty(\mathbb{R}^{2n}) \) have the property that the system \( p_i = \frac{\partial}{\partial q_i} \Phi (q, Q) \) has a unique smooth solution \( Q = Q(q,p) \).

Define \( P_i(q,p) = - \frac{\partial}{\partial Q_i} \Phi (q, Q) | _{Q= Q(q,p)} \)

Let \( \{\cdot,\cdot\} \) be the Poisson bracket, such that \( \{f,g\} = \sum_{j=1}^n \frac{\partial f}{\partial q_j} \frac{\partial g}{\partial p_j} - \frac{\partial f}{\partial p_j} \frac{\partial g}{\partial q_j} \)

Show that:

I) \( \{Q_i(q,p), Q_j(q,p)\} = 0 \)

II) \( \{Q_i(q,p), P_j(q,p)\} = \delta_{ij} \)

III) \( \{P_i (q,p), P_j(q,p)\} = 0 \)

Solution

Important equations

$$ \underbrace{\frac{\partial p_i}{\partial q_k}}_{0} = \frac{\partial}{\partial q_k}\left(\frac{\partial \Phi}{\partial q_i}(q, Q)\right) $$ $$ = \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial q_k}}_{\delta_{sk}} + \frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) $$ $$ \Rightarrow 0 = \frac{\partial^2 \Phi}{\partial q_i \partial q_k}(q, Q) + \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) $$ $$ \tag{1} \Rightarrow \boxed{ -\frac{\partial^2 \Phi}{\partial q_i \partial q_k}(q, Q) = \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) } $$

$$ \underbrace{\frac{\partial p_i}{\partial p_k}}_{\delta_{ik}} = \frac{\partial}{\partial p_k}\left(\frac{\partial \Phi}{\partial q_i}(q, Q)\right) $$ $$ = \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial p_k}}_{0} + \frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) $$ $$ \tag{2} \boxed{ \Rightarrow \delta_{ik} = \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) } $$

$$ \frac{\partial P_i}{\partial q_k} = \frac{\partial}{\partial q_k}\left(-\frac{\partial \Phi}{\partial Q_i}(q, Q)\right) $$ $$ = \sum_{s=1}^{n}\left(-\frac{\partial^2 \Phi}{\partial Q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial q_k}}_{\delta_{sk}} - \frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) $$ $$ \tag{3} \boxed{ \Rightarrow \frac{\partial P_i}{\partial q_k} = -\frac{\partial^2 \Phi}{\partial Q_i \partial q_k}(q, Q) - \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) } $$

$$ \frac{\partial P_i}{\partial p_k} = \frac{\partial}{\partial p_k}\left(-\frac{\partial \Phi}{\partial Q_i}(q, Q)\right) $$ $$ = \sum_{s=1}^{n}\left(-\frac{\partial^2 \Phi}{\partial Q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial p_k}}_{0} - \frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) $$ $$ \tag{4} \Rightarrow \boxed{ \frac{\partial P_i}{\partial p_k} = -\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) } $$

Solution I)

This is the most complicated problem but we calculate it first because we use it in II) und III) First we write equations (1) and (2) again an shift our eyeballs between them and look at it as matrix multiplication: $$ \underbrace{-\frac{\partial^2 \Phi}{\partial q_i \partial q_k}(q, Q)}_{D_{ik}} = \sum_{s=1}^{n}\left(\underbrace{\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)}_{A_{is}}\underbrace{\frac{\partial Q_s}{\partial q_k}}_{C_{sk}}\right) $$ $$ \underbrace{\delta_{ik}}_{\mathbb{1}} = \sum_{s=1}^{n}\left(\underbrace{\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)}_{A_{is}}\underbrace{\frac{\partial Q_s}{\partial p_k}}_{B_{sk}}\right) $$ Now let's do some linear algebra: We know that \( \mathbb{1} = AB \Rightarrow A^{-1} = B \text{ and } D = AC \) $$ \Rightarrow A^{-1}D = C $$ $$\Rightarrow BD = C $$ With that we can now calculate C which we need to calculate the Poisson bracket $$ C_{sk} = - \sum_{e=1}^{n}\left(\frac{\partial Q_s}{\partial p_e}\frac{\partial^2 \Phi}{\partial q_k \partial q_e}\right) $$ Now we change the indices such that it fits into the equation $$ \frac{\partial Q_i}{\partial q_k} = - \sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial p_s}\frac{\partial^2 \Phi}{\partial q_k \partial q_s}\right) $$ Finally we are prepared to calculate the Poisson bracket $$ \{Q_i, Q_j\} = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\frac{\partial Q_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial Q_j}{\partial q_k}\right) $$ Put in the sum from above $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_j}{\partial p_k}\left(-\sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial p_s}\frac{\partial^2 \Phi}{\partial q_k \partial q_s}(q, Q)\right)\right) - \frac{\partial Q_i}{\partial p_k}\left(-\sum_{s=1}^{n}\left(\frac{\partial Q_j}{\partial p_s}\frac{\partial^2 \Phi}{\partial q_k \partial q_s}(q, Q)\right)\right)\right) $$ Take the sum out $$ \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial Q_j}{\partial p_s}\frac{\partial^2 \Phi}{\partial q_k \partial q_s}(q, Q) - \frac{\partial Q_j}{\partial p_k}\frac{\partial Q_i}{\partial p_s}\frac{\partial^2 \Phi}{\partial q_k \partial q_s}(q, Q) \right) = 0 $$ Since we sum with \(s\) and \(k\) completely from \(1\) to \(n\) the two parts are equal and therefore the difference is zero.

Solution II)

$$ \{Q_i, P_j\} = \sum_{k=1}^{n}\left( \frac{\partial Q_i}{\partial q_k}\frac{\partial P_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial P_j}{\partial q_k} \right) $$ Now we put in the results from equation (3) and (4) $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\left(-\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right)\right) - \frac{\partial Q_i}{\partial p_k}\left(-\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q) - \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right)\right)\right) $$ Now we multiply out and separate all the sums. We can rearrange it like this because all the parts we put inside are independent on the running variable. $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right) + \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) - \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) $$ Now we rearrange again such that we can build a new Poisson bracket $$ = \underbrace{\sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right)}_{\color{red}{\delta_{ij}}} + \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q) \underbrace{\sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial Q_s}{\partial q_k} - \frac{\partial Q_i}{\partial q_k}\frac{\partial Q_s}{\partial p_k}\right)}_{=\{Q_s, Q_i\}=0}\right) $$ Now we use the definition from the problem and then the chain rule $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial p_k}{\partial Q_j} \right) = \frac{\partial Q_i}{\partial Q_j} = \delta_{ij} $$ The chain rule is here not correct. It would imply that \( Q \) is independent of \( q \) and that is not correct. Since I got no better solution I would write \( \delta_{ij} \) directly in the sum above.

Solution III)

$$ \{P_i, P_j\} = \sum_{k=1}^{n}\left(\frac{\partial P_i}{\partial q_k}\frac{\partial P_j}{\partial p_k} - \frac{\partial P_i}{\partial p_k}\frac{\partial P_j}{\partial q_k}\right) $$ Now we put in the results from equation (3) and (4) in all 4 parts of the sum $$ = \sum_{k=1}^{n}\left(\left(-\frac{\partial^2 \Phi}{\partial Q_i \partial q_k}(q, Q) - \underbrace{\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right)}_{(a)}\right)\left(\underbrace{-\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right)}_{(b)}\right) \\ - \left(\underbrace{-\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right)}_{(c)}\right)\left(-\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q) - \underbrace{\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right)}_{(d)}\right)\right) $$ If we multiply out we see that (a) * (b) = (c) * (d) and because of the sign they cancel out. So we multiply out the remaining parts $$ = \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_i \partial q_k}(q, Q)\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k} - \frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right) $$ And use the definition from the problem set $$ = \sum_{k=1}^{n}\sum_{s=1}^{n}\left(-\frac{\partial^2 \Phi}{\partial Q_i \partial q_k}(q, Q)\underbrace{\frac{\partial P_s}{\partial Q_j}}_{=0}\frac{\partial Q_s}{\partial p_k} + \underbrace{\frac{\partial P_s}{\partial Q_i}}_{=0}\frac{\partial Q_s}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right) $$

$$ = 0 $$

Alternative Solution

Alternative solution:

Here's a proof I've found on my usual source of knowledge (meaning a youtube-lecture):

Claim: For a generating function of type one (that's the smartman's name for the Phi in our problem) it holds: \( \{ Q, P \} = 1 \).

Proof:

We first fix the \( q \) coordinates in the momentum equation and derive by \( p \), denoted: \( \Phi_p |_q\).

So we get:

Equation 1:

$$ 1 = \Phi_{q Q} Q_p |_q \Rightarrow Q_p |_q = \frac{1}{\Phi_{q Q}} $$

Equation 2:

Holding \( p \) fixed and derive by \( q \):

$$ 0 = \frac{\partial p}{\partial q} = \Phi_{qq} + \Phi_{qQ}Q_{q}|_p \Rightarrow Q_{q}|_p = - \frac{\Phi_{qq}}{\Phi_{qQ}} $$

Equation 3:

Holding \( q \) fixed and derive by \( p \):

$$ P_p |_q = - \Phi_{QQ}Q_p |_q \Rightarrow P_p |_q = - \frac{\Phi_{QQ}}{\Phi_{qQ}} $$

where we used equation 1 again.

Equation 4:

Holding \( p \) fixed and derive by \( q \):

$$ P_q |_p = - \Phi_{Qq} - \Phi_{QQ} Q_q |_p \Rightarrow P_q |_p = - \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} $$

using equation 2.

We are now prepared to calculate:

$$ \{ Q, P \} = Q_qP_p - Q_pP_q = \frac{\Phi_{QQ}}{\Phi_{qQ}} \frac{\Phi_{qq}}{\Phi_{qQ}} - \frac{1}{\Phi_{q Q}} \left( - \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} \right) = 1 $$ by just cancelling out the terms. \( \square \)

( My first idea would have been:

By the script:

$$ \{ Q, P \} = \langle \nabla Q , J \nabla P \rangle_{\mathbb{R}^{2n}} = 1 $$

with \( \nabla = \left( \frac{\partial}{\partial q_1}, ..., \frac{\partial}{\partial p_n} \right) \) and

$$ J = \begin{bmatrix} 0 & I_n \\ -I_n & 0 \\ \end{bmatrix} $$

and this should result in the things we want to show (by some symplectic argument or so...).)

But it might be better to just show this (the property is surely right, found it in a classical mechanics textbook):

\( \vdash : \{ f, g \}_{q, p} = \{ Q, P \}_{q, p} \{ f, g \}_{Q, P} \)

Proof

With simplified notation:

\( \{ f, g \}_{p, q} = \sum_k \left( \frac{\partial f}{\partial q_k} \frac{\partial g}{\partial p_k} - \frac{\partial f}{\partial p_k} \frac{\partial g}{\partial q_k} \right) = \sum_k \left( \frac{\partial f}{\partial Q} \frac{\partial Q}{\partial q_k} + \frac{\partial f}{\partial P} \frac{\partial P}{\partial q_k} \right) \left( \frac{\partial g}{\partial Q} \frac{\partial Q}{\partial p_k} + \frac{\partial g}{\partial P} \frac{\partial P}{\partial p_k} \right) - \sum_k \left( \frac{\partial f}{\partial Q} \frac{\partial Q}{\partial p_k} + \frac{\partial f}{\partial P} \frac{\partial P}{\partial p_k} \right) \) \(\left( \frac{\partial g}{\partial Q} \frac{\partial Q}{\partial q_k} + \frac{\partial g}{\partial P} \frac{\partial P}{\partial q_k} \right) \)

\( = \sum_k \left( \frac{\partial Q}{\partial q_k} \frac{\partial P}{\partial p_k} - \frac{\partial Q}{\partial p_k} \frac{\partial P}{\partial q_k} \right) \left( \frac{\partial f}{\partial Q} \frac{\partial g}{\partial P} - \frac{\partial f}{\partial P} \frac{\partial g}{\partial Q} \right) = \{ Q, P \}_{q, p} \cdot \{ f, g \}_{Q, P} \)

Kicking out the ambiguity of the last term: By the script: \( \{ f, g \}_{Q, P} = \langle \nabla_{Q, P} \cdot f , J \nabla_{Q, P} \cdot g \rangle_{\mathbb{R}^{2n}} = \sum_k \left( \frac{\partial f}{\partial Q_k} \frac{\partial g}{\partial P_k} - \frac{\partial f}{\partial P_k} \frac{\partial g}{\partial Q_k} \right) \)

with \( \nabla_{Q,P} \) \( = \left( \nabla Q, \nabla P \right) \) \(= \left( \frac{\partial}{\partial Q_1}, ..., \frac{\partial}{\partial P_n} \right) \) and

\( J = \begin{bmatrix} 0 & I_n \\ -I_n & 0 \\ \end{bmatrix} \).

\(\square\)

Now compute: \( \{ Q, P \}_{q, p} \cdot \{ Q_i, Q_j \}_{Q, P} = \{ Q, P \}_{q, p} \cdot \{ P_i, P_j \}_{Q, P} = 0 \) which follows from the properties of the Poisson-bracket.

Furthermore: \( \{ Q, P \}_{q, p} \cdot \{ Q_i, P_j \}_{Q, P} = \{ Q, P \}_{q, p} \cdot \delta_{i j} = \delta_{i j} \)

which solves the problem.