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Chapter 8: Problem 4

Show that the Poisson bracket of a product can be calculated by Leibniz's rule: $$ \left(F_{1} F_{2}, H\right)=F_{1}\left(F_{2}, H\right)+F_{2}\left(F_{1}, H\right) . $$

Short Answer

Expert verified
Question: Prove the Poisson bracket of a product of two functions, F1 and F2, using Leibniz's rule. Answer: By applying Leibniz's rule for differentiation in combination with the definition of the Poisson bracket, we derived the following result: $$ \{F_1F_2, H\} = F_1\{F_2, H\} + F_2\{F_1, H\} $$ This result demonstrates that the Poisson bracket of a product can be calculated using Leibniz's rule.

Step by step solution

01

Recall the Definition of the Poisson Bracket and Leibniz's Rule

In classical mechanics, the Poisson bracket of two functions, F1 and F2, with respect to generalized coordinates (\(q_i\)) and momenta (\(p_i\)) is defined as follows: $$ \{F_1, F_2\} = \sum_{i=1}^n \left( \frac{\partial F_1}{\partial q_i} \frac{\partial F_2}{\partial p_i} - \frac{\partial F_1}{\partial p_i} \frac{\partial F_2}{\partial q_i}\right) $$ Leibniz's rule states that the derivative of the product of two functions is given by: $$ \frac{d(F_1F_2)}{dx} = F_1\frac{dF_2}{dx} + F_2\frac{dF_1}{dx} $$
02

Compute the Poisson Bracket of F1F2 and H

Let's compute the Poisson bracket of the product (F1F2) and H. We will first differentiate F1F2 with respect to the generalized coordinates \(q_i\) and momenta \(p_i\), and then apply the definition of the Poisson bracket to these derivatives. $$ \begin{aligned} \{F_1F_2, H\} &= \sum_{i=1}^n \left( \frac{\partial (F_1F_2)}{\partial q_i} \frac{\partial H}{\partial p_i} - \frac{\partial (F_1F_2)}{\partial p_i} \frac{\partial H}{\partial q_i}\right) \\ &= \sum_{i=1}^n \left( \left(F_1\frac{\partial F_2}{\partial q_i} + F_2\frac{\partial F_1}{\partial q_i}\right) \frac{\partial H}{\partial p_i} - \left(F_1\frac{\partial F_2}{\partial p_i} + F_2\frac{\partial F_1}{\partial p_i}\right) \frac{\partial H}{\partial q_i}\right) \end{aligned} $$ We applied Leibniz's rule for differentiation to obtain these expressions.
03

Apply the Definition of the Poisson Bracket to Simplify the Expression

Now we apply the definition of the Poisson bracket again to simplify the expression we obtained in Step 2: $$ \begin{aligned} \{F_1F_2, H\} &= \sum_{i=1}^n \left( F_1\frac{\partial F_2}{\partial q_i}\frac{\partial H}{\partial p_i} + F_2\frac{\partial F_1}{\partial q_i}\frac{\partial H}{\partial p_i} - F_1\frac{\partial F_2}{\partial p_i}\frac{\partial H}{\partial q_i} - F_2\frac{\partial F_1}{\partial p_i}\frac{\partial H}{\partial q_i}\right) \\ &= \sum_{i=1}^n F_1\left(\frac{\partial F_2}{\partial q_i}\frac{\partial H}{\partial p_i} - \frac{\partial F_2}{\partial p_i}\frac{\partial H}{\partial q_i}\right) + \sum_{i=1}^n F_2\left(\frac{\partial F_1}{\partial q_i}\frac{\partial H}{\partial p_i} - \frac{\partial F_1}{\partial p_i}\frac{\partial H}{\partial q_i}\right) \\ &= F_1\{F_2, H\} + F_2\{F_1, H\} \end{aligned} $$
04

Conclusion

We have shown that the Poisson bracket of a product can be calculated using Leibniz's rule: $$ \{F_1F_2, H\} = F_1\{F_2, H\} + F_2\{F_1, H\} $$

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Most popular questions from this chapter

Show that if at least one of the eigenvalues of a symplectic transformation \(S\) does not lie on the unit circle, then \(S\) is unstable.

Show that if all the eigenvalues of a linear transformation are distinct and lie on the unit circle, then the transformation is stable.

Show that the map \(A: \mathbb{R}^{2 n} \rightarrow \mathbb{R}^{2 n}\) sending \((\mathbf{p}, \mathbf{q}) \rightarrow(\mathbf{P}(\mathbf{p}, \mathbf{q}), \mathbf{Q}(\mathbf{p}, \mathbf{q}))\) is canonical if and only if the Poisson brackets of any two functions in the variables \((\mathbf{p}, \mathbf{q})\) and \((\mathbf{P}, \mathbf{Q})\) coincide: $$ (F, H)_{\mathbf{p}, \mathbf{q}}=\frac{\partial H}{\partial \mathbf{p}} \frac{\partial F}{\partial \mathbf{q}}-\frac{\partial H}{\partial \mathbf{q}} \frac{\partial F}{\partial \mathbf{p}}=\frac{\partial H}{\partial \mathbf{P}} \frac{\partial F}{\partial \mathbf{Q}}-\frac{\partial H}{\partial \mathbf{Q}} \frac{\partial F}{\partial \mathbf{P}}=(F, H)_{\mathbf{p}, \mathbf{q}} $$ Solution. Let \(A\) be canonical. Then the symplectic structures \(d \mathbf{p} \wedge d \mathbf{q}\) and \(d \mathbf{P} \wedge d \mathbf{Q}\) coincide. But the definition of the Poisson bracket \((F, H)\) was given invariantly in terms of the symplectic structure; it did not involve the coordinates. Therefore, $$ (F, H)_{\mathbf{p}, \mathbf{q}}=(F, H)=(F, H)_{\mathbf{P}, \mathbf{Q}} . $$ Conversely, suppose that the Poisson brackets \(\left(P_{i}, Q_{j}\right)_{\mathbf{p}, \mathbf{q}}\) have the standard form of Problem \(2 .\) Then, clearly, \(d \mathbf{P} \wedge d \mathbf{Q}=d \mathbf{p} \wedge d \mathbf{q}\), i.e., the map \(A\) is canonical.

Verify that \(\left(\mathbb{R}^{2 n}, \omega^{2}\right)\) is a symplectic manifold. For \(n=1\) the pair \(\left(\mathbb{R}^{2}, \omega^{2}\right)\) is the pair (the plane, area).

Let \(\pi_{1}\) and \(\pi_{2}\) be two \(k\)-dimensional planes in symplectic \(\mathbb{R}^{2 n}\). Is it always possible to carry \(\pi_{1}\) to \(\pi_{2}\) by a symplectic transformation? How many classes of planes are there which cannot be carried one into another?
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