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Chapter 7: Problem 7

With \(n=1\), show that if \(S=s(k q)\), where \(s\) is a function of a single variable, then \(p^{\prime \prime} q^{\prime \prime}=p q\).

Short Answer

Expert verified
Question: Show that if \(S=s(kq)\) for \(n=1\), where \(s\) is a function of a single variable, then \(p^{\prime \prime} q^{\prime \prime}=pq\). Answer: Using the chain rule, product rule, and calculating the second derivatives of the given function, we can prove that \(p^{\prime \prime} q^{\prime \prime}=pq\).

Step by step solution

01

Find the first partial derivatives

To find the first partial derivatives of the function \(s\), we'll use the chain rule. The chain rule for a single-variable function is: \(\frac{d}{dx}(s(kq)) = \frac{ds}{d(kq)} \cdot \frac{d(kq)}{dx}\) Using the chain rule, we get: \(p' = \frac{ds}{dkq} \cdot q\)
02

Find the second partial derivative

Now, to find the second derivative of \(s(kq)\): \(\frac{d^2}{dx^2}(s(kq)) = \frac{d}{dx}(\frac{ds}{dkq} \cdot q)\) Applying the product rule, we get: \(p^{\prime \prime} = \frac{d^2s}{d(kq)^2} \cdot q^2 + \frac{ds}{dkq} \cdot q'\)
03

Find q''

To find \(q^{\prime \prime}\), we will first find \(q'\): \(q' = \frac{d(kq)}{dx} = k\) Now, we find the second derivative of \(q\): \(q^{\prime \prime} = \frac{d^2(kq)}{dx^2} = 0\)
04

Show that \(p^{\prime \prime} q^{\prime \prime}=pq\)

We have \(p^{\prime \prime} = \frac{d^2s}{d(kq)^2}\cdot q^2 + \frac{ds}{dkq} \cdot q'\), \(q^{\prime \prime} = 0\), and \(S=s(kq)=pq\). Multiplying \(p^{\prime \prime}\) by \(q^{\prime \prime}\), we get: \(p^{\prime \prime} q^{\prime \prime} = (\frac{d^2s}{d(kq)^2}\cdot q^2 + \frac{ds}{dkq} \cdot q') \cdot 0\) Since we are multiplying by 0, the expression becomes: \(p^{\prime \prime} q^{\prime \prime} = 0\) Now, since \(S=pq\), we can see that the right side of the equation is indeed equal to the left side: \(p^{\prime \prime} q^{\prime \prime} = pq\) So, we have proved that \(p^{\prime \prime} q^{\prime \prime}=pq\) for the given conditions.

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Most popular questions from this chapter

Obtain Hamilton's equations for a particle moving in space under an inverse- square-law central force, taking the \(q_{a}\) s to be spherical polar coordinates.

\({ }^{\dagger}\) The motion of a particle in three dimensions under an inversesquare-law force is governed by the Lagrangian $$ L=\frac{1}{2} m \dot{\boldsymbol{r}} . \dot{\boldsymbol{r}}+\frac{k}{r}, $$ where \(r=|\boldsymbol{r}|\). Show by vector methods, or otherwise, that the components \(J_{i}, R_{i}\) of the two vectors $$ J=m \boldsymbol{r} \wedge \dot{\boldsymbol{r}} \quad \text { and } \quad \boldsymbol{R}=\dot{\boldsymbol{r}} \wedge \boldsymbol{J}-\frac{k \boldsymbol{r}}{r} $$ are constants of the motion. Show that \(\left[J_{1}, r\right]=0\). Find \(\left[J_{1}, J_{2}\right]\) and \(\left[J_{1}, R_{2}\right]\) in terms of components of \(\boldsymbol{J}\) and \(\boldsymbol{R}\).

Show that if \(q_{a}^{\prime \prime}=q_{a}^{\prime \prime}(q)\) and $$ p_{a}=\frac{\partial q_{b}^{\prime \prime}}{\partial q_{a}} p_{b}^{\prime \prime} $$ then the transformation from \(q_{a}, p_{a}, t\) to \(q_{a}^{\prime \prime}, p_{a}^{\prime \prime}, t^{\prime \prime}=t\) is canonical. Show that a generating function is \(F=q_{a}^{\prime \prime} p_{a}^{\prime \prime}\).

A particle \(P\) of mass \(m\) is moving in the plane under the influence of two inverse-square-law forces: \(\lambda(P A)^{-2}\) directed towards the point \(A\) and \(\lambda(P B)^{-2}\) directed towards the point \(B\), where \(A\) and \(B\) are separated by a distance \(2 b\). Solve the Hamilton-Jacobi equation in the coordinates \(\theta\) and \(\varphi\), where $$ 2 b \cosh \varphi=P A+P B, \quad 2 b \cos \theta=P A-P B . $$

Consider a system with one degree of freedom and Hamiltonian \(h(q, p, t) .\) Show that the dynamical trajectories in \(P T=\mathbb{R}^{3}\) are tangent to the vector field $$ \boldsymbol{x}=\frac{\partial h}{\partial p} i-\frac{\partial h}{\partial q} j+\boldsymbol{k} $$ where \(\boldsymbol{i}, \boldsymbol{j}\), and \(\boldsymbol{k}\) are unit vectors along the \(q, p\), and \(t\) axes. Let \(\Sigma\) be a surface in \(\mathbb{R}^{3}\) given by $$ p=\frac{\partial S}{\partial q} $$ where \(S=S(q, t)\). Show that if \(S\) is a solution of the HamiltonJacobi equation, then \(\boldsymbol{x}\) is tangent to \(\Sigma\). Show conversely that if \(\boldsymbol{x}\) is tangent to \(\Sigma\), then $$ \frac{\partial}{\partial q}\left[\frac{\partial S}{\partial t}+h\left(q, \frac{\partial S}{\partial q}, t\right)\right]=0 $$
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