13.6: Problems
- Page ID
- 19025
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)13.1. Generalized momentum is defined as
\[\mathbb{M} = \frac{\partial \mathcal{L}}{\partial \left(\frac{dV}{dr}\right)}.\nonumber \]
(a) Find the generalized momentum for the system described by the Lagrangian of Equation 13.3.51.
(b) The generalized momentum does not have the units of momentum. Identify the units of this generalized momentum.
(c) Write the Hamiltonian of Equation 13.3.50 as a function of \(r\), \(V\), and \(\mathbb{M}\) but not as a function of \(\frac{dV}{dr}\).
(d) Write the Lagrangian of Equation 13.3.51 as a function of \(r\), \(V\), and \(\mathbb{M}\) but not as a function of dV dr .
(e) Show that the Hamiltonian and Lagrangian found above satisfy the equation \(H = \mathbb{M} \frac{dV}{dr} − \mathcal{L}\).
13.2. In the analysis of this chapter, the generalized path was chosen as \(V\) and the generalized potential was chosen as \(\rho_{ch}\). The opposite choice is also possible where the generalized path is \(\rho_{ch}\) and the generalized potential is \(V\).
(a) Write the Hamiltonian of Equation 13.3.50 as functions of \(\rho_{ch}\) instead of \(V\), so it has the form \(H (r, \rho_{ch}, \frac{d\rho_{ch}}{dr})\).
(b) Repeat the above for the Lagrangian of Equation 13.3.51.
(c) Find the Euler-Lagrange equation using \(\rho_{ch}\) as the generalized path.
13.3. Verify that \[y = \frac{144}{\mathrm{t}^3}\nonumber \] is a solution of the Thomas Fermi equation [46].
(While this solution satisfies the Thomas Fermi equation, it is not useful in describing the energy of an atom. In the \(\mathrm{t} \rightarrow 0\) limit, this solution approaches infinity, \(y(0) \rightarrow \infty\). However, in the \(\mathrm{t} \rightarrow 0\) limit, the solution should approach a constant, \(y(0) \rightarrow 1\), to correctly describe the physical behavior of an atom [180].)
13.4. The previous problem discussed that \[y = \frac{144}{\mathrm{t}^3}\nonumber \] is a solution of the Thomas Fermi equation. Show that \[y = \frac{72}{\mathrm{t}^3}\nonumber \] is not a solution.
13.5. Prove that the Thomas Fermi equation is nonlinear.