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I'm studying Landau, Lifshitz - Mechanics. Could someone help me with this problem ? =)

Problem 2 (Page 27 3rd Edition) Determine the period of oscillation, as a function of the energy, when a particle of mass $m$ moves in fields for which the potential energy is

$(b)$ $U=-U_{0}/\cosh^{2}\alpha x$


The period is given by

$ T=4 \sqrt{\frac{m}{2}}\int\frac{dx}{\sqrt{E+\frac{U_{0}}{\cosh^{2}\alpha x}}}$

How can I evaluate this integral? I know that the answer is $T=(\pi/\alpha)\sqrt{2m/|E|}$

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closed as too localized by Manishearth Jan 26 '13 at 8:26


I removed my last comment, because it was incomplete. You should start with two things: 1) determine the domain of integration and 2) rewrite the integrand in the most compact way possible (i.e. use only dimensionless parameters). One way to see that you should not brute-force the integral is that $U_0$ does not appear in the final answer. –  Vibert Jan 24 '13 at 23:56

2 Answers 2

Hints following Vibert's advices:

  1. From energy conservation $$\frac{1}{2}m\dot{x}^2 ~=~ E-U,$$ with symmetric potential $$U ~=~ - \frac{U_0}{c^2},\qquad c~:=~\cosh(\alpha x), \qquad 0<-E<U_0,$$ one gets the quarter period $$\frac{T}{4}~=~ \sqrt{\frac{m}{2}} \int_0^{x_1}\frac{dx}{\sqrt{E-U}},$$ where $x_1>0$ is the upper turning point determined by the condition $E=U(x_1)$.

  2. Prove that $$\int_0^{x_1}\frac{dx}{\sqrt{E-U}}~=~\frac{1}{\alpha\sqrt{|E|}}\int_0^{\sqrt{a}}\frac{ds}{\sqrt{(a-s^2)}}, $$ where $$ s~:=~\sinh(\alpha x),\qquad ~a:=~ -U_0/E-1~>~0.$$

  3. Show that the last integral does not depend on $a>0$. Choose $a=1$, and perform a substitution $s=\sin(t)$.

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Remark: By a symmetric potential is meant an even potential $U(x)=U(-x)$. –  Qmechanic Jan 27 '13 at 11:49

Some idea to evaluate the integral: $$\int\frac{dx}{\sqrt{E+\frac{U_{0}}{\cosh^{2}\alpha x}}}=\int \cosh\alpha x\frac{dx}{\sqrt{E \cosh^{2}\alpha x+U_{0}}} = \int \frac{d(\sinh\alpha x)/\alpha}{\sqrt{(E+U_0) + E \sinh^{2}\alpha }}$$ The last equality use the identity $\cosh^2 x - \sinh^2 x = 1$. Using $y=\sinh \alpha x$, you should then able see it as a standard integral of the form $\int dy/\sqrt{a^2+y^2} = \sinh^{-1}(y/a)$

As what Vibert said, to get the correct answer, you need the correct domain of integration

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