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A period determination problem from Landau and Lifshitz

Posted by peeterjoot on January 23, 2013

A guest post by Alexandre Léonard

[Click here for a PDF of this post with nicer formatting]

Question: A period problem from [1] section 11 (problem 2b)

Determine the period of oscillation, as a function of the energy, when a particle of mass m moves in a field for which the potential energy is

\begin{aligned}U = U_0 \tan^2\alpha x\end{aligned} \hspace{\stretch{1}}(1.0.1)


The answer is:

\begin{aligned}T=\frac{\pi}{\alpha}\sqrt{\frac{2m}{E+U_0}}\end{aligned} \hspace{\stretch{1}}(1.0.2)

My attempt:
I start from the formula

\begin{aligned}T(E) = \sqrt{2m}\int_{x_1(E)}^{x_2(E)}\frac{dx}{\sqrt{E-U(x)}},\end{aligned} \hspace{\stretch{1}}(1.0.3)

where x_1(E) and x_2(E) are the limits of the motion. From the symmetry of our potential, it clear that we have:

\begin{aligned}T(E) = 2\sqrt{2m}\int_{0}^{x_2(E)}\frac{dx}{\sqrt{E-U(x)}}.\end{aligned} \hspace{\stretch{1}}(1.0.4)

Now we can easily find x_2(E):

\begin{aligned}E = U_0\tan^2\alpha x_2 \quad \rightarrow \quad x_2 = \frac{1}{\alpha}\text{arctan}{\sqrt{\frac{E}{U_0}}},\end{aligned} \hspace{\stretch{1}}(1.0.5)

and we are left with the following integral:

\begin{aligned}T(E) = 2\sqrt{2m}\int_{0}^{\frac{1}{\alpha}\text{arctan}{\sqrt{\frac{E}{U_0}}}}\frac{dx}{\sqrt{E-U_0\tan^2\alpha x}}.\end{aligned} \hspace{\stretch{1}}(1.0.6)

First obvious change of variable is y = \alpha x ~ \rightarrow ~ dy = \alpha dx which gives

\begin{aligned}T(E) = \frac{2}{\alpha}\sqrt{2m}\int_{0}^{\text{arctan}{\sqrt{\frac{E}{U_0}}}}\frac{dy}{\sqrt{E-U_0\tan^2 y}}.\end{aligned} \hspace{\stretch{1}}(1.0.7)

Factorizing E under the square root leads to:

\begin{aligned}T(E) = \frac{2}{\alpha}\sqrt{\frac{2m}{E}}\int_{0}^{\text{arctan}{\sqrt{\frac{E}{U_0}}}}\frac{dy}{\sqrt{1 -\frac{U_0}{E}\tan^2 y}}.\end{aligned} \hspace{\stretch{1}}(1.0.8)

Now write a^2=U_0/E and make a change of variables that will eliminate the square root:

\begin{aligned}a\tan y = \sin z.\end{aligned} \hspace{\stretch{1}}(1.0.9)

\begin{aligned}\cos z dz = a \sec^2y dy\end{aligned} \hspace{\stretch{1}}(1.0.10)

\begin{aligned}\int_{0}^{\text{arctan}{\sqrt{\frac{E}{U_0}}}} \frac{dy}{\sqrt{1 -\frac{U_0}{E  }\tan^2 y}} = \frac{1}{a} \int_{0}^{\pi/2} \frac{\cos ^2 y \cos z dz}{\cos z} = a \int_{0}^{\pi/2} \frac{dz}  {a^2 + \sin^2 z} \end{aligned} \hspace{\stretch{1}}(1.0.11x)

where we have used the square of the relation between y and z :

\begin{aligned} a^2 (1 + \cos^2 y) = \sin^2z \cos^2y \quad\Leftrightarrow \quad \frac{a^2}{a^2+\sin^2z} = \cos^2y.\end{aligned} \hspace{\stretch{1}}(1.0.11)

Let us now calculate the definite integral:

\begin{aligned}\int_{0}^{\pi/2} \frac{dx}{a^2 + \sin^2 x}\end{aligned} \hspace{\stretch{1}}(1.0.12)

First method: brute force
Here are the big lines to solve this integral (writing everything would really take some time. The only thing to guess is the proper change of variable. In our case the good one is the following:

\begin{aligned}u = \tan \frac{x}{2} \quad \rightarrow \quad du = \frac{1}{2}(1+u^2) dx\end{aligned} \hspace{\stretch{1}}(1.0.13)

With a bit of not too complicated trigonometry, we obtain:

\begin{aligned}\sin x = \frac{2u}{1+u^2}, \quad \cos x = \frac{1-u^2}{1+u^2}\end{aligned} \hspace{\stretch{1}}(1.0.14)

Injecting these results into the integral yields to:

\begin{aligned}\int_{0}^{\pi/2} \frac{dx}{a^2 + \sin^2 x} = \int_{0}^{1} \frac{2}{(a^2 + \frac{4u^2}{(1+u^2)^2}) (1+u^2)}du = \int_{0}^{1} \frac{2(1+u^2)}{a^2(1+(2+\frac{4}{a^2})u^2 + u^4)}\end{aligned} \hspace{\stretch{1}}(1.0.15)

We have to compute the integral of the ratio between two polynomials, and we just have to follow the recipe from kinder garden. First we find the roots of the denominator, I call them a_+ and a_-:

\begin{aligned}a_+ = \frac{-2-a^2+2\sqrt{1+a}}{a^2}\end{aligned} \hspace{\stretch{1}}(1.0.16)

\begin{aligned}a_- = \frac{-2-a^2-2\sqrt{1+a}}{a^2}\end{aligned} \hspace{\stretch{1}}(1.0.17)

Note that a being positive, a_- and a_+ are negative (not hard to see …)
and we get for the integral:

\begin{aligned}\int_{0}^{1} \frac{2(1+u^2)}{a^2(1+(2+\frac{4}{a^2})u^2 + u^4)} = \frac{2}{a^2}\int_{0}^{1} \frac{(1+u^2)}{(u^2-a_+)(u^2-a_-)}\end{aligned} \hspace{\stretch{1}}(1.0.18)

Rewrite the ratio as:

\begin{aligned}\frac{(1+u^2)}{(u^2-a_+)(u^2-a_-)} = \frac{A}{(u^2-a_+)}+ \frac{B}{(u^2-a_-)}\end{aligned} \hspace{\stretch{1}}(1.0.19)

we find A = \frac{1+a_+}{a_+-a_-} and B=\frac{1+a_-}{a_--a_+}. Injecting these results into the integral, we get:

\begin{aligned}\frac{2}{a^2}\int_{0}^{1} \frac{(1+u^2)}{(u^2-a_+)(u^2-a_-)} = \frac{2}{a^2}\left[ \frac{1+a_+}{a_+-a_-}\int_0^1 \frac{1}{u^2+|a_+|}du + \frac{1+a_+}{a_--a+}\int_0^1\frac{1}{u^2+|a_-|}du\right]\end{aligned} \hspace{\stretch{1}}(1.0.20)

It is now possible, to integrate:

\begin{aligned}\frac{2}{a^2}\int_{0}^{1} \frac{(1+u^2)}{(u^2-a_+)(u^2-a_-)} = \frac{2}{a^2}\left[ \frac{1+a_+}{a_+-a_-}\frac{1}{\sqrt{|a_+|}}\text{arctan} \frac{1}{\sqrt{|a_+|}}+ \frac{1+a_+}{a_--a+}\frac{1}{\sqrt{|a_-|}}\text{arctan} \frac{1}{\sqrt{|a_-|}} \right]\end{aligned} \hspace{\stretch{1}}(1.0.21)

This result is correct, but I wouldn’t put all the pieces together to get the answer even if it must be possible…

Second method: smart but uses deep results from mathematics
This method uses the Residue Theorem. I am not sure it is taught to many students but physicists, mathematicians, maybe engineers and I think that is all. Let me recall the problem: find the value of the definite integral:

\begin{aligned}\int_{0}^{\pi/2} \frac{dx}{a^2 + \sin^2 x}\end{aligned} \hspace{\stretch{1}}(1.0.22)

From the symmetry of the integrand we see that it is equivalent to calculate:

\begin{aligned}\frac{1}{4}\int_{0}^{2\pi} \frac{dx}{a^2 + \sin^2 x}\end{aligned} \hspace{\stretch{1}}(1.0.23)

Let me write the \sin as a sum of complex numbers:

\begin{aligned} \sin x = \frac{e^{ix}-e^{-ix}}{2i}\end{aligned} \hspace{\stretch{1}}(1.0.24)

Setting z=e^{ix} we have dz = izdx. Now, thinking of this integral as a contour integral in the complex plane, we see that the path is just the unit circle: e^{ix} with 0\leq x \leq 2\pi. So we must compute:

\begin{aligned}\frac{1}{4}\int_{0}^{2\pi} \frac{dx}{a^2 + \sin^2 x} =\frac{1}{4} \int_{C} \frac{-i}{a^2 + \left(\frac{z-z^{-1}}{2i}\right)^2} \frac{dz}{z} = i \int_C \frac{z}{z^4-(2+4a^2)z^2+1}dz\end{aligned} \hspace{\stretch{1}}(1.0.25)

Here we set u=z^2. Now we have to loop twice on the unit circle because u=e^{2ix}. But it is the same than looping only once and multiply the result by 2. However, this factor 2 cancels with the factor 2 from: du=2zdz and we get:

\begin{aligned}i \int_C \frac{z}{z^4-(2+4a^2)z^2+1}dz = i \int_C \frac{1}{u^2-(2+4a^2)u+1}du\end{aligned} \hspace{\stretch{1}}(1.0.26)

Here again we look for the roots of the denominators and find out which of those is contained inside the unit circle. The roots are given by:

\begin{aligned}u_+ = 1+2a^2+2a\sqrt{1+a^2}\end{aligned} \hspace{\stretch{1}}(1.0.27)

\begin{aligned}u_- = 1+2a^2-2a\sqrt{1+a^2}\end{aligned} \hspace{\stretch{1}}(1.0.28)

From the expression of u_+ it is clear that u_+ lies on the real axis and is bigger than 1. Now, a>0 and by looking carefully to u_- we can figure out that 0 < u_- < 1.
We have to find the residue of the integrand at the pole (the root) inside the unit circle. This is given by:

\begin{aligned}\text{Res}_{u_-} = \lim_{u\rightarrow u_-} \frac{(u-u_-)}{(u-u_+)(u-u_-)} = \frac{1}{u_- - u_+} = \frac{-1}{4a\sqrt{1+a^2}}\end{aligned} \hspace{\stretch{1}}(1.0.29)

And finally, the residue theorem tells us that the integral is equal to: 2\pi i \sum \text{Res}, where the sum is over all the poles contained inside the path. So we get:

\begin{aligned}\int_{0}^{\pi/2} \frac{dx}{a^2 + \sin^2 x} =i \int_C \frac{1}{u^2-(2+4a^2)u+1}du = i 2\pi i  \frac{-1}{4a\sqrt{1+a^2}} = \frac{\pi}{2a\sqrt{1+a^2}}\end{aligned} \hspace{\stretch{1}}(1.0.30)

Putting everything together leads to the final correct answer:

\begin{aligned}T(E) = \frac{2}{\alpha}\sqrt{\frac{2m}{E}} \sqrt{\frac{U_0}{E}}\frac{\pi}{2\sqrt{\frac{U_0}{E}}\sqrt{1+\frac{U_0}{E}}} =\frac{\pi}{\alpha}\sqrt{\frac{2m}{E+U_0}} \end{aligned} \hspace{\stretch{1}}(1.0.31)


[1] LD Landau and EM Lifshitz. Classical mechanics. 1960.

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