Spherical Pendulum - elliptic integral

[szczepan]

Hello everyone. In the 3rd edition of Mechanics by Landau and Lifshitz, paragraph 14, there is a problem concerning spherical pendulum. Calculations leading to the integral $$ t=\int \frac {d \Theta} {\sqrt{\frac{2}{ml^2}[E-U_{ef}(\Theta)]}},$$ $$ U_{ef}(\Theta)=\frac{M_z^2}{2ml^2sin^2\Theta}-mglcos\Theta, $$ are straightforward. The problem is that the forementioned integral should lead to the elliptic integral of the first kind, which is: $$
\int \frac{d \Theta}{\sqrt{1-k^2 sin^2 \Theta}}. $$ By factoring out the sine, i managed to obtain
$$ \frac{ml^2}{M_z} \int - \frac {sin \Theta d \Theta}{\sqrt{1- k^2sin^2\Theta}}, \ \ k= \sqrt{\frac{(E+mgl cos \Theta)2ml^2}{M_z^2}}. $$ As you can see, the numerator is wrong, and $k$ contains $\Theta$, so it's not what it should be. I've also tried juggling with some substitutions and trigonometric identities, but it just complicated the equations and the outcomes weren't even close to the elliptic integral form. Any ideas on how to do it?

 

[Twigg]

I'm pretty sure you misread the text on the effective potential. I looked at the print for problem 1 under paragraph 14 in the third edition, and what I see is $$U_{ef}(\Theta) = \frac{M_{z}^{2}}{2ml^{2}}sin^{2}\Theta - mglcos\Theta$$ They could've been a bit more clear with some parentheses.

 

[szczepan]

That's what I thought for a moment. I'm actually using the polish translation of the book, and this expression for effective potential energy is written as a fraction, just as in my first post. In the english version (e.g. this pdf: http://renaissance.ucsd.edu/courses/mae207/mech.pdf) it appears to be
$$ U_{ef}(\Theta)=\frac{1}{2}M_z^{2'}ml^2sin^2\Theta-mglcos\Theta. $$ But I'm almost sure that this $" \ ' \ "$ next to $M_z^2$ is really a poorly printed $" \ / \ "$. To really convince you, here's the derivation. In the Lagrange function for this spherical pendulum, we have a term $$ \frac{ml^2}{2} \dot{\phi}^2sin^2\Theta .$$ Then, we see that $$ ml^2 \dot{\phi} sin^2 \Theta = M_z, $$ from which we have $$ \dot{\phi}^2=\frac{M_z^2}{m^2l^4sin^4\Theta}. $$ So when we insert $\dot{\phi}^2$ into the Lagrange function, the effective potential energy will have $sin^2\Theta$ in the denominator, so I think the question still stands, unless I'm completely wrong.

 

[mpresic]

I think part of your confusion may be caused by using the same symbol theta in the definition of the elliptic integral and associating it with the theta in Landau and Liv. For example, multiply the numerator and denominator in the definition of the elliptic integral by cos theta. Then you will transform the definition of the elliptic integral with u = cos theta. The numerator becomes du. The denominator is the product of square root of 1-u squared and a square root of 1 - k squared u squared.

You now have the definition of the elliptic integral in terms of u. Now try to transform your equation for the spherical pendulum in such a way that you get a square root in the denominator which will be a product of 1-u squared times 1 - k squared u squared.

 

[mpresic]

Now I am thinking my last suggestion may not help

 

[szczepan]

I know that $\Theta$ in the definition can be whatever I like. As I said I tried various substitutions, even weird ones like $\frac{cos\Theta}{sin\Theta}=cos\phi$, or similar to those which where used by Landau in previous problems concerning elliptic integrals. I just can't seem to find the right one.

 

[mpresic]

I think I can help but Hoo Boy is it complicated. The expression within the integral is a cubic equation, after you perform a suitable transformation u = cosine theta in your integral. Now you go to Abramowitz and Stegun (for example) and look st their table 17.4.61 on page 597. That table shows 12 different forms for your integral that all have solutions in the form of a elliptic function of the first kind (called F in their notation). find the specific form necessary for your cubic equation, you need to determine the roots to the cubic equation and order them appropriately. This will be extraordinarily complicated to do and would probably require a computer doing symbolic algebra for the general solution. At least with Abramowitz and Stegun table 17.4.61 and knowint the integral shows a cubic polynomial in the square root in the denominator, the answer will be an elliptic function of the first kind.

 

[vanhees71]

I have a very short section on it in my mechanics lecture notes (in German). Maybe it helps:

http://theory.gsi.de/~vanhees/faq/mech/node67.html
http://theory.gsi.de/~vanhees/faq-pdf/mech.pdf

 

[szczepan]

Thank you for your answers. As I understood from equations only ( those fascinating long words in German are amazing but I can understand only a few words :) ), there is a way to solve the spherical pendulum problem without using the elliptic integrals. I think the table from Abramowitz and Stegun will be helpful, I was able to reduce the problem to a 3rd degree polynomial in the denominator, but I couldn't find a proper integral. Again thank you, I think the thread can be closed now.

 

[vanhees71]

...but in the end the integrals are precisely elliptic integrals of various kinds. They are pretty common in all kinds of pendula.

 

[Nohora Hernandez]

I think I can help but Hoo Boy is it complicated. The expression within the integral is a cubic equation, after you perform a suitable transformation u = cosine theta in your integral. Now you go to Abramowitz and Stegun (for example) and look st their table 17.4.61 on page 597. That table shows 12 different forms for your integral that all have solutions in the form of a elliptic function of the first kind (called F in their notation). find the specific form necessary for your cubic equation, you need to determine the roots to the cubic equation and order them appropriately. This will be extraordinarily complicated to do and would probably require a computer doing symbolic algebra for the general solution. At least with Abramowitz and Stegun table 17.4.61 and knowint the integral shows a cubic polynomial in the square root in the denominator, the answer will be an elliptic function of the first kind.

Hello. I follow your instruction and I found the following integral

[t-t_o=-l\sqrt{\frac{m}{2}} \int \frac{du}{\sqrt{(1-u^2)(E+mglu)-\frac{M_z^2}{2ml^2}}}\]

After of that I was checking the book that you said, nontheless its not clear for me how this integral has the same form that you mentioned? or perhaps, could you give me another advice to obtain the form as in the book says.

 

[TOAsh2004]

Hello together,

I don't know if anybody is still interested in this thread but i recently did these calculations myself, inspired by the small scene at the blackboard with the elliptic integrals in the film "The man who knew infinity". I stumbled across this thread and wanted to share my solution with you.

Greetings,

TOAsh2004