# Re: Susskind and Hrabovsky: Should the Lagrangian of a particle referred to a rotating frame omit the velocity dependent “potential”?

My question pertains to Lecture 6: Exercise 4 in The Theoretical Minimum by Leonard Susskind and George Hrabovsky. A suggested solution has been posted here: http://www.madscitech.org/tm/slns/

The problem is this: In a rotating frame find the Lagrangian and Euler-Lagrange equations of a particle subject to no (real) force in polar coordinates.

Previously the book introduced the Lagrangian and Euler-Lagrange equations for a particle which is unaccelerated in an inertial frame, using rectangular Cartesian coordinates of a reference frame rotating with constant angular velocity, $\omega.$ I have briefly reproduced that derivation using my own notation.

The origins of all coordinate systems are at the center of rotation. I have used complex numbers and a different sense of rotation (because I believe the book's choice is non-standard).

The inertial coordinates of the particle are

$$\mathfrak{r}=re^{i\rho}=x+iy.$$

The coordinates fixed in the rotating frame are

$$\overline{\mathfrak{r}}=re^{i\overline{\rho}}=\overline{x}+i\overline{y},$$

and their time derivatives are

$$\dot{\overline{\mathfrak{r}}}=\left(\dot{r}+i\dot{\overline{\rho}}r\right)e^{i\overline{\rho}}=\dot{\overline{x}}+i\dot{\overline{y}}.$$

These coordinate systems are related by

$$\mathfrak{r}=\overline{\mathfrak{r}}e^{i\omega t}=re^{i\left(\overline{\rho}+\omega t\right)}.$$

Since there is no (real) force, the Lagrangian in the inertial system is simply the kinetic energy:

$$L=\frac{m}{2}\dot{\mathfrak{r}}\dot{\mathfrak{r}}^{*}.$$

The square of the magnitude of the velocity transforms as follows:

$$\dot{\mathfrak{r}}\dot{\mathfrak{r}}^{*}=\left(\dot{\overline{\mathfrak{r}}}+i\omega\overline{\mathfrak{r}}\right)e^{i\omega t}\left(\dot{\overline{\mathfrak{r}}}^{*}-i\omega\overline{\mathfrak{r}}^{*}\right)e^{-i\omega t}$$

$$=\dot{\overline{\mathfrak{r}}}\dot{\overline{\mathfrak{r}}}^{*}+\omega^{2}\overline{\mathfrak{r}}\overline{\mathfrak{r}}^{*}+i\omega\left(\overline{\mathfrak{r}}\dot{\overline{\mathfrak{r}}}^{*}-\dot{\overline{\mathfrak{r}}}\overline{\mathfrak{r}}^{*}\right)$$

$$=\dot{\overline{\mathfrak{r}}}\dot{\overline{\mathfrak{r}}}^{*}+\omega^{2}\overline{\mathfrak{r}}\overline{\mathfrak{r}}^{*}+i\omega\left(\left(\overline{x}+i\overline{y}\right)\left(\dot{\overline{x}}-i\dot{\overline{y}}\right)-\left(\overline{x}-i\overline{y}\right)\left(\dot{\overline{x}}+i\dot{\overline{y}}\right)\right)$$

$$=\dot{\overline{x}}^{2}+\dot{\overline{y}}^{2}+\omega^{2}\left(\overline{x}^{2}+\overline{y}^{2}\right)+2\omega\left(\overline{x}\dot{\overline{y}}-\overline{y}\dot{\overline{x}}\right).$$

So the Lagrangian is

$$L=\frac{2}{m}\left(\dot{\overline{x}}^{2}+\dot{\overline{y}}^{2}+\omega^{2}\left(\overline{x}^{2}+\overline{y}^{2}\right)+2\omega\left(\overline{x}\dot{\overline{y}}-\overline{y}\dot{\overline{x}}\right)\right).$$

Set up the Euler Lagrange equations in the following steps

$$\begin{bmatrix}\dot{p}_{\overline{x}}\\ \dot{p}_{\overline{y}} \end{bmatrix}=\frac{\partial L}{\partial\overline{\mathfrak{r}}}=m\begin{bmatrix}\omega^{2}\overline{x}+\omega\dot{\overline{y}}\\ \omega^{2}\overline{y}-\omega\dot{\overline{x}} \end{bmatrix},$$

$$\begin{bmatrix}p_{\overline{x}}\\ p_{\overline{y}} \end{bmatrix}=\frac{\partial L}{\partial\dot{\overline{\mathfrak{r}}}}=m\begin{bmatrix}\dot{\overline{x}}-\omega\overline{y}\\ \dot{\overline{y}}+\omega\overline{x} \end{bmatrix},$$

$$m\begin{bmatrix}\ddot{\overline{x}}-\omega\dot{\overline{y}}\\ \ddot{\overline{y}}+\omega\dot{\overline{x}} \end{bmatrix}=m\begin{bmatrix}\omega^{2}\overline{x}+\omega\dot{\overline{y}}\\ \omega^{2}\overline{y}-\omega\dot{\overline{x}} \end{bmatrix}.$$

Put this in the form of Newton's equations of motion with pseudo-forces,

$$m\begin{bmatrix}\ddot{\overline{x}}\\ \ddot{\overline{y}} \end{bmatrix}=m\begin{bmatrix}\omega^{2}\overline{x}+2\omega\dot{\overline{y}}\\ \omega^{2}\overline{y}-2\omega\dot{\overline{x}} \end{bmatrix}.$$

The pseudo-forces consist of a centripetal force, and the negative of a Coriolis force. The book suggests treating the term in the Lagrangian producing the centripetal force as a potential energy:

$$V\left[r\right]=\omega^{2}\left(\overline{x}^{2}+\overline{y}^{2}\right).$$

The solution to Exercise 6-4 sets up the Lagrangian using the kinetic energy relative to the rotating coordinates, and a potential energy dependent on the radius. In my notation this is:

$$L=\frac{m}{2}\left(\dot{\overline{x}}^{2}+\dot{\overline{y}}^{2}\right)+V\left[r\right]=\frac{m}{2}\left(\dot{r}^{2}+\dot{\overline{\rho}}^{2}r^{2}\right)+V\left[r\right].$$

As I have written, the solution has the potential energy added to the kinetic energy, which I'm pretty sure is wrong. But that is not what I am asking about.

My question is this: is it appropriate to set up the Lagrangian in a non-inertial reference frame in a way that omits velocity-dependent forces?

The final result of the posted solution is the claim that $mr^{2}\ddot{\overline{\rho}}=0$. (He actually writes $mR^{2}\ddot{\theta}=0.$) If $\overline{\rho}=\theta$ were measured relative to an inertial frame, I wouldn't question that result. As it stands, I don't believe it is correct.

• Yup, you are right and the book is wrong. You can check your result against any standard mechanics textbook. – knzhou Sep 15 '18 at 15:10
• I get the appeal of trying to learn physics from a book like Susskind's, which promises to bring the layperson up to the level of a theoretical physicist in just a few hundred pages, but I've never seen anybody actually succeed. These kinds of books are incredibly incomplete and riddled with typos; even your average professor's casual handwritten lecture notes would be better. – knzhou Sep 15 '18 at 15:11
• The worst offenses aren't in the book, but rather in the online suggested solution. Working the problem has been valuable to me. There are plenty of imperfections in the book, but reading it certainly hasn't been a waste of time thus far. – Steven Thomas Hatton Sep 15 '18 at 16:34
• Best of luck then! But if you eventually want a change, you definitely have the drive and ability to tackle lecture notes. There are many good sets of notes out there with the same fun, conversational tone as a popular book. – knzhou Sep 15 '18 at 16:57