A planet of mass \(M\) is surrounded by a cloud of small particles in orbits around it. Their mutual gravitational attraction is negligible. Due to collisions between the particles, the energy will gradually decrease from its initial value, but the angular momentum will remain fixed, \(\boldsymbol{J}=\boldsymbol{J}_{0}\), say. The system will thus evolve towards a state of minimum energy, subject to this constraint. Show that the particles will tend to form a ring around the planet. [As in Problem 15 , the constraint may be imposed by the method of Lagrange multipliers. In this case, because there are three components of the constraint equation, we need three Lagrange multipliers, say \(\omega_{x}, \omega_{y}, \omega_{z} .\) We have to minimize the function \(E-\boldsymbol{\omega} \cdot\left(\boldsymbol{J}-\boldsymbol{J}_{0}\right)\) with respect to variations of the positions \(\boldsymbol{r}_{i}\) and velocities \(\dot{\boldsymbol{r}}_{i}\), and with respect to \(\boldsymbol{\omega}\). Show by minimizing with respect to \(\dot{\boldsymbol{r}}_{i}\) that once equilibrium has been reached the cloud rotates as a rigid body, and by minimizing with respect to \(\boldsymbol{r}_{i}\) that all particles occupy the same orbit.] What happens to the energy lost? Why does the argument not necessarily apply to a cloud of particles around a hot star?

Let's begin by setting up the Lagrange function, which we will minimize with respect to the variations of positions \(\boldsymbol{r}_{i}\) and velocities \(\dot{\boldsymbol{r}}_{i}\), and with respect to \(\boldsymbol{\omega}\). We have the following Lagrange function: $$ L=E-\boldsymbol{\omega}\cdot\left(\boldsymbol{J}-\boldsymbol{J}_{0}\right), $$ where E is the total energy of the system.

To find the equilibrium, we first minimize the Lagrange function with respect to the velocities \(\dot{\boldsymbol{r}}_i\). The derivative of L with respect to \(\dot{\boldsymbol{r}}_i\) is: $$ \frac{\partial L}{\partial\dot{\boldsymbol{r}}_i}=\frac{\partial E}{\partial\dot{\boldsymbol{r}}_i}-\frac{\partial\boldsymbol{\omega}}{\partial\dot{\boldsymbol{r}}_i}\cdot\left(\boldsymbol{J}-\boldsymbol{J}_0\right)-\boldsymbol{\omega}\cdot\frac{\partial\left(\boldsymbol{J}-\boldsymbol{J}_0\right)}{\partial\dot{\boldsymbol{r}}_i}. $$ We know that the energy of a particle is given by: $$ E_i = \frac{1}{2}m_i\dot{\boldsymbol{r}}_i^2 - G\frac{m_i M}{|\boldsymbol{r}_i|}, $$ where \(m_i\) is the mass of the \(i\)-th particle, and \(G\) is the gravitational constant. The derivative of the energy with respect to \(\dot{\boldsymbol{r}}_i\) is: $$ \frac{\partial E}{\partial\dot{\boldsymbol{r}}_i}=m_i\dot{\boldsymbol{r}}_i. $$ Now we take the derivative of the element inside the bracket of the Lagrange function with respect to \(\dot{\boldsymbol{r}}_i\): $$ \frac{\partial\left(\boldsymbol{J}-\boldsymbol{J}_0\right)}{\partial\dot{\boldsymbol{r}}_i}=m_i\boldsymbol{r}_i. $$ So the first term in the partial derivative of the Lagrange function vanishes: $$ \frac{\partial L}{\partial\dot{\boldsymbol{r}}_i}=m_i\dot{\boldsymbol{r}}_i-\boldsymbol{\omega}\cdot\left(m_i\boldsymbol{r}_i\right)=0. $$ This result shows that when equilibrium has been reached, the cloud rotates as a rigid body.

Now let's minimize the Lagrange function with respect to the positions \(\boldsymbol{r}_i\). The derivative of L with respect to \(\boldsymbol{r}_i\) is: $$ \frac{\partial L}{\partial\boldsymbol{r}_i}=\frac{\partial E}{\partial\boldsymbol{r}_i}-\frac{\partial\boldsymbol{\omega}}{\partial\boldsymbol{r}_i}\cdot\left(\boldsymbol{J}-\boldsymbol{J}_0\right)-\boldsymbol{\omega}\cdot\frac{\partial\left(\boldsymbol{J}-\boldsymbol{J}_0\right)}{\partial\boldsymbol{r}_i}. $$ The derivative of the energy with respect to \(\boldsymbol{r}_i\) is: $$ \frac{\partial E}{\partial\boldsymbol{r}_i}=G\frac{m_i M}{|\boldsymbol{r}_i|^3}\boldsymbol{r}_i. $$ We take the derivative of the element inside the bracket of the Lagrange function with respect to \(\boldsymbol{r}_i\): $$ \frac{\partial\left(\boldsymbol{J}-\boldsymbol{J}_0\right)}{\partial\boldsymbol{r}_i}=m_i\dot{\boldsymbol{r}}_i\times\boldsymbol{\hat{z}}, $$ where \(\boldsymbol{\hat{z}}\) is the direction perpendicular to the plane containing \(\boldsymbol{r}_i\). We now substitute the results and equate the derivative with respect to \(\boldsymbol{r}_i\) to zero: $$ G\frac{m_i M}{|\boldsymbol{r}_i|^3}\boldsymbol{r}_i-\boldsymbol{\omega}\times\left(m_i\dot{\boldsymbol{r}}_i\times\boldsymbol{\hat{z}}\right)=0. $$ This equation is only valid for lying in the same plane as the direction perpendicular to \(\boldsymbol{\hat{z}}\), meaning all particles will occupy the same orbit.

The energy lost in the system is primarily due to collisions between particles, which is radiated away as heat. This means the initial energy of the system decreases. The primary assumption in this problem is that the mutual gravitational attraction between the particles is negligible. In the case of a cloud of particles around a hot star, the assumption may not hold true, as the particles may also be affected by the radiation pressure originating from the star, which might cause the system to have a more complicated behavior.