\({ }^{\dagger}\) The motion of a particle in three dimensions under an inversesquare-law force is governed by the Lagrangian $$ L=\frac{1}{2} m \dot{\boldsymbol{r}} . \dot{\boldsymbol{r}}+\frac{k}{r}, $$ where \(r=|\boldsymbol{r}|\). Show by vector methods, or otherwise, that the components \(J_{i}, R_{i}\) of the two vectors $$ J=m \boldsymbol{r} \wedge \dot{\boldsymbol{r}} \quad \text { and } \quad \boldsymbol{R}=\dot{\boldsymbol{r}} \wedge \boldsymbol{J}-\frac{k \boldsymbol{r}}{r} $$ are constants of the motion. Show that \(\left[J_{1}, r\right]=0\). Find \(\left[J_{1}, J_{2}\right]\) and \(\left[J_{1}, R_{2}\right]\) in terms of components of \(\boldsymbol{J}\) and \(\boldsymbol{R}\).

First, we must find the time derivatives of the components \(J_i\) and \(R_i\). If their time derivatives are zero, then these components are constants of motion. The given vectors \({\boldsymbol{J}}\) and \({\boldsymbol{R}}\) are: $$ \boldsymbol{J}=m \boldsymbol{r} \wedge \dot{\boldsymbol{r}} \quad \text { and } \quad \boldsymbol{R}=\dot{\boldsymbol{r}} \wedge \boldsymbol{J}-\frac{k \boldsymbol{r}}{r} $$ Now, take time derivatives: $$ \frac{d \boldsymbol{J}}{d t} = m {\boldsymbol{\dot{r}}} \wedge {\boldsymbol{\ddot{r}}} \quad \text { and } \quad \frac{d \boldsymbol{R}}{d t}=\ddot{\boldsymbol{r}} \wedge \boldsymbol{J} +\dot{\boldsymbol{r}} \wedge \frac{d\boldsymbol{J}}{d t} - \frac{k \dot{\boldsymbol{r}}}{r} $$

Now we need to use the Euler-Lagrange equations to simplify the time derivatives. Remember that the Euler-Lagrange equations are expressed as follows: \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)=\frac{\partial L}{\partial q_i}\), where \(q_i\) are the generalized coordinates, in this case, the components of the position vector \(\boldsymbol{r}\). The Euler-Lagrange equations for each component \(r_i\) will give us the equation of motion under the inverse square-law force: $$ \frac{d^2 r_i}{d t^2}=-\frac{k r_i}{m r^3}. $$ Now substitute the Euler-Lagrange equations in the time derivatives of \(\boldsymbol{J}\) and \(\boldsymbol{R}\): $$ \frac{d \boldsymbol{J}}{d t} = m {\boldsymbol{\dot{r}}} \wedge \left(-\frac{k \boldsymbol{r}}{m r^3}\right) = 0 \quad \text { and } \quad \frac{d \boldsymbol{R}}{d t}=\left(-\frac{k \boldsymbol{r}}{m r^3}\right) \wedge \boldsymbol{J} +\dot{\boldsymbol{r}} \wedge (0) - \frac{k \dot{\boldsymbol{r}}}{r} $$ From the result above, we can see that \(\frac{d \boldsymbol{J}}{d t} = 0\), which means the components of the vector \(\boldsymbol{J}\) are constants of the motion. For \(\frac{d \boldsymbol{R}}{d t}\), note that the first two terms cancel out, leaving: $$ \frac{d \boldsymbol{R}}{d t} = - \frac{k \dot{\boldsymbol{r}}}{r} $$

To demonstrate that the components of the \(\boldsymbol{R}\) vector are constants of motion, we need to show that \(\boldsymbol{R}\) is parallel to \(\boldsymbol{\dot{r}}\). $$ \boldsymbol{R} \cdot \boldsymbol{\dot{r}} = \left(\dot{\boldsymbol{r}} \wedge \boldsymbol{J}-\frac{k \boldsymbol{r}}{r}\right) \cdot \boldsymbol{\dot{r}} = \boldsymbol{J} \cdot (\boldsymbol{\dot{r}}\wedge\boldsymbol{\dot{r}})-\frac{k \boldsymbol{r}}{r} \cdot \boldsymbol{\dot{r}} = 0 $$ Since the dot product is zero, \(\boldsymbol{R}\) is parallel to \(\boldsymbol{\dot{r}}\), and thus, the components of \(\boldsymbol{R}\) are also constants of motion.

Now, we are asked to find the Poisson brackets \(\left[J_{1}, r\right]\), \(\left[J_{1}, J_{2}\right]\), and \(\left[J_{1}, R_{2}\right]\) given the components of the vectors \(\boldsymbol{J}\) and \(\boldsymbol{R}\). The Poisson bracket of two variables \(A\) and \(B\) is defined as: $$ \left[A, B\right]=\frac{\partial A}{\partial q_i}\frac{\partial B}{\partial p_i}-\frac{\partial A}{\partial p_i}\frac{\partial B}{\partial q_i}, $$ where \(p_i\) are the generalized momenta conjugate to the generalized coordinates \(q_i\). Now, we can calculate the Poisson brackets: 1. \(\left[J_{1}, r\right] = 0\) since \(J_1\) doesn't depend on \(r\). 2. \(\left[J_{1}, J_{2}\right] = \frac{\partial J_{1}}{\partial r_i}\frac{\partial J_{2}}{\partial p_i}-\frac{\partial J_{1}}{\partial p_i}\frac{\partial J_{2}}{\partial r_i} = J_{3}\) (since the only non-zero terms involve the derivatives of \(J_1\) with respect to \(p_2\) and \(J_2\) with respect to \(p_1\)). 3. \(\left[J_{1}, R_{2}\right] = \frac{\partial J_{1}}{\partial r_i}\frac{\partial R_{2}}{\partial p_i}-\frac{\partial J_{1}}{\partial p_i}\frac{\partial R_{2}}{\partial r_i} = -R_{3}\) (similarly, the only non-zero terms involve the derivatives of \(J_1\) with respect to \(p_2\) and \(R_2\) with respect to \(p_1\)). Now, we've found that \(\left[J_{1}, r\right]=0\), \(\left[J_{1}, J_{2}\right]=J_{3}\), and \(\left[J_{1}, R_{2}\right]=-R_{3}\) using the components of the vectors \(\boldsymbol{J}\) and \(\boldsymbol{R}\). We also showed that the components of both these vectors are constants of motion, which was our main task.