Talk:Aufgaben:Problem 12

Exercise

Consider a mass \(m>0\) in the Newtonian gravitational field. Show that the gravitational force strength on the mass can be written as \begin{equation*} |\vec{F}| = \frac{{L_3}^2}{mr^4} \bigg| \frac{d^2 r}{d\phi^2} - \frac{2}{r} \left( \frac{dr}{d\phi} \right)^2 - r \bigg| \end{equation*} where \( \vec{L} = (L_1,L_2,L_3)\) is the angular momentum and \((r,\phi)\) are the polar coordinates.\\

Hint: show first that the total energy can be written as \(H = \frac{p_r^2}{2m} + \frac{p_\phi^2}{2mr^2} + U(r)\), where \(p_r,p_\phi\) are the generalized momenta defined as \(p_r = m\dot{r}, \; p_\phi = mr^2\dot{\phi}\) and \(U(r)\) is the potential. Then obtain an expression for \(-\partial U / \partial r\) from the equations of motion.

Solution

In polar coordinates is \( \vec{x} = \begin{pmatrix} r \cos \phi \\ r \sin \phi \end{pmatrix} \). Then we can calculate \( \dot{\vec{x}} \) depending on \( \dot{r} \) and \( \dot{\phi} \): \begin{equation*} \dot{\vec{x}} = \frac{\partial \vec{x}}{\partial r} \dot{r} + \frac{\partial \vec{x}}{\partial \phi} \dot{\phi} = \begin{pmatrix} \cos \phi \\ \sin \phi \end{pmatrix}\dot{r} + \begin{pmatrix}- r \sin \phi \\ r \cos \phi\end{pmatrix} \dot{\phi} \end{equation*}

The kinetic energy is then: \(T = \frac{1}{2} m |\dot{\vec{x}}|^2 = \frac{1}{2} m \left( \dot{r}^2 + r^2 \dot{\phi}^2 \right)\)

The total energy is \( H = T + U(r) \) where \( U(r) \) is the potential caused by the gravitational force \( \vec{F} \) with the relation: \begin{equation*} \vec{F} = -\vec{\nabla} U(r) = - \begin{pmatrix} \frac{\partial}{\partial r} \\ \frac{1}{r} \frac{\partial}{\partial \phi} \end{pmatrix} U(r) = \begin{pmatrix} - \frac{\partial U(r)}{\partial r} \\ 0 \end{pmatrix} \end{equation*}

We define the generalized momenta \( p_r = m \dot{r} \) and \(p_\phi = m r^2 \dot{\phi} \) and get: \begin{equation*} H = \frac{p_r^2}{2m} + \frac{p_\phi^2}{2mr^2} + U(r) \end{equation*}

Consider the Hamliton equations: \begin{equation*} \dot{r} = \frac{\partial H}{\partial p_r} = \frac{p_r}{m} \end{equation*} \begin{equation*} \dot{\phi} = \frac{\partial H}{\partial p_\phi} = \frac{p_\phi}{mr^2} \end{equation*} \begin{equation} \dot{p}_r = -\frac{\partial H}{\partial r} = \frac{p_\phi^2}{mr^3} - \frac{\partial U(r)}{\partial r} \end{equation} \begin{equation*} \dot{p}_\phi = -\frac{\partial H}{\partial \phi} = 0 \end{equation*}

From the third equation we get: \begin{equation} - \frac{\partial U(r)}{\partial r} = \dot{p}_r - \frac{p_\phi^2}{mr^3} = m \ddot{r} - m r \dot{\phi}^2 \end{equation}

We know that for Newtonian gravitational fields the angular momentum \( \vec{L} \) is conserved. If we expand our two-dimensional system by a third axis the angular momentum would have the form \( \vec{L} = \begin{pmatrix} 0 \\ 0 \\ L_3 \end{pmatrix} \) because the masspoint only moves on the xy-plane. Also we can identify \( L_3 \) with the generalized momentum \( p_\phi \) of the azimuthal angle. The equation turns to:

\begin{equation} - \frac{\partial U(r)}{\partial r} = m \ddot{r} - \frac{{L_3}^2}{mr^3} \end{equation}

First a bit preparing: \begin{equation} \dot{\phi} = \frac{L_3}{m r^2} \end{equation} \begin{equation} \dot{\phi}^2 = \frac{{L_3}^2}{m^2 r^4} \end{equation} \begin{equation} \ddot{\phi} = (-2) \frac{L_3}{m r^3} \cdot \frac{dr}{dt} \end{equation}

Now we say \(r\) is depending on \( \phi(t) \Rightarrow r\equiv r(\phi(t)) \) Consider: \begin{equation} \frac{dr}{dt}(\phi(t)) \overset{\text{[chain rule]}}{=} \frac{dr}{d\phi}(\phi(t)) \underbrace{ \frac{d\phi}{dt}(t) }_{ \dot{\phi} } = \frac{L_3}{m r^2} \frac{dr}{d\phi} \end{equation} with product and chain rule we get: $$\begin{align} \frac{d^2r}{dt^2} &=\frac{d}{dt}\left(\frac{dr}{d\phi}\right)\cdot\frac{d\phi}{dt}+\frac{dr}{d\phi}\cdot\frac{d}{dt}\left(\frac{d\phi}{dt}\right)\\ &=\bigg(\frac{d^2r}{d\phi^2}\underbrace{\frac{d\phi}{dt}\bigg)\cdot\frac{d\phi}{dt}}_{\dot{\phi}^2} +\frac{dr}{d\phi}\underbrace{\frac{d^2\phi}{dt^2}}_{\ddot{\phi}}\\ &=\frac{{L_3}^2}{m^2r^4}\frac{d^2r}{d\phi^2} +\frac{dr}{d\phi}\bigg((-2)\frac{L_3}{mr^3}\cdot\underbrace{\frac{dr}{dt}}_{\frac{dr}{dt}(\phi(t))}\bigg)\\ &=\frac{{L_3}^2}{m^2r^4}\frac{d^2r}{d\phi^2} +(-2)\frac{L_3}{mr^3}\cdot\frac{L_3}{mr^2}\left(\frac{dr}{d\phi}\right)^2\\ &=\frac{{L_3}^2}{m^2 r^4} \left(\frac{d^2r}{d\phi^2} -\frac{2}{r}\left(\frac{dr}{d\phi}\right)^2\right) \end{align}$$

We can now replace this in the equation: $$\begin{align} -\frac{\partial U(r)}{\partial r} & = m\ddot{r} -\frac{{L_3}^2}{mr^3}\\ & = \frac{{L_3}^2}{m r^4} \left( \frac{d^2 r}{d\phi ^2} -\frac{2}{r} \left( \frac{dr}{d\phi} \right)^2 -r \right) \end{align}$$

We notice that \( \frac{{L_3}^2}{m r^4} >0 \) and can proof the statement: $$\begin{align} |\vec{F}| &= \bigg|-\frac{\partial U(r)}{\partial r} \bigg| \\ &= \frac{{L_3}^2}{m r^4} \bigg| \frac{d^2 r}{d\phi ^2} -\frac{2}{r} \left( \frac{dr}{d\phi} \right)^2 -r \bigg| \end{align}$$