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| − | =Exercise=
| + | In the second solution. For proofing the bilinearity of anti-commutator it is sufficient to show linearity in the first argument. Then use that it is symmetric. |
| − | | + | [[User:Beni|Beni]] ([[User talk:Beni|talk]]) 16:42, 24 June 2015 (CEST) |
| − | Consider a mass \(m>0\) in the Newtonian gravitational field. Show that the gravitational force strength on the mass can be written as
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| − | \begin{equation*}
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| − | |\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|
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| − | \end{equation*}
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| − | where \( \vec{L} = (L_1,L_2,L_3)\) is the angular momentum and \((r,\phi)\) are the polar coordinates.\\
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| − | 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.
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| − | =Solution=
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| − | In polar coordinates is \( \vec{x} = \begin{pmatrix} r \cos \phi \\ r \sin \phi \end{pmatrix} \).
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| − | Then we can calculate \( \dot{\vec{x}} \) depending on \( \dot{r} \) and \( \dot{\phi} \):
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| − | \begin{equation*}
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| − | \dot{\vec{x}} = \frac{\partial \vec{x}}{\partial r} \dot{r} + \frac{\partial \vec{x}}{\partial \phi} \dot{\phi} =
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| − | \begin{pmatrix} \cos \phi \\ \sin \phi \end{pmatrix}\dot{r} + \begin{pmatrix}- r \sin \phi \\ r \cos \phi\end{pmatrix} \dot{\phi}
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| − | \end{equation*}
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| − | The kinetic energy is then: \(T = \frac{1}{2} m |\dot{\vec{x}}|^2
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| − | = \frac{1}{2} m \left( \dot{r}^2 + r^2 \dot{\phi}^2 \right)\)
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| − | The total energy is \( H = T + U(r) \) where \( U(r) \) is the potential caused by the gravitational force \( \vec{F} \) with the relation:
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| − | \begin{equation*}
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| − | \vec{F} = -\vec{\nabla} U(r)
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| − | = - \begin{pmatrix} \frac{\partial}{\partial r} \\
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| − | \frac{1}{r} \frac{\partial}{\partial \phi} \end{pmatrix} U(r)
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| − | = \begin{pmatrix} - \frac{\partial U(r)}{\partial r} \\ 0 \end{pmatrix}
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| − | \end{equation*}
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| − | We define the generalized momenta \( p_r = m \dot{r} \) and \(p_\phi = m r^2 \dot{\phi} \) and get:
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| − | \begin{equation*}
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| − | H = \frac{p_r^2}{2m} + \frac{p_\phi^2}{2mr^2} + U(r)
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| − | \end{equation*}
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| − | Consider the Hamliton equations:
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| − | \begin{equation*}
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| − | \dot{r} = \frac{\partial H}{\partial p_r} = \frac{p_r}{m}
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| − | \end{equation*}
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| − | \begin{equation*}
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| − | \dot{\phi} = \frac{\partial H}{\partial p_\phi} = \frac{p_\phi}{mr^2}
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| − | \end{equation*}
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| − | \begin{equation}
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| − | \dot{p}_r = -\frac{\partial H}{\partial r} = \frac{p_\phi^2}{mr^3} - \frac{\partial U(r)}{\partial r}
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| − | \end{equation}
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| − | \begin{equation*}
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| − | \dot{p}_\phi = -\frac{\partial H}{\partial \phi} = 0
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| − | \end{equation*}
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| − | From the third equation we get:
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| − | \begin{equation}
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| − | - \frac{\partial U(r)}{\partial r} = \dot{p}_r - \frac{p_\phi^2}{mr^3} =
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| − | m \ddot{r} - m r \dot{\phi}^2
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| − | \end{equation}
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| − | 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.
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| − | The equation turns to:
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| − | \begin{equation}
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| − | - \frac{\partial U(r)}{\partial r} =
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| − | m \ddot{r} - \frac{{L_3}^2}{mr^3}
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| − | \end{equation}
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| − | First a bit preparing:
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| − | \begin{equation}
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| − | \dot{\phi} = \frac{L_3}{m r^2}
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| − | \end{equation}
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| − | \begin{equation}
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| − | \dot{\phi}^2 = \frac{{L_3}^2}{m^2 r^4}
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| − | \end{equation}
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| − | \begin{equation}
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| − | \ddot{\phi} = (-2) \frac{L_3}{m r^3} \cdot \frac{dr}{dt}
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| − | \end{equation}
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| − | Now we say \(r\) is depending on \( \phi(t) \Rightarrow r\equiv r(\phi(t)) \)
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| − | Consider:
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| − | \begin{equation}
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| − | \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}
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| − | \end{equation}
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| − | with product and chain rule we get:
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| − | $$\begin{align}
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| − | \frac{d^2r}{dt^2}
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| − | &=\frac{d}{dt}\left(\frac{dr}{d\phi}\right)\cdot\frac{d\phi}{dt}+\frac{dr}{d\phi}\cdot\frac{d}{dt}\left(\frac{d\phi}{dt}\right)\\
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| − | &=\bigg(\frac{d^2r}{d\phi^2}\underbrace{\frac{d\phi}{dt}\bigg)\cdot\frac{d\phi}{dt}}_{\dot{\phi}^2}
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| − | +\frac{dr}{d\phi}\underbrace{\frac{d^2\phi}{dt^2}}_{\ddot{\phi}}\\
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| − | &=\frac{{L_3}^2}{m^2r^4}\frac{d^2r}{d\phi^2}
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| − | +\frac{dr}{d\phi}\bigg((-2)\frac{L_3}{mr^3}\cdot\underbrace{\frac{dr}{dt}}_{\frac{dr}{dt}(\phi(t))}\bigg)\\
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| − | &=\frac{{L_3}^2}{m^2r^4}\frac{d^2r}{d\phi^2}
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| − | +(-2)\frac{L_3}{mr^3}\cdot\frac{L_3}{mr^2}\left(\frac{dr}{d\phi}\right)^2\\
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| − | &=\frac{{L_3}^2}{m^2 r^4} \left(\frac{d^2r}{d\phi^2}
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| − | -\frac{2}{r}\left(\frac{dr}{d\phi}\right)^2\right)
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| − | \end{align}$$
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| − | We can now replace this in the equation:
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| − | $$\begin{align}
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| − | -\frac{\partial U(r)}{\partial r} & =
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| − | m\ddot{r} -\frac{{L_3}^2}{mr^3}\\ & =
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| − | \frac{{L_3}^2}{m r^4} \left( \frac{d^2 r}{d\phi ^2}
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| − | -\frac{2}{r} \left( \frac{dr}{d\phi} \right)^2 -r \right)
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| − | \end{align}$$
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| − | We notice that \( \frac{{L_3}^2}{m r^4} >0 \) and can proof the statement:
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| − | $$\begin{align}
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| − | |\vec{F}|
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| − | &= \bigg|-\frac{\partial U(r)}{\partial r} \bigg| \\
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| − | &= \frac{{L_3}^2}{m r^4} \bigg| \frac{d^2 r}{d\phi ^2}
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| − | -\frac{2}{r} \left( \frac{dr}{d\phi} \right)^2 -r \bigg|
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| − | \end{align}$$
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In the second solution. For proofing the bilinearity of anti-commutator it is sufficient to show linearity in the first argument. Then use that it is symmetric.
