A more elegant way of finding $\dot \phi$

Question

Is there a simple way of obtaining an expression for $\dot{\phi}$ from the equation $$r{d^2\over dt^2}\begin{pmatrix} r\cos \phi\\r\sin \phi\end{pmatrix}=F'(r){d\over dt}\begin{pmatrix} r\cos \phi\\r\sin \phi\end{pmatrix}$$ where $':=d/dr$ and $\dot:=d/dt$. without actually going through the messy explicitly calculating all the time derivatives? Thank you.

Answer 1

This may not be exactly what you are looking for, since I can only get $\dot\phi$ in terms of $r$ and $\dot{r}$ (not just $r$). But maybe you can work something else further.

Denote $\vec{x} = \begin{pmatrix} r\cos \phi \\ r\sin\phi\end{pmatrix}$ and $\vec{v} = \dot{\vec{x}}$. Your equation states that $$ r \dot{\vec{v}} = F'(r) \vec{v}\tag{1}$$

Step 1: taking the dot product against $\vec{v}$ you get $$ \frac12 r \frac{d}{dt} |\vec{v}|^2 = F'(r) |\vec{v}|^2 $$ or $$ \frac12 \frac{d}{dt} \ln( \dot{r}^2 + r^2 \dot{\phi}^2) = \frac{F'}{r} \tag{2} $$

Step 2: taking the dot product against $\vec{x}/r$ we note that $ \vec{x}\vec{v}/r = \dot{r}$. So equation (1) becomes $$ \vec{x} \dot{\vec{r}} = F'(r) \dot{r} $$ which we re-write as $$ r \ddot{r} - r^2 \dot{\theta}^2 = \dot{F} \tag{3} $$ where on the RHS we used the chain rule, and on the LHS we just computed.

Step 3: taking the dog product against $\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix} \vec{v}$, we see that the right hand side evaluates to 0. The left-hand side after a little bit of computation (and dividing by $r$) becomes $$ (r^2\dot{\phi}^2 - \ddot{r} r) \dot{\phi} + 2 \dot{r}^2 \dot{\phi} + \dot{r}r \ddot{\phi} = 0 \tag{4}$$

Step 4: plug equation (3) into (4), we get $$ \left( \dot{F} - 2\dot{r}^2 \right) \dot\phi = \dot{r} r \ddot\phi $$ which using the chain rule and reorganizing we get $$ \frac{F'}{r} = \frac{\ddot\phi}{\dot\phi} + 2 \frac{\dot{r}}{r} = \frac{d}{dt}\left( \ln \dot\phi + 2 \ln r\right) = \frac{d}{dt} \ln r^2\dot\phi \tag{5}$$

Step 5: now insert (2) into (5) you get $$ \ln\left( \dot{r}^2 + r^2 \dot{\phi}^2\right) + C = 2 \ln r^2\dot{\phi} $$ which implies that for some constant $C$ to be fixed by the initial conditions we have $$ \dot{r}^2 + r^2\dot{\phi}^2 = C r^4 \dot{\phi}^2 $$ which we can solve as $$ \frac{\dot{r}^2}{Cr^4 - r^2} = \dot{\phi}^2 \tag{*}$$

This gives an expression of $\dot\phi$ in terms or $r,\dot{r}$. Notably taking the square root of this expression (fixing the sign via the initial data) we see that in principle we can integrate both sides in time and get an expression of $\phi$ in terms of $r$. (In fact, a trigonometric substitution gives that the solution should be of the form $ \phi = A\sec^{-1} (B r) + D$ where $A$ and $B$ are determined from the constant $C$ in equation (*).)

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The gist of the above computation is exploiting "conservation laws". Equations (2) and (3) form some analogue of the "Work-energy" theorem for your dynamical system. Equation (4) is an analogue of "conservation of angular momentum".