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

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.

-
Does $r$ depend on $t$? Does $\phi$ depend on $r$? –  Willie Wong Jul 10 '12 at 10:55
@WillieWong: $r=r(t)$ ,$\phi=\phi(t)$ –  Valentin Jul 10 '12 at 11:56
@WillieWong: This actually results from a particle in the $x,y$ plane moving in a magnetic field $F'(r)/r$ in the z-direction. –  Valentin Jul 10 '12 at 11:59
What is the equation for $\phi$? You should have a couple of equations of motion. –  Jon Jul 10 '12 at 12:23
Second question: what do you want $\dot{\phi}$ in terms of? Are you asking to isolate the equation for $\phi$? Or are you asking for whether we can solve $\dot{\phi}$ directly without solving also for $r$? –  Willie Wong Jul 10 '12 at 13:39

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 (*).)

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".

-
BTW, from equation (1) one sees that the solution curve $\vec{x}(t)$ has no normal acceleration, hence by a change of time parametrisation must be a straightline (since it is a geodesic of the plane). Which automatically tells you that $\phi = A\sec^{-1}(Br) + D$... –  Willie Wong Jul 11 '12 at 10:28
Thank you, Willie!! –  Valentin Jul 11 '12 at 23:18