homework and exercises - Electron in the proximity of a magnetic monopole

I am puzzling about an exercise in the book "Electromagnetic Theory" by Ferraro (p.543).

An electron (mass $m$, charge $-e$) in a monopole magnetic field $\vec{B}\left(\vec{r}\right)=g\frac{\vec{r}}{r^3}$ has due to the Lorentz force $\vec{F}_L=-e\vec{v}\times\vec{B}$ the following equation of motion:

$$m\ddot{\vec{r}}=-\frac{eg}{r^3}\dot{\vec{r}}\times\vec{r}.$$

Since $\vec{J}=\vec{r}\times\vec{p}+eg\frac{\vec{r}}{r}$ is constant, the electron moves on the surface of a cone, which tip is in the origin at $\vec{r}=0$. Mathematically, $\vec{r}\cdot\vec{J}=egr=rJ\cos\vartheta$, thus the angle $\vartheta$ between $\vec{r}$ and $\vec{J}$ is constant.

Now I use spherical coordinates:

$$\vec{r}=r\mathbf{e}_{r}$$ $$\vec{v}=\dot{\vec{r}}=\dot{r}\mathbf{e}_{r}+r\dot{\vartheta}\mathbf{e}_{\vartheta}+r\dot{\varphi}\sin\vartheta\mathbf{e}_{\varphi}$$ $$\vec{a}=\ddot{\vec{r}}=\left(\ddot{r}-r\dot{\vartheta}^{2}-r\dot{\varphi}^{2}\sin^{2}\vartheta\right)\mathbf{e}_{r}+\left(2\dot{r}\dot{\vartheta}+r\ddot{\vartheta}-r\dot{\varphi}^{2}\sin\vartheta\cos\vartheta\right)\mathbf{e}_{\vartheta}+\left(2\dot{r}\dot{\varphi}\sin\vartheta+2r\dot{\vartheta}\dot{\varphi}\cos\vartheta+r\ddot{\varphi}\sin\vartheta\right)\mathbf{e}_{\varphi}$$

and obtain the following equations of motion (using $\dot{\vartheta}=0$):

$$\ddot{r}=r\dot{\varphi}^2\sin^2\vartheta \qquad(1)$$ $$r\dot{\varphi}^2\cos\vartheta=\frac{eg}{mr}\dot{\varphi} \qquad(2)$$ $$2\dot{r}\dot{\varphi}+r\ddot{\varphi}=0 \qquad(3)$$ with the boundary conditions: $$r\left(0\right)=r_0\qquad\varphi\left(0\right)=0.$$

In order to solve (1) for $r\left(t\right)$, one has to use the fact that the kinetic energy $\frac{1}{2}m\vec{v}^2$ is conserved. This holds due to the vector potential $\vec{A}=\frac{1}{2}\vec{B}\times\vec{r}=0$. Therefore,

$$\vec{v}^2=\dot{r}^2+r^2\dot{\varphi}^2\sin^2\vartheta\equiv u^2=\text{const}$$

Using this in (1) one gets the simplified differential equation:

$$r\ddot{r}=u^2-\dot{r}^2\qquad(4)$$

Does anyone know how to derive the following solution of (4)?

$$r\left(t\right)=r_0\sqrt{1+\left(\frac{ut}{r_0}\right)^2}\qquad(5)$$

Solving (2) for $\dot{\varphi}$:

$$\dot{\varphi}=\frac{eg}{mr^2\cos\vartheta}$$

and inserting (5) I obtain:

$$\varphi\left(t\right)=\frac{eg}{mr_0u\cos\vartheta}\arctan{\frac{ut}{r_0}}.$$

Now the solutions says:

$$r\left(\varphi\right)=\frac{r_0}{\cos\left(\varphi\sin\vartheta\right)}.$$

But I get something similar, but not equal:

$$r\left(\varphi\right)=\frac{r_0}{\cos\left(\frac{mr_0u\cos\vartheta}{eg}\varphi\right)}.$$

Has anyone an idea where I could have done something wrong?

Answer

Actually, I don't have to use the conservation of the kinetic energy. With a slightly different procedure, one gets another simplified differential equation.

Inserting (2) in (1):

$$\ddot{r}=\frac{e^2 g^2 \tan^2\vartheta}{m^2} \frac{1}{r^3}$$

Now look at:

$$\frac{\text{d}}{\text{dt}}\dot{r}^2=2\dot{r}\ddot{r}=\frac{e^2 g^2 \tan^2\vartheta}{m^2} \frac{2\dot{r}}{r^3}=\frac{\text{d}}{\text{dt}}\left(\frac{e^2 g^2 \tan^2\vartheta}{m^2}\frac{-1}{r^2}\right)$$

Thus, a simpler differential equation is obtained ($C=\text{const}$):

$$\dot{r}^2=\frac{e^2 g^2 \tan^2\vartheta}{m^2}\frac{-1}{r^2}+C$$

and can be solved by separation of variables:

$$\text{dt}=\frac{r\text{ dr}}{\sqrt{Cr^2-\frac{e^2 g^2 \tan^2\vartheta}{m^2}}}$$

$$t=\frac{1}{C}\sqrt{Cr^2-\frac{e^2 g^2 \tan^2\vartheta}{m^2}}+t_0$$

With $t_0=0$ and $r\left(0\right)=r_0$:

$$r\left(t\right)=\sqrt{\frac{e^2 g^2 \tan^2\vartheta}{m^2r_0^2}t^2+r_0^2}$$

Now solving for $\varphi\left(t\right)$ from (2):

$$\dot{\varphi}=\frac{eg}{m\cos\vartheta}\frac{1}{r^2}$$

$$\text{d}\varphi=\frac{m\cos\vartheta r_0^2}{eg\sin^2\vartheta}\frac{\text{dt}}{t^2+\left(\frac{m r_0^2}{e g \tan\vartheta}\right)^2}$$

Using the arctangent integral:

$$\int\frac{\text{dt}}{t^2+b^2}=\frac{1}{b}\arctan\frac{t}{b}$$

and $\varphi\left(0\right)=0$, one obtains:

$$\varphi\left(t\right)=\frac{1}{\sin\vartheta}\arctan\frac{e g \tan\vartheta\ t}{m r_0^2}$$

Finally,

$$r\left(\varphi\right)=r_0 \sqrt{\tan^2\left(\sin\vartheta \varphi\right)+1}=\frac{r_0}{\cos\left(\sin\vartheta\varphi\right)}$$