# Classical electron in magnetic monopole

Consider a mass $$m$$ with electrical charge $$+e$$ that is moving on a sphere of radius $$r$$ around the magnetic monopole $$\vec{B}=e_m\frac{\vec{r}}{r^3}.$$

The I get the following equation of motion $$m\ddot{\vec{r}}=\frac{ee_m}{cr^3}\,\dot{\vec{r}}\,\times\,\vec{r}$$ using the Lorentz force.

When I multiply both sides by $$\dot{\vec{r}}$$ I get that $$\dot{\vec{r}}^2=v^2=const$$. And from that I get $$r(t)=\sqrt{r_0^2+(v-v_0)^2t^2}$$ where $$r_0=r(0)$$ and $$v_0=v(0)$$. But that does not look like the mass keeps moving on the sphere.

Can anyone show me where I went wrong? Any help or advice is very much appreciated!

One has to be very careful in the interpretation of the found result $$\dot{\vec{r}}^2$$=const. If the charged particle moves on a sphere the best coordinates for the description of the motion of the particle are spherical coordinates ($$x=r\sin\theta\cos\phi,\, y=r\sin\theta\sin\phi,\,z=r\cos\theta$$). We will put the origin of the coordinate system at the same position as the magnetic monopole. Then the position vector of the charged particle is:

$$\vec{r} = r \vec{e}_r$$

where $$\vec{e}_r$$ is one of the unit vectors in spherical coordinates which points along the radial direction. Keeping this in mind we will write velocity in spherical coordinates:

$$\vec{\dot{r}} = \dot{r}\vec{e}_r + r \vec{\dot{e}_r}$$

The restriction of the motion of the charged particle to a sphere can simply expressed by $$\dot{r}=0$$. However, the unit vectors in spherical coordinates are not constant, so we have to compute $$\vec{\dot{e}_r}$$:

$$\vec{\dot{e}_r}= \frac{d\vec{e}_r}{dr}\frac{dr}{dt} + \frac{d\vec{e}_r}{d\theta}\frac{d\theta}{dt} + \frac{d\vec{e}_r}{d\phi}\frac{d\phi}{dt}$$

In order to compute the derivatives we recall ($$\vec{r}=r \vec{e}_r = x\vec{e}_x + y\vec{e}_y + z\vec{e}_z$$):

$$\vec{e}_r= \sin\theta \cos\phi \vec{e}_x +\sin\theta \sin\phi \vec{e}_y + \cos\theta \vec{e}_z$$

With the last formula the derivatives $$\frac{d \vec{e}_r}{d\theta}$$ and $$\frac{d \vec{e}_r}{d\phi}$$ can be computed ($$\frac{d\vec{e}_r}{dr}=0)$$. The final result is:

$$\vec{\dot{r}}= r \vec{\dot{e}_r}= r\left(\dot{\theta}\vec{e}_\theta + \dot{\phi}\sin\theta \vec{e}_\phi \right)$$

The found result would implicate:

$$const = \dot{\vec{r}}^2 = r^2 \left(\dot{\theta}\vec{e}_\theta + \dot{\phi}\sin\theta \vec{e}_\phi \right)^2 = r^2 \left(\dot{\theta}^2 + \dot{\phi}^2\sin^2\theta\right)$$

as $$\vec{e}^2_\theta= \vec{e}^2_\phi=1$$ and $$\vec{e}_\theta\cdot \vec{e}_\phi=0$$

These are some particular curves on the sphere, the best is to display them with a graphics software.

• Ah, of course! Now I see my mistake. Can you give me any hint on how to actually solve the differential equation? I've never cames across one with a cross product in it. – TwoStones Nov 19 '19 at 16:46
• Sorry if this question is just stupid (I am still new to theoretical physics), but what exactly is the solution then? – TwoStones Nov 19 '19 at 16:57
• @TwoStones. All curves which fulfill $\dot\theta^2 + \dot\phi^2 \sin\theta^2 = const$. They depend on the initial conditions. I guess, these are geodesics on the sphere, but I have no proof for this assumption, so take that with particular care. – Frederic Thomas Nov 19 '19 at 17:01
• Thank you very much! – TwoStones Nov 19 '19 at 17:03
• @TwoStones. You already solved the DE. And you even got a first integral of it, d.h. $\dot{\vec{r}}^2=const$. In CM there is not more required. – Frederic Thomas Nov 19 '19 at 17:05