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.

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – TwoStones Nov 19 '19 at 16:46
  • $\begingroup$ Sorry if this question is just stupid (I am still new to theoretical physics), but what exactly is the solution then? $\endgroup$ – TwoStones Nov 19 '19 at 16:57
  • 1
    $\begingroup$ @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. $\endgroup$ – Frederic Thomas Nov 19 '19 at 17:01
  • $\begingroup$ Thank you very much! $\endgroup$ – TwoStones Nov 19 '19 at 17:03
  • $\begingroup$ @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. $\endgroup$ – Frederic Thomas Nov 19 '19 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.