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I am having a lot of difficulty finding an approach to solving the following question:

A dyon is a particle with both electric and magnetic charge; in suitable units

$$\mathbf{E} = \frac{-Q}{4\pi\varepsilon_0 r^2 } \mathbf{e_r}$$

$$\mathbf{B} = \frac{-P}{4\pi\mu_0 r^2 } \mathbf{e_r}$$

A particle of mass $m$ and charge $e$ moves in the field of the dyon. Using the equation of motion for a charged particle

$$\mathbf{F}= e (\mathbf{E} + \mathbf{v} \times \mathbf{B})$$

and the expression for velocity in spherical polars

$$\dot{\mathbf{r}} = \dot{r} \mathbf{e_r} + r \dot\theta \mathbf{e_\theta} + r \dot\varphi \sin\theta \,\mathbf{e_\varphi}$$

find the vector equation of motion in the spherical polar basis $\{\mathbf{e_r}, \mathbf{e_\theta}, \mathbf{e_\varphi} \}$

I have attempted to solve the problem by first finding the cross product of $\mathbf{v} \times\mathbf{B}$

Since $\mathbf{v}= \dot{\mathbf{r}}$

Then just following through with the given equation of motion by adding vectors $\mathbf{E}+(\mathbf{v} \times \mathbf{B})$

Then finally multiplying everything through with the scalar $e$.

However I am not completely sure that this is the correct way to go about this problem, all help is very much greatly appreciated.

David K
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mnmakrets
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  • Friendly reminder that this page has a basic primer on how to properly type and format mathematics on this site so your equations are more easily understood. – JMoravitz Jul 15 '15 at 00:08
  • I just edited it, could you please next time not use whatever it was for the dots? I somehow wasn't able to delete them...the $LaTeX$ for the $\dot{x}$ can be seen by you if you make a right click and choose Tex commands – user190080 Jul 15 '15 at 00:21
  • Thank you so much, so sorry about the formatting errors. – mnmakrets Jul 15 '15 at 00:40
  • You might want to look at this answer to another question about cross products using spherical coordinates: http://math.stackexchange.com/a/243303/139123 – David K Jul 15 '15 at 02:39
  • I don't think there is anything fundamentally wrong in what you are doing. There is nothing special about spherical coordinates as far as the cross product is concerned. If $(r,\theta,\phi)$ is a right-handed coordinate system, then you will have $\mathbf{e_r} \times \mathbf{e_\theta} = \mathbf{e_\phi}$, etc.). The expression for $\ddot{\mathbf{r}}$ is a bit tricky, however. Just ensure that you stick with the "physics" spherical coordinate system as you have it - e.g. https://en.wikipedia.org/wiki/Spherical_coordinate_system - rather than the "mathematics" one. – Marconius Jul 15 '15 at 03:27

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