# how to express the result of triple cross product of two vectors in spherical coordinates (unit vector)

I have a system of differential equations like below: (real problem is more complex, here is just an example)

$$\frac{\mathrm{d}\textbf{M}_1}{\mathrm{d}t}=\frac{A}{M_{1}M_{2}}[\textbf{M}_1\times(\textbf{M}_1\times\textbf{M}_2)]+\frac{B}{M_1}(\textbf{M}_1\times\textbf{M}_2) \\ \frac{\mathrm{d}\textbf{M}_2}{\mathrm{d}t}=\frac{C}{M_{2}M_{1}}[\textbf{M}_2\times(\textbf{M}_2\times\textbf{M}_1)]+\frac{D}{M_2}(\textbf{M}_2\times\textbf{M}_1)$$ A, B, C, and D are just constants. $\textbf{M}_1$ and $\textbf{M}_2$ are two vectors and their derivative over time (t) has such relation. Since the magnitude of $\textbf{M}_1$ and $\textbf{M}_2$ is unchanged over time, and I am only interested in the change of $\theta$ and $\phi$ of both vectors, so I want to express the equations in spherical coordinates and use unit vectors of $\theta$ and $\phi$. Thus, the equations can be separated to 4 equations, each of which deals with only one parameter: $\hat{\theta}_1$, $\hat{\phi}_1$, $\hat{\theta}_2$ and $\hat{\phi}_2$. (In my case, $\theta_{1,2}$ is the angle between $\textbf{M}_{1,2}$ and +z direction, ${\phi}_{1,2}$ is the angle of $\textbf{M}_{1,2}$ with +x direction on x-y plane.)

The left side of the equations can thus be re-written as:

$$\frac{\mathrm{d}\textbf{M}_1}{\mathrm{d}t}=M_1(\frac{\mathrm{d}\theta_1}{\mathrm{d}t}\hat{\theta}_1+\frac{\mathrm{d}\phi_1}{\mathrm{d}t}\sin\theta_1\hat{\phi}_1) \\ \frac{\mathrm{d}\textbf{M}_2}{\mathrm{d}t}=M_2(\frac{\mathrm{d}\theta_2}{\mathrm{d}t}\hat{\theta}_2+\frac{\mathrm{d}\phi_2}{\mathrm{d}t}\sin\theta_2\hat{\phi}_2)$$

But I can’t express the right side of the equations to similar expression with $\hat{\theta}_1$, $\hat{\phi}_1$, $\hat{\theta}_2$ and $\hat{\phi}_2$.

So my question is, can we express the items on the right side as similar as the left side of the differential equations? I’ve tried this equation: $\textbf{a}\times(\textbf{b}\times\textbf{c})=\textbf{b}(\textbf{a}\cdot\textbf{c})-\textbf{c}(\textbf{a}\cdot{\textbf{b}})$ for that triple cross product. But it results in an equation with $\textbf{M}_1$ and $\textbf{M}_2$, and seems not to be expressed by $\hat{\theta}_1$, $\hat{\phi}_1$, $\hat{\theta}_2$ and $\hat{\phi}_2$ further. For $\textbf{M}_1\times\textbf{M}_2$, I have no clue...

Thanks a lot for your help!

Best regards,

You can write ${\bf M}_{1,2}=M_{1,2}\hat{r}_{1,2}$. Using the relationship between spherical and polar coordinates such as in https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates#Unit_vector_conversions, the second table, it is trivial to show that $\hat{r}=\hat{\theta}\times\hat{\phi}$, and $$\frac{d\hat{r}}{dt}=\frac{d\theta}{dt}\frac{\partial \hat{r}}{\partial \theta}+\frac{d\phi}{dt}\frac{\partial \hat{r}}{\partial \phi}=\frac{d\theta}{dt}\hat{\theta}+\frac{d\phi}{dt}\sin\theta\hat{\phi}$$ You already used this equation for the left side.
The cross product ${\bf M}_1\times{\bf M}_2$ now can be written as $${\bf M}_1\times{\bf M}_2=M_1M_2\hat{r}_1\times\hat{r}_2=M_1M_2(\hat{\theta}_1\times\hat{\phi}_1)\times(\hat{\theta}_2\times\hat{\phi}_2)$$ You can now use the following forms of the quadruple cross product $$(A\times B)\times(C\times D)=(D\cdot(A\times B))C-(C\cdot(A\times B))D\\=(A\cdot(C\times D))B-(B\cdot(C\times D))A$$ to obtain equations in terms of $\hat{\theta}_1$, $\hat{\theta}_2$, $\hat{\phi}_1$, $\hat{\phi}_2$.
• Thanks! I think maybe the triple cross product is easier, since I calculated by using the equation a×(b×c)=b(a⋅c)−c(a⋅b), and it turns out to be just a sum of $\textbf{M}_1$ and $\textbf{M}_2$. But for the next step, how to express them in spherical unit vector $\hat{\theta}$ and $\hat{\phi}$, I have no idea. Is it still not separable? – Enlong Liu Aug 17 '16 at 14:33