potential - How to cast dipole point charge force expression from cylindrical to Cartesian coordinates

I am currently building simulations of molecular dynamics and one thing I want to model is dipole interactions. I recently came across this post about calculating the force between a point charge and a dipole.

The libraries that I'm using only work in Cartesian coordinates, so I wanted to cast my calculations down to (x,y,z) vectors to control the visualization.

I'm fairly confident that what I have is fine until I do the coordinate transformation (though please point out problems!). In particular, I worry that the theta that I'm defining between my dipole and r-hat is not the same as the theta in the basis matrix.

Would appreciate some guidance on my work!

Answer

In my opinion you are going about this the wrong way. It is simpler to write the expressions in terms of the appropriate vectors, and use the chain rule, the product rule, and the known expressions for the gradient of functions involving the position vector. You can then easily evaluate the expressions in a Cartesian coordinate system.

The potential energy of interaction between the charge and the dipole is $$ V = -k \frac{\mu_1 Q_2}{r^3} \vec{e}\cdot\vec{r} $$ where the cosine of the angle is written as a scalar product of the unit vector $\vec{e}$ defining the direction of the dipole, and the unit vector $\vec{r}/r$ pointing towards the dipole from the charge. Notice that this has introduced an extra power of $r$ in the denominator. Then you know that \begin{gather*} \vec{\nabla} (\vec{e}\cdot\vec{r}) = \vec{e} \\ \text{and}\quad \vec{\nabla} r = \frac{\vec{r}}{r} , \quad\text{hence}\quad \vec{\nabla} r^{-3} = -3r^{-4} \vec{\nabla} r = -3r^{-5} \vec{r}. \end{gather*} So $$ \vec{F} = -\vec{\nabla} V = k \frac{\mu_1 Q_2}{r^3} \left[ \vec{e} - 3\frac{\vec{e}\cdot\vec{r}}{r^2}\vec{r} \right]. $$ Obviously equal and opposite forces act on the two particles. You'll need to take care with the signs.

You don't mention this, but if your dipolar molecule is moving, you will also need to evaluate the torque on it. This is easily done from the formula $$ \vec{\tau} = -\vec{e} \times \vec{\nabla}_e V = k \frac{\mu_1 Q_2}{r^3} \vec{e}\times\vec{r} $$ where $\vec{\nabla}_e$ means the gradient with respect to $\vec{e}$ rather than $\vec{r}$, and $\times$ is the vector product. There is no torque acting on the charge (it is a point object).