Force on a magnetic dipole in an external magnetic field

I want to find an expression for the force acting upon a magnetic dipole with dipole moment $$\mathbf{m}$$ if that dipole is positioned in a stationary, external magnetic field $$\mathbf{B}$$. The expression given for the force is the following (assuming that $$\nabla \times\mathbf{B}=0$$):

$$\mathbf{F}=(\mathbf{m}\cdot\nabla)\mathbf{B}\quad(1)$$

My question is mostly whether the expression above is equivalent to:

$$\begin{bmatrix} \frac{\partial \mathbf{B}}{\partial x} & \frac{\partial \mathbf{B}}{\partial y}& \frac{\partial \mathbf{B}}{\partial z} \end{bmatrix}\mathbf{m} \quad (2)$$

or equivalent to:

$$\begin{bmatrix} \frac{\partial \mathbf{B}}{\partial x} & \frac{\partial \mathbf{B}}{\partial y}& \frac{\partial \mathbf{B}}{\partial z} \end{bmatrix}^T\mathbf{m} \quad (3)$$

I basically found these two expressions ($$(2)$$ $$(3)$$) for the force from two different sources, so one of them must be wrong. I derived the first expression in the following way:

$$(\mathbf{m}\cdot\nabla)\mathbf{B}=(m_1\frac{\partial }{\partial x}+m_2\frac{\partial }{\partial y}+m_3\frac{\partial }{\partial z})\begin{bmatrix} B_1\\ B_2\\ B_3 \end{bmatrix}=\begin{bmatrix} m_1\frac{\partial B_1}{\partial x} + m_2\frac{\partial B_1}{\partial y} + m_3\frac{\partial B_1}{\partial z} \\ m_1\frac{\partial B_2}{\partial x} + m_2\frac{\partial B_2}{\partial y} + m_3\frac{\partial B_2}{\partial z} \\ m_1\frac{\partial B_3}{\partial x} + m_2\frac{\partial B_3}{\partial y} + m_3\frac{\partial B_3}{\partial z} \end{bmatrix}$$

The last expressions can be interpreted as the Matrix product $$(2)$$. Is that correct or am I missing something obvious?

Thanks!

When in doubt, use coordinates and index notation; the expression

$$\mathbf{F}=(\mathbf{m}\cdot\nabla)\mathbf{B}~~~(*)$$

can be written in cartesian coordinates in this way:

$$F_i = m_k \partial_k B_i$$

If you're wondering how do we know that the right-hand side of (*) expands this way, it is actually the definition of the shorthand $$\mathbf m \cdot \nabla$$.

If force coordinates are put into a row $$\mathbf F^T$$, then this row can be obtained as left multiplication of the matrix $$\mathbf G$$ with coordinates $$G_{ki} = \partial_k B_i$$ by the magnetic moment row $$\mathbf m^T$$:

$$\mathbf F^T = \mathbf m^T \cdot \mathbf G.$$

• Thanks for your response but I think your result might have an incorrect dimension since $\mathbf{m}^T\mathbf{G} \in \mathbb{R}^{1,3}$ and $\mathbf{F}\in\mathbb{R}^{3,1}$. – Mantabit Feb 13 '19 at 7:39
• @Mantabit you are right, the force on the left-hand side also has to be written as a row, because $i$ is a column index. It is fixed now. – Ján Lalinský Feb 13 '19 at 20:48
• Okay thanks, I checked your solution and it agrees with mine. – Mantabit Feb 13 '19 at 21:38