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I was reading Electromagnetism by David Tong. In Chapter 3 (Page 60), he deduced $$\mathbf{F} = ∇(\mathbf{B} \cdot \mathbf{m})\tag{3.28}$$ by Taylor expanding $\mathbf{B}$ around a certain point $\mathbf{r} = \mathbf{R}$. So I would think of this as the force of a current distribution with magnetic dipole $m$ would approximately experience in the space.

Then I got confused in the next section with this understanding. In the next chapter, Tong used this formula he derived to calculate the force between two dipoles. He first got, $$\mathbf{F} = \frac{\mu_0}{4\pi} \nabla \left[ \frac{3(\mathbf{m}_1 \cdot \hat{\mathbf{r}})(\mathbf{m}_2 \cdot \hat{\mathbf{r}}) - \mathbf{m}_1 \cdot \mathbf{m}_2}{r^3} \right]\tag{p.61}$$ and then $$\mathbf{F} = \frac{3\mu_0}{4\pi r^4} \left[ (\mathbf{m}_1 \cdot \hat{\mathbf{r}})\mathbf{m}_2 + (\mathbf{m}_2 \cdot \hat{\mathbf{r}})\mathbf{m}_1 + (\mathbf{m}_1 \cdot \mathbf{m}_2)\hat{\mathbf{r}} - 5(\mathbf{m}_1 \cdot \hat{\mathbf{r}})(\mathbf{m}_2 \cdot \hat{\mathbf{r}})\hat{\mathbf{r}} \right].\tag{3.31}$$ He still kept this $r$, saying that the dipole force drop off as $1/r^4$. But I don't understand how does this $r$ still mean anything here. It is not the distance between two dipoles. In Tong's own calculation, $$\mathbf{F} = \int_V d^3r \, \mathbf{J}(\mathbf{r}) \times \left[ (\mathbf{r} \cdot \nabla') \mathbf{B}(\mathbf{r}') \right] \Bigg|_{\mathbf{r}'=\mathbf{R}}.\tag{p.60}$$ He had this idea of evaluating at where $\mathbf{B}$ was expanded. But in the example, I cannot understand what does $r$ mean but a reference point you choose in space.

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    $\begingroup$ What makes you think it is not the distance between the two dipoles? In the derivation of the gradient expression, r is the distance from the point where the current density is being evaluated. $\endgroup$ Commented Mar 19 at 23:24
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    $\begingroup$ @Ruffolo but what does it mean for a dipole to be located somewhere? Isn't it a current distribution in the space? I don't quite get what you mean, a distance need two endpoints to measure the distance between them, if in this case one is R (the point where the 1st current distribution is distributed around), what is the other? We have already integrated along the 2nd current distribution, there is no 2nd point explicitly written in the notes. $\endgroup$ Commented Mar 20 at 0:07
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    $\begingroup$ Well then what is the force on? In your original expression for a force on a dipole in a field, where is the force applied, if the dipole is not located somewhere? $\endgroup$ Commented Mar 20 at 0:13
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    $\begingroup$ In the treatment by Tong you are supposed to consider the current distributions to be “localized” even in the original expression. Then in the dipole-dipole case you are supposed to consider r to be much larger than the size of the localized current distributions that make up the two dipoles. It’s not spelled out, but that’s the way to think of it. $\endgroup$ Commented Mar 20 at 0:33
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    $\begingroup$ For the general case you’d have varying forces at every current element. It wouldn’t look clean like in Tong. $\endgroup$ Commented Mar 20 at 0:34

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I am not sure I fully understand your question. I exclude the case of strictly point-like distributions and only consider distributions of finite extent, with a characteristic length denoted by $H$.

To determine the force acting on the distribution, one performs a Taylor expansion around a point denoted $Q$, with a remainder that is not written explicitly but whose order of magnitude can be estimated:

$$\vec{B}(P) = \vec{B}(Q) + \left( \overrightarrow{QP} \cdot \nabla \right)\vec{B} + \mathcal{O}\!\left( B(Q)\,\frac{QP^2}{L^2} \right)$$

Here, $\mathcal{O}$ denotes Landau notation, and $L$ is the characteristic length scale over which the field varies.

In general, one can neglect the remainder $$\mathcal{O}\!\left( B(Q)\, \frac{QP^2}{L^2} \right)$$ provided that $$\frac{QP^2}{L^2} \ll 1$$.

If one imposes $H \ll L$ and chooses the point $Q$ such that $QP$ is of order $H$, then the remainder in the Taylor expansion can indeed be neglected. This shows that the choice of the point $Q$ is not critical, as long as it does not lie too far from the center of the distribution.

If the point $Q$ is poorly chosen, such that $QP > H$, then neglecting the remainder is no longer justified and the formula loses its meaning.

Note that the same issue arises when stating that the torque on a dipole is $$\vec{\Gamma} = \vec{m} \times \vec{B}(Q)$$: the point $Q$ must be chosen according to the same criteria.

In the general case of an arbitrary point $Q'$, one must write:

$$\vec{\Gamma}(Q') = \vec{m} \times \vec{B}(Q) + \overrightarrow{Q'Q} \times \vec{R}$$

with

$$\vec{R} = \nabla \left( \vec{m} \cdot \vec{B} \right)$$

the resultant force acting on the dipole.

Hope it can help !

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