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$$ \boldsymbol{M}:=\text{magnetic moment vector} $$

$$ \boldsymbol{H} := \text{magnetic field vector which is generated at }~ \boldsymbol{r} ~ \text{by} ~ \boldsymbol{M} ~ $$

$$ \boldsymbol{r}:=\text{displacement vector which is composed of }~r~\text{and}~\theta~$$

$$ \phi_{m}:=\frac{ \boldsymbol{M} \cdot \boldsymbol{r} }{ 4\pi\mu_{0} r ^{3} } ~~ \leftarrow~~ \text{magnetic scalar potential} $$

$$ \boldsymbol{H}:= - \nabla \phi_{m} $$

$$ = - \nabla \left( \frac{ \boldsymbol{M} \cdot \boldsymbol{r} }{ 4\pi\mu_{0} r ^{3} } \right) $$

$$ = - \frac{ 1 }{ 4\pi\mu_{0} } \nabla \left( \frac{ \boldsymbol{M}\cdot \boldsymbol{r} }{ r ^{3} } \right)  \tag{1} $$

$$ = -\frac{ 1 }{ 4\pi\mu_{0} } \left\{ \left( \boldsymbol{M}\cdot \boldsymbol{r} \right) \nabla \left( \frac{ 1 }{ r ^{3} } \right) + \frac{ 1 }{ r ^{3} } \nabla \left( \boldsymbol{M}\cdot \boldsymbol{r} \right) \right\} \tag{2}$$

$$ = -\frac{ 1 }{ 4\pi\mu_{0} } \left\{ \left( \boldsymbol{M}\cdot \boldsymbol{r} \right) \left( -\frac{ 3 }{ r ^{4} } \frac{ \boldsymbol{r} }{ r } \right) + \frac{ 1 }{ r ^{3} } \boldsymbol{M} \right\} \tag{3} $$

$$ = \frac{1}{4\pi\mu_{0}} \left\{ -\frac{ \boldsymbol{M} }{ r ^{3} } + \frac{ 3 \left( \boldsymbol{M} \cdot \boldsymbol{r} \right) \boldsymbol{r} }{ r ^{5} } \right\} \tag{4} $$

About the first tag, I can easily get that the constant can be got out from the argument of nabla operator.

About the second tag, I can get that the operation similar to $~ \left( f(x)g(x) \right)'=f(x)'g(x)+ f(x)g(x)' ~$ is done .

About the third tag, what are going on.

First things to first,

$$ \nabla\left( \frac{ 1 }{ r ^{3} } \right) = -\frac{ 3 }{ r ^{4} } \frac{ \boldsymbol{r} }{ r } $$

Why the right term can gained from the left term?

Second thing is

$$ \nabla\left( \boldsymbol{M}\cdot\boldsymbol{r_{}} \right) = \boldsymbol{M} $$

Which website(s) should I refer?

Added. $~ 1/r ^{3} ~$ is not a bold form but of course it can be assumed that $~ 1 / r ^{3} ~$ has a direction.

So,

$$ \nabla\left( 1/r ^{3} \right) $$

$$ = -\frac{ 3 }{ r ^{4} } \cdot \frac{ \boldsymbol{r} }{ r } +0 $$

About the left term is the form of differentiated by the scalar $~r~$ and the right term is the form of differentiated by the direction.

About the right fraction of the left term,

I can guess that this fraction represents the unit vector with some direction.

But why the differentiation of direction? is to be zero?

ps. I will back after about 10 hours.

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  • $\begingroup$ Have you tried evaluating that $\nabla(1/r^3)$ componentwise? $\endgroup$
    – J. Murray
    Commented Jun 29, 2021 at 13:07
  • $\begingroup$ $1/r^3$ is a scalar and does not have a direction, see my answer below. $\endgroup$ Commented Jun 29, 2021 at 13:41

1 Answer 1

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Regarding your first question, I agree with J. Murray's comment that you should try evaluating the gradient component-wise.

For example for the $x$ component you'll have

$$\frac{\partial}{\partial x} \left(\frac{1}{r^3}\right) = \frac{-3}{r^4}\frac{\partial r}{\partial x} = \frac{-3}{r^4}\frac{\partial}{\partial x}\sqrt{x^2 + y^2 + z^2} \\ = \frac{-3}{r^4}\frac{x}{r}$$ if you do this for all components you'll get the expression you need.

Regarding your second point, I think $\mathbf{M}$ is constant, in which case $$\nabla (\mathbf{M}\cdot\mathbf{r}) = \nabla (M_1x + M_2 y + M_3 z) = \mathbf{M}$$

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  • $\begingroup$ Sorry, your formulas are too difficult for me currently.. $\endgroup$ Commented Jun 29, 2021 at 14:21
  • $\begingroup$ Do you understand what the nabla operator is and what derivatives are? Maybe you should start by asking a question (or friends/teachers) about multidimensional calculus/derivatives. $\endgroup$ Commented Jun 29, 2021 at 14:30

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