# One of the definitions of magnetic field which is given by using nabla operators

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

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

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

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

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

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

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

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

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

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

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.

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

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 $$\nabla (\mathbf{M}\cdot\mathbf{r}) = \nabla (M_1x + M_2 y + M_3 z) = \mathbf{M}$$