I'm solving some excercises on magnetostatics, and encounterded this on which i'm having some trouble. Given a particle of magnetic dipole moment $\mathbf{m}$, show that its current density is given as $\mathbf{J} = \left( \mathbf{m} \times \nabla \right) \delta \left( \mathbf{r} - \mathbf{r}_{0} \right)$, where $\mathbf{r}_{0}$ is the vector position of the particle.

I started from the stationary Ampere's law, given that the magnetic field $\mathbf{B}_{m}$ is due only by the magnetic dipole moment

$$ \nabla \times \mathbf{B}_{m} \left( \mathbf{r} - \mathbf{r}_{0} \right) = \mu_{0} \; \mathbf{J} \left( \mathbf{r} \right)$$

So that

$$ \boxed{\mathbf{J} \left( \mathbf{r} \right) = \frac{1}{\mu_{0}} \nabla \times \mathbf{B}_{m} \left( \mathbf{r} - \mathbf{r}_{0} \right)} \; \; \; \; (1)$$

Now, it terms of the magnetic dipole moment, the corresponding magnetic field is given as

$$ \boxed{\mathbf{B}_{m} \left( \mathbf{r} - \mathbf{r}_{0} \right) = \frac{\mu_{0}}{4\pi} \left[ \frac{\mathbf{m}\cdot\left(\mathbf{r} - \mathbf{r}_{0}\right)}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{5}} \left( \mathbf{r} - \mathbf{r}_{0} \right) - \frac{\mathbf{m}}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{3}} \right]} \; \; \; \; (2) $$

Replacing, then, (2) into (1)

$$ \mathbf{J}\left( \mathbf{r} \right) = \frac{1}{4 \pi} \nabla \times \left( \frac{\mu_{0}}{4\pi} \left[ \frac{\mathbf{m}\cdot\left(\mathbf{r} - \mathbf{r}_{0}\right)}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{5}} \left( \mathbf{r} - \mathbf{r}_{0} \right) - \frac{\mathbf{m}}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{3}} \right] \right) $$

$$ = \frac{1}{4 \pi} \left[ \nabla \times \left( \frac{\mathbf{m}\cdot\left(\mathbf{r} - \mathbf{r}_{0}\right)}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{5}} \left( \mathbf{r} - \mathbf{r}_{0} \right) \right) - \nabla \times \left( \frac{\mathbf{m}}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{3}} \right) \right] $$

$$ = \frac{1}{4 \pi} \left[ \nabla \left( \frac{\mathbf{m}\cdot\left(\mathbf{r} - \mathbf{r}_{0}\right)}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{5}} \right) \times \left( \mathbf{r} - \mathbf{r}_{0} \right) + \frac{\mathbf{m}\cdot\left(\mathbf{r} - \mathbf{r}_{0}\right)}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{5}} \nabla \times \left( \mathbf{r} - \mathbf{r}_{0} \right) \right. $$

$$ \left. - \nabla \left( \frac{\mathbf{1}}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{3}} \right) \times \mathbf{m} - \frac{\mathbf{1}}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{3}} \nabla \times \mathbf{m} \right] $$

Now, as $\nabla \times \left( \mathbf{r} - \mathbf{r}_{0} \right)=0$ and $\nabla \times \mathbf{m}=0$

$$ \boxed{\mathbf{J}\left( \mathbf{r} \right) = \frac{1}{4 \pi} \left[ \nabla \left( \frac{\mathbf{m}\cdot\left(\mathbf{r} - \mathbf{r}_{0}\right)}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{5}} \right) \times \left( \mathbf{r} - \mathbf{r}_{0} \right) - \nabla \left( \frac{\mathbf{1}}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{3}} \right) \times \mathbf{m} \right]} \; \; \; \; (3) $$

Now, by components

$$ \left( \nabla \left[ \frac{\mathbf{m}\cdot\left(\mathbf{r} - \mathbf{r}_{0}\right)}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{5}} \right] \right)_{i} = \partial_{i} \left( \frac{m_{j} (x_{j} - x^{0}_{j}) }{ [(x_{l} - x^{0}_{l})(x_{l} - x^{0}_{l})]^{5/2} } \right)$$

$$ = m_{j} \left\{ \frac{1}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {5}} \partial_{i} (x_{j} - x^{0}_{j}) + (x_{j} - x^{0}_{j}) \partial_{i} [(x_{l} - x^{0}_{l})(x_{l} - x^{0}_{l})]^{-5/2} \right\} $$

$$ = m_{j} \left\{ \frac{\delta_{ij}}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {5}} - \frac{5}{2}\frac{1}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {7}}2(x_{l} - x^{0}_{l})\delta_{il}(x_{j} - x^{0}_{j}) \right\} $$

$$ = m_{j} \left\{ \frac{\delta_{ij}}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {5}} - 5\frac{(x_{i} - x^{0}_{i})(x_{j} - x^{0}_{j})}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {7}} \right\} $$

$$ = \frac{m_{i}}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {5}} - 5 \frac{m_{j} (x_{j} - x^{0}_{j})}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {7}}(x_{i} - x^{0}_{i}) $$

$$ \boxed{\left( \nabla \left[ \frac{\mathbf{m}\cdot\left(\mathbf{r} - \mathbf{r}_{0}\right)}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{5}} \right] \right)_{i} = \left[ \frac{\mathbf{m}}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {5}} - 5 \frac{\mathbf{m} \cdot (\mathbf{r} - \mathbf{r}_{0}) }{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {7}}(\mathbf{r} - \mathbf{r}_{0}) \right]_{i}} \; \; \; \; (4) $$


$$ \left( \nabla \left[ \frac{1}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{3}} \right] \right)_{i} = \partial_{i} [(x_{j} - x_{j}^{0})(x_{j} - x_{j}^{0})]^{-3/2} $$

$$ = -\frac{3}{2} \frac{1}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {5}}2(x_{j} - x_{j}^{0})\delta_{ij} $$

$$ = -3 \frac{1}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {5}} (x_{i} - x_{i}^{0}) $$

$$ \boxed{\left( \nabla \left[ \frac{1}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{3}} \right] \right)_{i} = \left( -3 \frac{1}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {5}} (\mathbf{r} - \mathbf{r}_{0}) \right)_{i}} \; \; \; \; (5) $$

Replacing (4) and (5) into (3)

$$ \mathbf{J}\left( \mathbf{r} \right) = \frac{1}{4 \pi} \left[ \left( \frac{\mathbf{m}}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {5}} - 5 \frac{\mathbf{m} \cdot (\mathbf{r} - \mathbf{r}_{0}) }{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {7}}(\mathbf{r} - \mathbf{r}_{0}) \right) \times \left( \mathbf{r} - \mathbf{r}_{0} \right) - \left( -3 \frac{1}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {5}} (\mathbf{r} - \mathbf{r}_{0}) \right) \times \mathbf{m} \right] $$

And, because $\left( \mathbf{r} - \mathbf{r}_{0} \right) \times \left( \mathbf{r} - \mathbf{r}_{0} \right) = 0$

$$ \mathbf{J}\left( \mathbf{r} \right) = \frac{1}{4 \pi} \left[ \left( \frac{\mathbf{m}}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {5}} \right) \times \left( \mathbf{r} - \mathbf{r}_{0} \right) + 3 \frac{1}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {5}} (\mathbf{r} - \mathbf{r}_{0}) \times \mathbf{m} \right] $$

$$ \mathbf{J}\left( \mathbf{r} \right) = \frac{1}{4 \pi} \left(-2 \left( \frac{\mathbf{m}}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {5}} \right) \times \left( \mathbf{r} - \mathbf{r}_{0} \right) \right)$$

$$ \boxed{\mathbf{J}\left( \mathbf{r} \right) = -\frac{1}{2 \pi} \mathbf{m} \times \frac{\left( \mathbf{r} - \mathbf{r}_{0} \right)}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {5}}} $$

Here is where i'm stuck. I'm aware of the identity

$$ \nabla \cdot \left[ \frac{\mathbf{r} - \mathbf{r}_{0}}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^{3}} \right] = 4 \pi \delta (\mathbf{r} - \mathbf{r}_{0})$$

So, assuming that the answer presented in the statement is true (and that my calculations were correct), then it must be true that

$$ -\frac{1}{2 \pi} \frac{\left( \mathbf{r} - \mathbf{r}_{0} \right)}{\left| \mathbf{r} - \mathbf{r}_{0} \right|^ {5}} = \nabla \delta \left( \mathbf{r} - \mathbf{r}_{0} \right) $$

but, if this is indeed true, i'm not sure how to prove this.

Thanks in advance for any help!


A Herculean effort. Well done. It looks correct though I had only a brief look. Griffiths (Introduction to Electrodynamics) has a quite nice section on 3d delta functions in Sec. 1.5

The trick with delta functions is that they only make sense inside the integral. You want to analyse $\boldsymbol{\nabla}.\left(\frac{\mathbf{r}-\mathbf{r}_0}{\left|\mathbf{r}-\mathbf{r}_0\right|^3}\right)$. First, simplify it to:


Note, that since $\mathbf{r}_0$ is constant, there is no difficulty in switching from $\boldsymbol{\nabla}$ with respect to $\mathbf{r}$, to $\boldsymbol{\nabla}$ with respect to $\boldsymbol{r}$. Next, evaluate expression for $\boldsymbol{r}\neq\mathbf{0}$, you will find

$$\boldsymbol{\nabla}.\left(\frac{\boldsymbol{r}}{\left|\boldsymbol{r}\right|^3}\right)=0 \mbox{ for } \boldsymbol{r}\neq\mathbf{0}$$

This is a good start, if you are aiming to end up with a delta function. Next stick your candidate into an integral. Let $V$ be a sphere with radius $R$, and centre at $\mathbf{r}_0$

$$\int_V d^3 r \boldsymbol{\nabla}.\left(\frac{\boldsymbol{r}}{\left|\boldsymbol{r}\right|^3}\right)=\oint_{\partial V} d^2\, r\boldsymbol{\hat{r}}.\frac{\boldsymbol{r}}{\left|\boldsymbol{r}\right|^3}=\oint_{\partial V} d^2\, r \frac{1}{r^2}=\int_\mbox{solid angle} R^2 d^2\Omega \frac{1}{R^2}=4\pi$$

So you have a situation where your quantity is zero everywhere except one point, and at that one point, if you integrate your quantity, it gives a finite constant value = that's a delta function. You can do it with more rigour, but basically all you need is here, so:


Having said it, your approach is strange. You are using magnetic field to find the magnetic dipole. I would go in the other direction. I would define magnetic dipole to be either delta-function of magnetization, or a current due to point-like loop of current (yet another way to go is to consider low-order series expansion of the relevant Greens function - will ignore here). The second approach is more fun, so let us go with it.

Assume the current is circulating in anti-clockwise direction around the z-axis, in $z=0$ plane, and that loop radius is $R$. The current density is then (using cyllindrical coordinates $\rho,\phi,z$):


If the current through the loop is $I$, by integrating in $\mathcal{P}=\{y=0,x>0\}$ plane we can find:

$$I=\int_\mathcal{P}dx\,dz\,\mathbf{\hat{y}}.\mathbf{J}=\alpha \int_0^\infty d\rho\, \delta\left(\rho-R\right)=\alpha$$

Thus $\alpha=I$. As before, delta functions only make sense inside an integral, so let us stick our expression inside an integral with a arbitrary well-behaved vector field $\mathbf{V}$:

$$\int_V d^3 r \mathbf{V}.\mathbf{J}=I\int dz\, \int_0^\infty d\rho\int_0^{2\pi}\rho\,d\phi V_\phi \delta\left(\rho-R\right)\delta\left(z\right) = I R\int_0^{2\pi}d\phi \,V_\phi\bigg|_{\rho=R,\,z=0}=I \oint_{\partial \mathcal{D}} dl \mathbf{\hat{l}}.\mathbf{V}=I \int_{\mathcal{D}} d^2\rho \mathbf{\hat{z}}.\boldsymbol{\nabla}\times\mathbf{V}$$

In the penultimate step I have introduced a disk region ($\mathcal{D}$) of radius $R$, lying in $z=0$ plane at the origin, and interpreted the integral as the integral around the boundary of that disk. We can then use the inverse of the Stokes theorem (which works in simply-connected space) and deduce the expression for the integral. We can now take the limit $R\to 0$ to get:

$$\lim_{R\to 0}\int_V d^3 r \mathbf{V}.\mathbf{J}=I\pi R^2 \mathbf{\hat{z}}.\left(\boldsymbol{\nabla}\times\mathbf{V}\right)\bigg|_{\mathbf{r}=0}$$

All that remains to show, by integration by parts, is that:

$$\int_V d^3 r \mathbf{V}.\boldsymbol{\nabla}\times\mathbf{\hat{z}}\delta\left(\mathbf{r}\right)=\mathbf{\hat{z}}.\left(\boldsymbol{\nabla}\times\mathbf{V}\right)\bigg|_{\mathbf{r}=0}$$

Thus, setting $\mathbf{m}=\mathbf{\hat{z}}I\pi R^2$:


where $\mathbf{m}$ is a constant so it can be outside the derivative, there still seems to be sign mismatch. Because I have seen such problems before, I am inclined to think that my sign is correct, but I could be wrong.

| cite | improve this answer | |
  • $\begingroup$ Wow, I was absolutetly oblivius on how to treat the dirac delta identities, thanks a lot! Also, thanks for your answer, it looks a lot clearer than my approach! $\endgroup$ – Jpmarulandas Mar 19 at 0:59

Not the answer you're looking for? Browse other questions tagged or ask your own question.