Potential of a magnetic dipole

Tags:
1. Jan 7, 2016

AwesomeTrains

Hello everyone!
I'm stuck at an Electrodynamics problem and would be happy for some guidance

1. The problem statement, all variables and given/known data

A magnetic dipole $\vec{m}(t)=\vec{m}_0cos(\omega t)$ at the origin can be described by the current density $\vec{j}(\vec{r},t)=-\vec{m}(t)\times\vec{\nabla}\delta(\vec{r})$. Calculate the retarded potentials:
$\Phi(\vec{r},t)$ and $\vec{A}(\vec{r},t)$.
Given hint: "It is easier to first do the integration by parts and then transform the derivative with right to $\vec{r}'$ into the derivative with right to $\vec{r}$".
2. Relevant equations
$\vec{A}(\vec{r},t)=\int_V\frac{\vec{j}(\vec{r},t)}{|\vec{r}-\vec{r}'|}d^3\vec{r}'$
$\Phi(\vec{r},t)=\int_V\frac{\rho(\vec{r},t)}{|\vec{r}-\vec{r}'|}d^3\vec{r}'$

3. The attempt at a solution

I tried calculating the x-component of the vector potential and got this:
$A_x(\vec{r},t)=\int_V\frac{(m_y\partial_z-m_z\partial_y)}{|\vec{r}-\vec{r}'|}\delta(\vec{r}')d^3\vec{r}' =\frac{(m_y\partial_z-m_z\partial_y)}{|\vec{r}-0|}$
Where eg. $m_y$ ist the y-component of the dipole vector. I get the last equation because of the delta distribution property $\int^{\infty}_{\infty}f(x)\delta(x)dx=f(0)$.
But this doesn't seem right, I didn't do any integration like the hint was suggesting.
Is this the right approach? I don't understand when I'm supposed to make use of the hint.

Kind regards
Alex

2. Jan 7, 2016

TSny

In the integrands, the numerators should be evaluated at $\vec{r}^{\,'}$ rather than at $\vec{r}$.

Should the time in the integrand be the same as the time on the left side of the equations?

Note that the partial derivatives should be with respect to the primed coordinates. Therefore, these derivatives act on the delta function. Before you can integrate, you need to rewrite the integrand so that the derivatives no longer act on the delta-function. This is where integration by parts comes in. After that, follow the rest of the hint that was given.

3. Jan 8, 2016

AwesomeTrains

Thanks for the help
Do you mean like this:
$\vec{A}(\vec{r},t)=\int_V\frac{\vec{j}(\vec{r},t')}{|\vec{r}'-\vec{r}|}d^3\vec{r}'$ and $\Phi(\vec{r},t)=\int_V\frac{\rho(\vec{r},t')}{|\vec{r}'-\vec{r}|}d^3\vec{r}'$?
Sorry, I forgot the primes on t on the right side. The fields propagate with c which gives the retarded time $t'=t-\frac{|\vec{r}'-\vec{r}|}{c}$.
Does this mean that $\vec{m}(t')$ depends on $\vec{r}'$?
Can I do it like this:
$\int_V\frac{m_y\partial_z}{|\vec{r}-\vec{r}'|}\delta(\vec{r}')d^3\vec{r}'=\left[\frac{m_y(\vec{r}')\delta(\vec{r}')}{|\vec{r}'-\vec{r}|}\right]^V-\int_V\frac{\partial_z\vec{m}_y(\vec{r}')}{|\vec{r}-\vec{r}'|}\delta(\vec{r}')d^3\vec{r}'$ ? I used the integration by parts formula for one dimension but that doesn't seem right. Should I use some vector calculus formula on the cross product in the beginning instead?
Also how do I convert the derivative so that it is with right to $\vec{r}$?

4. Jan 8, 2016

TSny

In the numerators the current and charge densities should be evaluated at the primed position. That is, you should have $\vec{j}(\vec{r}\, ',t')$ and $\rho(\vec{r}\, ',t')$

Yes. $\vec{m}(t')$ depends on $\vec{r \, '}$ as well as on $t$ and $\vec{r}$.
I would continue to write $m$ as $m(t \, ')$, rather than $m(\vec{r} \, ')$ because $t \, '$ depends on $t$ and $\vec{r}$, as well as $\vec{r \, '}$. You can substitute for $t \, '$ a little later.

In the integral $\int_V\frac{m_y\partial_z \, }{|\vec{r}-\vec{r}'|}\delta(\vec{r}')d^3\vec{r}'$, the partial derivative is with respect to $z \, '$, not $z$. So, it should be $\int_V\frac{m_y}{|\vec{r}-\vec{r}'|} \partial_{z \, '} \delta(\vec{r}')d^3\vec{r}'$

When integrating by parts with respect to $z \, '$, you got a term of the form $\left[\frac{m_y(t \, ')\delta(\vec{r}')}{|\vec{r}'-\vec{r}|}\right]^V$. But this term should still contain integrations over $x \, '$ and $y \, '$. Nevertheless, you can argue that this boundary term in $z \, '$ should be zero. The other term that you got when integrating by parts is - $\int_V\frac{\partial_z\vec{m}_y(t \, ')}{|\vec{r}-\vec{r}'|}\delta(\vec{r}')d^3\vec{r}'$. The partial derivative should be with respect to $z \, '$ and $\partial_{z \, '}$ should act on the entire expression $\frac{\vec{m}_y(t \, ')}{|\vec{r}-\vec{r} \,'|}$. It is at this point that you can finish the hint and rewrite $\partial_{z \, '}$ in terms of $\partial_{z}$; i.e., in terms of the unprimed $z$. To do this, note that $\frac{\vec{m}_y(t \, ')}{|\vec{r}-\vec{r} \,'|}$ is a function only of the quantity $|\vec{r}-\vec{r} \,'|$ and $t$.
You could work with vector calculus identities to make the whole derivation more elegant and compact. In particular, the second identity listed here would be useful:
https://en.wikipedia.org/wiki/Vector_calculus_identities#Product_of_a_scalar_and_a_vector.

But the way you are doing it will also get you there.