# Jefimenko B-field approximation

1. Feb 26, 2010

### Pengwuino

I'm looking to evaluate the magnetic field using Jefimenko's equations. There is two parts to it but I'm just looking at the first. The approximation is r>>r' where r' is localized about the origin. The Jefimenko's equation for the magnetic field (the first term that I'm having trouble with) has:

$$B(\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} ,t) = \frac{{\mu _0 }}{{4\pi }}\int\limits_v {d^3 x'\{ [J(\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x}' ,t')]_{ret} \times \frac{{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over R} }}{{R^3 }}} \}$$

$$R = |\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} - \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} '|$$

using the taylor expansion:

$$\frac{1}{{|\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} - \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} '|^3 }} \approx \frac{1}{{|\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} |^3 }} + \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} ' \cdot [\nabla '(\frac{1}{{|\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} - \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} '|^3 }})]_{\vec x' = 0}$$

Now, 4 terms are generated. One will go away, one is easily solved, but what I'm having trouble with are terms like this:

$$\frac{{ - \mu _0 }}{{4\pi }}\int\limits_v {(\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} ' \cdot [\nabla '(\frac{1}{{R^3 }})]_{\vec x' = 0} )\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} ' \times [J(\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} ,t')]_{ret} d^3 x}$$

I want to pull the derivative operator off the 1/R^3 through an integration by parts but since it's evaluated at r'=0, I'm not exactly confident on how to do that. I want the derivative operator on things such as $$\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} ' \times [J(\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} ',t')]_{ret}$$ which are somewhat like the magnetic moment once integrated but wonder if I still evaluate the original part at r'=0 or does the evaluation switch over to the part I'm now using the derivative operator on? Or does it go onto both now?

Signed,
Confused in Antarctica

Last edited: Feb 26, 2010
2. Feb 26, 2010

### gabbagabbahey

You can't. As you said, $$\left.\mathbf{\nabla}'\left[\frac{1}{|\textbf{x}-\textbf{x}'|^3}\right]\right|_{\texbf{x}'=0}$$ is evaluated at $\textbf{x}'=0$, making the result a function only of $\textbf{x}$. So, just calculate that function and pull it out of the integral since it has no dependence on the primed coordinates.

3. Feb 26, 2010

### gabbagabbahey

Also, you seem to be missing a term involving $$\mathbf{\dot{J}}(\textbf{x}',t_r)$$ in your original equation.

4. Feb 26, 2010

### Pengwuino

Yes, I left out that term becuase I haven't looked at it yet, I was just wanting to post what part I was having problem with. After talking to a few people we came to the same conclusion however, doing the evaulation makes it impossible to pull off the gradiant I guess. I found out the hopefuly correct way of doing it and I shall give it a shot.

5. Feb 28, 2010

### Pengwuino

So now I'm stuck with doing these 3-dimensional integration by parts. I need to work with this integral. I'm trying to do it by integration by parts but since it's in 3 dimensions, I'm not sure how it's done. I'm quite amazed that I don't think I've ever run across a 3-dimensional IVP that didnt use Gauss' law or anything. The steps so far are:

$$\begin{array}{l} \int_V {(\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} \cdot \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} ')(} \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} ' \times \left. {[J(\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} ',t')]} \right|_{ret} )d^3 x' \\ u = (\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} \cdot \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} ') \\ dw = \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} \times \left. {[J(\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} ',t')]} \right|_{ret} )d^3 x' \\ w = 2\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over m} \\ du = ? \\ \end{array}$$

Now I figure I can't simply naively say that $$du = \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} d^3 x'$$ because then I have a vector multiplying a vector. Doing the actual computation seems to give me:

$$du = xdx' + ydy' + zdz'$$

and now I have just a sum of 3 integrations. Is this all correct mathematically? It's worrisome because the integral over prime coordinates diverge since the integration is over an arbitrary volume.

6. Mar 1, 2010

### gabbagabbahey

In one dimesion, integration by parts is derived from the product rule:

$$(fg)'=f'g+fg'\implies \int f'g dx=\int(fg)'dx-\int fg' dx= fg-\int fg' dx$$

In 3 dimensions, things are not so simple; there are actually 8 product rules and each leads to a different variation of integration by parts. So, in order to use IBP in vector calculus, you need to select an appropriate product rule.

For example, if I wanted to calculate $\int f(\textbf{r})(\mathbf{\nabla}g(\textbf{r}))\cdot d\textbf{r}$ over some curve, I might find it useful to use the product rule $\mathbf{\nabla}(fg)=(\mathbf{\nabla}f)g+f(\mathbf{\nabla}g)$ to transfer the derivative to $g$ instead.