How to Evaluate the Magnetic Field Using Jefimenko's Equations?

[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{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} ,t) = \frac{{\mu _0 }}{{4\pi }}\int\limits_v {d^3 x&#039;\{ [J(\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x}&#039; ,t&#039;)]_{ret} \times \frac{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over R} }}{{R^3 }}} \}

R = |\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} - \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} &#039;|

using the taylor expansion:

\frac{1}{{|\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} - \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} &#039;|^3 }} \approx \frac{1}{{|\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} |^3 }} + \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} &#039; \cdot [\nabla &#039;(\frac{1}{{|\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} - \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} &#039;|^3 }})]_{\vec x&#039; = 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{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} &#039; \cdot [\nabla &#039;(\frac{1}{{R^3 }})]_{\vec x&#039; = 0} )\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} &#039; \times [J(\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} ,t&#039;)]_{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{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} &#039; \times [J(\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} &#039;,t&#039;)]_{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

 

[gabbagabbahey]

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.

You can't. As you said, \left.\mathbf{\nabla}&#039;\left[\frac{1}{|\textbf{x}-\textbf{x}&#039;|^3}\right]\right|_{\texbf{x}&#039;=0} is evaluated at \textbf{x}&#039;=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.

 

[gabbagabbahey]

Also, you seem to be missing a term involving \mathbf{\dot{J}}(\textbf{x}&#039;,t_r) in your original equation.

 

[Pengwuino]

Yes, I left out that term because 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.

 

[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}<br /> \int_V {(\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} \cdot \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} &#039;)(} \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} &#039; \times \left. {[J(\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} &#039;,t&#039;)]} \right|_{ret} )d^3 x&#039; \\ <br /> u = (\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} \cdot \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} &#039;) \\ <br /> dw = \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} \times \left. {[J(\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} &#039;,t&#039;)]} \right|_{ret} )d^3 x&#039; \\ <br /> w = 2\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over m} \\ <br /> du = ? \\ <br /> \end{array}

Now I figure I can't simply naively say that du = \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over x} d^3 x&#039; because then I have a vector multiplying a vector. Doing the actual computation seems to give me:

du = xdx&#039; + ydy&#039; + zdz&#039;

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.

 

[gabbagabbahey]

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

(fg)&#039;=f&#039;g+fg&#039;\implies \int f&#039;g dx=\int(fg)&#039;dx-\int fg&#039; dx= fg-\int fg&#039; 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.