# Jefimenko B-field approximation

by Pengwuino
Tags: approximation, bfield, jefimenko
 PF Gold P: 7,120 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: $$B(\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup }} \over x} ,t) = \frac{{\mu _0 }}{{4\pi }}\int\limits_v {d^3 x'\{ [J(\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup }} \over x}' ,t')]_{ret} \times \frac{{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpo onup}} \over R} }}{{R^3 }}} \}$$ $$R = |\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup} } \over x} - \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} '|$$ using the taylor expansion: $$\frac{1}{{|\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\righth arpoonup}} \over x} - \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} '|^3 }} \approx \frac{1}{{|\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\righth arpoonup}} \over x} |^3 }} + \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} ' \cdot [\nabla '(\frac{1}{{|\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\righ tharpoonup}} \over x} - \mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\rightharpoonup}} \over x} '|^3 }})]_{\vec x' = 0}$$ 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
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.
 HW Helper P: 5,003 Also, you seem to be missing a term involving $$\mathbf{\dot{J}}(\textbf{x}',t_r)$$ in your original equation.
 PF Gold P: 7,120 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.
 HW Helper P: 5,003 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.