Jefimenko B-field approximation


by Pengwuino
Tags: approximation, bfield, jefimenko
Pengwuino
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#1
Feb26-10, 01:42 AM
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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:

[tex]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 }}} \} [/tex]

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

using the taylor expansion:

[tex]\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} [/tex]

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

[tex]\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} [/tex]

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 [tex]\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}
\over x} ' \times [J(\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$ }}
\over x} ',t')]_{ret} [/tex] 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,
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gabbagabbahey
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#2
Feb26-10, 10:37 AM
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Quote Quote by Pengwuino View Post
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, [tex]\left.\mathbf{\nabla}'\left[\frac{1}{|\textbf{x}-\textbf{x}'|^3}\right]\right|_{\texbf{x}'=0}[/tex] is evaluated at [itex]\textbf{x}'=0[/itex], making the result a function only of [itex]\textbf{x}[/itex]. So, just calculate that function and pull it out of the integral since it has no dependence on the primed coordinates.
gabbagabbahey
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Feb26-10, 10:45 AM
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Also, you seem to be missing a term involving [tex]\mathbf{\dot{J}}(\textbf{x}',t_r)[/tex] in your original equation.

Pengwuino
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#4
Feb26-10, 02:58 PM
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Jefimenko B-field approximation


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.
Pengwuino
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#5
Feb28-10, 11:36 PM
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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:

[tex]\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}[/tex]

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

[tex]du = xdx' + ydy' + zdz'[/tex]

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
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#6
Mar1-10, 11:40 AM
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In one dimesion, integration by parts is derived from the product rule:

[tex](fg)'=f'g+fg'\implies \int f'g dx=\int(fg)'dx-\int fg' dx= fg-\int fg' dx[/tex]

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 [itex]\int f(\textbf{r})(\mathbf{\nabla}g(\textbf{r}))\cdot d\textbf{r}[/itex] over some curve, I might find it useful to use the product rule [itex]\mathbf{\nabla}(fg)=(\mathbf{\nabla}f)g+f(\mathbf{\nabla}g)[/itex] to transfer the derivative to [itex]g[/itex] instead.


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