Apr 24th 2011, 02:13 PM #1 Member   Join Date: Nov 2010 Posts: 40 Please help with this derivation on part of the retarded potential I cannot do it with Latex, so I have to up load with image!! I have to actually go to physics forum, type in the Latex there and then copy on to Woed, then scan and then upload onto Tinypic to put it on here. If you have any better idea to do Latex direct here, that would save a lot of time. Please help me on this, Thanks Alan
Apr 24th 2011, 05:05 PM   #2

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 Originally Posted by yungman I cannot do it with Latex...
I'll look at the rest of the problem (no promises) but I can quickly address the LaTeX problem. There is a sticky about this in the LaTeX Help forum, but briefly we can use the "command line"

[img]http://latex.codecogs.com/png.latex?*LaTeX commands go here*[/img]

-Dan
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Apr 24th 2011, 05:28 PM   #3
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Posts: 40
 Originally Posted by topsquark I'll look at the rest of the problem (no promises) but I can quickly address the LaTeX problem. There is a sticky about this in the LaTeX Help forum, but briefly we can use the "command line" [img]http://latex.codecogs.com/png.latex?*LaTeX commands go here*[/img] -Dan
I read your other post and I'll look into the Latex stuff. How about the current question? I am getting quite despirate!!! Sorry!!

Thanks

 Apr 24th 2011, 05:42 PM #4 Forum Admin     Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,442 Okay, first an admission. I am somewhat familiar with advanced and retarded Green's functions, not advanced and retarded potentials. That they are related is clear, but I don't know just how to relate one to the other. I noticed an interesting mistake in the equation below (1). You are correct that there are similarities with (1), but you have missed the details of the primes. Here is what the equation should be: $\nabla ^{\prime} \cdot \vec{J}(\vec{r}~',t_r) = \frac{\partial J_{x'}}{\partial t_r} \frac{\partial t_r}{\partial x'} + \frac{\partial J_{y'}}{\partial t_r} \frac{\partial t_r}{\partial y'} + \frac{\partial J_{z'}}{\partial t_r} \frac{\partial t_r}{\partial z'}$ $= \frac{1}{c} \left ( \frac{\partial J_{x'}}{\partial t_r} \frac{\partial \eta }{\partial x'} + \frac{\partial J_{y'}}{\partial t_r} \frac{\partial \eta }{\partial y'} + \frac{\partial J_{z'}}{\partial t_r} \frac{\partial \eta }{\partial z'} \right ) = -\frac{1}{c} \frac{\partial \vec{J}(\vec{r}~')}{\partial t_r} \cdot \left ( \nabla ' \eta \right )$ That, at least, puts that term in the form you need it. -Dan __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.
 Apr 24th 2011, 09:33 PM #5 Member   Join Date: Nov 2010 Posts: 40 Thanks, actually that's the second part of my question, $\vec J_{(\vec r\;',t_r)}$ is at the source point (x',y',z') so It is should be $\vec J_{(\vec r\;',t_r)} =\hat x J_{x'} + \hat y J_{y'} +\hat z J_{z'}$ instead of $\vec J_{(\vec r\;',t_r)} =\hat x J_{x} + \hat y J_{y} +\hat z J_{z}$ in all the equation. But as show in (A), there is the charge density term that I really don't get: $\nabla'\cdot\vec J_{(\vec r\;',t_r)} =-\frac {\partial \rho_{(\vec r\;',t_r)}}{\partial t} -\frac 1 c \frac {\partial \vec J_{(\vec r\;',t_r)}}{\partial t_r}\cdot(\nabla'\eta)$ I guess I reverse engineering your latex and got it now!!! Now that I know how to post Latex, I am alive!!! Last edited by yungman; Apr 24th 2011 at 09:46 PM.

 Tags derivation, part, potential, retarded

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