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Griffiths Electrodynamics gradient of charge distribution

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Sparky_
#1
Apr7-13, 08:34 PM
P: 166
I do not understand the following from Griffiths’ Electrodynamics – page 424 Equation 10.21.

[tex]
\nabla p = \dot{p} \nabla {tr} = …
[/tex]

I’m not sure how much of this applies (I think my question is on the math) but p is the charge distribution, tr is the retarded time.

Is this an application of the chain rule?

With the gradient being a derivative with respect to spatial location (x,y,z), why is the time derivative showing up in the gradient? I initially want to say if something is dependent upon t but not on x, then its derivative with respect to x is zero.

The result looks like the chain rule applied – I don’t see why the time dependent portion shows up.

Can you help clear this up for me?

Thanks
Sparky_
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HomogenousCow
#2
Apr7-13, 08:54 PM
P: 366
continuity equation?
Sparky_
#3
Apr7-13, 09:45 PM
P: 166
I do not see it yet.

I see later on the same page

[tex]
\nabla \dot{p} = \ddot{p} \nabla {tr} = …
[/tex]

Can you explain further?

Somehow the gradient is giving an additional time derivative.

Thanks
Sparky_

MisterX
#4
Apr8-13, 12:41 AM
P: 584
Griffiths Electrodynamics gradient of charge distribution

[itex]\rho[/itex] has arguments like this:
[itex]\rho (\vec{r}', t_r(\vec{r}, \vec{r}', t)) [/itex]

The gradient is being applied w.r.t to the coordinates of [itex]\vec{r}[/itex] ( not [itex]\vec{r}'[/itex] which gets integrated away). The coordinates that we would be taking the derivative with respect to in order to obtain the gradient are only found in the parameters of [itex]t_r[/itex]. So this result is from the chain rule. Here is one component of the gradient, for example.
[itex](\nabla \rho)_x = \frac{\partial \rho (\vec{r}', t_r(x, y, z, \vec{r}', t))}{\partial x} = \dot{\rho}\frac{\partial t_r}{\partial x} [/itex]
HomogenousCow
#5
Apr8-13, 12:44 AM
P: 366
Oh I see, did he specify that the dot derivative is with respective to retarded time?
WannabeNewton
#6
Apr8-13, 01:21 AM
C. Spirit
Sci Advisor
Thanks
WannabeNewton's Avatar
P: 5,661
It won't matter. ##\partial _{t_{r}} = \frac{\partial t_{r}}{\partial t}\partial _{t} = \frac{\partial }{\partial t}(t - \frac{\mathfrak{r}}{c})\partial_{t} = \partial_{t}##.

Anyways, as noted above ##\rho = \rho(r',t_{r}) ## and ##r'## is no longer a variable after integration but ##t_{r} = t_{r}(t,x,y,z,r')## so ##\nabla \rho = \partial _{t_{r}}\rho \nabla t_{r} = \partial_{t}\rho \nabla t_{r}##. Not sure what that has to do with the conservation of 4-current (continuity equation) ##\partial_{a}j^{a} = 0##.
Sparky_
#7
Apr8-13, 09:56 AM
P: 166
Thank you!!

I went back in this section of the text and reread. I see that p (charge density) is specified p(r’, tr). That is actually the point of this topic (the nonstatic case).

You confirmed that this is an application of the chain rule and p is a function of position and tr.

Thank you for the help!
Sparky_


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