Griffiths Electrodynamics gradient of charge distribution

[Sparky_]

I do not understand the following from Griffiths’ Electrodynamics – page 424 Equation 10.21.

$$\nabla p = \dot{p} \nabla {tr} = …$$

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_

 

[HomogenousCow]

continuity equation?

 

[Sparky_]

I do not see it yet.

I see later on the same page

$$\nabla \dot{p} = \ddot{p} \nabla {tr} = …$$

Can you explain further?

Somehow the gradient is giving an additional time derivative.

Thanks
Sparky_

 

[MisterX]

$\rho$ has arguments like this:
$\rho (\vec{r}', t_r(\vec{r}, \vec{r}', t))$

The gradient is being applied w.r.t to the coordinates of $\vec{r}$ ( not $\vec{r}'$ 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 $t_r$. So this result is from the chain rule. Here is one component of the gradient, for example.
$(\nabla \rho)_x = \frac{\partial \rho (\vec{r}', t_r(x, y, z, \vec{r}', t))}{\partial x} = \dot{\rho}\frac{\partial t_r}{\partial x}$

 

[HomogenousCow]

Oh I see, did he specify that the dot derivative is with respective to retarded time?

 

[WannabeNewton]

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_]

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_