How to compute the divergence of retarded scalar potential

AI Thread Summary
The discussion centers on computing the divergence of the retarded scalar potential in time-dependent Maxwell's Equations. The user is confused about the derivative of the function f(r, r', t) and the application of the product rule in their calculations. They initially derive two different expressions for the gradient of the function, leading to a contradiction regarding the treatment of the charge density ρ. Ultimately, the user resolves their confusion by realizing that they incorrectly assumed ρ was a function of r instead of r'. This highlights the importance of correctly identifying the variables in mathematical expressions when applying derivatives.
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I'm learning time-dependent Maxwell's Equations and having difficulty understanding the following derivative:

Given f(\textbf{r}, \textbf{r}', t) = \frac{[\rho(\textbf{r}, t)]}{|\textbf{r} - \textbf{r}'|}

where

\textbf{r} = x \cdot \textbf{i} + y \cdot \textbf{j} + z \cdot \textbf{k}, in Cartesian Coordinates

\textbf{r}' = x' \cdot \textbf{i} + y' \cdot \textbf{j} + z' \cdot \textbf{k}

[\rho(\textbf{r}, t)] \stackrel{\Delta}{=} \rho(\textbf{r}, t_r) with t_r = t-\frac{|\textbf{r} - \textbf{r}'|}{c} and c is a non-zero constant(speed of EM wave indeed)

The tutorial I'm reading "infers" that

\nabla f(\textbf{r}, \textbf{r}', t) = \nabla \frac{1}{|\textbf{r} - \textbf{r}'|} \cdot [\rho(\textbf{r}, t)] + \frac{1}{|\textbf{r} - \textbf{r}'|} \cdot [\frac{\partial \rho(\textbf{r}, t)}{\partial t}] \cdot \nabla t_r -- (a)

where \nabla \stackrel{\Delta}{=} \frac{\partial}{\partial x} \cdot \textbf{i} + \frac{\partial}{\partial y} \cdot \textbf{j} + \frac{\partial}{\partial z} \cdot \textbf{k}

I'm confused by the latter part of the equation above. By applying the identity \nabla (g_1 \cdot g_2) = g_2 \cdot \nabla g_1 + g_1 \cdot \nabla g_2 to f(\textbf{r}, \textbf{r}', t) I get

\nabla f(\textbf{r}, \textbf{r}', t) = \nabla \frac{1}{|\textbf{r} - \textbf{r}'|} \cdot [\rho(\textbf{r}, t)] + \frac{1}{|\textbf{r} - \textbf{r}'|} \cdot \nabla [\rho(\textbf{r}, t)] -- (b)

then if (a) is correct I'll have

\nabla [\rho(\textbf{r}, t)] = [\frac{\partial \rho(\textbf{r}, t)}{\partial t}] \cdot \nabla t_r -- (c)

However, though trivial, \textbf{r}(x, y, z) = x \cdot \textbf{i} + y \cdot \textbf{j} + z \cdot \textbf{k} is still a function of x, y \, \text{and} \, z, so in my calculation

\nabla [\rho(\textbf{r}, t)] = \frac{\partial \rho(\textbf{r}, t_r)}{\partial \textbf{r}} \cdot \nabla \textbf{r} + \frac{\partial \rho(\textbf{r}, t_r)}{\partial t_r} \cdot \nabla t_r = \frac{\partial \rho(\textbf{r}, t_r)}{\partial \textbf{r}} \cdot \nabla \textbf{r} + [\frac{\partial \rho(\textbf{r}, t)}{\partial t}] \cdot \nabla t_r -- (d)

Obviously (d) contradicts (c) but unfortunately I can't figure out where I went wrong in my calculation.

Can someone help to point out my mistakes or guide me to some references? Any help is appreciated :)
 
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I have solved my problem. The mistake I made was that I took [\rho] = [\rho(\textbf{r}, t)] while it should be [\rho] = [\rho(\textbf{r}', t)].
 
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