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How to compute the divergence of retarded scalar potential

  1. Dec 2, 2014 #1
    I'm learning time-dependent Maxwell's Equations and having difficulty understanding the following derivative:

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


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

    [itex]\textbf{r}' = x' \cdot \textbf{i} + y' \cdot \textbf{j} + z' \cdot \textbf{k}[/itex]

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

    The tutorial I'm reading "infers" that

    [itex]\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[/itex] -- (a)

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

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

    [itex]\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)] [/itex] -- (b)

    then if (a) is correct I'll have

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

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

    [itex]\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[/itex] -- (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 :)
  2. jcsd
  3. Dec 6, 2014 #2
    I have solved my problem. The mistake I made was that I took [itex][\rho] = [\rho(\textbf{r}, t)][/itex] while it should be [itex][\rho] = [\rho(\textbf{r}', t)][/itex].
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