Doubt about the derivation of Liénard-Wiechert Potentials?

Question

When deriving the Lienard-Wiechert Potentials, there is one step that you need to perform: $$ \nabla_{\mathbf r}|\mathbf r - \mathbf r_s(t_r)| $$ Where $t_r$ is: $$ t_r = t - \frac{|\mathbf r - \mathbf r_s(t_r)|}{c} $$ $\mathbf r$, is the position where you are calculating the field and $\mathbf r_s$ is the position of a particle. According to the wikipedia page, the result is the following: $$ \nabla_{\mathbf r}|\mathbf r - \mathbf r_s(t_r)| = \nabla_{\mathbf r} t_r \frac{d|\mathbf r - \mathbf r_s(t_r)| }{d t_r} + \mathbf n_s = \frac{\mathbf n_s}{1-\boldsymbol \beta_s \cdot \mathbf n_s} $$ Where: $$ \mathbf n_s = \frac{\mathbf r - \mathbf r_s}{|\mathbf r - \mathbf r_s|} $$ $$ \boldsymbol{\beta_s} = \frac{\mathbf v_s}{c} $$ I don't quite understand this result. How do you obtain this $\nabla_{\mathbf r}|\mathbf r - \mathbf r_s|$ from the retarded time? I assume it cames from the chain rule, but I cannot derive it by myself. Any hint?

Answer 1

You're right that some problems could arise for a kind of circular dependence of the variables involved. Let's get some partial results first and then put everything together:

• gradient of the retarded time $\nabla t_r(\mathbf{r}) = \nabla \left( t - \frac{|\mathbf{r} - \mathbf{r}_s(t_r)|}{c} \right)$, that from direct computation reads $$\nabla t_r = -\frac{1}{c} \nabla|\mathbf{r} - \mathbf{r}_s(t_r)| \ ,$$ we don't need to manipulate this result further, so far, since the unknwon has appeared;

• gradient of the absolute value $|\mathbf{r} - \mathbf{r}_s(t_r)|$, that by direct computation reads $$\begin{aligned} \nabla |\mathbf{r} - \mathbf{r}_s(t_r)| & = \frac{1}{2 |\mathbf{r} - \mathbf{r}_s(t_r)|} \nabla |\mathbf{r} - \mathbf{r}_s(t_r)|^2 = \\ & = \frac{1}{2|\mathbf{r} - \mathbf{r}_s(t_r)|} \, 2 \left[ \mathbf{r} - \mathbf{r}_s(t_r) - ( \mathbf{r} - \mathbf{r}_s(t_r) ) \cdot \frac{d \mathbf{r}_s(t_r)}{d t_r} \nabla t_r \right] = \\ \end{aligned} \ .$$

• Using the epxression of the first point for $\nabla t_r$and defining $\mathbf{\hat{n}} = \frac{\mathbf{r} - \mathbf{r}_s(t_r)}{|\mathbf{r} - \mathbf{r}_s(t_r)|} $, and $\boldsymbol{\beta} = \frac{1}{c}\frac{d \mathbf{r}_s(t_r)}{d t_r}$, it should be not so hard to recast the last equation as $$\nabla |\mathbf{r} - \mathbf{r}_s(t_r)| = \mathbf{\hat{n}} + \mathbf{\hat{n}} \cdot \boldsymbol{\beta} \, \nabla |\mathbf{r} - \mathbf{r}_s(t_r)|$$

• Eventually, solving for $\nabla |\mathbf{r} - \mathbf{r}_s(t_r)|$, we get the desired expression $$\nabla |\mathbf{r} - \mathbf{r}_s(t_r)| = \frac{\mathbf{\hat{n}}}{1 - \mathbf{\hat{n}} \cdot \boldsymbol{\beta}} \ .$$

That should provide a way to get the desired result.

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Edit 1. Required details about $\nabla |\mathbf{r} - \mathbf{r}_s(t_r)|^2.$ Direct calculation using Cartesian coordinates (for simplicity) gives

$$\begin{aligned} \left\{ \nabla |\mathbf{r} - \mathbf{r}_s(t_r)|^2 \right\}_i & = \partial_i \sum_k \left( r_k - r_k^s(t_r(r_j)) \right)^2 = \\ & = 2 \sum_k \left( \partial_i r_k - \partial_i r_k^s \right) \left( r_k - r_k^s(t_r(r_j)) \right) = \\ & = 2 \sum_k \left( \delta_{ik} - \partial_{t_r} r_k^s \partial_i t_r(r_j) \right) \left( r_k - r_k^s(t_r(r_j)) \right) = \\ & = 2 \left( r_i - r_k^s(t_i(r_j)) \right) - 2 \partial_i t_r(r_j) \sum_k \partial_{t_r} r_k^s \left( r_k - r_k^s(t_r(r_j)) \right) = \\ & = 2 \left\{ \mathbf{r} - \mathbf{r}^s - \nabla t_r \, \dfrac{d \mathbf{r}^s}{dt_r} \cdot ( \mathbf{r} - \mathbf{r}^s ) \right\}_i \end{aligned}$$