# Doubt about the derivation of Liénard-Wiechert Potentials?

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?

• The term $\nabla_{\mathbf r} t_r$ is equal to $-1/c$ times $\nabla_{\mathbf r}|\mathbf r - \mathbf r_s(t_r)|$. This gives you a linear equation for such a gradient. Commented Feb 20 at 18:06
• @GiorgioP-DoomsdayClockIsAt-90 But $|\mathbf r - \mathbf r_s(t_r)|$, also depends on the retarded time. How would I do this? Commented Feb 20 at 18:21
• basics's answer made explicit my hint. Commented Feb 21 at 7:57
• You have to study this in a textbook. You have found out that Wikipedia is not much help. Commented Feb 22 at 17:36

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.

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}

• Sorry But I do not understand why $\nabla |\mathbf r - \mathbf r_s|^2 = 2 \left[ \mathbf{r} - \mathbf{r}_s(t_r) - ( \mathbf{r} - \mathbf{r}_s(t_r) ) \cdot \frac{d \mathbf{r}_s(t_r)}{d t_r} \nabla t_r \right]$? Could you explain it in depth please? Commented Feb 20 at 19:09
• editing........ Commented Feb 20 at 19:19
• you're welcome. Glad to manage to be helpful sometimes Commented Feb 20 at 19:29