# Retarder potential strange derivation

I have an old French textbook (the author died a few years ago) that develops in quite a very very detailed way the relativist Larmor formula on more than 35 pages.

However, I've been stuck for a few days on a step that may be obvious (...). Here is the figure used by the author:

And at a given moment he writes ($$\vec{\nabla}$$ is the nabla operator):

$$\vec{\nabla}\bigg(z-\frac{\vec{z}\cdot\vec{v}}{c}\bigg)=\frac{\vec{z}}{z}-\frac{\vec{z}}{z}\frac{v}{c}=\frac{\vec{z}}{z}-\hat{r}\frac{v}{s}=\frac{\vec{z}}{z}-\frac{\vec{v}}{c}$$

The thing is:

1. I don't understand the steps (and I'm not able to identify the missing steps too).
2. Sadly he never defined earlier what is $$\hat{r}$$

So I have to guess but I've been stuck for two weeks. If maybe someone has an idea...?

• What is $s$? ... – G. Smith Feb 28 at 21:00
• It’s $s=z-(\vec{z}\vec{v})/c$ – Vincent ISOZ Feb 28 at 21:10
• That belongs in the question. – G. Smith Feb 28 at 21:27
• No. I know where the $s$ comes from however.... i also don’t understand how he makes it appear in the above equalities.... – Vincent ISOZ Feb 28 at 22:39
• Without an explanation of $s$, the question lacks clarity. Your question is supposed to be understandable by anyone reading it. This site is not here to answer questions for you. It is here to be a Q&A resource for all. Since you have not clarified the question, I have voted to close it as unclear. – G. Smith Feb 28 at 23:01

To answer question 2: $$\hat r$$ is a common notation for unit vectors (vectors with length 1). They are defined as $$\hat r=\frac{1}{r}\vec r$$ To answer your first question consider what the nabla operator does. In this case it's a gradient so it takes in a function and spits out a vector containing the derivatives of that function: $$\nabla f(x,y,z)=\pmatrix{\frac{\partial f}{\partial x}\\\frac{\partial f}{\partial y}\\\frac{\partial f}{\partial z}}$$ Because our function only depends on $$z$$ we can see that the $$x$$, $$y$$ compontents of the gradient will be zero and we can just look at the $$z$$ component. So you can probably calculate $$\frac{\partial}{\partial z}\left(z-\frac{\vec z\cdot \vec v}{c}\right)=\frac{\partial}{\partial z}\left(z-\frac{z\, v_z}{c}\right)$$ We can write a vector generally as follows $$\vec v=v_x\hat x+v_y\hat y+v_z\hat z$$ So since the $$x,y$$ components of our gradient are zero the expression becomes $$\nabla \left(z-\frac{\vec z\cdot \vec v}{c}\right)=\hat z\left[\frac{\partial}{\partial z}\left(z-\frac{\vec z\cdot \vec v}{c}\right)\right]$$
• « The function depends only on $z$ ». How did you deduce that from the figure ? O_o – Vincent ISOZ Feb 28 at 21:13
• @VincentISOZ Actually I didn't look at the figure. I just looked at the first equation you typed. The term that nabla acts on only depends on $z$. – AccidentalTaylorExpansion Feb 28 at 21:20