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:

enter image description here

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


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...?

  • 1
    $\begingroup$ What is $s$? ... $\endgroup$ – G. Smith Feb 28 at 21:00
  • $\begingroup$ It’s $s=z-(\vec{z}\vec{v})/c$ $\endgroup$ – Vincent ISOZ Feb 28 at 21:10
  • 1
    $\begingroup$ That belongs in the question. $\endgroup$ – G. Smith Feb 28 at 21:27
  • $\begingroup$ No. I know where the $s$ comes from however.... i also don’t understand how he makes it appear in the above equalities.... $\endgroup$ – Vincent ISOZ Feb 28 at 22:39
  • $\begingroup$ 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. $\endgroup$ – 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 $$\nabla \left(z-\frac{\vec z\cdot \vec v}{c}\right)=\hat z\left[\frac{\partial}{\partial z}\left(z-\frac{\vec z\cdot \vec v}{c}\right)\right]$$

  • $\begingroup$ « The function depends only on $z$ ». How did you deduce that from the figure ? O_o $\endgroup$ – Vincent ISOZ Feb 28 at 21:13
  • $\begingroup$ @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 $z$. $\endgroup$ – AccidentalTaylorExpansion Feb 28 at 21:20
  • $\begingroup$ @VincentISOZ It's not possible for me to follow the steps after the first equal sign either. It seems like the author uses some approximations/simplifications but without the book I have no idea what he did. $\endgroup$ – AccidentalTaylorExpansion Feb 28 at 21:24
  • $\begingroup$ Thx for trying anyway. It’s also a pain in the ass for me a friend mathematician. We were not able to guess what he is doing here... sad that the author died.... $\endgroup$ – Vincent ISOZ Feb 28 at 22:40

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