0
$\begingroup$

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):

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

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...?

$\endgroup$
10
  • 1
    $\begingroup$ What is $s$? ... $\endgroup$
    – G. Smith
    Feb 28 at 21:00
  • $\begingroup$ It’s $s=z-(\vec{z}\vec{v})/c$ $\endgroup$ 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.... $\endgroup$ 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
0
$\begingroup$

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 $$\nabla \left(z-\frac{\vec z\cdot \vec v}{c}\right)=\hat z\left[\frac{\partial}{\partial z}\left(z-\frac{\vec z\cdot \vec v}{c}\right)\right]$$

$\endgroup$
4
  • $\begingroup$ « The function depends only on $z$ ». How did you deduce that from the figure ? O_o $\endgroup$ 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 $z$. $\endgroup$ 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. $\endgroup$ 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.... $\endgroup$ Feb 28 at 22:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.