# Relativistic generalization of Larmor fomula?

1. Apr 29, 2015

### genxium

While reading an online tutorial (http://www.phy.duke.edu/~rgb/Class/phy319/phy319/node146.html) about deriving the relativistic generalization of Larmor Formula, I got some problems with the steps.

Basically with an assumption $\beta \ll 1$ the author gets

$Power = \frac{e^2}{6 \pi \epsilon_0 c^3} |\frac{d\textbf{v}}{dt}|^2$ -- (1)

then by replacing $\textbf{v} = \frac{\textbf{p}}{m}$ he/she gets

$Power = - \frac{e^2}{6 \pi \epsilon_0 c^3} \frac{dp_{\mu} dp^{\mu}}{d\tau d\tau}$ -- (2)

where $p^{\mu}, p_{\mu}$ are contravariant and covariant forms of the momentum 4-vector respectively. To my understanding it's using Minkowski metric.

According to wikipedia (http://en.wikipedia.org/wiki/Larmor_formula#Covariant_form), result (2) makes sense because when $\beta$ again goes to $\beta \ll 1$ (2) reduces to (1).

So here comes a problem, the reasoning for (1) makes use of the assumption $\beta \ll 1$, thus the WHOLE CONTEXT is already non-relativistic, how come one can derive (2) from (1) in this context?

By the way there might be a mistake in the tutorial: when starting with (1), the author takes $\textbf{v} = \frac{\textbf{p}}{m}$, then he/she directly applies

$|\frac{d\textbf{v}}{dt}|^2 = \frac{1}{m^2} |\frac{d\textbf{p}}{dt}|^2$

which doesn't seems right, in $\frac{d(\textbf{p}/m)}{dt}$ both $\textbf{p}$ and $m$ are functions of $t$.

I did check other tutorials about Larmor formula like http://farside.ph.utexas.edu/teaching/em/lectures/node130.html, but the maths is taking much time to understand there :(

Any help will be appreciated.

2. Apr 29, 2015

### Mentz114

To get the covariant form the author has introduced Lorentz scalar $\frac{dp_{\mu} dp^{\mu}}{d\tau d\tau}$ making the expression manifestly covariant. This is not affected by the restriction on $\beta$.

You say that $m$ is a function of $t$. If $m$ is the rest mass then this is not true.

3. Apr 29, 2015

### genxium

about the mass, I'm still confused by the author's notation. I did notice that he/she uses $m$ to represent rest mass some chapters before. However it's not mentioned that the derivation is carried out in the particle's frame, hence $\textbf{v} = \frac{\textbf{p}}{m}$ should apply to any frame here -- or, say that the observer is measuring in frame $S$, then $S$ is not necessarily the particle's frame and all $\textbf{v}, \textbf{p}, m$ are with respect to $S$.

The introduction of Lorentz scalar is the last step of the derivation which is another way of saying "$(\frac{d\textbf{p}}{d\tau})^2 - (\frac{1}{c^2}\frac{dE}{d\tau})$ in Minkowski metric" to me. It's not clear to me when it jumps out of the $\beta \ll 1$ context.

4. Apr 29, 2015

### Mentz114

Rest mass is frame invariant in the $\beta<<1$ regime, however it is defined. But if you are using $m'=\gamma m$ or some kind of 'relativistic mass', I think that is wrong.

To be covariant the quantity must be a Lorentz scalar. $p^\mu p_\mu$ is obviously invariant under coordinate transformation because $\lambda p^\mu \lambda^{-1} p_\mu=p^\mu p_\nu = -mc^2$. ( note that $m$ here must be invariant ).

5. Apr 29, 2015

### genxium

I'd like to clarify my opinion for the mass first. Surely the rest mass is frame invariant, my point is that beginning with $|\frac{d\textbf{v}}{dt}|^2$, the author did

$\textbf{v} = \frac{\textbf{p}}{m}$ -- (3)

now what I argued was that the $m$ in (3) is NOT rest mass, thus it should be followed by

$|\frac{d\textbf{v}}{dt}|^2 = |\frac{d(\textbf{p}/m)}{dt}|^2$ -- (4)

and the $m$ in (4) is not rest mass either so it's time dependent.

6. Apr 30, 2015

### genxium

I'm afraid I was totally wrong from the very beginning, problem solved now.

@Mentz114 you're alright in your answers, thanks a lot but they didn't hit the key point of my confusion :)

Here's what solves my problem (from http://www.cv.nrao.edu/course/astr534/LarmorRad.html):