Relativistic generalization of Larmor fomula?

[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.

 

[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.

 

[genxium]

Thanks for the reply!

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

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 $S$.

To get the covariant form the author has introduced Lorentz scalar...

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.

 

[Mentz114]

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 $S$.

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.

The introduction of the 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.

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 ).

 

[genxium]

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.

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.

 

[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):

To treat particles moving at nearly the speed of light in the observer's frame, we must use Larmor's equation to calculate the radiation in the particle's rest frame and then transform the result to the observer's frame in a relativistically correct way.