Angular distribution of radiation in relativistic limit

[genxium]

While linear accelerating an electron, with direction of acceleration being the $z+$ axis of the spherical coordinates, its radiation in angular distribution form is(according to this tutorial: http://farside.ph.utexas.edu/teaching/em/lectures/node132.html)

$\frac{dP(t')}{d\Omega} = \frac{e^2\dot{u}}{16\pi^2\epsilon_0c^3}\frac{sin^2\theta}{[1-(u/c)cos\theta]^5}$

where $t'$ is retarded time, $\theta$ is the polar angle of measuring point in current spherical coordinate, $u$ is the value of velocity, $\dot{u}$ is the value of acceleration.

By differentiating wrt $cos\theta$ I can verify equation (1662) in the tutorial saying that

$\theta_{max} = arccos [\frac{1}{3(u/c)}(\sqrt{1+15u^2/c^2} - 1)]$

is the angle of maximum radiation.

However I don't get why taking $u/c \rightarrow 1$ results in $\theta_{max} \rightarrow 1/(2\gamma)$ where $\gamma$ should be $\frac{1}{\sqrt{1-u^2/c^2}}$.

It seems quite straight forward for me to get just $\theta_{max} \rightarrow arccos(1) \, = \, 0$ as $u/c \rightarrow 1$.

Where did I go wrong?

Any help is appreciated :)

 

[Orodruin]

The gamma factor goes to infinity in this limit so the angle does go to zero. The quoted expression describes how it goes to zero and to derive it you need to keep higher order terms. I suggest writing your expression in terms of $\gamma$ and making a Taylor expansion in $1/\gamma$.

 

[genxium]

I suggest writing your expression in terms of $\gamma$ and making a Taylor expansion in $1/\gamma$.

Thanks this sounds on the right track. I'll try it :)

 

[genxium]

OMG the calculation is taking me a long time and I'm still stuck in all the trials. By far I've done:

Denote $h = \frac{\gamma}{3\sqrt{\gamma^2-1}}(4\sqrt{1-\frac{15}{16\gamma^2}} - 1)$ thus $\theta_{max} = arccos(h) \approx \frac{\pi}{2} - (h + \frac{1}{6}h^3 + \frac{3}{40}h^5 + ...)$

And now the high order terms become complicated if one goes on in a straight forward calculation. Is there any workaround for this?

Btw I also doubt the necessity of such a hassle. If the importance of investigating $\frac{dP(t')}{d\Omega}$ is to divide the space into "lobes" where radiations strengthen/attenuate w.r.t $\theta$, then finding local extrema like $\theta_{max}(u, \dot{u}, ...), \theta_{min}(u, \dot{u}, ...)$ should be a good stop to the calculation.

 

[Orodruin]

OMG the calculation is taking me a long time and I'm still stuck in all the trials. By far I've done:

Denote $h = \frac{\gamma}{3\sqrt{\gamma^2-1}}(4\sqrt{1-\frac{15}{16\gamma^2}} - 1)$ thus $\theta_{max} = arccos(h) \approx \frac{\pi}{2} - (h + \frac{1}{6}h^3 + \frac{3}{40}h^5 + ...)$

And now the high order terms become complicated if one goes on in a straight forward calculation. Is there any workaround for this?

Btw I also doubt the necessity of such a hassle. If the importance of investigating $\frac{dP(t')}{d\Omega}$ is to divide the space into "lobes" where radiations strengthen/attenuate w.r.t $\theta$, then finding local extrema like $\theta_{max}(u, \dot{u}, ...), \theta_{min}(u, \dot{u}, ...)$ should be a good stop to the calculation.

You are attempting to taylor expand $\arccos(x)$ instead of $\arccos(1-x)$. You will have more success if you do the second.

 

[theodoros.mihos]

When set u/c -->1 for lower values the c/u --> 1 for larger values. So (c/u)^2 --> 1 much stronger than (c/u) and rewriting your equation as:
$$ \cos{\theta_{max}} = \frac{1}{3}\left(\sqrt{15+(c/u)^2}-(c/u)\right) $$ the limit for (c/u)-->1 is 1 so θ_max=0.

 

[genxium]

You are attempting to taylor expand $\arccos(x)$ instead of $\arccos(1-x)$. You will have more success if you do the second.

Hi Orodruin, the hint is cool however I still can't find out how to use it :(

It's not obvious to me how to get to the form $1-x$ from either $\frac{1}{3(u/c)}(\sqrt{1+15u^2/c^2} - 1)$ or $\frac{\gamma}{3\sqrt{\gamma^2 - 1}}(4\sqrt{1-\frac{15}{16\gamma^2}} - 1)$.

 

[Orodruin]

I suggest you rewrite it as
$$
\frac{1}{3\sqrt{1-1/\gamma^2}} (4\sqrt{1- \frac{15}{16\gamma^2}} - 1)
$$
and start expanding it in $1/\gamma^2$.

 

[genxium]

I suggest you rewrite it as
$$
\frac{1}{3\sqrt{1-1/\gamma^2}} (4\sqrt{1- \frac{15}{16\gamma^2}} - 1)
$$
and start expanding it in $1/\gamma^2$.

Do you mean Taylor expanding $\frac{1}{3\sqrt{1-1/\gamma^2}} (4\sqrt{1- \frac{15}{16\gamma^2}} - 1)$ first to get something like $f(\frac{1}{\gamma^2})$, then Taylor expanding $acos(f(\frac{1}{\gamma^2}))$?

 

[Orodruin]

Do you mean Taylor expanding $\frac{1}{3\sqrt{1-1/\gamma^2}} (4\sqrt{1- \frac{15}{16\gamma^2}} - 1)$ first to get something like $f(\frac{1}{\gamma^2})$, then Taylor expanding $acos(f(\frac{1}{\gamma^2}))$?

You already have a function which is a function $f(1/\gamma^2)$ so you can series expand it. It is easier to take the cosine of everything and expand the cosine for small angles.