1
$\begingroup$

For the last few hours, I tried to solve an exercise about synchrotron radiation but can't get to a solution. I think there are some concepts about special relativity that I didn't understand during the lecture. The exercise is the following:

It is given that the power P radiated by an accelerated charge e is: $$P = \frac{2}{3} \hbar \alpha \gamma^{6}(\dot{\vec{\beta}}^2-(\vec{\beta} \times \dot{\vec{\beta}})^2)$$ I am asked to express this in terms of the bending radius $\rho$ and the particle energy. I can assume highly relativistic particles.

It is clear to me that the cross product is a simple multiplication since the velocity and acceleration is perpendicular. But how can I express $\beta$ in a way that the term only depends on the particle energy?

I started with $\vec{\beta} = \frac{\vec{p}c}{E}$ and with that $\dot{\vec{\beta}} = \frac{\dot{\vec{p}}c}{E}$

Now $\dot{p}$ has to be the force (Lorentz force) so $\dot{p} = qvB$ and the magnetic field can be determined by the centripetal force: $qvB = \frac{mv^2}{\rho}$ $\Leftrightarrow$ $B = \frac{mv}{q\rho}$

$\rightarrow \quad$ $\dot{\vec{p}} = \frac{mv^2}{\rho}$

And now I'm stuck. Because if I continue like that it will end up with an expression dependent on the velocity, mass, energy of the particle and bending radius. And without a given mass I can't determine the radiation power P.

I think I made a mistake when saying $\dot{\vec{p}} = Force$ because I didn't consider special relativity effects at all but I have no idea what to use instead.

Thank you very much for your help!

$\endgroup$
1
  • $\begingroup$ Oh well. If your reasoning leads to a result that you doubt, do not just stop, but check your reasoning and write down its unescapable result. If the reasoning checks out, the result is a correct conclusion from the assumptions and the doubt is ungrounded. In this case, if $P$ turns out to be a function of both energy and mass, that is a fine result. The Larmor power for ultrarelativistic particle is a function of Lorentz factor and radius, and the Lorentz factor is not determined solely by energy. One has to know the mass. $\endgroup$ Commented Feb 5, 2019 at 0:21

1 Answer 1

0
$\begingroup$

$\def\vb{\vec\beta}$ First hint. You noted that $\vb$ and $\dot\vb$ are orthogonal. Then the expression in parentheses is $$\Bigl|\dot\vb\Bigr|^2\,(1 - \beta^2).$$

Second hint. For circular motion acceleration = $v^2/\rho$ so $$\Bigl|\dot\vb\Bigr| = {c\,\beta^2 \over \rho} \simeq {c \over \rho}.$$

And without a given mass I can't determine the radiation power $P$.

Indeed you can't. You have a $\gamma$ which must be converted in energy and mass is required.

$\endgroup$
1
  • $\begingroup$ Thank you very much! I got the right solution with this starting point $\endgroup$
    – Manuel
    Commented Feb 9, 2019 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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