Hi! I try to construct the emission spectrum from relativistic electron rotating in homogeneous magnetic field - synchrotron. In my lecture notes a found out one really easy derivation using the invariance of [itex]\frac{I'}{(\nu')^3}=\frac{I}{\nu^3}[/itex], where [itex] I [/itex] is the specific intensity and [itex] \nu[/itex] is the frequency. So the radiated intensity from inertial observe frame is [itex]I'=\frac{I (\nu')^3}{\nu^3}[/itex], using Doppler effect fromula [itex]I'=\frac{I}{\gamma^3\left(1-\frac{v}{c}\cos{\theta}\right)^3}[/itex], where [itex] \theta [/itex] is angle possition on the circular trajectory. I used [itex] \theta = \frac{\omega_{cyclotron} t}{\gamma} [/itex] . Then I plotted the resulting intensity, which looked ok (at least similar to some I found on the internet). I also did the fourier transformation (picture uploded). But the critical frequency is too hight, also the peaks are to widt - here (http://farside.ph.utexas.edu/teaching/em/lectures/node133.html) I found, that the maximum radiadion should be emmited at frequency [itex] \propto \gamma^2 \omega_{cyclotron} [/itex] , blue line at the picture. Thanks a lot for each advice!
Your first spectrum shows the intensity over time. The fourier transformation of this shows the frequency of intensity peaks, but not the frequency of the radiation itself. To get this, you would need the complex amplitude in the first plot, and you don't get this with the simple formula you used. If you want the frequency of the emitted radiation, I think you can neglect the overall motion of the electron around the circle and focus on the parts where you get the intensity peaks. This won't work with your formula for the total intensity, however.
Ok, thanks a lot! I thought that it is quite weird, that the frequency dependence came only from the periodic motion. So could I try to correct this formula by adding proper intensity I in the rest frame - the correct frequency dependence? (I tried to change some parameteres and if the period of peaks remain the same, but they would be thinner, I will get more shorter frequences - the correct spectrum)