<p>Synchrotron radiation is emitted by charged particles (mostly electrons) executing helical motion, accelerated by the Lorentz force exerted by the vector product of their velocity and the magnetic field. The frequency of the radiation depends how fast the electrons orbit, which in turn depends on the magnetic field strength.</p>
<p>The acceleration thus depends on their velocity <em>and</em> the strength of the magnetic field. For non-relativistic electrons the frequencies that the electromagnetic radiation appears would simply be the orbital frequency and would be
<span class="math-container">$$\omega = \frac{qB}{mc}\ ,$$</span>
where <span class="math-container">$q$</span> and <span class="math-container">$m$</span> are the charge and mass of the electron and <span class="math-container">$B$</span> is the magnetic field strength.</p>
<p>For Relativistic electrons, this frequency becomes <em>even</em> smaller (by the Lorentz factor <span class="math-container">$\gamma$</span>). If you work out what these frequencies are for the typical magnetic field strengths in galaxies then they are at very, very long radio wavelengths and at very small frequencies, well below the plasma frequency of the interstellar medium.</p>
<p>However, there are two effects that boost the spectrum back up to MHz, GHz, or in the cases of some very strong magnetic fields (around pulsars for example), even optical frequencies. These are that radiation is Doppler shifted to higher frequencies when the electrons move towards the observer. Secondly, the radiation from a relativistic electron is beamed into a narrow cone in the forward direction, so that a distant observer would see a set of narrow, short pulses as the electron spirals around the field lines. The spectrum (Fourier transform) of this pulsed emission leads to power at much higher frequencies than the simple orbital frequency of the electron, but still not in the visible range unless the magnetic fields are very strong.</p>
<p>Similar arguments would apply to synchrotron radiation from relativistic charged particles in orbit around stellar sized black hole. In this case, the relevant orbital frequency is the actual orbital frequency around the black hole. It is highest at the innermost stable circular orbit of a Schwarzschild black hole but is only <span class="math-container">$2200 (M/M_\odot)^{-1}$</span> Hz (I.e. just hundreds of Hz for typical black holes). So the same argument applies here. It is only the Doppler beaming effects that even boosts these frequencies up to observable radio wavelengths.</p>

CC BY-SA 4.0
2021-12-17T13:30:06.340Z
2021-12-17T13:30:06.340Z
ProfRob
<p>Why do populations of relativistic (high energy) electrons emitting synchrotron radiation emit at mostly radio wavelengths? The fact that they are high energy makes me think they would emit high energy photons. As a particle moves in the magnetic field of say, an accretion disk, is it constantly emitting synchrotron radiation as it goes around?</p>

<radiation><accretion-discs><synchrotron-radiation>
Why is synchrotron radiation from relativistic electrons low in energy?
theta
# Why is synchrotron radiation from relativistic electrons low in energy?

(Asked by: username_0 on 2021-12-17)

Why do populations of relativistic (high energy) electrons emitting synchrotron radiation emit at mostly radio wavelengths? The fact that they are high energy makes me think they would emit high energy photons. As a particle moves in the magnetic field of say, an accretion disk, is it constantly emitting synchrotron radiation as it goes around?

**Tags:** radiation, accretion-discs, synchrotron-radiation


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## 1 answers

### Answer by username_1 on 2021-12-17. Score: 4 | Accepted answer

Synchrotron radiation is emitted by charged particles (mostly electrons) executing helical motion, accelerated by the Lorentz force exerted by the vector product of their velocity and the magnetic field. The frequency of the radiation depends how fast the electrons orbit, which in turn depends on the magnetic field strength.

The acceleration thus depends on their velocity *and* the strength of the magnetic field. For non-relativistic electrons the frequencies that the electromagnetic radiation appears would simply be the orbital frequency and would be
$$\omega = \frac{qB}{mc}\ ,$$
where $q$ and $m$ are the charge and mass of the electron and $B$ is the magnetic field strength.

For Relativistic electrons, this frequency becomes *even* smaller (by the Lorentz factor $\gamma$). If you work out what these frequencies are for the typical magnetic field strengths in galaxies then they are at very, very long radio wavelengths and at very small frequencies, well below the plasma frequency of the interstellar medium.

However, there are two effects that boost the spectrum back up to MHz, GHz, or in the cases of some very strong magnetic fields (around pulsars for example), even optical frequencies. These are that radiation is Doppler shifted to higher frequencies when the electrons move towards the observer. Secondly, the radiation from a relativistic electron is beamed into a narrow cone in the forward direction, so that a distant observer would see a set of narrow, short pulses as the electron spirals around the field lines. The spectrum (Fourier transform) of this pulsed emission leads to power at much higher frequencies than the simple orbital frequency of the electron, but still not in the visible range unless the magnetic fields are very strong.

Similar arguments would apply to synchrotron radiation from relativistic charged particles in orbit around stellar sized black hole. In this case, the relevant orbital frequency is the actual orbital frequency around the black hole. It is highest at the innermost stable circular orbit of a Schwarzschild black hole but is only $2200 (M/M\_\odot)^{-1}$ Hz (I.e. just hundreds of Hz for typical black holes). So the same argument applies here. It is only the Doppler beaming effects that even boosts these frequencies up to observable radio wavelengths.


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