6
$\begingroup$

For a radio spectrum, when is it called thermal emission, synchrotron emission, self-absorbed synchrotron emission and inverted spectrum? They are all power-law and their difference is power-law index?

For a power-law $\nu^{\alpha}$, as long as $\alpha$ is positive, the spectrum is an inverted spectrum, right?

Please recommend reference books which introduce these models and their corresponding astrophysical phenomena.

$\endgroup$

1 Answer 1

8
+100
$\begingroup$

Terminology

The terminology of thermal and non-thermal emission is somewhat unfortunate. Formally, non-thermal emission is continuum radiation from particles with non-Maxwellian energy spectra. Practically, we generally consider it to be emission which is not described by Bremsstrahlung or black body radiation (source).1 Note that this is true for the entire electromagnetic spectrum, not just the radio region.

Synchrotron emission is a form of non-thermal emission which occurs when particles are accelerated perpendicular to their translational motion, specifically when the particles are relativistic (otherwise it is called cyclotron emission).

Self-absorbed synchrotron emission is described well in the introduction of this paper, and occurs when the synchrotron emission becomes a non-negligible source of heating. It occurs only when the self-absorption frequency is larger than the cooling frequency, i.e.

$$ \nu_{c} < \nu_{a} $$

The spectral index, $\alpha$, describes the dependence of the radiative flux density on the frequency of the emission. With frequency $\nu$ and radiative flux density $S_{\nu}$,

$$ S_{\nu} \propto \nu^{\alpha} $$

An inverted spectrum results from $0 < \alpha < 2.5$,2 and this is generally indicative of thermal emission. However, the observed emission can be depressed by absorption processes, so a positive spectral index (i.e. an inverted spectrum) is not robustly indicative of thermal emission on its own.


Radio Astronomy

Most radio astronomy uses radio interferometry instead of direct observation, there are a number of reasons for this: most emitters are very weak in the radio portion of the electromagnetic spectrum, radio telescopes for direct imaging have to be much larger than their shorter-wavelength counterparts, and radio interferometry allows the highest angular resolution of any technique (it is possible to create a radio interferometery array the size of the Earth, which we have done).

That being said, there are a number of phenomena and objects whose activity in the radio region lends them to direct imaging:

  • Masers - stimulated spectral line emission, usually in the range of $3$mm to $0.3$m. This phenomena can occur in the atmospheres of gas giants, the expanded fuel sheath of late-type stars, protosolar disks and nebulae, supernovae remnants, and black holes. The produced spectrum is characterized by high brightness, extraordinarily high black-body temperature equivalent typically on the magnitude of $10^9$ K, but as high as $10^{14}$ K, and very high polarization that is predominantly circular. A spectra which contains a sharp peak that would correspond to a black-body temperature above $10^9$ K would likely be the product of a maser.

  • Quasars - The supermassive black holes in the centers of galaxies sometimes have accretion disks or clouds which produce immensely luminous emissions as the material falls into the black hole. The sheer scale of the energy involved (often orders of magnitude larger than the total output of the rest of the galaxy) leads to a relatively uniform emission across much of the electromagnetic spectrum, from X-rays to far infrared.

  • Pulsars - The mechanisms for emission here are poorly understood, but in the formation of a neutron star (or even a white dwarf occasionally), the conservation of angular momentum results in a very rapid rotation. A strong magnetic field causes protons and electrons to accelerate on the star's surface which produces cyclotron and synchrotron emission. The emission exits the pulsar in a beam.

  • Radio galaxies - Active galaxy nuclei which are exceptionally luminous in the $30$ m - $300$ km region. The radio emission in these nuclei is generated by synchrotron action characterized by a smooth, broad-band radio spectra with strong polarization.


1 Not a published source, but does a pretty good job of explaining this in the introduction.

2 $2.5$ is the rough upper limit for $\alpha$ in the power law description.


$\endgroup$
13
  • 1
    $\begingroup$ 1 as long as α is positive, the spectrum is an inverted spectrum, right? 2 The difference of radio spectra is just spectral index? How to tell a radio spectrum is synchrotron emission? $\endgroup$ Aug 15, 2020 at 3:33
  • 1
    $\begingroup$ @questionhang 1 - Yes, positive spectral index means an inverted spectrum and usually that indicates thermal emission is dominating. 2 - No, the only difference is not spectral index, the spectra generated by different kinds of emission will vary. As to the question of how to tell if a spectrum is due to synchrotron emission, the answer to that is too long for the comments. I will add it to my answer when I get a chance $\endgroup$ Aug 15, 2020 at 3:40
  • 1
    $\begingroup$ @questionhang That is not enough spectral information to determine conclusively whether the source is synchrotron emission. I have described in my answer the primary objects and phenomena which produce emissions in the radio region. Without seeing a complete spectra (or at the very least more of a spectra) there is no way to determine the source of the emission. $\endgroup$ Aug 16, 2020 at 8:48
  • 1
    $\begingroup$ @questionhang Yes, preferably an entire spectra with a reasonable (log-scale linearity would suffice) sampling granularity. But at the very least a full radio spectra. $\endgroup$ Aug 16, 2020 at 9:48
  • 1
    $\begingroup$ @questionhang If you want to know how to determine if a spectrum contains synchrotron emissions, you should ask a fresh question. If you want help with a specific spectrum, you can include it (with all the relevant & necessary data) as an example in your new question. That way, it may be helpful for future readers, not just yourself. $\endgroup$
    – PM 2Ring
    Aug 16, 2020 at 13:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .