Comments below What is the maximum distance measurable with parallax? discuss challenges associated with parallax measurements of Betelgeuse and link to Wikipedia’s Betelgeuse; Distance measurements which contains the intriguing paragraph:

In 2008, using the Very Large Array (VLA), produced a radio solution of 5.07±1.10 mas, equaling a distance of 197±45 pc or 643±146 ly.83 As the researcher, Harper, points out: "The revised Hipparcos parallax leads to a larger distance (152±20 pc) than the original; however, the astrometric solution still requires a significant cosmic noise of 2.4 mas. Given these results it is clear that the Hipparcos data still contain systematic errors of unknown origin." Although the radio data also have systematic errors, the Harper solution combines the datasets in the hope of mitigating such errors.83 An updated result from further observations with ALMA and e-Merlin gives a parallax of 4.51±0.8 mas and a distance of 222 (+34/−48) pc.[10] Further observations have resulted in a slightly revised parallax of 4.51±0.80.10

83Harper et al. (2008) A New VLA-Hoppocaros Distance to Betelgeuse and its Implications

10Harper et al. (2017)A Updated 2017 Astrometric Solution for Betelgeuse

When radio astrometry is used to measure positions of Betelgeuse in an effort to determine its parallax, I am guessing (see below) that it is the thermal black body radiation from the star's "radio photosphere" rather than maser radiation from a cloud well outside of the star itself.

(This answer to How far have individual stars been seen by radio telescopes? discusses maser radiation from stars for example)

Harper et al. (2017) contains:

For the purpose of this work, the channels containing line emission were excluded from the analysis and a single continuum data set centered at ≃338 GHz with a ∼5.9 GHz bandwidth was used.

and Harper et al. (2008) mentions:

We have used the highest spatial resolutions available with the VLA, i.e., A-configuration with the Pie Town VLBA antenna, and these allow us to obtain positional uncertainties comparable to Hipparcos. Good u–v coverage was obtained for six frequency bands (Q, K, U, X, C, L)7 at five epochs. For each band we used two 50 MHz continuum channels recording full Stokes polarizations.

7These bands have nominal wavelengths: Q = 0.7 cm, K = 1.3 cm, U = 2.0 cm, X = 3.6 cm, C = 6 cm, and L = 20 cm.


  1. Would these then be from the star's "radio photosphere"?
  2. Is there such a thing as a "radio photosphere" that differs substantially from a star's optical photosphere?

1 Answer 1


Stars certainly do have radio emission from their photospheres, since blackbodies emit at all wavelengths. But that’s not usually what is detected, because it’s so faint. Doing some simple numbers, emission from a star like the Sun peaks at a wavelength of about 500 nm (= $5 \times 10^{-7}$ m), and the long-wavelength side of the Planck function, in the Rayleigh-Jeans approximation, goes as $\lambda^{-4}$, so at a wavelength of 5 cm (= $5 \times 10^{-2}$ m) the emission will be $10^{20}$ times fainter. With nearby or very luminous stars it is possible to detect this in some cases (e.g. with ALMA which is both very sensitive at operates at somewhat shorter wavelengths), but more typically the radio emission detected is from other processes like synchrotron emission.

Whether the radio photosphere is at a substantially different radius depends on the opacity at that wavelength, but Reid & Menten (1996) Radio Photospheres of Red Giant Stars suggests that for some giants it may be a lot bigger than the optical photosphere.

From the abstract:

[...]These observations suggest that long period variables have a “radio photosphere” at about two stellar radii, where the stellar radius is defined by line-free regions at optical wavelengths. The radio photosphere is just inside the SiO maser shell, and the limited variability of the radio emission suggests that stellar shocks are mostly damped within this region. The density and temperature of the radio photosphere are estimated to be >~ 1012 cm-3 and $\approx$ 1400 K, respectively. For these physical conditions, free electrons, obtained predominantly from the ionization of potassium and sodium atoms, provide the opacity through interactions with neutral H atoms and H2 molecules.

  • $\begingroup$ thank you for the concise yet surprisingly thorough answer! $\endgroup$
    – uhoh
    Commented Jul 10, 2020 at 13:25
  • $\begingroup$ I found this interesting radio2space.com/the-radio-sun $\endgroup$
    – uhoh
    Commented Jul 12, 2020 at 6:16
  • 1
    $\begingroup$ Cool - some great figures there. Along the same lines, ALMA was upgraded a few years ago to be able to observe the Sun (different set of challenges than observing faint sources). Technical paper here, news release here. $\endgroup$ Commented Jul 12, 2020 at 13:25
  • 1
    $\begingroup$ Sorry! It’s here: aanda.org/articles/aa/full_html/2018/11/aa34113-18/… $\endgroup$ Commented Jul 12, 2020 at 13:54
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    $\begingroup$ The Sun isn't a blackbody and certainly not at radio wavelengths. $\endgroup$
    – ProfRob
    Commented May 17, 2021 at 7:32

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