4 ms·

Please note that this is not a temperature measurement in the sense of a thermodynamic temperature. It's not made clear in the article, but the ten trillion deg

Please note that this is not a temperature measurement in the sense of a thermodynamic temperature. It's not made clear in the article, but the ten trillion degrees is a "brightness temperature". This is a convention in radio astronomy where you take a measured specific intensity (power / unit area / unit solid angle / frequency) and compute the temperature a blackbody would need to have to emit the same specific intensity. When the brightness temperatures are this high, it indicates that the emission process is non-thermal (in this case, synchrotron emission from relativistic electrons). See equation 2.32 of: https://www.cv.nrao.edu/~sransom/web/Ch2.html https://www.cv.nrao.edu/~sransom/web/Ch2.html

zyxzevn 8y ago
Or due to plasma double layers?
czbond 8y ago
It was not clear to me, thank you for clearing that up. I am more of the "look at the nice visuals" side of astronomy - so I'm sharing what I found for synchrotron emission of relativistic electrons. ;) http://www.astro.utu.fi/~cflynn/astroII/l4.html http://www.astro.utu.fi/~cflynn/astroII/l4.html
frob 8y ago
Ok, that makes so much more sense. Thank you. If it were in the 10-trillion K range, we would probably be in the regime of hadron dissociation. I say probably since the quark-hadron phase diagram is still less a precise piece of work and more a ... rough sketch.
adrianratnapala 8y ago
Thank you for that, it seemed a bit implausible that they could measure such a temperature using radio astronomy since the emission peak should be in the gamma rays. While I get it that we can't always be persnicketty about the definition of (thermodynamic) temperature I find these alternate, phenomenological temperatures often do more harm than good. It's not just that it results in misleadingly juicy press releases. In my experience, physicists themselves forget that they are talking about what is little more than a fiction and start sticking their "temperature" into equations where it doesn't belong.
- deleted 8y ago
  [deleted]
- SiempreViernes 8y ago
  Also: temperatures of things that are ill defined, like the ”temperature of the surface of the sun”.
  - privong 8y ago
    > Also: temperatures of things that are ill defined, like the ”temperature of the surface of the sun”. This has more to do with phrasing it as the "surface" of the Sun rather than the temperature being ill-defined. The proper term would be the temperature of the photosphere of the Sun, which is where the physical conditions finally become such that optical light can escape. The temperature at that layer sets the blackbody emission that we observe.
    - SiempreViernes 8y ago
      The photosphere is no more well defined unless you specify a particular wavelength, or you admit defeat and accept averages as perfectly fine physical properties.
      - privong 8y ago
        > The photosphere is no more well defined unless you specify a particular wavelength, or you admit defeat and accept averages as perfectly fine physical properties. That stellar spectra (ignoring absorption lines) are broadly consistent with blackbody emission suggests that the size of the photosphere changes slowly with wavelength (i.e., that the opacity isn't a strong function of wavelength, otherwise there'd be a strong temperature gradient in the optically emitting regions). So from the standpoint of optical emission it's within the precision of measurements to think about it as as a single radius/surface. Or to put it another way, the change in photosphere size across the UV/optical/NIR is, for most stars, a small fraction of the radius. Thus the relative change in the size is small compared to the overall size. So while you're technically correct, "accepting defeat" won't be of significant practical importance to our understanding of the basic properties of stars (at least when discussing the continuum; absorption lines may have larger "sized" photospheres due to their increased optical depth at larger distances). Thus while talking about a single photosphere is not technically correct statement, it's nonetheless a useful way to describe the physical system.

zyxzevn 8y ago

Or due to plasma double layers?

czbond 8y ago

It was not clear to me, thank you for clearing that up. I am more of the "look at the nice visuals" side of astronomy - so I'm sharing what I found for synchrotron emission of relativistic electrons. ;) http://www.astro.utu.fi/~cflynn/astroII/l4.html http://www.astro.utu.fi/~cflynn/astroII/l4.html

frob 8y ago

Ok, that makes so much more sense. Thank you. If it were in the 10-trillion K range, we would probably be in the regime of hadron dissociation. I say probably since the quark-hadron phase diagram is still less a precise piece of work and more a ... rough sketch.

adrianratnapala 8y ago

Thank you for that, it seemed a bit implausible that they could measure such a temperature using radio astronomy since the emission peak should be in the gamma rays. While I get it that we can't always be persnicketty about the definition of (thermodynamic) temperature I find these alternate, phenomenological temperatures often do more harm than good. It's not just that it results in misleadingly juicy press releases. In my experience, physicists themselves forget that they are talking about what is little more than a fiction and start sticking their "temperature" into equations where it doesn't belong.

deleted 8y ago
[deleted]
SiempreViernes 8y ago
Also: temperatures of things that are ill defined, like the ”temperature of the surface of the sun”.
- privong 8y ago
  > Also: temperatures of things that are ill defined, like the ”temperature of the surface of the sun”. This has more to do with phrasing it as the "surface" of the Sun rather than the temperature being ill-defined. The proper term would be the temperature of the photosphere of the Sun, which is where the physical conditions finally become such that optical light can escape. The temperature at that layer sets the blackbody emission that we observe.
  - SiempreViernes 8y ago
    The photosphere is no more well defined unless you specify a particular wavelength, or you admit defeat and accept averages as perfectly fine physical properties.
    - privong 8y ago
      > The photosphere is no more well defined unless you specify a particular wavelength, or you admit defeat and accept averages as perfectly fine physical properties. That stellar spectra (ignoring absorption lines) are broadly consistent with blackbody emission suggests that the size of the photosphere changes slowly with wavelength (i.e., that the opacity isn't a strong function of wavelength, otherwise there'd be a strong temperature gradient in the optically emitting regions). So from the standpoint of optical emission it's within the precision of measurements to think about it as as a single radius/surface. Or to put it another way, the change in photosphere size across the UV/optical/NIR is, for most stars, a small fraction of the radius. Thus the relative change in the size is small compared to the overall size. So while you're technically correct, "accepting defeat" won't be of significant practical importance to our understanding of the basic properties of stars (at least when discussing the continuum; absorption lines may have larger "sized" photospheres due to their increased optical depth at larger distances). Thus while talking about a single photosphere is not technically correct statement, it's nonetheless a useful way to describe the physical system.