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).
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
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.
> 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.
My understanding is that there's a historical reason for this convention. Many telescope observations are calibrated by comparing the sky measurements against measurements of thermal loads with known temperatures. Since the calibration sources are specified with a temperature, it's straightforward to measure the source brightness in units of a pseudo-temperature. But I agree that it's not intuitive for those familiar with it.
Another layer on this is fact that doppler broadening of emission lines from molecules means they cover a finite width. Since typical velocities in astronomy range from fractions of a km/s to hundreds of kilometers per second, line widths are typically given in km/s. Since these line widths originate due to doppler motions, a width in km/s is equivalent to a width in Hz (for a given reference frequency for the spectral line).
The combination of these two things means people often report integrated flux measurements in units of K * km / s! It seems bizarre at first, but this is equivalent to specifying it as power / unit area (again, when there's a reference frequency specified). But these units are often convenient in their own way.
Well I am not an astronomer, but the story you tell feels very similar to the how other parts of physics sometimes use weird kinds of temperatures for similar reasons.
In my field we sometimes liked to quote frequencies in Kelvins. What's and Planck-over-Boltzman between friends?
> But I agree that it's not intuitive for those familiar with it.
It's worse than that. The convention of using temperature usually arises from some such instrument-calibration where the temperature more-or-less exists, and is accurate. Then the same instrument gets used for things where temperature doesn't exist at all and sometimes doesn't even give a fair description of the energy scales. Yet people keep blithely reporting these things as temperatures.
> 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.
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.
> 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.
I am certainly not prepared to ignore lines in this pointless internet debate! I did my master thesis on lines, how dare you?!
Furthermore, this is a discussion of the sun and not some far away "star" nobody ever even resolved to a disk.
On the sun we see the convection zones plainly, and thus speaking of a single temperature of what is clearly a feature with structure could be wery misleading (in certain specialised circumstances).
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.
> That's hotter than the Hagedorn temperature, there could be some very interesting stuff going on.
Unfortunately it's not made clear in the article, but the temperature they quote is not an actual temperature, but instead a way of describing the specific intensity of the radiation. In this case the emission they're measuring is non-thermal (synchrotron emission), so the "brightness temperature" of the radiation field is not the same as the thermodynamic temperature. I'd put in a top-level comment about it too: https://news.ycombinator.com/item?id=17898224
However, any matter in the way of the emissions would still reach something like that temperature, right? There's got to be some dust out there that's over the hagedorn temperature.
Not likely, since the intervening material probably lets most of the radio emission through. So the radiative coupling between the synchrotron emission and that material is too weak to raise the temperature by much (if any).
> There's got to be some dust out there that's over the hagedorn temperature.
Interstellar dust grains are thought to be destroyed at temperatures of 1500-2000 K (depending on composition). And their absorption cross-section for absorbing radio waves is fairly small.
I don't understand a lot of that article. I recently read about the Planck temperature, beyond which the material would emit radiation smaller than Planck length, and which is therefore sort of an upper limit of temperature beyond which we don't know what would happen. For the Hagedorn temperature, it says that beyond this temperature, things will turn into "quark matter" (linking to another article explaining that this means "any of a number of theorized phases of matter whose degrees of freedom include quarks and gluons.", which isn't very helpful), but not why that would happen. I don't understand why beyond this point, matter should suddenly turn into other 'theorized' particles.
Basically, as particles approach the Hagedorn temperature they bleed off energy in the form of matter (hadrons). This process prevents the temperature from increasing, unless it is a quark plasma - in which case there is a second and much higher temperature. The plasma would arise due to degrees of freedom (which is what temperature actually measures) - matter would have an overwhelming freedom to break apart (which it does normally all the time: see nuclear radiation).
I don't understand a lot of it either, however my intuition tells me that it can be interpreted like just another state change.
My understanding of the transitions of matter between solid, liquid, and gaseous states is that it is primarily due to the amount of energy in the system. The more energy a particle has, the more likely it is to break its bonds with other particles around it and transition to another state. We tend to think of individual atoms being in some way the 'final' state once we reach some gaseous form, however it is not that great a leap to consider putting so much energy into the nucleus of an atom that its nucleus 'evaporates' into protons and neutrons (hadrons), and from there putting so much energy into those hadrons that they too evaporate into their constituent quarks. The difference in each case being that enough energy must be present to overcome each necessary nuclear force.
To be fair though, this is all very theoretical from what I've just read, and we don't truly know what happens yet.
Not sure why you were downvoted, I guess someone thought you were wrong? Anyway, I think it was a helpful explanation -- if anyone thinks it's wrong, they should (briefly) comment instead of downvoting.
See equation 2.32 of: https://www.cv.nrao.edu/~sransom/web/Ch2.html