# The Coronal Heating Myth Resolved

1. Jan 10, 2004

### Thomas2

The fact that the solar corona has a temperature of a few million degrees has puzzled solar physicists for a long time, considering the comparatively low temperature of about 6000oK at the sun's apparent surface (the photosphere). Clearly, the laws of thermodynamics seem to rule out that a cool gas volume (the photosphere) should be able to heat another gas volume (the corona) to a temperature of almost 1000 times its own. Various elaborate plasma processes have been proposed that would enable charged particles in the photosphere to be accelerated to such high temperatures, but all these can still not explain how unordered thermal energy of many particles should be transformed into ordered high energy of a few particles.
However, in the course of the 'coronal heating' discussion it has apparently not been recognized that a temperature of several million degrees is in fact the 'natural' temperature of the solar plasma (as determined by the gravitational energy), whereas the photospheric temperature is the 'abnormal' one.
The further discussion involves a number of mathematical formulae as well as illustrative figures (see my Coronal Heating page for this)

Last edited: Jan 26, 2004
2. Jan 11, 2004

### Orion1

Corona Temperature...

$$E = K_o T = \frac{ G M_o m_p}{ 2 r_c }$$

$$T_c = \frac{ G M_o m_p}{ 2 K_o r_c }$$

Tc = 1.148*10^7 K
Tc = 2*10^6 K (measured)

$$T_s = \frac{ G M_o m_e}{ 2 K_o r_s}$$

Ts = 6290.615 K
Ts = 5820 K (measured)

Mass-Luminosity Equasion:
$$r_s = \frac{}{2T_s ^2} \sqrt{ \frac{L_o}{\pi \sigma} \left( \frac{M_s}{M_o} \right)^n}$$

Orion-Thomas Integral:
$$r_s = \frac{ G M_s m_e}{ 2 K_o T_s} = \frac{}{2T_s ^2} \sqrt{ \frac{L_o}{ \pi \sigma} \left( \frac{M_s}{M_o} \right)^n}$$

Orion-Thomas Thermodynamic Equasion:
$$T_s = \frac{K_o}{G M_s M_e} \sqrt{ \frac{L_o}{ \pi \sigma} \left( \frac{M_s}{M_o} \right)^n}$$

n - Dynamic Stellar Property
n = -8.244 -> -16.769 - White Dwarf
n = 3.321 -> 3.376 - Main Sequence
n = 3.661 -> 3.763 - Giant, Supergiant

White Dwarf:
Md = .8*Mo
rd = 8769 km
Ts = 99857.879 K
Ts = 100000 K (measured)

Tc = 9.253*10^8 K
Tc = unknown (no known measure)

Giant: Arcturus
Mg = 4*Mo
rg = 1.74*10^10 m
Ts = 1006.498 K
Ts = 4290 K (measured)

Tc = 1.848*10^6 K
Tc = unknown (no known measure)

Supergiant (Red Giant): Betelgeuse
Msg = 18*Mo
rsg = 4.872*10^11 m
Ts = 161.758 K
Ts = 3100 K (measured)

Tc = 297.013 K
Tc = unknown (no known measure)

Mo = 1.989*10^30 kg - Sol/Stellar Mass
rc = 7*10^8 m - Sol/Stellar Coronal Radius
rs = 6.96*10^8 m - Sol/Stellar Radius
Lo - Sol Luminosity
Ms - Stellar Mass
mp = 1.672*10^-27 kg - Proton Mass (Plasma)
me = 9.109*10^-31 kg - Electron Mass (Plasma)
Ko - Boltzmann's Thermal Constant
Tc - Coronal Thermodynamic Gravitational Temperature.
Ts - Surface Thermodynamic Gravitational Temperature.
$$\sigma$$ - Stefan-Boltzmann Constant

Reference:
http://www.plasmaphysics.org.uk/research/sun.htm

Last edited: Jan 19, 2004
3. Jan 11, 2004

### Nereid

Staff Emeritus
Taking the same approach that you used for the Sun, and applying it to two of the solar system gas giants, Jupiter and Neptune, we get the following:

For Jupiter, Ts = 58 K; Tc = 107,000 K
For Neptune, Ts = 8.9 K; Tc = 16,300 K.

Similarly, the application of your approach to a 1 solar mass red giant, with a radius equal to that of the Earth's orbit (150 million km), gives:
Ts = 29 K; Tc = 53,800 K.

Finally, a white dwarf of mass 1 sol, radius 6950 km would have:
Ts = 63,000 K; Tc > 1 billion K

AFAIK, none of these temperatures match what's observed. Unless I've misunderstood your approach (or it applies uniquely to the Sun), there would seem to be strong observational evidence to show your explanation of coronal heating in the Sun is wrong.

4. Jan 12, 2004

### Thomas2

First of all, the main point I am making is that, according to classical mechanics, an atom in the solar atmosphere should actually have a temperature (i.e. kinetic energy) according to half it's gravitational potential energy (this is similar to the requirement that the centrifugal force is identical to gravitational force in a stable circular orbit). It is easy to work out that this temperature is orders of magnitude higher than the surface temperature of the sun, i.e. it is the latter which needs to be explained.

In a quantitative sense, I am referring my analysis primarily to the sun, simply because this is the only star where the mass and radius can be measured exactly. But even for the other examples mentioned above, the resultant surface temperature is not orders of magnitude off the measured values, apart from the temperature for Red Giants. The discrepancy for the latter is not surprising as the average volume number density is very small for these. Most of a Red Giant can in fact be considered to be just an expanded atmosphere. The 'surface' of the Red Giant (as given by the requirement that the density reaches about 3.1023cm-3, i.e. atoms cease to exist) is actually much closer to it's center but is invisible because it's radiation is absorbed by the expanded atmosphere. Since the latter provides most of the mass (the role of star and atmosphere are practically reversed compared to the sun), one should therefore actually observe T=53800 K as the temperature of the red giant, but although the cooling due to inelastic collisions is greatly reduced (because of the low plasma density spectral lines are not broadened to a continuum anymore and only protons in a small energy range can excite transitions), it is apparently still efficient enough to bring the temperature down to the observed value.

In any case, one has to take into account that I am referring to 'temperature' here as the kinetic energy of particles. This is usually not strictly identical to the 'radiation temperature' derived from stellar and planetary spectra. For the latter, relative intensities in certain spectral bands are used to derive a 'temperature' under the assumption of a Local Thermodynamic Equilibrium. However, the latter condition is often far from being fulfilled (see my website http://www.plasmaphysics.org.uk/#lte for more details).

Although it is only of secondary interest here, it is remarkable that the analysis on my http://www.plasmaphysics.org.uk/research/sun.htm" [Broken] ).

Last edited by a moderator: May 1, 2017
5. Jan 18, 2004

### Nereid

Staff Emeritus
This reference gives absolute radii of several thousand stars, many determined with an accuracy of ~<1%, which is more than sufficient for you to do quantitative analyses:
http://arxiv.org/abs/astro-ph/0012289

Similarly, masses have been determined equally accurately for several thousand (tens of thousands?) of stars.

For many (most?) of the stars, the photospheric temperatures are also well determined.

BTW, I didn't see any links to peer-reviewed papers on your website; could you give us some, to papers you've published?

6. Jan 19, 2004

### Thomas2

Nereid,

Obviously, the more accurate results will be associated with giant stars for which however my analysis does not strictly apply (as mentioned in my previous post).
If you look at Table 3 of your reference, the majority of the observations have actually an accuracy between 5% and 10% and this is obviously excluding any potential unknown systematic errors.
Anyway, these observations do not provide an alternative theoretical explanation for the photospheric and coronal temperatures considering the arguments outlined on my page http://www.plasmaphysics.org.uk/research/sun.htm.

If readers want to understand the basic physics in this context, I don't actually think it is a good idea to apply my analysis to more examples at this stage, in particular because (as indicated already above) the use of the 'spectral' temperature instead of the the kinetic temperature is generally questionable (it would require very complex model calculations to establish a connection between the two).

My work is summarized on my webpage http://www.plasmaphysics.org.uk/research/ where you can also find links to the major papers.

Last edited: Jan 20, 2004
7. Jan 21, 2004

### Orion1

Solar Muon Flux...

$$T_c = \frac{ G M_o m_ \mu}{ 2 K_o r_c }$$
Tc = 1.301*10^6 K
Tc = 2*10^6 K (measured)

The Coronal Temperature may be explained if it was heated by a Solar Muon Flux.