Electron beam heating/Laboratory

This laboratory is an activity for you to create a method of heating the solar corona or that of a star of your choice. While it is part of the radiation astronomy course principles of radiation astronomy, it is also independent.

Some suggested entities to consider are electromagnetic radiation, electrons, positrons, neutrinos, gravity, time, Euclidean space, Non-Euclidean space, magnetic reconnection, or spacetime.

More importantly, there are your entities.

Please define your entities or use available definitions.

Usually, research follows someone else's ideas of how to do something. But, in this laboratory you can create these too.

Okay, this is an astrophysics coronal heating laboratory.

Yes, this laboratory is structured.

I will provide an example of electron beam heating calculations. The rest is up to you.

Questions, if any, are best placed on the discussion page.

Control groups
The findings demonstrate a statistically systematic change from the status quo or the control group.

For creating a heating method, what would make an acceptable control group? Think about a control group to compare your universe or your process of creating a universe to. Or, at least a control group to compare your heating technique to.

I will provide one example.

Electron beam heating
The lecture on electron beam heating describes at least one example of the use of this technology to heat objects.

Flux
Def. the rate of transfer of energy (or another physical quantity) through a given surface, specifically electric flux, magnetic flux is called a flux.

"Mass flux of species α [is in units of] g/cm2 s" or, in dimensions [quantity of species α]·[area]−1·[time]−1, where an area may be cm2.

Detector distance
Let Voyager 1 be 17,932,000,000 km (119.9 AU) from the Sun at RA 17.163h Dec +12.44°, ecliptic latitude of 34.9°.

Let Voyager 1, now headed out of the solar system, be traveling at 17 km/s (38,000 mph).

At this distance, the Voyager 1 electron detector has detected both the number and direction of extraheliosphere electrons. For this laboratory example, let the electron flux be 2 e- cm-2 s-1 diffusing into our solar system from elsewhere in the galaxy. Each of these electrons has an energy of 10 MeV.

Electron beams
As of December 5, 2011, "Voyager 1 is about ... 18 billion kilometers ... from the [S]un [but] the direction of the magnetic field lines has not changed, indicating Voyager is still within the heliosphere ... the outward speed of the solar wind had diminished to zero in April 2010 ... inward pressure from interstellar space is compacting [the magnetic field] ... Voyager has detected a 100-fold increase in the intensity of high-energy electrons from elsewhere in the galaxy diffusing into our solar system from outside ... [while] the [solar] wind even blows back at us."

Photospheres
From the image at the top of this laboratory it appears that the coronal clouds about the Sun cover a partial shell from say 80° N to about 80° S that follows the curvature of the photosphere. There does seem to be a physical separation of the shell of coronal clouds from the concentric shell of the photosphere. The coronal holes perhaps being void of incoming electron flux.

Photosphere volumes
R⊙eq ≈ 6.955 x 105 km. The thickness of the photosphere is about 400 km. R⊙p ≈ 6.951 x 105 km.


 * $$V_{\odot p} = \frac{4\pi}{3} [R_{\odot eq}^3 - R_{\odot p}^3] km^3,$$


 * $$V_{\odot p} = \frac{4\pi}{3} [6.955^3 - 6.951^3] \times 10^{15} km^3,$$


 * $$V_{\odot p} = \frac{4\pi}{3} (0.580) \times 10^{15} km^3,$$


 * $$V_{\odot p} = 7.288 \times 10^{15} km^3.$$

Photosphere hydrogens
The density of the Sun is about 2 x 10-4 kg m-3. Or,


 * $$\rho_{\odot p} = 2 \times 10^{-4} kg \cdot m^{-3},$$


 * $$\rho_{\odot p} = 2 \times 10^{-4} kg \cdot [10^{-3} km]^{-3},$$


 * $$\rho_{\odot p} = 2 \times 10^5 kg \cdot km^{-3}.$$

One mole of H2 (gas) has a mass of 2.016 x 10-3 kg. The molar density of the photosphere may be


 * $$\rho_{\odot p} = \frac{2 \times 10^5 kg}{2.016 \times 10^{-3} kg/mole} km^{-3},$$


 * $$\rho_{\odot p} = \frac{2}{2.016} \frac{10^5}{10^{-3}} \frac{kg}{kg/mole} km^{-3},$$


 * $$\rho_{\odot p} = 0.992 \times 10^8 moles \cdot km^{-3},$$


 * $$\rho_{\odot p} = 10^8 moles \cdot km^{-3}.$$


 * $$V_{\odot p} = 7.288 \times 10^{15} km^3.$$


 * $$H_{2 \odot p} = (10^8 moles \cdot km^{-3}) \cdot (7.288 \times 10^{15} km^3),$$


 * $$H_{2 \odot p} = 7.288 \times 10^{23} moles.$$

Photosphere heating
For H2 (gas) the molar constant-volume heat capacity at 298 K is 20.18 J/(mol · K). At 2000 K it is about 25 J/(mol · K). Using a linear extrapolation,


 * $$C_{V,m} = (2.83 \times 10^{-3})T J/(mol \cdot K^2) + 19.3 J/(mol \cdot K),$$

for 5777 K, yields


 * $$C_{V,3,m} = (2.83 \times 10^{-3}) (5777) J/(mol \cdot K) + 19.3 J/(mol \cdot K),$$


 * $$C_{V,m} = 35.6 J/(mol \cdot K).$$

Before calculating the amount of energy or power necessary to heat the coronal clouds around the Sun, let's see if the influx of electrons from outside the heliosphere may be able to heat the surface of the photosphere (p) to 5777 K from 100 K.


 * $$\Delta Q = [35.6 J/(mol \cdot K)] \cdot (5777 - 100) K,$$


 * $$\Delta Q = (35.6) \cdot (5677) \frac{J}{mole \cdot K}{K},$$


 * $$\Delta Q = 2.02 \times 10^5 J/mole.$$


 * $$1 J = 6.24 \times 10^{18} eV.$$


 * $$\Delta Q = (2.02 \times 10^5 J/mole) \cdot (6.24 \times 10^{18} eV/J),$$


 * $$\Delta Q = (2.02) \cdot (6.24) \times 10^5 \times 10^{18} eV/mole,$$


 * $$\Delta Q = 12.6 \times 10^{23} eV/mole,$$


 * $$\Delta Q = 1.26 \times 10^{24} eV/mole.$$

Photosphere electron influx

 * $$\Delta Q = 1.26 \times 10^{24} eV/mole.$$


 * $$H_{2 \odot p} = 7.288 \times 10^{23} moles.$$

Voyager 1 is 17,932,000,000 km (119.9 AU) from the Sun at RA 17.163h Dec +12.44°, ecliptic latitude of 34.9°.

For this laboratory example, let the electron flux be 2 e- cm-2 s-1 diffusing into our solar system from elsewhere in the galaxy. Each of these electrons has an energy of 10 MeV.


 * $$\Phi_{e^-} = 2 e^- \cdot cm^{-2} \cdot s^{-1} \cdot \frac{(10^{-2} \cdot m \times 10^{-3} \cdot km/m)^{-2}}{cm^{-2}},$$


 * $$\Phi_{e^-} = 2 e^- \cdot cm^{-2} \cdot s^{-1} \cdot \frac{10^{10} \cdot km^{-2}}{cm^{-2}},$$


 * $$\Phi_{e^-} = 2 \times 10^{10} e^- \cdot km^{-2} \cdot s^{-1}.$$

If the electron flux measured by Voyager 1 is close to 2 e- cm-2 s-1 where each electron averages 10 MeV and these electrons are heading for the Sun, then each electron may strike the photosphere from anywhere in a sphere around the Sun.

To heat the photosphere to 5777 K takes


 * $$\Delta Q = (1.26 \times 10^{24} eV/mole) \cdot (7.288 \times 10^{23} moles),$$


 * $$\Delta Q = (1.26) \cdot (7.288) \times (10^{24} \times 10^{23}) \cdot eV,$$


 * $$\Delta Q = 9.18 \times 10^{47} eV.$$

The power (P) that may be deposited on the photospheric surface of the Sun is


 * $$P_{e^-} = 4 \pi R_{Voyager 1}^2 \cdot \Phi_{e^-} \cdot (10 MeV/e^-),$$


 * $$P_{e^-} = 4 \pi (1.7932 \times 10^{10} km)^2 \cdot (2 \times 10^{10} e^- \cdot km^{-2} \cdot s^{-1}) \cdot (10 MeV/e^-),$$


 * $$P_{e^-} = (4 \pi) \cdot (1.7932)^2 \cdot 2 \times (10^{20} \times 10^{10} \times 10^7) \cdot (km^2 \cdot e^- \cdot km^{-2} \cdot s^{-1} \cdot eV/e^-),$$


 * $$P_{e^-} = 80.8 \times 10^{37} \cdot eV \cdot s^{-1},$$


 * $$P_{e^-} = 8.08 \times 10^{38} eV \cdot s^{-1}.$$

The luminosity (in Watts, W) of the Sun is 3.846 x 1026 W. In eV/s this is


 * $$L_{\odot} = 3.846 \times 10^{26} W \cdot (10^7 erg/(s \cdot W)) \cdot 6.24 \times 10^{11} eV/erg,$$


 * $$L_{\odot} = (3.846) \cdot (6.24) \times (10^{26} \times 10^7 \times 10^{11}) \cdot (W \cdot erg/(s \cdot W)) \cdot eV/erg),$$


 * $$L_{\odot} = 24.0 \times 10^{44} \cdot eV \cdot s^{-1},$$


 * $$L_{\odot} = 2.40 \times 10^{45} \cdot eV \cdot s^{-1}.$$

If the energy of the incoming electrons is 700 MeV and the flux is 8.48 x 104 e- cm-2 s-1, then the power from the incoming electrons would be


 * $$P_{e^-} = (8.08 \times 10^{38} eV \cdot s^{-1}) \cdot (70) \cdot (8.48/2 \times 10^4),$$


 * $$P_{e^-} = (8.08) \cdot (70) \cdot (4.24) \times (10^4 \times 10^{38}) eV \cdot s^{-1},$$


 * $$P_{e^-} = 2400 \times 10^{42} eV \cdot s^{-1},$$


 * $$P_{e^-} = 2.40 \times 10^{45} eV \cdot s^{-1}.$$

Coronal heating
The power to heat the solar corona is on the order of 1039 eV s-1. While this is within range of the estimated electron influx beam heating, X-rays are usually generated between 0.1 and 120 keV. 10 to 700 MeV electrons would likely produce gamma rays rather than X-rays.

Report
Title: Corona and photosphere heating by interstellar electron influx.

Author: Marshallsumter (discuss • contribs) 01:08, 31 January 2014 (UTC)

Abstract:

By using estimates of the interstellar electron influx that may have been measured by Voyager 1, a back-of-the-envelope calculation shows that the electron influx may be sufficient to heat the solar corona to MK and the photosphere to 5777 K. These estimates put the coronal heating problem in perspective of the overall external heating of the outer Sun including the photosphere.

Introduction:

The news report of Voyager 1 reaching some 120 AU from the Sun contain interesting discoveries. The most significant may be a flux of electrons diffusing into our solar system from elsewhere in the galaxy. The solar wind has ceased earlier and has even been turned backward. These results suggest a net influx of electrons perhaps toward the Sun that may be involved in heating the solar corona. The possibility is explored using some "back-of-the-envelope" calculations.

Experimentation:

The experiment consists of two parts: the amount of power to either heat the photosphere to an effective temperature of 5777 K or to at least match the Sun's luminosity and to heat the solar corona into the MK range so that X-rays are emitted.

To accomplish the first experiment the volume of the photosphere is calculated followed by the number of moles of diatomic hydrogen gas, the principal component of the photosphere. Using an estimated constant volume specific heat capacity the amount of heat needed to warm the Sun up from 100 K (a likely pre-luminescent temperature of the proto-Sun before fusion to 5777 K) was calculated.

The second experiment is accomplished by estimating the power and influx of electrons from elsewhere in the galaxy.

Discussion:

The power calculated for the electron influx is compared with the first experiment and with the known luminosity of the Sun.

The initial guess put the power at about six orders of magnitude too low to match the Sun's current luminosity. However, increasing the electron energy by 70 times and the influx by about 4 x 104 brings the estimate into agreement. It is likely that either increase may be too much.

Other concerns focus on whether any of these electrons can reach the Sun. The Sun has a net surface negative charge. The solar wind contains electrons, protons and heavier nuclei that may alter the energy or influx significantly.

Conclusions:

While likely factors exist that may decrease the estimated electron energy or influx, it appears that such an influx may not only heat the solar corona but also the photosphere. Actual numbers from Voyager 1 may shed more light on this conclusion.

Evaluation
To assess your method of heating the solar corona, including your justification, analysis and discussion, I will provide such an assessment of my example for comparison.

If you chose another star, even of a different spectral type, be sure to include likely scale factors or other adjustable parameters.

Evaluation

Unless the influx is significantly higher at lower energies it seems that 10 to 700 MeV is way too high. The corona of the Sun would be producing gamma-rays not X-rays unless the negative surface charge diverts the energy loss into photospheric heating. No justification is given for the 104th increase in the influx estimate unless it is a focusing effect. Actual data on interstellar electrons does suggest MeV levels.

Hypotheses

 * 1) Energy expended in increasing gas or plasma temperatures up into the MeV category is where some of the 700 MeV electron energy is going.
 * 2) The influx of interstellar electrons is necessary and sufficient to heat the solar photosphere and corona to the observed properties and temperatures.
 * 3) As the photosphere of the Sun has a net negative charge, this repulsion may be enough to reduce the energy of incoming electrons below MeV levels.