REVISION LECTURE I SOLAR PHYSICS

PHAS M314/SPCEG012

April 23, 2012

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KJHP part of Revision Lecture

See http://www.mssl.ucl.ac.uk/~kjhp/

Go to directory SOLAR_PHYSICS_COURSE_2012/REVISION_LECTURE/

File is called PHASM314_revisionlecture2012.ppt

See PHASM314 Moodle for details.

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Recap: the Sun as a star

Sun formed by gravitational collapse of an interstellar gas cloud: now a main sequence star.

Age: from oldest meteorites ~ 4.7x109 yearsPossible sources of its energy are:

– thermal– gravitational– chemical– nuclear

If gravitational energy were sufficient to power the Sun, Sun’s lifetime = the Kelvin-Helmholtz time = 3 ×107 years.

Virial theorem: negative gravitational energy = 2 x thermal energy.So thermal energy also inadequate to power the Sun. Only nuclear reactions can provide the required energy.

Reactions are fusion of light elements because only they are abundant. In fact net reaction is 4H → He. Time scale is tnuclear where:

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Three possible ways for energy to be transported out:

1. Conduction2. Convection3. Radiation

Mean free path of the photons >> mean free path of electrons, so thermal conduction is negligible in the Sun.

Radiation is the more important mechanism at inner part (0.25 – 0.71 Ro) of solar interior. Convection predominates in the outer part (0.71 – 1 Ro).

Photon transport in solar interior

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Standard solar model

The standard solar model describes the physical conditions in the solar interior:

1) Element Abundances: mass fractions are X (hydrogen) = 0.71,Y(helium) = 0.27, Z(“metals”) = 0.02.

2) Hydrostatic equilibrium.3) Energy transport: Radiation in deep interior (photon

diffusion moderated by Rosseland opacity). Outer regions: convection.

4) Energy generation: Nuclear fusion reactions (p-p chain) are the source of energy.

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Neutrinos

Neutrinos produced in the core by nuclear reactions (mostly associated with p-p chain).

Predicted neutrino flux at Earth = 6.6x1014 m-2 s-1. Measured flux is between 1.9 and 3.6 times less than this.This was the Solar Neutrino Problem.

Solar neutrino detectors: Homestake, Gallex & SAGE, Kamiokande .

Explanation of discrepancy: neutrino flavour “oscillations”: the MSW effect.

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Photosphere

Photosphere: where energy generated at solar core escapes to space.

Temperature varies from 6400K to 4400K (temperature minimum)

Radiation black-body-like. Total solar irradiance (“solar constant”) = total amount of

radiation over all wavelengths per unit time per unit area reaching the top of the Earth’s atmosphere.

Spacecraft measurements give 1.368 kW m-2, varying by ± 0.3 % (over 11-year cycle; also sunspots).

ChromosphereChromosphere controlled by supergranular motions:

supergranules about 30,000 km across. Convection sweeps material & mag. fields to edges of cells. EUV and Hα emission enhanced at these locations.

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TE, LTE, NLTE and coronal equilibria

Thermodynamic equilibrium (TE): matter in complete equilibrium with radiation – a black body. Holds in solar interior.

Local thermodynamic equilibrium (LTE): thermodynamic equilibrium defined by the local value of T. Holds in the solar photosphere.

Non-LTE (NLTE): Kinetic temperature T different from the temperature characterizing the radiation passing through it. Applies in the solar chromosphere.

Coronal equilibrium: Equilibrium controlled entirely by the interactions of ions and electrons of corona.

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The solar corona: properties

Corona is much hotter than chromosphere or photosphere: 106 K -- 2×106 K (1—2 MK).

Corona is much less dense than chromosphere or photosphere: n (particles m-3) ~ 1013 – 3×1014 m-3.

General magnetic field ( ~1mT) in form of loop structures/streamers.

Temperature, density, and magnetic field enhanced over active regions, lower over coronal holes.

Corona is a fully ionized plasma, with electrons, protons, and He nuclei performing helices about magnetic field lines.

Ionization equilibrium: ionizations (collisional + autoionizations) balanced by recombinations (radiative + dielectronic).

Coronal heating: either by the dissipation of MHD waves or by lots of tiny flares. Evidence for both still evenly balanced.

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Solar flaresA solar flare is a sudden release of energy during which magnetic energy is converted to kinetic energy of fast particles, mass motions, and radiation.

Energy released up to 1025 J in the largest flares.

Energy release almost certainly due to magnetic reconnection, when oppositely directed field lines approach one another.

This can only happen when the diffusion term in the induction equation dominates, but theoretically this is extremely hard to come about.

Several schemes by which flares may occur in practice.

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Sun-like stars

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Sun-like stars are generally late (spectral types K, M) stars with characteristics like the Sun’s:

Starspots (visible from light curve modulations)Chromospheres (Ca II H and K lines in emission)Coronae (X-ray emission)Flare activity

Some non-solar-type stars:RS CVn stars: binary systems – very energetic flares (stars have interacting magnetic fields). T Tauri stars: very young stars with blobs of infalling material & powerful stellar wind.

Worked exam question: 2009 Qu. 1

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What are the mechanisms by which energy is transported from the boundary of the Sun’s core region to the photosphere?

Draw a simple diagram showing the way the temperature decreases from the core to the photosphere, and indicate on this diagram which mechanism dominates.

What sets the boundaries for each process? [6]

The mean free path of a photon in the solar interior is approximately 10-2 m. Assuming that photons travel by a random walk process, calculate the approximate time for a photon generated in the solar core to reach the photosphere. [4]

By considering an element of gas in the solar interior rising adiabatically, discuss the physical conditions leading to the onset of convection. [5]Derive an expression for the critical value of the radial temperature gradient at which convection will occur. [5]

Question: What are the mechanisms by which energy is

transported from the Sun’s core to the photosphere? Draw a simple diagram showing the way the temperature decreases from the core to the photosphere, and indicate on this diagram which mechanism dominates. What sets the boundaries for each process?

Answer: Energy is transported by radiation from a lower boundary of 0.25 Rʘ to 0.71 Rʘ .

Diagram:

The lower boundary 0.25 Rʘ is set by the nuclear reactions taking place within this limit, and the upper boundary 0.71 Rʘ is set by the onset of convection. 13

Question: The mean free path of a photon in the solar interior is

approximately 10-2 m. Assuming that photons travel by a random walk process, calculate the approximate time for a photon generated in the solar core to reach the photosphere.

 

Answer: Total number of random walk steps is (Rʘ /λ)2 where λ is the mean free path. The time taken to cross λ is λ/c, so total time taken for photons to reach photosphere is Rʘ

2 / (λc) :

Time = (7.108)2 / (10-2 x 3.108) = 1.63.1011s

No. of seconds in a year ~ 31.106 (i.e. 365.25 x 24 x 60 x 60)

So time = 5300 years.

 

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Question: By considering an element of gas in the solar interior rising adiabatically, discuss the physical conditions leading to the onset of convection.

Answer: Element of gas has initial density ρ1’, pressure P1’, and after rising has density ρ2’, pressure P2’ (outside densities and pressures are ρ1, P1 and ρ2, P2). If ρ2’ < ρ2 element keeps rising towards photosphere, where it radiates, cools and sinks. A convective cycle is thus set up and convection begins (so is convectively unstable).

 

The gas element is assumed to rise adiabatically, so PV γ = constant. In terms of density this is

So condition for stability is 15

/1

'1

'2'

1'22

P

P

γ

ρρ/1

'1

'2'

1'2

P

P

Question: Derive an expression for the critical value of the radial

temperature gradient at which convection will occur.

 Answer: The condition for stability can be written

or considering radial gradients

Perfect gas law is P = ρ kB T / (μ mH), so eliminate ρ to give

Or expanding or

Using magnitudes of gradients, criterion for stability is

If the temperature gradient dT/dr is greater than the RHS, there will be instability and convection will occur.

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dr

dP

P

T

dr

dT r

r

rr 1

dr

dP

PT

P

dr

d

rP

T r

rr

r

r

r

11

dr

dP

Pdr

d r

r

r

r

11

dr

dP

Pdr

dT

T

P

dr

dP

TP

T r

r

r

r

rr

rr

r

112

dr

dP

Pdr

dT

Tr

r

r

r

11

11

P

dP

P

dPPd

γρ

ργ

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/1

Worked exam question: 2010 Qu. 4

Starting with Ohm’s law for a plasma, derive the induction equation for a solar flare plasma. Indicate which is the diffusive and which the advective term. [5]

Give order-of-magnitude estimates for the diffusive and advective time scales for a pre-flare plasma with temperature 2 × 106 K, length scale 10,000 km, and velocity 100 km s-1, and calculate the magnetic Reynolds number. Compare the solar pre-flare plasma with the atmosphere separating the two components of an RS CVn binary system with temperature 107K and distance scale 106 km. [5]

 Give the reason for the magnetic buoyancy of a flux tube which is in thermal balance with its surroundings. Calculate the density difference between a flux tube of 10 Tesla intensity and the surrounding plasma. Both are in pressure equilibrium and located near the bottom of the convection zone. For the bottom of the convection zone use values of pressure p = 7·1012 Pascal, density = 2·102 kg m-3, and temperature T = 2·106 K. [5]

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Give a brief account of the effect of electric currents (non-potentiality) on the shape and connectivity of coronal loops. The Figure below shows two sets of magnetic extrapolations. The thick lines indicate magnetic field lines extrapolated from photospheric magnetic fields connecting positive (full contours, right-hand-side foot points in each figure) and negative (dashed contours, left-hand-side foot points in each figure) magnetic concentrations. Identify which extrapolation is computed with current-free (potential) and which one with current-carrying (non-potential) conditions. [5]

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Question: Starting with Ohm’s law for a plasma, derive the induction equation for a solar flare plasma. Indicate which is the diffusive and which the advective term.

Answer:

Ohm’s law for a plasma: η J = J/(σ μ0) = E + v × B (1)

where J = current density, E = electric field, B = magnetic flux density; η = magnetic diffusivity.

Take the curl of both sides:

η curl J = curl E + curl (v × B) (2)

or (1/ η) × L.H.S. of (2) is, by Maxwell’s equations,

curl J = curl curl B = grad div B - ▼2 B = - ▼2 B

R.H.S. of (2) is - ∂B/∂t + curl (v × B)

So ∂B/∂t = η ▼2 B + curl (v × B) (3)

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Eq. (3): ∂B/∂t = η ▼2 B + curl (v × B)

can be expressed by:

Rate of change of magnetic field in a flare volume = diffusive term + advective term. [Bookwork, 5 marks]

Question: Give order-of-magnitude estimates for the diffusive and advective time scales for a pre-flare plasma with temperature 2 × 106 K, length scale 10,000 km, and velocity 100 km s-1, and calculate the magnetic Reynolds number. Compare the solar pre-flare plasma with the atmosphere separating the two components of an RS CVn binary system with temperature 107K and distance scale 106 km.

Answer: Get an order-of-magnitude estimate of quantities by approximating ∂/∂t = 1 / t, curl = 1 / L,

▼2 = 1 / L2:

If there is no advective term, B / τD = η B / L2

or diffusion time, τD = L2 / η. 20

The classical value of the magnetic diffusivity (Spitzer) is

η = 109 Te-3/2 m2 s-1

(see rubric) where Te = electron temperature. For the quiet corona, Te = 2 MK, so η = 0.35 m2 s-1.

 

So the diffusive time scale τD for the pre-flare volume (take this to be L = 10,000 km = 107 m) is possibly as high as ~3.1014 s (10 million years).

If there is no diffusion, only advection, then approximately

B / τC = (v B) / L or advective time scale, τC = L / v and with v ~ 100 km/s, L = 107 m, τC = 100 seconds.

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Magnetic Reynolds number (dimensionless) is defined by Rm = τD / τC which is therefore ~ 3.1012 for coronal material.

 

For an RS CVn binary system, τD is approximately 3.1019 s. τC is 104 s, so Rm = 3.1015. [Partly unseen, 3 marks]

 

Question: Give the reason for the magnetic buoyancy of a flux tube which is in thermal balance with its surroundings. Calculate the density difference between a flux tube of 10 Tesla intensity and the surrounding plasma. Both are in pressure equilibrium and located near the bottom of the convection zone. For the bottom of the convection zone use values of pressure p = 7·1012 Pascal, density = 2·102 kg m-3, and temperature T = 2·106 K.

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Answer: Since magnetic pressure adds to the gas pressure pi inside the flux tube, local hydrostatic equilibrium requires:

p = pi + B2/20

If there is thermal equilibrium between the slender flux tube and its surroundings, then using the ideal gas law (p = RT):

i/pi = /p

Then it can be shown that - i = B2/ 20p

Using the given values for B (10 Tesla), p (7·1012 Pascal) and (2·102 kg m-3), and 0 = 4π x 4x10-7 H m-1 (rubric), one gets:

- i = 2·102 x (10 x 10) / [ 2 x (4 x 3.14 x 10-7) x (7x1012) ]

= 2·104 / 1.76·107

= 1.1·10-3 kg m-3.

[Part bookwork, part unseen – 5 marks]

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Question: Give a brief account of the effect of electric currents (non-potentiality) on the shape and connectivity of coronal loops. The Figure below shows two sets of magnetic extrapolations. The thick lines indicate magnetic field lines extrapolated from photospheric magnetic fields connecting positive (full contours, right-hand-side foot points in each figure) and negative (dashed contours, left-hand-side foot points in each figure) magnetic concentrations. Identify which extrapolation is computed with current-free (potential) and which one with current-carrying (non-potential) conditions.

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Answer: Both the shape and connectivity of magnetic field lines are different when they carry currents:

In potential (current-free) magnetic configurations field lines connect opposite polarities at right angles across the magnetic inversion line and have a planar shape.

Current-carrying (non-potential) configurations are characterized by sheared magnetic fields, where field lines assume a non-planar, sigmoidal shape and connect opposite polarities such way that their tops become increasingly parallel with the underlying magnetic inversion line.

In the Figure the magnetic extrapolation on the left shows non-potential current-carrying magnetic field lines, while the one on the right is a potential, current-free extrapolation.

[Part bookwork, part unseen – 5 marks]

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Exam etiquette

Make sure you read the exam paper rubric carefully, and apportion your time to the number of questions. Be guided by the number of marks in square brackets.

Please write clearly!

Put equations on fresh lines and define all the terms.

If you do a calculation, and need to do a “side” calculation, put it clearly in the right-hand margin, then when finished put a single diagonal line through it – you might get marks for being on the right track!

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