Lecture 11

The Sun

University of Naples Federico II, Academic Year 2011-2012Istituzioni di Astrofisica, read by prof. Massimo Capaccioli

EIT - Extreme-ultraviolet Imaging Telescope of SOHO

Learning outcomes

The student will see:

• xxx

Review: static stellar structure equations

2

2 2

4 4

Hydrostatic equilibrium: Equation of state:

Mass conservation: Energy generation:

Radiati o n:

r

H

r r

dP GM kTP

dr r m

dM dLr r

dr dr

ρ ρµ

π ρ π ρε

= − =

= =

2 3 2

3 1

pConvection (

olitr ) :op

e

164

r H r

P

dT L dT m GM

dr r T dr k r

γρκρ µπσ γ

∝

= − = −

The Solar modelBy solving the equations of the previous slide we can build up a model of the interior structure of the Sun.In general the differential equations are solved .

Instead of assuming a

numeric

polyt

ally

rope temperature gradient mode of energy transport

Boundary condition 0

, choose the depending on the

at :

.

......., ,: in the si

.........mplest case

at :

0 0

, , a

.

0 0 nd 0.s

r M L

r R P Tρ= = == = = =

⊙

Convection zones in the Sun

For the Solar model we can plot as a function of radius.

Where radiation most effecti this is , ve form is the o ener

ln ln

2.5 gy trans rf po t.

d P d T

>

The Solar interior

The interior can be divided into three regions:

1. Core: site of nuclear reactions

2. Radiative zone

3. Convective zone

Abundance distribution

32

H depleted in the core

He

He

pp chain

most abu

is ,

where is produced.

is an intermediate species

in the .

It is atndant top

H-burnin

the

of the ,

where

g region

Abundances

is lower.

a

T

⊳

⊳

i

i

i

i

i

⊳

within the ,

since

re homogeneou

the plasma is

s

convective zone

effectively

mi e .

x d

i

i

i

The solar model: evolutionAs the abundances in the corechange, the nuclear reaction rates change accordingly, and the luminosity L, temperature Te , and radiusR of the star are affected.

time

R R

L L⊙

⊙eT

Energy production

nuclear reaction rates higher temperature higher

is not

Although are where the is ,

most of the energy of produced center

amount of mass

the Sun because:

1. the in a

a

shell

radiu

at the

4s t is r dM π= 2 ;

. . there is more mass per unit

volume at

(assuming constant density);

2. the of

( ) at the center has b

large radius

mass fraction Hydrogen

depleteeen

due to

d

fusio , and the rat

X

n

i e

r drρ

2

e equations

depend o Xn .

rate of change of L

Recall: proton-proton chain

1 41 2

7 1 44 1 2

3 4 1 42 2 1 2

The net reactions are:

4 2 2 2

PPI

PPII

PP

2

2

In all cases,

III

neutr are produced ino w i hs h c

e

e

e

H He e

Be H He e

He He H e He

γ

γ

γ

υ

υ

υ

+

+

−

→ + + +

+ → + + +

+ + + → + +

have a .

So they leave the Sun , telling us what the physical conditi

negligig

ons are

le cross section

untouched

where they are produced.

Expected Solar neutrino flux

33 1bolometric Solar fluxHe n

3.864 10 erg s26.7

Given the , ,the energy produced by the production of a , ,and carried by

ucleusaverage energy neut

MeVq 0.6 the , ,

the rino

total Soa

lar nMe

uV

e trin

LQ

ν

−= ×=

=

( )( )33

38

2 1

6

13

102

is:

.

At the distance of the Earth, , the

o flux

flux per

3.864 102 1.851 10 neutrin

cm

osq 1.609 10 26

is:

, apparentl

.7 0.6

1.495 10 cm

6.588 10 neutrin y enorm

s

us!s oo4

L

Q

d

d

νν

ννφ

π

−

−

×Φ = × = = ×− × −

= ×

Φ= = ×

3717One type of neutrino detector on Earth uses an isotope of ( Cl), which

will (rarely) int

chlorine

radioactiveeract with a neutrino to produce a o isotope argon

f :

37 3717 18

This reaction requires the neutrino to have an energy

of or more, and can only detect

neutrinos from the "side-reactions" in the PP chain:

0.814 MeV

PPII

eCl Ar eυ −+ ⇔ +

3 4 7 7 1 82 2 4 4 1 5

7 7 8 84 3 5 4

7 1 4 8 43 1 2 4 2

The Homes

2

take (South

PP II

2

I

e e

He He Be Be H B

Be e Li B Be e

Li H He Be He

γ γυ υ− +

+ → + + → +

+ → + → + +

+ → →

3

2 430

Dakota) detector contains ~ 400,000 cm of perchloroethylene (C Cl ), a cleaning flu

id.

2 10 atoms of Cl isoto pe.

× o ne Arg on ato Det m eve ry ect 2 3 .s days÷

Direct observations of the core: neutrinos

Ray Davis

Direct observations of the core: neutrinos

7131

GALLEX SAGE natural

gallium aqueous gallium chloride solutio

More recently, the (also ) experiments uses of

in a to detect neutrinos via:

3

0

tons

100 to

n n 7132

1 1 21 1 1

2 1 31 1 2

3 3 4 12 2 2 1

This is sensitive to ( ) and can detect

neutrinos fro

lower neutrino energies

main branch PP chain

0.233 MeV

m the of the :

2

e

e

Ga Ge e

H H H e

H H He

He He He H

υ

υγ

−

+

+ ⇔ +

+ → + +

+ → +

+ → +

Both the and experiments detected (by a

factor 2 3) than were expected from the PP-chain reactions.

This problem exist 3

Homestake GALLEX fe

ed for .

The soluti

wer neutr

on was su

0

gg

i

e

nos

ste

y

d

ear

r

s

by

÷∼

esults from the in

Japan.

Results showed that

produced in the

can change into

Super-Kamiokande detector

electron neutrinos

upper atmosphere

t or .

This means neutrinos

au- muo

must h

n-neutrinos

so

ave

me mass

oscillate

and can t

between fl

heref

avo

ore

urs.

The Solar neutrino: problem and solution

The Solar neutrino problem: final solution

The Sudbury Neutrino Observatory (Ontario, Canada) uses heavy water, and was able to directly detect the flux of all types of neutrinos from the Sun.

The results are now completely consistent with the standard Solar model.

The Solar atmosphere

T ~ 106 K

T ~ 25000 K

T ~ 5770 K

The solar atmosphere extends thousands of km above the photosphere(from which the optical radiation is emitted)

It is of much lower density and higher temperature than the photosphere

T ~ 107 KCore

The extended solar spectrum

While the solar radiation is similar to a blackbody prediction at optical wavelengths, there is excess radiation at very short wavelengths. This light is also highly variable.

The radio Sun

Radio waves penetrate through the chromosphere and corona.

The image here shows the "transition region" between the chromosphere and the corona.

The infrared Sun

Infrared images show some features of the Sun's chromosphere, and some features in the corona.

Dark markings are caused by absorption of the infrared light by regions of high density.

The chromosphere

• UV (30.4 nm) images reveal the chromosphere.• Can sometimes see large prominences rising high above the surface of

the Sun. • At the north and south poles of the Sun, less EUV light is emitted -these

regions often end up looking dark in the pictures, giving rise to the term coronal holes.� These are low density regions extending above the surface where

the solar magnetic field opens up.

HeI emission

The X-ray Sun

The X-rays we see all come from the corona.

The corona is a very stormy place, constantly changing and erupting.

Movie from http://www.lmsal.com/SXT/sxt_movie.html

Sunspots

Dark (cool) regions of the photosphere.

Number of spots changes on a 11 year cycle.

Concentrations of magnetic field lines.

The Sun’s magnetic field

By studying sunspots we can learn about the nature of the Sun’s magnetic field.

Switches polarity every 11 years.

Recap: electromagnetic radiation

Region Wavelength [Å] Wavelength [cm] Frequency [Hz] Energy [eV]

Radio > 109 > 10 < 3×109 < 10−5

Microwave 109 ÷ 106 10 ÷ 0.01 3×109 ÷ 3×1012 10−5 ÷ 0.01

Infrared 106 ÷ 7000 0.01 ÷ 7×10−5 3×1012 ÷ 4.3×1014 0.01 ÷ 2

Visible 7000 ÷ 4000 7×10−5 ÷ 4×10−5 4.3×1014 ÷ 7.5×1014 2 ÷ 3

Ultraviolet 4000 ÷ 10 4×10−5 ÷ 10−7 7.5×1014 ÷ 3×1017 3 ÷ 103

X-Ray 10 ÷ 0.1 10−7 ÷ 10−9 3×1017 ÷ 3×1019 103 ÷ 105

Gamma Ray < 0.1 < 10−9 > 3×1019 > 105

h = 6.625×10−27 erg s k = 1.38×10-16 erg K−1

Thermal energy at ambient temperature 0.04 eV

Visible light photons 1.5 ÷ 3.5 eV

Dissociation energy of a molecule of NaCl in ions Na+ and Cl- 4.2 eV

Hydrogen atom dissociation energy 13.6 eV

Energy of an electron shooting the screen of a color TV ∼0.2 keV

X photons for medial diagnostics 0.2 MeV

Nuclear decay energies:

(1) gamma 0 ÷ 3 MeV

(2) beta 0 ÷ 3 MeV

(3) alpha 2 ÷ 10 MeV

Cosmic rays energy 1 MeV ÷ 1000 TeV

Recap: elettronVolt

1 eV/c² = 1.783×10−33 g1 keV/c² = 1.783×10−30 g1 MeV/c² = 1.783×10−27 g = 2×mel

1 GeV/c² = 1.783×10−24 g = mprot

Recap: energy scales