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					                                     Chapter 1
                      Physics of Thermodynamic Equilibrium


1.1 Thermodynamic Equilibrium (T.E.)


                                             Any system, which is left alone for
                                             sufficiently long time, will reach an
                                             equilibrium state known as thermal
                                             equilibrium.           Its   properties          then   are
                                             characterized by a single parameter called
                                             temperature. For example, when a system
                                             of gas atoms and photons is said to be in
                                             T.E., then the particles satisfy the following
                                             properties:




(1) The velocity distribution is Maxwellian velocity distribution


                                                                             4m 3                 mv 2 
                                                           f M (v )                               2kT  ,
                                                                                          v 2 exp      
                                                                          2mkT  2
                                                                                      3
                                                                                                        
                                                               
                                                                  f M v dv  1 , or
                                                           0


                                                                             2                E 
                                                                                          1
                                                           f M (E)                   E 2 exp    ,
                                                                           kT  2            kT 
                                                                                  3



                                                               
                                                                  f M E dE  1
                                                           0




where   v is the speed (magnitude of velocity) of particle, in non-relativistic limit
   mv 2
E      and f (v)dv  f ( E )dE .
    2


                                         1
(2) The distribution of atoms in each energy level obeys the Boltzmann
    law:
                                                         nj       gj E j  Ei 
                                                                exp         
                                                         ni   gi         kT 
                                                         gj     h j   i 
                                                                   
                                                            exp            
                                                         gi         kT 

                                                         Boltzman law
                                                                      E 
                                                         ni  g i exp  i  only valid when
                                                                      kT 
                                                         the temperature is very high




where n j ni  , g j  g i  and h j  i  are the population, statistical weight factor

(internal degree of freedom), and energy of the j (i)th energy level respectively,
                                             13.6eV
e.g. Hydrogen atom g l  2l  1 , h n  
                                               n2
(3) The distribution of ionized and neutral atoms is given by the Saha

equation




                                                      ne n g 0 22me kT  2
                                                                            3
                                                                                    
                                                              
                                                                             exp   I 
       A0             A  e                         n0    g0     h 3
                                                                                   kT 

       n0              n      ne                  where  I is the ionization potential , g 0
                                                                                             



                                                   and g 0 are the statistical weights for the
                                                   ion and the atom in ground states
                                                   respectively.




                                               2
We can prove the Saha equation as follows:
(a) According to the Boltzmann law ,


                                    1         
                                I  me v     
                                           2
            dN 0 (v)   g
                          exp     2                  (note: dN 0 and N 0 are per unit volume)
              N0        g0          kT        
                                              
                                              


where dN 0 (v ) is the differential number of ions in the ground state per unit volume,

with the free electron in velocity regime between v, v  dv , and N 0 is the number of
                                             
atoms in the ground state per unit volume. g 0 is the statistical weight factor of the

ion in its ground state and the differential electron statistical factor is g e .We have


                 
            g  g0 ge


     By definition


                                1
            p e  me v , dV                           (Ne: electron number density)
                                Ne
                                                       We have chosen the volume element dV
                                                       s.t. dVNe =1.
where N e is the electron density and dV is the volume consisting of one electron. We
also have

                                                                     p = mv
                   dVd 3 p   2 4me v 2 dv
                                  3
                                                                   d3p = dp2dp
            ge  2         
                     h3      Ne   h3                                   = 4(mv)2d(mv)


     Hence we have




                                                   3
                                        1                       
                                    I  me v                   
                                               2
            dN 0 (v) 8me g 0
                         3   
                      3       exp     2                       v 2 dv
              N0       h Ne g0          kT                      
                                                                
                                                                

                                                                                                   
             
            N0 Ne   dN  N
                   0 e 
                           8me g 0
                               3 
                                         
                                              
                                                    m v2 
                                    exp  I  exp  e v 2 dv
                                                                                         Not   
                                                                                               
                                                                                                        v is the absolute
                                                    2kT 
             N0       N0    h3 g0       kT  0          
                                                                                                               
                                                                                         magnitude,          0



                                                           3                                               1
                       8m g                               2
                                   2kT 
                              3    
                                                                                            m v2        2
                             exp  I                    e
                                                                  x2
                             e    0
                                          
                                                                         2
                                                                        x dx             x e           
                         3
                        h g0     kT  me                                                 2kT         
                                                            0                                           


                                        3
                         2me kT  2 2 g 0      
                                          
                                         exp  I                           Q.E.D.
                         h        g0          kT 
                             2




        
                          
where  e  x x 2 dx 
             2
                                  has been used. In fact this is derived in the class.
        0
                          4


Example. Some neutral hydrogen is first injected into a container at Ti  0 . If the

container is heated up to T f , what is the ratio of ionized hydrogen to neutral hydrogen

 Np   

N      ? N.B. g p  1 , g H  2 . (In ground state there are two states  and  ).
       
 H    


                  Np
       Let           . Using the Saha equation
                  NH


                                               3
            N p Ne             2me kT  2 2 g p      
                      N e                    exp  I 
                               h        gH          kT 
                                   2
             NH


       Recall



                                                       4
           N H mH  N e me  N p m p  N H  N p mH  mH N p 1   1   mH N e 1   1 


     (N.B. ρ is constant if the size of the container does not change, N e  N p , m p  m H
                                        1
                                    
                         2me kT  2
     , m p  m e )             is called the thermal de Broglie wavelength.
                         h       
                             2




     Thus


                   2                
         N e              3 exp  I 
                  1   mH           kT 


              2   m        
                  H3 exp  I    , T 
             1         kT 


        As T  0 , we can see that   0 , and as T   , we find    too.
        For   1, we need to solve a quadratic equation to obtain the exact result.


(4) The radiation field emitted by a thermal object with temperature T is
   isotropic, homogeneous and stationary.
     The frequency distribution is called a blackbody form (Planck function).


                                             1
                  2h 3      h  
         I  B  2       exp kT   1
                   c               


where I  = specific intensity (power/area/freq/solid angle=erg/sec Hz cm2 ster)

and B is the source function (a specific emission mechanism).




                                                  5
(5) The principle of detailed balancing applies

                                                     N ANB gAgB      E A  E B  E C  
          A B C                                              exp                      
                                                      NC    gC              kT           

(6) Kirchoff’s law – which relates the emission and absorption
     coefficients to the Planck function.


            B


or


                        2 h 3        1
          S        B  2
                          c          h

                                   e   kT
                                            1


where  is the emission coefficient,  is the absorption coefficient

(    cm 1 ) and S is the source function (same as B ).


N.B. In most astronomical situations, the (global) T.E. cannot be reached. Instead,
T.E. could be achieved in a small region. In this case, this small region is said to
have local thermodynamic equilibrium (LTE).


Optical depth      ds


Line of sight ds




                                                 6
1.2 Connection between Maxwellian Distribution and the Fundamental
Distributions

      Spin 1/2 particles (e.g. electrons, protons, neutrons, etc.) satisfy
the Fermi-Dirac distribution



                                                                                                                     1
                                                                                               nj                  
                                                                                                                     / kT
                                                                                                                              1
                                                                                                               j
                                                                                                      e
                                                                         where n j is the occupation number in
                                                                         the energy level with energy  j ,  is the
                                                                         chemical potential and T is the
                                                                         temperature.




     Integer spin particles (e.g.                                                                                             1
photons, helium, etc.) satisfy the                                                           n j   j    / kT
                                                                                                                   1
Bose-Einstein distribution
                                                                                                  e
        j    / kT
If e                      1 (classical limit), both distributions reduce to


                                                  
                                          j   / kT
                             nj  e                         j .



       This is called Maxwell-Boltzman (Maxwellian) distribution. We can
defined the probability distribution for the Maxwellian distribution as

                                                  
                                          j   / kT                    j / kT
                                    e                               e
                             fj                                                      .
                                    e                            e
                                               j   / kT                 j / kT

                                    j                             j



              In the continuous limit,


                                                                                     7
                                                             g E e  E / kT
                            f j  f E                 
                                                        g E e
                                                                          E / kT
                                                                                    dE
                                                        0

where g E  
                            dN
                               is called the density of state. For non-relativistic particle
                            dE
( E  p 2 / 2m ),
                                d 3 xd 3 p V 4p 2 dp V 4 2m  E 1 / 2 dE
                                                                                         3/ 2
N.B.                        dN                      
                                    h3            h3          h3
                              dN 4V 2m  E 1 / 2
                                               3/ 2
                                                   .
                              dE            h3

Therefore, the Maxwellian distribution becomes

                                                    E 1 / 2 e  E / kT
                            f M E            
                                            0
                                                    E 1 / 2 e  E / kT dE
                                                       E 1 / 2 e  E / kT
                                                             
                                            kT 3 / 2 0         x 1 / 2 e  x dx
                                            2E 1 / 2 e  E / kT
                                                                        .
                                                    kT 3 / 2
Classical limit
           j    / kT
Since e                      1 must hold for all  j , it must be true for e  0    / kT  1 too. Let
 0  0 , we obtain e   / kT  1 . From statistical mechanics, we can obtain
                              V
                e  /  
                            N3

                                            is the thermal de-Broglie wavelength. Since mkT 
                             h
where  
                                                                                                1/ 2
                                                                                                       ~ p of
                 2mkT         1/ 2


                                                                                                           V
the particle, so 3 represents a minimum volume occupied by a particle by Q.M. and
                                                                                                           N
represents the average volume for each particle. So, the classical limit will be
satisfied when the gas is extremely dilute. Alternatively, as T   , the
classical limit is obviously satisfied. These two conditions ensure us when the
Maxwellian distribution can be used.




                                                                                     8
1.3 Blackbody Radiation


     A blackbody is an object that absorbs all radiation incident on it. Its radiation
satisfies the Planck function:


                                                 1
                   2h 3       h  
          I  B  2        exp kT   1
                    c                


     The followings are some useful properties of the Planck function.


(A) Stefan-Boltzmann Law


     The intensity of the total radiation emitted by a blackbody is:


                                                            4
                            2h 3            d       2h  kT  x 3 dx 2 4 k 4 4
          I   I d                               2   x                T
                             c2             h      c  h  0 e  1 15h 3 c 2
                        0
                                        exp     1           
                                            kT                 4
                                                                  15




            h
where x       , k  1.38  10 16 erg s K-1, h  6.6  10 27 erg s, c  3 1010 cm s-1 and
            kT

  x 3 dx  4
 e x  1  15
0

                             outgoing energy

     The total flux                                                                    
                                                                                area   A

                                                  2
                                                                                                 
                                                                                                 
                                                                                                Fk
                                                       2 5 k 4 4
                                    1
                                                                                           
          Fn   I cos  I  cosd cos  d   I 
                    
                         d                                     T   BT 4
                     ˆ ˆ
                    k n d  0            0            15h 3 c 2


with  B  5.67  10 5 erg cm 2 s 1 K 4 (the Stefan-Boltzmann constant).




                                                       9
     The total radiation energy density


                  1         4     8 5 k 4  4
           R       Id         15h 3 c 3 T  aT
                               I                    4
                                              
                  c          c               
(R  velocity = energy flux)
           4 B
with a          7.56  10 15 ergcm 3 K 4 .
            c
            Some books also call it Stefan-Boltzmann constant, but B is more
            popular to be called as Stefan-Boltzmann constant.


(B) Limiting cases:


                                    h
     (i) Wien’s distribution (          1 )-high frequency limit (low temperature case)
                                    kT
                        2 h 3  h 
           B ( Wien )  2 exp    
                          c     kT 
                                               h
     (ii) Rayleigh-Jeans distribution (            1 )- low frequency limit ( e x  1  x )
                                               kT


                                     1
                      2h 3  h 
           B ( RJ )  2        
                       c  kT 




(C) Monotonicity with temperature


           B T1   B T2  if T1  T2


    noticing that for a given frequency,


      B
           0 can prove this result.
      T




                                                  10
                                           h                                               B T1 
                                       exp    
         B 2h           2   4
                                           kT 
    N.B.     2 2                  0, 
         T  c kT   h   2
                  exp kT   1                                                         B T2 
                          
(D) Wiens’ law:
                                                                                            T1

             m axT  0.29 cmK
 first come out as an experimental law                                                                       

                               d      c  2hc 2          1                   T2
        Proof. B  B             B 2  5
                               d                    hc              B d  B d
                                                   exp      1
                                                       kT            wavelength     frequency
                                                                        distribution   distribution


                                          B
        The maximum occurs at                 0
                                           


                                         hc 1       hc  
                                                exp       
        B      2         5            kT 2      kT    0 , x  hc
             2hc                                          2 
                   6     hc         5     hc                kT
                     exp
                                  1   exp
                                                       1
                            kT                         
                   
                                               kT   


         5(e x  1)  xe x


                       x
         1  e x 
                       5

iteration method
    x          4.9              4.92        4.94   4.96     4.98
                                                                                                       5
                                                                                                       x
 1  ex    0.99255            0.9927     0.9928   0.993   0.9932   1
                                                                                                  
                                                                                                 1-e-x
   x 5        0.98             0.984       0.988   0.992   0.996

                                                                                                         x
         x m  4.96                                                                     xm 5
                                                                                         




                                                    11
                  hc
      maxT           0.2902cmK
                4.96k
H.W. Calculate  m ax , note that  m ax m ax  c . Don’t worry because B (T )  B (T ) .
                                                   peak position of the two graphs are not at the same point

(E) Mean photon energy and number density


                                            
                                                  x4
                      hB (T )d            exp(x)  1 dx          (5) (5)
            h     0
                                         kT 
                                             0
                                                                kT              3.8kT
                          
                                                  x   3
                                                                      (4) (4)
                           B (T )d
                          0
                                              exp(x)  1 dx
                                             0




                             
                       1      B       8k 3 
           n       d   d  d   3 3 (3) (3)T 3  20.42T 3
                                       h c 
                0
                  h     c    0
                                h            


       is the energy density per unit frequency and  (n) is called the Riemann zeta
function. (c.f. Mathematical Results in appendix)


1.4 Applications to Stellar Spectra


      The continuous spectrum of a star is conventionally compared with a blackbody of
intensity B (T ) in three different ways.


      (a) Brightness temperature:
       The brightness temperature Tb of an object at a given wavelength (frequency)

      is the temperature of a blackbody having the same brightness (specific
      intensity) at that wavelength (frequency):


           B Tb   I (the specific intensity observed at a specific frequency  )




                                                   12
                               h
In radio wave band, usually        1 , then
                               kT


                2 2 k              I c 2
    B (T )           T    Tb 
                 c2                 2k 2


   If the deduced brightness temperature is unreasonably high, then Tb will imply
that some non-thermal emission processes are taking place in the source, e.g.
Tb  10 30 K for pulsar in radio wave.

(b) Effective temperature ( Teff ):



L   L d  4R 2 BTeff
                        4




There are two main differences between Tb and Teff .


i) Tb depends on  but Teff does not.


ii) In order to determine Teff , the size of the source must be known

(If the spectrum also has a blackbody shape, Teff = Tbb  R)


(c) Color temperature ( Tc )
(usually, only a finite range of spectrum can be obtained)


   The observed spectrum of a source may often have a shape which is similar to
that of a blackbody, but not necessarily with the same absolute value, then the color
temperature is defined as




                                             13
                                                    hc 
                                                exp       1
          f i  Bi , Tc    j     
                                            5        kT 
                                                j c
          f  j  B  j , Tc   i
                                 
                                        
                                                   hc 
                                                  kT   1
                                                exp      
          observed
          spectra                                   i c


     Note that while both Tb and Teff depend on the magnitude of the source intensity,

Tc depends only on the shape of the observed spectrum. In general Tc varies from one

spectral regime to another.


1.5 The Magnitude System


     The magnitude system has its origin in the classification of stars
according to their apparent visual brightness used by the Greek astronomers.
In the 19th century, this classification was quantified as a logarithmic law .

Nowadays, magnitudes are relative measures of monochromatic flux of a

source.
     If f ( ) is the monochromatic flux due to a source and f 0 is a reference
monochromatic flux, then the magnitude is


                         f ( ) 
          m  2.5 log           2.5 log f ( )  q0
                         f0 


with q0  2.5 log f 0 . q 0 defines the zero point of the magnitude scale. Sometimes,

another magnitude system is used, it is the Bolometric Magnitude which gives

the flux integrated over all wavelengths:




                                                      14
                                   
                          f ( )d 
          mb  2.5 log  0          
                         Fb         
                                    
                        
                                    
                                     


and Fb  2.52 10 8 Wm 2  2.52 10 5 erg sec1 cm 2 .


      Since the Sun is the most visible stellar object in the sky, the sun’s luminosity is
often used as a reference


          mb  0.23  5 log D  2.5 log( L / L )


where D is the distance in parsec from the source to the Earth, L is the luminosity of the
source and L   3.8  10 33 ergs 1 is the solar luminosity. Remark : for a very distance
object the correction due to interstellar absorption is necessary to include in the above
expression.
      So far, we have just calculated the apparent magnitude that is a measure of the
flux received on Earth. To compare the intrinsic luminosity, we introduce the
absolute magnitude, which is the apparent magnitude the source would
have at a distance of 10 pc (N.B. 1 parsec  3  10 18 cm ), corrected for any effects
of interstellar absorption A :


          M  m  5  5 log D  A


          N.B.: A is a complicated correction function which depends on interstellar
medium (ISM)


      Our sun has an absolute magnitude 4.6 and visual or apparent magnitude –26.73.
(A for Sun is very small because there is very little ISM.)




                                               15
                Example Estimate the apparent magnitude of the moon. The moon light comes from
                                                                L
                reflecting the sunlight.                             power received at the surface of the moon
                                                               4D 2
                                                                L
                                   the Moon like a flat mirror        area of the moonthe power received by the whole moon
                         RMoon
                             2
                                                               4D 2
                 LMoon          L   1.2  10 23 ergs 1            Moon
                          4D  2

Total surface area at                                                d
distance D from the                                                             Earth                     Sun
Sun
                                   d 
                                   1 pc   2.5 log( LMoon / L )
                mb  0.23  5 log      
                                                                                       D
                         13.5


                where D  1AU  1.5  1013 cm , d  3.8  1010 cm and RMoon  1.7  10 8 cm have been
                used.
                         The real observed value is –12.7. The discrepancy is due to the fact that the Moon
                absorbs some sunlight and only reflects 60% of the incident light. The amount of sunlight
                being reflected depends on the surface properties of the planet.                                     reflected
                                                                                                                LR   luminosity
                                                                     percentage of sunlight reflected: Albedo 
                                                                                                                LI
                                                                                                                     incident
                                                                                                                     luminosity




                                                                16
Appendix: Mathematical Results


          A useful result for evaluating the frequency moment of the Planck function is


              
                               2k   4
               B( , T )d  c 2 h 3   4   4T
                                                            4

              0




where (n  1)  n! is the gamma function and


                              
                        1      u n1        
                                      du   m n
                       (n)  e u  1
               ( n) 
                            0              m 1




is the Riemann zeta function.
N.B.
(I)


                                                                 
             u n1        e u u n1                                                    

           eu  1 du          u
                                     du   u n1e mu du  m n  x n1e  x dx  (n) m n
                        0 1 e           m 1 0             m 1                         m 1
          0
                                                                    
                                                                   0
                                                                         
                                                                          n 




where the change of variable mu  x has been used.


(II)


                                                        4                           4
                     2h 3    d 2h  kT                     x 3       2h  kT 
 B( , T )d    h  c 2  c 2  h 
      
                                                              e x  1 dx  c 2  h           (  4)   4
                                                                               
0                0
                   exp     1                              0

                       kT 
                                         h
                                    x                                                         zero moment    =0
                                         kT
                                                                                                   l moment    = l



                              Some useful numerical values of zeta function


                                                    17
n                 (n)
2                2 /6
3                 1.2
4                4 / 90
5                1.04
6                1.02


     (n)  1 for n>4




           18