CURICULUM VITAE
A. DATA DIRI
01. N a m a
02. Tempat/Tanggal Lahir
03. Jenis Kelamin
04. Fakultas/Jurusan
05. Pangkat/Golongan/NIP
06. Bidang Keahlian
07. Alamat Rumah
08. Alamat Kantor
09. e-mail
10. Riwayat Pendidikan Tinggi
Jenis Pendidikan
Sarjana (S1)
Pra Magister (Pra S2)
Magister (S2)
Doktor (S3)
Tempat
IKIP Ujung Pandang
ITB Bandung
ITB Bandung
Université de la Méditerranée
Marseille, Prancis
: Dr. H. Muris, M.Si
: Tinggas, 1965
: Laki-laki
: FMIPA/Fisika
: Lektor Kepala/IV/a/131925820
: Fisika Material
: BTN Minasa Upa G20/14 Makassar.
90224.
Telp. (0411) 886307
HP. 081342403676
: Jurusan Fisika FMIPA UNM
Kampus Parangtambung Makassar
Tlp/Fax. (0411)840622, HP. 081342403676
: murisfmipaunm@yahoo.com
:
Tahun lulus
1989
1992
1994
2001
Spesialisasi
Pendidikan Fisika
Fisika
Fisika Material
Fisika Material
B. Riwayat Pekerjaan
1.Dosen Tetap Jurusan Fisika FMIPA Universitas Negeri Makassar, 1990 - sekarang.
2.Ketua Program Studi Fisika FMIPA Universitas Negeri Makassar, 2003 - 2004.
3.Pembantu Dekan Bidang Akademik FMIPA Universitas Negeri Makassar, 2004 - sekarang.
4.Dosen Program Pascasarjana UNM Makassar, 2006 - sekarang

Rujukan Utama :
Introdution to Statistical Physics for Students by
Pointon
Longman, England
Rujukan Tambahan :
Buku Buku Fisika Zat Padat, Fisika Kuantum dan Fisika
Modern yang relevan

1.
Pengantar
2.
Statistik Maxwell Boltzmann
3.
Aplikasi Statistik Maxwell Boltzmann
4.
Statistik Bose Einstein
5.
Statistik Fermi Dirac
6.
Temperatur dan Entropy
7.
Aplikasi Statistik Termodinamika
8.
Ensemble Kanonik
9.
Grand Ensemble Kanonik

1.
Pengantar
2.
Statistik Maxwell Boltzmann
3.
Aplikasi Statistik Maxwell Boltzmann
4.
Statistik Bose Einstein
5.
Statistik Fermi Dirac
6.
Temperatur dan Entropy
7.
Aplikasi Statistik Termodinamika
8.
Ensemble Kanonik
9.
Grand Ensemble Kanonik

Sistim Termodinamika, Parameter Makroskopik
Sistim terbuka dimana dimungkinkan terjadi pertukanan energi dan materi dengan lingkungan.
Sistim tertutup terjadi pertukaran energi maupun materi dengan lingkungannya
Isolated systems tidak memungkinkan terjadinya pertukaran energi maupu materi dengan lingkungannya
Paramater internal dan external : temperatur, volume, tekanan, energi, medan magnet, dll. (nilai rata-rata, fluktuasi diabaikan).

Pengertian Dasar Statistik
Mean : Rata-rata
Mode : yang paling mungkin
Median : Titik tengah
Varians : Ragam, Lebar Distribusi

Pengertian Dasar Statistik
Misalkan suatu variabel yang diselidiki : 3,4,4,3,5,3,4
X

3

4

4

3

6

3

5

7
28

4
7
X
 x
1
 x
2
 x
3
 x
4
 x
5
 x
6
 x
7
7
X

 i x i
N

Pengertian Dasar Statistik
Rata-rata dengan fungsi probabilitas
4
5 x i
3
3
1
7 f
3 f(x i
)
3/7
3/7
1/7
1 x i f(x i
)
9/7
12/7
5/7
28/7 = 4
Ternyata diperoleh hasil rata-rata yang sama yakni 4

Pengertian Dasar Statistik
Hasil ini diperoleh dari pengembangan bentuk
X

 i
 i f .( x i
).
x i f ( x i
)
  i f ( x i
).
x i
 f ( x i
)

1
Jika fungsinya kontinyu maka : X



  x .
f ( x ) dx
Bagaimana anda mengartikan parameter statistik berikut ?
kontinyu diskrit

Pengertian Dasar Statistik

Fungsi Gaussian
Fungsi seperti akan banyak dijumpai dalam pembahasan statistik partikel

x
Ruang Euclid dan Ruang Fase
Ruang Euclid dV
 dxdydz dV z y dz dy dx

Ruang Euclid dan Ruang Fase
  p x
2  p y
2  p z
2
2 m
 
x
y
z
Ruang fase Ruang momentum d

6 N
 dx
1 dy
1 dz
1 dp x 1 dp y 1 dp z 1
........
dx i dy i dz i dp xi dp yi dp zi
........
dx
N dy
N dz
N dp xN dp yN dp zn
 i
N 

1 dx i dy i dz i dp xi dp yi dp zi
 i
N 

1
  i

Rata Rata Sifat Assembly
Misalkan dalam assembly terdapat sejumlah N molekul dengan energi total E dan berada dalam volume V.
p(N) menyatakan koordinat momentum x(N) menyatakan koordinat posisi p(N) x(N)

Rata Rata Sifat Assembly
Jika X adalah perilaku yang ingin dicari rata-ratanya dalam ruang fase tersebut
X


6

N
X
 x ( N ), p ( N )
  x ( N ), p ( N )
 d

6 N
Normalisasi terhadap ruang
X
 
6

N
X
 x ( N ), p ( N )
  x ( N ),

6

N
P
 x ( N ), p ( N )
 d

6 N p ( N )
 d

6 N

Rata Rata Sifat Assembly
Jika X merupakan fungsi yang diskrit, maka perata-rataan fungsi X dapat dinyatakan dengan :
X

 i
 i p i p i
X i
Normalisasi probabilitas menghasilkan
X
 i p i

1
  i p i
X i

a.
Klasik
- Terbedakan antara satu dengan lainnya (distinguishable)
- Energi kontinu
- Tak memenuhi prinsip larangan Pauli b.
Kuantum : Terdapat dua tipe
Tipe I (fermion) :
- Tak terbedakan antara satu dengan lainnya (indistinguishable)
- Energi disktrit
- Memenuhi prinsip larangan Pauli
Misalnya : elektron dalam zat padat

Assembli Klasik dan Kuantum b.
Kuantum : Terdapat dua tipe
Tipe II (boson) :
- Tak terbedakan antara satu dengan lainnya (indistinguishable)
- Energi disktrit
- Tidak memenuhi prinsip larangan Pauli
Misalnya : foton atau partikel alpha

Statistik Maxwell Boltzmann
Distribusi Energi
Misalkan dalam sistim yang ditinjau terdapat N sistim :
Sistem 1 dengan energi ε
1
Sistem 2 dengan energi ε
2
…………………….
Sistem i dengan energi ε i
…………………….
Sistem N dengan energi ε
N

Statistik Maxwell Boltzmann
Distribusi Energi
Misalkan dalam sistim yang ditinjau terdapat N sistim :
Sistem 1 dengan energi ε
1
Sistem 2 dengan energi ε
2
…………………….
Sistem i dengan energi ε i
…………………….
Sistem N dengan energi ε
N

Statistik Maxwell Boltzmann
Prinsip Kekekalan

Statistik Maxwell Boltzmann
Jumlah pilihan jika memilih sejumlah N
1 di antara N partikel
Jika g
1 menyatakan bobot, maka jumlah pilihan yang ada adalah :

Statistik Maxwell Boltzmann
Perluas lagi dengan mengambil sejumlah N
2 dari N-N
1
Perluas lagi dengan mengambil sampai n kali

Statistik Maxwell Boltzmann
Secara umum dapat ditulis :

Contoh Pemakaian
Empat partikel dengan notasi a,b,c dan d didistribusi pada dua pita energi 2 pada pita 1 dan 2 pada sistim 2. Bobot masing-masing adalah 3 dan 4.
Jadi : N
1
= N
2
= 2 g
1
= 3 , g
2
= 4
W

N !
N
1
!.
N
2
!
 g
1
2
.
g
2
2
W

4 !
2 !.2!

3 2 .
4 2 
864

a c b d
Contoh Pemakaian a c d b d c,a b
Ini hanyalah 3 contoh gambar dari 864 kemungkinan yang ada.
Sekarang adalah giliran anda untuk melengkapinya.

Statistik Maxwell Boltzmann
Peluang terbesar diperoleh dengan mengambil dw/dn = 0
Rumus Stirling

Distribusi Maxwell Boltzmann n (

) d
 
2

N

 kT

3 / 2 e
 
/ kT

1 / 2 d

0 exp



 k
B
T




g(

) g
 

C

0

=
P(

)
0


2D
3D k y

L x k z k y k
Aplikasi Statistik Maxwell Boltzmann k x
Untuk partikel kuantum dalam kotak 2D (e.g., electron pd FET): k x

 n x
L x k y

 n y
L y k
 k x
2  k y
2
N

1
4



L x k
2

L y
 k
2
 area

4

G
 k
2
4

G

1
4

2 m

 2
# states within ¼ of a circle of radius k g
2 D
   2 s
2



1

2 m
- Tak bergantung pd

N

1
8 k x

 

L x
3



L y k
3

L z
 k
3
 volume

6

2
G
 
 k
3
6

2
g(

)
G g
3 D
   2 s
4


2
1

2 m

3 / 2

1 / 2
 2

1
6

2
2 m

 2
3 / 2
3D
2D
1D

Thus, for 3D electrons
(2 s +1=2): g
3 D 
1
2

2
2 m

3 / 2

1 / 2
 2

f
Distribusi Kecepatan Maxwell fv



2
 m k
B
T


3 / 2 exp


 mv
2
2 k
B
T

 4
 v
2 dv v y
Nampak bahwa persamaan ini merupakan perkalian antara faktor Boltzmann dengan sebuah tetapan .
Tetapan tersebut dapat diperoleh dari normalisasi

0
 f
  dv

1 C



2
 m k
B
T


3 / 2 v z
P(v) dN
 

NP
  d
 
N exp



 k
B
T

 d

Distribusi energi, N
– the total # of particles dN
 

NP
  dv

N


2
 m k
B
T


3 / 2
4
 v
2 exp


 mv
2
2 k
B
T

 dv speed distribution (distribusi kecepatan) dN
  x

NP
  x dv

N


2
 m k
B
T


1 / 2 exp


 mv
2
2 k
B
T

 dv
Distrbusi kecepatan dalam arah x, v x
P(v x
)
" volume"
 v
 v
 dv


4
 v
2 dv v v x v v x

Karakteristik Nilai Kecepatan
P
 



2
 m k
B
T


3 / 2
4
 v
2 exp


 mv
2
2 k
B
T


Lihat bahwa distribusi ini tidak simetrik, sehingga perlu dicari perata-rataan sebagai berikut
P(v)
The root-mean-square speed is proportional to the square root of the average energy:
E

1
2 m
  rms
2 v rms

2 E m

3 k
B
T m v max v v rms v
Harga kec.maksimum :
Kelajuan rata-rata : v



0 v

P

 dP dv

 v
 v max

0
 v max

2 k
B
T m dv



2
 m k
B
T




0
4
 v 3 exp


 mv
2
2 k
B
T

 dv

8 k
B
T
 m v max
 v
 v rms

2

8 /
 
3

1

1 .
13

1 .
22

Soal (Maxwell distr.)
Consider a mixture of Hydrogen and Helium at T=300 K. Find the speed at which the Maxwell distributions for these gases have the same value.
P
 v , T , m




2
 m k
B
T


3 / 2
4
 v
2 exp


 mv
2
2 k
B
T




2
 m
1 k
B
T


3 / 2
4
 v
2 exp


 m
1 v
2
2 k
B
T





2
 m
2 k
B
T


3 / 2
4
 v
2 exp


 m
2 v
2 k
B
T
2


3
2 ln m m
2
1 v
2

2 k
B
T
 m
1
 m
2
 m
1 v
2
3
2 ln m
1

2 k
B
T

3
2 ln m
2
 m
2 v
2 k
B
T
2 v

3 k

B
T m
1
 m
1 ln m
2 m
2


3

1 .
38

10

23 
300
 ln
2

1 .
7

10

27
2

1 .
6 km/s

Soal (Maxwell distr.)
Find the temperature at which the number of molecules in an ideal Boltzmann gas with the values of speed within the range v - v+dv is a maximum.
P
 v , T , m




2
 m k
B
T


3 / 2
4
 v
2 exp


 mv
2
2 k
B
T

 maximum:

P

T

0
3
2


2
 m k
B
T


1 / 2 


2
 m k
B
T
2

 exp


 mv
2
2 k
B
T





2
 m k
B
T


3 / 2 exp


 mv
2
2 k
B
T



 mv
2 k
B
T
2
2



0
3
2


 mv
2
2 k
B
T



0 T
 mv
2
3 k
B
At home:
Find the temperature T at which the rms speed of Hydrogen molecules exceeds their most probable speed by 400 m/s.
Answer: 380K

 o
Pelebaran Garis Spektrum Doppler
Doppler.
Misalkan molekul gas melakukan radiasi dengan panjang gelombang dalam arah x dengan kecepatan v x menuju kepada seorang pengamat. Pengamat akan menerima radiasi dengan panjang gelombang.

 o
Pelebaran Garis Spektrum Doppler
Karena efek Doppler, maka panjang gelombang yang diamati pengamat adalah :
   o

 v x c

 v
 c
   o
 
 o dv x
  c
 o d


 o
Pelebaran Garis Spektrum Doppler
Dari distribusi Maxwell Boltzamann dN
 

Nf
  dv

N


2
 m k
B
T


3 / 2
4
 v
2 exp


 mv
2
2 k
B
T

 dv
Ubah sebagai fungsi panjang gelombang f
  d
 


2
 m k
B
T


3 / 2 exp


 mc
2
2 k
B
T
  
 o
2
 o

2

 c
 o d


 o
Pelebaran Garis Spektrum Doppler
Intensitas radiasi :
I
  d
 
Cf (

) d
 
I
  o exp

 mc
2
2 k
B
T
  
 o
2
 o

2
 d

I (
 o
)
I (

)
Dengan mengukur intensitas radiasi maka dapat ditentukan temperatur gas emisi

 o

 o
  
 p x
2
/ 2 me e / KT d


 e
 e / kT dT
Prinsip Ekipartisi Energi
Jika energi sistem dinyatakan dalam bentuk kuadrat posisi dan momentum maka tiap bentuk kuadrat tersebut akan memberikan energi rata-rata ½ kT
Contoh molekul gas dengan massa m, energinya dapat dinyatakan dengan
 x
 p 2 x
2 m
Maka energi rata-ratanya adalah :
  
 p x
2
/ 2 m e e / KT d


 e
 e / kT dT

Prinsip Ekipartisi Energi
2 p x
2
 x

 exp


(


 exp


(
 p
2 x
2 m
) / kT
 dxdydzdp y dp z
 p
2 x
2 m
)

/ kT
 dxdydzdp y dp x x x

2 p
2 x m
 



 exp( exp(
 p x
2

/ p x
2
/ 2 mkT ) dp x
2 mkT ) dp x p
2
Misalkan = u 2 maka
2 mkT
 x
 kT


  e



 e
 u
 u
2
2 u du
2 du

Prinsip Ekipartisi Energi
Hasilnya memberikan :


  e
 u
2 u
2 du

1
2


  e
 u
2 du
Maka :  x

1
2 kT p
2 x
2 mkT
 u
Karena ada satu bentuk kuadrat maka memberikan energi rata-rata ½ kT
Contoh 2 : Osilator harmonik dengan dua jenis energi
 x
 p
2 x
2 m

1
2
 x
2

Prinsip Ekipartisi Energi
Maka :
 x


 p
2
/ x
2 m


1
2
 x
2
 e
 e / kT d

 e
 e / kT d

 x
  

 



 p x
2 
1
2
 x
2
 

 

  exp




 exp 

 x  dx



 p x
2
2 m
 p x
2
2 m
1
2

 x
2
1
2

 kT
 x
2

 / kT 

 dxdp x


 dxdp x
Ubah ke koordinat polar : p x
2
2 m
 r
2 sin
2

,
1
2
 x
2  r
2 cos
2
 dp x dp y

2 ( m /

)
1
2 rdrd


Prinsip Ekipartisi Energi
Maka :
 
2 x
 

0
2
0 x
 
0

0 e r
2 e
 r
2
/ kT
/ kT r
3 dr rdr
 kT
Karena terdiri dari dua bentuk kuadrat maka energinya adalah
2 x ½ kT = kT
Untuk osilator harmonik 3D maka :
 

 p
2 x
2 m
 p
2 y
2 m
 p
2 x
2 m



1
2 kT

1
2 kT

1
2 kT

3
2 kT

3
2 kT

Prinsip Ekipartisi Energi
Energi rata-rata untuk osilator harmonik 3 D.
  p
2 x
2 m

1
2

1 x
2  p 2 y
2 m

1
2

2 y
2  p
2 x
2 m

1
2

3 z
2

6 .
1
2 kT

3 kT
Jadi dalam hal ini ada 6 derajat kebebasan ( f = 6) dimana tiap derajat kebebasan memberikan kontribusi energi sebesar ½ kT

Prinsip Ekipartisi Energi
Jika terdapat N
A
(bil. Avogadro) molekul gas dan berlaku sebagai osilator harmonik 3D, maka, terdapat 6 derajat kebebasan,maka :
E

6 N
A
1
2 kT

3 RT
Panas jenis per gram atom zat padat :

E

T v

3 R

5 , 94 kal/ o
K/gr.atom

Panas jenis gas
Jika terdapat N
A
(bil. Avogadro) molekul gas dan berlaku sebagai osilator harmonik 3D, maka, terdapat 6 derajat kebebasan,maka :
E

6 N
A
1
2 kT

3 RT
Panas jenis per gram atom zat padat :

E

T v

3 R

5 , 94 kal/ o
K/gr.atom

  s
 
 g g s s


1

1


!
n n s
!
s

!
STATISTIK BOSE-EINSTEIN g s
  g s

1

 n s

!
w s

 g s


 g g s s


1
1

!

 n s
!
n s
!

 
 g g s s


1
1



!
n n s s
!
!
 w
  s w s

STATISTIK BOSE-EINSTEIN
  s
  g
 g s s


1
1



!
n n s s
!
!
 w
  s w s

STATISTIK BOSE-EINSTEIN
 s


 log
 n s w
 x
  s

 dn s

0
 log
 n s w
 x
  s

0 log
  s w

  g s
 s
 log
1
 w s n s g s

1
 n s
  g s

1
  g s

1

 n s log n s


STATISTIK BOSE-EINSTEIN
 log
 n s w
 log
 g s

1
 n s

 log n s
 log
 n s w
 log

 g s n
 s n s

 log

 g s n
 s n s


 x
  s

0 g s n s
 e

 x
  s
 
1

STATISTIK BOSE-EINSTEIN n s
 e

 x
 g
 s s
 
1 n s
 g s
1
A e
 s
/ kT

1 w s
 n s
!
 g g s s
!
 n s

!

STATISTIK BOSE-EINSTEIN n s
 e

 x
 g
 s s
 
1 n s
 g s
1
A e
 s
/ kT

1 w s
 n s
!
 g g s s
!
 n s

!

STATISTIK BOSE-EINSTEIN

STATISTIK FERMI-DIRAC
W
  s w s w s
 n s g g !
!
 
!
s s
 n s
Jumlah untuk semua kemungkinan susunan yang berbeda untuk satu tingkatan energi
W
  s n s
!
 g g s s
!
 n s

!
Jumlah untuk semua kemungkinan susunan yang berbeda

STATISTIK FERMI-DIRAC log W

  s
 s log
 g s n s log
!
 g g s g s
 s
!
 n s n s

!
log n s

 g s
 n s
  g s
 n s
 
 s



 log W
 n s
   
 s


 dn s

0 Gunakan rumus Stirling
 log

s
   
 s

0

STATISTIK FERMI-DIRAC
 log W
 n s
 log g s
 n s n s log g s
 n s n s
   
 s

0 g s n s
 e


  


1

T=0
STATISTIK FERMI-DIRAC
~ k
B
T f
  e
1

  
F
 kT 
1 n
  d
 f
    d


F
 
, f
 
 e
 
1

1

1

F
 
, f
 
 e

1

1

0
( with respect to

)

=


STATISTIK FERMI-DIRAC n s
  e

   g
 s s
 
1
Distribusi jumlah partikel partikel f
  e
1

  
F
 kT 
1
Melalui normalisasi g s
= 1 diperoleh fungsi distribusi. Maka f(e) merupakan probabilitas sebagai fungsi energi
Sebagai fungsi probabilitas maka harga fungsi ini maksimum 1 dan minimum 0

Radiasi Benda Hitam
Two types of bosons:
(a) Composite particles which contain an even number of fermions. These number of these particles is conserved if the energy does not exceed the dissociation energy (~ MeV in the case of the nucleus).
(b) particles associated with a field, of which the most important example is the photon. These particles are not conserved: if the total energy of the field changes, particles appear and disappear.
We’ll see that the chemical potential of such particles is zero in equilibrium, regardless of density.

Radiation in Equilibrium with Matter
Typically, radiation emitted by a hot body, or from a laser is not in equilibrium: energy is flowing outwards and must be replenished from some source. The first step towards understanding of radiation being in equilibrium with matter was made by Kirchhoff, who considered a cavity filled with radiation , the walls can be regarded as a heat bath for radiation.
The walls emit and absorb e.-m. waves. In equilibrium, the walls and radiation must have the same temperature T . The energy of radiation is spread over a range of frequencies, and we define u
S
(

,T) d
 as the energy density (per unit volume) of the radiation with frequencies between
 and

+d

.
u
S
(

,T) is the spectral energy density.
The internal energy of the photon gas: u
 

  u
  
S

, T d
0
In equilibrium, u
S
(

,T) is the same everywhere in the cavity, and is a function of frequency and temperature only. If the cavity volume increases at T =const, the internal energy U = u (T) V also increases. The essential difference between the photon gas and the ideal gas of molecules: for an ideal gas, an isothermal expansion would conserve the gas energy, whereas for the photon gas, it is the energy density which is unchanged, the number of photons is not conserved, but proportional to volume in an isothermal change.
A real surface absorbs only a fraction of the radiation falling on it. The absorptivity
 is a function of
 and T ; a surface for which

(

) =1 for all frequencies is called a black body.

Photons Apa Itu ?
The electromagnetic field has an infinite number of modes (standing waves) in the cavity.
Any radiation field is a superposition of plane waves of different frequencies. The characteristic feature of the radiation is that a mode may be excited only in units of the quantum of energy hf (similar to a harmonic oscillators) :
 i

 n i

1 / 2
 h

T
This fact leads to the concept of photons as quanta of the electromagnetic field . The state of the el.-mag. field is specified by the number n for each of the modes, or, in other words, by enumerating the number of photons with each frequency.
According to the quantum theory of radiation, photons are massless
bosons of spin 1 (in units ħ
). They move with the speed of light :
The linearity of Maxwell equations implies that the photons do not interact with each other . (Non-linear optical phenomena are observed when a large-intensity radiation interacts with matter).
E ph
 h

E ph
 cp ph p ph

E ph c
 h
 c
Presence of a small amount of matter is essential for establishing equilibrium in the photon gas.
We’ll treat a system of photons as an ideal photon gas , and, in particular, we’ll apply the BE statistics to this system.
The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter.

Potensial Kimia Foton = 0
The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter. The textbook suggests that N can be found from the equilibrium condition:



F

N


T , V

0
On the other hand,




F
N


T , V
  ph
Thus, in equilibrium, the chemical potential for a photon gas is zero:
 ph

0
However, we cannot use the usual expression for the chemical potential, because one cannot increase N (i.e., add photons to the system) at constant volume and at the same time keep the temperature constant: 



F
N


T , V
- does not exist for the photon gas
Instead, we can use G

N

G

F

PV P
 

F

V

T
 
F

T , V

V
- by increasing the volume at T =const, we proportionally scale F
Thus, G

F

F
V
V

0
- the Gibbs free energy of an equilibrium photon gas is 0 !
 ph

G
N

0
For

= 0 , the BE distribution reduces to the
Planck’s distribution
: n ph
 f ph

1 exp


 k
B
T



1

1 exp

 h
 k
B
T



1
Planck’s distribution provides the average number of photons in a single mode of frequency

=

/h .

The average energy in the mode:
In the classical (high temperature) limit:
  k
B
T
  n h
  h
 exp

 h
 k
B
T



1
In order to calculate the average number of photons per small energy interval d

, the average energy of photons per small energy interval d

, etc., as well as the total average number of photons in a photon gas and its total energy, we need to know the density of states for photons as a function of photon energy.
k y k z extra factor of 2: two polarizations k x g
Rapat Keadaan Foton
N

1
8
L

 
 x
3



L k y
3

L z
 k
3
 volume

6

2
G
 dG
  d

  cp
 c  k G

6

2

3
 
3 g
3 D ph
 k
3
6

2

2

2

2
 
3
3 D g ph
 
 g
3 D ph
  d
 d

 h

 
2 h
 c 
2
 
3

8
 c
3
2 g
3 D ph
 

8

2 c
3

Spektrum Radiasi Benda Hitam
Rata-rata jumlah foton per satuan volume denga frekwensi
 dan

+d

: g d
  u
S
  d
 u s

T
 h
 g

8
 h c
3 exp
 

3 h
 

1
Rapat Spektrum (hukum Radiasi Planck) u adalahfungsi energi: u
  d
  u
  d
 u
    d
 d

 u
 h

, T

 h
Radiasi spektrum benda hitam u

T

8

 
3


3 exp

 k
B
T



1 u(

,T) - the energy density per unit photon energy for a photon gas in equilibrium with a blackbody at temperature T .

Pendekatan Klasik (f kecil ,
 besar), Hkm Rayleigh-Jeans
Pd frekwensi rendah dan temp. tinggi
 h
 
1 exp
  h
 

1
  h
 u s
 
8
 h c
3 exp
 

3 h
 

1

8
 
2 c
3 k
B
T
Hukum Rayleigh-Jeans
- purely classical result (no h ), can be obtained directly from equipartition
This equation predicts the so-called ultraviolet catastrophe
– an infinite amount of energy being radiated at high frequencies wavelengths.
or short

Hukum Rayleigh-Jeans u sebagai fungsi dari panjang gelombang u
 
, T
 d
   u

T d


 d
 d

  hc

2

 u
 
, T


8

 
3 h c
 exp

 hc
 k
B
T
3



1 hc

2

8


5 hc 1 exp

 hc
 k
B
T



1
In the limit of large

: u
 
, T
 large


8
 k

4
B
T
1

4

 frekwensi tinggi , Hukum Pergeseran Wien’s
At high frequencies:
 h
 
1 exp
  h
 

1
 exp
  h
  u s
,

8
 c
3 h

3 exp

  h
 
- Ditemukan secara eksperimen oleh Wien
Wien
Nobel 1911
Maksimum u(

) berfeser ke frekwensi tinggi ketika temperatur naik.
 max

2 .
8 k
B
T h
 du d

 const
 d d

 h
 k
B
T









 exp

 h
 k
B
T k h
B

T


3



1







3
 x
 e x 
3

 const




3 e x x

2
1
 x

2 .
8 x x
3 e
 e x 
1
2





0 h k
 max
B
T

2 .
8
Hukum
Pergeseran
Wien
- the
“most likely” frequency of a photon in a blackbody radiation with temperature T
Numerous applications
(e.g., non-contact radiation thermometry)

u
  u

,
 max
  max h
 max

2 .
8 k
B
T k
B
T hc
 max
- does this mean that

2 .
8
? Wrong!
 max
 max u
 
, T
 d
 du df
  u

T d

 const
 d dx

 d
 d


 x
5
 exp

1
 hc

2

 x

1


 u
 
, T


8

 
3
 const
 


 h c
 exp

 hc
 k
B
T
3 x
6
 exp
5 x



1

1

 hc

2

8


5 hc
 x
5
 x

 exp
2
 exp x
 

1 x

2



1 exp

 hc
 k
B
T

0



1
5 x
 exp 1 /

1

 exp
    max
 hc
5 k
B
T
T = 300 K

 max

10
 m
“night vision” devices

Radiasi Sinar Matahari
Temperatur permukaan- 5800K
 max
 hc
5 k
B
T

0 .
5
 m
As a function of energy , the spectrum of sunlight peaks at a photon energy of u max
 h
 max

2 .
8 k
B
T

1 .
4 eV


(u max
)

0.88
 m, near infrared
- close to the energy gap in Si, 1.2 eV, which has been so far the best material for photovoltaic devices (solar cells)
Spectral sensitivity of the eye:

Hukum Radiasi Stefan-Boltzmann
Jumlah total foton persatuan volume : n

N
V
 
0
 n d
 
8
 c
3

0
 
2 exp

 h
 k
B
T



1 d
 
8
 c
3 k
B
T h

3

0
 e x
2 dx x 
1

8
 k
B hc

3
T
3 
2 .
4
- increases as T 3
Energi total foton per satuan volume : (apat energi gas foton)
 
2

5 k
B
4
15 h
3 c
2 u
Tetapan Stefan-Boltzmann u

U
V
 
0
  exp
 g

 
 

1 d
 
8

5
15
 
B
 
3
4

4
 c
T
4
Hukum Stefan-
Boltzmann
Energi rata-rata per foton :
  u
N

15
8
 hc
5
3
   
8

B
B
T
3 
3
2 .
4

4

15

2 .
4 k
B
T

2 .
7 k
B
T
(just slightly less than the “most” probable energy)

Daya yang dipancarkan oleh Benda Hitam
For the
“uni-directional” motion, the flux of energy per unit area
 c
 u energy density u
1m 2 c

1s
Integration over all angles provides a factor of ¼: power emitted by unit area

1
4 c
 u
(the hole size must be >> the wavelength)
 c
4

4
 c
T
4  
T
4
Thus, the power emitted by a unit-area surface at temperature T in all directions: power
 c
4 u
The total power emitted by a sphere of radius R : total power emitted by a sphere

4

R
2

T
4
T

Beberapa Contoh u
The value of the Stefan-Boltzmann constant:

4


T
4 c

5 .
76

10

8
W /

K
4 m
2

Consider a human body at 310K. The power emitted by the body:

T
4 
500 W / m
2
While the emissivity of skin is considerably less than 1, it emits sufficient infrared radiation to be easily detectable by modern techniques (night vision).
Radiative transfer:
Liquid nitrogen is stored in a vacuum or Dewar flask, a container surrounded by a thin evacuated jacket. While the thermal conductivity of gas at very low pressure is small, energy can still be transferred by radiation. Both surfaces, cold and warm, radiate at a rate:
J rad


1
 r
 
T i
4
Dewar
W / m
2 i=a for the outer (hot) wall, i=b for the inner (cold) wall, r
– the coefficient of reflection, (1r )
– the coefficient of emission
Let the total ingoing flux be J , and the total outgoing flux be
J’
:
J


1
 r
 
T a
4  r J

The net ingoing flux: J
J
 

1
 r
 
T b
4  rJ

J
 
1

1
 r r


T a
4 
T b
4

If r =0.98 (walls are covered with silver mirror), the net flux is reduced to
1% of the value it would have if the surfaces were black bodies ( r =0).

Efek Rumah Kaca
Absorption:
P ower in
 


R
E
2

  
Sun
4


R
Sun
R orbit


2 the flux of the solar radiation energy received by the Earth ~ 1370 W/m 2
Emission: Power out

4

R
E
2 
T
E
4
T
E





4


R
Sun
R orbit


2
 1


/ 4
T
Sun
Transmittance of the Earth atmosphere

= 1
–
T
Earth
= 280K However, in reality
R orbit
= 1.5
·10 11 m
R
Sun
= 7
·10 8 m

= 0.7
–
T
Earth
= 256K
To maintain a comfortable temperature on the Earth, we need the Greenhouse Effect !
The complicated issue of global worming : adding CO
2
(and other “greenhouse” gases) to the atmosphere tends in itself to raise the earth’s average temperature, but also may increase cloudiness, which lowers it. One thing is clear: since climate is largely determined by the heat balance in the atmosphere, anything that changes the atmospheric absorption is bound to have climatic consequences.

Pengurangan Massa Matahari
The spectrum of the Sun radiation is close to the black body spectrum with the maximum at a wavelength

= 0.5
 m. Find the mass loss for the Sun in one second.
How long it takes for the Sun to loose 1% of its mass due to radiation? Radius of the
Sun: 7 ·10 8 m, mass - 2 ·10 30 kg.
 max
= 0.5
 m
  max
 hc
5 k
B
T

T
 hc
5 k
B
 max



6 .
6

10

34
5

1 .
38

10

23

3

10
8

0 .
5

10

6
K

5 , 740 K


P
 power emitted by a sphere


4

R
2

T
4
 
2

5 k
B
4
15 h
3 c
2

5 .
7

10

8
W m
2
K
4
This result is consistent with the flux of the solar radiation energy received by the Earth
(1370 W/m 2 ) being multiplied by the area of a sphere with radius 1.5
·10 11 m (Sun-Earth distance).
P

4
 
R
Sun

2 

 hc
2 .
8 k
B
 max


4

4


7

10
8 m
2


5 .
7

10

8
W m
2
K
4


5 ,740K

4 
3 .
8

10
26
W the mass loss per one second
1% of Sun’s mass will be lost in dm
 dt
P c
2


3 .
8
3


10
10 8
26 m
W
2


4 .
2

10 9 kg/s
 t

0 .
01 M dm / dt

2

10
28 kg
4 .
2

10
9 kg/s

4 .
7

10
18 s

1.5

10
11 yr

Fungsi Distribusi untuk gas Fermi Ideal
The probability of the by n i i -state with energy
 i to be occupied particles (the total energy of this state n i
 i
) :
The grand partition function for all particles in the i th singleparticle state (the sum is taken over all possible values of n i
) :
P
  i
, n i


Z i
  n i
1
Z exp


 n i
 i
 k
B
T
 n i exp


 n i
  i
   
 k
B
T


The mean number of particles in this state: n
If the particles are fermions , n can only be 0 or 1 :

1 exp
 

 k
B
T
 


1 n i
  n i n i
P
  i

0

P
 

1

P
 

the Fermi-Dirac distribution
Z i

1
 exp



 k

B
T
 exp



 k

B
T

1
 exp



 k

B
T






1

1

 exp
 

 k
B
T

~ k
B
T


At T = 0 , all the states with

<
 have the average # of particles 1 (i.e., they are occupied with 100% probability), all the states with

>
 have the average
# of particles 0 (i.e., they are unoccupied). With increasing T , the step-like function is “smeared” over the energy range ~ k
B
T .
T=0
( with respect to

)

=


Fungsi Distribusi Gas Bose Ideal
The grand partition function for all particles in the i th single-particle state:
(the sum is taken over the possible values of n i
)
If the particles are bosons , n can any integer

0:
Z i
  n i exp


 n i
  i
   
 k
B
T
Z i

1
 exp



 k

B
T
 

 exp



2
     
 k
B
T
 exp



3
     
 k
B
T

....
Z i

1
 exp



 k

B
T
 

 exp



 k

B
T
 

2
 exp



 k

B
T
 

3

....

1
1
 exp



 k

B
T
 

The mean number of particles in this state: n i
  n i n i
P
  i

  n i n i
0

P
 

1

P exp


 n i
     
 k T
B
Z
 
1
Z

2

P
 n i

...


 x


 x exp

 n i x


   k
B
T



1
Z

Z
 x n i
 
1
Z

Z
 x
 

1
 e
 x
 
 x 


1

1 e
 x 


 e
 x
1
 e
 x
 e x
1

1 n i

1 exp
 

 k
B
T
 


1
Distribusi Bose
Einstein
The mean number of particles in a given state for the BEG can exceed unity, it diverges as
  
, and is nonexistent for

>

.

Probabilitas, Fungsi Distribusi, Rapat Keadaan ….
U(x)
 x
The probability that the system is in state s with energy E and N particles
P
  i

1
Z exp




 i
 k
  
B
T n i




1
 s
P
 

1
The macrostate of such system is completely defined if we know the mean occupancy for all energy levels, which is often called the distribution function : f E

While f(E) is often less than unity (much less in the case of an ideal gas), it is not a probability. (e.g., it can exceed unity in a Bose gas).
 i f
 
 n where n=N/V – the density of particles
If we can neglect the spectrum discreteness: n



0 g
    d
 where g(

) is the density of states

Kaitan Termodinamika, Potensial Kimia
Consider the grand potential
   k
B
T ln Z which is a generalization of F=-k
B
T lnZ d
  
SdT

PdV

Nd

- the appearance of
μ as a variable, while computationally very convenient for the grand canonical ensemble, is not natural. Thermodynamic properties of systems are eventually measured with a given density of particles. However, in the grand canonical ensemble, quantities like pressure or N are given as functions of the “natural” variables
T , V and
μ
. Thus, we need to use

 
/
, V
 
N to eliminate
μ in terms of T and n=N/V .
 

T
  
T

S

N

U , V


U

N
S , V


F

N

T , V
Boltzmann
Gas

Boltzmann
  k
B
T ln

 n
Q n



0
μ for an ideal gas is negative : when you add a particle to a system and want to keep S fixed, you typically have to remove some energy from the system.

MB
< 0: - the occupancy n
B
 exp
 

 k
B
T
 
 cannot be negative for any


Fermi
Gas
Potensial Kimia untuk Gas Fermi n
F
 
 f
F
 

1 exp
 

 k
B
T
 


1 n


0
 g
 
T , V , N


 n

N / V d
 


0

    exp
 

 g

 
  k
B
T



1 d

When the average number of fermions in a system (their density) is known, this equation can be considered as an implicit integral equation for

(T,n).
It also shows that
 determines the mean number of particles in the system just as T determines the mean energy. However, solving the eq. is a non-trivial task.

/E
F

E
F

1


2
12

 k
B
T
E
F


2

....
depending on n and T ,
 for fermions may be either positive or negative .
1
1 k
B
T/E
F
The limit T

0: adding one fermion to the system at T=0 increases its energy U by E
F
. In this situation F = U-TS = U ( S is also 0 : all the fermions are packed into the lowest-energy states), so that the chemical potential, which is the change in F produced by the addition of one particle, is E
F
:  
T

0


E
F
The change of sign of

(n,T) indicates the crossover from the degenerate Fermi system (low T , high n ) to the Boltzmann statistics.
The condition k
B
T << E
F is equivalent to n >> n
Q
:
The crossover occurs at n~n
Q becomes negative:
When n<<n
Q the chemical potential

Boltzmann
 n n
Q

3
4



E
F k
B
T
 k
B
T ln

 n
Q n


3 / 2



0

Potensial Kimia untuk Gas Bose
Bose
Gas n
BE

1 exp
 

 k
B
T
 


1 n



0 g d
 


0 g exp
 

 k
B
T
 


1 d

The occupancy cannot be negative for any

, thus, for bosons ,
 
0 (
 varies from 0 to

). Also, as T

0,
 
0

 
BE T

0
 exp
1
 

1

 
BE T

0


0

 1 ,
,

 

0
0
For bosons, the chemical potential is a non-trivial function of the density and temperature
(for details, see the lecture on BE condensation).
T

Pendekatan Klasik
The Fermi-Dirac and Bose-Einstein distributions must reduce to the Maxwell-
Boltzmann distribution in the classical limit , n i

1 for all i . Hence, exp
 

 k
B
T
 
 
1 and n i

1 exp



 

 k
B
T
 




 exp 



 

 k
B
T
 



 the Maxwell-
Boltzmann distribution
The same result, of course, we would get if we start from the equation for the average n k in Boltzmann statistics: n i

NP

N
Z
1 exp



 k
B
T






   k
B
T ln
Z
1
N

Z
1
N
 exp



 k
B
T





 exp


 k
B
T

 exp



 k
B
T


 exp

 
  k

B
T
  

Comparison of the MB, FD, and BE distributions plotted for the same value of

.
Note that the MB distribution makes no sense when the average # of particle in a given state becomes comparable to 1 (violation of the dilute limit).

=


Pendekatan Klasik (cont.)
In terms of the density, the classical limit corresponds to n << the quantum density: n
 n
Q



2
 mk
B
T h 2


3 / 2
We can also rewrite this condition as T>>T
C where T
C is the so-called degeneracy temperature of the gas, which corresponds to the condition n~ n
Q
. More accurately:
T
C

2
 h 2 mk
B

 n
2 .
6


2 / 3
For the FD gas, T
C
T
C
~ E
F
/k
B where E
F is the Fermi energy (Lect. 24) , for the BE gas is the temperature of BE condensation (Lect. 26).
Critical density for bosons: n



0 exp

  g

 
   

1 d
 


 g

2 s
4


1
2
2 m
 2
3 / 2

1 / 2




2 s
4


1
2
2 mk
B
T
 2
3 / 2 

0
2 x
1 / exp
 x
 


1 dx
Since
 
0, the maximum possible value of n is obtained when

= 0, and

0
 exp x
1 /
 
2

1 dx

1 .
3
 n cr

1 .
3

2 s
4


1
2
2 mk
B
T
 2
3 / 2

2 .
6 n
Q where n
Q is the quantum concentration, which varies as T 3/2

U k
B
T
C
3
Pendekatan Ketiga Distribusi
Fermi-Dirac
Maxwell-Boltzmann
Bose-Einstein
S
Nk
B
3
2
1
2
1 zero-point energy,
Pauli principle
1
1 2 3
T
T
C
2 3
T
T
C
T
C

2
 h 2 mk
B

 n
2 .
6


2 / 3

Comparison between Distributions
C
V
/Nk
B
Fermi-Dirac
Maxwell-Boltzmann
Bose-Einstein
2
1.5
0 1 T/T
C

Maxwell
Boltzmann n k

1 exp
 

 k
B
T
 

Comparison between Distributions
Bose
Einstein n k

1 exp
 

 k
B
T
 


1
Fermi
Dirac n k

1 exp
 

 k
B
T
 


1 distinguishable
Z=(Z
1
) N / N!
n
K
<<1 indistinguishable integer spin 0,1,2 … indistinguishable halfinteger spin 1/2,3/2,5/2 … spin doesn’t matter bosons fermions localized particles
 don’t overlap wavefunctions overlap total
 symmetric wavefunctions overlap total
 anti-symmetric photons
4 He atoms free electrons in metals electrons in white dwarfs gas molecules at low densities
“unlimited” number of particles per state n
K
<<1 unlimited number of particles per state never more than 1 particle per state

Aplikasi Statistik Termodinamika
Paramagnetism
Fungsi Partisi

Aplikasi Statistik Termodinamika
Momen magnet rata-rata
Fungsi Partisi

Aplikasi Statistik Termodinamika
Kapasitas panas magnetik

Aplikasi Statistik Termodinamika
Untuk temperatur rendah

Aplikasi Statistik Termodinamika
Jika dideferensial terhadap B

Aplikasi Statistik Termodinamika