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CURICULUM VITAE
A. DATA DIRI
01. N a m a : Dr. H. Muris, M.Si
02. Tempat/Tanggal Lahir : Tinggas, 1965
03. Jenis Kelamin : Laki-laki
04. Fakultas/Jurusan : FMIPA/Fisika
05. Pangkat/Golongan/NIP : Lektor Kepala/IV/a/13192582006. Bidang Keahlian : Fisika Material
07. Alamat Rumah : BTN Minasa Upa G20/14 Makassar.
90224.
Telp. (0411) 886307
HP. 081342403676
08. Alamat Kantor : Jurusan Fisika FMIPA UNM
Kampus Parangtambung Makassar
Tlp/Fax. (0411)840622, HP. 081342403676
09. e-mail : [email protected]
10. Riwayat Pendidikan Tinggi :
Jenis Pendidikan Tempat Tahun lulus Spesialisasi
Sarjana (S1)
Pra Magister (Pra S2)
Magister (S2)
Doktor (S3)
IKIP Ujung Pandang
ITB Bandung
ITB Bandung
Universit de la MditerraneMarseille, Prancis
1989
1992
1994
2001
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
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Fisika Statistik
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
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Pokok Bahasan
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
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Pokok Bahasan
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
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Sistim Termodinamika, Parameter Makroskopik
Sistim terbuka dimana dimungkinkan
terjadi pertukanan energi dan materidengan lingkungan.
Sistim tertutup terjadi pertukaran
energi maupun materi dengan
lingkungannya
Isolated systemstidak
memungkinkan terjadinya pertukaran
energi maupu materi dengan
lingkungannya
Paramater internal dan external : temperatur, volume, tekanan, energi,
medan magnet, dll. (nilai rata-rata, fluktuasi diabaikan).
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Pengertian Dasar Statistik
Mean : Rata-rata
Mode : yang paling mungkin
Median : Titik tengah
Varians : Ragam, Lebar Distribusi
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Pengertian Dasar Statistik
Misalkan suatu variabel yang diselidiki : 3,4,4,3,5,3,4
47
28
7
5363443
X
7
7654321 xxxxxxxX
N
xX i
i
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Pengertian Dasar Statistik
Rata-rata dengan fungsi probabilitasxi f f(xi) xi f(xi)
3 3 3/7 9/7
4 3 3/7 12/7
5 1 1/7 5/7
7 1 28/7 = 4
Ternyata diperoleh hasil rata-rata yang sama yakni 4
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Pengertian Dasar Statistik
Hasil ini diperoleh dari pengembangan bentuk
i
ii
i
i
i
ii
xxfxf
xxf
X ).()(
)..(
1)( ixf
Jika fungsinya kontinyu maka :
dxxfxX )(.
Bagaimana anda mengartikan parameter statistik berikut ?
kontinyu
diskrit
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Pengertian Dasar Statistik
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Fungsi Gaussian
Fungsi seperti akan banyak dijumpai dalam pembahasan statistikpartikel
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Ruang Euclid dan Ruang Fase
Ruang Euclid
dxdydzdV
z
xy
dy
dxdz
dV
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Ruang Euclid dan Ruang Fase
zyx dpdpdxdydzdpd
mppp zyx
2
222
znyNxNNNN
ziyixiiii
zyxN
dpdpdpdzdydx
dpdpdpdzdydx
dpdpdpdzdydxd
........
........
1111116
N
i
ziyixiiii dpdpdpdzdydx1
N
i
id1
Ruang fase Ruang momentum
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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)
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Rata Rata Sifat Assembly
Jika X adalah perilaku yang ingin dicari rata-ratanya dalam ruang
fase tersebut
N
NdNpNxPNpNxXX
6
6)(),()(),(
Normalisasi terhadap ruang
N
N
N
N
dNpNxP
dNpNxPNpNxXX
6
6
6
6
)(),(
)(),()(),(
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Rata Rata Sifat Assembly
Jika X merupakan fungsi yang diskrit, maka perata-rataan fungsi X
dapat dinyatakan dengan :
i
i
iii
p
XpX
Normalisasi probabilitas menghasilkan
1i
ip
i
iiXpX
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Assembli Klasik dan Kuantum
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
a. Klasik
- Terbedakan antara satu dengan lainnya (distinguishable)
- Energi kontinu
- Tak memenuhi prinsip larangan Pauli
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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
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Statistik Maxwell Boltzmann
Distribusi Energi
Misalkan dalam sistim yang ditinjau terdapat N sistim :
Sistem 1 dengan energi 1
Sistem 2 dengan energi 2.
Sistem i dengan energi
i
.
Sistem N dengan energi N
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Statistik Maxwell Boltzmann
Distribusi Energi
Misalkan dalam sistim yang ditinjau terdapat N sistim :
Sistem 1 dengan energi 1
Sistem 2 dengan energi 2.
Sistem i dengan energi
i
.
Sistem N dengan energi N
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Statistik Maxwell Boltzmann
Prinsip Kekekalan
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Statistik Maxwell Boltzmann
Jumlah pilihan jika memilih sejumlah N1 di antara N partikel
Jika g1 menyatakan bobot, maka jumlah pilihan yang ada adalah :
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Statistik Maxwell Boltzmann
Perluas lagi dengan mengambil sejumlah N2 dari N-N1
Perluas lagi dengan mengambil sampai n kali
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Statistik Maxwell Boltzmann
Secara umum dapat ditulis :
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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 : N1 = N2 = 2 g1 = 3 , g2 = 4
8644.3!.2!2
!4 22 W
22
21
21
.!!.
! ggNN
NW
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Contoh Pemakaian
a b a b c,a
c d c d d b
Ini hanyalah 3 contoh gambar dari 864 kemungkinan yang ada.
Sekarang adalah giliran anda untuk melengkapinya.
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Statistik Maxwell Boltzmann
Peluang terbesar diperoleh dengan mengambil dw/dn = 0
Rumus Stirling
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Distribusi Maxwell Boltzmann
de
kT
Ndn kT 2/1/
2/3
2)(
0
TkB
exp
0
g() Cg
=
0
P()
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Aplikasi Statistik Maxwell Boltzmann
2D
kx
ky Untuk partikel kuantum dalam kotak 2D (e.g., electron pd FET):
22
yx
y
y
y
x
xx kkk
L
nk
L
nk
2
222 2
4
1
444
1
mG
kkG
areak
LL
kkN
yx
k
# states within of
a circle of radius kxL
3D
ky
kx
kz
2/3
222
3
2
332
6
1
66
3/4
8
1
mG
kkG
volumek
LLL
kkN
zyx
- Tak
bergantung pd
2/12/3
22
3 2412
msg D
2
2
2
12
msg D
g()
3D
2D
1D
Thus, for 3D electrons
(2s+1=2):
2/12/3
22
3 2
2
1
mg
D
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Distribusi Kecepatan Maxwell
vx
vy
vz
dvv
dvvv
24
volume""
v
10
dvvf
2/3
2
Tk
mC
B
dvvTk
mv
Tk
mfvvf
BB
22
2/3
42
exp2
Nampak bahwa persamaan ini merupakan perkalianantara faktor Boltzmann dengan sebuah tetapan.
Tetapan tersebut dapat diperoleh dari normalisasi
dvTkmv
vTk
m
NdvvNPvdN BB
2exp42
22
2/3
dTk
NdNPdNB
exp
dvTk
mv
Tk
mNdvvNPvdN
BB
xx
2exp
2
22/1
Distr ib usi energi, N the total # of particles
speed distr ibut ion (dist r ibus i kecepatan)
Distrb usi k ecepatan dalam arah x, vx
P(vx)
vx
v
P(v)
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Tk
mvv
Tk
mvP
BB 2exp4
2
22
2/3
Lihat bahwa distribusi ini tidak simetrik, sehingga
perlu dicari perata-rataan sebagai berikut
Karakteristik Nilai Kecepatan
0
23
0
8
2exp4
2 m
Tkdv
Tk
mvv
Tk
mdvvPvv B
BB
Harga kec.maksim um :
m
Tkv
dv
vdP B
vv
20 max
max
Kelaju an rata-rata :
P(v)
maxv rmsvv
22.113.113/82max rmsvvv
v
The roo t-mean-sq uare speedis proportional to
the square root of the average energy:
m
Tk
m
EvvmE Brmsrms
32
2
1 2
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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.
Tk
vmv
Tk
m
Tk
vmv
Tk
m
BBBB 2exp4
22exp4
2
2
22
2/3
2
2
12
2/3
1
Tk
vmm
Tk
vmm
BB 2ln2
3
2ln2
3 222
2
11
km/s6.1
107.12
2ln3001038.13ln3
2
ln
2
327
23
21
2
1
21
2
2
1
mm
m
mTk
vmm
Tk
v
m
mB
B
Tk
mvv
Tk
mmTvP
BB 2exp4
2,,
22
2/3
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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.
022
exp22
exp222
32
222/3
2
2
2/1
Tk
mv
Tk
mv
Tk
m
Tk
mv
Tk
m
Tk
m
BBBBBB
0,
T
TvPmaximum:
BB k
mvT
Tk
mv
30
22
322
Tk
mvv
Tk
mmTvP
BB 2exp4
2,,
22
2/3
Find the temperature Tat which the rms speed of Hydrogen molecules exceeds their
most probable speed by 400 m/s.
Answer: 380K
At home:
o
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Pelebaran Garis Spektrum Doppler
Bagian ini adalah salah satu contoh penerapan distribusi laju dari
statistik Maxwell Boltzmann, yakni pelebaran spektrum akibat efekDoppler.
Misalkan molekul gas melakukan radiasi dengan panjang
gelombang dalam arah x dengan kecepatan vx menuju kepada
seorang pengamat. Pengamat akan menerima radiasi denganpanjang gelombang.
o
o
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Pelebaran Garis Spektrum Doppler
Karena efek Doppler, maka panjang gelombang yang diamati
pengamat adalah :
c
vxo 1
o
ocv
dc
dvo
x
o
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Pelebaran Garis Spektrum Doppler
dvTk
mvv
Tk
mNdvvNfvdN
BB
2exp4
2
22
2/3
Dari distribusi Maxwell Boltzamann
Ubah sebagai fungsi panjang gelombang
d
c
Tk
mc
Tk
mdf
oo
o
BB
2
222/3
2exp
2
o
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Pelebaran Garis Spektrum Doppler
d
Tk
mcIdCfdI
o
o
B
o
2
22
2exp)(
)(I
Intensitas radiasi :
o
)( oI Dengan mengukur intensitas
radiasi maka dapat ditentukan
temperatur gas emisi
o
dmep KTex /2 2/
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Prinsip Ekipartisi Energi
m
pxx
2
2
Jika energi sistem dinyatakan dalam bentuk kuadrat posisi dan momentum maka tiap
bentuk kuadrat tersebut akan memberikan energi rata-rata kT
Contoh molekul gas dengan massa m, energinya dapat dinyatakan dengan
Maka energi rata-ratanya adalah :
dTe kTe /
dTe
demp
kTe
KTe
x
/
/2 2/
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Prinsip Ekipartisi Energi
mkT
px
2
2
m
px
2
2
Nyatakan energi sebagai dan
xxxy
p
m
xx
x
x
zy
x
x
dpmkTpdpdxdydzdpkT
dpmkTpm
pdpdxdydzdpkT
m
p
x
x )2/exp(/)(exp
)2/exp(2
/)2
(exp
2
2
222
2
Misalkan = u2
maka
due
duuekT
u
u
x2
2 2
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Prinsip Ekipartisi Energi
Hasilnya memberikan :
dueduue uu22
212
Maka : kTx
2
1
Karena ada satu bentuk kuadrat maka memberikan energi rata-rata kT
Contoh 2 : Osilator harmonik dengan dua jenis energi
22
2
1
2x
m
pxx
umkT
px 2
2
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Prinsip Ekipartisi Energi
Maka :
de
dexmx
p
kTe
kTe
x /
/2
2
12/
2
xx
xx
x
x
dxdpkTxm
p
dxdpkTxm
pxp
2
2
22
22
2
1
2exp
/2
1
2exp
2
1
Ubah ke koordinat polar :
,sin2
22
2
rm
px 222
cos2
1rx
dpxdx
rdrdmdpdp yx21
)/(2
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Prinsip Ekipartisi Energi
Maka :
kT
rdred
drred
x
kTr
x
kTr
2
0 0
/
2
0 0
3/
2
2
Karena terdiri dari dua bentuk kuadrat maka energinya adalah
2x kT = kT
Untuk osilator harmonik 3D maka :
kT
kTkTkTkTm
p
m
p
m
p xyx
2
3
2
3
2
1
2
1
2
1
222
222
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Prinsip Ekipartisi Energi
Energi rata-rata untuk osilator harmonik 3 D.
kT
kT
zm
py
m
px
m
p xyx
3
21.6
2
1
22
1
22
1
2
2
3
2
2
2
2
2
1
2
Jadi dalam hal ini ada 6 derajat kebebasan ( f = 6) dimana tiap
derajat kebebasan memberikan kontribusi energi sebesar kT
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Prinsip Ekipartisi Energi
Jika terdapat NA (bil. Avogadro) molekul gas dan berlaku sebagai
osilator harmonik 3D, maka, terdapat 6 derajat kebebasan,maka :
RTkTNE A 3
2
16
Panas jenis per gram atom zat padat :
K/gr.atomkal/94,53o
RT
E
v
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Panas jenis gas
Jika terdapat NA (bil. Avogadro) molekul gas dan berlaku sebagai
osilator harmonik 3D, maka, terdapat 6 derajat kebebasan,maka :
RTkTNE A 3
2
16
Panas jenis per gram atom zat padat :
K/gr.atomkal/94,53o
RT
E
v
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STATISTIK BOSE-EINSTEIN
!1 sss ngg
!!1
!1
ss
ssss
ng
nggw
!!1
!1
ss
ss
ng
ng
s
sww
s ss
ss
ng
ng
!!1
!1
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STATISTIK BOSE-EINSTEIN
s
sww
s ss
ss
ng
ng
!!1
!1
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STATISTIK BOSE-EINSTEIN
0log
ss
s
s
dnxn
w
0log
ss
xn
w
s
ssssssss
s
s
nnggngng
ww
log1log11log1
loglog
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STATISTIK BOSE-EINSTEIN
ssss
nngn
wlog1log
log
s
ss
s
n
ng
n
wlog
log
0log
s
s
ss xn
ng
1 sx
s
s en
g
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STATISTIK BOSE-EINSTEIN
1
sx
s
se
gn
1/1 kTs
sse
A
g
n
!!!
sss
ssngn
gw
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STATISTIK BOSE-EINSTEIN
1
sx
s
se
gn
1/1 kTs
sse
A
g
n
!!!
sss
ss
ngn
gw
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STATISTIK BOSE-EINSTEIN
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STATISTIK FERMI-DIRAC
s
swW
!!!
sss
s
sngn
g
w
s sss
s
ngn
gW
!!
!Jumlah untuk semua kemungkinan susunan
yang berbeda
Jumlah untuk semua kemungkinan susunanyang berbeda untuk satu tingkatan energi
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STATISTIK FERMI-DIRAC
s
ssssssss
s sss
s
ngngnngg
ngn
gW
logloglog
!!
!loglog
s
s
s
dnn
Ws
0log
0log
s
sn
W
Gunakan rumus Stirling
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STATISTIK FERMI-DIRAC
s
ss
s n
ng
n
W
loglog
0log ss
ss
nng
1 se
ng
s
s
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STATISTIK FERMI-DIRAC
T=0
~ kBT
= (with respect to)
1
1
kTFef
dgfdn
11
1,0
efF
011
,0 efF
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STATISTIK FERMI-DIRAC
1
se
gn ss
11
kTFef
Distribusi jumlah partikel partikel
Melalui normalisasi gs = 1 diperolehfungsi distribusi. Maka f(e) merupakan
probabilitas sebagai fungsi energi
Sebagai fungsi probabilitas maka harga fungsi ini maksimum 1dan minimum 0
R di i B d Hit
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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. Well see that the chemical
potential of such particles is zero in
equilibrium, regardless of density.
Radiation in Eq ilibri m ith Matter
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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 cavi ty f i l led with radiat ion, 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 uS(,T)d as the energy density (per unit volume) of theradiation with frequencies between and +d. uS(,T) is the spectral energy density.The internal energy of the photon gas:
dTuTu S
0
,
In equilibrium, uS(,T) is the same everywhere in the cavity, and is a function offrequency and temperature only. If the cavity volume increases at T=const, the
internal energy U=u(T)Valso increases. The essential difference between the
photon gas and the ideal gas of molecules: for an ideal gas, an isothermal expansionwould conserve the gas energy, whereas for the photon gas, it is the energy densi ty
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 ablack body.
Photons Apa Itu ?
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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 m ay be exc i ted on ly in un i ts of the quantumof energyh f(similar to a harmonic oscillators) :
hnii 2/1
This fact leads to the concept ofpho tons as qu anta of the electrom agnet ic f ield. The
state of the el.-mag. field is specified by the numbernfor 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 sp in1 (in units ). They move with the speed of light :
ch
c
Ep
cpE
hE
ph
ph
phph
ph
The linearity of Maxwell equations implies that the photons do no t
interact with each other. (Non-linear optical phenomena are
observed when a large-intensity radiation interacts with matter).
T
The mechanism of establishing equilibrium in a photon gas is absorp t ion and emiss ion
of photons by matter.
Presence of a small amount of matter is essential for establishing equilibrium in the
photon gas. Well treat a system of photons as an ideal photo n gas, and, in particular,
well apply the BE statistics to this system.
Potensial Kimia Foton = 0
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Potensial Kimia Foton 0The mechanism of establishing equilibrium in a photon gas is absorpt ion
and emiss ionof photons by matter. The textbook suggests that Ncan be
found from the equilibrium condition:
0phThus, in equilibrium, the chemicalpotent ia lfor a photon gas is zero:
0
,
VTN
F
On the other hand,ph
VTNF
,
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:
VTN
F
,
- does not exist for the photon gas
Instead, we can use NG PVFG V
VTF
V
FP
T
,
- by increasing the volume at T=const, we proportionally scale F
0 VV
F
FGThus,- the Gibbs free energy of an
equilibrium photon gas is 0 ! 0 N
Gph
For = 0, the BE distribution reduces to the Plancks distribution:
1exp
1
1exp
1,
Tk
h
Tk
Tfn
BB
phph
Plancks distribution provides the averagenumber of photons in a single mode of
frequency =/h.
h
hn
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Rapat Keadaan Foton
ky
kx
kz
2
3
2
33
66
3/4
8
1
kkG
volumek
LLL
kkN
zyx
extra factor of 2:
two polarizations
322
3
32
3
26
cg
cGkccp
d
dGg Dph
32
32
2
33 8
cc
hh
d
dgg Dph
D
ph
3
23 8
cg
D
ph
1exp
Tk
hhn
B
In the classical (high temperature) limit: TkB
The average energy in the mode:
In order to calculate the average number of photons per small energy interval d, theaverage energy of photons per small energy interval d, etc., as well as the totalaverage number of photons in a photon gas and its total energy, we need to know the
densi ty of states for photo nsas a function of photon energy.
Spektrum Radiasi Benda Hitam
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Spektrum Radiasi Benda Hitam
Radiasi s pektrum
benda h i tam
1exp
8,
3
3
Tk
hcTu
B
Rata-rata jumlah foton per satuan volume denga frekwensi dan +d:
u(,T)- the energy density per unit photonenergy for a photon gas in equilibrium with
a blackbody at temperature T.
dTudfgS
,
- Rapat Spektrum (hukum Radiasi Planck)
1exp
8
,
3
3
hc
h
fghTus
hThud
dTuTudTudTu ,,,,,
uadalahfungsi energi:
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Pendekatan Klasik (f keci l, besar), Hkm Rayleigh-Jeans
This equation predicts the so-called
ul t rav io let catastrop he an infinite
amount of energy being radiated at
high frequencies or short
wavelengths.
Hukum Rayleigh-Jeans
Pd frekwensi rendah dan temp. tinggi hhh 1exp1
Tkchc
hTu Bs 3
23
3
8
1exp
8,
- purely classical result (no h), can be
obtained directly from equipartition
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Hukum Rayleigh-Jeans
4
1
In the limit of large :
4large
8,
TkTu B
1exp
18
1exp
8,,,
52
3
32
Tk
hc
hchc
Tk
hc
ch
hcTu
hc
d
ddTudTu
BB
u sebagai fungsi dari panjang gelombang
frekwensi tinggi Hukum Pergeseran Wiens
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frekwensi tinggi , Hukum Pergeseran Wien s
Wien
Maksimum u() berfeser ke frekwensi tinggi ketika temperatur naik.
0
113
1exp
2
32
3
x
x
x
B
B
B
eex
exconst
Tk
hTk
h
Tk
hd
dconstddu
8.233 xex x
Hukum
Pergeseran
Wien
Numerous applications
(e.g., non-contact radiation thermometry)
- the mostlikely frequency of a photon in a
blackbody radiation with temperature Tu
(,
T)
8.2max Tkh
B
h
TkB8.2max
Nobel 1911
At high frequencies: hhh exp1exp1
hc
hTus exp
8, 3
3
- Ditemukan secara eksperimen oleh Wien
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max max
1exp
18
1exp
8,,,
52
3
32
Tk
hc
hchc
Tk
hc
ch
hcTu
hc
d
ddTudTu
BB
8.2max Tk
h
B
- does this mean that
8.2max
Tk
hc
B?
max maxWrong!
0
1/1exp
/1exp
1/1exp
5
1/1exp
125
2
65
xx
xx
xxconst
xxdx
dconst
df
du
xxx /1exp1/1exp5 Tk
hc
B
5max
night vision devices
T =300 K max10 m
Tu , Tu ,
Radiasi Sinar Matahari
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Radiasi Sinar Matahari
Temperatur permukaan- 5800K
As a funct ion o f energy, the spectrum of sunlight peaks at a photon energy of
m5.05
max Tk
hc
B
eVTkhu B 4.18.2maxmax (umax) 0.88 m, near infrared
Spectral sensitivity of the eye:
- close to the energy gap in Si, 1.2 eV,which has been so far the best material
for photovoltaic devices (solar cells)
Hukum Radiasi Stefan Boltzmann
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Hukum Radiasi Stefan-Boltzmann
Jumlah total foton persatuan volume :
Energi total foton per satuan volume : (apat
energi gas foton)
3
45
0 15
8
1exp hc
Tkd
g
V
UTu B
Hukum Stefan-
Boltzmann23
45
15
2
ch
kB
Tetapan Stefan-Boltzmann 4
4T
cTu
4.2818
1exp
8 33
0
23
30
2
30
Thc
k
e
dxx
h
Tk
cd
Tk
hcdgnV
N
n
B
x
B
B
- increases as T3
Energi rata-rata per foton :
TkTkTkhc
hcTk
N
TuBB
B
B 7.24.2154.2815
84
33
345
(just slightly less than the most
probable energy)
D di k l h B d Hit
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Daya yang dipancarkan oleh Benda Hitam
For the uni-directional motion, the flux of energy per unit area
c 1senergy densi tyu
1m2
T424sphereabyemittedpowertotal TR
uc
Integration over all angles provides a factor of :
uc4
1areaunitbyemittedpower
Thus, the power emitted by a unit-area
surface at temperature Tin all directions: 44
4
4
c
4
cpower TT
cTu
The total power emitted
by a sphere of radius R:
(the hole size must be >> the wavelength)
B b C t h 44
TT
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Beberapa Contoh
The value of the Stefan-Boltzmann constant:
4Tc
Tu
248 /1076.5 mKW
Consider a human body at 310K. The power emitted by the body: 24 /500 mWT
While the emissivity of skin is considerably less than 1, it emits sufficient infrared
radiation to be easily detectable by modern techniques (night vision).
Radiat ive transfer:
Dewar
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, energycan still be transferred by radiation. Both surfaces, cold and warm, radiate at a rate:
24 /1 mWTrJ irad
The net ingoing flux:
rJTrJJrTrJ ba 44 11
Let the total ingoing flux be J, and the total outgoing flux be J:
i=afor the outer (hot) wall, i=bfor the inner (cold) wall,
r the coefficient of reflection, (1-r) the coefficient of emission
441
1ba TT
r
rJJ
Ifr=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
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Transmittance of the Earth atmosphere
Absorpt ion:
Emiss ion:42
4outPower EE TR
the flux of the solar radiation energyreceived by the Earth ~ 1370 W/m2
2
42inower
orbit
SunSunE
R
RTRP
Sun
orbit
SunE T
R
RT
4/12
4
=1 TEarth = 280K
Rorbit= 1.51011 m
RSun= 7108 m
However, in reality =0.7 TEarth = 256KTo maintain a comfortable temperature on the Earth, we need the Greenhouse Effect !
The complicated issue ofgloba l worm ing: adding CO2 (and othergreenhouse gases)
to the atmosphere tends in itself to raise the earths 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
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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: 7108 m, mass - 2 1030 kg.
max = 0.5 m
KKk
hcT
Tk
hc
BB
740,5105.01038.15
103106.6
55 623
834
max
max
This result is consistent with the flux of the solar radiation energy received by the Earth
(1370 W/m2) being multiplied by the area of a sphere with radius 1.51011 m (Sun-Earth
distance).
42
8
23
45
Km
W107.5
15
2 ch
kB 424sphereabyemittedpower TRP
W108.3,740K5Km
W107.5m1074
8.24 26
4
42
828
4
max
2
B
Sunk
hcRP
kg/s102.4
m103
W108.3 928
26
2
c
P
dt
dmthe mass loss per one second
1% of Suns mass will be lost inyr101.5s107.4
kg/s102.4
kg102
/
01.0 11189
28
dtdm
Mt
Fungsi Distribusi untuk gas Fermi Ideal
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The probability of the i-state with energy ito be occupiedby niparticles (the total energy of this state nii) :
Tk
nn
ZnP
B
iiiii
exp
1,
If the particles are fermions, ncan only be 0 or1:
TkZ
B
i
exp1
The grand partition function for all particles in the ithsingle-
particle state (the sum is taken over all possible values ofni) :
in B
ii
i Tk
n
Z
exp
1exp
1
Tk
n
B
- the Ferm i-Dirac
distr ibut ion
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 ~ kBT.
The mean number of particles in this state:
TkTk
Tk
PPnPnn
BB
B
niii
i
exp1
1
exp1
exp
1100
T=0
~ kBT
= (with respect to)
Fungsi Distribusi Gas Bose Ideal
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The grand partition function for all particles in the
ithsingle-particle state:
(the sum is taken over the possible values ofni)
in B
iii
Tk
nZ
exp
If the particles are bosons,ncan any integer 0:
....3exp2expexp1
TkTkTkZ
BBB
i
Tk
TkTkTkZ
B
BBB
i
exp1
1....expexpexp1
32
The mean
number of
particles in this
state:
x
Z
Z
xn
xZZ
Tk
n
n
TkxPPPnPnn
ii
i
n
i
n
B
i
i
Bn
iii
1exp
1exp
...221100
1
1
11
11
1
xx
x
x
x
iee
e
exe
x
Z
Zn
1exp
1
Tk
n
B
i
Dis tr ibus i Bos eEinstein
The mean number of particles in a given state for the BEG can exceed unity, it diverges as
, and is nonexistent for > .
P b bilit F i Di t ib i R t K d
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Probabilitas, Fungsi Distribusi, Rapat Keadaan .
The macrostate of such system is completely defined if
we know the mean occupancy for all energy levels,
which is often called thedis t r ibu t ion funct ion
:
EnEf
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).
nEfi
1s
EP
x
U(x)
The probabi l i tythat the system is in state swith energy Eand Nparticles
1exp1
Tk
nT
ZP
B
iii
where n=N/V the density of particles
If we can neglect the
spectrum discreteness: dfgn
0
where g() is thedensi ty o f states
Kaitan Termodinamika, Potensial Kimia
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cannot be negative for any
Tkn
B
B
exp
for an ideal gas is negat ive: when you add a
particle to a system and want to keep Sfixed, youtypically have to removesome energy from the
system.
VTVSVU N
F
N
U
N
STTn
,,,
,
BoltzmannGas
0ln
n
nTk QBBoltzmann
Consider the grand p otent ial ZTkB ln which is a generalization ofF=-kBT lnZ
NdPdVSdTd
- the appearance of 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 orNare given as functions of the natural variables
T,Vand . Thus, we need to use to eliminate in terms ofTand
n=N/V. NVT ,/
MB < 0: - the occupancy
Potensial Kimia untuk Gas Fermi
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1exp
1
Tk
fn
B
FF
d
Tk
nT
gdfgn
B
00 1
,exp
Fermi
Gas
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 Tdetermines the mean
energy. However, solving the eq. is a non-trivial task.
The limit T0: adding one fermion to the system at T=0 increases its energy Uby EF. In
this situation F = U-TS = U(Sis also0: all the fermions are packed into the lowest-energy
states), so that the chemical potential, which is the change in Fproduced by the addition
of one particle, is EF: FET 0
depending on nand T, forfe rmionsmay be either
pos i t iveornegat ive.
....12
1
22
F
B
F E
Tk
E
kBT/EF1
1 /EF
The change of sign of (n,T) indicates the crossover from thedegenerate Fermi system (low T, high n) to the Boltzmann statistics.
The condition kBT > nQ:
2/3
3
4
Tk
E
n
n
B
F
Q
The crossover occurs at n~nQ When n
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Bose
Gas
1exp
1
Tk
n
B
BE
The occupancy cannot be negative for any , thus, forbosons, 0 (varies from 0 to ). Also, as T0, 0
0,1
0,0
10/0exp
100
TBETBE
nn
d
Tk
gdfgn
B
00
1exp
T
For bosons, the chemical potential is a non-trivial function of the density and temperature
(for details, see the lecture on BE condensation).
Pendekatan Klasik
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The Fermi-Dirac and Bose-Einstein distributions must reduce to the Maxwell-
Boltzmann distribution in the classical l imi t, for all i. Hence,
TkTkTkTkN
Z
N
ZTk
TkZ
NsNPn
BBBB
B
B
i
expexpexpexplnexp 11
1
1in
1exp
TkB
Tk
Tk
nB
B
i
exp
exp
1and
the Maxwell-Bol tzmann
distr ibut ion
The same result, of course, we would get if we start from the equation for the
average nkin Boltzmann statistics:
=
Comparison of the MB, FD, and BE
distributions plotted for the same
value of . Note that the MBdistribution makes no sense when the
average # of particle in a given state
becomes comparable to 1 (violation of
the dilute limit).
Pendekatan Klasik (cont.)
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2/3
2
2
h
Tmknn BQ
In terms of the density, the classical limitcorresponds to n>TCwhere TCis the so-called degeneracytemperatureof the gas, which corresponds to the condition n~ nQ. More accurately:
3/22
6.22
n
mk
hT
B
C
For the FD gas, TC
~ EF/k
Bwhere E
Fis the Fermi energy (Lect. 24) , for the BE gas
TCis the temperature of BE condensation (Lect. 26).
Cri t ical densi ty for boson s:
0
2/12/3
22
2/1
2/3
22
01exp
2
4
122
4
12
1expdx
x
xTmksmsgd
gn B
3.1
1exp0
2/1
dxx
xSince 0, the maximum p ossib le value ofnis obtained when = 0, and
QB
cr nTmks
n 6.22
4
123.1
2/3
22
where nQis the quantum concentration,
which varies as T3/2
Pendekatan Ketiga Distribusi
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CBTk
U
CT
T1 2 3
1
2
3
3/22
6.22
n
mk
hT
B
C
BNk
S
CT
T
1 2 3
1
2
3
Fermi-Dirac
Maxwell-Boltzmann
Bose-Einstein
zero-point
energy,
Pauli
principle
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Comparison between Distributions
T/TC
CV/NkB
1
1.5
2
0
Fermi-DiracMaxwell-Boltzmann
Bose-Einstein
Comparison between Distributions
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Maxwell
Boltzmann
Fermi
Dirac
Bose
Einstein
distinguishable
Z=(Z1)N/N!
nK
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Paramagnetism
Fungsi Partisi
Aplikasi Statistik Termodinamika
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Momen magnet rata-rata
Fungsi Partisi
Aplikasi Statistik Termodinamika
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Kapasitas panas magnetik
Aplikasi Statistik Termodinamika
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Untuk temperatur rendah
Aplikasi Statistik Termodinamika
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Jika dideferensial terhadap B
Aplikasi Statistik Termodinamika
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