fisika-statistik.ppt

7/27/2019 fisika-statistik.ppt

1/90

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


2/90

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


3/90

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


4/90

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


5/90

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).


6/90

Pengertian Dasar Statistik

Mean : Rata-rata

Mode : yang paling mungkin

Median : Titik tengah

Varians : Ragam, Lebar Distribusi


7/90


Misalkan suatu variabel yang diselidiki : 3,4,4,3,5,3,4

47

28

7

5363443

X

7

7654321 xxxxxxxX

N

xX i

i


8/90


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


9/90


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


10/90



11/90

Fungsi Gaussian

Fungsi seperti akan banyak dijumpai dalam pembahasan statistikpartikel


12/90

Ruang Euclid dan Ruang Fase

Ruang Euclid

dxdydzdV

z

xy

dy

dxdz

dV


13/90

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


14/90

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)


15/90


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

)(),(

)(),()(),(


16/90


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


17/90

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


18/90

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


19/90

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


20/90


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


21/90


Prinsip Kekekalan


22/90


Jumlah pilihan jika memilih sejumlah N1 di antara N partikel

Jika g1 menyatakan bobot, maka jumlah pilihan yang ada adalah :


23/90


Perluas lagi dengan mengambil sejumlah N2 dari N-N1

Perluas lagi dengan mengambil sampai n kali


24/90


Secara umum dapat ditulis :


25/90

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


26/90

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.


27/90


Peluang terbesar diperoleh dengan mengambil dw/dn = 0

Rumus Stirling


28/90

Distribusi Maxwell Boltzmann

de

kT

Ndn kT 2/1/

2/3

2)(

0

TkB

exp

0

g() Cg

=

0

P()


29/90

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


30/90

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)


31/90

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


32/90

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


33/90

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


34/90

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


35/90


Karena efek Doppler, maka panjang gelombang yang diamati

pengamat adalah :

c

vxo 1

o

ocv

dc

dvo

x

o


36/90


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


37/90


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/


38/90

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/


39/90


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


40/90


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


41/90


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


42/90


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


43/90


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


44/90


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


45/90

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


46/90

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


47/90


s

sww

s ss

ss

ng

ng

!!1

!1


48/90


0log

ss

s

s

dnxn

w

0log

ss

xn

w

s

ssssssss

s

s

nnggngng

ww

log1log11log1

loglog


49/90


ssss

nngn

wlog1log

log

s

ss

s

n

ng

n

wlog

log

0log

s

s

ss xn

ng

1 sx

s

s en

g


50/90


1

sx

s

se

gn

1/1 kTs

sse

A

g

n

!!!

sss

ssngn

gw


51/90


1

sx

s

se

gn

1/1 kTs

sse

A

g

n

!!!

sss

ss

ngn

gw


52/90



53/90

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


54/90


s

ssssssss

s sss

s

ngngnngg

ngn

gW

logloglog

!!

!loglog

s

s

s

dnn

Ws

0log

0log

s

sn

W

Gunakan rumus Stirling


55/90


s

ss

s n

ng

n

W

loglog

0log ss

ss

nng

1 se

ng

s

s


56/90


T=0

~ kBT

= (with respect to)

1

1

kTFef

dgfdn

11

1,0

efF

011

,0 efF


57/90


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


58/90

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


59/90

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 ?


60/90

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


61/90

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


62/90

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


63/90

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:


64/90

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


65/90

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


66/90

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


67/90

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


68/90

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


69/90

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


70/90

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


71/90

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


72/90

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


73/90

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


74/90

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


75/90

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


76/90

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


77/90

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


78/90

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


79/90

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


80/90

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.)


81/90

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


82/90

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


83/90

Comparison between Distributions

T/TC

CV/NkB

1

1.5

2

0

Fermi-DiracMaxwell-Boltzmann

Bose-Einstein

Comparison between Distributions


84/90

Maxwell

Boltzmann

Fermi

Dirac

Bose

Einstein

distinguishable

Z=(Z1)N/N!

nK


85/90

Paramagnetism

Fungsi Partisi

Aplikasi Statistik Termodinamika


86/90

Momen magnet rata-rata

Fungsi Partisi



87/90

Kapasitas panas magnetik



88/90

Untuk temperatur rendah



89/90

Jika dideferensial terhadap B



90/90

fisika-statistik.ppt

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Transcript of fisika-statistik.ppt