today’s outline - october 02, 2013segre/phys570/13f/lecture_13.pdf · 2013. 10. 3. · today’s...

Today’s Outline - October 02, 2013

• Inelastic Scattering

• Scattering from Molecules

• Scattering from Liquids & Glasses

• Small Angle Scattering

Reading assignment: Chapter 4.1-4.4

Homework Assignment #04:Chapter 4: 2, 4, 6, 7, 10due Monday, October 21, 2013

C. Segre (IIT) PHYS 570 - Fall 2013 October 02, 2013 1 / 11

Inelastic scattering

The form factors for all atoms dropto zero as Q →∞, however, otherprocesses continiue to scatter pho-tons.

In particular, Compton scatteringbecomes dominant.

Compton scattering is an inelasticprocess: |~k| 6= |~k ′| and it is alsoincoherent.

The Compton scattering containsinformation about the momentumdistribution of the electrons in thegroundstate of the atom.

0 5 10

Compton scattering is an inelasticprocess: |~k| 6= |~k ′|

and it is alsoincoherent.

0 5 10

10500 11000 11500 12000 12500Energy (eV)

ay Inte

nsity (

units) φ=160

Total atomic scattering

We can now write the total scat-tering from an atom as the sum oftwo components:

∼ r20 |f (Q)|2(

∼ r20S(Z ,Q)

Recall that f (Q)→ Z as Q → 0

For the incoherent scattering we ex-pect S(Z ,Q)→ Z as Q →∞

ZHe = 2 ZAr = 18

∼ r20 |f (Q)|2

∼ r20S(Z ,Q)

ZHe = 2 ZAr = 18

∼ r20 |f (Q)|2(

∼ r20S(Z ,Q)

ZHe = 2 ZAr = 18

∼ r20 |f (Q)|2(

∼ r20S(Z ,Q)

ZHe = 2 ZAr = 18

∼ r20 |f (Q)|2(

∼ r20S(Z ,Q)

ZHe = 2 ZAr = 18

∼ r20 |f (Q)|2(

∼ r20S(Z ,Q)

ZHe = 2 ZAr = 18

∼ r20 |f (Q)|2(

∼ r20S(Z ,Q)

ZHe = 2 ZAr = 18

Scattering from molecules

From the atomic form factor, wewould like to abstract to the nextlevel of complexity, a molecule (wewill leave crystals for Chapter 5).

Fmol(~Q) =∑j

fj(~Q)e i~Q·~r

As an example take the CF4

molecule

We have the following relation-ships:

= OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB

Fmol(~Q) =∑j

fj(~Q)e i~Q·~r

molecule

= OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB

Fmol(~Q) =∑j

fj(~Q)e i~Q·~r

molecule

= OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB

Fmol(~Q) =∑j

fj(~Q)e i~Q·~r

molecule

= OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB

Fmol(~Q) =∑j

fj(~Q)e i~Q·~r

molecule

= OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB

Fmol(~Q) =∑j

fj(~Q)e i~Q·~r

molecule

OA = OB

= OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB

Fmol(~Q) =∑j

fj(~Q)e i~Q·~r

molecule

OA = OB = OC

= OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB

Fmol(~Q) =∑j

fj(~Q)e i~Q·~r

molecule

OA = OB = OC = OD

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB

Fmol(~Q) =∑j

fj(~Q)e i~Q·~r

molecule

OA = OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB

Fmol(~Q) =∑j

fj(~Q)e i~Q·~r

molecule

OA = OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u

= −z= OA · OB

Fmol(~Q) =∑j

fj(~Q)e i~Q·~r

molecule

OA = OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z

= OA · OB

Fmol(~Q) =∑j

fj(~Q)e i~Q·~r

molecule

OA = OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB

Fmol(~Q) =∑j

fj(~Q)e i~Q·~r

molecule

OA = OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB

Fmol(~Q) =∑j

fj(~Q)e i~Q·~r

molecule

OA = OB = OC = OD = 1

OA = OO ′ + O ′A

OB = OO ′ + O ′B

OA · OD = 1 · 1 · cos u = −z= OA · OB

− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

3u = cos−1(−z) = 109.5

but from the triangle OO ′A

(O ′A)2 = 1− z2

Fmol± = f C (Q) + f F (Q)

[3e∓iQR/3 + e±iQR

]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B

= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

3u = cos−1(−z) = 109.5

(O ′A)2 = 1− z2

]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

3u = cos−1(−z) = 109.5

(O ′A)2 = 1− z2

]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

3u = cos−1(−z) = 109.5

(O ′A)2 = 1− z2

]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

3u = cos−1(−z) = 109.5

(O ′A)2 = 1− z2

]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

3u = cos−1(−z) = 109.5

(O ′A)2 = 1− z2

]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

3u = cos−1(−z) = 109.5

(O ′A)2 = 1− z2

]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

3u = cos−1(−z) = 109.5

(O ′A)2 = 1− z2

]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

u = cos−1(−z) = 109.5

(O ′A)2 = 1− z2

]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

3u = cos−1(−z) = 109.5

(O ′A)2 = 1− z2

]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

3u = cos−1(−z) = 109.5

(O ′A)2 = 1− z2

|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f Fsin(QR)

QR+ 12|f F |2

sin(Q√

− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

3u = cos−1(−z) = 109.5

(O ′A)2 = 1− z2

|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f Fsin(QR)

QR+ 12|f F |2

sin(Q√

− z = (OO ′ + O ′A) · (OO ′ + O ′B)

= z2 + 0 + 0 + O ′A · O ′B= z2 + (O ′A)2 cos(120)

= z2 + (1− z2) cos(120)

= z2 − 1

2(1− z2)

0 = 3z2 + 2z − 1

3u = cos−1(−z) = 109.5

(O ′A)2 = 1− z2

]|Fmol |2 = |f C |2 + 4|f F |2 + 8f C f F

sin(QR)

QR+ 12|f F |2

sin(Q√

8/3R)C. Segre (IIT) PHYS 570 - Fall 2013 October 02, 2013 5 / 11

The Radial Distribution Function

Ordered 2D crystal Amorphous solid or liquid

Total scattered intensity

I (~Q) = f (~Q)2∑n

e i~Q·~rn

e−i~Q· ~rm

= f (~Q)2∑n

e i~Q·(~rn− ~rm)

= Nf (~Q)2 + f (~Q)2∑n

∑m 6=n

e i~Q·(~rn− ~rm)

The sum over m 6= m is now replaced with an integral of the continuousatomic pair density function, ρn(~rnm

I (~Q) = Nf (~Q)2 + f (~Q)2∑n

∫Vρn(~rnm)e i

~Q·(~rn− ~rm) dVm

= Nf (~Q)2 + f (~Q)2∑n

[ρn(~rnm)− ρat ]e i~Q·(~rn− ~rm) dVm

+ f (~Q)2ρat∑n

∫Ve i

= I SRO(~Q)

+ I SAXS(~Q)

I (~Q) = f (~Q)2∑n

e i~Q·~rn

e−i~Q· ~rm = f (~Q)2

e i~Q·(~rn− ~rm)

= Nf (~Q)2 + f (~Q)2∑n

∑m 6=n

e i~Q·(~rn− ~rm)

I (~Q) = Nf (~Q)2 + f (~Q)2∑n

∫Vρn(~rnm)e i

= Nf (~Q)2 + f (~Q)2∑n

+ f (~Q)2ρat∑n

∫Ve i

= I SRO(~Q)

+ I SAXS(~Q)

I (~Q) = f (~Q)2∑n

e i~Q·~rn

e−i~Q· ~rm = f (~Q)2

e i~Q·(~rn− ~rm)

= Nf (~Q)2 + f (~Q)2∑n

∑m 6=n

e i~Q·(~rn− ~rm)

I (~Q) = Nf (~Q)2 + f (~Q)2∑n

∫Vρn(~rnm)e i

= Nf (~Q)2 + f (~Q)2∑n

+ f (~Q)2ρat∑n

∫Ve i

= I SRO(~Q)

+ I SAXS(~Q)

I (~Q) = f (~Q)2∑n

e i~Q·~rn

e−i~Q· ~rm = f (~Q)2

e i~Q·(~rn− ~rm)

= Nf (~Q)2 + f (~Q)2∑n

∑m 6=n

e i~Q·(~rn− ~rm)

I (~Q) = Nf (~Q)2 + f (~Q)2∑n

∫Vρn(~rnm)e i

= Nf (~Q)2 + f (~Q)2∑n

+ f (~Q)2ρat∑n

∫Ve i

= I SRO(~Q)

+ I SAXS(~Q)

I (~Q) = f (~Q)2∑n

e i~Q·~rn

e−i~Q· ~rm = f (~Q)2

e i~Q·(~rn− ~rm)

= Nf (~Q)2 + f (~Q)2∑n

∑m 6=n

e i~Q·(~rn− ~rm)

I (~Q) = Nf (~Q)2 + f (~Q)2∑n

∫Vρn(~rnm)e i

= Nf (~Q)2 + f (~Q)2∑n

+ f (~Q)2ρat∑n

∫Ve i

= I SRO(~Q)

+ I SAXS(~Q)

I (~Q) = f (~Q)2∑n

e i~Q·~rn

e−i~Q· ~rm = f (~Q)2

e i~Q·(~rn− ~rm)

= Nf (~Q)2 + f (~Q)2∑n

∑m 6=n

e i~Q·(~rn− ~rm)

I (~Q) = Nf (~Q)2 + f (~Q)2∑n

∫Vρn(~rnm)e i

= Nf (~Q)2 + f (~Q)2∑n

+ f (~Q)2ρat∑n

∫Ve i

= I SRO(~Q)

+ I SAXS(~Q)

I (~Q) = f (~Q)2∑n

e i~Q·~rn

e−i~Q· ~rm = f (~Q)2

e i~Q·(~rn− ~rm)

= Nf (~Q)2 + f (~Q)2∑n

∑m 6=n

e i~Q·(~rn− ~rm)

I (~Q) = Nf (~Q)2 + f (~Q)2∑n

∫Vρn(~rnm)e i

= Nf (~Q)2 + f (~Q)2∑n

+ f (~Q)2ρat∑n

∫Ve i

= I SRO(~Q) + I SAXS(~Q)

Liquid Scattering

For the moment, let us ignore the SAXS term and focus on the shortrange order term.

I SRO(~Q) = Nf (~Q)2 + f (~Q)2∑n

[ρn(~rnm)− ρat ] e i~Q·(~rn−~rm)dV

When we average over all choices of origin in the liquid, 〈ρn(~rnm)〉 → ρ(~r)and the sum simplifies to N giving:

I SRO(~Q) = Nf (~Q)2 + Nf (~Q)2

[ρ(~r)− ρat ] e i~Q·~rdV

Performing an orientational average results in

I SRO(~Q) = Nf (~Q)2 + Nf (~Q)2

∫ ∞0

4πr2 [ρ(~r)− ρat ]sinQr

Liquid Scattering

I SRO(~Q) = Nf (~Q)2 + f (~Q)2∑n

I SRO(~Q) = Nf (~Q)2 + Nf (~Q)2

∫ ∞0

Liquid Scattering

I SRO(~Q) = Nf (~Q)2 + f (~Q)2∑n

I SRO(~Q) = Nf (~Q)2 + Nf (~Q)2

∫ ∞0

Liquid Scattering

I SRO(~Q) = Nf (~Q)2 + f (~Q)2∑n

I SRO(~Q) = Nf (~Q)2 + Nf (~Q)2

∫ ∞0

Liquid Scattering

I SRO(~Q) = Nf (~Q)2 + f (~Q)2∑n

I SRO(~Q) = Nf (~Q)2 + Nf (~Q)2

∫ ∞0

Liquid Scattering

I SRO(~Q) = Nf (~Q)2 + f (~Q)2∑n

I SRO(~Q) = Nf (~Q)2 + Nf (~Q)2

∫ ∞0

S(Q) - The Liquid Structure Factor

A bit of rearrangment results in the expression for the liquid structurefactor, S(Q).

S(Q) =I SRO(~Q)

Nf (Q)2= 1 +

∫ ∞0

r [ρ(~r)− ρat ] sin(Qr)dr

When Q → ∞, the short wave-length limit, 1/Q → 0 eliminatesall dependence on the interparticlecorrelations and S(Q)→ 1.

When Q → 0, i.e. the long wave-length limit, sin(Qr)/Q → r andS(Q) is dominated by the densityfluctuations in the system

We can rewrite the structure factor equation

Q [S(Q)− 1] =

∫ ∞0

4πr [ρ(~r)− ρat ] sin(Qr)dr =

∫ ∞0H(r) sin(Qr)dr

Which is the sine Fourier Transform of the deviation of the atomic densityfrom its average, H(r) = 4πr [g(r)− 1]

S(Q) =I SRO(~Q)

Nf (Q)2= 1 +

∫ ∞0

Q [S(Q)− 1] =

∫ ∞0

S(Q) =I SRO(~Q)

Nf (Q)2= 1 +

∫ ∞0

Q [S(Q)− 1] =

∫ ∞0

S(Q) =I SRO(~Q)

Nf (Q)2= 1 +

∫ ∞0

Q [S(Q)− 1] =

∫ ∞0

S(Q) =I SRO(~Q)

Nf (Q)2= 1 +

∫ ∞0

Q [S(Q)− 1] =

∫ ∞0

S(Q) =I SRO(~Q)

Nf (Q)2= 1 +

∫ ∞0

Q [S(Q)− 1] =

∫ ∞0

4πr [ρ(~r)− ρat ] sin(Qr)dr

S(Q) =I SRO(~Q)

Nf (Q)2= 1 +

∫ ∞0

Q [S(Q)− 1] =

∫ ∞0

S(Q) =I SRO(~Q)

Nf (Q)2= 1 +

∫ ∞0

Q [S(Q)− 1] =

∫ ∞0

S(Q) =I SRO(~Q)

Nf (Q)2= 1 +

∫ ∞0

Q [S(Q)− 1] =

∫ ∞0

Radial Distribution Function

We can invert the Fourier Transform to obtain

H(r) =2

∫ ∞0

Q [S(Q)− 1] sin(Qr)dQ

and thus the radial distribution fuction can be obtained from the structurefactor (an experimentally measureable quantity).

g(r) = 1 +1

2π2rρat

∫ ∞0

This formalism holds for both non-crystalline solids and liquids, eventhough inelastic scattering dominates in the latter.

The relation between radial distribution function and structure factor canbe extended to multi-component systems where g(r) → gij(r) andS(Q) → Sij(Q).

H(r) =2

∫ ∞0

g(r) = 1 +1

2π2rρat

∫ ∞0

H(r) =2

∫ ∞0

g(r) = 1 +1

2π2rρat

∫ ∞0

H(r) =2

∫ ∞0

g(r) = 1 +1

2π2rρat

∫ ∞0

H(r) =2

∫ ∞0

g(r) = 1 +1

2π2rρat

∫ ∞0

H(r) =2

∫ ∞0

g(r) = 1 +1

2π2rρat

∫ ∞0

Structure in Supercooled Liquid Metals

Measurement of the liquid structure factor of molten metals have shownthat there is short range order which leads to the phenomenon ofsupercooling.

“Difference in Icosahedral Short-Range Order in Early and Late Transition Metal Liquids”,

G.W. Lee et al. Phys. Rev. Lett 93, 037802 (2004).

This indicates thepresence of icosahedralclusters which inhibitcrystallization.

today’s outline - october 02, 2013segre/phys570/13f/lecture_13.pdf · 2013. 10. 3. · today’s...

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