today’s outline - february 15, 2018csrri.iit.edu/~segre/phys570/18s/lecture_12.pdf · c. segre...

Today’s Outline - February 15, 2018

• Scattering from two electrons

• Scattering from atoms

• Scattering from molecules

• Radial distribution function

• Liquid scattering

Homework Assignment #03:Chapter 3:1,3,4,6,8due Thursday, February 22, 2018

C. Segre (IIT) PHYS 570 - Spring 2018 February 15, 2018 1 / 22

Scattering from two electrons

Consider systems where there is only weak scattering, with no multiplescattering effects. We begin with the scattering of x-rays from twoelectrons.

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

The scattering from the second electron willhave a phase shift of φ = ~Q ·~r .

A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

= 2r20

(1 + cos(~Q ·~r)

~Q = (~k − ~k ′)

|~Q| = 2k sin θ =4π

λsin θ

A(~Q) = −r0(

1 + e i~Q·~r)

I (~Q) = A(~Q)∗A(~Q)

(1 + e i

~Q·~r)(

1 + e−i~Q·~r)

I (~Q) = r20

(1 + e i

~Q·~r + e−i~Q·~r + 1

)= 2r20

(1 + cos(~Q ·~r)

)C. Segre (IIT) PHYS 570 - Spring 2018 February 15, 2018 2 / 22

Scattering from many electrons

for many electrons

generalizing to a crystal

A(~Q) = −r0∑j

e i~Q·~rj

A(~Q) = −r0∑N

e i~Q· ~RN

e i~Q·~rj

Since experiments measure I ∝ A2, the phase information is lost. This is aproblem if we don’t know the specific orientation of the scattering systemrelative to the x-ray beam.

We will now look at the consequences of this orientation and generalize tomore than two electrons

for many electrons

A(~Q) = −r0∑j

e i~Q·~rj

A(~Q) = −r0∑N

e i~Q· ~RN

e i~Q·~rj

for many electrons

A(~Q) = −r0∑j

e i~Q·~rj

A(~Q) = −r0∑N

e i~Q· ~RN

e i~Q·~rj

for many electrons

A(~Q) = −r0∑j

e i~Q·~rj

A(~Q) = −r0∑N

e i~Q· ~RN

e i~Q·~rj

for many electrons

A(~Q) = −r0∑j

e i~Q·~rj

A(~Q) = −r0∑N

e i~Q· ~RN

e i~Q·~rj

for many electrons

A(~Q) = −r0∑j

e i~Q·~rj

A(~Q) = −r0∑N

e i~Q· ~RN

e i~Q·~rj

Two electrons — fixed orientation

The expression

I (~Q) = 2r20

(1 + cos(~Q ·~r)

)assumes that the two electronshave a specific, fixed orienta-tion. In this case the intensityas a function of Q is.

Fixed orientation is not theusual case, particularly for solu-tion and small-angle scattering. 0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

The expression

I (~Q) = 2r20

(1 + cos(~Q ·~r)

Fixed orientation is not theusual case, particularly for solu-tion and small-angle scattering.

0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

The expression

I (~Q) = 2r20

(1 + cos(~Q ·~r)

Fixed orientation is not theusual case, particularly for solu-tion and small-angle scattering. 0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

Orientation averaging

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

∫e iQr cos θ sin θdθdφ∫

sin θdθdφ

4π2π

∫ π

0e iQr cos θ sin θdθ

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

Consider scattering from twoarbitrary electron distribu-tions, f1 and f2. A(~Q), isgiven by

and the intensity, I (~Q), is

if the distance between thescatterers,~r , remains constant(no vibrations) but is allowedto orient randomly in spaceand we take ~Q along the z-axis

substituting x = iQr cos θ anddx = −iQr sin θdθ⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r

⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

if the distance between thescatterers,~r , remains constant(no vibrations) but is allowedto orient randomly in space

and we take ~Q along the z-axis

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

if the distance between thescatterers,~r , remains constant(no vibrations) but is allowedto orient randomly in space

and we take ~Q along the z-axis

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

substituting x = iQr cos θ anddx = −iQr sin θdθ

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

=sin(Qr)

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

A(~Q) = f1 + f2ei ~Q·~r

I (~Q) = f 21 + f 22 + f1f2ei ~Q·~r + f1f2e

−i ~Q·~r⟨I (~Q)

⟩= f 21 + f 22 + 2f1f2

⟨e i~Q·~r⟩

sin θdθdφ

4π2π

∫ π

(− 1

)∫ −iQr

iQrexdx

sin(Qr)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Randomly oriented electrons

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

Recall that when we had a fixedorientation of the two electrons,we had and intensity variationI (~Q) = 2r20 (1 + cos(Qr)).

When we now replace the twoarbitrary scattering distributionswith electrons (f1, f2 → −r0),we change the intensity profilesignificantly.⟨I (~Q)

⟩= 2r20

sin(Qr)

)0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

⟩= 2r20

sin(Qr)

)0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

⟩= 2r20

sin(Qr)

0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

When we now replace the twoarbitrary scattering distributionswith electrons (f1, f2 → −r0),we change the intensity profilesignificantly.

⟨I (~Q)

⟩= 2r20

sin(Qr)

0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

⟩= 2r20

sin(Qr)

)0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

⟨I (~Q)

⟩= f 21 +f 22 +2f1f2

sin(Qr)

⟩= 2r20

sin(Qr)

)0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

Scattering from atoms

Single electrons are a good first example but a real system involvesscattering from atoms. We can use what we have already used to write anexpression for the scattering from an atom, ignoring the anomalous terms.

with limits in units of r0 of

The second limit can be understoodclassically as loss of phase coher-ence when Q is very large and asmall difference in position resultsin an arbitrary change in phase.

ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs

where zs is a screening correction

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

Z for Q→ 0

0 for Q→∞

However, it is better to usequantum mechanics to calculatethe form factor, starting with ahydrogen-like atom as an example.The wave function of the 1s elec-trons is just

ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

Z for Q→ 0

0 for Q→∞

ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

Z for Q→ 0

0 for Q→∞

ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

Z for Q→ 0

0 for Q→∞

ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

Z for Q→ 0

0 for Q→∞

ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

Z for Q→ 0

0 for Q→∞

ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

Z for Q→ 0

0 for Q→∞

ψ1s(r) =1√πa3

e−r/a, a =a0

Z − zs

f 0(~Q) =

∫ρ(~r)e i

~Q·~rd~r

Z for Q→ 0

0 for Q→∞

Hydrogen form factor calculation

Since ρ(r) = |ψ1s(r)|2, the form factor integral becomes

f 01s(~Q) =1

∫e−2r/ae i

~Q·~r r2 sin θdrdθdφ

the integral in φ gives 2π and if we choose ~Q to be along the z direction,~Q ·~r → Qr cos θ, so

f 01s(~Q) =1

∫ ∞0

2πr2e−2r/a∫ π

0e iQr cos θ sin θdθdr

∫ ∞0

2πr2e−2r/a1

[−e iQr cos θ

∣∣∣π0dr

∫ ∞0

2πr2e−2r/a1

[e iQr − e−iQr

∫ ∞0

2πr2e−2r/a2 sin(Qr)

f 01s(~Q) =1

∫e−2r/ae i

f 01s(~Q) =1

∫ ∞0

2πr2e−2r/a∫ π

∫ ∞0

2πr2e−2r/a1

[−e iQr cos θ

∣∣∣π0dr

∫ ∞0

2πr2e−2r/a1

[e iQr − e−iQr

∫ ∞0

f 01s(~Q) =1

∫e−2r/ae i

f 01s(~Q) =1

∫ ∞0

2πr2e−2r/a∫ π

∫ ∞0

2πr2e−2r/a1

[−e iQr cos θ

∣∣∣π0dr

∫ ∞0

2πr2e−2r/a1

[e iQr − e−iQr

∫ ∞0

f 01s(~Q) =1

∫e−2r/ae i

f 01s(~Q) =1

∫ ∞0

2πr2e−2r/a∫ π

∫ ∞0

2πr2e−2r/a1

[−e iQr cos θ

∣∣∣π0dr

∫ ∞0

2πr2e−2r/a1

[e iQr − e−iQr

∫ ∞0

f 01s(~Q) =1

∫e−2r/ae i

f 01s(~Q) =1

∫ ∞0

2πr2e−2r/a∫ π

∫ ∞0

2πr2e−2r/a1

[−e iQr cos θ

∣∣∣π0dr

∫ ∞0

2πr2e−2r/a1

[e iQr − e−iQr

∫ ∞0

f 01s(~Q) =1

∫e−2r/ae i

f 01s(~Q) =1

∫ ∞0

2πr2e−2r/a∫ π

∫ ∞0

2πr2e−2r/a1

[−e iQr cos θ

∣∣∣π0dr

∫ ∞0

2πr2e−2r/a1

[e iQr − e−iQr

∫ ∞0

f 01s(~Q) =1

∫e−2r/ae i

f 01s(~Q) =1

∫ ∞0

2πr2e−2r/a∫ π

∫ ∞0

2πr2e−2r/a1

[−e iQr cos θ

∣∣∣π0dr

∫ ∞0

2πr2e−2r/a1

[e iQr − e−iQr

∫ ∞0

Form factor calculation

f 01s(~Q) =1

∫ ∞0

If we write sin(Qr) = Im[e iQr

]then the integral becomes

f 01s(~Q) =4

∫ ∞0

re−2r/aIm[e iQr

[∫ ∞0

re−2r/ae iQrdr

this can be integrated by parts with

u = r dv = e−r(2/a−iQ)dr

du = dr v = −e−r(2/a−iQ)

(2/a− iQ)

f 01s(~Q) =4

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

f 01s(~Q) =1

∫ ∞0

f 01s(~Q) =4

∫ ∞0

re−2r/aIm[e iQr

[∫ ∞0

re−2r/ae iQrdr

(2/a− iQ)

f 01s(~Q) =4

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

f 01s(~Q) =1

∫ ∞0

f 01s(~Q) =4

∫ ∞0

re−2r/aIm[e iQr

[∫ ∞0

re−2r/ae iQrdr

]this can be integrated by parts with

(2/a− iQ)

f 01s(~Q) =4

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

f 01s(~Q) =1

∫ ∞0

f 01s(~Q) =4

∫ ∞0

re−2r/aIm[e iQr

[∫ ∞0

re−2r/ae iQrdr

(2/a− iQ)

f 01s(~Q) =4

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

f 01s(~Q) =1

∫ ∞0

f 01s(~Q) =4

∫ ∞0

re−2r/aIm[e iQr

[∫ ∞0

re−2r/ae iQrdr

(2/a− iQ)

f 01s(~Q) =4

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

f 01s(~Q) =1

∫ ∞0

f 01s(~Q) =4

∫ ∞0

re−2r/aIm[e iQr

[∫ ∞0

re−2r/ae iQrdr

(2/a− iQ)

f 01s(~Q) =4

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

f 01s(~Q) =1

∫ ∞0

f 01s(~Q) =4

∫ ∞0

re−2r/aIm[e iQr

[∫ ∞0

re−2r/ae iQrdr

(2/a− iQ)

f 01s(~Q) =4

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

]C. Segre (IIT) PHYS 570 - Spring 2018 February 15, 2018 9 / 22

f 01s(~Q) =4

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

the first term becomes zero because re−r(2/a−iQ) → 0 as r → 0, thesecond is easily integrated

f 01s(~Q) =4

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

(2/a− iQ)2

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

f 01s(~Q)

(a/2)4(4Q/a)

[1 + (Qa/2)2]2=

[1 + (Qa/2)2]2

f 01s(~Q) =4

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

]the first term becomes zero because re−r(2/a−iQ) → 0 as r → 0, thesecond is easily integrated

f 01s(~Q) =4

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

(2/a− iQ)2

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

f 01s(~Q)

(a/2)4(4Q/a)

[1 + (Qa/2)2]2=

[1 + (Qa/2)2]2

f 01s(~Q) =4

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

f 01s(~Q) =4

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

(2/a− iQ)2

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

f 01s(~Q)

(a/2)4(4Q/a)

[1 + (Qa/2)2]2=

[1 + (Qa/2)2]2

f 01s(~Q) =4

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

f 01s(~Q) =4

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

(2/a− iQ)2

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

f 01s(~Q)

(a/2)4(4Q/a)

[1 + (Qa/2)2]2=

[1 + (Qa/2)2]2

f 01s(~Q) =4

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

f 01s(~Q) =4

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

(2/a− iQ)2

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

f 01s(~Q)

(a/2)4(4Q/a)

[1 + (Qa/2)2]2=

[1 + (Qa/2)2]2

f 01s(~Q) =4

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

f 01s(~Q) =4

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

(2/a− iQ)2

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

f 01s(~Q)

(a/2)4(4Q/a)

[1 + (Qa/2)2]2=

[1 + (Qa/2)2]2

f 01s(~Q) =4

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

f 01s(~Q) =4

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

(2/a− iQ)2

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

f 01s(~Q)

(a/2)4(4Q/a)

[1 + (Qa/2)2]2

f 01s(~Q) =4

[re−r(2/a−iQ)

(2/a− iQ)

∣∣∣∣∣∞

∫ ∞0

e−r(2/a−iQ)

(2/a− iQ)dr

f 01s(~Q) =4

[− e−r(2/a−iQ)

(2/a− iQ)2

∣∣∣∣∣∞

(2/a− iQ)2

[(2/a + iQ)2

(2/a− iQ)2(2/a + iQ)2

[(2/a)2 + i(4Q/a)− Q2

[(2/a)2 + Q2]2

]f 01s(~Q) =

(a/2)4(4Q/a)

[1 + (Qa/2)2]2=

[1 + (Qa/2)2]2

1s and atomic form factors

f 01s(~Q) =1

[1 + (Qa/2)2]2

This partial form factor will varywith Z due to the Coulomb inter-action

In principle, one can compute thefull atomic form factors, however,it is more useful to tabulate the ex-perimentally measured form factors 0 5

f 0 (Q) =4∑

aje−bj sin2 θ/λ2 + c =

4∑j=1

aje−bj (Q/4π)2 + c

f 01s(~Q) =1

[1 + (Qa/2)2]2

f 0 (Q) =4∑

4∑j=1

f 01s(~Q) =1

[1 + (Qa/2)2]2

In principle, one can compute thefull atomic form factors, however,it is more useful to tabulate the ex-perimentally measured form factors

0 1 2 3 4 5

f0 1s(Q

f 0 (Q) =4∑

4∑j=1

f 01s(~Q) =1

[1 + (Qa/2)2]2

In principle, one can compute thefull atomic form factors, however,it is more useful to tabulate the ex-perimentally measured form factors 0 1 2 3 4 5

f0 1s(Q

f 0 (Q) =4∑

4∑j=1

f 01s(~Q) =1

[1 + (Qa/2)2]2

f0 1s(Q

f 0 (Q) =4∑

aje−bj sin2 θ/λ2 + c

f 01s(~Q) =1

[1 + (Qa/2)2]2

f0 1s(Q

f 0 (Q) =4∑

4∑j=1

f 01s(~Q) =1

[1 + (Qa/2)2]2

f 0 (Q) =4∑

4∑j=1

Two hydrogen atoms

Previously we derived the scat-tering intensity from two local-ized electrons both fixed andrandomly oriented to the x-rays

when we now replace the two lo-calized electrons with hydrogenatoms, we have, for fixed atoms

and if we allow the hydrogenatoms to be randomly orientedwe have

with no oscillating structure inthe form factor

0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

Two hydrogen atoms

0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

Two hydrogen atoms

0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

Two hydrogen atoms

0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

Two hydrogen atoms

0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

Two hydrogen atoms

0 0.5 1 1.5 2

Q (units of 2π/r)

nsity (

units o

f r2 o

Inelastic scattering

The form factors for all atoms dropto zero as Q →∞, however, otherprocesses continue 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 theground state 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 scatter-ing from an atom as the sum of twocomponents:

∼ r20 |f (Q)|2(dσ

∼ r20S(Z ,Q)

recall that f (Q)→ Z as Q → 0

so |f (Q)|2 → Z 2 as Q → 0

and for incoherent scattering we ex-pect S(Z ,Q)→ Z as Q →∞

ZHe = 2 ZAr = 18 0 5 10 15 20

Q [Å-1

f r02 )

We can now write the total scatter-ing from an atom as the sum of twocomponents:(

∼ r20 |f (Q)|2

∼ r20S(Z ,Q)

so |f (Q)|2 → Z 2 as Q → 0

ZHe = 2 ZAr = 18 0 5 10 15 20

Q [Å-1

f r02 )

∼ r20 |f (Q)|2(dσ

∼ r20S(Z ,Q)

so |f (Q)|2 → Z 2 as Q → 0

ZHe = 2 ZAr = 18 0 5 10 15 20

Q [Å-1

f r02 )

∼ r20 |f (Q)|2(dσ

∼ r20S(Z ,Q)

so |f (Q)|2 → Z 2 as Q → 0

ZHe = 2 ZAr = 18 0 5 10 15 20

Q [Å-1

f r02 )

∼ r20 |f (Q)|2(dσ

∼ r20S(Z ,Q)

so |f (Q)|2 → Z 2 as Q → 0

ZHe = 2 ZAr = 18 0 5 10 15 20

Q [Å-1

f r02 )

∼ r20 |f (Q)|2(dσ

∼ r20S(Z ,Q)

so |f (Q)|2 → Z 2 as Q → 0

ZHe = 2 ZAr = 18 0 5 10 15 20

Q [Å-1

f r02 )

∼ r20 |f (Q)|2(dσ

∼ r20S(Z ,Q)

so |f (Q)|2 → Z 2 as Q → 0

ZHe = 2 ZAr = 18

0 5 10 15 20

Q [Å-1

f r02 )

∼ r20 |f (Q)|2(dσ

∼ r20S(Z ,Q)

so |f (Q)|2 → Z 2 as Q → 0

ZHe = 2 ZAr = 18 0 5 10 15 20

Q [Å-1

f r02 )

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

The CF4 scattering factor

− 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

− 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

(O ′A)2 = 1− z2

− 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

(O ′A)2 = 1− z2

− 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

− 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

− 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

− 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

− 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

− 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

− 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

(O ′A)2 = 1− z2

− 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

(O ′A)2 = 1− z2

]C. Segre (IIT) PHYS 570 - Spring 2018 February 15, 2018 16 / 22

Fmol± = f C (Q)+f F (Q)

[3e∓iQR/3+e±iQR

|Fmol± |2 = |f C |2+10|f F |2+6|f F |2 cos(QR/3)+2f C f F [3 cos(QR/3)+cos(QR)]

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

QR+12|f F |2

sin(Q√

The plot shows the structure factorof CF4, its orientationally averagedstructure factor, and the form fac-tor factor of Mo which has the samenumber of electrons as CF4

The logarithmic plot shows thespherically averaged structure fac-tor compared to the inelastic scat-tering for CF4

[3e∓iQR/3+e±iQR

]|Fmol± |2 = |f C |2+10|f F |2+6|f F |2 cos(QR/3)+2f C f F [3 cos(QR/3)+cos(QR)]

QR+12|f F |2

sin(Q√

The plot shows the structure factorof CF4,

its orientationally averagedstructure factor, and the form fac-tor factor of Mo which has the samenumber of electrons as CF4

[3e∓iQR/3+e±iQR

QR+12|f F |2

sin(Q√

The plot shows the structure factorof CF4, its orientationally averagedstructure factor,

and the form fac-tor factor of Mo which has the samenumber of electrons as CF4

[3e∓iQR/3+e±iQR

QR+12|f F |2

sin(Q√

[3e∓iQR/3+e±iQR

QR+12|f F |2

sin(Q√

[3e∓iQR/3+e±iQR

QR+12|f F |2

sin(Q√

The radial distribution function

Ordered 2D crystal Amorphous solid or liquid

Take a circle (sphere) of radius r and thickness dr and count the numberof atom centers lying within the ring. Then expand the ring radius by drto map out the radial distribution function g(r)

Take a circle (sphere) of radius r and thickness dr and count the numberof atom centers lying within the ring.

Then expand the ring radius by drto map out the radial distribution function g(r)

Total scattered intensity

Consider a mono-atomic (-molecular) system where the total scatteredintensity is given by

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)

The sum over m 6= m is now replaced with an integral of the continuousatomic pair density function, ρn(~rnm) and adding and subtracting theaverage atomic density ρat

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~Q·(~rn− ~rm) dVm = I SRO(~Q)

+ I SAXS(~Q)

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)

I (~Q) = Nf (~Q)2 + f (~Q)2∑n

∫Vρn(~rnm)e i

= Nf (~Q)2 + f (~Q)2∑n

+ f (~Q)2ρat∑n

+ 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

+ 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

+ 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)

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

and adding and subtracting theaverage atomic density ρat

I (~Q) = Nf (~Q)2 + f (~Q)2∑n

∫Vρn(~rnm)e i

= Nf (~Q)2 + f (~Q)2∑n

+ f (~Q)2ρat∑n

+ 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~Q·(~rn− ~rm) dVm

= 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

+ 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~Q·(~rn− ~rm) dVm = 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∫V

[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∫V

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∫V

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∫V

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∫V

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∫V

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 function can be obtained from thestructure factor (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

today’s outline - february 15, 2018csrri.iit.edu/~segre/phys570/18s/lecture_12.pdf · c. segre...

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