Luiz
Leal
Oak
Ridge
National
Laboratory
Lectures
Presented
at
the
Nuclear
Engineering
Department
of
the
Massachusetts
Institute
of
Technology
(MIT)
Courtesy of Luiz Leal, Oak Ridge National Laboratory. Used with permission.
Thermal Neutron Scattering
Kernel Development
The
procedure
consists
of
finding
the
amplitude
of
the
scattering
wave
function
that
leads
to
the
differential
scattering
cross
section.
In
doing
so,
a
few
things
will
be
assumed:
(1) First
Born
approximation;
(2) Fermi
Pseudo‐potential;
(3) Time‐dependent
Schrödinger
equation
(TDSE)
The
first
two
assumptions
have
already
been
discussed.
Why
assume
the
TDSE
is
explained
as
follows:
the
energy
dependency
(explicitly)
in
the
Schrödinger
equation
implies
that
the
eigenstates
of
the
scattering
system
are
known,
i.e.,
initials
and
final
states.
Usually
they
are
not
known
and
even
if
they
are
known
there
will
a
large
number.
It
is
desired
to
“eliminate”
the
explicit
appearance
of
the
eigenstates.
In
so
doing
one
uses
∂ψ
E → i
∂t
or
 2 2

∂ψ ∇ + V ψ = i
−
∂t
 2m

Leo
Van
Hove
championed
this
idea.
The
time
dependence
is
subsequently
used
in
a
Fourier
transformation.
Assuming
that
the
interaction
occurred
at
t = t 0 .
The
wave
function
at
the
detector
at
the
€

position
r for
t > t 0 is
a
solution
of
the
equation
€
€
€


 2 2


∂ψ ( r , t)
∇ + V (r , t)ψ (r ,t) = i
−
∂t €
 2m

€
For
t < t 0 before
the
collision
has
occurred
the
incident‐neutron
wave
function
is


i(k .r − w 0 t )
ψ inc (r , t) = e
where

k → neutron momentum
w 0 → neutron energy

The
solution
at
the
detector
at
(r ,t) is
!
!
!
" ( r ,t) = " inc ( r ,t) + " scat (r ,t) €
Given
that
for
the
incident
wave
from
the
potential
region
€
ψ inc away
 
 2 2
∂ψ ( r ,t)

 ∇ + i
ψ inc (r ,t) = 0
∂t 
2m
then
 
 2 2
∂ψ ( r ,t)



 ∇ + i
ψ scat (r ,t) = V ( r ,t)ψ ( r ,t)
∂t 
2m
€
Approximations:
€
(a)
Born
approximation
In
the
right‐hand
side
of
the
above
equation:


ψ ( r , t ) → ψ inc ( r , t ) (b)
Fermi
Pseudo‐potential

2π 2
 
anδ[r − Rn (t)]
Vn (r , t) =
m
where

Rn → position of nucleus n at time t
an → bound scattering related to the bound cross sec tion as σ b = 4πan2
(an2 =
σb
)
4π
Hence
the
equation
to
be
solved
is
€


 2 2 mi ∂ 

 

a
(
,t)
(r
,
∇
+
ψ
t)
=
4
π
δ
[(r
−
R
(t
)]
ψ
r


∑
scat
n
n 0
inc


 ∂t 
n

€
€
Note
that
Rn (t 0 ) is
for
reactions
that
occurred
at
the
t = t 0 .
Making
use
of
the
Green’s
function
method
to
 
€
solve
the
equation
above.
Let
G(r − r ',t − t 0 ) be
the
Green’s
function,
hence:
 2 2 mi ∂   € '
 
∇ +  ∂t  G(r − r ,t − t 0 ) = 4 πδ (r − r ')δ (t − t 0 ) 
The
solution
for
ψ scat (r, t) is

ψ scat (r , t) =
∫

 

∫ dτ dt0G(r − r ',t − t0 )ψ inc (r ',t0 )∑ anδ[r − Rn (t0 )]
τ t0
€
€
€
€
€
€
n
where
τ is
the
volume.
 
G(r
− r ',t − t 0 ) and
Substituting
the
values
for
ψ inc we
have

m t

3/2
ψ scat ( r , t) = i
dt 0 (t − t 0 ) ∑ an ∫ dτ δ[r − Rn (t 0 )] ×
€
∫
2π −∞
n
 im r − r ' 2 
 
 × exp[i(k 0 .r − w 0 t)]
exp 
 2 (t − t 0 ) 
The
above
expression
is
the
solution
for
the
scattered
 wave
ψ scat at
the
detector
at
the
position r for
the
time
t > t 0 for
an
incident
neutron
of
energy
w 0 and
momentum
k0 .
It
is
not
clear
how
to
obtain
the
scattered
amplitude
from
th
€e
above
equation.
Leon
Van
Hove
came
up
with
a
cle€aver
idea
of
u€sing
a
Fourier
€
transform
of
the
type

 ' iw' t
ψ scat ( r , t) = ∑ f (r , w )e
w'
and
T
 ∫ dt e ψscat (r, t)

1
f (r , w') =
T
€
€
€

f (r, w) iw' t
0
relates
to
the
scattered
wave
amplitude.
After
LOTS
OF
ALGEBRA

1
f (r , w') =
Tr'
T
∫ dt
0
0
e
iw' t 0

 
iκ . r
∑ an ∫ dτδ[r − Rn (t0 )]e
n
For
derivation
of
the
above
equation
see:
Thermal
Neutron
Scattering
by
Engelstaff
pages
49‐52
The
scattered
neutron
has
  
κ = k0 − k → momentum change
w'= w 0 − w → energy change The
scattering
differential
cross
section
is
defined
as
€

d 2σ
2
2 T υ
= r'
f (r ,w)
dEdΩ
h υ0
υ 0 and υ initial and final neutron velocity
€
Read:
The
Elements
of
Neutron
Interaction
Theory
Anthony
Foderaro
page
555

2
f (r ,w) =
1 +∞

 iκ .( r − r ")
− iwτ
*
dτ e ∑ am an ∫ dr " ∫ dr e
×
2 2 ∫
T r' −∞
m,n
 
 
δ[r " −Rn (0)]δ[r − Rn (τ )]
The
bar
indicates
time
averaging.
€
The
double
differential
cross
section
is
finally
obtained
as
d 2σ
1 υ +∞

 iκ .r
−iwτ
*
=
dτ e
am an ∫ dr ' ∫ dr e ×
∑
∫
dEdΩ h υ 0 −∞
m,n
 
  
δ[r " − Rn (0) − r ]δ[r − Rn (τ )]
€
The
Space­Time
Correlation
Function
(Van
Hove)

1
 
  
G(r , τ ) = ∑ ∫ dr'δ[r ' −Rn (0) − r ]δ[r − Rn (τ )]
N m,n

G( r , τ ) is not the Green' s function!!
€
€
€

G(r
,τ ) Interpretation
of
the
Two
parts:

€
for
m = n (diagonal
terms)
Gs ( r , τ ) 
for
m ≠ n (off‐diagonal
terms) Gd (r , τ ) €
€
Interpretation:

Gs (r , τ ) → self − correlation function :
a
second
IDENTICAL
nucleus
is
present

€
€
€
€
€
Gd ( r , τ ) → distinct − correlation function :
second
DISTINCT
nucleus
is
present



G ( r , τ ) = Gs ( r , τ ) + Gd ( r , τ ) where
a
N

1
 
  
Gs (r , τ ) = ∑ ∫ dr'δ[r ' −Rn (0) − r ]δ[r − Rn (τ )]
N n=1
and
N

1
 
  
dr'δ[r ' −Rn (0) − r ]δ[ r − Rn (τ )]
Gd (r , τ ) =
∑
∫
N m≠n=1
Defining
1
< a >= ∑ a*m anδmn
N m,n
2
< a >2 =
d 2σ
d 2σ coh d 2σ incoh
=
+
dEdΩ dEdΩ dEdΩ €
€
€
1
*
a
∑
m an
2
N m≠n
Assuming
only
coherent
inelastic
scattering
present,
i.e.,
< a 2 >=< a > 2 = a d 2σ
d 2σ coh
=
€
dEdΩ dEdΩ d 2σ
a2 υ
=
dEdΩ h υ 0
Recall
that
∫



i(κ .r −w' τ )
G(r , τ )
∫ dr dτ e
€
€
σb
a =
4π
and
2
υ
=
υ0
E
E0
d 2σ
σb
=
dEdΩ 4 π
E ˜ 
S (κ,w')
E0
w h e re


1

i(
κ
˜S (κ, w ') = ∫ ∫ dr dτ e .r −w'τ )G(r, τ )
h
˜S (κ, w') → Scattering Law ˜S (κ, w')
can
be
obtained
directly
from
d 2σ
measurements
of
dEdΩ 
˜
Properties
of
S (κ ,w') (a) Depends
only
on
the
dynamics
of
the
scatter
center
€
(b)
Sum
ruler

κ
˜
∫ w S(κ,w) dw = 2M 2
(c)
Condition
derived
from
the
detailed
balance
˜S (−κ,w−) = e
−
w
KT
˜S (κ,w) €
€
€
Definitions:
β
2
˜S (κ,w) = e S(α, β ) −
Where
 2κ 2
α=
2mAKT
E − E0
β=
KT and
Since
 
  
2
2
2
κ = k0 − k → κ = k0 + k − 2k 0 . k cosθ
 
κ
 (k + k − 2k 0 . k cosθ )
α=
=
2mAKT
2mAKT
E 0 + E − 2 EE 0 cosθ )
α=
AKT
2
2
2
2
0
2
Hence
the
double
differential
cross
section
becomes
2
dσ
σb
=
dEdΩ 4 πKT
β
2
E
e S(α , β )
E0
−
Notations:
d 2σ
dEdΩ
or
σ s (E 0 → E,θ )
or
σ s (E 0 → E, µ)
where
µ = cosθ
€
Simple
Example:
(1) Scatterer
is
a
single
nucleus
of
mass
M.
Only
coherent
scattering
is
accounted
for;
(2)
Scatterer
is
free
and
at
rest
This
corresponds
to
the
situation
dealt
with
in
the
theory
of
neutron
moderation
where
the
chemical
region
effects
are
negligible.
σb
σ s (E 0 → E, µ) =
4π
µ = cosθ
E ˜ 
S (κ ,w)
E0
˜S (κ, w') is
a
delta
function
as
1/ 2

+
−
2
(EE
E
E
)
µ
0
˜S (κ,w) = δ ( E 0 − E + 0
)
A
M
µ = cosθ and A =
m
Recall
that
w → E 0 − E
 2κ 2
E 0 + E − 2(EE 0 )1/ 2 µ
→
2M
A
Also,
dΩ = 2π sin θ dθ
µ = cosθ
dµ = −sin θ dθ
dΩ = 2π (−dµ)
and
σ s (E 0 → E) =
∫ σ (E
s
0
→ E, µ) dΩ
4π
−1
σ s (E 0 → E) = 2π ∫ σ s (E 0 → E, µ) dµ
+1
1/ 2


σ E
σ s (E 0 → E) = 2π b  
4π  E 0 
€
−1
E 0 + E − 2(EE 0 )1/ 2 µ
)dµ
∫ δ (E 0 − E +
A
+1
Change
of
variables
E 0 + E − 2(EE 0 )1/ 2 µ
x = E0 − E +
A
such that
2(EE 0 )1/ 2
dx = −
dµ
A
€
(E  E )
µ = ±1 → x  = E 0 − E +
A
1/ 2
0
1/ 2 2
1/ 2


σb E
A
σ s (E 0 → E) = 2π
 
4 π  E 0  2(E 0 E)1/ 2
σb A
σ s (E 0 → E) =
4 E0
Recall
that
2
 A + 1
σb = 
 σ free
 A 
x+
∫ δ(x)dx
x−
2
( A + 1) 1
σ s (E 0 → E) =
σ free
4 A E0
 A −1
if α = 

 A + 1
hence
2
4A
→ 1− α =
2
(A + 1)
1
σ s (E 0 → E) =
σ free
(1− α )E 0
Energy
2
range?
A + 2Aµ + 1
E=
E0
2
(A + 1)
µ = 1 → E = E0
µ = −1 → E = (1− α )E 0 
1
 (1− α )E σ free
0

σ s (E 0 → E) = 


0

for (1− α )E 0 < E < E 0
otherwise
(Nuclear
Reactor
Analysis
Duderstadt
and
Hamilton
page
44)
€
MIT OpenCourseWare
http://ocw.mit.edu
22.106 Neutron Interactions and Applications
Spring 2010
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