Electromagnetic radiation
mass  acceleration  force=charge  field
Along this line one
does not observe
any acceleration
Here one observes the full acceleration,
but delayed in time by R/c !
m  acc  [e]Ein (t  R / c)  [e]E0e  i (t  R / c )
R
 [ e] E0 e i  t  ei ( / c ) R
 [ e] Ein (t )  ei kR
Erad (t )
ei kR
 r0
;
Ein (t )
R
e2
5
r0 

2.82

10
Ang
2
4 0 mc
1
Radiation from a dipole-antenna
due to the oscillating charge in the antenna
Guess that
Erad ( R ) 
1
R  E 2  1  energy  density  radiated energy  1  4 R 2
rad
R2
R2
Erad  charge q ; Erad ( R, t )  observed
acceleration (t  R / c)
1 e 1
Erad ( R, t )  
 2  acc(t  R / c) ;
1 e 2 acc
R
4

To get the dimension right !
0 c
e  Erad ( R, t ) 
 2 
R 4 0
Force
c
m/sec 2
= Energy
(m/sec) 2
OK
Polarisation
P=cos2(ψ)
P=1
Interference (mathematical)
k’
Q
Phase
0 .
2 .
4.
wavecrest
sc.ampl.  r0 (1  ei Qr )
r
k
sc.ampl.  r0  e
1
i Qr j
j
many
z
l
in 
1,2 … many
2
z
2  k  z  k  r
sc.ampl.  r0   (r ) ei Qr dr
Number density
as drawn #2 is behind #1 for ”in”
l
out 
 k 'r but ahead for ”out” , therefore ” – ”
res  ( k  k ' )  r  Q  r
The dream experiment
A2  f1ei Qr1  f 2ei Qr2
; I 2  A2  A2*
I 2  { f1ei Qr1  f 2ei Qr2 } { f1ei Qr1  f 2ei Qr2 }
 f12  f 22  f1 f 2ei Q(r1 r2 )  f1 f 2ei Q(r1 r2 )
I2
I
orient .average
sin(Q r12 )
 f  f  2 f1 f 2
Q r12
2
1
2
2
  fi  2 fi f j
2
orient .average
i
i j
sin(Q ri j )
Q ri j
1,2 … many
Measuring atomic and molecular formfactors from gas scattering
Intensity 
Q  2k sin q
1
2
 f (Q)
sin 2q
Detector
Viewing
Field
Kr
Gas cell
f atom   el (r)  ei Qr dr
f mol   f j  e
j
i Qr j
Detector
2q
X-ray beam
4
f (Q)   a j e
0
j 1
a=[15.2354 6.70060 4.35910 2.96230];
b=[3.0669 0.241200 10.7805 61.4135];
c=1.71890; % Ga
 b j ( Q / 4 )2
 cj
a=[16.6723 6.0710 3.4313 4.27790];
b=[2.63450 0.2647 12.9479 47.7972];
c=2.531; % As
Q
sin q
d sin q
q
r
e
i Qr
orientational
average
d   d sin q dq
dq
d

Unit sphere
 Qr
e
i Qr
orientational
average
e


i Q r cosq
sin q dq d
 sin q dq d
2 (Qr )

1

x  Qr
2  2
ei x dx

sin(Qr )
Qr
Fourier transform of a Gaussian
f ( x)  Ae
 a2 x2

Fourier transform
: F (q) 

f ( x) ei q x dx

A   q2 /(4 a2 )
F (q) 
e
a
1
 x 2 / 2 2
or with f ( x) 
e
2 
; F ( q)  e
 q 2 2 / 2
Gaussian( x)   ( x) when   0

 f ( x) ( x) dx  f (0)

F.T.Gaussian( x)   ( x)  1 (delocalized in q )
localized in x
Convolution of 2 Gaussians
h( x )   e
H ( q)  e
 x12 / 212
 q212 / 2
e
e
 ( x1 x )2 / 2  22
 q2 22 / 2
i.e. h(x) is also Gaussian with
dx1
e
 q2 2 / 2
 2   12   22
locally a plane wave
  Aei k r
Side View
Top View
ei k r
3D :   A
r

Energy density

2
Ring wave (2D)
or
Spherical wave (3D)
4 r 2 independent of r
Surface area
scattering length
Spherical wave
k’
i kR
A
e
R

Area is
2
R 
I scatt 
 in  ei k x
k
d
 A2
d
Perfect plane
wave
Flux  : c  
A2 2
ce e
R 
2
R
 c  A2  
i kR  i kR
Almost plane
wave when
Defines the
scattering
cross
section
c
1
A point scatterer in the beam
d
Intensity  Flux  Area I scattered  in   
d
thru 
Aperture d
d  l
2
2LL=Nl
No real beam is perfectly monochromatic
P(l)
l
ll
l
wavelength band
l
2LL=(N+1)(ll)
From the 2 equations, derive
1 l2
LL 
2 l
the longitudinal coherence length
No real beam is perfectly collimated
P(q)
source
With R being the distance from
observation point
to
LT 
show from the figure that
l R
q
2D
q
2 LT  q  l
A and B out-of-phase
A
D
B
q
q
LT
The transverse
coherence length
Here
A and B beams in phase
Absorption
 dI  I ( z )  dz   
I ( z)  I 0  e  z
 (atomic number Z ,  )
 1 
 ( Z ,  )  Z 4  
  
3
The experimental setup
0.7 x 0.7 mm
Sample
Scintillator
q < 2 mdeg
X-rays
E = 8-27 keV
20m
Rotation
CCD
Tomography
• Study the bulk
structures, 3D
• Nondestructive
• Small
lengthscales
(350 nm)
Single slice
100 microns
Galathea III
Fra http://www.unge-forskere.dk/
Compton Scattering
Energy and momentum conservation
l' 
  1  lC k (1  cos )
l '
lC 
mc
 3.86 103 Å