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8/8/2019 FEL laser induced heating of the electron gas in metal
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FEL laser induced heating of the electron gas in metal
Agnieszka [my maiden name :-)]
DESY September 04
I wish to thank my supervisor,dr Jacek Krzywinski,
for introducing me to the absolutely supercoolworld of plasma physics.
Abstract
The laser heating of plasma electrons via the inverse Bremsstrahlung process isconsidered. The theoretical model of plasma electrons assumes Fermi distributionfunctions and describes the above mentioned process by a collision integral. Thekinetic equation is derived, and the change in average kinetic energy of electrons isobtained.
1 Introduction
Plasma may absorb a significant part of the laser light by inverse Bremsstrahlung process.During that process, the plasma electron gains energy from the electromagnetic field ofthe laser beam by absorbing photons while colliding the nucleus. We consider the inverseBremsstrahlung absorption of Bremsstrahlung laser radiation. The beam is treated as aclassical plane electromagnetic wave. We consider a perfectly homogeneous and isotropiccrystal. There is no lattice deformation during the collision process. We neglect the energytransport since our sample is a thin film metal. We investigate a very rare gas so that weconsider only binary collisions and particles are completely uncorrelated before the collision(molecular chaos). According to these assumptions we can make use of the standardsemi-classical transport theory with its Boltzmann equation and calculate dynamics of theelectron gas with standard collision functions. The plasma electrons are described by thesolution of the Schrodinger equation for the electron in the electromagnetic field of the laserbeam. The scattering of the electrons by nuclei is treated using first-order perturbationtheory. A Fermi-Dirac distribution function for the electrons is assumed. The transitionprobabilities are used to write the kinetic equation for the electrons. The rate of changeof the kinetic energy is derived for a weak-field case. Asymptotic cases are considered.
1
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where
V(r) =Ze2
rer/a is the Yukawa potential ,
V( k) =4Z e2
1/a2 + k2is the Fourier transform of V(r) ,
= h2
(k22 k
21)
2m,
=eE0m
hk with respect to the z direction,
= tan1
kykx
k2 k1 = k is the exchanged momentum and k1 = k .
(4)
The transition probability per unit time is then
| a( k) |2
t =
2
h |V
k
|
2
n=J
2
n
h
( nh) , (5)
where Jn is the Bessel function of order n.The transition probability per unit time with the absorption or emission of n photons:
T(n, k) =2
h| V
k
|2 J2n
h
( nh) , (6)
The function implies the energy conservation in the electron-photon system. Since thenucleus carry of some momentum during the collision process, the momentum of the systemis not conserved.
The change in the number of the electrons N(k) can be expressed schematically as:
N(k)t
=
n=1
k
+
k+ k h
k
k + k
hk
k
hk + k
k h
k+ k
which means
N(k)
t
=
n=1
k
[T(n, k)N(k)(1 N(k + k)) + T(n, k)N(k1) [1 N(k2)]
T(n, k2 k1)N(k2) [1 N(k1)] T(n, k2 k1)N(k2) [1 N(k1)]
=
n=,n=0
k
[T(n, k)[N(k) N( k + k)] .
(7)
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We express the electron kinetic equation as a change of the number of electrons withthe momentum hk in time:
N(k)
t=
n=,n=0
k
[T(n, k)[N(k) N( k + k)] . (8)
Assuming a Fermi distribution for the electrons, letting the sum over k become anintegral, and using Eq. (6), we expand Eq. (8) to
f(k)
t=
2
h
1
83
|
4Ze2
1/a2 + k2|2
1
1 + e(EFh2k2/2m)/KT
1
1 + e(EFh2(k+k)2/2m)/KT
n=
J2n
h
( nh)d3k .
(9)
The electron-photon collision term for inverse Bremsstrahlung in a metal, describing the
free electrons absorption of the energy of laser light of angular frequency and amplitudeof electric field E0 reads
f(k)
t=
2
h
1
83
|
4Ze2
1/a2 + k2|2
1
1 + e(EFh2k2/2m)/KT
1
1 + e(EFh2(k+k)2/2m)/KT
n=
J2n
eE0hk
2mh2
( nh)d3k .
(10)
3 Average change of the electrons kinetic energyThe energy gain for a given electron in plasma will be found for the case of the weakelectromagnetic field.
For this case
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Using Eq. (9) with only first two terms of the sum of the Bessel retained, we have
f(k)
t=
2
h
1
83
|
4Ze2
1/a2 + k2|2
1
1 + e(EFh2k2/2m)/KT
1
e(EFh2(k+k)2/2m)/KT
2h2
(h2k2
2m+
2h2k k
2m+ h) (
h2k2
2m+
2h2k k
2m h) d
3k
=2
h
1
83
|
4Ze2
1/a2 + k2|2
1
1 + e(EFh2k2/2m)/KT
1
e(EFh2(k+k)2/2m)/KT
eE0hk2mh2
2 (
h2k2
2m+
2h2k k
2m+ h) (
h2k2
2m+
2h2k k
2m h)
d3k .
(12)
Let the stand for the angle between k and k, and z = cos . The unitary vectorn = (nx, ny, nz) points into the same direction as the electric field vector.
k described in spherical coordinates is
k = (k sin cos , k sin cos , k cos ) . (13)
The coordinate of k parallel to the n (the electromagnetic field direction) reads
k = (n k)n . (14)
The perpendicular coordinate is k = k k. The squared modulus of k is
( k k)2 = ( k)2 2 k k + ( k)
2
= ( k)2 2(n k)2 + (n k)2
= (k)2 1 (sin cos nx
+ sin sin ny
+ cos nz)2 .
(15)
In the next step we assign cos as z1 and z2 to appropriate parts of the integral
(k)2(1 (nx
1 z2i cos + ny
1 z2i sin + nzzi)
2)
= (k)2
1 n2x(1 z2i )cos
2 n2y(1 z2i )sin
2 n2zz2i 2nxnzzi
1 z2i cos
2nxny(1 z2i )sin cos 2nynzzi
1 z2i sin
.
(16)
2 defined by (4) expands to:
2
=
e2E20m22 h
2
(k)2
1 n2zz
2i (n
2x + n
2y)(1 z
2i )/2
. (17)
Since the variable appears in the integral only via expression (15), all we have to dois to integrate (15) over . Integrating from to 2 gives:
(k)22
1 n2zz2i (n
2x + n
2y)(1 z
2i )/2
. (18)
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We can write down the Dirac deltas in (12) as:
h2(k)2
2m+
2h2kkz
2m+ h
h2(k)2
2m+
2h2kkz
2m h
= (g1(z)) (g2(z)) ,
(19)
where
g1(z) = h2(k)2/(2m) + h22kkz/(2m) + h , (20)
g2(z) = h2(k)2/(2m) + h22kkz/(2m) h . (21)
The roots ofg1,2 are
z1 = h(k)2 + 2m
2hkk, (22)
z2 = h(k)2 2m
2hkk. (23)
The derivatives of g1,2 are given by the following equation:
g1(z) = g2(z) = 2h
2kk/(2m) . (24)
The Dirac delta can be written down as:
h2(k + k)2
2m
h2k2
2m+ h
(25)
h2(k + k)2
2m
h2k2
2m h . (26)
There are two distributions integrated. The first one does not depend on the variable
of integration and the other depend on it via h2(k+k)2
2m. The presence of the Dirac deltas is
to take into account the energy conservation during the collision process. Owning this thedistribution functions can be taken outside the integration over spherical angles. According
that we can divide the integral into to parts (one Dirac delta each). The we can put h2k2
2m h
or h2k2
2m+ h respectively for h
2(k+k)2
2mand finally transfer the distribution functions as
explained above.
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With the above tasks executed we can integrate over z and
f(k)
t=
2Z2e6E20mh34k
1
1 + e(EFh2k2/2m)/(KT)
1
1 + e(EF(h2k22mh)/2m)/(2mKT)
d(k)(k)3
(1/a2 + (k)2)2
1 n2z
(h(k)2 + 2m)2
4h2k2(k)2 (n2x + n
2y)
1
(h(k)2 + 2m)2
4h2k2(k)2
/2
+
1
1 + e(EFh2k2/2m)/(KT)
1
1 + e(EF(h2k2+2mh)/2m)/(2mKT)
d(k)
(k)3
(1/a2 + (k)2)2
1 n2z
(h(k)2 2m)2
4h2k2(k)2 (n2x + n
2y)
1
(h(k)2 2m)2
4h2k2(k)2
/2
.
(27)
It is easy to change the result given by Eq. (27) to suit a coordinate system describedby Fig. 2, in which the field vector is oriented along the z axis and k vector is arbitrary.We notice, that Eq. (27) depends on k only through k = |k|, and on n only through n2xand n2x + n
2y. Since n is unitary, we have n
2x + n
2y = 1 n
2z. However, nzk is equal to the
component of k parallel to the field vector. Switching to spherical coordinates, we havethe radial coordinate ofk equal to k (no substitution) and angular coordinates equal to and (according to Fig. 2).
Coordinate is an angle between the field vector and k, hence its cosine is equal to nz.To transform Eq. (27) to the new coordinate system, if suffices to substitute n2z with cos
2 and n2x + n
2y with 1 n
2z = sin
2 :
f(k)
t(k, ,) =
2Z2e6E20mh34k
1
1 + e(EFh2k2/2m)/(KT)
1
1 + e(EF(h2k22mh)/2m)/(KT)
d(k)
(k)3
(1/a2 + (k)2)2
1 cos2
(h(k)2 + 2m)2
4h2k2(k)2 sin2
1
(h(k)2 + 2m)2
4h2k2(k)2
/2
+
1
1 + e(EFh2k2/2m)/(KT)
1
1 + e(EF(h2k2+2mh)/2m)/(KT)
d(k) (k)
3
(1/a2 + (k)2)2
1 cos2
(h(k)2 2m)2
4h2k2(k)2 sin2
1
(h(k)2 2m)2
4h2k2(k)2
/2
.(28)
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After integrating over angles and
f(k)
t=
8
3
2Z2e6E20mh34k
1
1 + e(EFh2k2/2m)/(KT)
1
1 + e(EF(h2k22mh)/2m)/(KT) d(k)
(k)3
(1/a2 + (k)2)2
+
1
1 + e(EFh2k2/2m)/(KT)
1
1 + e(EF(h2k2+2mh/2m))/(KT)
d(k)
(k)3
(1/a2 + (k)2)2
=
162Z2e6E203mh34k
1
1 + e(EFh2k2/2m)/(KT)
1
1 + e(EF(h2k22mh)/2m)/(KT)
+
1
1 + e(EFh2k2/2m)/(KT)
1
1 + e(EF(h2k2+2mh/2m))/(KT)
d(k)(k)3
(1/a2 + (k)2)2.
(29)
We integrate over k and we obtain
f(k)
t=
162Z2e6E203mh34k
1
1 + e(EFh2k2/2m)/(KT)
1
1 + e(EF(h2k22mh)/2m)/(KT)
+
1
1 + e(EFh2k2/2m)/(KT)
1
1 + e(EF(h2k2+2mh)/2m)/(KT)
1
2
1
(1 + a2(k)2)+ log(1 + a2(k)2)
.
(30)
3.1 The integral over k limitsNow we have to find the upper and lower limits of the integral over k from the conditionthat 1 < z1,2 < 1.
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for z1:
1 = h(k)2 + 2m
2hkkh(k)2 2hkk + 2m = 0
(k)2 2kk + 2m/h = 0
= 4k2 8m/h
k1 = k +
k2 2m/h , k2 = k
k2 2m/h
1 = h(k)2 + 2m
2hkkh(k)2 + 2hkk + 2m = 0
(k)2 + 2kk + 2m/h = 0
= 4k2 8m/h
k3 = k + k2 2m/h , k4 = k k2 2m/h
(31)
for z2:
1 = h(k)2 2m
2hkkh(k)2 2hkk 2m = 0
(k)2 2kk 2m/h = 0
= 4k2 + 8m/h
k1 = k +
k2 + 2m/h , k2 = k
k2 + 2m/h
1 = h(k)2 2m
2hkkh(k)2 + 2hkk 2m = 0
(k)2 + 2kk 2m/h = 0
= 4k2 + 8m/h
k3 = k +
k2 + 2m/h , k4 = k
k2 + 2m/h
(32)
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We must take into account the fact the radial coordinate k has a positive value andchoose the limit from k > 0 range only.
Then the integration for k are
1. integral with z1
k [k
k2 2m/h, k +
k2 2m/h] (33)
2. integral with z2
k [k +
k2 + 2m/h, k +
k2 + 2m/h] (34)
We integrate over k in the respective limits
f(k)
t=
162Z2e6E203mh34k
1
1 + e(EFh2k2/2m)/(KT)
1
1 + e(EF(h2k22mh)/2m)/(KT)
4a2h2kk2 2m/h + (h2(1 + 4a2k2) 4a2hm + 4a4m22) (log
2a2m + h(1 + 2a2k(k
k2 2m/h))
h
log
2a2
m + h(1 + 2a2
k(k +
k2
2m/h))h
)
+
1
1 + e(EF+h2k2/2m)/(KT)
1
1 + e(EF(h2k22mh)/2m)/(KT)
4a2h2kk2 + 2m/h + (h2(1 + 4a2k2) + 4a2hm + 4a4m22) (log
2a2m + h(1 + 2a2k(k +
k2 2m/h))
h
log2a2m + h(1 + 2a2k(k + k2 + 2m/h))
h
) .
(35)
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3.2 Fermi energy temperature dependence
The Fermi energy temperature dependence can be derived from a Fermi-Dirac distributionfunction normalization. A Maxwell-Boltzmann distribution is a good approximation for aFermi-Dirac in a classical (high-temperature) limit. We use the approximation
f(r, k) = e
EFh2k2/(2m)
KT (36)
N =
0
f(r, k)d3rd3k
eEFKTV
eh2k2
2mKT
3= V e
EFKT
2mKT
h2
3/2 (37)
eEFKT =
N
V
h2
2mKT
3/2(38)
Finally we get
EF = KT log
NV
h2
2mKT
3/2(39)
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4 High temperature limit
We approximate the Fermi-Dirac distribution with the Boltzmann distribution for theelectrons
f(k)
t
=162Z2e6E20
3mh3
4
ke(EFh2k2/2m)/(KT) e(EF(h2k22mh)/2m)/(KT)
4a2h2kk2 2m/h + (h2(1 + 4a2k2) 4a2hm + 4a4m22) (log
2a2m + h(1 + 2a2k(k
k2 2m/h))
h
log
2a2m + h(1 + 2a2k(k +
k2 2m/h))
h
)
/(2(h2(1 + 4a2k2) 4a2hm + 4a2m22))
+
e(EFh2k2/2m)/(KT) e(EF(h2k2+2mh)/2m)/(KT)
4a2h2kk2 + 2m/h + (h2(1 + 4a2k2) + 4a2hm + 4a4m22) (log
2a2m + h(1 + 2a2k(k +
k2 2m/h))
h
log
2a2m + h(1 + 2a2k(k +
k2 + 2m/h))
h
)
/
(2(h2
(1 + 4a2
k2
) + 4a2
hm + 4a2
m2
2
)) .
(40)
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We substitute the Fermi energy calculated as (Eq. (39)) to above equation to get
f(k)
t=
162Z2e6E203mh34k
N
h2
2mKT
3/2
eh2k2/(2mKT) e(h
2k22mh)/(2mKT)
4a2h2kk2 2m/h + (h2(1 + 4a2k2) 4a2hm + 4a4m22) (log
2a2m + h(1 + 2a2k(k
k2 2m/h))
h
log
2a2m + h(1 + 2a2k(k +
k2 2m/h))
h
)
/(2(h2(1 + 4a2k2) 4a2hm + 4a2m22))
+
eh
2k2/(2mKT) e(h2k2+2mh)/(2mKT)
4a2
h2
k
k2 + 2m/h + (h2
(1 + 4a2
k2
) + 4a2
hm + 4a4
m2
2
)
(log
2a2m + h(1 + 2a2k(k +
k2 2m/h))
h
log
2a2m + h(1 + 2a2k(k +
k2 + 2m/h))
h
)
/(2(h2(1 + 4a2k2) + 4a2hm + 4a2m22))
.
(41)
The kinetic energy of the electron is assumed to be unchanged during the collisionprocess, which means the approximation h 0 is proper. In this case the integral overk limits are from 0 to 2k.
f(k)
t=
4
3
(2)1/2Z2e6E20N
m5/24(KT)3/2k
eh2k2/(2mKT) e(h
2k22mh)/(2mKT)
2a2k2
1 + 4a2k2+
1
2log[1 + 4a2k2]
+
eh
2k2/(2mKT) e(h2k2+2mh)/(2mKT)
2a2k2
1 + 4a2k2+
1
2log[1 + 4a2k2]
= 43
(2)1/2Z2e6E20Nm5/24(KT)3/2k
2a
2k2
1 + 4a2k2+ 1
2log[1 + 4a2k2]
eh2k2/(2mKT)
2
eh/(KT) + eh/(KT)
.
(42)
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Since we have1
2
ez + ez
= cosh z , (43)
we can expand cosh z into a Taylor series:
cosh z =
n=0
z2n
(2n)!= 1 +
1
2z2 +
1
24z4 +
1
170z6 +
1
40320z8 + . . . (44)
and write down
f(k)
t=
4
3
(2)1/2Z2e6E20N
m5/24(KT)3/2k
2a2k2
1 + 4a2k2+
1
2log[1 + 4a2k2]
eh2k2/(2mKT)
2 2cosh
h
KT
=
4
3
(2)1/2Z2e6E20N
m5/24K3/2T3/2k
2a2k2
1 + 4a2k2+
1
2log[1 + 4a2k2]
eh2k2/(2mKT)
2 2
1 + 1
2
h
KT
2
=4
3
(2)1/2Z2e6E20N
m5/24(KT)3/2k
2a2k2
1 + 4a2k2+
1
2log[1 + 4a2k2]
eh
2k2/(2mKT)
h
KT
2
=4
3
(2)1/2Z2e6E20Nh2
m5/22(KT)7/2k
2a2k2
1 + 4a2k2+
1
2log[1 + 4a2k2]
eh
2k2/(2mKT) .
(45)
The change in average kinetic energy of the electrons is
d
dt
= d3kk2h2
2m
f(k)
t
. (46)
We integrate two parts of the expression (45) separately
the first term0
h2k2
2mkdk
2a2k2
1 + 4a2k2
eh
2k2/(2mKT)
=1
128
8KT( 1
a2
8KmT
h2)
eh2
8a2KmT h2
0, h2
8a2KmT
a4m
, (47)
the other term0
h2k2
2mkdk
log
1 + 4a2k2
eh
2k2/(2mKT)
=KT
16a2h2
2(h2 4a2KmT) + e
h2
8a2KmT (h2 8a2KmT)
1
eh2
8a2KmT
.
(48)
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For a the (27) simplifies:
f(k)
t=
8
3
2Z2e6E20mh34k
1
1 + e(EFh2k2/2m)/(KT)
1
1 + e(EF(h2k22mh)/2m)/(KT)
d(k)
1
k
+
1
1 + e(EFh2k2/2m)/(KT)
1
1 + e(EF(h2k2+2mh/2m))/(KT)
d(k)
1
k
.(53)
For high-temperature approximation we get
f(k)
t=
82Z2e6E20N
3mh34kN
h2
2mKT
3/2
eh2k2/(2KmT) e(h
2k22mh)/(2KmT)
+
eh
2k2/(2KmT) eh2k2+2mh/(2KmT)
d(k)
1
k.
(54)
Assuming the infinitesimally small energy change during the collision process we takethe integral limits from 0 to 2k. To avoid log[0] term we take the cutoff (kmin) for lowerk limit
f(k)
t=
82Z2e6E20N
3mh34k
h2
2mKT
3/2
eh2k2/(2mKT) e(h
2k22mh)/(2mKT)
(log[2k] log[kmin])
+
eh
2k2/(2KmT) e(h2k2+2mh)/(2KmT)
(log[2k] log[kmin])
= 21/2Z2e6E20N3m5/24(KT)3/2k
(log[2k] log[kmin0]) eh2k2/(2mKT)
2
eh/(2KT) + eh/(2KT)
=21/2Z2e6E20N
3m5/24(KT)3/2k(log[2k] log[kmin]) e
h2k2/(2mKT)
h
KT
2
=21/2Z2e6E20Nh
2
3m5/22(KT)7/2k(log[2k] log[kmin]) e
h2k2/(2mKT) .
(55)
We integrate over k
d dt
= 21/2Z2e6E20Nh2
3m5/22(KT)7/2
0
h2k32m
(log[2k] log[kmin])eh2k2
2mKT dk
=21/2Z2e6E20Nh
2
3m5/22(KT)7/2
0
h2k3
2meh2k2
2mKT log[2k]
0
dkh2k3
2meh2k2
2mKT log[kmin]
dk .
(56)
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The result is
d
dt=
21/2Z2e6E20Nh2
3m5/22(KT)7/2
0
h2k3
2m(log[2k] log[kmin])e
h2k2
2mKTdk
=21/2Z2e6E20Nh
2
3m5/22(KT)7/2
0
h2k3
2meh2k2
2mKT log[2k]dk
0
h2k3
2meh2k2
2mKT log[kmin]
dk
= 21/2Z2e6E20Nh2
3m5/22(KT)7/2
K2mT2
2h(1 + log[4]) + K
2mT2
hlog[kmin]
dk ,
(57)
where the Euler gamma has its numerical value 0.577Finally we get
d
dt=
21/2Z2e6E20Nh
3m3/22(KT)3/2
1
2(1 + log[4]) + log[kmin])
. (58)
5 Conclusions
We have presented the theoretical model of the electron gas irradiated by a laser beam.We have considered absorption by inverse Bremsstrahlung to derive the electron kineticequation and the rate of change of the average kinetic energy of the electrons.
To be continued. . .
A
f(x)(g(x))dx integrating formula
My aim is to calculate the below mentioned integral:f(x)(g(x))dx . (A.59)
Converting variables
y = g(x)dy = g(x)dx ,
(A.60)
we obtain f(x)(g(x))dx =
f(g1(y))(y)
1
g(g1(y))dy . (A.61)
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Substituting x = g1(y), we havef(x)(g(x))dx = f(x0)
1
g(x0). (A.62)
If a function g1(y) is multi-valued, e.g.
g(x) = x2 1
g1(y) = 1 ,(A.63)
our integral becomes a sum of obtained expression evaluations for all the roots. If there isxi which is an i-th root of the g(x) function (i = 1, . . . , N ), then
f(x)(g(x))dx =
Ni=1
f(xi)1
g(xi). (A.64)
18
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