FEL laser induced heating of the electron gas in metal

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


2/18


3/18

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)

3


4/18

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


5/18

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)

5


6/18

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.

6


7/18

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)

7


8/18

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.

8


9/18

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)

9


10/18

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)

10


11/18

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)

11


12/18

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)


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)

12


13/18

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)


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)

13


14/18

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)

14


15/18


16/18

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)

16


17/18

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)

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FEL laser induced heating of the electron gas in metal