NEUTRON COLLISION THEORY M. Ragheb 3/30/2006

INTRODUCTION We wish to analyze the process by which neutrons scatter upon collision with the nuclei of different materials. The intended use is in shielding, dosimetry, and criticality calculations. The energy loss per collision would characterize the properties of different energy moderating materials such as graphite, light water, and heavy water. The kinematics of two-body collisions processes are best described using the Center of Mass system (CM), rather than the Laboratory (LAB) system of coordinates. The reason is that scattering is isotropic in the CM frame, and it is easier described in it. We then introduce the concept of a microscopic and macroscopic neutron cross section and describe the use of compiled cross sections data to estimate collision rate densities and reaction rates.

RELATIONSHIPS BETWEEN VELOCITIES AND ENERGIES IN CM AND LAB SYSTEMS The CM system is characterized by: 1. The total momentum in the CM system is zero. 2. The magnitudes of the CM velocities do not change in a collision. Their velocity vectors are rotated through the CM scattering angle. 3. Cross sections are calculated in the CM system, but are measured and used in the LAB system. 4. The total energy in the CM system is always less than in the LAB system. The energy difference is taken up by the center of mass’ motion itself. Let us consider: Mass of target nucleus = A Mass of neutron = 1 Target nucleus is stationary, implying that VL = 0 We can now deduce the relationships between velocities and energies in the CM and LAB systems. The collision coordinates in the LAB and CM system before and after a collision are shown in Figs.1 and 2, as well as the relationships between the scattering angles in the LAB and the CM frames. The Center of Mass velocity, is obtained by a momentum balance before and after a collision as: ( 1 + A)vCM = 1 ⋅ vL + A ⋅ VL

By taking VL = 0, we get:

vCM =

1 vL 1+ A

(1)

The neutron velocity in the center of mass system using Eqn. 1 is:

vC = vL - vCM = vL -

1 vL . 1+ A

From which:

vC =

A vL 1+ A

(2)

The target velocity in the Center of Mass system is from Eqn. 1:

VC = VL - vCM = −

1 vL 1+ A

(3)

where again we took VL = 0. The total momentum in CM frame is by using Eqns. 2 and 3:

1 ⋅ vC + A ⋅ VC = 1 ⋅

A A vL vL = 0 1+ A 1+ A

The total kinetic energy in the LAB system is:

EL =

1 1 1 ⋅ 1 ⋅ vL2 + AVL2 = vL2 2 2 2

where VL was taken as zero. The total kinetic energy in the CM system is:

EC =

1 1 ⋅ 1 ⋅ vC2 + AVC2 2 2

Using Eqns.2 and 3 we get:

1 A2 1 A ⋅1⋅ v2 + vL2 2 L 2 (A + 1 ) 2 (A + 1 )2 1 A(A + 1 ) 2 = vL 2 (A + 1 )2

EC =

(4)

Thus:

EC =

1 A 2 1 2 ⋅ vL = µ vL 2 A +1 2

(5)

where:

µ=

A.1 A +1

,

is the reduced mass. The relationship between the LAB and CM velocities from Eqns 4 and 5 is:

EC = µ EL =

A EL A +1

(6)

Thus EC < EL, since the center of mass motion itself takes the energy difference. Applying conservation of momentum in the CM system before and after a collision yields: Before collision 1 ⋅ v C + AV C

After collision =

'

'

AV C + 1 ⋅ v C

Rewriting this vector equation component-wise in the x and y directions:

vC − AVC = -AVC' cos θ c + vC' cos θc

(7)

0 = - AVC' sin θ c + vC' sin θc

(8)

vC' = AVC'

(9)

Equation 8 implies that:

Substituting in Eqn. 7 we get: vC = AVC

(10)

Then using Eqn. 3 we get:

vC =

A VL 1+ A

(11)

Applying conservation of energy in the CM system yields:

1 1 1 1 .1.vC2 + AVC2 = .1.v'C2 + AV'C2 2 2 2 2 Substituting for vc and v'c from Eqns.9 and 10: 1 2 2 1 1 1 A VC + AVC2 = A2V'C2 + AV'C2 2 2 2 2 2 2 (A + 1 )VC = (A + 1 )V'C

which yields: VC = V'C

(12)

Substituting in Eqns. 9 and 10: vC = v'C

(13)

Thus the velocities do not change in the CM frame. RELATIONSHIP BETWEEN SCATTERING CROSS SECTION IN LAB AND CM FRAMES

From Fig.3, we can write for the horizontal and vertical components: Horizontal:

v'L cos θ L = vCM + v'C cos θ C

(14)

Vertical:

v'L sin θ L = v'C sin θ C

(15)

Dividing the Left Hand Side (LHS) of both equations, we get: tan θ L =

Here we used from Eqn.1:

v'C sin θ c sin θC = vCM + v'C cos θC 1 + cos θ C A

vCM =

1 vL 1+ A

(16)

and from Eqns.2,13:

v 'C = vC =

A A vL vL v ' C = vC = 1+ A 1+ A

Now:

σ L (θ L )sinθ L ⋅ dθ L = σ CM (θ C )sinθ C ⋅ dθ C where:

σ L (θ L ) is the differential scattering cross section in the LAB system, σ CM (θC ) is the differential scattering cross section in the CM system.

Thus:

σ L (θ L ) = σ CM (θC ) ⋅ From Eqns.15 and 13:

sin θC v ' L v ' L = = sin θ L v 'C vC

sinθC dθC ⋅ sinθ L dθ L

(17)

Thus, from Eqn. 11: sin θC = sin θ L

v 'L A vL 1+ A

(18)

Substituting in Eqn.17:

σ L (θ L ) = σ CM (θC ) ⋅

1 + A v ' L dθ C ⋅ A vL dθ L

(17)'

Differentiating Eqn.16 with respect to θ L , we get:

sec2θ L =

1 = cos2 θ L

[(

1 + cos θC ) cos θ C + sin 2 θ C ] dθ A ⋅ C 1 dθ L 2 ( + cosθC ) A

Thus: 1 + cos θ C ) 2 dθC A = dθ L cos2 θ .( 1 cos θ + 1) L C A (

(19)

To get an expression for cos 2 θ L in terms of θ C , we use Eqn.14: cos2 θ L = ( =(

vCM v 'C cos θC ) 2 + v 'L v 'L 1 vL A vL cos θC ) 2 + 1 + A v 'L 1 + A v 'L

by use of Eqns. 1, 11 and 13. Thus: cos2 θ L =

1 v2L (1 + A cos θC ) 2 2 2 (1 + A) v ' L

Substituting Eqn.20 into Eqn.19, we get:

(20)

1 + cosθ C ) 2 dθ C A = 1 A2 vL2 1 dθ L ⋅ 2 ⋅ ( + cos θ C ) 2 ( cos θ C + 1) 2 (1 + A) v ' L A A (

(1 + A) 2 vL2 A2 v '2L = 1 ( cos θ C + 1) A

(21)

Substituting Eqn.21 into Eqn. (17)':

σ L (σ L ) = σ CM (σ C ) ⋅

To get the value of (

(1 + A)3 v '3L 1 ⋅ 3 ⋅ 3 A vL ( 1 cos θ + 1) C A

(17)''

v 'L 3 ) we use the triangular relationship from Fig. 3: vL 2 v '2L = vCM + v 'C2 − 2v 'C vCM cos(180o − θC ) 2 = vCM + v '2C + 2v 'C vCM cos θC

Substituting for v CM

from Eqn.1, using v 'C = vC , and substituting for vL from Eqn.11, we get: v '2L 1 + A2 + 2 A cos θC = vL2 (1 + A) 2

(22)

Substituting Eqn.22 into Eqn. (17)'', we get:

σ L (θ L ) = σ CM (θ C ) ⋅

(1 + A)3 1 (1 + A2 + 2 A cos θC )3/ 2 ⋅ ⋅ 1 A3 (1 + A)3 ( cos θC + 1) A

Finally: 1 2 + .cos θC + 1)3/ 2 2 A A σ L (θ L ) = σ CM (θC ) 1 ( cos θC + 1) A (

which relates the scattering cross sections in the LAB and CM systems.

(23)

RELATIONSHIP BETWEEN THE INITIAL AND FINAL ENERGIES To relate the final and initial particle energies in the LAB system, we use Eqn. 22: 1 2 E ' 2 .1.v ' L A2 + 1 + 2 A cos θC = = 1 E ( A + 1) 2 2 .1.vL 2

(24)

Defining the collision parameter: ⎛ A −1⎞ α =⎜ ⎟ ⎝ A +1⎠

2

(25)

Thus: 1 + α = 2.

A2 + 1 2A , 1-α = 2. 2 ( A + 1) ( A + 1) 2

And:

E' =[

(1 + α ) + (1 − α ) cos θC ]⋅ E 2

(26)

which relates the initial and final energies for a collision.

SPECIAL CHARACTERISTICS OF PARTICLES COLLISIONS ENERGY TRANSFER Equation 26 describing the relationship between the initial and final energy of a neutron after collision with a nucleus possesses several important characteristics: 1. It implies that the energy transfer from neutron to nucleus is related to the scattering angle in the CM system. For the case of no collision we have θC = 0 , then:

E' = E . 2. The maximum energy loss occurs in a back scattering collision: θC = 180o . Then:

E ' = αE .

The maximum energy loss would be: ∆Emax = E − E ' = E − α E = (1 − α )E 3. Not only a neutron cannot gain energy in an elastics collision with a stationary nucleus (E'

INTRODUCTION We wish to analyze the process by which neutrons scatter upon collision with the nuclei of different materials. The intended use is in shielding, dosimetry, and criticality calculations. The energy loss per collision would characterize the properties of different energy moderating materials such as graphite, light water, and heavy water. The kinematics of two-body collisions processes are best described using the Center of Mass system (CM), rather than the Laboratory (LAB) system of coordinates. The reason is that scattering is isotropic in the CM frame, and it is easier described in it. We then introduce the concept of a microscopic and macroscopic neutron cross section and describe the use of compiled cross sections data to estimate collision rate densities and reaction rates.

RELATIONSHIPS BETWEEN VELOCITIES AND ENERGIES IN CM AND LAB SYSTEMS The CM system is characterized by: 1. The total momentum in the CM system is zero. 2. The magnitudes of the CM velocities do not change in a collision. Their velocity vectors are rotated through the CM scattering angle. 3. Cross sections are calculated in the CM system, but are measured and used in the LAB system. 4. The total energy in the CM system is always less than in the LAB system. The energy difference is taken up by the center of mass’ motion itself. Let us consider: Mass of target nucleus = A Mass of neutron = 1 Target nucleus is stationary, implying that VL = 0 We can now deduce the relationships between velocities and energies in the CM and LAB systems. The collision coordinates in the LAB and CM system before and after a collision are shown in Figs.1 and 2, as well as the relationships between the scattering angles in the LAB and the CM frames. The Center of Mass velocity, is obtained by a momentum balance before and after a collision as: ( 1 + A)vCM = 1 ⋅ vL + A ⋅ VL

By taking VL = 0, we get:

vCM =

1 vL 1+ A

(1)

The neutron velocity in the center of mass system using Eqn. 1 is:

vC = vL - vCM = vL -

1 vL . 1+ A

From which:

vC =

A vL 1+ A

(2)

The target velocity in the Center of Mass system is from Eqn. 1:

VC = VL - vCM = −

1 vL 1+ A

(3)

where again we took VL = 0. The total momentum in CM frame is by using Eqns. 2 and 3:

1 ⋅ vC + A ⋅ VC = 1 ⋅

A A vL vL = 0 1+ A 1+ A

The total kinetic energy in the LAB system is:

EL =

1 1 1 ⋅ 1 ⋅ vL2 + AVL2 = vL2 2 2 2

where VL was taken as zero. The total kinetic energy in the CM system is:

EC =

1 1 ⋅ 1 ⋅ vC2 + AVC2 2 2

Using Eqns.2 and 3 we get:

1 A2 1 A ⋅1⋅ v2 + vL2 2 L 2 (A + 1 ) 2 (A + 1 )2 1 A(A + 1 ) 2 = vL 2 (A + 1 )2

EC =

(4)

Thus:

EC =

1 A 2 1 2 ⋅ vL = µ vL 2 A +1 2

(5)

where:

µ=

A.1 A +1

,

is the reduced mass. The relationship between the LAB and CM velocities from Eqns 4 and 5 is:

EC = µ EL =

A EL A +1

(6)

Thus EC < EL, since the center of mass motion itself takes the energy difference. Applying conservation of momentum in the CM system before and after a collision yields: Before collision 1 ⋅ v C + AV C

After collision =

'

'

AV C + 1 ⋅ v C

Rewriting this vector equation component-wise in the x and y directions:

vC − AVC = -AVC' cos θ c + vC' cos θc

(7)

0 = - AVC' sin θ c + vC' sin θc

(8)

vC' = AVC'

(9)

Equation 8 implies that:

Substituting in Eqn. 7 we get: vC = AVC

(10)

Then using Eqn. 3 we get:

vC =

A VL 1+ A

(11)

Applying conservation of energy in the CM system yields:

1 1 1 1 .1.vC2 + AVC2 = .1.v'C2 + AV'C2 2 2 2 2 Substituting for vc and v'c from Eqns.9 and 10: 1 2 2 1 1 1 A VC + AVC2 = A2V'C2 + AV'C2 2 2 2 2 2 2 (A + 1 )VC = (A + 1 )V'C

which yields: VC = V'C

(12)

Substituting in Eqns. 9 and 10: vC = v'C

(13)

Thus the velocities do not change in the CM frame. RELATIONSHIP BETWEEN SCATTERING CROSS SECTION IN LAB AND CM FRAMES

From Fig.3, we can write for the horizontal and vertical components: Horizontal:

v'L cos θ L = vCM + v'C cos θ C

(14)

Vertical:

v'L sin θ L = v'C sin θ C

(15)

Dividing the Left Hand Side (LHS) of both equations, we get: tan θ L =

Here we used from Eqn.1:

v'C sin θ c sin θC = vCM + v'C cos θC 1 + cos θ C A

vCM =

1 vL 1+ A

(16)

and from Eqns.2,13:

v 'C = vC =

A A vL vL v ' C = vC = 1+ A 1+ A

Now:

σ L (θ L )sinθ L ⋅ dθ L = σ CM (θ C )sinθ C ⋅ dθ C where:

σ L (θ L ) is the differential scattering cross section in the LAB system, σ CM (θC ) is the differential scattering cross section in the CM system.

Thus:

σ L (θ L ) = σ CM (θC ) ⋅ From Eqns.15 and 13:

sin θC v ' L v ' L = = sin θ L v 'C vC

sinθC dθC ⋅ sinθ L dθ L

(17)

Thus, from Eqn. 11: sin θC = sin θ L

v 'L A vL 1+ A

(18)

Substituting in Eqn.17:

σ L (θ L ) = σ CM (θC ) ⋅

1 + A v ' L dθ C ⋅ A vL dθ L

(17)'

Differentiating Eqn.16 with respect to θ L , we get:

sec2θ L =

1 = cos2 θ L

[(

1 + cos θC ) cos θ C + sin 2 θ C ] dθ A ⋅ C 1 dθ L 2 ( + cosθC ) A

Thus: 1 + cos θ C ) 2 dθC A = dθ L cos2 θ .( 1 cos θ + 1) L C A (

(19)

To get an expression for cos 2 θ L in terms of θ C , we use Eqn.14: cos2 θ L = ( =(

vCM v 'C cos θC ) 2 + v 'L v 'L 1 vL A vL cos θC ) 2 + 1 + A v 'L 1 + A v 'L

by use of Eqns. 1, 11 and 13. Thus: cos2 θ L =

1 v2L (1 + A cos θC ) 2 2 2 (1 + A) v ' L

Substituting Eqn.20 into Eqn.19, we get:

(20)

1 + cosθ C ) 2 dθ C A = 1 A2 vL2 1 dθ L ⋅ 2 ⋅ ( + cos θ C ) 2 ( cos θ C + 1) 2 (1 + A) v ' L A A (

(1 + A) 2 vL2 A2 v '2L = 1 ( cos θ C + 1) A

(21)

Substituting Eqn.21 into Eqn. (17)':

σ L (σ L ) = σ CM (σ C ) ⋅

To get the value of (

(1 + A)3 v '3L 1 ⋅ 3 ⋅ 3 A vL ( 1 cos θ + 1) C A

(17)''

v 'L 3 ) we use the triangular relationship from Fig. 3: vL 2 v '2L = vCM + v 'C2 − 2v 'C vCM cos(180o − θC ) 2 = vCM + v '2C + 2v 'C vCM cos θC

Substituting for v CM

from Eqn.1, using v 'C = vC , and substituting for vL from Eqn.11, we get: v '2L 1 + A2 + 2 A cos θC = vL2 (1 + A) 2

(22)

Substituting Eqn.22 into Eqn. (17)'', we get:

σ L (θ L ) = σ CM (θ C ) ⋅

(1 + A)3 1 (1 + A2 + 2 A cos θC )3/ 2 ⋅ ⋅ 1 A3 (1 + A)3 ( cos θC + 1) A

Finally: 1 2 + .cos θC + 1)3/ 2 2 A A σ L (θ L ) = σ CM (θC ) 1 ( cos θC + 1) A (

which relates the scattering cross sections in the LAB and CM systems.

(23)

RELATIONSHIP BETWEEN THE INITIAL AND FINAL ENERGIES To relate the final and initial particle energies in the LAB system, we use Eqn. 22: 1 2 E ' 2 .1.v ' L A2 + 1 + 2 A cos θC = = 1 E ( A + 1) 2 2 .1.vL 2

(24)

Defining the collision parameter: ⎛ A −1⎞ α =⎜ ⎟ ⎝ A +1⎠

2

(25)

Thus: 1 + α = 2.

A2 + 1 2A , 1-α = 2. 2 ( A + 1) ( A + 1) 2

And:

E' =[

(1 + α ) + (1 − α ) cos θC ]⋅ E 2

(26)

which relates the initial and final energies for a collision.

SPECIAL CHARACTERISTICS OF PARTICLES COLLISIONS ENERGY TRANSFER Equation 26 describing the relationship between the initial and final energy of a neutron after collision with a nucleus possesses several important characteristics: 1. It implies that the energy transfer from neutron to nucleus is related to the scattering angle in the CM system. For the case of no collision we have θC = 0 , then:

E' = E . 2. The maximum energy loss occurs in a back scattering collision: θC = 180o . Then:

E ' = αE .

The maximum energy loss would be: ∆Emax = E − E ' = E − α E = (1 − α )E 3. Not only a neutron cannot gain energy in an elastics collision with a stationary nucleus (E'