Page 314 - Intro to Tensor Calculus

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r
Figure 2.5-4. Volume element and solid angle about position ~.
0
depending upon the cross-sections, exit with energy E − ∆E = E and thus will contribute a gain to the
volume element. In terms of the flux φ the gains due to scattering into the volume element are denoted by
Z Z
0 ~
~
~
~ 0
~ 0
0
0
dΩ dE Σ(E → E, Ω → Ω)φ(~r, E , Ω,t) dτ dE dΩ
r
0
and represents the particles at position ~ experiencing a scattering collision with a particle of energy E and
~
~
direction Ω which causes the particle to end up with energy between E and E + dE and direction Ωin dΩ.
~ 0
The summations are over all possible initial energies.
In terms of φ the losses are due to those particles leaving the volume element because of scattering and
are
~
~
Σ s (E,~r)φ(~r, E, Ω,t)dτ dE dΩ.
The particles which are lost due to absorption processes are
~
~
Σ a (E,~r)φ(~r, E, Ω,t) dτ dE dΩ.
The total change to the number of particles in an element of phase space per unit of time is obtained by
summing all gains and losses. This total change is
Z Z
dN 0 ~
~
~
~ 0
0
dτ dE dΩ= dΩ dE Σ(E → E, Ω → Ω)φ(~r, E , Ω,t) dτ dE dΩ
~ 0
0
dt
~
− Σ s (E,~r)φ(~r, E, Ω,t)dτ dE dΩ
(2.5.72)
~
~
− Σ a (E,~r)φ(~r, E, Ω,t) dτ dE dΩ
~
~
+ S(~r, E, Ω,t)dτ dE dΩ.
The rate of change dN on the left-hand side of equation (2.5.72) expands to
dt
dN ∂N ∂N dx ∂N dy ∂N dz
= + + +
dt ∂t ∂x dt ∂y dt ∂z dt
∂N dV x ∂N dV y ∂N dV z
+ + +
∂V x dt ∂V y dt ∂V z dt