Momentum integrals evaluation

The procedure adopted here consists of taking a momentum balance on an element of fluid. The resulting Momentum Equation involves no assumptions concerning the nature of the flow. However, it includes an integral, the evaluation of which requires a knowledge of the velocity profile ux = f(y). At this stage assumptions must be made concerning the nature of the flow in order to obtain realistic expressions for the velocity profile. 

The integrals Ma a can be evaluated in direct space using ellipsoidal coordinates, or alternatively in momentum space using methods discussed in reference [17], The result is ... 

For maximimi mmierical efficiency, all derivative terms of the AWF are kept in momentum space so that one FFT is saved for each of their evaluations. For example, during the evaluation of the following general double integral, the first FFT can be avoided ... 

The spin-orbit operator LS given in Eq. (67) is expressed in terms of the individual electron-orbital and spin-momentum operators rather than the total momentum operators L and S. It can be shown (/, 5) that when evaluating integrals involving only LS functions of the same configuration, ls can be replaced by... 

The pressure is presumed to vary smoothly throughout the length of the channel, so it can be expanded in a first-order Taylor series. The wall shear stress is presumed to come from an empirical correlation, since, by assumption, the model does not consider radial variations in the axial velocity. The control-surface integrals can be evaluated simply to yield an equation for the axial momentum balance... 

In the last line we used Eqs. (3.10) (3.13), find we exploited the reflection symmetry of the chain the chain is invariant under a relabeling j — ft — j. To evaluate the result (3.20), for small values of the momentum q, such that q2f 2/2d = q2 2n = 0(1), we replace the summations by integrals. (This step will be discussed in more detail below.)... 

In order to evaluate the matrix element in Eq. (4.151), (p S p0), we must calculate three-dimensional integrals. In the following we show, however, that the matrix element can be reduced to a sum over one-dimensional S -matrix elements. This is obtained via an expansion of the momentum eigenstates (R p) in a basis where we can use that the angular momentum of the relative motion is conserved. 

The matrix elements in the integral are evaluated by introducing a unit operator using the momentum eigenfunctions between the operator and the coordinate eigenfunction ... 

In the first line we have introduced the unit operator using the momentum eigenstates because we may then evaluate the resulting matrix elements. In the third line we have introduced the momentum eigenfunctions in the coordinate representation, and in the last line we have used the standard integral f dx exp(— p2x2 qx) = exp(matrix element in Eq. (F.33) we find by analogy that... 

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