MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 6 pps

GEGENBAUER FUNCTIONS (TYPE A) 153 9.8 GEGENBAUER FUNCTIONS (TYPE A) The Gegenbauer equation in general is given as d2C;' (x) dC2' (x) dx2 dx - (2X' + 1)x- + n(n + 2X')C,"'(x) = 0. (9.250) (1 - x2) For X = 1/2 this equation reduces to the Legendre equation. values of n its solutions reduce to the Gegenbauer or Legendre polynomials: For integer (9.251) In the study of surface oscillations of a hypersphere one encounters the equa- tion - (2m + 3)x- dU,"(x) + Xurn A( >=o, (9.252) d2U," (x) dx2 dx (1 -x2) solutions of which could be expressed in terms of the Gegenbauer polynomials as where X = (1 - m)(l+ m + 2). (9.254) Using x = -cos6 UT(X) = Z?(Q)(sinQ)-rn-l we can put Equation (9.252) into the second canonical form as (9.255) (9.256) (9.257) m(m + 1) dQ2 On the introduction of A" = x + (m + I)~, (9.258) and comparing with, Equation (9.90), this is of type A with c = p = d = 0, a = 1, and z = 8, and its factorization is given by k(6, m) = m cot B p(m) = m2. (9.259) (9.260) 154 STURM-LIOUVILLE SYSTEMS AND THE FACTORIZATION METHOD The solutions are found by using and the formulas (9.261) (9.262) Note that Z[n is the eigenfunction corresponding to the eigenvalue A"= (2+112, 1-m=0,1,2 , , (9.263) that is, to x = (1 + 1)2 - (m + 1)2 = (2 - m)(l+ m + 2). (9.264) 9.9 SYMMETRIC TOP (TYPE A) The wave equation for a symmetric top is encountered in the study of simple molecules. If we separate the wave function as U = @(U) exp(iK4) exp(im+), (9.265) where 8,4, and + are the Euler angles and K and m are integers, O(8) satisfies the second-order ordinary differential equation dO(8) (m - KCOs8)2 @(e) + a@(U) = 0, (9.266) &O(U) de2 +cotu - dB sin2 u where (9.267) A, W, C, and h are other constants that come from the physics of the problem. With the substitution Y = O(U) sin'/2 U, (9.268) Equation (9.266) becomes Y + (a + K~ + 1/4)Y = 0. (9.269) (m- 1/2)(m+ 1/2)+rc2-2rn~cosU sin2 u BESSEL FUNCTIONS (TYPE C) 155 This equation is of type A, and we identify the parameters in Equation (9.90) as a = 1, c = -1/2, d = -6, p = 0. The factorization is now given by k(0, m) = (m - 1/2) cot e - K/ sin 8, p(m) = (m - 1/2)2. Eigenfunctions can be obtained from SinJ-n+l/2 COSJ+n+l/2 - e 2 2 by using 2 1r2 1 YE-’ = (J + -)2 - (m - -)2 L The corresponding eigenvalues are c7 + K + 1/4 = (J + 1/2)2 J - Iml and J - 1.1 = 0,1,2, so that (9.270) (9.271) (9.272) (9.273) (9.274) (9.275) (9.276) 9.10 BESSEL FUNCTIONS (TYPE C) Bessel’s equation is given as z2J$(z) +zJL(z) + (A22 - m2)Jm(.) = 0. (9.277) Multiplying this equation by l/z, we obtain the first canonical form as (9.278) where p(z) = z, and ~(z) = 2. (9.279) 156 STURM-LIOUVILLE SYSTEMS AND THE FACTORIZATION METHOD A second transformation, (9.280) (9.281) gives us the second canonical form 9 = 0. 1 d21k (m2 - 1/4) dx2 22 -+ [x- (9.282) This is type C, and its factorization is given as (9.283) p(m) = 0. (9.284) (m- f) k(x,m) = ~ x' Because p(m) is neither a decreasing nor an increasing function of m, we have no limit (upper or lower) to the ladder. We have only the recursion relations and where 9, = XWm(A1/2X). 9.11 HARMONIC OSCILLATOR (TYPE D) The Schrodinger equation for the harmonic oscillator is given as (9.285) (9.286) (9.287) (9.288) where equation can be written in either of the two forms (See Problem 9.14) = (h/p~)*/~x and A = 2E/hw in terms of the physical variables. This O-O,@x = (A + 1)Qx (9.289) and 0+0_9x = (A - l)Ikx, (9.290) PROBLEMS 157 where (9.291) Operating on Equation (9.289) with Of and on Equation (9.290) with 0- we obtain the analog of Theorem I as *A,, 0: o+*x (9.292) and QA-2 a o-*x. (9.293) Moreover, corresponding to Theorem IV, we find that we can not lower the eigenvalue X indefinitely. Thus we have a bottom of the ladder A = 2nf 1, n = 0,1,2 ,"' . (9.294) Thus the ground state must satisfy o-qo = 0, (9.295) (9.296) Now the other eigenfunctions can be obtained from \kn+l = [an + 2]-'/20+*11,, (9.297) *,_I = [2n]-"20_*,. (9.298) Problems 9.1 Starting from the first canonical form of the Sturm-Liouville equation: dx [P(Z)F] + q(2)9(2) + XW(Z)*(Z) = 0, x E [a, b] , derive the second canonical form: where d2p1 2 dpdw 1 d2w pdz dz w dz2 pdz2 I + + 158 STURM-LIOUVILLE SYSTEMS AND THE FACTORlZATlON METHOD by using the transformations Y(Z) = w [W(~)P(~)I~~~ and W(X) dz = dx [m] 9.2 Derive the normalization constants in W+I" fl(cos6) and 21 + 1(1 -m)! 1 y,-"(e,$> = [L-1" p,(cos6). J 2 (i+l)! 27r 9.3 Derive the normalization constant in 9.4 Derive Equation (9.195), which is given as 9.5 The general solution of the differential equation is given as the linear combination y(x) = C, sin fix + C, cos Ax. Show that factorization of this equation leads to the trivial result with k(x,m) = 0, p(m) = 0, and the corresponding ladder operators just produce other linear combinations of sin Ax and cos dz. 9.6 Show that taking k(z, m) = h(z) + kl(z)m + k2(z)m2 PROBLEMS 159 does not lead to any new categories, except the trivial solution given in Prob lem 9.5. A similar argument works for higher powers of m. 9.7 in k(z, m), no new factorization types appear. 9.8 Show that Show that as long as we admit a finite number of negative powers of m is a periodic function of m with the period one. Use this result to verify 9.9 Derive the stepdown operator in 9.10 the equation Follow the same procedure used in Path I in Section 9.6.5 to derive ct m 1 (1+1-m) yln,?’(8) = - C,+i,m (1 + 1 - m) J(l + m + 1)(1+ m + 2) COS~ 7- (1+;)sin6’}yY(B). d8 sin8 9.11 tions of the first kind: Use the factorization method to show that the spherical Hankel func- hp = j, + in, can be expressed as Hint: Introduce in yj‘ + [ 1 - - ”1 y1 = 0. 160 STURM-LIOUVILLE SYSTEMS AND THE FACTORIZATION METHOD 9.12 normalized eigenfunctions y(n, E, T) of the differential equation Using the factorization method, find a recursion relation relating the to the eigenfunctions with 1 f 1. Hint: First show that 1 = n - 1,n - 2, , 1 = integer and the normalization is 9.13 The harmonic oscillator equation d2 9 dx2 - + (E - x2)*(x) = 0 is a rather special case of the factorization method because the operators O& are independent of any parameter. i) Show that the above equation factorizes as d dx o,= x and d dx 0- = -x. ii) In particular, show that if 9&(z) is a solution for the energy eigenvalue E, then is a solution for E + 2, while is a solution for E - 2. iii) Show that E has a minimum with Emin = 1, En = 2n + 1, n = 0,1,2, . . . PROBLEMS 161 and show that the E < 0 eigenvalues are not allowed. to €,in and then use it to express all the remaining eigenfunctions. iv) Using the factorization technique, find the eigenfunction corresponding Hint: Use the identity 9.14 Show that the standard method for the harmonic oscillator problem leads to a single ladder with each function on the ladder corresponding to a different eigenvalue A. This follows from the fact that ~(z, m) is independent of m. The factorization we have introduced in Section 9.11 is simpler, and in fact the method of factorization originated from this treatment of the problem. 9.15 The spherical Bessel functions jl(2) are related to the solutions of d2Yl l(1 + 1) - &2+ [ I x2 ] Yd2) =o, (regular at x = 0) by Y1 (XI j&) = 2 Using the factorization technique, derive recursion formulae i) Relating j~(z) to j,+,(z) and j~-l(z). ii) Relating ji(x) to j,+,(x) and jl-l(z) . This Page Intentionally Left Blank [...]... cos8 0 -sin8 D(+) = [ 1 "1 1 sin+ 0 cosq5 0 cos+ -sin+ sin8 cos8 sin$ cos+ 0 o , 1 (10.53) , 0 0 1 (10.54) (10.55) In terms of the individual rotations, elements of the complete transformation matrix can be written as A = DCB, A= + + + + + cos+cos - cos8 sin +sin cos - cos 6sin +cos sin 8 sin - sin + (10. 56) (10.57) ++ + + cos+ sin cos8cos+sin sin cos 6cos +cos -sin 8cos q5 - sin + ++ The inverse of... -sin4 1 - sin $ sin4cos$ cos~cos$ Reversing the order we get R2R1 = [ (10.93) - sin $ cos Cp cos $ sin $ sin Cp 0 cos Cp sin 4 sin $ - cos sin q5 cos $ cos 4 (10.94) $ J It is clear that for finite rotations these two matrices are not equal: RiR2 # R2R1 (10.95) However, for small rotations, say by the amounts Sic, and approximations sin 6 $ cos6$ to find R1R2 = [ N N 1 6+ 6$ 64 ,we can use the S+, sin... use the S+, sin 6 p N 6 p C C (10. 96) 1, cos6Cp _N 1 0 1 -64 =R2R, (10.97) Note that in terms of the generators [Eq (10.84)] we can also write this a s R1R2 =[ 1 0 0 0 1 0 o 0 0 0 -1 + 6 $ 0 0 0 0 0 ] + S$Xl+ 64 x2 = I + 64 x2 + 6+ Xl =I = R2R1, +64 0 0 [o 0 0 -1 A] 0 (10.98) 178 COORDINATES AND TENSORS which again proves that infinitesimal rotations commute 10.7 CARTESIAN TENSORS Certain physical properties... A(t)r(O) (10.72) Differentiating and using Equation (10 .61 ) we obtain -= dr(t) dt A'(t)r(O) = A'(O)A(t)r(O) = Xr(t), (10.73) ( 10.74) ( 10.75) where X = A'(0) (10. 76) Differentiating Equation (10.75) we can now obtain the higher-order derivatives as (10.77) Using these in the Taylor series expansion of r ( t )about t = 0 we write (10.78) 1 76 COORDINATES AND TENSORS thus obtaining r(t) = (I + X t + z1X... their transformation properties 10.1 CARTESIAN COORDINATES In threedimensional Euclidean space a Cartesian coordinate system can be constructed by choosing three mutually orthogonal straight lines A point is defined by giving its coordinates, ( q , z 2 , q ) , or by using the position vector 163 164 COORDINATES AND TENSORS fig 10.1 Cartesian coordinate system + as r r - IL& -f- +z2G2 +~ 3 G 3 = (XI... Differentiating with respect t o t 2 and putting t2 = 0 and result that will be useful to us shortly as A'(t) = A'(O)A(t) (10.59) (10 .60 ) tl = t, we obtain a (10 .61 ) 1 174 COORDINATES AND TENSORS 2 + r -* /' f r Fig 10 .6 Passive and active views of the rotation matrix 10.5 ACTIVE A N D PASSIVE INTERPRETATIONS OF ROTATIONS It is possible t o view the rotation matrix A in F=Ar (10 .62 ) as an operator acting on... vectors 6; and &$, 2 will have the same form as 7 2 2 2 2 (10. 165 ) + where gLl = GI, thus proving its tensor character Because ?i’ and b are arbitrary vectors, we can take them as the infinitesimal displacement vector a;- as thus 2 2 (10. 167 ) gives the line element with the metric g;j = e - e j , i A A i , j = 1,2 (10. 168 ) 188 COORDINATES AND TENSORS Hence a and a are indeed the contravariant and the... Cartesian coordinates, their transformations, and Cartesian tensors We then generalize our discussion to generalized coordinates and general tensors The next stop in our discussion is the coordinate systems in Minkowski spacetime and their transformation properties We also introduce four-tensors in spacetime and discuss covariance of laws of nature We finally discuss Maxwell’s equations and their transformation... metric tensor are also given as (Gantmacher) (10.1 56) 1 86 COORDINATES AND TENSORS where A3'= cofactor of gji (10.157) g = det g i j (10.158) and 10.8.3 Geometric Interpretation of Covariant and Contravariant Components Covariant and contravariant indices can be geometrically interpreted in terms of oblique axis A vector 3 in the coordinate system shown in Figure 10.7 can be written as + = a%+ a + a%2,... system In the case of the active view, we also need to know how an operator A transforms under coordinate transformations Considering a transformation represented by the matrix B, we multiply both sides of Equation (10 .62 ) by B to write BI; = BAr (10 .63 ) Using BB-I = B - ~ B I, = (10 .64 ) we now write Equation (10 .63 ) as BF = BAB-lBr, - = A'r' r' (10 .65 ) (10 .66 ) In the new coordinate system T and r are . (1+;)sin6’}yY(B). d8 sin8 9.11 tions of the first kind: Use the factorization method to show that the spherical Hankel func- hp = j, + in, can be expressed as Hint: Introduce in. COORDINATES In threedimensional Euclidean space a Cartesian coordinate system can be constructed by choosing three mutually orthogonal straight lines. A point is defined by giving its coordinates,. found by the right-hand rule (Fig. 10.2). 166 COORDINATES AND TENSORS Fig. 10.3 Motion in Cartesian coordinates 10.1.2 Differentiation of Vectors In a Cartesian coordinate system motion