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Spherical-coordinates

# Spherical-coordinates - Physical Chemistry Course Number...

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Unformatted text preview: Physical Chemistry Course Number: C362 9 The rotational problem 9.1 Reduced Mass 1. For a diatomic molecules, we can write the Hamiltonian as: h2 H(r1, r2) = - 2M1 where Cartesian coordinates: 2 2 r1 h2 - 2M2 2 r2 + V (r1, r2) (9.8) 2 2 2 = 2+ 2+ 2 x y z (9.9) 2. What is the center of mass of this system? The position of the center of mass is given by the vector: RCM and let us introduce the vector: r r2 - r1 Or we can write the old coordinates in terms of the new as: r1 = RCM - and r2 = RCM + M2 r M1 + M2 (9.12) (9.11) M1 r1 + M2 r2 M1 + M2 (9.10) M1 r (9.13) M1 + M2 Homework: Show Eqs. (9.12) and (9.13) to be true. 3. So now we have introduced two new coordinates, the center of mass vector, and a vector denoting the internal motion. Let us see if we can write the kinetic energy of the two particle system in terms of these two new coordinates. pr2 2 pr1 2 + T = 2M1 2M2 dr1 2 1 dr2 1 + M2 = M1 2 dt 2 dt Chemistry, Indiana University 46 2 c 2011, Srinivasan S. Iyengar (instructor) Physical Chemistry 2 Course Number: C362 1 M2 dr dRCM = M1 - + 2 dt M1 + M2 dt M1 dr dRCM 1 M2 + 2 dt M1 + M2 dt dRCM 1 (M1 + M2 ) = 2 dt 2 2 1 M1 M2 + 2 M1 + M2 dr dt 2 (9.14) 4. Homework: Show Eq. (9.14) to be true. To achieve this realize first that |a|2 = a a. 5. Now if we represent the total mass, Mtot = (M1 + M2 ) and the reduced M mass, = M11 M22 then we have +M 1 dRCM T = Mtot 2 dt pCM 2 p 2 + = 2Mtot 2 9.2 Spherical coordinates 1. We will first introduce a concept called the metric tensor. These two words essentially mean "length element". And the length element has to be different in different coordinate systems. While in the Cartesian coordinate system the length elements would be along the x, y and z directions, in a spherical coordinate system the length elements would be along the r, and directions, which are obviously different. The metric tensor is a matrix that basically tells us how these length elements relate to each other when we go from one coordinate system to another. (In the present case we want to go from the Cartesian to the spherical coordinate system.) 2. The metric tensor is defined in the following fashion for a transformation from {x, y, z} {u1, u2, u3}. (Note that for a spherical coordinate system {u1, u2, u3} {r, , }.) gi,j = Chemistry, Indiana University 2 1 dr + 2 dt 2 (9.15) y y z z x x + + ui uj ui uj ui uj 47 (9.16) c 2011, Srinivasan S. Iyengar (instructor) Physical Chemistry Course Number: C362 where i, j can be 1, 2, 3, for the coordinate system transformation that we are interested in. 3. For illustration we can write down the metric tensor for the {x, y, z} {r, , }. But first we note that rsincos rsinsin rcos x2 + y 2 + z 2 z z cos = = 2 r x + y2 + z2 y tan = x Thus, g1,1 = g2,2 g3,3 x x y y z z + + =1 r r r r r r x x y y z z = + + = r2 x x y y z z + + = r2 sin2 = (9.18) And we also find that g1,2 = g2,1 = g1,3 = g3,1 = g2,3 = g3,2 = 0. 4. So how does the metric tensor help us? It turns out that the Laplacian for such orthogonal coordinate systems (where gi,j = 0 for i = j) can be written as: 2 2 2 1 g 1/2 2 = 2+ 2+ 2= (9.19) 1/2 u x y z i gi,i ui i g where g is the determinant of the metric tensor. Hence in our case g = g1,1 g2,2 g3,3 since the off-diagonal elements are zero. The gradient operator is given by: Hence the Laplacian for {x, y, z} {r, , } is: 2 x y z r2 = = = = (9.17) = 2 1 1 r sin + sin + r2 sin r r sin 48 (9.20) Chemistry, Indiana University c 2011, Srinivasan S. Iyengar (instructor) ...
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