04082010reducedmass

04082010reducedmass - magnitues ) as r = r 1 + r 2 , we can...

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Quantum Mechanics: Rotation of Diatomics: Reducing the Problem 7th April 2010 Consider two masses, m 1 and m 2 , separated by a constant distance r 0 . You can think of the masses as being ”connected” or ”bonded”, but this is in some loose sense. Mass m 1 is a distance r 1 from the center of mass of the system; mass m 2 is a distance r 2 from the center of masss. The system is now taken to be rotating in three dimensions about some axis of rotation. The angular frequency (in radians sec ) of rotation is ω . The kinetic energy of the system is: KE = 1 2 m 1 v 2 1 + 1 2 m 2 v 2 2 = 1 2 m 1 r 2 1 ω 2 + 1 2 m 2 r 2 2 ω 2 = 1 2 ± m 1 r 2 1 + m 2 r 2 2 ² ω 2 = 1 2 I ω 2 where I is deFned as the moment of inertia. If we consider the sum of r 1 and r 2 (considering only their

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Unformatted text preview: magnitues ) as r = r 1 + r 2 , we can write the moment of inertia in terms of the reduced mass as follows: I = m 1 r 2 1 + m 2 r 2 2 = r 2 where the reduced mass is deFned as: = m 1 m 2 m 1 + m 2 1 Thus, the kinetic energy in terms of the reduced mass becomes: KE = 1 2 r 2 2 = 1 2 v 2 The last equation shows us that in terms of the reduce mass, the dynamics of a rigid, diatomic rotor is equivalent to that of a particle of reduced mass (eFectively, we have reduced a two-body problem to a one-body problem). 2...
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This note was uploaded on 02/02/2012 for the course CHEM 444 taught by Professor Dybowski,c during the Fall '08 term at University of Delaware.

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04082010reducedmass - magnitues ) as r = r 1 + r 2 , we can...

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