Chapter_8_Notes

Classical Mechanics

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Unformatted text preview: 1 PHYS620 Fall 2009 Notes for Chapter 8 Central force motion The motion of two particles interacting by a force that has direction along the line joining the particles and strength that depends only on the separation of the two particles is equivalent to a central force motion. In the center-of-mass reference frame, the system is completely determined by the relative position of the two particles. Furthermore, because the Lagrangian is spherically symmetric about the center-of-mass, the angular momentum about the center-of-mass is conserved. Hence the relative motion lies in a plane that has its normal in the direction of the angular momentum vector. The system can then be described by two generalized co-ordinates. Let the two particles have masses, m 1 and m 2 , and position vectors r 1 and r 2 in the center-of-mass frame. The Lagrangian is ( ) 1 1 1 2 2 2 2 1 1 1 . 2 2 L m m U = ⋅ + ⋅-- r r r r r r ɺ ɺ ɺ ɺ (8.1.1) However, because the center-of-mass is at the co-ordinate origin, r 1 and r 2 are not independent but are related by 1 1 2 2 0. m m + = r r (8.1.2) In terms of the particle separation, 2 1 , =- r r r (8.1.3) we have 2 1 1 2 , m m m = - + r r (8.1.4) and 1 2 1 2 . m m m = + r r (8.1.5) The Lagrangian is then ( ) ( ) 2 2 2 1 1 2 1 2 1 2 1 2 1 2 1 1 2 2 1 . 2 m m L m m U r m m m m m m U r m m = ⋅ + ⋅ - + + = ⋅ - + r r r r r r ɺ ɺ ɺ ɺ ɺ ɺ (8.1.6) The quantity 1 2 1 2 , m m m m μ = + (8.1.7) is called the reduced mass of the system. 2 We have reduced the motion of two particles to an equivalent problem of a single particle of mass μ acted on by a force directed along the line from the particle to a fixed point. The generalized momentum conjugate to the position vector r has components , i i L p r ∂ = ∂ ɺ (8.1.8) and hence . μ = p r ɺ (8.1.9) The angular momentum , = × L r p (8.1.10) is orthogonal to both r and p . Hence the motion is in the plane through the origin with normal parallel to L . The motion has two degree of freedom and can be described by using polar co- ordinates ( r , θ ). In these co-ordinates, the Lagrangian is ( ) ( ) 2 2 2 1 . 2 L r r U r μ θ = +- ɺ ɺ (8.1.11) Since this does not depend explicitly on θ , i.e. θ is ignorable there is a first integral of motion 2 , L p r l θ μ θ θ ∂ = = = ∂ ɺ ɺ (8.1.12) where l is a constant. In other words, the angular momentum about an axis through the origin normal to the plane of motion is conserved. Since the rate at which the position vector sweeps out area is 2 1 , 2 2 dA d l r dt dt θ μ = = (8.1.13) conservation of angular momentum leads to Kepler’s second law of orbital motion. Also because the Lagrangian does not explicitly depend on time, the Hamiltonian is conserved. Furthermore, the two conditions for the Hamiltonian to be equal to the energy are satisfied. Hence r θ d θ O 3 ( ) ( ) 2 2 2 2 2 2 2 1 2 1 1 ....
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Chapter_8_Notes - 1 PHYS620 Fall 2009 Notes for Chapter 8...

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