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Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY 2.71 Optics Spring ’09 Problem Set #4 Posted March 16, 2009 — Due Wednesday, April 1, 2009 1. Particle in quadratic and linear potential Consider a particle whose Hamil- tonian is given by p 2 + p 2 1 H ( q x ,q z ; p x ,p z ) = x y + kq 2 + mgq x , (1) 2 m 2 x where m is the particle mass, k is a spring constant, and g is a constant with units of acceleration. For simplicity, we have neglected the y dimension, i.e. we have restricted the motion to take place on the xz plane. a) Write down the Hamiltonian equations as in Notes p. wk5–b–7, and reduce them to a 4 × 4 set of ordinary differential equations for the dynamical variables q x , q y , p x , p y by taking the appropriate partial derivatives on (1). b) Noting that the equations for ( q x ,p x ) and ( q z ,p z ) can be decoupled from each other, derive a single second–order differential equation for each pair and then solve these second order differential equations with initial conditions q x ( t = 0) = q , q z ( t = 0) = 0; p x ( t = 0) = p , p z ( t = 0) = p z . c) Give an example of a physical system that conforms to this model. d) Is the Hamiltonian conserved in this system? Hint: Compute the derivative d H/ d t using (1). 2. Quadratic GRIN Consider the quadratic GRIN profile that is most often im- plemented in practice, n ( x ) = n 2 − α q x 2 . (2) 2 Unlike the mechanical system of Problem 1, the Hamiltonian equations cannot be solved analytically in this case, so we will integrate them numerically. Before that, however, we will make some simplifications. Consider the diagram on the next page. According to the derivation that we did in class, the position vector q = ( q x ,q y ,q z ) and momentum vector p = ( p x ,p y ,p z ) are parameterized by the variable s . The “time” variable s indexes location along the ray in the sense that as we advance s from to s 1 to s 2 to s 3 and so on, we progress along successive locations on the ray path (see the figure.) plane input plane r a y p a t h p output q plane xz z y ’ x ’ ( ) z z ( ) [ ] (0) (0) q p s s s z z z x y 1 3 2 1 2 3 It turns out that we can eliminate the variable s and instead...
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- Spring '09
- Hamiltonian equations, Screen Hamiltonian, quadratic grin