Lecture29

Lecture29 - Physics 344 Foundations of 21st Century...

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Physics 344 Foundations of 21 st Century Physics: Relativistic and Quantum Systems Instructor: Dr. Mark Haugan Office: PHYS 282 haugan@purdue.edu TA: Dan Hartzler Office: PHYS 7 dhartzle@purdue.edu Grader: Fan Chen Office: PHYS 222 chen926@purdue.edu Office Hours: If you have questions, just email us to make an appointment. We enjoy talking about physics! Help Session: Thursdays 2:00 – 4:00 in PHYS 154 Reading: Sections 1.3 through 1.6 and Chapters 4 and 5 in Six Ideas that Shaped Physics, Unit Q Exam 2: Thursday, December 1 at 8:00pm in ARMS B061
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Electromagnetic Cavity Modes Studying pulses bouncing between mirrors allowed us to extend our study of interference far beyond the multi-slit cases we first studied. cos( ) n n n E α δ = and sin( ) n n n E β = 2 2 2 n n n E = + and tan( ) n n n = ( 29 ( 29 1 n=1 n=1 ( , ) sin( )sin( ) sin( ) sin( )cos( ) cos( )sin( ) sin( ) sin( ) cos( ) y n n n n n n n n n n n n n n n n E x t E k x t E k x t t k x t t ϖ = = + = + = + ( 29 ( 29 0 n=0 n=0 ( , ) cos( )cos( ) cos( ) cos( )cos( ) sin( )sin( ) 1 cos( ) cos( ) sin( ) n n z n n n n n n n n n n n n n n E E B x t k x t k x t t c c k x t t c = = + = - = - Using Fourier’s ideas, we are able to represent any state of the electromagnetic field within a cavity as a superposition of the standing wave fields (mode oscillations) possible within the cavity. For y-polarized fields, we found that and with k n = n π /a , ω n = ck n and equivalently
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We also found that we could determine the α n and β n coefficients representing specific fields within a cavity from the configuration of the fields at an initial time, t = 0, n=1 ( ,0) sin( ) y n n E x k x β n=0 ( ,0) cos( ) z n n cB x k x α = and ( 29 n=1 ( , ) sin( ) sin( ) cos( ) y n n n n n E x t k x t t ϖ = + ( 29 ( 29 n=0 ( , ) 1/ cos( ) cos( ) sin( ) z n n n n n B x t c k x t t = - We did this using Fourier’s Trick which exploits the orthogonality of the sine and cosine mode functions on the interval x = [0, a ], 0 sin sin 2 a mn n x m x a dx a a π δ = 0 cos cos 2 a mn n x m x a dx a a = and concluding that 0 0 n=1 n=1 2 2 2 sin( ) ( ,0) sin( )sin( ) 2 a a m y n m n n mn m a k x E x dx k x k x dx a a a = = = 0 0 n=1 n=1 2 2 2 cos( ) ( ,0) cos( )cos( ) 2 a a m z n m n n mn m c a k x B x dx k x k x dx a a a = = = and
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Lecture29 - Physics 344 Foundations of 21st Century...

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