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l03 - lecture 3 Topics Where are we Consequences of Time...

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lecture 3 Topics: Where are we? Consequences of Time Translation Invariance and Linearity Uniform circular motion Harmonic oscillation for more degrees of freedom The double pendulum The damped harmonic oscillator Where are we? Last time, we saw how initial conditions appeared in a number of different examples of force laws. The last of these, the harmonic oscillator, is a particularly important system because it has two general properties, time translation invariance and linearity, that appear in many many physical systems. Because linearity went by pretty quickly last time, let me briefly review how it works. The equation of motion, F = ma , for the harmonic oscillator is linear because there is a single x in each term. It can be written as m d 2 x dt 2 + K x = 0 (1) We can think of this as a single “operator” acting on x . = ˆ m d 2 dt 2 + K ! x = 0 (2) In this form, it may be more clear why you the solutions form a linear space. If you have two solutions, x 1 ( t ) and x 2 ( t ) , you can form an arbitrary linear combinations and still get solutions, because the same operator acts on both, and multiplying by a constant doesn’t affect the validity of the solution, ˆ m d 2 dt 2 + K ! x 1 = 0 ˆ m d 2 dt 2 + K ! x 2 = 0 ˆ m d 2 dt 2 + K ! ( a x 1 + b x 2 ) = 0 (3) This fact has remarkable consequences, as we will see shortly. Consequences of Time Translation Invariance and Linearity Time translation invariance is an example of a symmetry. The physics of the harmonic oscillator looks the same if all clocks are reset by the same amount. When a symmetry is combined with the property of linearity, the result is an extremely powerful tool for studying the solutions of the system’s equation of motion. The reason is that because of linearity, the solutions of the equation 1
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of motion form what mathematicians call a linear space. You can add them together and multiply them by constants and you still have solutions. Because of this, we can use the tools of linear algebra to understand them. In particular, we can choose a convenient set of basis solutions that behave as simply as possible under time translations. For the symmetry of time translation, it is a mathematical fact that the basis solutions are just exponentials. We can always find solutions of the form 1 z ( t ) = z (0) e Ht (4) What is special about this form (and I am not going to discuss this in detail - I hope that you will see this beautiful argument in more detail in Physics 15c) is that when you change the setting of your clock by taking t t + a , the exponential (4) is the only function that just changes by a multiplicative constant, z ( t + a ) = z (0) e H ( t + a ) = z (0) e Ht e Ha = e Ha z ( t ) (5) You can always use the linearity of the space of solutions to find particularly convenient solutions that behave in this simple way under time translations - and then the result has to be an exponential.
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