Difference between revisions of "CM Course Notes"

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==220 Notes==
 
==220 Notes==
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'''Lagrangian Mechanics'''
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Method for finding normal modes of system <div class="mw-collapsible mw-collapsed" style="width:750px">
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Since we expect oscillatory motion of a normal mode (where ω is the same for both masses), we try:
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\[
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x_1(t) = A_1 e^{i \omega t} \\
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x_2(t) = A_2 e^{i \omega t}
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  \]
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Euler-Lagrange Equation of system without constraints: <div class="mw-collapsible mw-collapsed" style="width:750px">
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:<math>\frac{\mathrm{d}}{\mathrm{d}t} \left ( \frac {\partial  L}{\partial \dot{q}_j} \right ) =  \frac {\partial L}{\partial q_j} </math>
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'''Hamiltonian Mechanics'''
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Hamiltonian i.t.o. Lagrangian:
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:<math>H = \dot{\mathbf{q}}  \frac{\partial \mathcal{L}}{\partial \dot{\mathbf{q}}} - \mathcal{L} </math>
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Transform from rotating to fixed frame:
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\[
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\vec{v}' = \vec{v}  - \vec{\omega} \times \vec{r}
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\]
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'''Relativity'''
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Write the product $\gamma\beta$ in terms of only $\gamma$:<div class="mw-collapsible mw-collapsed" style="width:750px">
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:<math>\gamma \beta = \sqrt{\frac{\beta^2}{1- \beta^2}}= \sqrt{\frac{A}{1-\beta^2} + B} \\
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\text{In the words of the notorious J.D. Jackson, ''we see'' that }A = 1 \text{ and }B = -1 \text{ , such that } \\ \gamma \beta = \sqrt{\frac{1}{1-\beta^2} - 1}=\sqrt{\gamma^2 - 1}  </math>
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E.g.: Lim#3021
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===''<small>Exercises</small>''===
 
===''<small>Exercises</small>''===

Revision as of 23:19, 14 April 2014

Here is a conglomerate of notes gathered for the graduate CM class.


Contents

220 Notes

Lagrangian Mechanics

Method for finding normal modes of system

Since we expect oscillatory motion of a normal mode (where ω is the same for both masses), we try: \[ x_1(t) = A_1 e^{i \omega t} \\ x_2(t) = A_2 e^{i \omega t} \]

Euler-Lagrange Equation of system without constraints:

\[\frac{\mathrm{d}}{\mathrm{d}t} \left ( \frac {\partial L}{\partial \dot{q}_j} \right ) = \frac {\partial L}{\partial q_j} \]

Hamiltonian Mechanics

Hamiltonian i.t.o. Lagrangian:

\[H = \dot{\mathbf{q}} \frac{\partial \mathcal{L}}{\partial \dot{\mathbf{q}}} - \mathcal{L} \]

Transform from rotating to fixed frame:

\[ \vec{v}' = \vec{v} - \vec{\omega} \times \vec{r} \]

Relativity

Write the product $\gamma\beta$ in terms of only $\gamma$:

\[\gamma \beta = \sqrt{\frac{\beta^2}{1- \beta^2}}= \sqrt{\frac{A}{1-\beta^2} + B} \\ \text{In the words of the notorious J.D. Jackson, ''we see'' that }A = 1 \text{ and }B = -1 \text{ , such that } \\ \gamma \beta = \sqrt{\frac{1}{1-\beta^2} - 1}=\sqrt{\gamma^2 - 1} \] E.g.: Lim#3021


Exercises

General Identities


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