Ex 1.28 Prequel 1

Structure and Interpretation of Classical Mechanics

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An analogous result holds when the f_\alpha‘s depend explicitly on time.

a. Show that in this case the kinetic energy contains terms that are linear in the generalized velocities.

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Eq. (1.141):

\displaystyle{  \textbf{v}_\alpha = \partial_0 f_\alpha (t,q) + \partial_1 f_\alpha (t, q) v  }

Eq. (1.142):

\displaystyle{T(t, q, v) = \frac{1}{2} \sum_{\alpha} m_\alpha v^2_\alpha}

Eq. (1.133):

\displaystyle{  \begin{aligned}  \mathcal{P}_i &= (\partial_2 L)_i \\   \end{aligned}}

where \displaystyle{\mathcal{P}} is called the generalized momentum.

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\displaystyle{  \begin{aligned}  \mathcal{P} \dot Q &= (\partial_2 L) \dot Q  \\   &= (\partial_2 (T-V)) \dot Q  \\   &= (\partial_2 T) \dot Q  \\   \end{aligned}}

where \displaystyle{\dot Q} is the velocity selector. And V has no velocity component?

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“… T is a homogeneous function of the generalized velocities of degree 2.”

\displaystyle{  \begin{aligned}  n &= 2 \\   (\partial_2 T) \dot Q &= ? \\   \end{aligned}}

\displaystyle{  \begin{aligned}  x Df(x) &= nf(x)  \\   \end{aligned}}

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What does the D actually mean?

\displaystyle{ \begin{aligned}    x Df(x) &= nf(x)  \\   x \frac{d}{dx} f(x) &= nf(x)  \\   \end{aligned}}

\displaystyle{ \begin{aligned}    \mathbf{v} D T (\mathbf{v}) &= 2 T (\mathbf{v}) ? \\     \end{aligned}}

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\displaystyle{ \begin{aligned}     \partial_{\vec 2}   = \frac{\partial}{\partial \vec{\dot q}}   &= \begin{bmatrix} \frac{\partial}{\partial \dot q_1} & \frac{\partial}{\partial \dot q_2} & ... \end{bmatrix} \\ \\      \mathcal{P} \dot Q &= (\partial_2 T) \dot Q \\ \\    \begin{bmatrix} \mathcal{P}_1 & \mathcal{P}_2 & ... \end{bmatrix}     \begin{bmatrix} \dot Q_1 \\ \dot Q_2 \\ \vdots \end{bmatrix}     &=    \begin{bmatrix} (\partial_2 T)_1 & (\partial_2 T)_2 & ... \end{bmatrix}     \begin{bmatrix} \dot Q_1 \\ \dot Q_2 \\ \vdots \end{bmatrix} \\     \\    &=    \begin{bmatrix} \frac{\partial T}{\partial \dot q_1} &  \frac{\partial T}{\partial \dot q_2}  & ... \end{bmatrix}     \begin{bmatrix} \dot Q_1 \\ \dot Q_2 \\ \vdots \end{bmatrix} \\     \\     &=    \frac{\partial T}{\partial \dot q_1} \dot Q_1 + \frac{\partial T}{\partial \dot q_2} \dot Q_2  + ...     \\       \end{aligned}}

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Euler’s Homogeneous Function Theorem:

If

\displaystyle{ \begin{aligned}      f(t x_1, t x_2, ...) &= t^n f( x_1, x_2 , ...),  \\     \end{aligned}}

then

\displaystyle{ \begin{aligned}      \sum_{i} x_i \frac{\partial f}{\partial x_i} &= n f(\mathbf{x}) \\    \end{aligned}}

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\displaystyle{T(t, q, v) = \frac{1}{2} \sum_{\alpha} m_\alpha v^2_\alpha}

So

\displaystyle{\begin{aligned}     T(t, q, uv)     &= \frac{1}{2} \sum_{\alpha} m_\alpha (uv)^2_\alpha \\    &= u^2 T(t, q, v) \\    \end{aligned}    }

Therefore,

\displaystyle{ \begin{aligned}     \frac{\partial T}{\partial \dot q_1} \dot Q_1 + \frac{\partial T}{\partial \dot q_2} \dot Q_2  + ...     &= 2 T \\     \end{aligned}}

— Me@2022.09.27 12:26:09 PM

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2022.09.27 Tuesday (c) All rights reserved by ACHK