# 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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