Ex 1.28 The energy function

Structure and Interpretation of Classical Mechanics

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

c. Show, using Euler’s theorem, that the energy function is $\mathcal{E} = A + B$.

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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}}

— Euler’s Homogeneous Function Theorem

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\displaystyle{\begin{aligned} L + D_t F &= A - B \\ \\ D_t F &= - \frac{1}{2} \sum_\alpha m_\alpha \left \{ g_\alpha(t,q) + [\partial_0 f_\alpha (t,q)]^2 + 2 \partial_0 f_\alpha (t,q) \partial_1 f_\alpha (t,q) v \right \} \\ \\ A &= \frac{1}{2} \sum_\alpha m_\alpha [\partial_1 f_\alpha (t,q) ]v^2 \\ - B &= - V(t, q) - \frac{1}{2} \sum_\alpha m_\alpha g_\alpha(t,q) \\ \end{aligned}}

This answer is not totally correct, since the generalized velocity, $v$, should be a vector.

— Me@2022-11-01 08:58:52 AM

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Eq. (1.133):

$\displaystyle{\mathcal{P}_i = (\partial_2 L)_i}$

Eq. (1.144):

$\displaystyle{\mathcal{E} = \mathcal{P} \dot Q - L}$,

where $\displaystyle{\mathcal{P}}$ is the momentum state function.

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\displaystyle{\begin{aligned} \mathcal{E} &= \mathcal{P} \dot Q - A + B \\ &= \left( \partial_2 L \right) \dot Q - A + B \\ &= \left( \partial_2 (A - B) \right) \dot Q - A + B \\ \end{aligned}}

Since $B$ has no velocity dependence,

\displaystyle{\begin{aligned} \mathcal{E} &= \left( \partial_2 A \right) \dot Q - A + B \\ &= \begin{bmatrix} (\partial_2 A)_1 & (\partial_2 A)_2 & ... \end{bmatrix} \begin{bmatrix} \dot Q_1 \\ \dot Q_2 \\ \vdots \end{bmatrix} - A + B \\ \\ \end{aligned}}

Since $A$ is a homogeneous function of the generalized velocities of degree 2, by Euler’s Homogeneous Function Theorem,

\displaystyle{ \begin{aligned} \sum_{i} v_i \frac{\partial A}{\partial v_i} &= 2 A(v) \\ \end{aligned}}

— Me@2023-04-06 12:49:49 PM

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