Ex 1.25 Properties of Dt

Structure and Interpretation of Classical Mechanics

The total time derivative $\displaystyle{D_t F}$ is not the derivative of the function $\displaystyle{F}$ . Nevertheless, the total time derivative shares many properties with the derivative. Demonstrate that $\displaystyle{D_t}$ has the following properties …

$\displaystyle{D_t (F + G) = D_t F + D_t G}$

~~~

Eq. (1.108):

$\displaystyle{ D(F \circ \Gamma[q]) = (DF\circ\Gamma[q])D\Gamma[q] }$

Eq. (1.109):

$\displaystyle{ DF \circ \Gamma[q] = \left[ \partial_0 F \circ \Gamma[q], \partial_1 F \circ \Gamma[q], \partial_2 F \circ \Gamma[q], ... \right] }$

Eq. (1.110):

$\displaystyle{ \left(D \Gamma[q] \right)(t) = \left( 1, Dq(t), D^2 q(t), ... \right) = \begin{bmatrix} 1 \\ Dq(t) \\ D^2 q(t) \\ ... \\ \end{bmatrix} \\ }$

$\displaystyle{ \begin{aligned} D(F \circ \Gamma[q])(t) &= (DF\circ\Gamma[q])D\Gamma[q](t) \\ &= \left[ \partial_0 F \circ \Gamma[q], \partial_1 F \circ \Gamma[q], \partial_2 F \circ \Gamma[q], ... \right] \begin{bmatrix} 1 \\ Dq(t) \\ D^2 q(t) \\ ... \\ \end{bmatrix} \\ &= \partial_0 F \circ \Gamma[q] + \partial_1 F \circ \Gamma[q] D q(t) + \partial_2 F \circ \Gamma[q] D^2 q(t) + ... \\ &= \partial_0 F \circ \Gamma[q] u(t) + \partial_1 F \circ \Gamma[q] D q(t) + \partial_2 F \circ \Gamma[q] D^2 q(t) + ... \\ \end{aligned} }$ ,

where $\displaystyle{u(t) \equiv 1}$ .

$\displaystyle{ \begin{aligned} D(F \circ \Gamma[q]) &= \partial_0 F \circ \Gamma[q] u + \partial_1 F \circ \Gamma[q] D q + \partial_2 F \circ \Gamma[q] D^2 q + ... \\ &= \partial_0 F \circ \Gamma[q] J_0 \circ \Gamma[q] + \partial_1 F \circ \Gamma[q] J_1 \circ \Gamma[q] + \partial_2 F \circ \Gamma[q] J_2 \circ \Gamma[q] + ... \\ \end{aligned} }$

where

$\displaystyle{ \begin{aligned} I_0 \circ \Gamma[q] &= t \\ \\ I_{n>0} \circ \Gamma[q] &= I_{n>0} (t, q, v, a, ...) \\ &= I_{n>0} (t, q, Dq, D^2 q, ...) \\ &= D^{(n-1)} q \\ \\ J_{n} \circ \Gamma[q] &= D(I_n (t, q, v, a, ...)) \\ \end{aligned} }$

The meaning of $\displaystyle{\delta_\eta (fg)[q]}$ is

$\displaystyle{\delta_\eta (f[q]g[q])}$

— Me@2019-04-27 07:02:38 PM

$\displaystyle{ \begin{aligned} D(F \circ \Gamma[q]) &= \partial_0 F \circ \Gamma[q] J_0 \circ \Gamma[q] + \partial_1 F \circ \Gamma[q] J_1 \circ \Gamma[q] + \partial_2 F \circ \Gamma[q] J_2 \circ \Gamma[q] + ... \\ &= \left[(\partial_0 F) J_0 + (\partial_1 F) J_1 + (\partial_2 F) J_2 + ... \right] \circ \Gamma[q] \\ \end{aligned} }$

.
Eq. (1.113):

$\displaystyle{ D_t F \circ \Gamma[q] = D(F \circ \Gamma[q]) }$

$\displaystyle{ \begin{aligned} D_t F \circ \Gamma[q] &= \left[(\partial_0 F) J_0 + (\partial_1 F) J_1 + (\partial_2 F) J_2 + ... \right] \circ \Gamma[q] \\ \\ D_t F &= (\partial_0 F) J_0 + (\partial_1 F) J_1 + (\partial_2 F) J_2 + ... \\ \end{aligned} }$

Eq. (1.114):

$\displaystyle{ \begin{aligned} D_t F (t, q, v, a, ...) &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v + \partial_2 F(t, q, v, a, ...) a + ... \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} D_t F \circ \Gamma[q] (t) &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v(t) + \partial_2 F(t, q, v, a, ...) a(t) + ... \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} &D_t (F + G) \circ \Gamma[q] (t) \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} &= \partial_0 \left[F(t, q, v, a, ...) + G(t, q, v, a, ...)\right] \\ &+ \partial_1 \left[F(t, q, v, a, ...) + G(t, q, v, a, ...)\right] v(t) \\ &+ \partial_2 \left[F(t, q, v, a, ...) + G(t, q, v, a, ...)\right] a(t) + ... \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} &= \left[ \partial_0 F(t, q, v, a, ...) + \partial_0 G(t, q, v, a, ...)\right] \\ &+ \left[ \partial_1 F(t, q, v, a, ...) + \partial_1 G(t, q, v, a, ...)\right] v(t) \\ &+ \left[ \partial_2 F(t, q, v, a, ...) + \partial_2 G(t, q, v, a, ...)\right] a(t) + ... \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v(t) + \partial_2 F(t, q, v, a, ...) a(t) + ... \\ &+ \partial_0 G(t, q, v, a, ...) + \partial_1 G(t, q, v, a, ...) v(t) + \partial_2 G(t, q, v, a, ...) a(t) + ... \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} &= D_t F \circ \Gamma[q] (t) + D_t G \circ \Gamma[q] (t) \\ \end{aligned} }$

In short,

$\displaystyle{ \begin{aligned} D_t (F + G) \circ \Gamma[q] (t) &= D_t F \circ \Gamma[q] (t) + D_t G \circ \Gamma[q] (t) \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} D_t (F + G) \circ \Gamma[q] (t) &= (D_t F + D_t G) \circ \Gamma[q] (t) \\ \\ D_t (F + G) \circ \Gamma[q] &= (D_t F + D_t G) \circ \Gamma[q] \\ \\ D_t (F + G) &= D_t F + D_t G \\ \end{aligned} }$

— Me@2022-04-20 11:42:52 AM

Physics Town

continues

Ex 1.25 Properties of Dt

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