Ex 1.25 Properties of Dt

Structure and Interpretation of Classical Mechanics

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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}

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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} \\   }

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\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}.

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\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}  }

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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}  }

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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}  }

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\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}  }

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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}  }

So

\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

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2022.04.21 Thursday (c) All rights reserved by ACHK