- #71
etotheipi
Apologies for a late reply; the pubs were open again today 
Given any frame ##F = P \zeta \eta \xi## one can define the time derivative with respect to ##F## of an arbitrary vector ##\mathbf{u}##$$\mathcal{D}_F \mathbf{u} := \frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} \big{|}_{F} = \frac{\mathrm{d}{u}^{\zeta}}{\mathrm{d}t} \mathbf{e}_{\zeta} + \frac{\mathrm{d}{u}^{\eta}}{\mathrm{d}t} \mathbf{e}_{\eta} + \frac{\mathrm{d}{u}^{\xi}}{\mathrm{d}t} \mathbf{e}_{\xi}$$It's easy to show, given another frame ##G = Q \rho \sigma \tau##, that ##\mathcal{D}_F \mathbf{u} = \mathcal{D}_G \mathbf{u} + \boldsymbol{\omega} \times \mathbf{u}## where ##\omega_i = \frac{1}{2} \epsilon_{ijk} \dot{\mathcal{R}}_{jl} \mathcal{R}_{kl}## and ##(\mathcal{R}_{ij})## is the orthogonal, time-dependent matrix such that ##\mathbf{e}^G_i = \mathcal{R}_{ij} \mathbf{e}^F_j##.
Now let us show how to obtain the equations of motion for a particle in an arbitrary reference system. Begin with an inertial reference system ##K = Oxyz##, in which the Lagrangian is simply$$\mathscr{L} = \frac{1}{2} m v^2 - \varphi$$where ##\mathbf{v} := \mathcal{D}_K \boldsymbol{x}##. Let's now consider a second reference system ##L = O'x'y'z'## and define ##\mathbf{R}(t) := \overrightarrow{O(t)O'(t)}##, such that ##\boldsymbol{x} = \mathbf{R} + \boldsymbol{x}'##. Operating with ##\mathcal{D}_K## yields, upon defining ## \boldsymbol{\mathcal{V}} := \mathcal{D}_K \mathbf{R}## and ##\mathbf{v}' := \mathcal{D}_L \boldsymbol{x}'##\begin{align*}
\mathcal{D}_K \boldsymbol{x} &= \mathcal{D}_K \mathbf{R} + \mathcal{D}_K \boldsymbol{x}' \\
\mathbf{v} &= \boldsymbol{\mathcal{V}} + \left[ \mathcal{D}_L \boldsymbol{x}' + \boldsymbol{\omega} \times \boldsymbol{x}' \right] \\
\mathbf{v} &= \boldsymbol{\mathcal{V}} + \mathbf{v}' + \boldsymbol{\omega} \times \boldsymbol{x}'
\end{align*}It follows that\begin{align*}v^2 = \mathcal{V}^2 + v'^2 + (\boldsymbol{\omega} \times \boldsymbol{x}')^2 + 2 \mathbf{v}' \cdot \boldsymbol{\omega} \times \boldsymbol{x}' + 2 \boldsymbol{\mathcal{V}} \cdot \mathbf{v}' + 2 \boldsymbol{\mathcal{V}} \cdot \boldsymbol{\omega} \times \boldsymbol{x}'
\end{align*}The new Lagrangian is then written\begin{align*}
\mathscr{L}' = \frac{1}{2} m \left( \mathcal{V}^2 + v'^2 + \boldsymbol{\omega} \cdot (\boldsymbol{x}' \times (\boldsymbol{\omega} \times \boldsymbol{x}')) + 2 \mathbf{v}' \cdot \boldsymbol{\omega} \times \boldsymbol{x}' + 2 \boldsymbol{\mathcal{V}} \cdot \mathbf{v}' + 2 \boldsymbol{\mathcal{V}} \cdot \boldsymbol{\omega} \times \boldsymbol{x}' \right) - \varphi
\end{align*}where we have re-written ##(\boldsymbol{\omega} \times \boldsymbol{x}')^2 = \boldsymbol{\omega} \cdot (\boldsymbol{x}' \times (\boldsymbol{\omega} \times \boldsymbol{x}'))##. Let us now compute the derivatives of the Lagrangian\begin{align*}
\frac{\partial \mathscr{L}'}{\partial x'^i} &= -m \epsilon_{ijk} \epsilon_{klm} \omega^j \omega^l x'^m -m \epsilon_{ijk}\omega^j (\mathcal{V}^k + \dot{x}'^k) - \partial_i \varphi \\ \\
\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial \mathscr{L}'}{\partial \dot{x}'^i} &= m \dot{\mathcal{V}}^i + m \ddot{x}'^i + m\epsilon_{ijk} ( \dot{\omega}^j x'^k + \omega^j \dot{x}'^k)
\end{align*}The Euler-Lagrange equation ##\partial \mathscr{L}' / \partial \boldsymbol{x}' = d_t \left(\partial \mathscr{L}' / \partial \dot{\boldsymbol{x}}' \right)## thus implies the following equation of motion in the ##L## reference system, after converting from suffix notation back into vector notation,$$m \mathbf{a}' = - \nabla \varphi - m \boldsymbol{\mathcal{A}} - 2m \boldsymbol{\omega} \times \mathbf{v}' - m \boldsymbol{\alpha} \times \boldsymbol{x}' -m \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{x}')$$where ##\boldsymbol{\mathcal{A}} := \mathcal{D}_K \boldsymbol{\mathcal{V}} = \mathcal{D}_L \boldsymbol{\mathcal{V}} + \boldsymbol{\omega} \times \boldsymbol{\mathcal{V}} ## is the translational acceleration of the ##L##-system with respect to the ##K##-system and ##\boldsymbol{\alpha} := \mathcal{D}_{K} \boldsymbol{\omega}## is the rotational acceleration of the ##L##-system with respect to the ##K##-system. And of course ##\mathbf{a}' = \mathcal{D}_L^2 \boldsymbol{x}'## is the acceleration of the particle with respect to the ##L##-system.
This is the equation of motion of a particle in an arbitrary reference system ##L##; the additional terms arising from the co-ordinate transformation are not forces, although they are sometimes affectionally referred to as 'inertial forces'.
Given any frame ##F = P \zeta \eta \xi## one can define the time derivative with respect to ##F## of an arbitrary vector ##\mathbf{u}##$$\mathcal{D}_F \mathbf{u} := \frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} \big{|}_{F} = \frac{\mathrm{d}{u}^{\zeta}}{\mathrm{d}t} \mathbf{e}_{\zeta} + \frac{\mathrm{d}{u}^{\eta}}{\mathrm{d}t} \mathbf{e}_{\eta} + \frac{\mathrm{d}{u}^{\xi}}{\mathrm{d}t} \mathbf{e}_{\xi}$$It's easy to show, given another frame ##G = Q \rho \sigma \tau##, that ##\mathcal{D}_F \mathbf{u} = \mathcal{D}_G \mathbf{u} + \boldsymbol{\omega} \times \mathbf{u}## where ##\omega_i = \frac{1}{2} \epsilon_{ijk} \dot{\mathcal{R}}_{jl} \mathcal{R}_{kl}## and ##(\mathcal{R}_{ij})## is the orthogonal, time-dependent matrix such that ##\mathbf{e}^G_i = \mathcal{R}_{ij} \mathbf{e}^F_j##.
Now let us show how to obtain the equations of motion for a particle in an arbitrary reference system. Begin with an inertial reference system ##K = Oxyz##, in which the Lagrangian is simply$$\mathscr{L} = \frac{1}{2} m v^2 - \varphi$$where ##\mathbf{v} := \mathcal{D}_K \boldsymbol{x}##. Let's now consider a second reference system ##L = O'x'y'z'## and define ##\mathbf{R}(t) := \overrightarrow{O(t)O'(t)}##, such that ##\boldsymbol{x} = \mathbf{R} + \boldsymbol{x}'##. Operating with ##\mathcal{D}_K## yields, upon defining ## \boldsymbol{\mathcal{V}} := \mathcal{D}_K \mathbf{R}## and ##\mathbf{v}' := \mathcal{D}_L \boldsymbol{x}'##\begin{align*}
\mathcal{D}_K \boldsymbol{x} &= \mathcal{D}_K \mathbf{R} + \mathcal{D}_K \boldsymbol{x}' \\
\mathbf{v} &= \boldsymbol{\mathcal{V}} + \left[ \mathcal{D}_L \boldsymbol{x}' + \boldsymbol{\omega} \times \boldsymbol{x}' \right] \\
\mathbf{v} &= \boldsymbol{\mathcal{V}} + \mathbf{v}' + \boldsymbol{\omega} \times \boldsymbol{x}'
\end{align*}It follows that\begin{align*}v^2 = \mathcal{V}^2 + v'^2 + (\boldsymbol{\omega} \times \boldsymbol{x}')^2 + 2 \mathbf{v}' \cdot \boldsymbol{\omega} \times \boldsymbol{x}' + 2 \boldsymbol{\mathcal{V}} \cdot \mathbf{v}' + 2 \boldsymbol{\mathcal{V}} \cdot \boldsymbol{\omega} \times \boldsymbol{x}'
\end{align*}The new Lagrangian is then written\begin{align*}
\mathscr{L}' = \frac{1}{2} m \left( \mathcal{V}^2 + v'^2 + \boldsymbol{\omega} \cdot (\boldsymbol{x}' \times (\boldsymbol{\omega} \times \boldsymbol{x}')) + 2 \mathbf{v}' \cdot \boldsymbol{\omega} \times \boldsymbol{x}' + 2 \boldsymbol{\mathcal{V}} \cdot \mathbf{v}' + 2 \boldsymbol{\mathcal{V}} \cdot \boldsymbol{\omega} \times \boldsymbol{x}' \right) - \varphi
\end{align*}where we have re-written ##(\boldsymbol{\omega} \times \boldsymbol{x}')^2 = \boldsymbol{\omega} \cdot (\boldsymbol{x}' \times (\boldsymbol{\omega} \times \boldsymbol{x}'))##. Let us now compute the derivatives of the Lagrangian\begin{align*}
\frac{\partial \mathscr{L}'}{\partial x'^i} &= -m \epsilon_{ijk} \epsilon_{klm} \omega^j \omega^l x'^m -m \epsilon_{ijk}\omega^j (\mathcal{V}^k + \dot{x}'^k) - \partial_i \varphi \\ \\
\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial \mathscr{L}'}{\partial \dot{x}'^i} &= m \dot{\mathcal{V}}^i + m \ddot{x}'^i + m\epsilon_{ijk} ( \dot{\omega}^j x'^k + \omega^j \dot{x}'^k)
\end{align*}The Euler-Lagrange equation ##\partial \mathscr{L}' / \partial \boldsymbol{x}' = d_t \left(\partial \mathscr{L}' / \partial \dot{\boldsymbol{x}}' \right)## thus implies the following equation of motion in the ##L## reference system, after converting from suffix notation back into vector notation,$$m \mathbf{a}' = - \nabla \varphi - m \boldsymbol{\mathcal{A}} - 2m \boldsymbol{\omega} \times \mathbf{v}' - m \boldsymbol{\alpha} \times \boldsymbol{x}' -m \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{x}')$$where ##\boldsymbol{\mathcal{A}} := \mathcal{D}_K \boldsymbol{\mathcal{V}} = \mathcal{D}_L \boldsymbol{\mathcal{V}} + \boldsymbol{\omega} \times \boldsymbol{\mathcal{V}} ## is the translational acceleration of the ##L##-system with respect to the ##K##-system and ##\boldsymbol{\alpha} := \mathcal{D}_{K} \boldsymbol{\omega}## is the rotational acceleration of the ##L##-system with respect to the ##K##-system. And of course ##\mathbf{a}' = \mathcal{D}_L^2 \boldsymbol{x}'## is the acceleration of the particle with respect to the ##L##-system.
This is the equation of motion of a particle in an arbitrary reference system ##L##; the additional terms arising from the co-ordinate transformation are not forces, although they are sometimes affectionally referred to as 'inertial forces'.
Last edited by a moderator:
