PhyslibAlpha.ClassicalMechanics.MomentMap.GalileanMass

The zero element of the Galilean Lie algebra is defined as the quadruple $(\omega, \beta, \gamma, \varepsilon) = (0, 0, 0, 0)$, where $\omega \in \mathbb{R}^3$ is the rotation (axial vector), $\beta \in \mathbb{R}^3$ is the change of velocity, $\gamma \in \mathbb{R}^3$ is the space translation, and $\varepsilon \in \mathbb{R}$ is the time translation.

The addition of two elements $Z = (\omega, \beta, \gamma, \epsilon)$ and $Z' = (\omega', \beta', \gamma', \epsilon')$ in the Galilean Lie algebra is defined component-wise as $Z + Z' = (\omega + \omega', \beta + \beta', \gamma + \gamma', \epsilon + \epsilon')$, where $\omega$ is the rotation (axial vector), $\beta$ is the change of velocity, $\gamma$ is the space translation, and $\epsilon$ is the time translation.

For any two elements $Z = (\omega, \beta, \gamma, \varepsilon)$ and $Z' = (\omega', \beta', \gamma', \varepsilon')$ in the Galilean Lie algebra, Souriau's bracket $[Z, Z']$ is defined as the element: $[Z, Z'] = (\omega' \times \omega, \omega' \times \beta - \omega \times \beta', \omega' \times \gamma - \omega \times \gamma' + \varepsilon \beta' - \varepsilon' \beta, 0)$ Here, $\omega, \omega' \in \mathbb{R}^3$ represent rotations (axial vectors), $\beta, \beta' \in \mathbb{R}^3$ represent changes of velocity, $\gamma, \gamma' \in \mathbb{R}^3$ represent space translations, and $\varepsilon, \varepsilon' \in \mathbb{R}$ represent time translations. The symbol $\times$ denotes the standard cross product in $\mathbb{R}^3$.

For any two elements $Z$ and $Z'$ in the Galilean Lie algebra, Souriau's bracket satisfies the antisymmetry property $[Z, Z'] + [Z', Z] = 0$, where $0$ denotes the zero element of the algebra.

For any three elements $Z, Z', Z''$ of the Galilean Lie algebra, Souriau's bracket $[\cdot, \cdot]$ satisfies the Jacobi identity: $$[Z, [Z', Z'']] + [Z', [Z'', Z]] + [Z'', [Z, Z']] = 0$$ where $0$ is the zero element $(0, 0, 0, 0)$ of the algebra.

For any vectors $\beta, \gamma \in \mathbb{R}^3$, the Lie bracket of a boost (change of velocity) and a space translation in the Galilean Lie algebra is zero: $$[(0, \beta, 0, 0), (0, 0, \gamma, 0)] = 0$$ where $( \omega, \beta, \gamma, \varepsilon )$ denotes an element of the Galilean Lie algebra consisting of a rotation $\omega$, a change of velocity $\beta$, a space translation $\gamma$, and a time translation $\varepsilon$. This result shows that a change of velocity and a space translation commute.

Given a Galilean torsor $\mu = (l, g, p, E)$ and an element of the Galilean Lie algebra $Z = (\omega, \beta, \gamma, \varepsilon)$, the pairing $\mu(Z)$ is defined as: $$\mu(Z) = l \cdot \omega - g \cdot \beta + p \cdot \gamma - E \varepsilon$$ where $l$ is the angular momentum, $g$ is the center-of-mass term, $p$ is the linear momentum, and $E$ is the energy of the torsor $\mu$; and $\omega$ is the rotation axial vector, $\beta$ is the velocity boost, $\gamma$ is the spatial translation, and $\varepsilon$ is the time translation of the algebra element $Z$.

The cocycle $f_0$ is a 2-form on the Galilean Lie algebra that maps two elements $Z$ and $Z'$ to the real value defined by $f_0(Z, Z') = \beta \cdot \gamma' - \beta' \cdot \gamma$, where $\beta, \beta' \in \mathbb{R}^3$ are the boost (change of velocity) vectors and $\gamma, \gamma' \in \mathbb{R}^3$ are the space translation vectors of $Z$ and $Z'$ respectively, and $\cdot$ denotes the standard dot product.

For any two elements $Z$ and $Z'$ of the Galilean Lie algebra, the cocycle $f_0$ satisfies the antisymmetry property: $$f_0(Z, Z') = -f_0(Z', Z)$$ where $f_0$ is the 2-form on the Galilean Lie algebra defined by $f_0(Z, Z') = \beta \cdot \gamma' - \beta' \cdot \gamma$, with $\beta, \beta'$ being the boost vectors and $\gamma, \gamma'$ being the space translation vectors of $Z$ and $Z'$, respectively.

For any three elements $Z, Z', Z''$ of the Galilean Lie algebra, the Galilean cocycle $f_0$ satisfies the cyclic identity: $$f_0(Z, [Z', Z'']) + f_0(Z', [Z'', Z]) + f_0(Z'', [Z, Z']) = 0$$ where $[ \cdot, \cdot ]$ denotes Souriau's bracket on the Galilean Lie algebra.

The Galilean Lie algebra cocycle $f_0$ evaluated on a pure boost along the $x$-axis (represented by the vector $(1, 0, 0)$) and a pure space translation along the same axis is equal to $1$. In the notation of the Galilean Lie algebra elements $Z = (\omega, \beta, \gamma, \epsilon)$, this is expressed as $f_0(\langle 0, (1, 0, 0), 0, 0 \rangle, \langle 0, 0, (1, 0, 0), 0 \rangle) = 1$.

Let $f_0$ be the 2-cocycle on the Galilean Lie algebra defined by $f_0(Z, Z') = \beta \cdot \gamma' - \beta' \cdot \gamma$, where $Z = (\omega, \beta, \gamma, \epsilon)$ and $Z' = (\omega', \beta', \gamma', \epsilon')$ are elements representing rotations, velocity boosts, space translations, and time translations respectively. Let $[Z, Z']$ denote Souriau's bracket on the Galilean Lie algebra and let $\mu_0(Z)$ denote the pairing of a Galilean torsor $\mu_0$ with an algebra element $Z$. For any real number $M$, the map $(Z, Z') \mapsto M f_0(Z, Z')$ is a coboundary of the Lie algebra (i.e., there exists a Galilean torsor $\mu_0$ such that $M f_0(Z, Z') = \mu_0([Z, Z'])$ for all $Z, Z'$) if and only if $M = 0$.

The cocycle $f_0$ on the Galilean Lie algebra, defined for elements $Z = (\omega, \beta, \gamma, \varepsilon)$ and $Z' = (\omega', \beta', \gamma', \varepsilon')$ by $$f_0(Z, Z') = \beta \cdot \gamma' - \beta' \cdot \gamma,$$ is not a coboundary of the algebra. That is, there does not exist a Galilean torsor $\mu_0$ such that for all $Z, Z'$ in the Galilean Lie algebra, the pairing of $\mu_0$ with the Lie bracket $[Z, Z']$ satisfies: $$f_0(Z, Z') = \mu_0([Z, Z']).$$

The subtraction operation on the evolution space for $N$ material points is defined component-wise. For two points $y = (t, r, v)$ and $y' = (t', r', v')$ in the evolution space, where $t, t' \in \mathbb{R}$ represent time, $r, r'$ represent the particle positions, and $v, v'$ represent the particle velocities, their difference is given by: $$y - y' = (t - t', r - r', v - v')$$

For an element $Z = (\omega, \beta, \gamma, \epsilon)$ of the Galilean Lie algebra and a point $y = (t, \{r_j\}, \{v_j\})$ in the evolution space of $N$ particles, the vector field $Z_V(y)$ is defined by the infinitesimal transformations: $$\delta t = \epsilon$$ $$\delta r_j = \omega \times r_j + t \beta + \gamma$$ $$\delta v_j = \omega \times v_j + \beta$$ where $\omega \in \mathbb{R}^3$ represents the rotation (axial vector), $\beta \in \mathbb{R}^3$ the velocity change (boost), $\gamma \in \mathbb{R}^3$ the space translation, and $\epsilon \in \mathbb{R}$ the time translation. The terms $r_j$ and $v_j$ denote the position and velocity of the $j$-th material point, respectively, and $\times$ denotes the cross product in $\mathbb{R}^3$.

The function computes the linear part of the affine Galilean vector field $Z_V$ applied to a tangent vector $w$. Given an infinitesimal Galilean transformation $Z = (\omega, \beta, \gamma, \epsilon)$ and a vector $w = (t, \mathbf{r}_j, \mathbf{v}_j)$ in the evolution space of $N$ particles, the result is the vector $(0, \omega \times \mathbf{r}_j + t \beta, \omega \times \mathbf{v}_j)$, where $\omega, \beta, \mathbf{r}_j, \mathbf{v}_j \in \mathbb{R}^3$ and $\times$ is the cross product.

For any elements $Z, Z'$ of the Galilean Lie algebra and any point $y$ in the evolution space of $N$ material points, the vector field associated with the Souriau bracket $[Z, Z']$ satisfies: $$([Z, Z'])_V(y) = L_{Z'}(Z_V(y)) - L_Z(Z'_V(y))$$ where $Z_V(y)$ is the Galilean vector field at $y$, and $L_Z$ is the linear part of the affine vector field $Z_V$. This identity identifies the bracket in the Galilean Lie algebra with the Lie bracket of the corresponding vector fields on the evolution space.

Given a system of $N$ material points with masses $m_j \in \mathbb{R}$, let $y$ be a point in the evolution space consisting of time $t \in \mathbb{R}$, positions $r_j \in \mathbb{R}^3$, and velocities $v_j \in \mathbb{R}^3$. For two tangent vectors $dy = (dt, dr_j, dv_j)$ and $\delta y = (\delta t, \delta r_j, \delta v_j)$ at $y$, the Lagrange form $\sigma$ is defined as the scalar: $$\sigma(dy, \delta y) = \sum_{j=1}^N m_j \left( dv_j \cdot (\delta r_j - v_j \delta t) - \delta v_j \cdot (dr_j - v_j dt) \right)$$ where $\cdot$ denotes the standard dot product in $\mathbb{R}^3$.

For a system of $N$ material points with masses $m_j \in \mathbb{R}$, let $y$ be a point in the evolution space. For any two tangent vectors $dy$ and $\delta y$ at $y$, the Lagrange form $\sigma$ satisfies the antisymmetry property: $$\sigma(dy, \delta y) = -\sigma(\delta y, dy)$$ where the Lagrange form is defined as $\sigma(dy, \delta y) = \sum_{j=1}^N m_j \left( dv_j \cdot (\delta r_j - v_j \delta t) - \delta v_j \cdot (dr_j - v_j dt) \right)$.

For a system of $N$ material points with masses $m_j \in \mathbb{R}$, let $y = (t, \mathbf{r}_j, \mathbf{v}_j)$ be a point in the evolution space. Let $dy = (1, \mathbf{v}_j, \mathbf{0})$ be the tangent vector corresponding to the free motion (i.e., constant velocity motion) through $y$. Then, for any tangent vector $\delta y$ at $y$, the Lagrange form $\sigma$ satisfies: $$\sigma(dy, \delta y) = 0$$ This expresses that the velocity vector of the free motion lies in the kernel of the Lagrange form $\sigma$.

For a system of $N$ material points with masses $m_j \in \mathbb{R}$, let $y$ be a point in the evolution space characterized by the time $t$, positions $\mathbf{r}_j \in \mathbb{R}^3$, and velocities $\mathbf{v}_j \in \mathbb{R}^3$. The moment is defined as the Galilean torsor $(\mathbf{l}, \mathbf{g}, \mathbf{p}, E)$ where: - $\mathbf{l} = \sum_{j=1}^N m_j (\mathbf{r}_j \times \mathbf{v}_j)$ is the total angular momentum; - $\mathbf{g} = \sum_{j=1}^N m_j (\mathbf{r}_j - t \mathbf{v}_j)$ is the boost-related component (static moment); - $\mathbf{p} = \sum_{j=1}^N m_j \mathbf{v}_j$ is the total linear momentum; - $E = \frac{1}{2} \sum_{j=1}^N m_j \|\mathbf{v}_j\|^2$ is the total kinetic energy.

Given an evolution space point $y = (t, \mathbf{r}_j, \mathbf{v}_j)$ and a displacement vector $dy = (dt, d\mathbf{r}_j, d\mathbf{v}_j)$, this function defines the affine line $s \mapsto y + s \, dy$ for $s \in \mathbb{R}$. The resulting point in the evolution space is given by the coordinates $(t + s \, dt, \mathbf{r}_j + s \, d\mathbf{r}_j, \mathbf{v}_j + s \, d\mathbf{v}_j)$, where $t$ is time, $\mathbf{r}_j$ are the positions, and $\mathbf{v}_j$ are the velocities of the $N$ material points.

Given a state $y = (t, \mathbf{r}_j, \mathbf{v}_j)$ in the evolution space of $N$ material points (where $t$ is the time, $\mathbf{r}_j$ are the positions, and $\mathbf{v}_j$ are the velocities) and a time interval $s \in \mathbb{R}$, the function returns the state $(t + s, \mathbf{r}_j + s \mathbf{v}_j, \mathbf{v}_j)$ corresponding to the free motion of the particles.

Consider a system of $N$ material points with masses $m_j \in \mathbb{R}$. Let $y = (t, \mathbf{r}_j, \mathbf{v}_j)$ be a point in the evolution space, where $t$ is the time, $\mathbf{r}_j \in \mathbb{R}^3$ are the positions, and $\mathbf{v}_j \in \mathbb{R}^3$ are the velocities. Let $Z = (\boldsymbol{\omega}, \boldsymbol{\beta}, \boldsymbol{\gamma}, \varepsilon)$ be an element of the Galilean Lie algebra, where $\boldsymbol{\omega}$ is the rotation axial vector, $\boldsymbol{\beta}$ is the velocity boost, $\boldsymbol{\gamma}$ is the spatial translation, and $\varepsilon$ is the time translation. The pairing of the moment $\mu$ of the system with $Z$ is given by: $$\mu(Z) = \sum_{j=1}^N m_j \left[ (\mathbf{r}_j \times \mathbf{v}_j) \cdot \boldsymbol{\omega} - (\mathbf{r}_j - t \mathbf{v}_j) \cdot \boldsymbol{\beta} + \mathbf{v}_j \cdot \boldsymbol{\gamma} - \frac{1}{2} \varepsilon |\mathbf{v}_j|^2 \right]$$

In the evolution space of $N$ free material points with masses $m_j \in \mathbb{R}$, let $\mu(y)$ be the moment map (represented as a Galilean torsor) and $\sigma(y)$ be the Lagrange form. For any point $y$ in the evolution space, any tangent vector $dy$ at $y$, and any element $Z$ of the Galilean Lie algebra, the derivative of the pairing between the moment and $Z$ along the path $y + s \, dy$ at $s = 0$ is equal to the Lagrange form evaluated at $dy$ and the Galilean vector field $Z_V(y)$ associated with $Z$: $$\left. \frac{d}{ds} \langle \mu(y + s \, dy), Z \rangle \right|_{s=0} = \sigma(y)(dy, Z_V(y))$$ where the pairing $\langle \mu, Z \rangle$ is defined as $l \cdot \omega - g \cdot \beta + p \cdot \gamma - E \epsilon$ for a torsor $\mu = (l, g, p, E)$ and algebra element $Z = (\omega, \beta, \gamma, \epsilon)$.

For a system of $N$ material points with masses $m_j \in \mathbb{R}$, let $y$ be a point in the evolution space representing the initial time, positions, and velocities. Let $\text{motion}(y, s)$ be the state of the system after it has evolved for a time interval $s \in \mathbb{R}$ under free motion (i.e., with constant velocities). The Galilean moment $\mu$ of the system is invariant under this evolution: $$\text{moment}(m, \text{motion}(y, s)) = \text{moment}(m, y)$$ where the moment $\mu = (\mathbf{l}, \mathbf{g}, \mathbf{p}, E)$ is the torsor composed of the total angular momentum $\mathbf{l}$, the static moment $\mathbf{g}$, the total linear momentum $\mathbf{p}$, and the total kinetic energy $E$.

Given a system of $N$ material points where $m: \{0, \dots, N-1\} \to \mathbb{R}$ is a function assigning a mass $m_j$ to each particle $j$, the total mass of the system is the sum of the individual masses, defined as: $$\sum_{j} m_j$$

For a system of $N$ material points with masses $m_j \in \mathbb{R}$ and positions $\mathbf{r}_j \in \mathbb{R}^3$ (given by the state $y$ in the evolution space), the center of mass $\mathbf{R}$ is defined as: $$\mathbf{R} = \frac{1}{M} \sum_{j} m_j \mathbf{r}_j$$ where $M = \sum_{j} m_j$ is the total mass of the system. According to the convention $0^{-1} = 0$, if the total mass $M$ is zero, the center of mass is the zero vector.

Consider a system of $N$ free material points with masses $m_j$ and total mass $M = \sum_{j=1}^N m_j$. Let $y$ be a point in the evolution space and $Z, Z'$ be elements of the Galilean Lie algebra. Let $Z_V(y)$ and $Z'_V(y)$ be the Galilean vector fields on the evolution space corresponding to $Z$ and $Z'$, respectively. The Lagrange form $\sigma$ evaluated on these vector fields satisfies the relation: $$\sigma(Z_V(y), Z'_V(y)) = \mu([Z, Z']) + M f_0(Z, Z')$$ where $\mu$ is the moment of the system at state $y$, $[Z, Z']$ is Souriau's bracket on the Galilean Lie algebra, and $f_0(Z, Z')$ is the Galilean algebra cocycle defined by $f_0(Z, Z') = \beta \cdot \gamma' - \beta' \cdot \gamma$.

For a system of $N$ material points with total mass $M = \sum_{j} m_j$, if $M \neq 0$, then the Galilean cocycle $M f_0$ is not a coboundary of the Lie algebra. Specifically, there does not exist a Galilean torsor $\mu_0$ such that for all elements $Z, Z'$ of the Galilean Lie algebra, the following equality holds: $$M f_0(Z, Z') = \langle \mu_0, [Z, Z'] \rangle$$ where $f_0(Z, Z') = \beta \cdot \gamma' - \beta' \cdot \gamma$ is the fundamental cocycle defined by velocity boosts $\beta, \beta'$ and spatial translations $\gamma, \gamma'$, and $[Z, Z']$ denotes the Souriau bracket of the Galilean Lie algebra.

For a system of $N$ material points with individual masses $m_j$ and a non-zero total mass $M = \sum_{j} m_j \neq 0$, let $y$ be a state in the evolution space with associated time $t$. The boost-related component $\mathbf{g}$ of the Galilean moment at $y$ is given by the relation: $$\mathbf{g} = M \mathbf{R} - t \mathbf{p}$$ where $\mathbf{R}$ is the center of mass and $\mathbf{p}$ is the total linear momentum of the system.

For a system of $N$ material points with masses $m_j \in \mathbb{R}$, let $y$ be a state in the evolution space and $y(s)$ be the state after a free motion of duration $s \in \mathbb{R}$. Let $M = \sum_{j} m_j$ be the total mass, $\mathbf{R}(y)$ be the center of mass at state $y$, and $\mathbf{p}(y)$ be the total linear momentum at state $y$. Then the center of mass at the new state is given by: $$\mathbf{R}(y(s)) = \mathbf{R}(y) + s \frac{\mathbf{p}(y)}{M}$$ where, by convention, $M^{-1} = 0$ if $M = 0$.