note-math

Non-relativistic spacetime $ℝ \times ℝ^{3}$

action-point-particle-non-relativity_(tag) Action of path $𝑡 ⇝ 𝑥 (𝑡)$

\int 𝑑 𝑡 (\frac{1}{2} 𝑚 {\dot{𝑥}}^{2} - 𝑈 (𝑥))

let $𝑋 (𝑡)$ be a time-varying vector field or time-varying δ diffeomorphism, or $(1, 𝑋 (𝑡, 𝑥_{1 . . 3}))$ is a special type of vector field of non-relativistic spacetime $ℝ \times ℝ^{3}$

let $𝑋 (𝑡)$ is zero at the boundary — fix the endpoints of the path

let the differential of the action be zero

0 = \partial 𝑆 \cdot 𝑋 (𝑡) = \int 𝑑 𝑡 (\frac{\partial}{\partial \dot{𝑥}} (\frac{1}{2} 𝑚 {\dot{𝑥}}^{2}) \cdot \dot{𝑋} - \frac{\partial 𝑈}{\partial 𝑥} (𝑥) \cdot 𝑋)

where $\frac{\partial}{\partial \dot{𝑥}} (\frac{1}{2} 𝑚 {\dot{𝑥}}^{2}) = 𝑚 \dot{𝑥}$

use product rule

\frac{𝑑}{𝑑 𝑡} ((𝑚 \dot{𝑥}) \cdot 𝑋) = 𝑚 \ddot{𝑥} \cdot 𝑋 + 𝑚 \dot{𝑥} \cdot \dot{𝑋}

and $𝑋 (𝑡)$ is zero at the boundary, such that

\int 𝑑 𝑡 (\frac{𝑑}{𝑑 𝑡} (𝑚 \dot{𝑥}) \cdot 𝑋) = 0

So the differential of the action is

0 = - \int 𝑑 𝑡 ((𝑚 \ddot{𝑥} + \frac{𝑑 𝑈}{𝑑 𝑥}) 𝑋)

holds for all δ diffeomorphism $𝑋 (𝑡)$ , thus giving the Lagrange-equation (alias Euler–Lagrange-equation), for non-relativistic point particles, Newton-equation_(tag)

𝑚 \ddot{𝑥} + \frac{\partial 𝑈}{\partial 𝑥} = 0

The momentum part of the action $\int 𝑑 𝑡 (\frac{1}{2} 𝑚 {\dot{𝑥}}^{2})$ does not use the volume-form of $ℝ^{3}$ , but instead uses the quadratic form $| \dot{𝑥} |^{2}$ of $ℝ^{3}$ and the volume-form $# 𝑑 𝑡$ of time $ℝ$

The Lagrangian can be written as a function $𝐿 (𝑥, \dot{𝑥}) = \frac{1}{2} 𝑚 {\dot{𝑥}}^{2} - 𝑈 (𝑥)$ (a function on the tangent bundle)

point-particle-Lagrange-equation_(tag)

For a general $𝐿$ , repeat the above process. Action

\int 𝑑 𝑡 (𝐿 (𝑥, \dot{𝑥}))

Differential of the action

\int 𝑑 𝑡 (\frac{\partial 𝐿}{\partial 𝑥} \cdot 𝑋 + \frac{\partial 𝐿}{\partial \dot{𝑥}} \cdot \dot{𝑋})

use

\frac{𝑑}{𝑑 𝑡} (\frac{\partial 𝐿}{\partial \dot{𝑥}} \cdot 𝑋) = (\frac{𝑑}{𝑑 𝑡} \frac{\partial 𝐿}{\partial \dot{𝑥}}) \cdot 𝑋 + \frac{\partial 𝐿}{\partial \dot{𝑥}} \cdot \dot{𝑋}

is zero at the boundary + integral is zero for all δ diffeomorphism $𝑋$ ==> point-particle-Lagrange-equation_(tag)

\frac{\partial 𝐿}{\partial 𝑥} - \frac{𝑑}{𝑑 𝑡} \frac{\partial 𝐿}{\partial \dot{𝑥}} = 0

Euclidean metric-manifold

Generalize to Euclidean metric-manifold

Needs #link(<metric-connection>)[]

Although the metric volume form is not used, due to the form of the kinetic energy part of the action, it is still related to the metric geodesic

Symmetry and conserved quantity

Handling the symmetry of non-relativistic spacetime $ℝ \times ℝ^{3}$ alias the Galileo group, generated from the translation of $ℝ \times ℝ^{3}$ , rotation, non-relativistic boost

let $𝑡 ⇝ 𝑥 (𝑡)$ be the solution to the action equation

Note that the variation of $𝑥 (𝑡)$ along the symmetry may make it no longer a solution to the equation, i.e., the symmetric δ diffeomorphism may not be zero at the boundary, i.e., it will change the endpoints of the path, so the relevant derivative of the action at the solution $𝑥 (𝑡)$ may not be zero

Time translation

In non-relativity, the mapping that preserves the measure and direction of time $ℝ$ is the time translation $𝑡 ⇝ 𝑡 + 𝑎$

δ variation of the integral area calculation-1-action-point-particle-non-relativity_(tag)

\begin{matrix} \frac{𝑑}{𝑑 𝑠} \int_{𝑡_{0} + 𝑠 𝑎}^{𝑡_{1} + 𝑠 𝑎} 𝑑 𝑡 (𝐿 (𝑡)) & = & (𝐿 (𝑡_{1}) - 𝐿 (𝑡_{0})) \cdot 𝑎 \\ = & \int_{𝑡_{0}}^{𝑡_{1}} 𝑑 𝑡 (\frac{𝑑}{𝑑 𝑡} 𝐿 (𝑡) \cdot 𝑎) \end{matrix}

The first equation can come from the fundamental theorem of calculus + derivative of composite functions

In general, changing the region $𝐼$ by $exp (𝑎 (𝑡))$ #link(<vector-field-as-δ-diffeomorphism>)[by δ diffeomorphism] $𝑎 (𝑡)$

\frac{𝑑}{𝑑 𝑠} \int_{exp (𝑠 𝑎 (𝑡)) 𝐼} 𝑑 𝑡 (𝑓 (𝑡)) = \int_{𝐼} 𝑑 𝑡 (\frac{𝑑}{𝑑 𝑡} 𝑓 (𝑡) \cdot 𝑎 (𝑡))

On the other side, use #link(<integral-change-of-variable-formula>)[change of variable formula]

\int_{𝑡_{0} + 𝑠 𝑎}^{𝑡_{1} + 𝑠 𝑎} 𝑑 𝑡 (𝐿 (𝑡)) = \int_{𝑡_{0}}^{𝑡_{1}} 𝑑 𝑡 (𝐿 (𝑡 + 𝑠 𝑎))

Apply it to

\begin{matrix} 𝐿 (𝑡) & = & 𝐿 (𝑥 (𝑡), \dot{𝑥} (𝑡)) \end{matrix}

Then use the exchange of differential and integral

\frac{𝑑}{𝑑 𝑠} \int_{𝑡_{0}}^{𝑡_{1}} 𝑑 𝑡 = \int_{𝑡_{0}}^{𝑡_{1}} 𝑑 𝑡 \frac{𝑑}{𝑑 𝑠}

Derivative of composite function $\frac{𝑑}{𝑑 𝑠} 𝐿 (𝑥 (𝑡 + 𝑠 𝑎), \dot{𝑥} (𝑡 + 𝑠 𝑎)) = \frac{\partial 𝐿}{\partial 𝑥} \dot{𝑥} 𝑎 + \frac{𝑑 𝐿}{𝑑 \dot{𝑥}} \ddot{𝑥} 𝑎$

is the variation of the action on the (changing endpoints) δ differentiation $\frac{𝑑}{𝑑 𝑠} (0) 𝑥 (𝑡 + 𝑠 𝑎) = 𝑎 \dot{𝑥} (𝑡) = 𝑋 (𝑡)$ at the solution $𝑥 (𝑡)$ calculation-2-action-point-particle-non-relativity_(tag)

\begin{matrix} \frac{𝑑}{𝑑 𝑠} \int_{𝑡_{0} + 𝑠 𝑎}^{𝑡_{1} + 𝑠 𝑎} 𝑑 𝑡 (𝐿 (𝑡)) & = & \int 𝑑 𝑡 (\frac{\partial 𝐿}{\partial 𝑥} \cdot 𝑋 + \frac{\partial 𝐿}{\partial \dot{𝑥}} \cdot \dot{𝑋}) \\ = & \int 𝑑 𝑡 ((\frac{\partial 𝐿}{\partial 𝑥} - \frac{𝑑}{𝑑 𝑡} \frac{\partial 𝐿}{\partial \dot{𝑥}}) \cdot 𝑋 + \frac{𝑑}{𝑑 𝑡} (\frac{\partial 𝐿}{\partial \dot{𝑥}} \cdot 𝑋)) \\ by product-rule of \frac{𝑑}{𝑑 𝑡} (\frac{\partial 𝐿}{\partial \dot{𝑥}} \cdot 𝑋) \\ = & \int 𝑑 𝑡 (\frac{𝑑}{𝑑 𝑡} (\frac{\partial 𝐿}{\partial \dot{𝑥}} \cdot 𝑋)) \\ Lagrange-equation \frac{\partial 𝐿}{\partial 𝑥} - \frac{𝑑}{𝑑 𝑡} \frac{\partial 𝐿}{\partial \dot{𝑥}} = 0 \end{matrix}

recall

\begin{matrix} \frac{𝑑}{𝑑 𝑠} \int_{𝑡_{0} + 𝑠 𝑎}^{𝑡_{1} + 𝑠 𝑎} 𝑑 𝑡 (𝐿 (𝑡)) & = & 𝐿 (𝑡) |_{𝑡_{0}}^{𝑡_{1}} \cdot 𝑎 \\ = & \int_{𝑡_{0}}^{𝑡_{1}} 𝑑 𝑡 (\frac{𝑑}{𝑑 𝑡} 𝐿 (𝑡) \cdot 𝑎) \end{matrix}

use $𝑋 = 𝑎 \dot{𝑥}$ , merge the previous item with the next item

\int_{𝑡_{0}}^{𝑡_{1}} 𝑑 𝑡 \frac{𝑑}{𝑑 𝑡} (\frac{\partial 𝐿}{\partial \dot{𝑥}} \cdot 𝑋) = (\frac{\partial 𝐿}{\partial \dot{𝑥}} \cdot \dot{𝑥} \cdot 𝑎) |_{𝑡_{0}}^{𝑡_{1}}

Get

(\frac{\partial 𝐿}{\partial \dot{𝑥}} \cdot \dot{𝑥} - 𝐿) |_{𝑡_{0}}^{𝑡_{1}} \cdot 𝑎 = 0

Quantity

𝐸 = \frac{\partial 𝐿}{\partial \dot{𝑥}} \cdot \dot{𝑥} - 𝐿

Called the energy of the action $𝐿$ , is invariant along time $𝑡$ , forall $𝑎 \in ℝ$ , i.e. conserved. This is true for $𝑡_{0} < 𝑡_{1}$ also imply $\frac{𝑑}{𝑑 𝑡} 𝐸 = 0$

For $𝐿 = \frac{1}{2} 𝑚 \dot{𝑥} {(𝑡)}^{2} - 𝑈 (𝑥 (𝑡))$ the energy is energy-point-particle-non-relativity_(tag)

\begin{matrix} 𝐸 & = & 𝑚 {\dot{𝑥}}^{2} - (\frac{1}{2} 𝑚 {\dot{𝑥}}^{2} - 𝑈) \\ = & \frac{1}{2} 𝑚 {\dot{𝑥}}^{2} + 𝑈 \end{matrix}

Homogeneity of Time ==> Conservation of Energy

Spatial translation

Kinetic energy part of the action

\int 𝑑 𝑡 (\frac{1}{2} 𝑚 {\dot{𝑥}}^{2})

Although the spatial translation δ diffeomorphism is not zero at the boundary or changes the path endpoints, the time endpoints remain unchanged, and spatial translation does not change kinetic energy. with $𝑥 + 𝑠 𝑎$ , $\frac{𝑑}{𝑑 𝑡} (𝑥 + 𝑠 𝑎) = \dot{𝑥}$

\begin{matrix} \frac{𝑑}{𝑑 𝑠} (𝑠 = 0) \int_{𝑡_{0}}^{𝑡_{1}} 𝑑 𝑡 (\frac{1}{2} 𝑚 {(\frac{𝑑}{𝑑 𝑡} (𝑥 + 𝑠 𝑎))}^{2}) \\ = & \frac{𝑑}{𝑑 𝑠} \int_{𝑡_{0}}^{𝑡_{1}} 𝑑 𝑡 (\frac{1}{2} 𝑚 {\dot{𝑥}}^{2}) \\ = & 0 \end{matrix}

So similar to #link(<calculation-2-action-point-particle-non-relativity>)[the case of energy], with δ diffeomorphism $\frac{𝑑}{𝑑 𝑠} (𝑥 + 𝑠 𝑎) = 𝑎 = 𝑋 (𝑡)$

0 = \int_{𝑡_{0}}^{𝑡_{1}} 𝑑 𝑡 (\frac{𝑑}{𝑑 𝑡} (\frac{\partial 𝐿}{\partial \dot{𝑥}} \cdot 𝑋)) = \frac{\partial 𝐿}{\partial \dot{𝑥}} |_{𝑡_{0}}^{𝑡_{1}} \cdot 𝑎

momentum-point-particle-non-relativity_(tag) The momentum of the action $\int 𝑑 𝑡 (\frac{1}{2} 𝑚 {\dot{𝑥}}^{2})$

𝑚 \dot{𝑥} \cdot 𝑎

is invariant along time $𝑡$ , forall $𝑎 \in ℝ^{3}$ , i.e. conserved

More generally, let the action $\int 𝑑 𝑡 (𝐿 (𝑥, \dot{𝑥}))$ with $\frac{\partial 𝐿}{\partial 𝑎} = 0$ such that the endpoints in this direction do not affect the action, then the momentum

\frac{\partial 𝐿}{\partial \dot{𝑥}} \cdot 𝑎

is conserved

Homogeneity of Space ==> Translation Invariance of $𝐿$ ==> Conservation of Total Linear Momentum

Lagrangian

𝐿 (𝑥, \dot{𝑥}) = \frac{1}{2} \sum 𝑚_{𝑖} {\dot{𝑥}}_{𝑖}^{2} - \sum_{𝑖 < 𝑖^{'}} 𝑈 (| 𝑥_{𝑖} - 𝑥_{𝑖^{'}} |)

$\frac{\partial 𝐿}{\partial 𝑎} = 0$ forall $𝑎 \in ℝ^{3}$ (by $(𝑥_{𝑖} + 𝑎) - (𝑥_{𝑖^{'}} + 𝑎) = 𝑥_{𝑖} - 𝑥_{𝑖^{'}}$ ) so the momentum

\sum 𝑚 {\dot{𝑥}}_{𝑖} \cdot 𝑎

is conserved

Spatial rotation

Choose an origin. Lagrangian

𝐿 (𝑥, \dot{𝑥}) = \frac{1}{2} \sum 𝑚 {\dot{𝑥}}^{2} - 𝑈 (| 𝑥 |)

$so (3)$ is represented as a rotation around axis $𝑛$ , cross product $𝑛 \times 𝑥$ is δ rotation

Rotation around axis ==> $(𝑛 \times 𝑥) ⟂ 𝑛$

in $ℂ$ , $𝑒^{𝜃 i} 𝑧 = 𝑧 + 𝜃 i 𝑧 + 𝑜 (𝑧)$ with $𝜃 i \in Im (ℂ) = u (1) = so (2)$ , $𝑧 ⟂ i 𝑧$ and magnitude $| 𝜃 i | = | 𝑛 |$

Length is invariant to direction $\frac{\partial | 𝑥 |}{\partial 𝑥} \cdot (𝑛 \times 𝑥) = 0$

Similar to the case of momentum, if the Lagrangian is invariant to rotation, the δ diffeomorphism (tangent vector field) is $𝑛 \times 𝑥$ , thus

rotation-momentum-point-particle-non-relativity_(tag) Rotation momentum rotation-momentum alias angular momentum angular-momentum

𝑚 \dot{𝑥} \cdot (𝑛 \times 𝑥) = 𝑛 \cdot (𝑥 \times 𝑚 \dot{𝑥})

is invariant along time $𝑡$ , forall $𝑛 \in so (3)$

quantity

𝑥 \times 𝑚 \dot{𝑥}

is also called rotation momentum

More generally, $𝐿 (𝑥, \dot{𝑥})$ with $𝑛 \cdot (𝑥 \times \partial) 𝑓 = 0$ such that the Lagrangian is invariant under rotation about $𝑛$ , then the rotational momentum about $𝑛$ is

𝑥 \times \frac{\partial 𝐿}{\partial \dot{𝑥}}

Isotropy of Space ==> Rotational Invariance of $𝐿$ ==> Conservation of Total Angular Momentum

$𝑎 \cdot (𝑏 \times 𝑐)$ = #link(<volume-of-parallelogram>)[parallelepiped directed volume] span by $𝑎, 𝑏, 𝑐$ in Euclidean $ℝ^{3}$

The rotational momentum $𝑥 \times 𝑚 \dot{𝑥}$ is constant with respect to time, so ${span (𝑥 \times \dot{𝑥})}^{⟂}$ is a constant 2d plane. Since $(𝑥 \times \dot{𝑥}) ⟂ 𝑥, \dot{𝑥}$ , $span (𝑥, \dot{𝑥}) \subset {span (𝑥 \times \dot{𝑥})}^{⟂}$ , $𝑥 (𝑡)$ is in the constant two-dimensional plane ${span (𝑥 \times \dot{𝑥})}^{⟂}$

For the Lagrangian of a system of point particles

𝐿 (𝑥, \dot{𝑥}) = \frac{1}{2} \sum 𝑚_{𝑖} {\dot{𝑥}}_{𝑖}^{2} - \sum_{𝑖 < 𝑖^{'}} 𝑈 (| 𝑥_{𝑖} - 𝑥_{𝑖^{'}} |)

Total rotational momentum

𝑛 \cdot \sum (𝑥_{𝑖} \times 𝑚 {\dot{𝑥}}_{𝑖})

is invariant along time $𝑡$

Non-relativistic boost

Non-relativistic boost $𝑥 ⇝ 𝑥 + 𝑡 \cdot 𝑣$

The conserved quantity of the action $\int \frac{1}{2} 𝑚 {\dot{𝑥}}^{2} 𝑑 𝑡$ is

𝑚 (𝑡 \cdot \dot{𝑥} - 𝑥) \cdot 𝑣

forall $𝑣 \in ℝ^{3}$

The action $\frac{1}{2} 𝑚 {\dot{𝑥}}^{2}$ has all δ symmetries of non-relativistic spacetime $ℝ \times ℝ^{3}$ , a 10 dimensional conserved quantity
The action $\frac{1}{2} 𝑚 {\dot{𝑥}}^{2} - 𝑈 (𝑥)$ has conserved energy
The action $\sum \frac{1}{2} 𝑚_{𝑖} {\dot{𝑥}}_{𝑖}^{2} - \sum_{𝑖 < 𝑖^{'}} 𝑈 (𝑥_{𝑖} - 𝑥_{𝑖^{'}})$ has conserved energy and momentum
The action $\frac{1}{2} 𝑚 {\dot{𝑥}}^{2} - 𝑈 (| 𝑥 |)$ has conserved energy and rotational momentum
The action $\sum \frac{1}{2} 𝑚_{𝑖} {\dot{𝑥}}_{𝑖}^{2} - \sum_{𝑖 < 𝑖^{'}} 𝑈 (| 𝑥_{𝑖} - 𝑥_{𝑖^{'}} |)$ has conserved energy, momentum, rotational momentum, 7 dimension

Non-relativistic potential $(𝑉, 𝐴) \in ℝ \times ℝ^{3}$

Rigid body

Parameterized by $SO (3)$ (or $SO (2)$ ), so it can be regarded as a non-relativistic particle on the Euclidean manifold $SO (3)$ . But the use of metric or the use of kinetic energy is not the #link(<Killing-form>)[] of $so (3)$ , because for objects that are not uniformly mass-distributed spheres, rotations in different directions have different inertias. Moment of inertia i.e. metric $𝑔$ may need to be calculated additionally

The moment of inertia can also be used as a symmetric operator under the Killing-form, with the characteristic basis being the principal axes of inertia and the eigenvalues being the moments of inertia.