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Free particles

Let's try to solve the physical problem of a roughly point-like neutral particle of (positive) mass $m$, subject to no force, in a number $n$ of spacelike dimensions, where we will set $n = 3$ if it is necessary to specify it (while many elements of this do not depend on the number of dimensions, this will be true for instance the detailed structure of the Galilean group).

There are several background structures we need to talk about. First is the structure of spacetime. There are two choices we can make : we can take only space itself (for instance $\mathbb{R}^3$ for classical Euclidian space), or spacetime (such as $\mathbb{R}^{(3+1)}$). This is unrelated to whether we are using classical mechanics or special relativity, as both can be done in either formalism.

There are a few different ways for which we can define a "particle". We can define either an actual point-like entity, or as some localized entity for which a single position through time is the main axis of analysis.

Overall, many of those methods can be categorized by the following attributes :

Various combinations of answers can give various different theories here. Not all possible combinations are done, and some of these do not fit with any of these criterias either.

Galilean version

The original correct version of the treatment of a free body was done by Galileo in "The Two New Sciences", in the chapter on uniform motions. While he does not actually solve the problem from the dynamics, merely describes what its solution is, it is still of some interest to see its original description.

He defines uniform motion to be one where the distances traversed by the moving particle during any equal intervals of time to be equal, or in modern terms,

\begin{equation} \forall t_1, t_2, T \in \mathbb{R},\ | \vec{x}(t_1 + T) - \vec{x}(t_1) |^2 = | \vec{x}(t_2 + T) - \vec{x}(t_2) |^2 \end{equation}

To treat this notion, he defines the following four axioms :

In modern terms, if we write this constant distance for a given $T$ as $d_T$, and we define the speed $v_T = d_T / T$ this means that

Newtonian mechanics

Free particles in Newtonian mechanics are simply point particles with no force applied to them, so that $\vec{F} = 0$. In other words,

\begin{eqnarray} \ddot{\vec{x}}(t) = 0 \end{eqnarray}

This is fairly easily solved for the velocity as

\begin{equation} \dot{\vec{x}}(t) = \vec{v}_0 \end{equation}

For some constant of integration $\vec{v}_0$ (corresponding to the initial velocity : $\dot{\vec{x}}(0) = \vec{v}_0$), and for the position

\begin{equation} \vec{x}(t) = \vec{v}_0 t + \vec{x}_0 \end{equation}

for some constant of integration $\vec{x}_0$, corresponding to the initial position ($\vec{x}(0) = \vec{x}_0$). For future use, it is also good to consider the equations in term of the momentum, $\vec{p} = m \dot{\vec{x}}$, with the equation of motion

\begin{equation} \frac{d\vec{p}(t)}{dt} = 0 \end{equation}

and similarly the solution

. \begin{equation} \vec{p}(t) = \vec{p}_0 \end{equation}

Original description

The original description of the Newtonian treatment of the free particle is given in Newton's Principia Mathematica, where the laws are given as follow :

Law I : Every body perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed thereon. Law II : The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in. which that force is impressed. Law III : To every action there is always opposed an equal reaction : or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.

As we are dealing with the free particle here, only the first law applies, meaning that as the Galilean case, its motion is a straight line. Few details are given for the free particle case beyond this in Newton, which focuses more on the gravitational case.

Conserved quantities

The free particles has the following list of conserved quantities. We will here just show that they are indeed conserved, and defer their theoretical underpinning for later on. As we will see, they are all fundamentally invariants due to being defined from the initial conditions of the system (ie the time dependence of those quantities drop out).

First is the momentum, which is by definition conserved in Newtonian mechanics (if $m$ is independent of time, anyway),

\begin{equation} \vec{p} = m \dot{\vec{x}}(t) = m \vec{v}_0 \end{equation}

which is obviously enough preserved in time. Likewise, the energy

\begin{equation} E = \frac{m}{2} |\dot{\vec{x}}(t)|^2 = \frac{m}{2} |\vec{v}_0|^2 \end{equation}

and the center of mass, given by the quantity

\begin{eqnarray} \mathrm{CM} &=& m (\vec{x} t - \vec{x})\\ &=& m (\vec{v}_0 t - \vec{v_0} t - \vec{x}_0)\\ &=& - m \vec{x}_0 \end{eqnarray}

Given a point $\vec{y}$, fixed within this coordinate system, the angular momentum with respect to that point is

\begin{eqnarray} \vec{J}_{\vec{y}} &=& (\vec{x} - \vec{y}) \times m \dot{\vec{x}}\\ &=& (\vec{x}_0 - \vec{y} + \vec{v}_0 t) \times m \vec{v}_0\\ &=& m (\vec{x}_0 - \vec{y}) \times \vec{v}_0 \end{eqnarray}

and simply being composed of constants, $\vec{J}$ has both its norm $|\vec{J}|^2$ and each individual components being constant.

and finally, free particles also admit the following two invariants : the dilation invariant,

\begin{eqnarray} D &=& \frac{d}{dt} (\frac{m}{2} |\vec{x}(t)|^2) - 2 t E\\ &=& \frac{m}{2} \frac{d}{dt} (|\vec{x}_0|^2 + t^2 |\vec{v}_0|^2 + 2 t \vec{x}_0 \cdot \vec{v}_0) - m t |\vec{v}_0|^2\\ &=& m (t |\vec{v}_0|^2 + \vec{x}_0 \cdot \vec{v}_0) - m t |\vec{v}_0|^2\\ &=& m \vec{x}_0 \cdot \vec{v}_0 \end{eqnarray}

and the special conformal transformation invariant,

\begin{eqnarray} K &=& \frac{m}{2} |\vec{x}|^2 - tD - t^2 E\\ &=& \frac{m}{2} (|\vec{x}_0|^2 + t^2 |\vec{v}_0|^2 + 2 t \vec{x}_0 \cdot \vec{v}_0) - t m \vec{x}_0 \cdot \vec{v}_0 - t^2 \frac{m}{2} |\vec{v}_0|^2\\ &=& \end{eqnarray}

Symmetries

For the Newtonian formalism, we consider our physical laws as a system of differential equations, in our case the differential operator of the form

\begin{eqnarray} F(t, \vec{x}(t), \dot{\vec{x}}(t), \ddot{\vec{x}}(t)) = 0 \end{eqnarray}

specifically only depending on $\ddot{\vec{x}}(t)$, with some initial conditions $\vec{x}(t_i) = \vec{x}_i$, $\dot{\vec{x}}(t_i) = \vec{v}_i$.

A symmetry group $G$ (a Lie group) acts on this system by point transformations, which are transformations on the variable $t$ and function $\vec{x}$

\begin{eqnarray} g : (t, \vec{x}) \mapsto (f_g(t), \phi_g(t, \vec{x})) \end{eqnarray}

which we could see alternatively as a change of variables,

\begin{eqnarray} t &\to& t' = f_g(t)\\ \vec{x} &\to& \vec{x}' = \phi_g(t, \vec{x}) \end{eqnarray}

The derivatives of those transformations are given by the chain rule. If we write the derivatives of our transformation function as $A_g = D \alpha_g(t)$

\begin{equation} \frac{d}{dt} (g \cdot \vec{x})(t) = \end{equation}

There are a few important groups of symmetries associated with the free particles. In terms of time symmetries, ie the symmetries decided by the function $f$, the free particle is invariant under time translation and reflection, which are the group $\mathbb{R}$ and $\mathbb{Z}_2$, or taken together, $\mathrm{E}(1) \cong \mathbb{Z}_2 \rtimes \mathbb{R}$ (the one-dimensional Euclidian group), which is a semidirect product as their actions do not commute. If we take the time reflection $k \in \mathbb{Z}_2$ and the time translation $T_a$ by $a$, the composition of two such transformations is

\begin{eqnarray} (k', T_{a'}) (k, T_a) t &=& (k', T_{a'}) (-1)^{k} (t + a)\\ &=& (-1)^{k'} ((-1)^{k} (t + a)) + a' \end{eqnarray}

So that the semidirect product is given by (using $\mathbb{Z}_2$ as the multiplicative group $\{ -1, +1 \}$)

\begin{eqnarray} T_a I_t t = T_a (-t) = -t + a \end{eqnarray}

Its action on the Newton equation is given as, for $t \to t' = - t + a$, with the Jacobian $dt'/dt = -1$,

\begin{eqnarray} \frac{d^2}{dt'} \vec{x}(t') &=& (-1^2) \frac{d^2}{dt^2} \vec{x}(-t + a) \end{eqnarray}

Hence if $\vec{x}(t)$ is a solution of the equation, so is $\vec{x}(t')$

The equation is also invariant under the rotation group, $\mathrm{O}(3)$. Given an orthogonal matrix $R$, such that $R^T R = R R^T = I$,

The Newton equation is also invariant under boosts, which are transformations of the type

\begin{equation} (t, \vec{x}) \mapsto (t, \vec{x} + \vec{v}t) \end{equation}

for $\vec{v} \in \mathbb{R}^3$. These form a group of the form $\mathbb{R}^3$, as any two such transformation commutes

The full symmetry group of these symmetries taken together is the inhomogeneous Galilean group, $\mathrm{IGal}(3)$ (which you can find descriptions here or here). As we will use this group quite a lot as we go, let's look into it for a bit.

The inhomogeneous Galilean group $\mathrm{IGal}(3)$, and its homogeneous version, the homogeneous Galilean group $\mathrm{Gal}(3)$, are two Lie groups which can be defined from the kinematics of space as the group which acts on $\mathbb{R} \times \mathbb{R}^3$, the pair of time and space $(t, \vec{x})$, by a series of parameters :

\begin{equation} \forall g \in \mathrm{IGal}(3),\ g = (s, \vec{a}, \vec{b}, R) \end{equation}

and two involutions, $P$ and $T$ (the time and space involutions), with the following standard affine transformation :

\begin{equation} g(t, \vec{x}) = (t + a, R \vec{x} + \vec{v} t + \vec{a}) \end{equation}

In its full form, the Galilean

the homogeneous Galilean group can be decomposed as

\begin{equation} \mathrm{Gal}(3) \cong \mathbb{R}^3 \rtimes \mathrm{O}(3) \end{equation}

where the semidirect product of boosts and rotations is given by

\begin{equation} (\vec{b}', R) (\vec{b}, R) = \end{equation}

while for the inhomogeneous Galilean group,

\begin{eqnarray} \mathrm{IGal}(3) \cong (\mathbb{R} \times \mathbb{R}^3) \rtimes \mathbb{R}^3 \rtimes \mathrm{O}(3) \end{eqnarray}

The homogeneous space associated to $(\mathrm{IGal}(3), \mathrm{Gal}(3))$

In addition to the Galilean group, which is the symmetry group of most (non-relativistic) equations of motion, the free particle admits a larger symmetry group, the Schrödinger group, $\mathrm{SL}(5)$, which in addition to the Galilean transformation, admits two additional types of symmetries. The first is that of dilations, where we simply rescale both time and distances to compensate each other (this is roughly the freedom to redefine our units)

\begin{eqnarray} (t, \vec{x}) \mapsto (\alpha t, \alpha^{-2} \vec{x}) \end{eqnarray}

Klein geometry

Other coordinates

As the free particle is fundamentally a straight line, Cartesian coordinates are the ideal setting for them. But of course any coordinate system can be worked out.

A common one is that of spherical coordinates. While the ideal setting in this case is that the particle's origin should be at the center of the sphere, then simply reducing to a one dimensional problem along a ray, let's look at the case of an arbitrary point (this could be for instance used to compute motion of objects in space in some common astronomical coordinate system). As usual with spherical coordinates, acceleration has to keep in mind the nonlinear time dependence of angular coordinates

\begin{eqnarray} \ddot{\vec{x}}(t) &=& (\ddot{r} - r \dot{\theta}^2 - r \dot{\varphi} \sin^2(\theta)) \vec{e}_r\\ && + (r \ddot{\theta} + 2 \dot{r} \dot{\varphi} - r \dot{\varphi}^2 \sin(\theta) \cos(\theta)) \vec{e}_\theta\\ && + (r \ddot{\varphi} \sin(\theta) + 2 \dot{r} \dot{\varphi} \sin(\theta) + 2 r \dot{\theta} \dot{\varphi} \cos(\theta)) \vec{e}_{\varphi} \end{eqnarray}

To simplify things a bit, parts of this equation are given by invariant quantities. The angular momentum defined by the center of the coordinate system is given by

\begin{eqnarray} \vec{J} &=& \vec{x} \times (m \dot{\vec{x}})\\ &=& r \vec{e}_r \times m (\dot{r} \vec{e}_r + r (\dot{\theta} \vec{e}_{\theta} + \dot{\varphi} \sin(\theta) \vec{e}_{\varphi}))\\ &=& mr^2 \vec{e}_r \times (\dot{\theta} \vec{e}_{\theta} + \dot{\varphi} \sin(\theta) \vec{e}_{\varphi})\\ &=& mr^2 (-\dot{\varphi} \sin(\theta) \vec{e}_{\theta} + \dot{\varphi} \vec{e}_{\varphi}) \end{eqnarray}

which is indeed a conserved quantity, as we have

\begin{eqnarray} \dot{\vec{J}} &=& m \frac{d}{dt} \left( r^2 (-\dot{\varphi} \sin(\theta) \vec{e}_{\theta} + \dot{\varphi} \vec{e}_{\varphi}) \right)\\ &=& m \left( 2 \dot{r} r (-\dot{\varphi} \sin(\theta) \vec{e}_{\theta} + \dot{\varphi} \vec{e}_{\varphi} + r^2 \frac{d}{dt} (-\dot{\varphi} \sin(\theta) \vec{e}_{\theta} + \dot{\varphi} \vec{e}_{\varphi}) ) \right) \end{eqnarray}

\begin{eqnarray} J^2 &=& \end{eqnarray}

As we saw earlier, every component of $\vec{J}$ is conserved for the free particle

\begin{eqnarray} J_z &=& m (x v_y - y v_x)\\ &=& m r^2 \dot{\varphi} \sin^2(\theta) \end{eqnarray}

and the energy is

\begin{eqnarray} E &=& \frac{m}{2} |\dot{\vec{x}}|^2\\ &=& \frac{m}{2} \left( \dot{r}^2 + r^2 (\dot{\theta}^2 + \dot{\varphi}^2 \sin^2(\theta)) \right) \end{eqnarray}

So that we can rewrite them as

\begin{eqnarray} \ddot{\vec{x}}(t) &=& \left( \ddot{r} - r \dot{\theta}^2 - \frac{J_z^2}{m^2 r^4} \right) \vec{e}_r\\ && + (r \ddot{\theta} + 2 \dot{r} \dot{\varphi} - r \dot{\varphi}^2 \sin(\theta) \cos(\theta)) \vec{e}_\theta\\ && + (r \ddot{\varphi} \sin(\theta) + 2 \dot{r} \dot{\varphi} \sin(\theta) + 2 r \dot{\theta} \dot{\varphi} \cos(\theta)) \vec{e}_{\varphi} \end{eqnarray}

Being free, this means that all three components must be zero, giving us the following system to solve :

\begin{eqnarray} 0 &=& \ddot{r} - r \dot{\theta}^2 - r \dot{\varphi} \sin^2(\theta)\\ 0 &=& r \ddot{\theta} + 2 \dot{r} \dot{\varphi} - r \dot{\varphi}^2 \sin(\theta) \cos(\theta))\\ 0 &=& r \ddot{\varphi} \sin(\theta) + 2 \dot{r} \dot{\varphi} \sin(\theta) + 2 r \dot{\theta} \dot{\varphi} \cos(\theta) \end{eqnarray}

If we consider an actual radial line, this means that all angular derivatives $\dot{\theta}$, $\dot{\varphi}$ vanish, leaving us only with $\ddot{r} = 0$, as expected. If we do not require this constraint however, we can look at the solution by converting the Cartesian solution,

\begin{eqnarray} r(t) &=& \sqrt{(x_0 + v_x t)^2 + (y_0 + v_y t)^2 + (z_0 + v_z)^2}\\ \theta(t) &=& \arccos(\frac{z_0 + v_z t}{r(t)})\\ \varphi(t) &=& \mathrm{atan2}(y_0 + v_y t, x_0 + v_x t) \end{eqnarray}

We now need to convert our initial conditions into spherical ones, given by the trio $(r_0, \theta_0, \varphi_0)$ and $(v_{r, 0}, v_{\theta, 0}, v_{\varphi, 0})$, as

\begin{eqnarray} r_0 &=& \sqrt{x_0 + y_0 + z_0 }\\ \theta_0 &=& \arccos(z_0){r_0})\\ \varphi_0 &=& \mathrm{atan2}(y_0, x_0) \end{eqnarray}

and conversely,

\begin{eqnarray} x_0 &=& \\ y_0 &=& \\ z_0 &=& \end{eqnarray}

Cylindrical coordinates

Analysis

Those equations will occur quite a lot in what will follow, as they are the core of what the motion of a free particle is, so let's look at them in a bit more detail. We are considering the equation of a map $\mathbb{R} \to \mathbb{R}^n$, which, unless we consider weaker senses of derivatives (this will not be necessary here), are at least $C^2$ functions.

\begin{eqnarray} \ddot{\vec{x}}(t) = 0 \end{eqnarray}

This is a second order ordinary differential equation with constant coefficients, a map from $C^2(\mathbb{R}, \mathbb{R}^n)$ to $C^0(\mathbb{R}, \mathbb{R}^n)$.

If we consider instead the equation of the momentum, $\dot{p} = 0$, this is a differential equation of the form $y'(t) = f(t, y(t))$, with $f$ the constant zero map. Pretty obviously $f$ is smooth in both variables and in particular is globally Lipschitz, since for any two $t_1, t_2 \in \mathbb{R}$, and any two points $\vec{x}_1, \vec{x}_2 \in \mathbb{R}^n$

\begin{eqnarray} |f(t_1, \vec{x}_1) - f(t_2, \vec{x}_2)| = 0 \leq M \|(t_2, \vec{x}_2) - (t_1, \vec{x}_1)\| \end{eqnarray}

By the Picard–Lindelöf theorem, the solution to this equation given some initial condition $\vec{p}_0 = \vec{p}(0)$ is unique, so that we can indeed say that this is the solution, and furthermore, using the equation

\begin{eqnarray} \dot{x}(t) = \frac{\vec{p}(t)}{m} \end{eqnarray}

with $\vec{p}(t) = \vec{p}_0$, we have the same situation for the constant function $f(t, \vec{x}(t)) = p_0 / m$, also globally Lipschitz, with a unique solution. Therefore, given an initial position and velocity, this equation has a unique solution.

Its Green function $G(t, t')$ is defined by considering the linear differential operator $\partial_t^2$ acting on the space of distributions $\mathcal{D}'$

the solution of the distributional equation

\begin{eqnarray} L G(t, t') = \delta(t' - t) \end{eqnarray}

The Green's function is

\begin{equation} G(t, t') = \theta(t - t') \frac{(t - t')}{2} + \theta(t' - t) \frac{(t' - t)}{2} \end{equation}

Decomposed into "retarded" and "advanced" solutions :

\begin{eqnarray} G^-(t, t') &=& \theta(t - t') \frac{(t - t')}{2} G^+(t, t') &=& \theta(t' - t) \frac{(t' - t)}{2} \end{eqnarray}

Rigid body mechanics

If we consider the motion of a rigid body, of mass $m$ and moment of inertia $I$, its equations of motion are given by

\begin{eqnarray} \dot{p} &=& 0\\ \dot{L} &=& 0 \end{eqnarray}

As both the force $\vec{F}$ and the torque $\vec{T} = \vec{x} \times \vec{F}$ are zero. The momentum solution is the same as usual (corresponding to the motion of the center of mass of the object)

\begin{equation} \vec{p} = \vec{p}_0 \end{equation}

While its angular momentum will similarly be a constant

\begin{equation} \vec{L} = \vec{L}_0 = I \vec{\omega} \end{equation}

giving the object a constant angular velocity around its axis, the object is simply freely rotating as it moves.

An alternative formalism for this is to deal with it in the context of screw theory. In screw theory, we consider pairs of vectors $(\vec{S}, \vec{V})$

Wrench :

\begin{eqnarray} R = (\vec{F}, \vec{P} \times \vec{F}) \end{eqnarray}

Newton's law :

\begin{equation} \frac{d}{dt} \{ K \} = \{ T \} \end{equation}

As we have no force involved, the wrench here is the zero screw $(0, 0)$

As there is no force to worry about here, the screw formalism will not differ significantly from the usual Newtonian one.

Continuum mechanics

If we consider furthermore the motion of some localized object, described by a mass distribution $\rho$ and a local velocity vector field $\vec{u}$, a standard equation of motion for such a case is given by Cauchy's momentum equation

\begin{equation} \frac{D}{Dt} \vec{u} = \frac{1}{\rho} \vec{\nabla} \cdot \mathbf{\sigma} + \vec{f} \end{equation}

with $\sigma$ its stress tensor and $\vec{f}$ the body force. In the free case, $\vec{f}$ vanishes, so that the equation reduces to

\begin{equation} \rho \frac{D}{Dt} \vec{u} = \vec{\nabla} \cdot \mathbf{\sigma} \end{equation}

or, expanding the material derivative,

\begin{equation} \rho (\frac{\partial \vec{u}}{\partial t} + \vec{u} \cdot \vec{\nabla} \vec{u}) = \vec{\nabla} \cdot \mathbf{\sigma} \end{equation}

In addition to this, we also have the law of conservation of mass and of energy,

\begin{equation} \frac{\partial \rho}{\partial t} + \vec{\nabla} \cdot (\rho \vec{u}) = 0 \end{equation}

which implies using the product rule

\begin{equation} \frac{\partial \rho}{\partial t} = -((\vec{\nabla} \rho) \cdot \vec{u} + \rho \vec{\nabla} \cdot \vec{u} ) \end{equation}

The global motion of the body is given by the averaged mass density and velocity,

\begin{equation} \vec{x} = \int_{\mathbb{R}^3} \vec{y} \rho(\vec{y}) d\vec{y} \end{equation}

By the Reynolds transport theorem, we have that

\begin{equation} \frac{d}{dt} \int_{U} \vec{u}(x) d^3x = \int_{U} \frac{\partial}{\partial t} \vec{u}(x) d^3x + \int_{\partial U} (\vec{v}_b \cdot \vec{n}) \vec{u} dA \end{equation} \begin{equation} \ddot{\rho} = \end{equation}

Lagrangian mechanics

One of the main tool for mechanics is of course the Lagrangian perspective. This section will not do every possible perspective on the Lagrangian for a free particle, as many other cases are also relating to it, but simply the "traditional" Lagrangian formalism (what Segal calls the "mythological picture" of the Lagrangian). We will look at a few of the potential formal issues, but not define it fully formally yet.

Dynamics

The Lagrangian of a free particle is typically given by

\begin{equation} L = \frac{m}{2} \dot{\vec{x}} \cdot \dot{\vec{x}} \end{equation}

with action

\begin{equation} S[\vec{x}; [t_i, t_f]] = \int_{t_i}^{t_f} dt\ \frac{m}{2} \dot{\vec{x}} \cdot \dot{\vec{x}} \end{equation}

This action is defined for some function $\vec{x}$, on the space of $C^1$ functions mapping $\vec{x} : [t_a, t_b] \to \mathbb{R}^n$, with appropriate boundary/initial value conditions. The variation of the action within that function space gives (using for instance standard methods like Gateaux derivatives)

\begin{eqnarray} \delta S &=& \int_{t_i}^{t_f} dt \left[\frac{\partial L}{\partial \vec{x}} \delta \vec{x} + \frac{\partial L}{\partial \dot{\vec{x}}} \delta \dot{\vec{x}} \right] \end{eqnarray}

which is, integrating by parts,

\begin{eqnarray} \delta S &=& \int_{t_i}^{t_f} dt \left[ (\frac{\partial L}{\partial \vec{x}} - \frac{d}{dt}\frac{\partial L}{\partial \dot{\vec{x}}}) \delta \vec{x} \right] + \left[ \frac{\partial L}{\partial \dot{\vec{x}}} \cdot \delta \vec{x} \right]^{t_2}_{t_1} \end{eqnarray}

As the perturbation $\delta \vec{x}$ is always $0$ on the boundaries, this leads us, by the fundamental lemma of the calculus of variation, to

\begin{equation} \frac{\partial L}{\partial \vec{x}} - \frac{d}{dt}\frac{\partial L}{\partial \dot{\vec{x}}} = 0 \end{equation}

\begin{equation} - \frac{d}{dt} (m \dot{\vec{x}}(t)) = 0 \end{equation} \begin{equation} \ddot{\vec{x}}(t) = 0 \end{equation}

We can solve this in two ways. As an initial value problem, $\vec{x}(0) = \vec{x}_0$, $\dot{\vec{x}}(0) = \vec{v}_0$, the solution is the same as classically,

\begin{equation} \vec{x}(t) = \vec{v}_0 t + \vec{x}_0 \end{equation}

As a boundary value problem, $\vec{x}(t_i) = \vec{x}_i$, $\vec{x}(t_f) = \vec{x}_f$, we get

\begin{equation} \vec{x}(t) = \frac{\vec{x}_f - \vec{x}_i}{t_f - t_i} t + \frac{1}{2} \left[ \vec{x}_i + \vec{x}_f - \frac{\vec{x}_f - \vec{x}_i}{t_f - t_i} (t_i + t_f)\right] \end{equation}

with the velocity

\begin{equation} \dot{\vec{x}}(t) = \frac{\vec{x}_f - \vec{x}_i}{t_f - t_i}\right] \end{equation}

Once we have the actual solution, we can also compute the actual extremal action if we wish to.

\begin{eqnarray} S &=& \int_{t_i}^{t_f} dt \frac{m}{2} \vec{v}_0 \cdot \vec{v}_0\\ &=& \frac{m}{2} \vec{v}_0 \cdot \vec{v}_0 [t_f - t_i]\\ \end{eqnarray}

or, in terms of the boundaries,

\begin{equation} S = \frac{m}{2} \frac{|\vec{x}_b - \vec{x}_a|^2}{t_f - t_i}\\ \end{equation}

Having actually computed the extremal action, we can also recast it as a function of the boundary conditions, to obtain the Dirichlet on-shell action :

\begin{equation} S[t_i, \vec{x}_i; t_f, \vec{x}_f] = \frac{m}{2} \frac{|\vec{x}_b - \vec{x}_a|^2}{t_f - t_i} \end{equation}

From this solution we can also define the Lagrangian flow, where given the tuple $(\vec{x}_0, \vec{v}_0)$, its flow by the Lagrangian with time $T$ will be its time evolution after $T$ :

\begin{equation} \Phi_T(\vec{x}_0, \vec{v}_0) = (\vec{x}_0 + \vec{v}_0 T, \vec{v}_0) \end{equation}

This function is invertible,

\begin{eqnarray} \Phi^{-1}_T(\vec{x}, \vec{v}) &=& (\vec{x} - \vec{v} T, \vec{v})\\ &=& \Phi_{-T}(\vec{x}, \vec{v}) \end{eqnarray}

and furthermore, for any two points $\vec{x}_1$, $\vec{x}_2$, there always exists some initial velocity $\vec{v}$ for any value of $T$ evolving one into the other, which is simply

\begin{equation} \vec{v} = \frac{\vec{x}_2 - \vec{x}_1}{T} \end{equation}

Symmetries

The symmetries of the Lagrangian for the free particles are the following ones, and can be derived using Noether's identities. In the case of a classical point particle, the transformations are typically given by two types of transformations, those on the time, given by some diffeomorphism

\begin{eqnarray} \alpha : \mathbb{R} &\to& \mathbb{R}\\ t &\mapsto& t' = \alpha(t) \end{eqnarray}

and those on the position, possibly depending on the time (such as boosts), given by a diffeomorphism

\begin{eqnarray} \phi_{(-)} : \mathbb{R}^3 &\to& (\mathbb{R} \to \mathbb{R}^3)\\ \vec{x} &\mapsto& \vec{y} = \phi_{(-)} \circ \vec{x} \end{eqnarray}

Our new configuration $\vec{y} = \phi_{(-)} \circ \vec{x}$ is still a function of the original time $t$ so that if we wish to evaluate it at $t'$, we typically define the new configuration $\vec{x}' = \phi \circ \vec{x} \circ \alpha^{-1}$,

\begin{equation} \vec{x}'(t') = (\phi_{(-)} \circ \vec{x} \circ \alpha^{-1} \circ \alpha) (t) = (\phi_{(-)} \circ \vec{x}) (t) \end{equation}

We then say that the action is invariant under such transformations if the transformation acting on both the configuration and the domain of integration is invariant,

\begin{equation} S[\vec{x}; U] = S[(\phi_{(-)} \circ \vec{x}) \circ \alpha^{-1}; \alpha(U)] \end{equation}

To look at the effect of such transformations on the action, we can first recast it by using the integration by substitution. Using the variable $s = \alpha(t)$,

\begin{eqnarray} S[(\phi_{(-)} \circ \vec{x}) \circ \alpha^{-1}; \alpha(U)] &=& \int_{\alpha(U)} L(s, \vec{y}(s), \frac{d}{ds} \vec{y}(s)) ds \end{eqnarray}

The simplest case is the time symmetries, where $\phi$ is the identity map.

If we perform the affine coordinate transform $t \to t' = \alpha t + T$, with $\alpha$ some scaling factor and $T$ a time translation, let's redefine the action by substitution in the integral,

\begin{eqnarray} S[x; [t_a, t_b]] &=& \int_{t_a}^{t_b} \frac{m}{2}\dot{\vec{x}} dt \end{eqnarray}

[...]

Boundary conditions

When we say that the configuration space of the solution has "boundary conditions", what does it mean exactly?

First we have to figure out which boundary conditions we are referring to exactly. In typical Lagrangian mechanics, we usually seem to imply that the boundary conditions are given by the configuration at the bounaries, that is

\begin{eqnarray} \vec{x}(t_i) &=& \vec{x}_i\\ \vec{x}(t_f) &=& \vec{x}_f \end{eqnarray}

a type of Dirichlet boundary condition, which is why we say that the variation is zero at those points. But of course in real physics problems, that is not the actual boundary conditions we have, as we typically pick initial conditions, which is a type of mixed boundary conditions.

This stems from a duality within some systems between the boundary values of the configuration and the initial value problem. As we saw earlier, for the equation of motion of the free particle, there is exactly one solution for any given initial condition of the position and velocity. This is formalised by the notion of path structures on jet manifolds (cf here). This notion is typically done using fiber bundles, but we will delay the bundle description for now to the section on bundle formalisms. As we are using the manifold $\mathbb{R}^3$, we can simply use tuples to work with here.

A path structure on $\mathbb{R}^n$ is a way to describe the solutions of a system of second order differential equations. If we describe our underlying variable and function as a tuple $\mathbb{R} \times \mathbb{R}^n$, $(t, \vec{x})$, we define the first jet space of this tuple by a tuple including its first derivatives,

\begin{equation} J^1 \mathbb{R} \times \mathbb{R}^n \cong \{ (t, \vec{x}, \dot{\vec{x}})\ |\ t \in \mathbb{R},\ \vec{x}, \dot{\vec{x}} \in \mathbb{R}^n \} \end{equation}

And its second jet space with the second derivative

\begin{equation} J^2 \mathbb{R} \times \mathbb{R}^n \cong \{ (t, \vec{x}, \dot{\vec{x}}, \ddot{\vec{x}})\ |\ t \in \mathbb{R},\ \vec{x}, \dot{\vec{x}}, \ddot{\vec{x}} \in \mathbb{R}^n \} \end{equation}

A path structure can be described among other definitions by a family of curves such that, for any two values $(x, p)$, there exists exactly one curve in this family such that, for some $t$,

\begin{eqnarray} \vec{x}(t) &=& \vec{x}\\ \dot{\vec{x}}(t) &=& \vec{p} \end{eqnarray}

In other words, we have exactly one curve passing through a given point in a given direction. Such path structures correspond to a system of ordinary differential equations,

\begin{equation} \ddot{\vec{x}}(t) = \vec{H}(\vec{x}, \dot{\vec{x}}) \end{equation}

where $H$ is homogeneous of degree two in $\dot{\vec{x}}$, ie

\begin{equation} \vec{H}(\vec{x}, \alpha \dot{\vec{x}}) = \alpha^2 \vec{H}(\vec{x}, \dot{\vec{x}}) \end{equation}

Projectivized :

\begin{equation} P (J^1 \mathbb{R} \times \mathbb{R}^n) \cong \end{equation}

[...]

The free particle is in fact what is called the "flat model" of a path structure, where the underlying ODE is merely $\ddot{\vec{x}} = 0$

Double fibration :

\begin{equation} P^{n+1} \overset{\pi}{\leftarrow} F_{0,1}^{n+1} \overset{\rho}{\rightarrow} G_1(P^{n+1}) \end{equation}

Once we have this duality, what does it mean to impose the boundary condition?

As we've seen, the boundary conditions are of fundamental importance to deal with the boundary term of the variation,

\begin{eqnarray} \left[ \frac{\partial L}{\partial \dot{\vec{x}}} \cdot \delta \vec{x} \right]^{t_2}_{t_1} \end{eqnarray}

meaning that if we wish to have the actual Euler-Lagrange equations as a condition to make the variation vanish, we have to find a way to make this part vanish. This term will only vanish for arbitrary variations if $\partial_{v} L$ itself vanishes at the boundary, which means that it will only be true in our case for the static particle, a notion which is not even frame invariant.

The usual notion is then that we do not permit arbitrary variations of the action, but only variations vanishing on the boundary, $\delta \vec{x}(\partial [t_i, t_g]) = 0$.

van Vleck determinant

In the Lagrangian formalism, there is commonly some duality between the expression of the action in terms of its boundary conditions at initial and final time, and its initial conditions in terms of its initial position and velocity. This is given by the van Vleck determinant.

The extremal action for the boundary conditions $\vec{x}(t_i) = \vec{x}_i$, $\vec{x}(t_i) = \vec{x}_i$

\begin{eqnarray} S(\vec{x}_i, t_i; \vec{x}_f, t_f) &=& \frac{m}{2} \frac{\| \vec{x}_f - \vec{x}_i \|^2}{t_f - t_i} \end{eqnarray}

There is only a single extremal path for free particles, therefore there is a unique van Vleck determinant, which is

\begin{eqnarray} \Delta(\vec{x}_i, t_i; \vec{x}_f, t_f) &=& (-1)^n \det (\frac{\partial^2 S}{\partial q_i \partial q_f})\\ &=& (-1)^n \frac{m}{2} \frac{\det( 2(\vec{x}_f + \vec{x}_i - 1) )}{t_f - t_i} \end{eqnarray}

Global aspects

In this Lagrangian formalism, the configuration space is typically something like $C^\infty(\mathbb{R}, \mathbb{R}^n)$ (if we only assume smooth trajectories here, which will be helpful to simplify things), or using some subset of its domain like $C^\infty(\I, \mathbb{R}^n)$, where we pick $I$ to be some connected subset of $\mathbb{R}$. As a space it has many important properties. It is of course a vector space as $\mathbb{R}^n$ is itself a vector space, where the linearity manifests as

\begin{equation} (\vec{x_1} + \vec{x_1})(t) = \vec{x_1}(t) + \vec{x_1}(t) \end{equation}

As a vector space it is infinite dimensional, which a detailed treatment thereof can be found in here. Furthermore, to give it better properties for variational calculus, it is a locally convex vector space. A subset of a vector space $A \subset V$ is called convex if for any two non-negative real numbers $a, b$, such that $a + b = 1$, we have

\begin{equation} v_1, v_2 \in A \to a v_1 + b v_2 \in A \end{equation}

which in the case of a finite dimensional vector space translates to the actual notion of convexity, ie any segment between two points belongs to the subset.

For more details on this space, we have to define a topology on it, so that we can give it the structure of a topological vector space (TVS), where the underlying vector space as a set is given a topology $\tau$, and the vector space operations $(+, -, \cdot)$ are all continuous in that topology.

$C^\infty(\I, \mathbb{R}^n)$ can be given a variety of topologies, coming from different perspectives. As a space of mappings between topological spaces, we can give it the compact-open topology, where for every compact subset $K$ of $\mathbb{R}$ (such as closed intervals $[a,b]$) and open subset $Y$ of $\mathbb{R}^n$, we can associate the set $V(K, U)$ of all functions $f$ such that $f(K) \subseteq U$. The collection of all such $V(K, U)$ is a subbase of the compact open topology, ie every such collection of functions is an open set of $C^\infty(\mathbb{R}, \mathbb{R}^n)$, and the compact-open topology is the smallest topology having those sets open.

\begin{equation} B = \bigcup_{K \Subset \mathbb{R}} \bigcup_{U \in \tau(\mathbb{R}^n)} V(K, U) \end{equation}

Another common topology used is the Whitney topology. For $J^k(\mathbb{R}, \mathbb{R}^n)$ the jet space of our functions, ie for $j^k f \in J^k(\mathbb{R}, \mathbb{R}^n)$, we have

\begin{equation} j^k f = (t, \vec{x}(t), \dot{\vec{x}}(t), \ldots, \vec{x}^{(k)}(t)) \end{equation}

so that $J^k(\mathbb{R}, \mathbb{R}^n) \cong \mathbb{R} \times \prod_{i = 1}^k \mathbb{R}^n$, define an open subset of the jet space $U$ and the map

\begin{equation} S^k(U) = \{ f \in C^\infty(\mathbb{R}, \mathbb{R}^n) | (j^k f)(\mathbb{R}) \subseteq U \} \end{equation}

Both of those common topologies can be derived from the notion of locally convex topological vector spaces. A TVS is locally convex if it has a neighbourhood basis of $0$ composed of balanced convex sets, ie there's a family of balanced convex open sets $B \subseteq \tau$ such that $0$ belongs to every open set of $B$ and if $U \in \tau$, $0 \in U$, then there is an open set $U' \in B$ such that $U \subseteq U$.

This definition is equivalent to the existence of a family of seminorms on $V$. A seminorm $p : V \to \mathbb{R}$ is such that

and given any family of seminorms on $V$, this implies that there exists a topology which makes it a locally convex TVS.

In the case of $C^\infty(I, \mathbb{R}^n)$, a family of seminorms is given by, for every compact subset $K$ of $I$

\begin{equation} p_K(\vec{x}) = \sup_{t \in K} |\vec{x}(t)| \end{equation}

Those seminorms generate the compact-open topology, as can be seen by considering

[...]

Another family of seminorms is given by considering in addition the bounds of the derivatives.

\begin{equation} p_{K, n}(\vec{x}) = \sup_{t \in K} |\vec{x}^{(n)}(t)| \end{equation}

As we are considering typically the action on some compact interval of time, $C^\infty([t_i, t_f], \mathbb{R}^n)$, our space is furthermore a Fréchet space, with the family of seminorms

\begin{equation} \| \vec{x} \|_k = \sup(\{ |\vec{x}^{(k)}(t)|\ | t \in [t_i, t_f]\}) \end{equation}

In other words, the greatest distance from zero, velocity, acceleration, etc, of that trajectory.

In addition to being a Fréchet space, $C^\infty(I, \mathbb{R}^n)$ is a Banach space : it is a complete normed space given some norm $\| \cdot \|$

Given those structures, we can finally define derivatives on this space, the variations we've been using. For a space of functions like $C^\infty(I, \mathbb{R}^n)$, functions on that space are functionals, ie maps of the form

\begin{eqnarray} F : C^\infty(I, \mathbb{R}^n) &\to& \mathbb{R}\\ f &\mapsto& F(f) \end{eqnarray}

Common such functionals are for instance distributions and integrals. In this case, there are two common derivatives used in such configuration spaces. First is the Gateaux derivative, which is similar to the standard notion of a derivative on $\mathbb{R}^n$ : given a vector $h \in C^\infty(I, \mathbb{R}^n)$,

\begin{equation} \frac{\delta F(f)}{\delta_h f} = \lim_{t \to 0} \frac{F(f + t h) - F(f)}{t} \end{equation}

The second standard derivative is the Fréchet derivative, which parallels the Taylor theorem on real functions : Given the Fréchet norm on $C^\infty(I, \mathbb{R}^n)$, the Fréchet derivative in the $h$ direction is given by the existence of another function $A : \mathbb{R} \to \mathbb{R}^n$ such that

\begin{equation} \lim_{\|h\|_{F}} \frac{\| F(f + h) - F(f) - Ah\|_{\mathbb{R}}}{\|h\|_F} = 0 \end{equation}

The topology allows us to define limits in the usual way,

Additional spaces that will be of importance here is the tangent space $T \mathrm{Conf}$, corresponding to the space of variations of the configuration, what we commonly write as $\delta \vec{x}$. As $\mathrm{Conf}$ is a vector space, we have $T \mathrm{Conf} \cong \mathrm{Conf}$

Our action is a functional on the configuration space, $S : \mathrm{Conf} \to \mathbb{R}$, with however a few subtleties. The action is typically defined on a specific region of the parameter space (here $\mathbb{R}$), which is here the interval $I = [t_i, t_f] \subset \mathbb{R}$

Given our action $S : C^\infty(\mathbb{R}, \mathbb{R}^n)$, its variation $\delta S$ has the critical points

\begin{equation} \mathrm{Crit}(\delta S) = \{ X \in C^\infty(\mathbb{R}, \mathbb{R}^n)\ |\ \delta S|_{X} = 0 \} \end{equation}

Its Hessian $\mathrm{Hess}(S)$ is the second variation of the action,

\begin{equation} \mathrm{Hess}(S) = \frac{\delta S}{\delta x \delta x} \end{equation}

The action is additive, in the sense that the configuration space has the structure of a vector space and for any three solutions $x_1, x_2, x_3$, such that $x_1$ and $x_3$ have disjoint support,

\begin{equation} \mathrm{supp}(x_1) \cap \mathrm{supp}(x_3) = \varnothing \end{equation}

ie if $\vec{x}_1$ is $0$ outside of the interval $[t_{i,1}, t_{f,1}]$ and likewise for $\vec{x}_3$ and $[t_{i,3}, t_{f,3}]$, and those intervals do not overlap, then we have the additivity rule,

\begin{equation} S[x_1 + x_2 + x_3] = S[x_1 + x_2] + S[x_2 + x_3] - S[x_2] \end{equation}

as can be shown

\begin{eqnarray} S[x_1 + x_2 + x_3] &=& \int_{\mathbb{R}} \frac{m}{2} | \vec{x}_1 + \vec{x}_2 + \vec{x}_3|^2 dt\\ &=& \int_{\mathbb{R}} \frac{m}{2} \left( | \vec{x}_1 + \vec{x}_2|^2 + |\vec{x}_3|^2 + 2 \vec{x}_1 \cdot \vec{x}_3 + 2 \vec{x}_2 \cdot \vec{x}_3 \right) dt\\ &=& S[x_1 + x_2] + \int_{\mathbb{R}} 2 \vec{x}_1 \cdot \vec{x}_3 dt + \int_{\mathbb{R}} \left[ |\vec{x}_3|^2 + 2 \vec{x}_2 \cdot \vec{x}_3 + |\vec{x}_2|^2 - |\vec{x}_2|^2 \right] dt\\ &=& S[x_1 + x_2] + S[x_2 + x_3] - S[x_2] + \int_{\mathbb{R}} 2 \vec{x}_1 \cdot \vec{x}_3 dt \end{eqnarray}

As $\vec{x}_1 \cdot \vec{x}_3$ is $0$ for any value of $t$ from their disjointness, the additivity is verified. It is in addition partially additive, where given two disjoint solutions $x_1$, $x_2$,

\begin{equation} S[x_1 + x_2] = S[x_1] + S[x_2] \end{equation}

which can be derived from additivity and the fact that $S[0] = 0$.

Observables

The general set of observables on such a theory is simply the set of all functionals on whatever configuration space we have decided for the theory, for instance for smooth paths,

\begin{equation} \mathrm{Obs} = [C^\infty([t_a, t_b], \mathbb{R}^3), \mathbb{R}] \end{equation}

As for any Lagrangian theory, we can split our observables into various types. Most fundamental being the "field" observables, which are point-wise observables on the configuration itself , which only depend on $\vec{x}$ at a given time $t$, and extending the configuration space to its derivatives, its various derivatives. These include observables such as position, velocity and acceleration. In terms of functionals, those would be Dirac delta functionals and their derivatives on the configuration space.

Linear and affine combinations of such field observables are the linear and affine observables, including quantities like center of mass and momentum. If we do not require the various elements of this linear sum to be at the same time, these can also be quantities such as positions and velocities averaged over a few different points in time.

Allowing the product of field observables gives us the polynomial observables,

\begin{equation} f(\vec{x}, t) = \sum_{k} \alpha_k \prod_{} x^{(i_k)}_{j_k}(t) \end{equation}

where each monomial is given by a product of components of a derivative of $\vec{x}$ at $t$. A specific case of polynomial observable is that of microcausal variables,

Hamiltonian mechanics

The momentum associated with the Lagrangian is given by the Legendre transformation,

\begin{eqnarray} \vec{p} &=& \frac{\partial L}{\partial \dot{\vec{x}}}\\ &=& m \dot{\vec{x}} \end{eqnarray}

Or more precisely, the dual vector defined by its action on a vector $\vec{v}$ as

\begin{equation} \vec{p}[\vec{v}] = m \vec{x} \cdot \vec{v} \end{equation}

The map from momentum to velocity is invertible, since the Hessian of the Lagrangian with respect to the velocities is simply $\mathrm{Hess}(L) = m \delta$, with $\delta$ the inner product, which is of full rank. This map inversion is simply $m^{-1} \delta$, so that the velocity expressed from the momentum is

\begin{equation} \vec{v}(\vec{p}) = \frac{\vec{p}}{m} \end{equation}

or more specifically the vector obtained from $\vec{p}$ (which is a dual vector) by the musical isomorphism with the metric $\delta$.

So that we can write the Hamiltonian out as the Legendre transform

\begin{eqnarray} H &=& \vec{v}(\vec{p}) \cdot \vec{p} - L\\ &=& \frac{\vec{p}\cdot\vec{p}}{m} - \frac{m}{2} \vec{v}(\vec{p}) \cdot \vec{v}(\vec{p}) \\ &=& \frac{\vec{p}\cdot\vec{p}}{2m} \end{eqnarray}

We can therefore use the Hamiltonian as usual to define the time evolution of any observable : for any time-independent observable, $\dot{O} = \{ H, O \}$, as $H$ is independent from $t$, with $\{ -, - \}$ the Poisson bracket.

The equations of motion are given by the time evolution of our phase space variables,

\begin{eqnarray} \dot{x} &=& \{ H, \vec{x} \}\\ &=& \frac{\partial H}{\partial \vec{x}} \frac{\partial \vec{x}}{\partial \vec{p}} - \frac{\partial \vec{x}}{\partial \vec{x}} \frac{\partial H}{\partial \vec{p}}\\ &=& - \frac{\partial H}{\partial \vec{p}}\\ \dot{p} &=& \{ H, \vec{p} \}\\ &=& \frac{\partial H}{\partial \vec{x}} \frac{\partial \vec{p}}{\partial \vec{p}} - \frac{\partial \vec{p}}{\partial \vec{x}} \frac{\partial H}{\partial \vec{p}}\\ &=& \frac{\partial H}{\partial \vec{x}}\\ \end{eqnarray}

as usual

\begin{eqnarray} \frac{\partial H}{\partial \vec{x}} &=& -\dot{\vec{p}}\\ &=& 0\\ \frac{\partial H}{\partial \vec{p}} &=& \dot{\vec{x}}\\ &=& \frac{\vec{p}}{m} \end{eqnarray}

with the solution

\begin{eqnarray} \vec{p}(t) &=& \vec{p}_0\\ \vec{x}(t) &=& \frac{\vec{\vec{p}_0}}{m} t + \vec{x}_0 \end{eqnarray}

For later purpose, it may be useful to compute our basic Poisson brackets. With our canonical coordinates, this is

\begin{eqnarray} \left\{ f, g \right\} &=& \frac{\partial f}{\partial \vec{x}} \cdot \frac{\partial g}{\partial \vec{p}} - \frac{\partial g}{\partial \vec{x}} \cdot \frac{\partial f}{\partial \vec{p}} \end{eqnarray}

It then follows that

\begin{eqnarray} \left\{ x^i, p^j \right\} &=& \delta^{ij}\\ \left\{ x^i, x^j \right\} &=& 0\\ \left\{ p^i, p^j \right\} &=& 0 \end{eqnarray}

Other useful quantities within the formalism is to look also at the Poisson brackets of our various Noether currents, $\vec{J}$ and $Q_{\mathrm{CoM}}$

Routhian mechanics

There exists a middle point between Lagrangian and Hamiltonian mechanics, called Routhian mechanics, where we only perform a Legendre transform on a specific set of degrees of freedom. There is fairly little change involved for the case of the free particle. If we pick one set of coordinates $\{ x_i \}_{i \in I}$ as Lagrangian variables and another $\{ x_j \}_{j \in J}$ as Hamiltonian, $I$, $J$ disjoint ang $I \cup J$ being all the variables, the Routhian is given by

\begin{equation} R(t, \vec{x}_i, \dot{\vec{x}}_i, \vec{x}_j, \vec{p}_j) = \sum_{j = 1}^{|J|} p_i v_i(p_i) - L(t, \vec{x}_i, \dot{\vec{x}}_i, \vec{x}_j, \vec{p}_j) \end{equation}

So in our case,

\begin{eqnarray} R(t, \vec{x}_i, \dot{\vec{x}}_i, \vec{x}_j, \vec{p}_j) &=& \sum_{j = 1}^{|J|} \frac{p_i^2}{m} - \frac{m}{2} \sum_{k = 1}^n \dot{x}_k^2\\ &=& \frac{1}{2m} \sum_{j = 1}^{|J|} p_i^2 - \frac{m}{2} \sum_{i = 1}^{|I|} \dot{x}_i^2 \end{eqnarray}

The equations of motion of the Routhian are simply the ones of Lagrangian mechanics on $I$ and Hamiltonian mechanics on $J$.

Invariance methods

From the symmetries of the systems we have, we can look at the conserved quantities. The four Noether currents of the free particle are the energy,

\begin{equation} E = \frac{1}{2} mv^2 \end{equation}

the momentum

\begin{equation} \vec{p} = m\vec{v} \end{equation}

the angular momentum

\begin{equation} \vec{J} = m \vec{x} \times \vec{v} \end{equation}

and the center of mass

\begin{equation} Q_{\text{CM}} = m (\vec{v} t - \vec{x}) \end{equation}

From their invariance and the boundary condition that $\vec{x}(0) = \vec{x}_0$, $\vec{v}(0) = \vec{v}_0$, we can infer that

\begin{eqnarray} E(t_f) = E(t_i) = \frac{1}{2} m v_i^2 = \frac{1}{2} m v_f^2 \to v_f^2 = v_i^2 \end{eqnarray} \begin{eqnarray} \vec{p}(t_f) = \vec{p}(t_i) = m\vec{v}_i = m\vec{v}_f \to \vec{v}_f = \vec{v}_i \end{eqnarray} \begin{eqnarray} Q_{\text{CM}}(t_i) &=& Q_{\text{CM}}(t_f) \\ &=& m (\vec{v}_i t_i - \vec{x}_i)\\ &=& m (\vec{v}_f t_f - \vec{x}_f)\\ &\to& \vec{x}_f - \vec{x}_i = \vec{v}_i (t_f - t_i) \end{eqnarray}

which is the solution we've previously seen.

Liouville methods

If we consider our system to not be defined by a sharp set of boundary conditions, but by some probability distribution, we must use the Liouville formalism.

If we have the phase space $P$, the probability density $\rho$ is a probability density on $P$ such that the probability to measure the free particle as within some subset $U \subset P$ is given by

\begin{equation} \int_{U} \rho(q, p) d^nq d^np \end{equation}

As a function on a phase space, its time evolution is as usual given by the Poisson bracket of with the Hamiltonian :

\begin{eqnarray} \dot{\rho} &=& \{ H, \rho \} \end{eqnarray}

Giving us the Liouville equation, which, in our case, as $H$ is independent from $q$,

\begin{eqnarray} \dot{\rho} + \frac{1}{m} \vec{p} \cdot \vec{\nabla}_{q} \rho = 0 \end{eqnarray}

As we have that $\vec{p}$ is independent from $q$, we can rewrite this as

\begin{eqnarray} \dot{\rho} + \frac{1}{m} \vec{\nabla}_{q} \cdot (\rho \vec{p}) = 0 \end{eqnarray}

This is a continuity equation for $\vec{u} = \rho \vec{p}$, the flux of the momentum. Using the method of characteristics, the characteristic curves are

\begin{eqnarray} \frac{dt}{ds} &=& 1\\ \frac{d\vec{p}}{ds} &=& 0\\ \frac{d\vec{q}}{ds} &=& \frac{1}{m} \vec{p}\\ \frac{dz}{ds} &=& 0 \end{eqnarray}

with solutions

\begin{eqnarray} t(s) &=& s + t_0\\ \vec{p}(s) &=& \vec{p}_0\\ \vec{q}(s) &=& \frac{s}{m} \vec{p}_0 + \vec{q}_0\\ z(s) &=& z_0 \end{eqnarray}

with the boudnary conditions

\begin{eqnarray} t(0) &=& 0\\ \vec{p}(0) &=& \vec{p}_0\\ \vec{q}(0) &=& \vec{q}_0\\ z(0) &=& \rho(\vec{q}_0, \vec{p}_0) \end{eqnarray} \begin{equation} \rho(q(t), p(t), t) = \rho_0(q_0, p_0) \end{equation} \begin{equation} (q_0, p_0) = (q - \frac{p}{m} t, p) \end{equation} \begin{equation} \rho(q, p, t) = \rho_0(q - \frac{p}{m} t, p) \end{equation}

A few basic examples we can apply this to is the sharp case, where $\rho(0)$ is the Dirac distribution for a given point in phase space, $\rho_0 = \delta_{(q, p)}$. This case simply reduces to the standard case,

\begin{equation} \rho(q, p, t) = \delta_{(q - \frac{p}{m} t, p)} \end{equation}

where $p$ remains sharply defined and $q$ simply changes position as usual. Another case we can look at is the multivariate Gaussian

\begin{equation} \rho_0(q, p, t) = \frac{1}{2\pi \sigma_q \sigma_p} \exp(- \frac{1}{2 \sigma_q^2} (\vec{q} - \vec{q}_0)^2) \exp(- \frac{1}{2 \sigma_p^2} (\vec{p} - \vec{p}_0)^2) \end{equation}

Dirac parametrized system

As an example of a simple parametrized system, it is possible to turn the point particle into a parametrization invariant Lagrangian, by considering the degrees of freedom $(t, \vec{x})$, parametrized by $\tau$, in which case the action is

\begin{equation} S = \int_{t_i}^{t_f} d\tau\ \frac{m}{2} \frac{\dot{\vec{x}}(\tau) \cdot \dot{\vec{x}}(\tau)}{\dot{t}(\tau)} \end{equation}

This action is invariant under the reparametrization $\tau \to \tau' = f(\tau)$, for some diffeomorphism $f$, in which case, we get via change of variables

\begin{eqnarray} S' &=& \int_{\tau'_1}^{\tau'_2} d\tau' \frac{m}{2} \frac{\dot{\vec{x}}(\tau') \cdot \dot{\vec{x}}(\tau')}{\dot{t}(\tau')}\\ &=& \int_{\tau_1}^{\tau_2} d\tau' f'(\tau) \frac{m}{2} \frac{\dot{\vec{x}}(f(\tau)) \cdot \dot{\vec{x}}(f(\tau))}{\dot{t}(f(\tau))}\\ \end{eqnarray}

We can verify that our system is underdetermined by computing the equations of motion :

\begin{eqnarray} \frac{\partial L}{\partial \vec{x}} &=& 0\\ \frac{\partial L}{\partial t} &=& 0\\ \frac{\partial L}{\partial \dot{\vec{x}}} &=& m \frac{\dot{\vec{x}}(\tau)}{\dot{t}(\tau)}\\ \frac{\partial L}{\partial \dot{t}} &=& -\frac{m}{2} \frac{\dot{\vec{x}}(\tau) \cdot \dot{\vec{x}}(\tau)}{\dot{t}^2(\tau)} \end{eqnarray}

So that the Euler-Lagrange equations become

\begin{eqnarray} \frac{d}{dt}\left(\frac{\dot{\vec{x}}(\tau)}{\dot{t}(\tau)}\right) &=& 0\\ \frac{d}{dt}\left(\frac{\dot{\vec{x}}(\tau) \cdot \dot{\vec{x}}(\tau)}{\dot{t}^2(\tau)} \right) &=& 0 \end{eqnarray}

which can be integrated into

\begin{eqnarray} \frac{\dot{\vec{x}}(\tau)}{\dot{t}(\tau)} &=& \vec{C}_1\\ \frac{\dot{\vec{x}}(\tau) \cdot \dot{\vec{x}}(\tau)}{\dot{t}^2(\tau)} &=& C_2 \end{eqnarray}

We'll note that the first equation reduces to the second by taking its norm :

\begin{eqnarray} \frac{\dot{\vec{x}}(\tau)}{\dot{t}(\tau)} \cdot \frac{\dot{\vec{x}}(\tau)}{\dot{t}(\tau)} &=& \vec{C}_1 \cdot \vec{C}_1\\ &=& C_2 \end{eqnarray}

We therefore have that for $4$ degrees of freedom, we only have three independent equations.

\begin{equation} \dot{\vec{x}}(\tau) = \dot{t}(\tau) \vec{C}_1 \end{equation}

Our solution is therefore

\begin{equation} \vec{x}(\tau) = \vec{x}_0 + \int d\tau\ \dot{t}(\tau) \vec{C}_1 \end{equation}

The basic gauge we can select here is $t = \tau$, in which case we simply get our previous case back. Now let's see the Hamiltonian method applied here. Our momenta are

\begin{eqnarray} \vec{p} &=& m \frac{\dot{\vec{x}}(\tau)}{\dot{t}(\tau)}\\ p_t &=& -\frac{m}{2} \frac{\dot{\vec{x}}(\tau) \cdot \dot{\vec{x}}(\tau)}{\dot{t}^2(\tau)} \end{eqnarray}

As our problem is under-determined, we cannot invert our momenta back to velocities. We are going to need to apply some constraints to our system. The important quantity to consider here is the rank of the matrix

\begin{equation} \frac{\partial^2 L}{\partial \dot{q}^a \partial \dot{q}^b} \end{equation}

with $q = (t, \vec{x})$. This matrix turns out to be

\begin{equation} \frac{\partial^2 L}{\partial \dot{q}^a \partial \dot{q}^b} = \begin{pmatrix} m \frac{\dot{\vec{x}}(\tau) \cdot \dot{\vec{x}}(\tau)}{\dot{t}^3(\tau)} & -\frac{m}{2} \frac{\dot{\vec{x}}(\tau)}{\dot{t}^2(\tau)} \\ m \frac{1}{\dot{t}(\tau)} & \frac{m}{\dot{t}(\tau)} \delta_{ij} \end{pmatrix} \end{equation}

with determinant

\begin{equation} \mathrm{det} = \end{equation}

...

The obvious constraint we get from the form of our momenta is

\begin{equation} \Phi = \frac{\vec{p} \cdot \vec{p}}{2m} + p_t = 0 \end{equation}

...

The Hamiltonian vanishes (as they typically do for parametrization invariant Lagrangians), and all that remains is the constraint Lagrangian

\begin{equation} H = \vec{p} \cdot \dot{\vec{x}} + p_t \dot{t} - L = 0 \end{equation}

meaning that only the constraint term will remain in the extended Hamiltonian

\begin{equation} H_E = v \Phi \end{equation}

The Dirac brackets of our theory are

\begin{eqnarray} \{ \vec{x}, \vec{p} \} &= & \delta\\ \{ t, p_t \} &=& 0\\ \{ \vec{x}, p_t \} &= & \delta\\ \end{eqnarray}

Bundle methods I : Sprays

For more abstract ways in classical mechanics, we can define the configuration space of a point particles by the tangent bundle. Given Euclidian space $E = \mathbb{R}^n$, the path of a particle $\gamma : \mathbb{R} \to E$ is described by the section of the trivial bundle $\pi_{\gamma} : \mathbb{R} \times E \to \mathbb{R}$, so that a section of this is

\begin{eqnarray} s : \mathbb{R} &\to& \mathbb{R} \times E\\ t &\mapsto& (t, \vec{x}(t)) \end{eqnarray}

which as a trivial vector bundle is simply a function given by

\begin{equation} \gamma = \mathrm{pr}_2 \circ s : \mathbb{R} \to E \end{equation}

The associated first jet bundle $J^1\pi_\gamma$ is then simply a slight extension of the tangent bundle, with its two projnections

\begin{eqnarray} \pi_0 : T \mathbb{R} \times TE \to \mathbb{R}\\ \pi_{1, 0} : T \mathbb{R} \times TE \to \mathbb{R} \times E \end{eqnarray}

The closest equivalent of the Newtonian formalism in this context is done via the notion of sprays, which are roughly speaking sources for the acceleration of a curve. In this context, a free particle is a curve $\gamma : L \to E$, with a lift to the tangent bundle $\gamma^\uparrow : L \to TE$, such that the curve is a flow line of the spray defining the dynamic of the theory. The spray $\xi$ is a vector field on $TE$, ie $\xi : TE \to TTE$, obeying <\p>

Semi-sprays can be used to define any dynamics (they are in fact the geometric way to define Newtonian mechanics without any restriction). As we have an action, we will more specifically use the semi-spray defined by the action functional

...

In addition to being a semispray, this is also a spray

Nonlinear connections?

Let's look at the infinite jet bundle case. $\pi : E = \mathbb{R} \times Q \to \mathbb{R}$

\begin{equation} J^{2k - 1} E = T\mathbb{R} \oplus T^{2k - 2} M \approx \mathbb{R} \oplus T^{2k - 2} M \end{equation}

Cartan form :

\begin{equation} \Theta_L = L dt + \sum_{s = 0}^{k -1} \left( \sum_{r = s}^{k - 1} (-1)^{r - s} \frac{d^{r - s}}{dt^{r - s}} \frac{\partial L}{\partial q^A_{(r + 1)}} \right) \psi_{(s)}^A \end{equation}

As we have $k = 1$ here, the sum is done only over $r = s = 0$, giving us

\begin{equation} \Theta_L = L dt + \frac{\partial L}{\partial v^A} (dq^A - v^A dt) \end{equation}

Extremal if for any $\xi \in \mathfrak{X}(J^1 Y)$, so $\xi = f dt + g dq + h dv$,

\begin{equation} (j^1 X)^* (\iota_\xi) d \Theta_L = 0 \end{equation} \begin{equation} d \Theta_L = d(L dt) + d(\frac{\partial L}{\partial v^A} (dq^A - v^A dt)) \end{equation}

link

Path structures

where if we have our equation $\ddot{x}(t) = 0$, this is a system generated on the jet bundle of our configuration space $\pi : \mathbb{R} \times \mathbb{R}^3 \to \mathbb{R}$, $J^1 \pi$, with coordinates $(t, x, v)$ by the two $1$-forms

\begin{eqnarray} \theta &=& dx - v dt\\ \omega &=& dv \end{eqnarray}

where $\theta$ is the contact form of the jet bundle (enforcing that the variable $v$ corresponds to derivatives of $x$), and $\omega$ is the $1$-form associated with the ODE, saying that the derivative of $v$ is zero. These constraints are given for sections of our bundles which are integral curves, ie if we have a curve $\gamma : \mathbb{R} \to \mathbb{R}^3$, with its lift to the jet bundle $j^1 \gamma$, then [...]

The paths that inhabit our space are defined by their direction, which is a point of the projectivized tangent bundle of our underlying space, $PT\mathbb{R}^3$.

Flag variety of pairs of incident lines and $2$-planes :

\begin{equation} F_{1,2}(\mathbb{R}^4) = \{ (\ell, P) \ |\ \ell \subset P,\ [\ell] \in P^{3},\ [P] \in \mathrm{Gr}(\mathbb{R}^{4})\} \cong G / P_{1,2} \end{equation}

The duality is given by the double fibraton on where every

Bundle methods II : Lagrangian

The Lagrangian method using the bundle formulation is that a Lagrangian is a map from the jet bundle of the bundle involved (technically the infinite jet bundle $J^\infty \pi$, but here we will only consider the first jet $J^1 \pi$) to the bundle of top forms $\bigwedge^n M$. In our case, this means that the Lagrangian is a function

\begin{equation} L : J^1 \pi \cong T \mathbb{R} \times TE \to \Lambda^n E \end{equation}

It is also common to replace this instead by picking some section of the bundle directly and applying its jet to the Lagrangian. If we disregard the $\mathbb{R}$ component of the bundle for now,

\begin{equation} S[\vec{x}; U] = \int_U (j^1 \vec{x})^* L \end{equation}

In this context, the Lagrangian is then the map

The variational bicomplex

Symmetries

Just like the case of the traditional Lagrangian method, let's look at how symmetries affect the bundle Lagrangian description.

In the bundle treatment of mechanics, the symmetries of the Lagrangian are given by bundle automorphisms, that is, diffeomorphisms on the bundle which respect the bundle structure

\begin{eqnarray} \mathrm{Aut}(E) = \{ f \in \mathrm{Diff}(E)\ | \ \} \end{eqnarray}

In which case the symmetries are those bundle automorphisms which leave the action invariant. This leads to the same notions of Noether currents and charges. This can be written by considering two types of transformations see here, which are, considering our theory as happening on the fiber bundle $\mathbb{R} \times \mathbb{R}^3 \to \mathbb{R}$, first transformations happening on the base space (ie transformations on time in our case), and transformations on the fiber.

In the case of point particles, we can find the full set of symmetries of the action by hand. The set of all automorphisms of the $\pi : \mathbb{R}^n \to \mathbb{R}$ bundle is

\begin{equation} \Phi(t, \vec{x}) = (\alpha(t), \vec{f} \circ \vec{x}(t)) \end{equation}

which acts on sections as

\begin{eqnarray} \Phi(t, \vec{x}(t)) &=& (\alpha(t), \vec{f}(\vec{x}(t), t))\\ &=& (s, \vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s))) \end{eqnarray}

with the new variable $s = \alpha(t)$ (the reparametrized time variable). Then the action becomes, with the help of some variable change,

\begin{eqnarray} S(\Phi(\vec{x})) &=& \int_{\alpha(t_a)}^{\alpha(t_b)} (\frac{d}{dt}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s)) \cdot (\frac{d}{dt}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s))) \frac{dt}{ds} ds\\ &=& \int_{\alpha(t_a)}^{\alpha(t_b)} (\frac{ds}{dt} \frac{d}{ds}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s)) \cdot (\frac{ds}{dt} \frac{d}{ds}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s))) \frac{dt}{ds} ds\\ &=& \int_{s_a}^{s_b} \dot{\alpha}(\alpha(s)) \left[ \dot{\vec{x}}(t) \frac{\partial}{\partial \vec{x}}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s)) + \frac{\partial}{\partial s}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s)) \right] \cdot (\frac{d}{ds}\vec{f}(\vec{x}(\alpha^{-1}(s)), \alpha^{-1}(s))) ds\\ &=& \end{eqnarray}

The action is invariant under the inhomogeneous Galilean group, $\mathrm{Gal}(n)$ cf. This group acts as an affine transformation on the $(n+1)$-dimensional space and time of classical mechanics as

\begin{equation} g \in \mathrm{Gal}(n),\ g \cdot (t, \vec{x}) = (t + T, \mathbf{R} \vec{x} + \vec{v} t + \vec{a}) \end{equation}

each group element $g$ beging defined by the elements $(T, \mathbf{R}, \vec{v}, \vec{a})$, of a time translation $T \in \mathbb{R}$, a rotation $\mathbf{R} \in \mathrm{O}(n)$, a boost $\vec{v} \in \mathbb{R}^n$, and a spatial translation $\vec{a} \in \mathbb{R}^n$. There are in addition two discrete transformations, the spatial reversal, already included in $\mathrm{O}$, $\vec{x} \to \vec{x}$, and the time reversal $t \to -t$.

Composition

This can be shown by considering the various transformations :

Time translations : For the transformation $t \to t' = t + T$, the action transforms as

\begin{eqnarray} S &=& \int_{t_i}^{t_f} dt'\ \frac{m}{2} \dot{\vec{x}}(t') \cdot \dot{\vec{x}}(t')\\ &=& \int_{t_i}^{t_f} dt\ \frac{m}{2} \dot{\vec{x}}(t + T) \cdot \dot{\vec{x}}(t + T)\\ &=& \int_{t_i}^{t_f} dt\ \frac{m}{2} \dot{\vec{x}}(t + T) \cdot \dot{\vec{x}}(t + T) \end{eqnarray}

Spatial translations : For the transformation $\vec{x} \to \vec{x}' = \vec{x} + \vec{a}$

Bundle methods III : Hamiltonian

If we consider our configuration space $TE$, the equivalent of the Legendre transform in terms of bundle is given by [...]

Legendre bundle etc

Bundle methods for constraints

The method of constraints can be better understood in the more geometric sense of bundle theory. The basic issue of a system with constraints is that rather than having a symplectic structure as Lagrangian theories typically do, it has a presymplectic structure, ie if we consider the symplectic form $\omega$ generated by its Lagrangian, there may exists vectors $X$ in the tangent bundle of the phase space such that, for any vector $Y$,

\begin{equation} \forall Y \in \mathfrak{X}(M),\ \omega(X, Y) = 0 \end{equation}

As a manifold, our phase space is rather simple, as it is simply the cotangent bundle of our configuration space, $\mathbb{R} \times \mathbb{R}^{n + 1}$.

\begin{equation} TQ = T\mathbb{R} \oplus T \mathbb{R}^{n + 1} \end{equation}

The symplectic form on this

Co-adjoint orbit method

Given a Klein pair of two groups, $(H, G)$, we wish to define an action that corresponds to a free particle of the kinematic group $G$, with the homogeneous group $H$. The groups involved for classical mechanics are the Galilean group $\mathrm{Gal}(n)$, and the inhomogeneous Galilean group $\mathrm{IGal}(n)$. The Galilean group can be defined as a linear group on $\mathbb{R}^{n+1}$, acting as

\begin{eqnarray} \mathrm{IGal}(n) \times \mathbb{R}^{n+1} &\to& \mathbb{R}^{n+1}\\ \left( (s, \vec{a}, \vec{v}, \mathbf{R}), (t, \vec{x}) \right) &\mapsto& (t + s, \mathbf{R} \vec{x} + \vec{v} t + \vec{a}) \end{eqnarray}

with a time translation by $s$, a space translation by $\vec{a}$, a boost by $\vec{v}$, and a rotation by $\mathbf{R}$. The homogeneous Galilean group is simply its homogeneous action, where the translation $\vec{a}$ is set to zero. As a manifold, we have

\begin{eqnarray} \mathrm{IGal}(n) \cong \mathbb{R}^{2n + 1} \times \mathrm{SO}(3) \end{eqnarray}

while as a group,

\begin{eqnarray} \mathrm{IGal}(n) \cong \mathbb{R}^{n + 1} \rtimes_\rho \mathrm{SE}(3) \end{eqnarray}

The method of co-adjoint orbits is done by considering the principal bundle $\pi : \mathrm{IGal}(n) \to \mathrm{IGal}(n) / \mathrm{Gal}(n)$, which is the bundle of total space $\mathrm{IGal}(n)$, base space $\mathbb{R}^{n+1}$, with fiber $\mathrm{Gal}(n)$.

The adjoint representation of a group is given by the pullback of the conjugation representation $(g, h) \mapsto g h g^{-1}$ by the tangent map $T : G \to TG$, defined at the identity. For $X \in \mathfrak{g}$,

\begin{equation} \mathrm{Ad}_g(X) = (d \rho_g)_e(X) = (\rho_g \circ \exp(tX))' (0) = (g \exp(tX)g^{-1})'(0) = g X g^{-1} \end{equation}

The co-adjoint representation is the one given similarly by acting on the dual algebra $\mathfrak{g}^*$.

\begin{eqnarray} \mathrm{Ad}^* : G \to \mathrm{GL}(\mathfrak{g}^*) \end{eqnarray}

The dual algebra $\mathfrak{gal}^*(3)$ is given by the linear maps from $\mathfrak{gal}(3)$ to $\mathbb{R}$. Such dual algebra elements can be represented by a matrix whose action on a Lie algebra element is via the Killing form $B$,

\begin{equation} B : \mathfrak{g} \times \mathfrak{g} \to \mathbb{R} \end{equation}

So that we can simply represent such duals by the lifting $X \in \mathfrak{g} \to B(X, -) \in \mathfrak{g}^*$. Within our context of representing elements of $\mathfrak{g}$ by a column vector, we can represent dual elements by a row vector, where the dual basis of $\mathfrak{gal}^*$ is given by the transpose of the original basis, in which case

\begin{eqnarray} \alpha(X) &=& B(X_\alpha, X)\\ &=& \mathrm{Tr}(\mathrm{ad}_{X_\alpha} \mathrm{ad}_{X}) &=& \ldots &=& u_i v_i \mathrm{Tr}((T^i)^T T^j)\\ &=& u_i v_i \delta^{ij}\\ &=& u_i v^i \end{eqnarray}

So that an adjoint member of the algebra is represented by a row vector.

A co-adjoint action is simply the action of $G$ on $\mathfrak{g}^*$, as $g \alpha g^{-1}$. Like the adjoint case, if we represent our dual algebra element as an $\mathbb{R}^{10}$ vector, the corresponding co-adjoint action is then a $10 \times 10$ matrix.

The co-adjoint orbits are the orbits of the co-adjoint action in $\mathfrak{g}^*$, so that in our case, we can represent such orbits by subspaces of $\mathbb{R}^{10}$.

To construct the co-adjoint representation of the Galilean group, let's first consider the case of $\mathrm{SO}(3)$.

Given the standard representation of the Galilean group on $\mathbb{R}^{n+1}$, we can define the dual algebra with the space

\begin{equation} \mathfrak{gal}^*(3) = \mathbb{R}^4 \oplus \mathfrak{se}^*(3) \cong \mathbb{R}^4 \oplus ((\mathbb{R}^3)^* \oplus (\mathfrak{so}(3))^*) \cong \mathbb{R}^4 \oplus (\mathbb{R}^3* \oplus \mathbb{R}^3) \end{equation}

We denote an element of $\mathfrak{gal}(3)$ as $((E, \vec{p}), (\vec{q} \oplus \vec{F}))$

The orbits can be classified by three parameters $(k, r, s)$, $k > 0$, $r \geq 0$ and $s \geq 0$. With

The physical interest of the coadjoint method is that a curve on $G$ can be projected down to its homogeneous space $G / H$, and our action can therefore be defined as a curve on $G$ itself before being projected down. We therefore consider a curve $\gamma : L \to G$, the action corresponding to the coadjoint orbit $\alpha \in \mathfrak{g}^*$ is given by

\begin{equation} S = \int_I \langle \alpha, \gamma^{-1} \dot{\gamma} \rangle d\tau \end{equation}

(show that it's invariant under coadjoint action)

$\dot{g}$ is the pullback of the left-invariant Maurer-Cartan form

The coadjoint orbit of interest in our case will be the spin $0$ and mass $m > 0$, with representative $M(m_0, E_0, 0, 0, 0)$.

Non-linear representation : $\mathrm{SO}(3)$

\begin{equation} g = \end{equation}

Symplectic method

Most physical systems admit a description in terms of a symplectic manifold. The two main cases are done using either the Lagrangian or Hamiltonian. First let's see the general case of symplectic manifolds.

Symplectic manifolds

A symplectic manifold $(M, \omega)$ is a pair of a manifold $M$ (in our case the phase space $\mathbb{R}^{2n}$) along with a closed non-degenerate $2$-form $\omega$, ie a map $\omega : TM \times TM \to \mathbb{R}$ obeying, for any two vector fields $X, Y \in \mathfrak{X}(M)$

\begin{eqnarray} \omega(X, X) &=& 0\\ d\omega &=& 0 \end{eqnarray}

and

\begin{equation} \forall Y,\ \omega(X, Y) = 0 \leftrightarrow X = 0 \end{equation}

As we are simply using the space $\mathbb{R}^{2n}$, we can use the Darboux coordinates globally and write $\omega$ as

\begin{equation} \omega = \begin{pmatrix} 0 & I_n\\ -I_n & 0 \end{pmatrix} \end{equation}

and since $\mathbb{R}^{2n}$ has trivial cohomology, furthermore we can define $\omega$ in terms of a symplectic potential $\theta$, as it is a closed form, $\omega = d \theta$, where using the coordinates $(q, p)$,

\begin{equation} \theta = \langle p, dq \rangle \end{equation}

Lagrangian symplectic manifolds

If we consider our Lagrangian theory, with our Lagrangian

\begin{equation} L : TQ \to \bigwedge^n Q \end{equation}

We can define a symplectic structure on $TQ$ by using the almost-tangent structure. An almost-tangent structure is a $G$-structure on the manifold for matrices of the form

\begin{equation} g = \begin{pmatrix} A & 0 \\ B & A \end{pmatrix} \end{equation}

with $A, B$ $n \times n$ matrices with $A$ non-singular. This structure manifests itself the following way : if a $2n$-dimensional manifold $M$ admits an almost-tangent structure, then its tangent bundle $TM$ admits an endomorphism $J$ such that $J^2 = $ and $J$ has rank $n$.

\begin{equation} J = \begin{pmatrix} 0 & 0 \\ I_n & 0 \end{pmatrix} \end{equation}

In our case, our underlying manifold is $TE$, and its tangent bundle is the double tangent bundle $T^2 E$, where this endomorphism is defined by

\begin{eqnarray} J = \iota \circ d\pi \end{eqnarray}

with $\iota$ the canonical vertical lift, where given a point $u \in TE$, $\pi(u) = p$,

\begin{eqnarray} \iota_u : T_{x} E &\to& V_u(TE) = \mathrm{ker}(d_u \pi)\\ v &\mapsto& \frac{d}{dt} (u + t v) \end{eqnarray}

and $d \pi$ is the tangent map of the tangent projection $\pi : TE \to M$. We therefore have

\begin{equation} \mathrm{Im}(J) = V(TN) = \mathbb{ker}(d\pi) \begin{equation}

making it rank $n$, and as its image is in the kernel, $J^2 = \iota \circ d\pi (J) = 0$.

Bundle coordinates :

\begin{equation} J = dx^i \otimes \partial_{v^i} \end{equation}

Any manifold with an almost tangent structure admits a symplectic structure,

\begin{equation} \omega = - d (dL \cdot J) \end{equation}

Hamiltonian symplectic manifold

The Hamiltonian formalism here also can lead to a symplectic structure. If we consider our phase space (here the cotangent bundle), we would like to send our various physical quantities to it. A way to do this is by using the momentum $1$-form,

\begin{equation} p = d L = \frac{\partial L}{\partial q^i} dx^i \end{equation} \begin{equation} \pi : T^* M \to M \end{equation} \begin{equation} d\pi : TT^* M \to TM \end{equation}

a point $m \in T^* M$

With the Hamiltonian that we computed, we can work out the case in symplectic geometry. Picking the symplectic vector space $(V, \omega)$ for $\mathrm{dim}(V) = 2n$ and $\omega$ the canonical symplectic form, we define the two subspaces $V = Q \oplus P$ of the underlying space and momentum space, with coordinates $(q, p) \in \mathbb{R}^n \times \mathbb{R}^n$. In those coordinates,

\begin{equation} \omega = \begin{pmatrix} 0 & -I\\ I & 0 \end{pmatrix} \end{equation}

The Hamiltonian vector $X_H$ is defined as the unique vector field on $V$ obeying

\begin{equation} dH(Y) = \omega (X_H, Y) \end{equation}

which is, in our coordinates,

\begin{eqnarray} \frac{\partial H}{\partial \vec{x}} \vec{Y}_x &=& - \vec{X}_{H, x} \vec{Y}_{x} \\ \frac{\partial H}{\partial \vec{p}} \vec{Y}_p &=& \vec{X}_{H, p} \vec{Y}_{p} \end{eqnarray}

The Hamiltonian vector field is then given by

\begin{eqnarray} X_{H} &=& \omega dH\\ &=& (\frac{\partial H}{\partial p_i}, - \frac{\partial H}{\partial q_i})\\ &=& (\vec{p}, 0)\\ \end{eqnarray}

The flow of the position is then given by $\vec{p}$, while there is no flow for the momentum, which remains fixed.

Symplectic symmetries

Dynamic symmetry

Poisson manifolds

As with any symplectic structure, this also generates a Poisson structure, a bivector $\Pi \in \bigwedge^2 V$ with the Schouten–Nijenhuis bracket

\begin{equation} [\Pi, \Pi] = 0 \end{equation} \begin{equation} \{ f, g \} = \Pi(df \wedge dg) \end{equation}

Lagrangian correspondence

In the Lagrangian correspondence method, as described here, we

Action $S : [\mathbb{R}, \mathbb{R}^n] \to \mathbb{R}$,

\begin{equation} S = \int_{t_i}^{t_f} dt\ \frac{m}{2} \dot{\vec{x}} \cdot \dot{\vec{x}} \end{equation}

Inverse Legendre transform

\begin{eqnarray} \phi^{-1}_L : T^*M &\to& TM\\ (p,q) &\mapsto& (v(p, q), q) \end{eqnarray} \begin{equation} L_{[t_i, t_f]} = (\phi_L^{-1} \times \phi_L^{-1}) (\mathrm{graph}(\Phi_{H_L}|^{t_f}_{t_i})) \end{equation} \begin{equation} L_{[t_1, t_2]} \circ L_{[t_0, t_1]} = L_{[t_0, t_2]},\ \lim_{t_1 \to t_0} L_{[t_0, t_1]} = \mathrm{graph}(\mathrm{Id}) \end{equation} \begin{eqnarray} \pi_{[t_i, t_f]} : M^{[t_i, t_f]} &\to& TM \times TM\\ x(t) &\mapsto& ((x(t_i), \dot{x}(t_i)), (x(t_f), \dot{x}(t_f))) \end{eqnarray}

$\mathrm{EL}_{[t_i, t_f]} \subset M^{[t_i, t_f]}$ the space of solutions to the EL equations.

\begin{equation} L_{[t_i, t_f]} = \pi_{[t_i, t_f]} (\mathrm{EL}_{[t_i, t_f]}) \end{equation}

Galilean manifold

A Galilean manifold is a manifold $M$ (representing here the classical spacetime) along with a Galilean structure, which is given by the appropriate reduction of the Klein pair we've seen, $(\mathrm{IGal}(n), \mathrm{Gal}(n))$, which is composed of the clock one-form $\omega$ and the spatial metric $\delta$, a rank $(2,0)$ tensor, such that the kernel of of $\delta$ is spanned by $\omega$ :

\begin{equation} \forall \alpha \in \mathfrak{\Omega}^1M,\ h(\omega, \alpha) = 0 \end{equation}

It is equipped with a Koszul connection $\nabla$ which is parallel to the clock and metric, in that

\begin{equation} \nabla \omega = 0,\ \nabla \delta = 0 \end{equation}

In terms of what we have seen so far, this is the bundle space $\mathbb{R} \times E$, where $\delta$ is the inner product metric on Euclidian space, and the clock $\omega$ is the structure giving us the splitting structure that makes every slice $\mathbb{R}^3$ simultaneous. On $\mathbb{R} \times E$, the kernel $\mathrm{ker}(\omega)$ is ...

The Newtonian case has a few additional conditions which make it the "classical spacetime" :

The case of a point particle is given by a timelike curve with the Dirac action (expressed here in terms of the structures on the Galilean space)

\begin{equation} S = \int_I \frac{m}{2} \frac{\delta(\dot{X}, \dot{X})}{\omega(\dot{X})} d\tau \end{equation}

or, equivalently using our action from the coadjoint orbit method,

...

The equation of motion is given by the usual solution for the Dirac constrained Lagrangian.

The BV-BRST method

The Dirac lagrangian can be used to illustrate the BV-BRST method. Consider the action that we've seen for it.

\begin{equation} S = \int_{t_i}^{t_f} d\tau\ \frac{m}{2} \frac{\dot{\vec{x}}(\tau) \cdot \dot{\vec{x}}(\tau)}{\dot{t}(\tau)} \end{equation}

As an action functional, this is the map from the parameter $\tau \in L$ into the configuration space $(t, \vec{x}) \in \mathbb{R}^{n+1}$. As a fiber bundle, this is the bundle $\pi : L \times \mathbb{R}^{n + 1} \to L$, and our configuration space is the space of sections $\Gamma(L \times \mathbb{R}^{n + 1})$,

\begin{eqnarray} \sigma : L &\to& L \times \mathbb{R}^{n + 1}\\ \tau &\mapsto& \sigma(\tau, t, \vec{x}) \end{eqnarray}

We are dealing with some element of the mapping space $[L, \mathbb{R}^4]$. There is a split of this configuration into a timelike and spacelike part, defined by two projection maps $\mathrm{pr}_t$ and $\mathrm{pr}_s$ (this is the projectors defined by the clock $1$-form on a Galilean space). Its time derivative will be the first jet of that configuration, with the product defined on the jet bundle by the spatial metric $\delta$.

\begin{equation} S[X] = \int_I d\tau\ \frac{m}{2} \frac{\langle \mathrm{pr}_s(j X), \mathrm{pr}_s(j X) \rangle}{\mathrm{pr}_t j X)} \end{equation}

From its form we know easily enough that it is a local action functional. Its generalized Lagrangian is

\begin{equation} S[X]_f = \int_I d\mathrm{vol}_L\ \frac{m}{2} \frac{\dot{\vec{x}}(\tau) \cdot \dot{\vec{x}}(\tau)}{\dot{t}(\tau)} \end{equation}

The evolutionary vectors we want are given by the differential map of the configuration

\begin{equation} \delta X = (\delta t, \delta x) \end{equation}

Those are our fields and antifields.

Master equation : for any evolutionary vector field $V \in T\mathrm{Conf}$,

\begin{equation} \langle \delta S[X], V \rangle = 0 \end{equation}

To solve this, we need to find first the kernel of the $1$-form $dS$. This is the classic Noether equation

\begin{eqnarray} \frac{\delta S}{\delta X} &=& \end{eqnarray}

As expected, the symmetry group is the group of diffeomorphism on the line $\mathrm{Diff}(L)$, which is just $\mathrm{Aut}(L)$, locally represented by its Lie algebra of vector fields on the line, $\mathrm{Lie}(\mathrm{Diff}(L)) = \mathfrak{X}(L)$. It acts on the configuration space $X \in \mathrm{Conf}$ as

\begin{eqnarray} \forall g \in \mathrm{Diff}(L),\ g^* X &=& \end{eqnarray}

To find the set of gauge-invariant observables, we need to perform the Koszul-Tate resolution of this action.

\begin{equation} 0 \to \mathrm{Ker}(dS) \hookrightarrow \mathfrak{X}(L) \overset{dS(-)}{\longrightarrow} [L, \mathbb{R}^n] \overset{\rho}{ \longrightarrow} [L, \mathbb{R}^n] \otimes \mathfrak{g}^* \to 0 \end{equation}

Derived intersection?

BRST complex

The Chevalley-Eilenberg algebra of the gauge group is the one given by the tangent Lie algebroid of the line, $\mathrm{CE}(TL)$.

\begin{equation} \mathrm{CE}(TL) = (\bigwedge_{C^\infty}(L) \Gamma(T^*L), d_{\mathrm{dR}}) \end{equation} \begin{equation} \mathrm{CE}(TL) = \mathbb{R} \overset{d_{\mathrm{dR}}}{\longrightarrow} \mathbb{R} \to 0 \to 0 \to \ldots \end{equation}

In terms of sections :

\begin{equation} C^\infty(L) \overset{d_{\mathrm{dR}}}{\longrightarrow} C^\infty(L) \times \bigwedge^1 L \to 0 \to 0 \to \ldots \end{equation}

The Lagrangian gerbe

For a more formally correct theory, we need to use a gerbe. Rather than use the action $S$ as an integral (ie a map from the homology on the space and the cohomology of $n$-forms to the real numbers), we instead take a collection of open sets (for instance a cover by compact open sets) of the underlying space

\begin{equation} \coprod_i U_i = M \end{equation}

Let's consider the Cech cohomology of this cover. The $n$-form intersections of those open sets can be describe by the $n$-fold fiber product

\begin{equation} U_{ijk\ldots} = \coprod_{i,j,k} \end{equation}

The cohomological Lagrangian

As we hinted at earlier, some modifications must be made to deal properly with the boundary conditions of Lagrangians to deal with them.

The cohomological perspective on Lagrangian mechanics (as described here) is given by the variational bicomplex associated with it.

Group cohomology

An important aspect of the symmetries of a space in mechanics will be given by the cohomology of the group. This will give us quite a lot of important properties later on, and in particular goes towards a lot to explain differences between relativistic and non-relativistic dynamics. So let's take a moment to compute the cohomology of the Galilean group.

Categorical method

cf. Lawvere

In the differential cohesive Grothendieck $\infty$-topos $\mathbf{H} = \mathbf{Smooth}$ of sheaves over the site of Cartesian spaces

\begin{equation} \mathbf{CartSp}_{\mathrm{Smooth}} \end{equation}

of open sets of Cartesian spaces $\mathbb{R}^n$ with the subcanonical coverage.

The relevant entities we need for classical mechanics are all located within this topos, which is the real line $L \cong \mathbb{R}$, meant to represent the time and any observable quantity, physical space and spacetime $\mathbb{R}^3$, $\mathbb{R}^4$, the space of all trajectories $\mathrm{Conf} = [L, \mathbb{R}^3]$, in the internal hom, and the moduli space of symplectic manifold, $\Omega$, for which any morphism into it represents a given symplectic structure over said manifold. For instance, the standard symplectic manifold on our phase space $P$ is some manifold

\begin{eqnarray} P_\omega : P \to \Omega \end{eqnarray}

or alternatively, represented by an element of the internal hom

\begin{eqnarray} P_\omega : 1 \to [P, \Omega] \end{eqnarray}

Special relativity

The special relativistic version of free particles can be done in any number of formalisms, similarly to the non-relativistic case. The simplest one being simply the relativistic version of Newton's second law

\begin{equation} f^\mu = mc^2 \ddot{x}^\mu(\tau) \end{equation}

which is here simply $\ddot{x}^\mu(\tau) = 0$. Similarly as before, this has the solution

\begin{equation} x^\mu(\tau) = v^\mu_0 \tau + x_0^\mu \end{equation}

Our particle is a standard slower than light particle if our initial velocity $v$ is timelike : $g(v,v) < 0$, in which case the velocity itself will be timelike. Similarly, it will be a null particle if $g(v,v) = 0$, and a tachyonic particle if $g(v,v) > 0$.

The action associated with a timelike point particle is the Nambu-Goto action, which is proportional to the length functional :

\begin{equation} S[X; [\tau_1, \tau_2]] = -m \int_{\tau_1}^{\tau_2} ds = -m \int_{\tau_1}^{\tau_2} d\tau \sqrt{\dot{x}^\mu(\tau) \dot{x}_\mu(\tau)} \end{equation}

with the usual variation

\begin{eqnarray} \delta S &=& \int_{\tau_1}^{\tau_2} d\tau \left[ \frac{\partial L}{\partial x^\mu} \delta x^\mu + \frac{\partial L}{\partial \dot{x}^\mu} \delta \dot{x}^\mu \right]\\ &=& \int_{\tau_1}^{\tau_2} d\tau \frac{\partial L}{\partial \dot{x}^\mu} \delta \dot{x}^\mu \end{eqnarray} \begin{eqnarray} \frac{\partial L}{\partial \dot{x}^\mu} &=& -m \frac{\partial \sqrt{l}}{\partial l} \frac{\partial l}{\partial \dot{x}^\mu}\\ &=& -m \frac{1}{2 \sqrt{\dot{x}^\mu(\tau) \dot{x}_\mu(\tau)}} \dot{x}_\mu(\tau) \end{eqnarray}

Equation of motion :

\begin{eqnarray} -\frac{m}{2} \frac{d}{d\tau} \left[\frac{1}{\sqrt{\dot{x}^\mu(\tau) \dot{x}_\mu(\tau)}} \dot{x}_\mu(\tau) \right] &=& 0\\ &=& -\frac{m}{2} \left[ \frac{d}{d\tau} (\frac{1}{\sqrt{\dot{x}^\mu(\tau) \dot{x}_\mu(\tau)}}) \dot{x}_\mu(\tau) + \frac{1}{\sqrt{\dot{x}^\mu(\tau) \dot{x}_\mu(\tau)}} \ddot{x}_\mu(\tau) \right]\\ &=& -\frac{m}{2} \left[ - \frac{1}{(\dot{x}^\mu(\tau) \dot{x}_\mu(\tau))^{3/2}} (\ddot{x}^\mu(\tau) \dot{x}_\mu(\tau)) \dot{x}_\mu(\tau) + \frac{1}{\sqrt{\dot{x}^\mu(\tau) \dot{x}_\mu(\tau)}} \ddot{x}_\mu(\tau) \right] \end{eqnarray}

Moving along a few terms, this gives us

\begin{eqnarray} \left[ - \frac{\ddot{x}^\mu(\tau) \dot{x}_\mu(\tau)}{(\dot{x}^\mu(\tau) \dot{x}_\mu(\tau))} \dot{x}_\mu(\tau) + \ddot{x}_\mu(\tau) \right] = 0 \end{eqnarray}

This is not quite what we expected, compared to the usual form of a free particle. This is because we forgot to take into account the parametrization invariance. Let's consider for instance the curve $x^\mu(\tau) = (\tau, 0, 0, 0)$. This is obviously a solution to both equations. By parametrization invariance, then, so should it be under the new parametrization $\tau' = \tau^3 + \tau$, with the curve $x^\mu(\tau) = (\tau^3 + \tau, 0, 0, 0)$. But here, our original equations of motion become

\begin{equation} \ddot{x}^\mu(\tau) = (6 \tau, 0, 0, 0) \end{equation}

On the other hand, we do have

\begin{eqnarray} - \frac{\ddot{x}^\mu(\tau) \dot{x}_\mu(\tau)}{(\dot{x}^\mu(\tau) \dot{x}_\mu(\tau))} \dot{x}_\mu(\tau) &=& - \frac{6 \tau \times 3 \tau^2}{3 \tau^2 * 3 \tau^2} \dot{x}_\mu(\tau)\\ &=& - 2 \tau^{-1} \dot{x}_\mu(\tau)\\ &=& (- 6 \tau, 0, 0, 0) \end{eqnarray}

which does compensate this term. To obtain a gauge-fixed equation, we can simply fix $x^0 = $

Parametrization invariance

\begin{eqnarray} \frac{\partial^2 L}{\partial \dot{x}^\mu \dot{x}^\nu} &=& -\frac{m}{2} \frac{\partial}{\partial \dot{x}^\nu} \frac{1}{ \sqrt{\dot{x}^\sigma(\tau) \dot{x}_\sigma(\tau)}} \dot{x}_\mu(\tau)\\ &=& \end{eqnarray}

Fixed gauge action

Once the gauge has been sorted out, we can recast our Nambu-Goto action in a few different ways.

Polyakov action

Another action we can use for the relativistic point particle is the Polyakov action, which contains the gauge term $e$ due to reparametrization invariance.

\begin{equation} S[X, e] = \int_\Sigma (\frac{1}{e(\tau)} \dot{X}^\mu(\tau) \dot{X}_\mu\nu(\tau) - e(\tau) m^2) d\tau \end{equation} \begin{eqnarray} \frac{\partial L}{\partial X^\mu} &=& 0\\ \frac{\partial L}{\partial \dot{X}^\mu} &=& \frac{2}{e(\tau)}\dot{X}^\mu(\tau)\\ \frac{\partial L}{\partial e} &=& -\frac{1}{e^2(\tau)} \dot{X}^\mu(\tau) \dot{X}_\mu(\tau) - m^2 \\ \frac{\partial L}{\partial \dot{e}} &=& 0 \end{eqnarray}

Symmetry group

BV-BRST of the relativistic particle

Just like the case of the Dirac parametrized system, we can solve the point particle relativistic Lagrangian via BV-BRST methods. Once again the mapping space is simply $[L, \mathbb{R}^{n+1}]$, but with no canonical splitting between time and space, and the inner product of the metric tensor $\eta$ :

General relativity

There are a few actions we can use in general relativity, as well as the choice of whether or not we consider a test particle or if our particle reacts to our spacetime. Obviously it is well known that the motion of a point particle in a spacetime is a geodesic, obeying the geodesic equation

\begin{equation} \ddot{x}^\mu(\tau) + {\Gamma^\mu}_{\alpha\beta}(x(\tau)) \dot{x}^\alpha(\tau) \dot{x}^\beta(\tau) = 0 \end{equation}

but we'll see how to derive these.

Nambu-Goto action

The Nambu-Goto action is given by the embedding $X$ of a curve $\gamma$ in our (fixed) spacetime $(M, g)$, such that we minimize the arc length of our curve.

\begin{eqnarray} S[[\tau_1, \tau_2]; X] &=& m \int_{\tau_1}^{\tau_2} d\mu[X_* g]\\ &=& m \int_{\tau_1}^{\tau_2} d\tau \sqrt{g_{\mu\nu}(X) \dot{X}^\mu(\tau) \dot{X}^\nu(\tau)} \end{eqnarray}

This action's Hessian is degenerate, as can be seen

\begin{equation} \frac{\delta S}{\delta \dot{X} \delta X} \end{equation}

Polyakov action

The Polyakov action generalizes easily enough to curved space,

\begin{equation} S[X, e; U] = \int_U \end{equation}

Backreaction

In a spacetime with otherwise no other sources of matter (and disregarding other potential sources of curvature like gravitational waves), the resulting spacetime is the Schwarzschild metric. This is not trivial to prove rigorously, as the stress-energy tensor is here a distribution and has its singular support on a $1$-dimensional submanifold, meaning that the Geroch-Traschen condition doesn't apply : the metric cannot be expressed by a tensor distribution. We'll have to do with the Colombeau algebra method for this.

First let's consider a point particle's stress energy tensor. If we express it as a distribution, it will be

\begin{equation} T^{\mu\nu} = \mu \int d^\tau \dot{X}^\mu \dot{X}^\nu \delta^{(n)}(x - X(\tau)) \end{equation}

There's a variety of ways we can consider what this expression means exactly,

Real objects

As usual, we may not have to consider the case of perfectly point-like objects, which is not particularly realistic. A real "free" object will not actually move perfectly freely, but will have its backreaction effects by the small local gravitational perturbation it does.

The usual framework for such cases is given by the MiSaTaQuWa equation

The geodesic spray

The general relativistic cases can be solved via the method of sprays with what is called the geodesic spray, which is the spray whose flow generates geodesics.

Non-relativistic quantum mechanics (canonical quantization)

The canonical quantization of non-relativistic free particles is fairly straightforward (ignoring some subtelties like Groenewold-Van Hove's theorem), as we have the basic Poisson brackets

\begin{eqnarray} \left\{ x_i, p_j \right\} &=& \delta_{ij}\\ \left\{ x_i, x_j \right\} &=& 0 \\ \left\{ p_i, p_j \right\} &=& 0 \end{eqnarray}

The canonical quantization map $Q$ simply associates those to operators

\begin{equation} \forall A, B \in C^{\infty}(\mathrm{Phase}),\ \exists \hat{A}, \hat{B} \in \mathcal{L}(\mathcal{H}),\ \{ A, B \} \mapsto \frac{1}{i \hbar} \left[ \hat{A}, \hat{B} \right] \end{equation}

then gives us

\begin{eqnarray} \left[ \hat{x}_i, \hat{p}_j \right] &=& i \hbar \delta_{ij} \hat{I}\\ \left[ \hat{x}_i, \hat{x}_j \right] &=& 0 \\ \left[ \hat{p}_i, \hat{p}_j \right] &=& 0 \end{eqnarray}

By the Stone-von Neumann theorem, for finite dimensions, all sets of operators satisfying those conditions are unitarily equivalent, and in particular for the Hilbert space of square integrable functions on $\mathbb{R}^n$, $\mathcal{H} = L^2(\mathbb{R}^n, d\mu)$, with $\psi(\vec{x}) \in L^2(\mathbb{R}^n, d\mu)$, we can use

\begin{eqnarray} \hat{\vec{x}} \psi(\vec{x}) &=& \vec{x} \psi(\vec{x})\\ \hat{\vec{p}}_j \psi(\vec{x}) &=& -i\hbar \vec{\nabla} \psi(\vec{x}) \end{eqnarray}

Then we can convert our Hamiltonian according to this quantization :

\begin{eqnarray} \hat{H} &=& \frac{\hat{\vec{p}} \cdot \hat{\vec{p}}}{2m}\\ &=& -\frac{\hbar^2}{2m} \Delta \end{eqnarray}

From Schrödinger's equation, we get therefore

\begin{equation} -\frac{\hbar^2}{2m} \Delta \psi(t, \vec{x}) = i \hbar \frac{\partial}{\partial t} \psi(t, \vec{x}) \end{equation}

which is the typical form of the Schrödinger equation for a free particle. This is a variant of the heat equation with complex arguments. As we're in the space of square integrable functions over $\mathbb{R}^n$, we can use the Fourier transform to obtain

\begin{equation} -\frac{\hbar^2}{2m} \| \vec{k} \|^2\tilde{\psi}(t, \vec{k}) = i \hbar \frac{\partial}{\partial t} \tilde{\psi}(t, \vec{k}) \end{equation}

or, slightly rearranged

\begin{equation} \frac{\partial}{\partial t} \tilde{\psi}(t, \vec{k}) = i\frac{\hbar}{2m} \| \vec{k} \|^2\tilde{\psi}(t, \vec{k}) \end{equation}

which is the defining equation of an exponential function, with the solution

\begin{equation} \tilde{\psi}(t, \vec{k}) = C(\vec{k}) e^{-i\frac{\hbar}{2m} \| \vec{k} \|^2} \end{equation}

with $C(\vec{k})$ an integration constant.

Another simple enough way to get solutions for this equation is to consider the generalized eigenvectors associated with our operators. In our case, we'll take the space on which $\hat{p}$ is defined, which is the dense subset of $C^1$ functions of $L^2(\mathbb{R}^n, d\mu)$ where the derivative is itself an $L^2$ function, which is the $H^1(\mathbb{R}^n) = W^{1,2}(\mathbb{R}^n)$ Sobolev space.

\begin{equation} \psi(t, \vec{x}) = \int f(p) e^ipx \end{equation}

For more specific cases, let's consider the closest states to a classical free particle, which is a coherent state. A coherent state is a state of minimal uncertainty, that is,

\begin{equation} \Delta x \cdot \Delta p = \frac{\hbar}{2} \end{equation}

Such that $\Delta x \approx \Delta p$, up to some dimensional factor. This makes it a wavepacket with the least spread in position and momentum space.

Symmetries

Just as in the classical case, we can investigate the set of symmetries on our system and their action on it. As in general for the quantization of a system, we are looking for the projective representation of the classical group, in our case the Galilean group

Non-relativistic quantum mechanics (geometric quantization)

To have the geometric quantization of the free particle, let's first find the prequantization of its phase space. Given our phase space $\mathbb{R}^{2n}$, the prequantization bundle associated to it is a circle bundle (with typical fiber $\mathrm{U}(1)$) over it. As $\mathbb{R}^{2n}$ is contractible, there is only one such bundle which is the trivial one,

\begin{eqnarray} \pi : \mathrm{U}(1) \times P &\to& P\\ (e^{i\theta}, \vec{x}, \vec{p}) &\mapsto& (\vec{x}, \vec{p}) \end{eqnarray}

This circle bundle is required to be equipped with a connection such that its curvature is equal to the symplectic form,

\begin{equation} F_{\nabla} = \omega \end{equation}

Non-relativistic quantum mechanics (path integral quantization)

For the path integral quantization, we just consider the Wick-rotated Feynman integral

\begin{equation} W(\vec{x}_i, \vec{x}_f) = \int_{x_i}^{x_f} d\mu[x(t)] \end{equation}

with $d\mu[x(t)]$ the Gaussian measure. We take $\mathbb{X}$ to be the space of continuous functions from the time interval $I = [t_i, t_f]$ to our physical space $\mathbb{R}^n$, $C(I, \mathbb{R}^n)$, with $\mathbb{X}'$ its dual as the space of bounded measures on $I$ :

\begin{equation} \forall x \in \mathbb{X}, x' \in \mathbb{X}',\ \langle x', x \rangle = \int_{I} dx'(t) x(t) \end{equation}

Cylinder sets : For each sequence $(t_j)_{j}$ of distinct moments on $[t_i, t_f]$, with $t_0 = t_i$ and $t_N = t_f$

Non-relativistic quantum mechanics (deformation quantization)

To deform the classical phase space formulation of our system into a quantum system, let's consider the algebra defined by the Moyal product

\begin{equation} f \star g = fg + \sum_{i = 1}^\infty \hbar^n C_n(f,g) \end{equation}

with $f,g$ sections of the Poisson manifold $(M, \Pi)$, and $f \star g$ a section of the deformed Poisson manifold $M\left[\!\left[\hbar\right]\!\right]$.

Groenewold-Moyal star product :

\begin{equation} f \star g = \exp (\frac{i\hbar}{2} \Pi) (f,g) \end{equation} \begin{equation} f \star g = fg + \hbar \sum_{i,j} \Pi^{ij} \langle \nabla_{i} f, \nabla_{j} g \rangle + \hbar^2 \ldots \end{equation}

Non-relativistic quantum mechanics (BRST quantization)

By using our parametrized version of our free particle, we can use the BRST quantization to hopefully find the same quantum system as before.

Bohmian mechanics

While quite possibly not correct in some aspects, Bohmian mechanics is entirely fit to describe the motion of point particles. There are many variants of it, but let's consider the basic one.

The system is composed of two parts : the particle itself, $\vec{x}$, and the guiding wave $\psi$. The guiding wave obeys the Schrödinger equation,

\begin{equation} i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \Delta \psi \end{equation}

while the particle obeys the guidance equation,

\begin{equation} \dot{\vec{x}} = \frac{\hbar}{m} \Im(\frac{\vec{\nabla}\psi}{\psi}) \end{equation}

Relativistic quantum mechanics

Relativistic quantum mechanics is famed for working out poorly, but this is mainly due to the non-rigorous way in which it is usually quantized. Fortunately, using constraint quantization it is possible to work it out, at least in the free case.

For the Nambu-Goto action, the associated Poisson algebra is given by the CCR

\begin{equation} \{ X^\mu, P^\nu \} = \delta \end{equation}

Fourier methods

One method we can use to derive the extremal action of an action is to consider our configuration space as the map $[t_i, t_f] \to \mathbb{R}^n$. Interpreting it as the mapping space $[S^1, \mathbb{R}^n]$, where we considered a preferred point $\theta_0 \in S^1$ and the discontinuous kind of functions such that $\vec{x}(\theta_0^+) = \vec{x}_i$, $\vec{x}(\theta_0^-) = \vec{x}_f$, this space is spanned by the Fourier series

\begin{equation} \vec{x}(\theta) = \sum_{n \in \mathbb{Z}} \vec{c}_n e^{i\frac{2\pi n \theta}{t_f - t_i}} \end{equation}

with time derivative

\begin{equation} \dot{\vec{x}}(\theta) = \sum_{n \in \mathbb{Z}} \frac{2\pi n}{t_f - t_i} \vec{c}_n e^{i\frac{2\pi n \theta}{t_f - t_i}} \end{equation}

The corresponding action

\begin{eqnarray} S[x] &=& \int_{t_i}^{t_f} \frac{m}{2} (\sum_{j,k \in \mathbb{Z}} \frac{4\pi^2 jk}{(t_f - t_i)^2} \vec{c}_j \cdot \vec{c}_k e^{i\frac{2\pi (j + k) \theta}{t_f - t_i}}) d\theta \end{eqnarray}

Prove the commutation of the integrals (Fubini's theorem)

\begin{eqnarray} S[x] &=& \frac{m}{2} \sum_{j,k \in \mathbb{Z}} \frac{4\pi^2 jk}{(t_f - t_i)^2} \vec{c}_j \cdot \vec{c}_k (\int_{t_i}^{t_f} e^{i\frac{2\pi (j + k) \theta}{t_f - t_i}} d\theta) \end{eqnarray}

Synthetic mechanics

Using the axiomatic system of Hartry Field's Science without Numbers, we can deduce the time evolution of a system of free particles.

First we need to set an origin and scale, given by a point $o$ in Euclidian space and three segments $oe_1$, $oe_2$ and $oe_3$, such that they form an orthonormal basis, ie :

\begin{equation} \vdash \forall i, j,\ \mathrm{Cong}(oe_i, oe_j) \wedge \mathrm{Cong}(\widehat{e_i o e_j}, \widehat{e_i o e_k}) \end{equation}

ie they are each of identical length and of identical angles between them. For those angles to be perpendicular, we need to add the following conditions : For each $e_i$, we define the line $X_i$ by the unique line which intersects $o$ and $e_i$. We then define another point $\overline{e}_i$ (the opposite vector) by being a point on the other side of $o$ (ie we have $\mathrm{Btw}(\overline{e}_i, o, e_i)$) such that those two segments are congruents, $\mathrm{Cong}(o\overline{e}_i, oe_i)$, as such a point can be given by axiom III.1 of Hilbert's axioms. The perpendicularity of any two such segments is then simply expressed by

\begin{equation} \vdash \forall i, j,\ \mathrm{Cong}(\widehat{e_i o e_j}, \widehat{e_i o \overline{e}_j}) \end{equation}

every vector's angle with another is equal to the angle with its opposite, the definition of a right angle.

We also define an initial instant $t_0$, as well as a time scale $\Delta_t$ given by $t_0$ and some other time $t_1$.

For any line $AB$, we define a new line obtained by some $k$-fold extension of the line by the distance $AB$ in the direction of $B$ (using a ruler and compass for instance) as

\begin{equation} A + kB \end{equation}

We will say that for such a line $AB$ extended in that fashion, if the point $X$ is incident to the extension of $AB$, we will say that if $X$ is between

\begin{equation} \vdash \mathrm{Betw}(X, A + kB, A + (k+1)B) \end{equation}

that $X$ is in the interval $(k, k+1)$, or just the interval $k$ for short.

For points in space, we will construct 6 different lines. Given three integers $k_1, k_2, k_3 \in \mathbb{N}$

\begin{equation} o + k_1 e_1, o + k_2 e_2, o + k_3 e_3, o + (k_1 + 1) e_1, o + (k_2 + 1) e_2, o + (k_3 + 1) e_3 \end{equation}

Given a point $X$, we demand that those three numbers are such that for each $o + k_i e_i, o + (k_i + 1) e_i$, given the two orthogonal plane to the two endpoints $S_i, S_i'$, the orthogonal projection of $X$ on those two planes, $X_i$, $X_i'$ are such that

\begin{equation} \vdash \mathrm{Betw}(X, X_i, X_i') \end{equation}

We say that $X$ is at coordinate interval $(k_i, k_i + 1)$ (or just $k_i$) for that $i$. If we have the point $X$ at coordinate interval $k_1$, $k_2$ and $k_3$, we can define 8 points from them. We can define the points at coordinate $(k_1,k_2,k_3)$ by the parallel transport of $o + k_1 e_1$ along $o + k_2 e_2$, giving a new line $\ell$, and then parallel transport $o + k_3 e_3$ along $o + k_2 e_2$ and then along $\ell'$, forming the point at coordinate $(k_1,k_2,k_3)$ at its end.

ex ey ez O p--- p--+ p-+- p-++ p+-- p+-+ p++- p+++ X

The $8$ points defined are the points

\begin{eqnarray} p_{---} &=& (k_1, k_2, k_3)\\ p_{+--} &=& ((k_1 + 1), k_2, k_3)\\ p_{-+-} &=& (k_1, (k_2 + 1), k_3)\\ p_{--+} &=& (k_1, k_2, (k_3 + 1))\\ p_{++-} &=& ((k_1 + 1), (k_2 + 1), k_3)\\ p_{-++} &=& (k_1, (k_2 + 1), (k_3 + 1))\\ p_{+-+} &=& ((k_1 + 1), k_2, (k_3 + 1))\\ p_{+++} &=& ((k_1 + 1), (k_2 + 1), (k_3 + 1)) \end{eqnarray}

Those $8$ points form a parallepiped $P$ (in fact a cube), that we will say is the cube at $(k_1, k_2, k_3)$, which contains the point $X$.

Given this system, we define our particle's original position and time as follow : the time $t_i$ is in the interval $(t_0 + k \Delta_t, t_0 + (k+1) \Delta_t)$

while the initial position $x_i$ is given by the cube $P_i$ containing it.

The initial velocity is given by considering some small interval of time $\delta_t$ and observing the position of the particle at $t_i + \delta t$, denoted by $x_{i}'$. $x_i'$ is contained within another cube $P'$, and $t_i + \delta_t$ is within some interval $(...)$.

The question we now wish to solve is within what cube $P_f$ will the particle be contained after an interval of time placing it within the time interval $(..)$

Jet methods

The most general action one can write for a particle moving in a spacetime (of unknown kinematic), assuming no terms beyond the second order, is some acceleration field $A$ which is a section from the first to the second jet bundle of the bundle for curves, $\pi : \mathbb{R} \times \mathbb{R}^n \to \mathbb{R}$

\begin{eqnarray} A : J^1 \pi &\to& J^2\pi\\ (t, x, \dot{x}) &\mapsto& (t, x, \dot{x}, A(t, x, \dot{x})) \end{eqnarray}

Dynamic groups

The time evolution of a point particle can also be described as the action of the Galilean group on the symplectic space with respect to time translation.

WZW model

The point particle can be described as a one dimensional Wess-Zumino-Witten model.

Given a Lie group $G$, a Wess-Zumino-Witten model with parameter space $\Sigma$ and target space $G$ is given by an action on the mapping space $[\Sigma, G]$,

\begin{equation} S_{\mathrm{WZW}} = S_{\mathrm{kin}} + S_{\mathrm{Top}} + S_{\partial} \end{equation}

where the kinetic term is the standard kinetic term for a sigma model with the metric being the unique bi-invariant metric on that Lie group

\begin{equation} S_{\mathrm{kin}}[X] = \int_{\Sigma} d \mu(X^* B) \end{equation}

The topological terms

\begin{equation} S_{\mathrm{Top}} = \int_{\partial \Sigma} \end{equation}

And the boundary term

\begin{equation} S_{\partial} = \int_{\partial \Sigma} \end{equation}

for the case of a point particle, we will consider the one dimensional WZW model of the Galilean group.

The Galilean group can be decomposed in a variety of ways,

\begin{equation} \mathrm{IGal}(n) = ((Tr_n \otimes B_3) \circ T) \circ \mathcal{R} = H \circ \mathcal{R} \end{equation}

The space $H$ is the evolution space of the particle, given by the triplet of position, speed and time $(t, \vec{x}, \vec{v})$.

Equivalent to the jet bundle

\begin{equation} \omega^A = dx - v dt,\ \omega^V = dv \end{equation} \begin{equation} \omega^{(2)} = \omega^A \wedge \omega^V \end{equation}

Discrete methods

If we discretize our configuration space by taking $N + 1$ equidistant points within $[t_a, t_b]$,

\begin{equation} \{ t_i \}_{0 \leq i \leq N},\ t_0 < t_1 < \ldots < t_{N - 1} < t_{N} \end{equation}

with $(t_{i+1} - t_{i}) = \Delta t$, our configuration space is then for maps of the $N+1$ components $\{ t_i \}$ to $E$, or equivalently, an element of the vector space $\mathbb{R}^{(N + 1) \times 3}$, which are the $N+1$ samples of $\vec{x}$

\begin{equation} \{ \vec{x}_i \} \end{equation}

The derivatives of those points are then defined by the difference equations

\begin{equation} \vec{v}_{i} = \frac{1}{\Delta t} (\vec{x}_{i + 1} - \vec{x}_i) \end{equation}

meaning that $\vec{v}$ is defined for $0 \leq i \leq N$.

\begin{eqnarray} \dot{\vec{v}} &=& \vec{0}\\ \dot{\vec{x}} &=& \vec{v} \end{eqnarray}

Last point approximation :

\begin{eqnarray} \vec{v}_{n+1} &=& \vec{v}_n\\ \vec{x}_{n+1} &=& \vec{x}_n + \vec{v}_{n+1} \Delta t \end{eqnarray}

The solution for the velocity is simply $v_n = v_0$ for any value of $\Delta t$, as can simply be shown by recursion ($v_1 = v_0$, if $v_{n} = v_0$, $v_{n+1} = v_n = v_0$), corresponding to the constant speed. For the position, we have to solve

\begin{eqnarray} \vec{x}_{n+1} &=& \vec{x}_n + \vec{v}_0 \Delta t \end{eqnarray}

This is an arithmetic sequence, with constant term $\vec{v}_0 \Delta t$. Its solution is therefore

\begin{eqnarray} \vec{x}_{n} &=& \vec{x}_0 + \vec{v}_0 n \Delta t \end{eqnarray}

In the limit $n \Delta t \to t$,

\begin{eqnarray} \vec{x}(t) &=& \vec{x}(0) + \vec{v}(0) t \end{eqnarray}

Second-order Taylor approximation :

\begin{eqnarray} \vec{v}_{n+1} &=& \vec{v}_n\\ \vec{x}_{n+1} &=& \vec{x}_n + \vec{v}_{n} \Delta t \end{eqnarray}

Polymer quantization

Polymer quantization is a variant of quantization used among other things in loop quantum gravity which drops some constraints of the usual types of quantization. Taking up what we found in canonical quantization previously, we used the Stone-von Neumann theorem to work out which representation of the canonical commutation relations generated by the Poisson algebra we could use, but part of the assumptions of Stone-von Neumann rests on the continuity of the representation. Allowing the extension of such operators to the discontinuous case is what allows the polymer quantization.

Machian mechanics

There were a few attempts by various physicists to implement the ideas of Ernst Mach regarding the use of only relative quantities for mechanics.

Reissner's theory : Two points with gravitating mass $m_1, m_2$ separated by a distance $r$ possess a kinetic energy

\begin{equation} T = m_1 m_2 \dot{r}^2 f(r) \end{equation}

A particle cannot be meaningfully "free" in the context of Mach, as an individual isolated particle does not possess a position or momentum. But if we consider a particle in a homogeneous universe and compare it to another particle that we consider its center (plus however many other particles to form a reference frame), we can compute it.

First let's consider an infinity of particles, all aligned on a lattice. That is, there are particles labelled $(i,j,k)$ such that for any two particles of neighbouring labels, the distance is the same $a$ for all such particles, and all distances match the Euclidian expectations for a cubical grid for further away points. All those particles are at a constant relative distance of each other, with a relative speed of 0.

We also consider another particle outside of these. We will place it at what we would call in a frame as being at position $(0, 0, 0)$ (it is at a distance $0$ of the central particle $(0,0,0)$). It is also at an initial velocity of $\dot{r}_{(0,0,0)} = (1,0,0)$, moving at a speed of one length of the lattice for every unit of time.

To do our further calculations, we need to figure out the relative position of this free particle with respect to every point of the lattice. If the free particle is at position $(x,y,z)$, ie at a distance of $\sqrt{x^2 + y^2 + z^2}$ of $(0,0,0)$, its relative distance from $(i,j,k)$ is

\begin{equation} r_{(i,j,k)} = \| (x - i, y - j, z - k) \| \end{equation}

As the other particles are at rest with respect to the origin, this does not change the relative velocity.

Its kinetic energy is then

\begin{equation} T = \sum_{i,j,k} m_i m \dot{r}_{(i,j,k)} f(r_{(i,j,k)}) \end{equation}

Semiclassical mechanics

Quantum mechanics has a few semiclassical approximations.

WKB approximation

Stochastic

Local nets

An alternative way to define a quantum theory on an underlying space is via the theory of local nets. Given a kinematic space $M$ with some kinematic group $K$,

Path formalism

An alternative solution to the problem of reparametrization is to use instead of curves going through spacetime a path, which is the equivalence class of curves up to reparametrization :

\begin{equation} [\gamma] = \left\{ \gamma \in [L, M] | \forall \gamma_1, \gamma_2,\ \exists g \in \mathrm{Diff}(L),\ \gamma_1 = \gamma_2 \circ g \right\} \end{equation}

Experimental perspective

A final notion to keep in mind is that of the actual experimental perspective. What do we actually do when we actually perform such measurements.

First issue is that of course, free particles do not exist, as we are not expecting anything in current physical theories to act as either free or as particles. Every object interacts however slightly with its environment, and no object is actually point-like with well-defined velocities. In a typical experiment of free motion, we first have to define what our object is, which depending on the underlying theory we assume may not be as trivial as it seems, despite the process being fairly easy to do on an intuitive level. If we consider the matter underlying the theory as a continuum (as it is likely to be if we are to trust the most common modern theories, like quantum field theory), it becomes difficult to properly differentiate the boundary between an object and its environment, and likewise in the other direction if we consider it as a collection of elementary particles. While those are concerns on an epistemologic level, however, they are not typically very problematic experimentally for the most part, especially at the level of the motion of free particles.

Approximation of a free particle

As we said, no actual object we can experiment on is truly free. The exact nature of the interactions that will weigh on the path of such objects will vary a lot depending on the type of object, their scale or the setting we use, but they are as varied as air resistence, contact forces, electromagnetic forces, gravitational forces, radiative pressures, Brownian forces and so on.

If we were doing everything properly, the best way to do the analysis of the free particle problem would be to consider the real forces involved and compute the error (or at least bounds upon it) that they entail. This is very rarely done, especially in any detail greater than asymptotic analysis, which only give us the scaling of errors and not their bounds. For a very brief example, let's try to compute it for a solvable case, the motion of a particle with friction. A rather simple model is that of linear drag, where we simply add a term proportional to the object's speed, with

\begin{eqnarray} \ddot{\vec{x}}(t) + \frac{k}{m} \dot{\vec{x}}(t) = 0 \end{eqnarray}

The exact solution is given by the integration of the speed, $\vec{v} = \dot{\vec{x}}$, as

\begin{eqnarray} \frac{\dot{\vec{v}}(t)}{\vec{v}(t)} = -\frac{k}{m} \end{eqnarray}

which, using some handwaving,

\begin{eqnarray} \frac{d\vec{v}(t)}{\vec{v}(t)} = -\frac{k}{m} dt \end{eqnarray}

Can be integrated to

\begin{eqnarray} \ln(\vec{v}) = -\frac{k}{m} t + C_0 \end{eqnarray}

where $C_0$ is an integration constant, such that at $t = 0$, $C_0 = \ln(\vec{v}_0)$. Our final form is given by exponentiation,

\begin{eqnarray} \vec{v} = \vec{v}_0 \exp(-\frac{k}{m} t) \end{eqnarray}

This can be checked to reduce to the free case in the limit $k \to 0$, and to have a steeper decrease of velocity depending on $k$. The position is then given by

\begin{eqnarray} \vec{x} &=& C_1+ \vec{v}_0 \int \exp(-\frac{k}{m} t) dt\\ &=& C_1 - \vec{v}_0 \frac{m}{k} \exp(-\frac{k}{m} t) \end{eqnarray}

with $C_1$ given by

\begin{eqnarray} \vec{x}_0 &=& C_1 - \vec{v}_0 \frac{m}{k} \end{eqnarray}

so that

\begin{eqnarray} \vec{x} &=& \vec{x}_0 + \vec{v}_0 \frac{m}{k} (1 - \exp(-\frac{k}{m} t)) \end{eqnarray}

which also reduce to the free case given that

\begin{eqnarray} -\frac{m}{k} (1 - \exp(-\frac{k}{m} t)) \approx t + \mathcal{O}(k) \end{eqnarray}

Given this, we can find the exact error produced by assuming that $k = 0$ :

\begin{eqnarray} | \vec{x}_{\text{free}} - \vec{x} | &=& |v_0| (t + \vec{v}_0 \frac{m}{k} (1 - \exp(-\frac{k}{m} t)))\\ &=& |v_0| \frac{k}{m} t^2 + \mathcal{O}(t^3) \end{eqnarray}

meaning that the error increases over time, and gets steeper for higher drag coefficients and smaller masses. Typically, the drag coefficient of a small spherical object in air will be given by something like Stokes' law,

\begin{equation} k \approx 6 \pi \mu R \end{equation}

with $\mu$ the dynamic viscosity, which in air is about

\begin{equation} \mu \approx 3.178\ \times 10^{-5}\ \mathrm{kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-1} \end{equation}

and $r$ the radius. For a small object $R < 0.1\ \mathrm{m}$, $m < 1\ \mathrm{kg}$, the coefficient would therefore be somewhere in the range

\begin{equation} \frac{k}{m} \approx 5.9903\ \times 10^{-5}\ \mathrm{s}^{-1} \end{equation}

meaning that the linear part of the error will be something like

\begin{eqnarray} e &\approx& |v_0| \frac{k}{m} t^2\\ &\approx& 5.9903\ \times 10^{-5} |v_0| \frac{k}{m} t^2 \end{eqnarray}

So that for a speed in the magnitude of $1\ \mathrm{m} \cdot \mathrm{s}^{-1}$, the error will be of the scale of $10^{-5}$ after one second and increase quadratically after that.

Approximation of free motion

As it is typically difficult to have even a good approximation of a free object, there are a variety of compromises that are usually employed.

The equivalence principle

First is of course that on Earth, it is very difficult to have an experiment free of gravity. There are a few ways to work around this issue. One is to consider the case of restricting the motion to that of planar motion, and have the gravitational force compensated by another force, typically reaction forces. In such cases we can consider the non-vertical degrees of freedom to be free, as can be shown easily enough by some constraint analysis of the Lagrangian.

Another type of experiment exploits the equivalence principle. There are a few different ways to state it, and different such principles, but roughly speaking it states that within a small enough neighbourhood, an object subjet to no other forces than gravitational moves as if subject to no forces. This stems from the nature of gravity in general relativity (and other geometric gravitational theories) as locally integrable $G$-structures : within a small enough neighbourhood (formally speaking, infinitesimal, but in practical term, such that any error would be at least linear in the radius of the region), spacetime reduces to its local Klein geometry, which in the case of general relativity is that of Minkowski space, and for Newtonian gravity as a Newton-Cartan theory, as classical mechanics. Hence for an experiment of small enough scale and duration (typically any experiment given the average spacetime curvature on Earth), it is perfectly fine to consider a particle in freefall as free, assuming the rest of the experimental apparatus is also in freefall and it is subject to no other forces (major anyway).

Air friction

As we saw, air friction can deviate from free motion quite a lot cumulatively. Therefore it is best to try to reduce it as much as possible.

The first obvious factor is simply the parameters of the object itself going into it. If we were to look at the air drag in more details, more aerodynamic shapes, higher mass and smaller size would go towards reducing air drag. Furthermore, reducing the drag of the fluid itself can also help, by picking less viscous fluids or use lesser pressure.

Another important class of forces is that of friction. Both in the case of air friction and, if we consider the case of motion on a surface, dry friction on the surface.

Operational measurements

For an experimental setup we must also have a set of observables to measure. For a mechanics problem like the free particle, this will typically be measurements of position at various moments, or measurements of trajectories directly. The method for those measurements will vary depending on the exact type of object we experiment on and the scales of time and distance involved.