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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.

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.

1. 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

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.

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, 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

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}

2. 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.

3. Continuum mechanics

If we consider furthermore the motion of some localized object, say a mass distribution $\rho$, its equation of motion is given by the integral version of this

\begin{equation} \ddot{\rho} = \end{equation}

4. Lagrangian mechanics

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[[t_i, t_f]; \vec{x}] = \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

\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, $\vec{x}(t_a) = \vec{x}_a$, $\vec{x}(t_b) = \vec{x}_b$, we get

\begin{equation} \vec{x}(t) = \frac{\vec{x}_b - \vec{x}_a}{t_b - t_a} t + \frac{1}{2} \left[ \vec{x}_a + \vec{x}_b - \frac{\vec{x}_b - \vec{x}_a}{t_b - t_a} (t_a + t_b)\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}

Symmetries

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}(\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, $\mathrm{Gal}(n)$. 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 $\vec{x} \to \vec{x}' = \vec{x} + \vec{a}$

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 $C^\infty(\mathbb{R}, \mathbb{R})$ (if we only assume smooth trajectories here) can be given a variety of topologies. As a space of continous maps between two 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 $f(K) \subseteq U$.

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}

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.

Banach space, Morse theorem

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 $x_2$

Observables

5. Hamiltonian mechanics

The momentum associated with the Lagrangian is

\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 $v$ as

\begin{equation} \vec{p}[v] = m \vec{x} \cdot 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), via the function

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

or more specifically the vector obtained from $\vec{p}$ 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, $\dot{O} = \{ H, O \}$, 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 $Q_{\mathrm{CoM}}$

6. 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.

7. Lagrangian correspondence

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}

8. 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}

9. Bundle methods

For more abstract ways in classical mechanics, we can define the configuration space of a point particles by the tangent bundle. Given Euclidian space $E = \mathbb{R}^n$, our tangent bundle is $TE \cong \mathbb{R}^n \times \mathbb{R}^n$. Then the particle is described by the map $\gamma : \mathbb{R} \to E$, and our action is a function $S : TE \to \mathbb{R}$,

A free particle is a curve $\gamma : L \to E$, with a lift to $\gamma^\uparrow : L \to TE$, 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>

• $(\pi_{TM})_* \xi_{X} = X$
• For the tangent structure $J$ on $TE$, we have $J\xi = V$, $V$ the canonical vector field on $TE$.
• For the canonical flip $j : TTE \to TTE$, $j \xi = \xi$

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

10. 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

11. 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}

For the Galilean group, the adjoint

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}

A co-adjoint action is simply the action of $G$ on $\mathfrak{g}^*$

The co-adoint orbits are the orbits of the co-adjoint action in $\mathfrak{g}^*$.

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}

12. 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.

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\\ \forall Y,\ \omega(X, Y) = 0 &\leftrightarrow& X = 0 \end{eqnarray}

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

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}

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}

13. 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}

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

• The curvature of the connection is zero (spacelike only?)
• The connection has a zero torsion
• Additional non-uniqueness parameters?
• The clock form obeys $\alpha \wedge d\alpha = 0$ (causality)
• The spacetime is topologically trivial : $M = \mathbb{R}^{(n+1)}$

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

14. 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}

15. 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} \bigsqucup_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} = \bigsqucup_{i,j,k} \end{equation}

16. 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

17. 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 = -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}

Polyakov action

Another action we can use for the relativistic point particle is the Polyakov action

\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}

18. 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 $\eta$ :

19. 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}

Polyakov action

The Polyakov action generalizes easily enough to curved space,

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

MisQ etc

The geodesic spray

20. 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 $Q$

\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} \[ \hat{A}, \hat{B} \] \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 $\mathcal{H} = L^2(\mathbb{R}^n, d\mu)$, 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}

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}

with the solution

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

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

21. 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 $t_N = t_f$

22. 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}

23. 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.

24. 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}

25. 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.

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

26. 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 $\vec{x}(\theta^+) = \vec{x}_i$, $\vec{x}(\theta^-) = \vec{x}_f$, 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}

27. 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}

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.

[diagram]

The $8$ points defined are the points $(k_1, k_2, k_3)$, $((k_1 + 1), k_2, k_3)$, $(k_1, (k_2 + 1), k_3)$, $(k_1, k_2, (k_3 + 1))$, $((k_1 + 1), (k_2 + 1), k_3)$, $(k_1, (k_2 + 1), (k_3 + 1))$, $((k_1 + 1), k_2, (k_3 + 1))$ and $((k_1 + 1), (k_2 + 1), (k_3 + 1))$. Those $8$ points form a parallepiped $P$, that we will say is the parallepiped 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 parallepiped $P$ 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 parallepiped $P'$, and $t_i + \delta_t$ is within some interval $(...)$.

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

28. 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}

29. 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.

30. WZW model

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

Given a Lie group $G$,

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}

31. Discrete methods

\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}

33. Polymer quantization

34. 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}

34. Semiclassical mechanics

35. Local nets

36. 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] = \{ \gamma \in [L, M] | \forall \gamma_1, \gamma_2,\ \exists g \in \mathrm{Diff}(L),\ \gamma_1 = \gamma_2 \circ g