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# Coordinate-free derivation of the point particle

In general relativity, the point particle is defined by a one-dimensional submanifold of the spacetime. In other words, an inclusion map

\begin{eqnarray} X : I &\hookrightarrow& M\\ \lambda &\mapsto& X(\lambda) \end{eqnarray}

$I$ being some $1$-manifold, usually either $\mathbb{R}$ or $S$, but in most realistic cases simply $\mathbb{R}$.

As with most extended objects, we can either the Nambu-Goto or the Polyakov action, or some supersymmetric variant thereof, but for now let's keep it generic. Our action is simply a linear functional from the space of such maps to $\mathbb{R}$. As we are not quite dealing with a simple vector space here, it is best to look a bit more into our configuration space.

As with all branes/extended objects, the configuration space $\mathrm{Conf}$ of our point particle is the mapping space $\mathrm{Map}(I, M)$, which is the space of continuous maps from $I$ to $M$. As a mapping space, it is equipped with the compact-open topology. For any compact subset $X$ of $I$ (typically some closed interval $[a, b]$ in our case), and an open subset $U \subset M$, then every other continuous function $Y \in \mathrm{Map}(\mathbb{R}, M)$ such that $Y(X) \subset U$, is part of the subbase of the compact-open topology. If we only consider closed intervals as our compact subsets, this will roughly correspond to all the curves entirely contained within $U$. If we consider every compact subsets $X$ of $I$ and every open subsets $U$ of $M$, and note the subset of our mapping space obeying those properties as $V(X,U)$, we can define the topology of our mapping space the usual way, as the union of finite intersections of all possible $V(X,U)$.

This will therefore give us a notion of how to define two nearby trajectories, but this isn't enough for our purpose, because we would also like to consider the set of curves with fixed endpoints.

[...] Doing this for fields and for point particles will be somewhat different here, as one will be a map over curves while the other will be a map over fields. As you seem to go for curves here, let's see how it works out The Lagrangian (density) here is a map from the local informations of the curve (in our case, its position and velocity at time $\lambda$) to a $1$-form, and the action will then map it to $\mathbb{R}$. So to recapitulate, we have curves : $$X : \mathbb{R} \to M$$ You have the space of curves, $C^\infty(\mathbb{R}, M)$, which could be of any regularity but let's keep it smooth here, the action maps those curves to $\mathbb{R}$, $$S : C^\infty(\mathbb{R}, M) \to \mathbb{R}$$ And then we have the Lagrangian, which maps informations of that curve at specific points (the *jet extension*, as it is called) to $1$-forms, $$L : M \times T^*M \to \mathbb{R}$$ All assembled, this is something like \begin{eqnarray} S[X] = \int L(X(\lambda), \dot{X}(\lambda)) d\lambda \end{eqnarray} Now we want to find out the variation of this action. The variation of the trajectory of our particle here will just reassign points of the manifold in some smooth way, such that the boundaries do not move. For two given curves, we have some diffeomorphism $X' = \phi(X)$, such that $\phi$ is the identity on the boundaries. Since we are only concerned about small variations, we can just consider families of diffeomorphisms connected to the identity, ie $$\lim_{\varepsilon \to 0} \phi_\varepsilon(X) = X$$ Such diffeomorphisms, at least locally (which we can always do since we can just focus our action to some compact region), are equivalent to the flow of some vector field, ie for any diffeomorphism $\phi$, there exists a vector field $V_\phi$ such that : $$\Phi^{V_\phi}_1 = \phi$$ with $\Phi$ the vector flow. This is roughly what corresponds to our local variation, except with some parameter $\varepsilon$ instead of $1$. Our variation is therefore something of the form \begin{eqnarray} \frac{\delta S[X]}{\delta X_V} &=& \lim_{\varepsilon \to 0} \frac{S[\Phi^{V_\phi}_\varepsilon (X)] - S[X]}{\varepsilon}\\ &=& \lim_{\varepsilon \to 0} \frac{1}{\varepsilon} \int \left[ L(\Phi^{V_\phi}_\varepsilon X(\lambda), \frac{d}{d\lambda} \Phi^{V_\phi}_\varepsilon X(\lambda)) - L(X(\lambda), \dot{X}(\lambda))\right] d\lambda \end{eqnarray} As usual in such matters, our variation is done in a specific direction, in this case indicated by our vector field $V$, and hopefully the final result will not depend on this specific vector field. As $L$ is a local quantity that is just a $1$-form, we can Taylor expand it (carefully), we are just varying parameters on a vector space. Let's consider \begin{eqnarray} L(\Phi^{V_\phi}_{0 + \varepsilon} X(\lambda), \frac{d}{d\lambda} \Phi^{V_\phi}_{0 + \varepsilon} X(\lambda)) &=& L(X(\lambda), \dot{X}(\lambda)) + \varepsilon \left[\frac{d}{d\varepsilon} L(\Phi^{V_\phi}_{\varepsilon} X(\lambda), \frac{d}{d\lambda} \Phi^{V_\phi}_{\varepsilon} X(\lambda))\right] \biggr\rvert_{\varepsilon = 0} + \mathcal{O}(\varepsilon^2) \end{eqnarray} So our variation is \begin{eqnarray} \frac{\delta S[X]}{\delta X_V} &=& \int \left[\frac{d}{d\varepsilon} L(\Phi^{V_\phi}_{\varepsilon} X(\lambda), \frac{d}{d\lambda} \Phi^{V_\phi}_{\varepsilon} X(\lambda))\right] \biggr\rvert_{\varepsilon = 0} d\lambda \end{eqnarray} This is what we need to expand to get our EL equations. For this we need to work out the derivative of a tensor field along a curve. In our case, every point $X(\lambda)$ on that curve will travel along along some curve $X_\varepsilon(\lambda) = \Phi^{V_\phi}_\varepsilon X(\lambda)$, and this curve will have a velocity of simply $V$ at $\varepsilon = 0$.
Last updated : 2021-08-30 23:52:01
Tags : physics , general relativity