I have a fairly specific question. My intuition says the answer is "yes", but there is a natural generalizations in which I take out all the "physics", and then I think the answer is "no".
<h3>Background and definitions</h3>
On $\mathbb R^d$ (with its usual metric), pick a differential one-form $b$ and a smooth function $c$. The tangent bundle $T\mathbb R^d$ is just $\mathbb R^{2d}$; define the **Lagrangian** $L: T\mathbb R^d \to \mathbb R$ by $L(v,q) = \frac12 |v|^2 + b(q)\cdot v + c(q)$, where $v$ is the fiber coordinate on $T\mathbb R^d$, $q$ is the base coordinate on $\mathbb R^d$, and $\cdot$ is the canonical pairing of a one-form with a vector. A **path of length $t$** is a smooth map $\gamma: [0,t] \to \mathbb R^d$; it has a canonical lift $(\dot\gamma,\gamma): [0,t] \to T\mathbb R^d$. The **action** of a path $\gamma$ of length $t$ is the integral $A[\gamma] = \int_0^t L(\dot\gamma(\tau),\gamma(\tau))d\tau$. By adjusting signs, one can include paths of negative length; a path of length $0$ is a point in $T\mathbb R^d$ and has zero action.
Consider the set $P$ of all paths (of arbitrary length); it is an infinite-dimensional smooth manifold. There are various natural projections from $P$ to finite dimensions. The "initial-value map" $P \to T\mathbb R^d \times \mathbb R$ takes a path $\gamma: [0,t]\to \mathbb R^d$ to the triple $(\dot\gamma(0),\gamma(0),t)$. I will be more interested in the "boundary-value map" $P \to \mathbb R^d \times \mathbb R^d \times \mathbb R$ taking $\gamma \mapsto (\gamma(0),\gamma(t),t)$. The fiber over a point in $\mathbb R^d \times \mathbb R^d \times \mathbb R$ is an affine space modeled on the space of **Dirichlet** paths $\gamma: [0,t] \to \mathbb R^d$ with $\gamma(0) = 0 = \gamma(t)$.
I like to think of the action $A$ as a Morse function on fibers of the boundary-value map. Let $C \subset P$ be the set of **classical** paths, i.e. paths $\gamma$ so that  if $\xi$ is Dirichlet ( is the differential of the action at $\gamma$; $\cdot$ is the canonical pairing). Equivalently, $\gamma \in C$ if $\gamma$ satisfies the **Euler-Lagrange equations** $\frac{\partial L}{\partial q}(\dot\gamma,\gamma) = \frac{d}{d\tau}\bigl[ \frac{\partial L}{\partial v}(\dot\gamma,\gamma) \bigr]$. Since the Euler-Lagrange equations are second-order nondegenerate, the initial-value map restricts to a diffeomorphism of $C$ to an open subset of $T\mathbb R^d \times \mathbb R$ containing $T\mathbb R^d \times \{0\}$.
If I really want to think of $A$ as a Morse function, I should require that its critical points (the classical paths) be nondegenerate. Let $\gamma$ be a (classical) path of length $t$, and $V$ the vector space of Dirichlet paths of length $t$. Then the second derivative or Hessian of $A$ is well-defined as a map $H : V \to V^*$. In fact, the Hessian makes sense as a second-order linear differential operator on the space of all paths of length $t$. Let's say that a classical path is **nondegenerate** if $\ker H = 0$ (or, rather, does not intersect the space $V$ of Dirichlet paths). The set $C'$ of nondegenerate classical paths is an open (I'm pretty sure) subset of $C$.
<h3>My question</h3>
Is the space $C'$ (path) connected?
Bonus question: what happens if you change the signature of the metric on $\mathbb R^d$?
<h3>Edit</h3>
The answer to my original question is "no". Let $d = 1$, $b = 0$, and $c(q) = \frac12 q^2$. Then a classical path of length $t$ is nondegenerate if and only if $t$ is not an integer multiple of $\pi$. This is a very nongeneric Lagrangian (it is the **harmonic oscillator**, and is exactly solvable). Also, I think with my definitions, paths of length $0$ are always degenerate.
So let me ask a more restricted question. Let's suppose that $b$ and $c$ are only supported in a compact neighborhood. Then classical paths that do not enter this neighborhood are precisely the straight lines, and they are all generic (provided $t \neq 0$). Is it true that the space of classical nondegenerate paths with positive length is connected with the restriction that $b,c$ have compact support?