# What do conjugate points mean in lagrangian mechanics?

Good evening, my mathematical physics professor explained conjugate points to us in a way that just didn't make much sense to me, so I looked for a deeper more geometric definition online. I found some, but I'm having trouble understanding the connection between the "geometric" definition of conjugate points and the "lagrangian" one. I'll start by sharing my understanding of it in the geometric sense (which may well be wrong):

Take a point $$p$$ on a riemannian manifold and a smooth family $$\gamma_s(t)$$ of geodesics passing through it, say at $$t=0$$. Let $$\gamma = \gamma_0.$$ We define the Jacobi field associated to $$\gamma$$ as $$J(t) = \left.\frac{d}{ds}\gamma_s(t)\right|_{s=0}$$ which intuitively describes the separation between the points on $$\gamma$$ and an infinitely close other geodesic in the family at time $$t$$.

From what I've read, an equivalent definition is to say that a Jacobi field is any solution of the equation $$\frac{D^2}{dt^2}J + R(J, \dot \gamma)\dot \gamma = 0 \ \ \ (\star_1)$$ where $$\frac{D^2}{dt^2}$$ denotes the derivative along $$\gamma$$ and $$R$$ is the Riemann curvature tensor.

A point $$q = \gamma(\tau)$$ is said to be a conjugate of $$p$$ if $$J(\tau) = 0$$, meaning that our family "merges" into this point (at least in the limit). For example, any two opposite points of $$S^2$$ are conjugate.

Now for the lagrangian definition: the setting is 1-d lagrangian mechanics, so we have some functions $$\gamma, \eta: \Bbb{R} \to \Bbb{R}$$, an action functional $$\mathcal{A}_\mathcal{L}(\gamma) \triangleq \int_{t_0}^{t_1} \mathcal{L}(\gamma(t), \dot\gamma(t), t)dt,$$ its first variation (let's call $$\phi(\lambda) = \mathcal{A}_\mathcal{L}(\gamma+\lambda\eta)$$) $$\delta\mathcal{A}_\mathcal{L}(\gamma; \eta) \triangleq \phi'(0)$$ and its second variation $$\delta^2\mathcal{A}_\mathcal{L}(\gamma; \eta) \triangleq \phi''(0).$$ It turns out that if we freeze $$\gamma$$ we can write $$\delta^2\mathcal{A}_\mathcal{L}(\gamma; \eta)$$ as an action functional $$\mathcal{A}_\mathcal{Q}(\eta)$$ ($$\mathcal{Q}$$ is called "auxiliary lagrangian"), so we can consider the Euler-Lagrange equation associated with it. Any solution to this equation is called a Jacobi field, and if we couple it with the initial conditions $$\eta(\tau) = 0, \dot\eta(\tau) = 1$$ for some time $$\tau$$, we have a unique solution $$\tilde\eta(t)$$. Finally, for every solution $$\tau_\star$$ of $$\tilde\eta(\tau_\star) = 0$$ we say that $$(\tau, \gamma(\tau))$$ and $$(\tau_\star, \gamma(\tau_\star))$$ are conjugate points.

For completeness, the mentioned Euler-Lagrange equation turns out to be $$-\left(a \dot \eta\right)' + (c-\dot b)\eta=0 \ \ \ (\star_2)$$ where $$a(t) = \mathcal{L}_{\dot\gamma\dot\gamma}(\gamma(t), \dot\gamma(t), t), b(t) = \mathcal{L}_{\gamma\dot\gamma}(\gamma(t), \dot\gamma(t), t), c(t) = \mathcal{L}_{\gamma\gamma}(\gamma(t), \dot\gamma(t), t).$$

Now, several questions:

1. Do $$\star_1$$ and $$\star_2$$ correspond in some way? They do look sort of similar but I can't really figure out what $$J$$ and $$R$$ would be in the lagrangian formulation. It seems like $$J$$ would correspond to $$\eta$$, but the substitution doesn't quite turn out right.
2. What manifold am I considering in the lagrangian setting? What is the "hidden geometry" of this context? A while ago we defined the Jacobi metric in the mechanical context (where $$L = T-V$$ and $$E = T+V$$ is conserved) as $$ds^2 = (E - V(q))\sum_{h, k}A_{hk}(q)dq_hdq_k$$ where $$q_i$$ are the lagrangian coordinates and $$A$$ is the kinetic matrix, but I can't really see the connection with what I described until now. I guess the manifold we are considering now, since we are in the 1-dimensional case, would be $$(t, \gamma(t)) \in \Bbb{R} \times \Bbb{R}$$ with the euclidean metric on the time coordinate and some metric dependent on the lagrangian on the space coordinate, but I can't really pull everything together.
3. What do conjugate points represent in the mechanical context? Really, I can't figure this out. This auxiliary lagrangian doesn't seem to have some neat physical or geometric meaning to me, so I'm just a bit confused about this.

The physics lesson looked like a nonsensical gibberish of derivatives to me, so any help is appreciated.

Here are some comments that may help.

First note that a geodesic flow is a special case of a mechanical system (`free particles' ie zero potential). So upon making sense of $$*_2$$ in the mechanical setting one can view the differential geometry concepts as a special case.

The trajectories of the mechanical system are extremals (~critical points) of the action functional, and the second variation (~Hessian) is meaningful when taken along a trajectory: one considers only $$\delta^2\mathcal{A}_{\mathcal{L}}(\gamma;\eta)$$ for $$\gamma$$ a trajectory.

The meaning of $$*_1$$ or $$*_2$$ is in both cases the linearized equations of motion along a given trajectory. That is vector fields along a given trajectory $$\gamma$$ obtained by differentiating one parameter families of trajectories.

You can check this directly, by considering a solution $$\gamma_\epsilon = \gamma + \epsilon \eta$$ of the Euler-Lagrange equations, and sending $$\epsilon \to 0$$:

$$\frac{d}{dt}\partial_{\dot q}L(\gamma_\epsilon, \dot\gamma_\epsilon) = \partial_qL(\gamma_\epsilon, \dot\gamma_\epsilon)$$

The left side expands as $$\frac{d}{dt} \left( \partial_{\dot q}L(\gamma, \dot\gamma) + \epsilon (\partial_q\partial_{\dot q}L(\gamma, \dot\gamma)\eta + \partial_{\dot q}^2L(\gamma, \dot\gamma) \dot\eta\right) + O(\epsilon^2)$$, while the right side expands to $$\partial_qL(\gamma, \dot\gamma) + \epsilon \left(\partial_q^2L(\gamma, \dot\gamma)\eta + \partial_{\dot q}\partial_qL(\gamma, \dot\gamma)\dot\eta\right) + O(\epsilon)^2$$. Since $$\gamma$$ satisfies the Euler-Lagrange equations, one may divide by $$\epsilon$$ and take the limit arriving at $$\frac{d}{dt}\left( b\eta + a\dot\eta \right) = b\dot \eta + c\eta$$ for the linearized equations of motion around the given trajectory $$\gamma$$. Note that upon cancelling a $$b\dot\eta$$ from both sides, these are exactly the equations $$*_2$$: the Euler-Lagrange equations of the second variation.

Hence vector fields satisfying $$*_2$$ have the same interpretation as Jacobi fields in the geodesic case: the vector fields obtained by differentiating a one-parameter family of trajectories (extremals) or equivalently as the vector fields satisfying the Euler-Lagrange equations of the second variation along a given trajectory.

As well, conjugate points in the mechanical context have the same interpretation as in differential geometry; one only replaces geodesic with trajectory or extremal.