To define the monodromy matrix, assume that the period for one cycle of a halo orbit is denoted as $T$, the initial time as $t_0$, and the state-transition matrix is defined as $\phi$. Then, the monodromy matrix is defined as the state-transition matrix for one full orbital cycle, $\phi(t_0,T)$. A good reference for a more detailed definition can be found here.

The monodromy matrix has 6 eigenvalues (three pairs) which are $\lambda_1>1, \lambda_2<1,\lambda_3=\lambda_4,\lambda_5=\lambda_6$.

My question is what is the physical interpretation of the eigenvalues of the monodromy matrix specifically and how does their corresponding eigenvectors define the hyperbolic invariant manifolds in the CR3BP.


1 Answer 1


Eigenvalues and Eigenvectors

Before specifically addressing the monodromy matrix, it's important to make sure you have a physical understanding of what eigenvalues and eigenvectors represent in general. I highly recommend 3blue1brown's youtube video on this topic:

I will distill the important points below.

Consider the two-dimensional $x-y$ plane for simplicity, with a general vector denoted by $\overrightarrow{a} = a_x \hat{i} + a_y \hat{j}$. We can view a matrix $A$, \begin{align} A = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} \end{align} as a linear transformation which can operate on the likes of $\overrightarrow{a}$. To visualize the effect $A$ would have on $\overrightarrow{a}$, consider the effect it has on the coordinate system's unit vectors. In the original system, the unit vectors are given by a line connecting the origin with $\hat{i} = [1 \ 0]$ and $\hat{j} = [0 \ 1]$, respectively. The first column of $A$ tells us the coordinates of where the unit vector $\hat{i}$ would land if we applied the linear transformation, and the second column tells us the same coordinates associated with transforming $\hat{j}$. (As a sanity check, verify mentally that this is consistent with the identity matrix returning the original vector, and how any scalar $k$ multiplied into the identity matrix has the simple effect of stretching the original vector.)

So why is this important? Well, it gives us a way to visualize what happens to any arbitrary vector such as $\overrightarrow{a}$ when we apply the linear transformation $A$. The $x$ and $y$ coordinates of $\overrightarrow{a}$ will be rotated and/or stretched corresponding to how $A$ rotates and stretches the unit vectors of the system (you can visualize this as the 2D space being rotated/stretched itself). Now, lets consider a special case of what could happen. Due to the particular way a linear transformation rotates and/or contracts the 2D space, specific vectors may exist that end up only expanding or contracting without changing their direction. An obvious example is the case mentioned earlier: the identity matrix multiplied by a scalar, which we can easily visualize as stretching whichever vector it operates on. Other cases are harder to visualize, but they may all be described by the following relationship \begin{align} A\overrightarrow{v} = \lambda \overrightarrow{v} \end{align} What this says is that when the linear transformation operates on some vector$\overrightarrow{v}$, the vector is stretched along the direction it is already pointing by some factor $\lambda$. For a particular transformation, we call the set of vectors $\overrightarrow{v}$ that exhibit this behavior eigenvectors, and we call the factor they are stretched by $(\lambda)$ the eigenvalues corresponding to the vectors.

The State Transition Matrix

You correctly define the Monodromy matrix . However, it's worth spending a minute to recall where the State Transition Matrix (STM) comes from and what it represents. Namely, nonlinear ODEs can be expensive to compute. If we solve for the state over time, $\overrightarrow{X}(t)$, resulting from a particular initial condition, $\overrightarrow{X}_0$, it would be nice if it were possible to approximately solve for how $\overrightarrow{X}(t)$ changes due to perturbations to $\overrightarrow{X}_0$, without re-solving the nonlinear ODE. (Here you can assume $\overrightarrow{X}$ denotes the spacecraft position+velocity). If we take a Taylor series for some future point, $\overrightarrow{X}_f$, as a function of the initial point subject to a small perturbation, denoted by $\delta \overrightarrow{X}_0$, we have \begin{align} \overrightarrow{X}_f(\overrightarrow{X}_0 + \delta \overrightarrow{X}_0) = \overrightarrow{X}_f(\overrightarrow{X}_0) + \frac{\partial \overrightarrow{X}_f}{\partial \overrightarrow{X}_0} \delta \overrightarrow{X}_0 + \frac{1}{2} \delta \overrightarrow{X}_0^T \frac{\partial^2 \overrightarrow{X}_f}{\partial \overrightarrow{X}_0^2} \delta \overrightarrow{X}_0 + \cdots \end{align} To define a linearization we truncate all terms after the first. Then, the perturbed final state is given by \begin{align} \overrightarrow{X}_f(\overrightarrow{X}_0 + \delta \overrightarrow{X}_0) - \overrightarrow{X}_f(\overrightarrow{X}_0) = \frac{\partial \overrightarrow{X}_f}{\partial \overrightarrow{X}_0} \delta \overrightarrow{X}_0 \\ \delta \overrightarrow{X}_F = \frac{\partial \overrightarrow{X}_f}{\partial \overrightarrow{X}_0} \delta \overrightarrow{X}_0 \end{align} In other words, a perturbation at the initial state is propagated to a perturbation at the final state by this matrix which we call the State Transition Matrix, \begin{align} \Phi(t_f,t_0) = \frac{\partial \overrightarrow{X}_f}{\partial \overrightarrow{X}_0} \end{align} whose name now makes intuitive sense. If we model a small perturbation at time $t=t_0$, then we can propagate it forward and approximate the resulting perturbation at $t=t_f$ (for small changes, this is a good approximation).

The Monodromy Matrix

As you identify in the question, the Monodromy matrix $M$ is the State Transition Matrix (STM) $\Phi$ after one orbit period. In other words, given a perturbation at some point of an orbit, the Monodromy matrix tells us the effect of that perturbation one period later. Now, we can start to actually answer your question. Let's first address the following: ``What is the physical interpretation of the eigenvectors and eigenvalues of the monodromy matrix?''

As discussed earlier for the general linear transformation $A$, its eigenvectors tell us which vectors/directions, if any, are purely expanded or contracted under $A$, and the eigenvalues indicate by how much. In the context of the Monodromy matrix, the eigenvectors represent the same thing. In other words, the eigenvectors describe directions along which an applied perturbation will scale. Depending on the eigenvalue, a perturbation applied in the direction given by the eigenvector will either grow ($\lvert \lambda \rvert > 1$), dampen ($\lvert \lambda \rvert < 1$), or stay the same ($\lvert \lambda \rvert = 1$). While there are 6 total eigenvectors/eigenvalues, it may be shown for the CRTBP that two are complex, and of the remaining four real-valued ones, the eigenvalues exist in reciprocal pairs, where one pair is simply unity. Therefore, we care primarily about the remaining pair of eigenvalues, $\lambda_1$ and $\lambda_2 = 1/\lambda_1$. Since it is a reciprocal pair, if the eigenvalue is not unity, then there will be both a direction of growth and a direction of contraction. If the perturbation dampens over time, then we refer to the direction as stable; whereas if the perturbation grows, then we consider the direction to be unstable. Stable and unstable directions naturally exist in pairs in the CRTBP.

Invariant Manifolds

The next part of your question asks how eigenvectors lead to the invariant manifolds in a Halo orbit. Now that we understand how the eigenvectors correspond to stable/unstable directions, we are in a position to address this part of your question. Instead of giving a "textbook definition" of invariant manifolds, let's state a couple of facts that we now understand about any example Halo orbit, and we will see that the meaning of invariant manifolds naturally falls out.

Consider a spacecraft traversing any reference unstable Halo orbit (by unstable, we only mean that the eigenvalues of the Monodromy matrix are not all unity). We now know that at each point on this orbit, there exists a stable and an unstable direction. If we applied a perturbation in the unstable direction at each point on the orbit, we could generate a ``family'' of trajectories that exponentially depart from the reference orbit by propagating forward in time. Note that traveling "backwards in time" along any of these unstable trajectories would approach the reference halo orbit.

What if we wanted to identify the trajectories that would approach the reference orbit forward in time? In that case, we could consider a perturbation in the stable direction, and then propagate backwards in time. Take a moment to think about this if it sounds strange. The reason we propagate backwards in time is that it shows us example paths, upon which a spacecraft could be placed, to approach the reference orbit in a stable sense (recall the perturbation is being dampened when applied in this direction).

We call the set of all trajectories which arrive forward in time to the reference orbit its stable manifold, and we call the set of all trajectories that arrive backwards in time its unstable manifold (as they depart forward in time). Therefore, we see that the eigenvectors of the Monodromy matrix may be used to compute the invariant manifolds for a particular periodic orbit. Additionally, we can start to better understand the motivation behind using stable/unstable manifolds for trajectory design in the CRTBP. Unstable and stable manifolds provide natural mechanisms for transferring to and away from periodic orbits, suggesting a philosophy for designing efficient trajectories between different orbits.

  • $\begingroup$ Is that the case for any closed orbit in the CRTBP? Can you direct me to sources about that? $\endgroup$
    – user449911
    Mar 4, 2022 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.