Springs in Hyperbolic Space

Unlike in Euclidean geometry, springs oscillate in hyperbolic space.

In this note we derive the differential equations governing a spring in hyperbolic space. In particular, we study the case where a spring is pushed perpendicularly to its length, and show that this causes the spring to oscillate. This is in contrast to Euclidean geometry, where a spring moving at constant speed through space remains at rest.

Phase Space

Consider a spring where half the rest-length is denoted $\ell_o$.

If the spring is massless and there is a single point mass on each end, the entire configuration of the system is uniquely specified by giving these two endpoints of the spring, or a point in $\mathbb{H}^3\times\mathbb{H}^3$. Thus, a state of the system is given by an element of the tangent bundle $Q=T(\mathbb{H}^3\times\mathbb{H}^3)\cong T\mathbb{H}^3\times T\mathbb{H}^3$

As such we represent configurations of the system as pairs $((p_1,v_1),(p_2,v_2))$ for $p_i\in\mathbb{H}^3$ and $v_i\in T_{p_i}\mathbb{H}^3$. These are the four things the figure above lets you drag: two masses and two velocities.

The Lagrangian

Kinetic energy $K$ determines a norm on the phase space $Q$ by

$K\left((p_1,v_1),(p_2,v_2)\right)=\frac{1}{2}m\langle v_1,v_1\rangle_{p_1}+\frac{1}{2}m\langle v_2,v_2\rangle_{p_2}$

where $\langle\cdot,\cdot\rangle$ is the Riemannian metric on $\mathbb{H}^3$. In Newtonian mechanics, the force from a spring is proportional to the difference from its rest length, by some proportionality constant $k$. In terms of potential energy, this is equivalent to the potential being proportional to the squared difference between the spring’s actual length and rest length, with proportionality constant $\tfrac{1}{2}k$:

$V\left((p_1,v_1),(p_2,v_2)\right)=\frac{1}{2}k\left(\mathrm{dist}(p_1,p_2)-2\ell_o\right)^2$

Where $\mathrm{dist}$ is the geodesic distance in $\mathbb{H}^3$. The difference $K-V$ between these gives the Lagrangian of classical mechanics for the spring:

$\mathcal{L}=\frac{m}{2}\left(\langle v_1,v_1\rangle_{p_1}+\langle v_2,v_2\rangle_{p_2}\right)-\frac{k}{2}\left(\mathrm{dist}(p_1,p_2)-2\ell_o\right)^2$

The Initial Condition

The particular motion we want to study starts from the spring at rest at its equilibrium length, given a sideways shove: the masses begin a distance $2\ell_o$ apart, and each is launched with speed $v_o$ perpendicular to the spring, both in the same direction. Setting the figure above to the push imposes exactly this.

The clause that will do the real work is the last one. Saying that the two velocities point “the same way” means that $v_2$ is the parallel translate of $v_1$ along the geodesic segment joining the masses — and it is this that makes the initial state symmetric enough for the arguments below.

Symmetries Constraining Evolution

The symmetries of this system greatly constrain its future dynamics; a fact which allows us to choose reasonable coordinates in which to perform the calculation.

• The Evolution of the spring is confined to a hyperbolic plane: Because the initial velocities are parallel translates of one another along the geodesic segment representing the initial spring, the initial configuration’s state lies in the tangent space to the hyperbolic plane containing the spring and the geodesics defined by these tangent vectors. Let $R$ be the reflection in this plane, and note that $R$ is a symmetry of the laws of physics (it’s an isometry) and simultaneously fixes the initial condition. Thus $R$ fixes the entire evolution of the system, so all future states lie in the set of fixed points of $R$. But $R$ fixes only the aforementioned hyperbolic plane. Thus the evolution of the system is confined to this plane.

• The midpoint of the spring is confined to a geodesic: Restricting ourselves to the hyperbolic plane in which all evolution occurs, let $g$ be the geodesic which is perpendicular to the spring at its midpoint, and let $G$ be the reflection of the hyperbolic plane across this geodesic. Let $c$ denote the initial center point of the geodesic, and $c(t)$ denote its evolution over time. We repeat the same argument as above. $G$ is a symmetry of the laws of physics (it’s an isometry), and $G$ fixes the initial conditions. Thus, the evolution of the system and $G(\mathrm{system})$ are equal, differing only by which mass is called ‘1’ and which is called ‘2’. But, the center $c(t)$ of the spring lies on the fixed set of $G$, and so must remain there for all time. This fixed set is the geodesic through the initial center, so the mass is confined to this geodesic.

• The state of each mass is a reflection of the other across the geodesic on which the spring’s center travels: Let $g$ be the geodesic tracking the spring’s center, and $G$ the reflection across it, as above. By the reasoning there, the system and $G(\mathrm{system})$ have the same evolution, save switching which is called mass 1 and mass 2. Thus, the state of mass 1 is precisely the reflection in $g$ of the state of mass 2, and vice versa. This is built into the figure above: only mass 1 is yours to move, and mass 2 is drawn as its mirror image.

Simplifying the Lagrangian

The symmetries above allow us to massively simplify our analysis, by careful choice of coordinates. First, we can make do with the state of only one of the endpoint masses, as the other is recoverable by application of the isometry $G$. This lets us rewrite the Lagrangian as

$$\mathcal{L}=\frac{m}{2}\left(\langle v,v\rangle_{p}+\langle G_\ast v,G_\ast v\rangle_{G(p)}\right)-\frac{k}{2}\left(\mathrm{dist}(p,G(p))-2\ell_o\right)^2$$

The first term here simplifies as $G$ is an isometry, so it does not change the length of tangent vectors: $\langle G_\ast v,G_\ast v\rangle_{G(p)}=\langle v,v\rangle_{p}$ so the overall term is just twice the hyperbolic length of $v$. The distance term in the potential energy simplifies as well. Letting $c$ denote the point on the spring lying on the central geodesic,

$$\begin{align} \mathrm{dist}(p,G(p))&=\mathrm{dist}(p,c)+\mathrm{dist}(c,G(p))\\ &=\mathrm{dist}(p,c)+\mathrm{dist}(G(c),G(p))\\ &=\mathrm{dist}(p,c)+\mathrm{dist}(c,p)\\ &=2\mathrm{dist}(p,c) \end{align}$$

These equalities are justified as (i) the points $p,c$, and $G(p)$ all lie on a common geodesic, (ii) $c=G(c)$ is fixed by the reflection as it lies on the central geodesic, and (iii) $G$ is an isometry so it does not change distances between pairs of points. Together this gives

$$\begin{align} \mathcal{L}&=\frac{m}{2}\left(\langle v,v\rangle_p+\langle v,v\rangle_p\right)-\frac{k}{2}\left(2\mathrm{dist}(c,p)-2\ell_o\right)^2\\ &=m\langle v,v\rangle_p-2k\left(\mathrm{dist}(c,p)-\ell_o\right)^2 \end{align}$$

Useful Coordinates

Now we turn to choosing useful coordinates on the hyperbolic plane containing the system’s evolution, for measuring the quantities which appear in this Lagrangian. Two useful constraints:

• (i) Because $\mathrm{dist}(c,p)$ shows up, we want one of the coordinates to measure the distance from the central geodesic $g$. We will call this distance $\ell$.
• (ii) because $\langle v,v\rangle_p$ shows up, we would like an orthogonal coordinate system so that we can compute this quantity by ‘the Pythagorean theorem’.

In fact, these two simple constraints already fully fix our coordinates up to a real valued function. To nail down something uniquely we further ask that the coordinate orthogonal to the central geodesic measures the actual hyperbolic distance along this geodesic, which we will call $h$. A picture of these coordinates on the Poincaré disk are drawn below.

The Lagrangian in Coordinates

Now it remains only to write down our Lagrangian in these chosen coordinates, where $\ell$ measures the distance of one of the spring’s endpoints from the central geodesic, and $h$ measures the geodesic distance along the central geodesic $g$ from the starting point. The Lagrangian contains two terms, one of which directly contains the coordinate $\ell=\mathrm{dist}(c,p)$, and the other which contains the magnitude square of the velocity of (either) mass. This is the only quantity we must compute in coordinates.

Because the coordinate system is orthogonal, we may decompose the velocity into components as $v=v_h+v_\ell$, which satisfy $\langle v,v\rangle = v_h^2+v_\ell^2$

As the $\ell$-coordinate accurately measures distance along geodesics perpendicular to the trajectory of the center (by construction), $\dot{\ell}=v_\ell$. The $h$-coordinate measures geodesic distance along $g$ (that is, where $\ell=0$), for positive values of $\ell$, distance increases quicker than $h$-coordinate value, by a factor which we compute below:

The factor is easiest to see by writing the coordinates down explicitly on the hyperboloid. Take the ambient form $\langle x,y\rangle=x_0y_0+x_1y_1-x_2y_2$ and let $g$ be the geodesic $h\mapsto(\sinh h,0,\cosh h)$. Walking a distance $\ell$ along the perpendicular to $g$ at parameter $h$ lands at

$P(h,\ell)=\left(\sinh h\cosh\ell,\ \sinh\ell,\ \cosh h\cosh\ell\right)$

which does lie on the hyperboloid, since $\langle P,P\rangle=\cosh^2\ell(\sinh^2h-\cosh^2h)+\sinh^2\ell=-1$. Differentiating,

$$\begin{align} \partial_h P&=\left(\cosh h\cosh\ell,\ 0,\ \sinh h\cosh\ell\right)\\ \partial_\ell P&=\left(\sinh h\sinh\ell,\ \cosh\ell,\ \cosh h\sinh\ell\right) \end{align}$$

and these three inner products are all that is needed:

$\langle\partial_h P,\partial_h P\rangle=\cosh^2\ell,\qquad \langle\partial_\ell P,\partial_\ell P\rangle=1,\qquad\langle \partial_h P,\partial_\ell P\rangle=0$

The second confirms that $\ell$ measures distance, the third that the coordinates are orthogonal — the two things we asked for — and the first is the factor we were after: the metric in these coordinates is

$ds^2=\cosh^2\ell\, dh^2+d\ell^2$

Thus, we see that $v_h=\dot{h}\cosh(\ell)$ and we can write out the Lagrangian in coordinates

$\mathcal{L}=m\left(\dot{\ell}^2+\dot{h}^2\cosh^2(\ell)\right)-2k\left(\ell-\ell_o\right)^2$

The Equations of Motion

Now we have arrived at a system that can be treated with ordinary classical mechanics: all of the geometry has been incorporated into the choice of coordinates and resulting form of the Lagrangian. To find the trajectory $q(t)=(h(t),\ell(t))$ the system will follow, we must solve

$$\frac{\partial\mathcal{L}}{\partial h}=\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial{\dot{h}}}$$ $$\frac{\partial\mathcal{L}}{\partial \ell}=\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial{\dot{\ell}}}$$

For the first of these, note that $\mathcal{L}$ is actually independent of $h$, so the left side is zero. Thus, the right side says the time derivative of $\frac{\partial \mathcal{L}}{\partial \dot{h}}$ is zero, so this quantity is constant - it’s conserved along all solutions. Computing,

$$\begin{align} \frac{\partial \mathcal{L}}{\partial \dot{h}} &= \frac{\partial}{\partial\dot{h}}\left[m\left(\dot{\ell}^2+\dot{h}^2\cosh^2(\ell)\right)-2k\left(\ell-\ell_o\right)^2\right]\\ &= m\frac{\partial}{\partial\dot{h}}\dot{h}^2\cosh^2(\ell)\\ &=2m\cosh^2(\ell)\dot{h} \end{align}$$

Since we already know $m$ to be constant, we can simplify this and conclude

Moving to the second Euler Lagrange equation for the system, we start with the left side

$$\begin{align} \frac{\partial\mathcal{L}}{\partial \ell} &=\frac{\partial}{\partial \ell}\left[m\left(\dot{\ell}^2+\dot{h}^2\cosh^2(\ell)\right)-2k\left(\ell-\ell_o\right)^2\right]\\ &= m\frac{\partial}{\partial \ell}\dot{h}^2\cosh^2(\ell)- \frac{\partial}{\partial \ell}2k\left(\ell-\ell_o\right)^2\\ &= 2m\dot{h}^2\cosh(\ell)\sinh(\ell)-4k(\ell-\ell_o) \end{align}$$

And then compute the right:

$$\begin{align} \frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{\ell}} &= \frac{d}{dt}\frac{\partial}{\partial \dot{\ell}}\left[m\left(\dot{\ell}^2+\dot{h}^2\cosh^2(\ell)\right)-2k\left(\ell-\ell_o\right)^2\right]\\ &=\frac{d}{dt}\left[m\frac{\partial}{\partial\dot{\ell}}\dot{\ell}^2\right]\\ &=\frac{d}{dt}\left[2m\dot{\ell}\right]\\ &=2m\ddot{\ell} \end{align}$$

Equating these and dividing through by the constant $2m$ yields an equation of motion for $\ell$:

$\ddot{\ell}=\dot{h}^2\cosh(\ell)\sinh(\ell)-2\frac{k}{m}(\ell-\ell_o)$

This equation is coupled to $h$ via the appearance of $\dot{h}$, but this can be removed using the conserved quantity discovered previously. For some constant $H$ along the entire trajectory of the system, we have $\dot{h}=H/\cosh^2\ell$, and substituting this in yields

$$\begin{align} \ddot{\ell} &= \dot{h}^2\cosh(\ell)\sinh(\ell)-2\frac{k}{m}(\ell-\ell_o)\\ &=\left(\frac{H}{\cosh^2\ell}\right)^2\cosh(\ell)\sinh(\ell)-2\frac{k}{m}(\ell-\ell_o)\\ &=H^2\operatorname{sech}^2\ell\tanh\ell -2\frac{k}{m}(\ell-\ell_o) \end{align}$$

This gives a full description of the hyperbolic spring system in terms of a pair of differential equations:

Initial Conditions

As a final step, it’s useful to rewrite the constant of motion $H$ in terms of properties of the initial condition. Say the initial velocity of each mass is $v_o$, and it’s pushed from rest, at equilibrium length $2\ell_o$. Using the relation of velocity of a mass at distance $\ell$ to velocity at the center, we see that $\dot{h}(0)=v_o/\cosh(\ell_o)$. But $H=\dot{h}\cosh^2\ell$ is a constant of motion, so

$$\begin{align} H&=\dot{h}\cosh^2\ell\\ &= \dot{h}(0)\cosh^2\ell_o\\ &=\frac{v_o}{\cosh\ell_o}\cosh^2\ell_o\\ &=v_o\cosh\ell_o \end{align}$$

If we are primarily concerned with the oscillation of the spring (as opposed to how far it’s traveled along the geodesic) this reduces our entire problem to a single ordinary differential equation written explicitly in terms of its initial conditions.