Geodesics

A geodesic is the correct replacement for a straight line on a curved space. If a curve lies in ordinary Euclidean space, being straight means that its acceleration vanishes: \displaystyle \ddot{x}=0. But if the curve is constrained to lie on a curved surface, its ambient acceleration need not vanish. For example, a great circle on a sphere bends in \displaystyle \mathbb{R}^3 , but it is still intrinsically straight on the sphere. The right condition is not that the full acceleration vanish, but that its tangential component vanish. Equivalently, any acceleration that remains is only the acceleration forced by the embedding. Thus, intrinsically, a geodesic is a curve whose velocity is transported parallel to itself. In local coordinates this becomes a second-order differential equation. The cleanest way to derive it is from an action.

Let \displaystyle M be a manifold with metric \displaystyle ds^2=g_{ij}(x)dx^i dx^j. For a curve \displaystyle x^i=x^i(\lambda), define the energy functional \displaystyle E[\gamma]=\frac12\int g_{ij}(x)\dot{x}^i\dot{x}^jd\lambda. The integrand is the kinetic energy determined by the metric: \displaystyle L=\frac12 g_{ij}(x)\dot{x}^i\dot{x}^j. Geodesics are critical points of this functional, hence satisfy the Euler–Lagrange equations

\displaystyle \frac{d}{d\lambda}\left(\frac{\partial L}{\partial\dot{x}^k}\right)-\frac{\partial L}{\partial x^k}=0.

Compute \displaystyle \frac{\partial L}{\partial\dot{x}^k}=\frac12 g_{ij}\frac{\partial}{\partial\dot{x}^k}(\dot{x}^i\dot{x}^j)=\frac12 g_{ij}(\delta^i_k\dot{x}^j+\dot{x}^i\delta^j_k)=\frac12(g_{kj}\dot{x}^j+g_{ik}\dot{x}^i)=g_{kj}\dot{x}^j,
using \displaystyle g_{ik}=g_{ki}. Therefore

\displaystyle \frac{d}{d\lambda}(g_{kj}\dot{x}^j)=\frac{dg_{kj}}{d\lambda}\dot{x}^j+g_{kj}\ddot{x}^j=\partial_\ell g_{kj}\dot{x}^\ell\dot{x}^j+g_{kj}\ddot{x}^j.

Also, \displaystyle \frac{\partial L}{\partial x^k}=\frac12\partial_k g_{ij}\dot{x}^i\dot{x}^j. Substituting into Euler–Lagrange gives

\displaystyle g_{kj}\ddot{x}^j+\partial_\ell g_{kj}\dot{x}^\ell\dot{x}^j-\frac12\partial_k g_{ij}\dot{x}^i\dot{x}^j=0.

Relabel \displaystyle \ell=i and use the symmetry of \displaystyle \dot{x}^i\dot{x}^j :

\displaystyle \partial_i g_{kj}\dot{x}^i\dot{x}^j=\frac12(\partial_i g_{kj}+\partial_j g_{ki})\dot{x}^i\dot{x}^j.

Hence \displaystyle g_{kj}\ddot{x}^j+\frac12(\partial_i g_{kj}+\partial_j g_{ki}-\partial_k g_{ij})\dot{x}^i\dot{x}^j=0. Multiplying by the inverse metric \displaystyle g^{mk} gives

\displaystyle \ddot{x}^m+\frac12 g^{mk}(\partial_i g_{kj}+\partial_j g_{ki}-\partial_k g_{ij})\dot{x}^i\dot{x}^j=0.


Define the Christoffel symbols by \displaystyle \Gamma^m_{ij}=\frac12 g^{mk}(\partial_i g_{jk}+\partial_j g_{ik}-\partial_k g_{ij}). Then

\displaystyle {\ddot{x}^m+\Gamma^m_{ij}\dot{x}^i\dot{x}^j=0.}

This is the geodesic equation. The point of the computation is that the correction term is not added artificially; it is forced by the variation of the metric. In flat Cartesian coordinates, \displaystyle \partial_k g_{ij}=0, so \displaystyle \Gamma^m_{ij}=0, and the equation reduces to \displaystyle \ddot{x}^m=0. Thus the geodesic equation is simply “zero acceleration” rewritten so that it makes intrinsic sense on a curved space.

Note: We use the energy functional \displaystyle \frac12\int g_{ij}\dot x^i\dot x^jd\lambda rather than the length functional \displaystyle \int \sqrt{g_{ij}\dot x^i\dot x^j}d\lambda because it produces the same geodesic paths when the curve is parametrized at constant speed, but the Euler–Lagrange equations are much cleaner. Length measures only the geometric image of the curve; energy also measures how the curve is parametrized. Thus critical points of energy are geodesics with an affine, constant-speed parameter. In arclength \displaystyle s , where \displaystyle g_{ij}\frac{dx^i}{ds}\frac{dx^j}{ds}=1, , the energy reduces to \displaystyle E=\frac12\int ds , so it is just one half of the length.

Euclidean plane in polar coordinates

Start with the Euclidean plane \displaystyle \mathbb{R}^2 , but write it in polar coordinates \displaystyle (r,\theta) . Then \displaystyle ds^2=dr^2+r^2d\theta^2, so \displaystyle g_{rr}=1,\quad g_{\theta\theta}=r^2,\quad g_{r\theta}=0. For a curve \displaystyle r=r(\lambda),\ \theta=\theta(\lambda), the energy Lagrangian is \displaystyle L=\frac12(\dot r^2+r^2\dot\theta^2).

For \displaystyle \theta ~~, \displaystyle \frac{\partial L}{\partial\dot\theta}=r^2\dot\theta,\quad \frac{\partial L}{\partial\theta}=0. Hence \displaystyle \frac{d}{d\lambda}(r^2\dot\theta)=0, so \displaystyle r^2\dot\theta=\ell. This is conservation of angular momentum, coming from the fact that the metric does not depend on \displaystyle \theta. For \displaystyle r~~ , \displaystyle \frac{\partial L}{\partial\dot r}=\dot r,\quad \frac{\partial L}{\partial r}=r\dot\theta^2. Thus \displaystyle \ddot r-r\dot\theta^2=0. Expanding the conserved quantity gives \displaystyle \frac{d}{d\lambda}(r^2\dot\theta)=2r\dot r\dot\theta+r^2\ddot\theta=0, hence \displaystyle \ddot\theta+\frac{2}{r}\dot r\dot\theta=0.

Therefore the geodesic equations are \displaystyle {\ddot r-r\dot\theta^2=0,\qquad \ddot\theta+\frac{2}{r}\dot r\dot\theta=0.}

These are not curved-space effects. The plane is still flat; the apparent force terms come from using curved coordinates. In Cartesian coordinates, the same geodesics are ordinary straight lines. A line can be written as \displaystyle x\cos\theta_0+y\sin\theta_0=a. Since \displaystyle x=r\cos\theta,\ y=r\sin\theta, this becomes \displaystyle r\cos(\theta-\theta_0)=a. Thus even flat space can produce nontrivial-looking geodesic equations when the coordinate system itself bends. From \displaystyle {\ddot r-r\dot\theta^2=0,\quad \ddot\theta+\frac{2}{r}\dot r\dot\theta=0} , the second equation is \displaystyle \frac{d}{d\lambda}(r^2\dot\theta)=0 , hence \displaystyle r^2\dot\theta=\ell. Put \displaystyle u=1/r and use \displaystyle \theta as the parameter. Since \displaystyle \dot\theta=\ell u^2 , we get \displaystyle \dot r=-\ell u' and \displaystyle \ddot r=-\ell^2u^2u''. Also \displaystyle r\dot\theta^2=\ell^2u^3. Substituting into \displaystyle \ddot r-r\dot\theta^2=0 gives \displaystyle -\ell^2u^2u''-\ell^2u^3=0, so \displaystyle u''+u=0. Therefore \displaystyle u=A\cos\theta+B\sin\theta=C\cos(\theta-\theta_0). Since \displaystyle u=1/r , the geodesics satisfy \displaystyle {r\cos(\theta-\theta_0)=a.} This derives the straight-line form directly from the polar geodesic equations.

Round Sphere \displaystyle S^2

On the unit sphere, use colatitude \displaystyle \theta and longitude \displaystyle \phi . The metric is \displaystyle ds^2=d\theta^2+\sin^2\theta,d\phi^2, so for a curve \displaystyle \theta=\theta(\lambda),\ \phi=\phi(\lambda) the Lagrangian is \displaystyle L=\frac12(\dot\theta^2+\sin^2\theta\dot\phi^2). Since \displaystyle L is independent of \displaystyle \phi , we get the conserved quantity \displaystyle \frac{\partial L}{\partial\dot\phi}=\sin^2\theta\dot\phi=\ell. Thus \displaystyle \dot\phi=\ell/\sin^2\theta. The \displaystyle \theta equation gives \displaystyle \frac{d}{d\lambda}\dot\theta-\sin\theta\cos\theta\dot\phi^2=0, hence \displaystyle \ddot\theta-\sin\theta\cos\theta\dot\phi^2=0. Therefore the geodesic equations are \displaystyle {\ddot\theta-\sin\theta\cos\theta\dot\phi^2=0,\quad \ddot\phi+2\cot\theta\dot\theta\dot\phi=0.}

To solve them directly, use \displaystyle \phi as the parameter and set \displaystyle u=\cot\theta. From \displaystyle \sin^2\theta\dot\phi=\ell , one obtains after substitution into the \displaystyle \theta equation the simple harmonic equation \displaystyle u''+u=0, where primes mean \displaystyle d/d\phi. Hence \displaystyle u=A\cos\phi+B\sin\phi=C\cos(\phi-\phi_0). Since \displaystyle u=\cot\theta, the geodesics satisfy \displaystyle {\cot\theta=C\cos(\phi-\phi_0).} This is the spherical-coordinate equation of a great circle.

The same conclusion is clearer from the extrinsic viewpoint. Embed \displaystyle S^2\subset\mathbb{R}^3 and let \displaystyle x(s)\in S^2 be unit speed. Since \displaystyle x\cdot x=1, differentiating gives \displaystyle x\cdot\dot x=0, and differentiating again gives \displaystyle x\cdot\ddot x+\dot x\cdot\dot x=0. For unit speed, \displaystyle \dot x\cdot\dot x=1, so \displaystyle x\cdot\ddot x=-1. A geodesic has no tangential acceleration, so its acceleration is purely normal to the sphere. The normal direction is \displaystyle x, hence \displaystyle \ddot x=-x. Therefore \displaystyle x(s)=a\cos s+b\sin s, with \displaystyle a,b\in\mathbb{R}^3 orthonormal. The curve lies in the plane spanned by \displaystyle a,b , and that plane passes through the origin. Thus geodesics on the round sphere are exactly great circles.

Upper Half Plane

Now take the upper half-plane \displaystyle H^2=\{(x,y):y>0\} with metric \displaystyle ds^2=\frac{dx^2+dy^2}{y^2}. The Lagrangian is \displaystyle L=\frac12\frac{\dot x^2+\dot y^2}{y^2}.

Since \displaystyle L is independent of \displaystyle x , we get a conserved quantity: \displaystyle \frac{\partial L}{\partial \dot x}=\frac{\dot x}{y^2}=a, so \displaystyle \dot x=ay^2. Assume unit speed, so \displaystyle \frac{\dot x^2+\dot y^2}{y^2}=1. Substituting \displaystyle \dot x=ay^2 gives \displaystyle a^2y^2+\frac{\dot y^2}{y^2}=1, hence \displaystyle \dot y^2=y^2-a^2y^4=y^2(1-a^2y^2). Therefore \displaystyle \frac{dx}{dy}=\frac{\dot x}{\dot y}=\pm \frac{ay}{\sqrt{1-a^2y^2}}. Integrating, \displaystyle x-c=\mp \frac1a\sqrt{1-a^2y^2}. Squaring gives \displaystyle (x-c)^2=\frac1{a^2}(1-a^2y^2), so \displaystyle {(x-c)^2+y^2=\frac1{a^2}.}

Thus the nonvertical geodesics are Euclidean semicircles centered on the boundary line \displaystyle y=0 , hence orthogonal to it. The remaining case is \displaystyle a=0 , which gives \displaystyle \dot x=0 , so \displaystyle x=\text{constant}. These are vertical lines. Therefore the geodesics in the upper half-plane are exactly vertical lines and semicircles orthogonal to the boundary.

Poincare Disk

Use polar coordinates on the disk. Since \displaystyle dx^2+dy^2=dr^2+r^2d\theta^2 , the Poincaré metric is \displaystyle ds^2=\frac{4}{(1-r^2)^2}(dr^2+r^2d\theta^2). Thus the energy Lagrangian is \displaystyle L=\frac12\frac{4}{(1-r^2)^2}(\dot r^2+r^2\dot\theta^2). The coordinate \displaystyle \theta is cyclic, so \displaystyle \ell=\frac{\partial L}{\partial\dot\theta}=\frac{4r^2\dot\theta}{(1-r^2)^2} is conserved. For unit speed, \displaystyle \frac{4}{(1-r^2)^2}(\dot r^2+r^2\dot\theta^2)=1. Using \displaystyle \dot\theta=\frac{\ell(1-r^2)^2}{4r^2} and writing \displaystyle r'=\frac{dr}{d\theta} , this becomes \displaystyle (r')^2=\frac{4r^4}{\ell^2(1-r^2)^2}-r^2.

The useful substitution is \displaystyle u=\frac{1+r^2}{2r}. Then \displaystyle u'=\frac{r^2-1}{2r^2}r', so \displaystyle (u')^2=\frac1{\ell^2}-\frac{(1-r^2)^2}{4r^2}. But \displaystyle u^2-1=\frac{(1-r^2)^2}{4r^2}, hence \displaystyle (u')^2=\left(1+\frac1{\ell^2}\right)-u^2. Differentiating gives \displaystyle u''+u=0. Therefore \displaystyle u=A\cos(\theta-\theta_0). Returning to \displaystyle r , \displaystyle \frac{1+r^2}{2r}=A\cos(\theta-\theta_0), or equivalently \displaystyle r^2-2Ar\cos(\theta-\theta_0)+1=0.

This is the equation of a Euclidean circle. A circle centered on the ray \displaystyle \theta=\theta_0 with center distance \displaystyle A and radius \displaystyle \sqrt{A^2-1} has precisely this equation. Since \displaystyle A^2=1+(A^2-1), it meets the unit boundary circle \displaystyle r=1 orthogonally. The exceptional case \displaystyle \ell=0 gives \displaystyle \dot\theta=0 , hence radial diameters. Thus the geodesics of the Poincaré disk are precisely Euclidean circles and lines orthogonal to \displaystyle |z|=1.

Cylinder

Take the cylinder of radius \displaystyle R parametrized by \displaystyle X(\theta,z)=(R\cos\theta,R\sin\theta,z). Its induced metric is \displaystyle ds^2=R^2d\theta^2+dz^2. Hence the geodesic Lagrangian is \displaystyle L=\frac12(R^2\dot\theta^2+\dot z^2). Both coordinates are cyclic, so \displaystyle R^2\dot\theta=a,\quad \dot z=b. Therefore \displaystyle \theta(\lambda)=\theta_0+\frac{a}{R^2}\lambda,\quad z(\lambda)=z_0+b\lambda.

Putting \displaystyle \omega=\frac{a}{R^2} , the geodesics are \displaystyle X(\lambda)=(R\cos(\theta_0+\omega\lambda),R\sin(\theta_0+\omega\lambda),z_0+b\lambda). Thus the geodesics are helices. The special cases are also included: if \displaystyle b=0 , we get horizontal circles; if \displaystyle \omega=0 , we get vertical straight lines; if both are nonzero, we get genuine helices.

If the cylinder is cut and unrolled into a plane, these geodesics become ordinary straight lines. The reason is that the cylinder is intrinsically flat. Its metric has constant coefficients, so the geodesic equations are just \displaystyle \ddot\theta=0,\quad \ddot z=0.

Surfaces of Revolution

Consider a surface of revolution \displaystyle X(u,v)=(f(u)\cos v,f(u)\sin v,h(u)). If \displaystyle u is arclength along the generating curve, then \displaystyle f'(u)^2+h'(u)^2=1, and the induced metric is \displaystyle ds^2=du^2+f(u)^2dv^2. Hence \displaystyle L=\frac12(\dot u^2+f(u)^2\dot v^2). Since \displaystyle v is cyclic,
\displaystyle \ell=\frac{\partial L}{\partial\dot v}=f(u)^2\dot v is conserved. For unit speed, \displaystyle \dot u^2+f(u)^2\dot v^2=1. Substituting \displaystyle \dot v=\ell/f(u)^2 gives \displaystyle \dot u^2=1-\frac{\ell^2}{f(u)^2}. Thus \displaystyle \frac{du}{dv}=\frac{\dot u}{\dot v}=\pm\frac{f(u)^2}{\ell}\sqrt{1-\frac{\ell^2}{f(u)^2}}. Geometrically, if \displaystyle \alpha is the angle the geodesic makes with a meridian, then the angular component of unit velocity is \displaystyle f(u)\dot v, so \displaystyle \sin\alpha=f(u)\dot v. Therefore \displaystyle f(u)\sin\alpha=f(u)^2\dot v=\ell. This is Clairaut’s relation: along a geodesic on a surface of revolution, \displaystyle f(u)\sin\alpha is constant.

For the catenoid, \displaystyle X(u,v)=(a\cosh(u/a)\cos v,\ a\cosh(u/a)\sin v,\ u).
Its metric is \displaystyle ds^2=\cosh^2(u/a)(du^2+a^2dv^2). Hence \displaystyle L=\frac12\cosh^2(u/a)(\dot u^2+a^2\dot v^2). Again \displaystyle v is cyclic, so \displaystyle \ell=\frac{\partial L}{\partial\dot v}=a^2\cosh^2(u/a)\dot v. For arclength parameter, \displaystyle \cosh^2(u/a)(\dot u^2+a^2\dot v^2)=1. Using \displaystyle \dot v=\ell/(a^2\cosh^2(u/a)) , we get \displaystyle \cosh^2(u/a)\dot u^2+\frac{\ell^2}{a^2\cosh^2(u/a)}=1. Thus \displaystyle \dot u^2=\frac{1}{\cosh^2(u/a)}\left(1-\frac{\ell^2}{a^2\cosh^2(u/a)}\right). Therefore the geodesic shape is given by the quadrature
\displaystyle \frac{du}{dv}=\pm\frac{a^2\cosh(u/a)}{\ell}\sqrt{1-\frac{\ell^2}{a^2\cosh^2(u/a)}}.
This is the catenoid version of Clairaut’s relation. Since the radius of the parallel circle is \displaystyle a\cosh(u/a), the conserved quantity is \displaystyle a\cosh(u/a)\sin\alpha=\ell.

Lie Groups

A Lie group is a space whose points are transformations, and where transformations can be multiplied. Examples are translations, rotations, scalings, and rigid motions. A curve \displaystyle g(t) in a Lie group is therefore a continuously moving transformation. To measure the velocity of \displaystyle g(t) , we move the tangent vector \displaystyle \dot g(t) back to the identity of the group. This gives the body velocity \displaystyle \xi(t)=g(t)^{-1}\dot g(t). So instead of describing velocity at every different point \displaystyle g(t) , we describe all velocities inside one fixed vector space: the tangent space at the identity, called the Lie algebra. A left-invariant metric means that the kinetic energy of a velocity does not depend on where the group element is; it depends only on the body velocity \displaystyle \xi. Thus the Lagrangian has the form \displaystyle L=L(\xi). For a quadratic metric, choose coordinates \displaystyle \xi=(\xi^1,\ldots,\xi^n). Then the most general kinetic energy is \displaystyle L(\xi)=\frac12\sum_{i,j}G_{ij}\xi^i\xi^j, where \displaystyle G_{ij} is a constant symmetric positive matrix. The momentum variables are \displaystyle p_i=\frac{\partial L}{\partial \xi^i} =\sum_j G_{ij}\xi^j. The only extra ingredient is that the group multiplication may be noncommutative. Infinitesimally this is recorded by constants \displaystyle c^k_{ij} , defined by \displaystyle [e_i,e_j]=\sum_k c^k_{ij}e_k, where \displaystyle e_1,\ldots,e_n is a basis for the Lie algebra. These constants measure how the basic infinitesimal motions fail to commute. For a left-invariant metric, the geodesic equations reduce to the following Euler–Lagrange form on the variables \displaystyle \xi^i and \displaystyle p_i : \displaystyle \dot p_k=\sum_{i,j}c^j_{ki}\xi^i p_j. Together with \displaystyle p_i=\sum_jG_{ij}\xi^j, this gives a closed system of ordinary differential equations for the body velocity.

So the practical recipe is: Choose coordinates \displaystyle \xi^i for velocity at the identity. Write the kinetic energy \displaystyle L=\frac12G_{ij}\xi^i\xi^j. Compute \displaystyle p_i=\partial L/\partial\xi^i. Use the commutation rules \displaystyle [e_i,e_j]=c^k_{ij}e_k to get \displaystyle \dot p_k=\sum_{i,j}c^j_{ki}\xi^i p_j. Finally reconstruct the actual curve from \displaystyle \dot g=g\xi. This is the Lie-group version of the geodesic equation. In ordinary coordinates, geodesics satisfy \displaystyle \ddot x^m+\Gamma^m_{ij}\dot x^i\dot x^j=0. On a Lie group with a left-invariant metric, the same equation becomes an equation for the moving-frame velocity \displaystyle \xi(t). The geometry is the same; only the coordinates have changed.

Translations:

The simplest Lie group is translation space \displaystyle \mathbb R^n. A point \displaystyle q\in\mathbb R^n represents the translation \displaystyle x\mapsto x+q. Since translations commute, there is no moving-frame correction: the body velocity is just the ordinary velocity \displaystyle \dot q. With constant metric matrix \displaystyle M , the Lagrangian is \displaystyle L=\frac12\dot q^T M\dot q. Euler–Lagrange gives \displaystyle \frac{d}{dt}\left(\frac{\partial L}{\partial\dot q}\right)-\frac{\partial L}{\partial q}=0. Since \displaystyle L does not depend on \displaystyle q , and \displaystyle \frac{\partial L}{\partial\dot q}=M\dot q, we get \displaystyle \frac{d}{dt}(M\dot q)=0. Hence \displaystyle p=M\dot q is constant, so \displaystyle \dot q=M^{-1}p=v is constant, and therefore \displaystyle q(t)=q_0+vt. Thus geodesics in the translation group are straight lines: momentum is conserved because the kinetic energy is the same at every point.

Rotations in the Plane:

A rotation in the plane is described by a single angle \displaystyle \theta(t). Thus \displaystyle SO(2) is a one-dimensional configuration space, and its natural kinetic energy has the form \displaystyle L=\frac12 I\dot\theta^2, where \displaystyle I>0 is a constant weight for angular motion. Euler–Lagrange gives \displaystyle \frac{d}{dt}\left(\frac{\partial L}{\partial\dot\theta}\right)-\frac{\partial L}{\partial\theta}=0. Since \displaystyle \frac{\partial L}{\partial\dot\theta}=I\dot\theta and \displaystyle \frac{\partial L}{\partial\theta}=0, we obtain \displaystyle \frac{d}{dt}(I\dot\theta)=0. Hence \displaystyle I\dot\theta=\ell is conserved, so \displaystyle \dot\theta=\ell/I=\omega is constant. Therefore \displaystyle \theta(t)=\theta_0+\omega t. In matrix form, the geodesic is \displaystyle R(t)=\begin{pmatrix}\cos(\theta_0+\omega t)&-\sin(\theta_0+\omega t)\\ \sin(\theta_0+\omega t)&\cos(\theta_0+\omega t)\end{pmatrix}. Thus geodesics on \displaystyle SO(2) are uniform rotations: the angular momentum is conserved because the kinetic energy does not depend on the angle itself.

Rotations in three dimensions

For rotations in space, the configuration is a matrix \displaystyle R(t)\in SO(3) , so \displaystyle R^TR=I and \displaystyle \det R=1. A curve \displaystyle R(t) has velocity \displaystyle \dot R(t) , and left translation brings this velocity back to the identity by \displaystyle \Omega=R^{-1}\dot R=R^T\dot R. . Define the body angular velocity by \displaystyle \Omega=R^{-1}\dot R=R^T\dot R. Differentiating \displaystyle R^TR=I gives \displaystyle \dot R^TR+R^T\dot R=0, so \displaystyle \Omega is skew-symmetric. Hence \displaystyle \Omega=\begin{pmatrix}0&-\omega_3&\omega_2\\ \omega_3&0&-\omega_1\\ -\omega_2&\omega_1&0\end{pmatrix}, which we identify with \displaystyle \omega=(\omega_1,\omega_2,\omega_3). Write \displaystyle \Omega=\omega_iE_i, where \displaystyle [E_j,E_k]=\varepsilon_{jki}E_i. Choose local coordinates \displaystyle q=(q^1,q^2,q^3) on \displaystyle SO(3). Since \displaystyle R=R(q), \displaystyle \dot R=\frac{\partial R}{\partial q^a}\dot q^a. Hence \displaystyle \Omega=R^{-1}\dot R=R^{-1}\frac{\partial R}{\partial q^a}\dot q^a. Write \displaystyle R^{-1}\frac{\partial R}{\partial q^a}=A_{ia}(q)E_i. Then \displaystyle \omega_i=A_{ia}(q)\dot q^a.

The kinetic-energy Lagrangian is \displaystyle L(q,\dot q)=\frac12(I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2). Define \displaystyle M_i=\frac{\partial L}{\partial\omega_i}=I_i\omega_i. Since \displaystyle \omega_i=A_{ia}\dot q^a, \displaystyle \frac{\partial L}{\partial\dot q^a}=M_iA_{ia}. Therefore \displaystyle \frac{d}{dt}\left(\frac{\partial L}{\partial\dot q^a}\right)=\dot M_iA_{ia}+M_i\frac{\partial A_{ia}}{\partial q^b}\dot q^b. Also, \displaystyle \frac{\partial L}{\partial q^a}=M_i\frac{\partial A_{ib}}{\partial q^a}\dot q^b.

Substituting into Euler–Lagrange gives \displaystyle \dot M_iA_{ia}+M_i\left(\frac{\partial A_{ia}}{\partial q^b}-\frac{\partial A_{ib}}{\partial q^a}\right)\dot q^b=0. Now set \displaystyle B_a=R^{-1}\frac{\partial R}{\partial q^a}=A_{ia}E_i. These matrices satisfy the identity \displaystyle \frac{\partial B_a}{\partial q^b}-\frac{\partial B_b}{\partial q^a}=-[B_a,B_b]. Thus \displaystyle \frac{\partial A_{ia}}{\partial q^b}-\frac{\partial A_{ib}}{\partial q^a}=-\varepsilon_{jki}A_{ja}A_{kb}. Hence \displaystyle \dot M_iA_{ia}-M_i\varepsilon_{jki}A_{ja}A_{kb}\dot q^b=0. Since \displaystyle A_{kb}\dot q^b=\omega_k, this becomes \displaystyle A_{ja}\left(\dot M_j-M_i\varepsilon_{jki}\omega_k\right)=0. Because \displaystyle A_{ja} is invertible, \displaystyle \dot M_j=M_i\varepsilon_{jki}\omega_k. This is exactly \displaystyle \dot M=M\times\omega.

Finally, with \displaystyle M=(I_1\omega_1,I_2\omega_2,I_3\omega_3), \displaystyle M\times\omega=((I_2-I_3)\omega_2\omega_3,\ (I_3-I_1)\omega_3\omega_1,\ (I_1-I_2)\omega_1\omega_2). Therefore \displaystyle I_1\dot\omega_1=(I_2-I_3)\omega_2\omega_3,\qquad I_2\dot\omega_2=(I_3-I_1)\omega_3\omega_1,\qquad I_3\dot\omega_3=(I_1-I_2)\omega_1\omega_2. These are Euler’s equations, derived directly from the ordinary Euler–Lagrange equations in local coordinates on \displaystyle SO(3).

Planar Motions:

For planar rigid motions, the group is \displaystyle SE(2) . A configuration is \displaystyle (x,y,\theta), where \displaystyle (x,y) is position and \displaystyle \theta is rotation angle. Use velocity variables \displaystyle \xi=(\omega,u,v), where \displaystyle \omega is angular velocity and \displaystyle u,v are translation velocities measured in the moving coordinates. Take the left-invariant kinetic energy \displaystyle L=\frac12(I\omega^2+mu^2+mv^2). Then the momenta are \displaystyle M=\frac{\partial L}{\partial\omega}=I\omega,\quad p_u=\frac{\partial L}{\partial u}=mu,\quad p_v=\frac{\partial L}{\partial v}=mv.

For \displaystyle SE(2) , the Euler–Lagrange equations in these velocity variables are
\displaystyle \dot M=u p_v-v p_u,\qquad \dot p_u=\omega p_v,\qquad \dot p_v=-\omega p_u.
Substituting \displaystyle p_u=mu,\ p_v=mv gives \displaystyle \dot M=umv-vmu=0, so \displaystyle I\dot\omega=0 , hence \displaystyle \dot\omega=0. The other two equations become \displaystyle m\dot u=\omega mv,\qquad m\dot v=-\omega mu, so \displaystyle \dot u=\omega v,\qquad \dot v=-\omega u. Thus \displaystyle \omega is constant, while \displaystyle (u,v) rotates with angular speed \displaystyle \omega. If \displaystyle \omega=0, then \displaystyle u,v are constant and the motion is a straight translation. If \displaystyle \omega\neq0, then \displaystyle u(t)=A\cos(\omega t)+B\sin(\omega t),\qquad v(t)=-A\sin(\omega t)+B\cos(\omega t). The actual group curve is recovered from \displaystyle \dot\theta=\omega,\qquad \dot x=u\cos\theta-v\sin\theta,\qquad \dot y=u\sin\theta+v\cos\theta. So the geodesics of this left-invariant metric on \displaystyle SE(2) are planar rigid motions with constant angular speed and rotating translation velocity.

Heisenberg Group

For the Heisenberg group, use velocity variables \displaystyle \xi=uX+vY+wZ with the only nonzero bracket \displaystyle [X,Y]=Z. Take the left-invariant kinetic energy \displaystyle L=\frac12(u^2+v^2+w^2). Then the momenta are \displaystyle p_u=\frac{\partial L}{\partial u}=u,\quad p_v=\frac{\partial L}{\partial v}=v,\quad p_w=\frac{\partial L}{\partial w}=w.

The Euler–Lagrange equations in these left-invariant velocity variables are determined by the bracket. Since \displaystyle [X,Y]=Z , the only coupling is between \displaystyle u,v and the central momentum \displaystyle w. The equations are \displaystyle \dot u=-vw,\qquad \dot v=uw,\qquad \dot w=0. Thus \displaystyle w is constant. The first two equations say that \displaystyle (u,v) rotates with angular speed \displaystyle w: \displaystyle u(t)=A\cos(wt)+B\sin(wt),\qquad v(t)=-A\sin(wt)+B\cos(wt). The energy is conserved: \displaystyle \frac{d}{dt}(u^2+v^2+w^2)=2u(-vw)+2v(uw)+0=0. So the speed stays fixed, but the direction of the horizontal velocity turns. This turning is not caused by an external force; it comes from the noncommutativity \displaystyle [X,Y]=Z.

SL_2

For \displaystyle SL(2,\mathbb R) , the group consists of \displaystyle 2\times2 real matrices with determinant \displaystyle 1. Its infinitesimal velocities are traceless matrices. Use the standard basis
\displaystyle H=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad E=\begin{pmatrix}0&1\\0&0\end{pmatrix},\quad F=\begin{pmatrix}0&0\\1&0\end{pmatrix}.
Write the velocity as \displaystyle \xi=aH+bE+cF. The commutation rules are \displaystyle [H,E]=2E,\quad [H,F]=-2F,\quad [E,F]=H. Take the left-invariant kinetic energy \displaystyle L=\frac12(Aa^2+Bb^2+Cc^2). Then the momenta are \displaystyle p_a=Aa,\quad p_b=Bb,\quad p_c=Cc. The Euler–Lagrange equations in left-invariant velocity coordinates give
\displaystyle \dot p_a=bp_c-cp_b,\qquad \dot p_b=-2ap_b+cp_a,\qquad \dot p_c=2ap_c-bp_a.
Substituting \displaystyle p_a=Aa,\ p_b=Bb,\ p_c=Cc gives
\displaystyle A\dot a=(C-B)bc,\qquad B\dot b=-2AB,ab+AC,ac,\qquad C\dot c=2AC,ac-AB,ab.
More cleanly, \displaystyle A\dot a=(C-B)bc,\qquad \dot b=-2ab+\frac{A}{B}ac,\qquad \dot c=2ac-\frac{A}{C}ab. After solving for \displaystyle a,b,c , the group curve is recovered from \displaystyle \dot g=g\xi. Thus geodesics on \displaystyle SL(2,\mathbb R) become a three-dimensional nonlinear system reflecting the noncommutativity of the generators \displaystyle H,E,F.

Sp_2

For the symplectic group \displaystyle Sp(2,\mathbb R) , one has \displaystyle Sp(2,\mathbb R)\cong SL(2,\mathbb R). Indeed, in dimension two, preserving area is the same as preserving the standard symplectic form. Thus the same computation applies. A matrix \displaystyle g\in Sp(2,\mathbb R) satisfies \displaystyle g^T Jg=J, where \displaystyle J=\begin{pmatrix}0&1\\-1&0\end{pmatrix}. The velocity \displaystyle \xi=g^{-1}\dot g satisfies \displaystyle \xi^T J+J\xi=0, which forces \displaystyle \xi to be traceless. Therefore \displaystyle \xi=aH+bE+cF , and the same equations above describe its geodesics.

For the higher symplectic group \displaystyle Sp(2n,\mathbb R) , the definition is \displaystyle g^T Jg=J. Differentiating gives the Lie algebra condition \displaystyle \xi^T J+J\xi=0. A convenient way to write such velocities is \displaystyle \xi=JK, where \displaystyle K=K^T is symmetric. Choose a left-invariant kinetic energy such as \displaystyle L=\frac{1}{2}{tr}(K^2). Then the geodesic equation has the same form: compute the momentum from \displaystyle L , use the commutator \displaystyle [\xi,\eta]=\xi\eta-\eta\xi, and evolve the momentum by \displaystyle \dot M=[M,\xi]
for the standard trace pairing. The curve is then reconstructed from \displaystyle \dot g=g\xi.


Schwarzschild Spacetime

The Schwarzschild metric is \displaystyle ds^2=-\left(1-\frac{2M}{r}\right)dt^2+\left(1-\frac{2M}{r}\right)^{-1}dr^2+r^2d\Omega^2. By spherical symmetry, restrict to the equatorial plane \displaystyle \theta=\pi/2. Write \displaystyle A(r)=1-\frac{2M}{r}. Then \displaystyle ds^2=-Adt^2+A^{-1}dr^2+r^2d\phi^2, and the geodesic Lagrangian is \displaystyle L=\frac12\left(-A\dot t^2+A^{-1}\dot r^2+r^2\dot\phi^2\right).

The coordinates \displaystyle t and \displaystyle \phi are cyclic, so their momenta are conserved: \displaystyle \frac{\partial L}{\partial\dot t}=-A\dot t=-E, hence \displaystyle \dot t=\frac{E}{A}; and \displaystyle \frac{\partial L}{\partial\dot\phi}=r^2\dot\phi=\ell, hence \displaystyle \dot\phi=\frac{\ell}{r^2}. For timelike geodesics, normalize \displaystyle g_{\mu\nu}\dot x^\mu\dot x^\nu=-1. Thus \displaystyle -A\dot t^2+A^{-1}\dot r^2+r^2\dot\phi^2=-1. Substituting the conserved quantities gives \displaystyle -\frac{E^2}{A}+\frac{\dot r^2}{A}+\frac{\ell^2}{r^2}=-1. Multiplying by \displaystyle A gives \displaystyle -E^2+\dot r^2+A\frac{\ell^2}{r^2}=-A, so \displaystyle \dot r^2=E^2-A\left(1+\frac{\ell^2}{r^2}\right). Hence \displaystyle \dot r^2+V_{\mathrm{eff}}(r)=E^2, \qquad V_{\mathrm{eff}}(r)=\left(1-\frac{2M}{r}\right)\left(1+\frac{\ell^2}{r^2}\right). This reduces the geodesic problem to one-dimensional radial motion in an effective potential.

For null geodesics, the normalization is instead \displaystyle g_{\mu\nu}\dot x^\mu\dot x^\nu=0. The same conserved quantities give \displaystyle -\frac{E^2}{A}+\frac{\dot r^2}{A}+\frac{\ell^2}{r^2}=0. Multiplying by \displaystyle A , \displaystyle \dot r^2=E^2-A\frac{\ell^2}{r^2}. Thus \displaystyle \dot r^2+V_{\mathrm{null}}(r)=E^2, \qquad V_{\mathrm{null}}(r)=\left(1-\frac{2M}{r}\right)\frac{\ell^2}{r^2}. If \displaystyle \ell=0 , the motion is radial. If \displaystyle \ell\neq0 , the angular equation \displaystyle \dot\phi=\ell/r^2 couples radial motion to bending around the black hole.

A circular null orbit has constant \displaystyle r , so it must sit at a critical point of \displaystyle V_{\mathrm{null}}. Since \displaystyle V_{\mathrm{null}}=\ell^2\left(\frac1{r^2}-\frac{2M}{r^3}\right), we compute \displaystyle V_{\mathrm{null}}'=\ell^2\left(-\frac2{r^3}+\frac{6M}{r^4}\right). Setting this equal to zero gives \displaystyle -2r+6M=0, hence \displaystyle r=3M. Therefore Schwarzschild spacetime has a circular photon orbit at \displaystyle r=3M , the photon sphere.

de Sitter Spacetime

In \displaystyle 1+1 dimensions, take \displaystyle ds^2=-dt^2+\cosh^2(Ht)d\chi^2. The Lagrangian is \displaystyle L=\frac12(-\dot t^2+\cosh^2(Ht)\dot\chi^2). Since \displaystyle \chi is cyclic, \displaystyle p=\frac{\partial L}{\partial\dot\chi}=\cosh^2(Ht)\dot\chi is conserved, so \displaystyle \dot\chi=\frac{p}{\cosh^2(Ht)}.

The \displaystyle t equation is \displaystyle \frac{d}{d\lambda}(-\dot t)-H\cosh(Ht)\sinh(Ht)\dot\chi^2=0, hence \displaystyle \ddot t+H\cosh(Ht)\sinh(Ht)\dot\chi^2=0. Substituting \displaystyle \dot\chi=\frac{p}{\cosh^2(Ht)} gives \displaystyle \ddot t+\frac{Hp^2\sinh(Ht)}{\cosh^3(Ht)}=0.

For timelike geodesics, choose proper time so \displaystyle -\dot t^2+\cosh^2(Ht)\dot\chi^2=-1. Then \displaystyle -\dot t^2+\frac{p^2}{\cosh^2(Ht)}=-1, so \displaystyle \dot t^2=1+\frac{p^2}{\cosh^2(Ht)}. Therefore \displaystyle \frac{d\chi}{dt}=\frac{\dot\chi}{\dot t}=\frac{p/\cosh^2(Ht)}{\sqrt{1+p^2/\cosh^2(Ht)}}. Thus as \displaystyle |t| grows, \displaystyle \cosh(Ht) grows and \displaystyle \dot\chi decays: freely moving particles become nearly fixed in the expanding coordinate \displaystyle \chi.

For null geodesics, \displaystyle -\dot t^2+\cosh^2(Ht)\dot\chi^2=0, so \displaystyle \frac{d\chi}{dt}=\pm\frac1{\cosh(Ht)}. Hence \displaystyle \chi-\chi_0=\pm\int\frac{dt}{\cosh(Ht)}=\pm\frac{2}{H}\arctan(e^{Ht})+C. Light travels only a finite coordinate distance as \displaystyle t\to\infty , reflecting the de Sitter horizon.

Anti-de Sitter Spacetime

In global \displaystyle 1+1 anti-de Sitter spacetime, take \displaystyle ds^2=-\left(1+\frac{r^2}{L^2}\right)dt^2+\left(1+\frac{r^2}{L^2}\right)^{-1}dr^2. Put \displaystyle A(r)=1+\frac{r^2}{L^2}. Then the Lagrangian is \displaystyle L_{\mathrm{geo}}=\frac12(-A\dot t^2+A^{-1}\dot r^2). Since \displaystyle t is cyclic, \displaystyle \frac{\partial L_{\mathrm{geo}}}{\partial\dot t}=-A\dot t=-E, so \displaystyle A\dot t=E.

For timelike geodesics, choose proper time \displaystyle \tau , so \displaystyle -A\dot t^2+A^{-1}\dot r^2=-1. Substituting \displaystyle \dot t=E/A gives \displaystyle -\frac{E^2}{A}+\frac{\dot r^2}{A}=-1. Multiplying by \displaystyle A , \displaystyle -E^2+\dot r^2=-A, hence \displaystyle \dot r^2=E^2-1-\frac{r^2}{L^2}. Therefore \displaystyle \dot r^2+\frac{r^2}{L^2}=E^2-1.

Differentiate this first integral with respect to \displaystyle \tau : \displaystyle 2\dot r\ddot r+\frac{2r\dot r}{L^2}=0. Hence, away from turning points, \displaystyle \ddot r+\frac1{L^2}r=0, and by continuity this holds everywhere. Thus \displaystyle r(\tau)=R\sin\left(\frac{\tau}{L}+\delta\right),\qquad R^2=E^2-1. So timelike geodesics in global AdS oscillate with frequency \displaystyle 1/L.

This is the key contrast with de Sitter: de Sitter expansion tends to separate freely moving particles in the spatial coordinate, while global anti-de Sitter geometry acts like a confining potential. Timelike geodesics fall inward and return periodically rather than escaping to infinity.

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