Take the 2-minute tour ×

I'm trying to read Lambourne's Relativity, Gravitation and Cosmology, but as this seems more of a maths question I've posted it here rather than in the physics forum.

The author talks about affinely parameterized geodesics and then, in Exercise 3.9, shows that the equator on the surface of a sphere is a geodesic.

My question is this:

How do you find the geodesic of a simple curved surface that isn't a sphere (for example, $z = x^2 + y^2$)?

I'm guessing that because the geodesic equation contains Christoffel symbols I would first need to find the metric for that particular surface, but I don't know how to do that.

I'm therefore guessing that my second question would need to be how do you find the metric for a simple curved surface (again for example $z = x^2 + y^2$).

Apologies if I've asked anything ridiculous.

Thank you.

share|improve this question
Do you understand the spherical case? –  M.B. Jul 23 '11 at 21:29
Also, understand that you can very rarely get an explicit formula for geodesics on a surface. I suppose you could solve the differential equations numerically to get a good approximation though. –  Vhailor Jul 23 '11 at 21:33
you solve differential equations. the simplest case would be finding the geodesics of the plane (and hope you get straight lines!) then the sphere and hyperbolic plane if you want to understand how it is done (in some cases where you can get nice solutions). you can try solving them using variational techniques (solving the euler-lagrange equations for the appropriate functional) or solving the geodesic equation $\nabla_{\dot{\gamma}}\dot{\gamma}=0$ –  yoyo Jul 24 '11 at 14:09
Thanks. I think I can follow the explanation of the plane and spherical cases (exercises 3.8 and 3.9 in Lambourne - very helpful as these are worked examples). However, I confess I can't quite see why a geodesic curve is defined as one that parallel transports its own tangent vector. –  Peter4075 Jul 24 '11 at 19:06
add comment

3 Answers

up vote 4 down vote accepted

As others have mentioned, it is usually not easy to find the geodesics of a given surface, so let me address some of your other questions.

Let's start with a simple case. If a surface $S$ in $\mathbb{R}^3$ is the graph of a function $z = f(x,y)$, then the metric $g$ can be represented in coordinates by the matrix $$g = \begin{pmatrix} 1 + (f_x)^2 & f_xf_y \\ f_xf_y & 1 + (f_y)^2 \end{pmatrix},$$ where subscripts denote partial derivatives.

In the case of $z = x^2 + y^2$, for example, we have $$g = \begin{pmatrix} 1 + 4x^2 & 4xy \\ 4xy & 1 + 4y^2 \end{pmatrix}.$$

More generally, if a surface $S$ in $\mathbb{R}^3$ is given by a local parametrization $\phi(x,y) = (\phi^1(x,y), \phi^2(x,y), \phi^3(x,y))$, then the metric $g$ can be represented by the matrix $$g = \begin{pmatrix} \langle \phi_x, \phi_x\rangle & \langle\phi_x, \phi_y\rangle \\ \langle\phi_x, \phi_x\rangle & \langle\phi_y, \phi_y\rangle \end{pmatrix},$$ where I am using the notation $\langle v, w\rangle = v \cdot w$ to denote the inner product (or "dot product") in $\mathbb{R}^3$. For example, $$\langle \phi_x, \phi_y\rangle = \langle (\phi^1_x, \phi^2_x, \phi^3_x), (\phi^1_y, \phi^2_y, \phi^3_y)\rangle = \frac{\partial\phi^1}{\partial x}\frac{\partial\phi^1}{\partial y} + \frac{\partial\phi^2}{\partial x}\frac{\partial\phi^2}{\partial y} + \frac{\partial\phi^3}{\partial x}\frac{\partial\phi^3}{\partial y}.$$ The case above where the surface is given as the graph of the function $z = f(x,y)$ is the special case where $\phi(x,y) = (x,y, f(x,y)).$

Note that the metric (tensor) was classically called the first fundamental form.

Now, I personally like to think of geodesics as unit speed curves that have zero geodesic curvature (rather than as curves which satisfy the geodesic equation).

To say that a curve $\alpha(t) = (x(t), y(t), z(t))$ is unit speed means that $|\alpha'(t)| \equiv 1$. (It is a fact that all geodesics have constant speed.) To say that a curve $\alpha$ has zero geodesic curvature means, well, that the geodesic curvature $\kappa_g \equiv 0$, where $$\kappa_g = \langle N \times \alpha', \alpha'' \rangle,$$ where $N$ is the unit normal vector to the surface. A nice apparatus for thinking about this definition is the Darboux frame.

Anyway, here are three more geometric facts that might help you in your search for geodesics, or at least your intuition for them:

Fact 1 (Corollary to Meusnier's Theorem): Let $\Pi$ be a plane that intersects a surface $S$. If at each point of intersection the plane $\Pi$ is perpendicular to the tangent plane to $S$, then the curve of intersection is a geodesic. (Such a curve is called a normal section.)

Fact 2: On a surface of revolution, every meridian is a geodesic.

Fact 3 (Clairaut's Theorem): Let $S$ be a surface of revolution, let $\alpha$ be a curve on $S$ with unit speed, let $\rho\colon S \to \mathbb{R}$ be the distance of a point of $S$ to the axis of rotation, and let $\psi$ be the angle between $\alpha'$ and the meridians of $S$.

Conclusion: If $\alpha$ is a geodesic, then $\rho \sin \psi$ is constant along $\alpha$. Conversely, if $\rho \sin\psi$ is constant along $\alpha$, and if no part of $\alpha$ is part of some parallel of $S$, then $\alpha$ is a geodesic. (cf. Clairaut's Relation.)

Remarks: I usually visualize Fact 1 by considering a right circular cylinder (try this!). Fact 2 should give you some of the geodesics on the elliptic paraboloid $z = x^2 + y^2$. Fact 3 can be used to determine the geodesics on the pseudosphere.

share|improve this answer
Jesse - thank you. There's a huge amount to take in there, but I now have the metric of my function, plus a general formula for the metric. I clicked on your link to "first fundemental form" and that looked interesting as well. –  Peter4075 Jul 29 '11 at 10:57
add comment

First of all, as Vhailor said, it is difficult to find an explicit formula for geodesics on a surface in general. This is related to the fact that by definition geodesics are solutions to a differential equation which can be quite ugly depending on the shape of the surface. In this case you can follow Vhailor's advice and solve the equation numerically using a computer.

I will not give you a complete answer for which I would need several pages, but I will try to give you some ideas. Consider also the notes by Misha Rudnev which seem to solve your question.

Concerning your second question, you are right. You need the Christoffel symbols of your surface which depends on the metric. In general, if you have an embedding of a surface into $\mathbb{R}^n$, your metric is the restriction of the Euclidean metric of $\mathbb{R}^n$. In the case of the cone $z=x^2+y^2$, such an embedding is for example given by $$ (\theta,z)\in[0,2\pi[\times\mathbb{R}^+\stackrel{\phi}{\longrightarrow}(\sqrt{z}\cdot \sin(\theta),\sqrt{z}\cdot \cos(\theta),z)\in\mathbb{R}^n $$ Now you can deduce the metric of the cone by the general formula $$ g_{c}(X,Y)=g_0\big(\phi_{*}(X),\phi_{*}(Y)\big)\;\;\;\;\;\;\;(1) $$ where $\phi_*$ denotes the differential of $\phi$, $g_c$ the metric of the cone, $g_0$ the Euclidean metric, (i.e. $g_0(e_i,e_j)=\delta_{i,j}$) and $X$ and $Y$ are tangent vectors to $[0,2\pi[\times\mathbb{R}^+$. Try to do the explicit computation as an exercise. ($\it{Hint}$: If you write a vectors $X$ in coordinates, $\phi_*(X)$ is just the Jacobian matrix of $\phi$ applied to $X=(X_1,X_2,X_3)^t$).

Now the Christoffel symbols are given in general by $$ \Gamma ^k_{ij}=\frac{1}{2}g^{kl}\big(-\partial_lg_{ij}+\partial_ig_{jl}+\partial_jg_{il}\big) $$

You see that this becomes a lot of computation if you want to solve the differential equation for geodesics afterwards, even for an "easy" shape. However, sometimes you can use tricks to find geodesics explicitely without solving any equation. For example a geodesic $\gamma:[a,b]\rightarrow M$ is a curve whose length is less then the lenght of every other curve that joins $\gamma(a)$ and $\gamma(b)$. If you haven't done it already, a good exercise is to try to apply this to the sphere in order to find geodesics. (Hint: The image of a minimal geodesic joining two points $x$ and $y$ on the sphere by the orthogonal reflection relatively to a plane cutting the sphère into two equal hemispheres and passing by $x$ and $y$ is a geodesic. More general, the image of a geodesic by a (local) isometry is a geodesic).

share|improve this answer
Julien - thank you, but I'm afraid this is way above my head. Eg, what does "restriction of the Euclidean metric" mean, and why is z = x^2 + y^2 a cone? I thought it was some sort of 3d parabola (paraboloid?). I'm happy to accept that the answer to the geodesic bit of my question is beyond me but is there a simpler explanation of how to find the metric of a curved surface? –  Peter4075 Jul 24 '11 at 19:40
Of course you are right! In my enthusiasm I mixed up the equations of the cone and the paraboloid:/ Nevertheless, the approach remains the same. –  Julien Meyer Jul 24 '11 at 21:53
"restriction of the Euclidean metric" is exactly what states formula (1). $\phi_*(X)$ and $\phi_*(Y)$ are tangent vectors to your surface and you use the Euclidean metric $g_0$ to multiply them. One can rephrase all this using differential forms. Do you know a bit about them? –  Julien Meyer Jul 24 '11 at 22:03
Julien - you're talking to someone who thinks 'Jacobian' relates to 17th century Scottish politics! Can we do this in baby steps? If I take partial derivatives of my precious surface z = x^2 + y^2 (which I can do), is there then some way to find the metric of the surface by using the Euclidean metric? I could then (theoretically at least) find the Christoffel symbols for the surface. Do I need to parametize the surface first before taking partial derivatives? Am I on the right track? Thanks for your patience. –  Peter4075 Jul 25 '11 at 14:09
add comment

How about this: for $z=F(x,y)$, a curve is given by

$$ X(t)= ( x(t), y(t), F(x(t),y(t)) )$$ The surface normal is $$ N = \left( \frac{dF}{dx}, \frac{dF}{dy}, -1 \right)$$

Want a unit speed curve:

$$ |X'|^2=1$$ Want curve acceleration parallel to the normal $ ( \frac{dF}{dx}, \frac{dF}{dy}, -1 )$, so

$$ \operatorname{cross}( X'', N )=0$$ That get's messy quickly - four non-linear PDE's. But it is fun to play with eg $ z=x^2+y^2$

share|improve this answer
add comment

Your Answer


