Geodesic of a curved surface

Question

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.

Answer 1

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 $$g = \begin{pmatrix} 1 + 4x^2 & 4xy \\ 4xy & 1 + 4y^2 \end{pmatrix}.$$

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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 $$g = \begin{pmatrix} \langle \phi_x, \phi_x\rangle & \langle\phi_x, \phi_y\rangle \\ \langle\phi_y, \phi_x\rangle & \langle\phi_y, \phi_y\rangle \end{pmatrix},$$ 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.

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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.

Answer 2

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).

Answer 3

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$