[Math] How to find geodesics on a one sheet hyperboloid

Question

I am looking at the following exercise:

Describe four different geodesics on the hyperboloid of one sheet
$$x^2+y^2-z^2=1$$ passing through the point $(1, 0, 0)$.

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We have that a curve $\gamma$ on a surface $S$ is called a geodesic if $\ddot\gamma(t)$ is zero or perpendicular to the tangent plane of the surface at the point $\gamma (t)$, i.e., parallel to its unit normal, for all values of the parameter $t$.

Equivalently, $\gamma$ is a geodesic if and only if its tangent vector $\dot\gamma$ is parallel along $\gamma$.

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Could you give me some hints how we can find in this case the geodesics?

Answer 1

First, look at some pictures of hyperboloids, to get a feeling for their shape and symmetry.

There are two ways to think of your hyperboloid. Firstly, it's a surface of revolution. You can form it by drawing the hyperbola $x^2 - z^2 = 1$ in the plane $y=0$, and then rotating this around the $z$-axis.

Another way to get your hyperboloid is as a "ruled" surface. Take two circles of radius $\sqrt2$. One circle, $C_1$, lies in the plane $z=1$ and has center at the point $(0,0,1)$. The other one, $C_2$, lies in the plane $z=-1$ and has center at the point $(0,0,-1)$. As you can see, $C_1$ lies vertically above $C_2$. Their parametric equations are: \begin{align} C_1(\theta) &= (\sqrt2\cos\theta, \sqrt2\sin\theta, 1) \\ C_2(\theta) &= (\sqrt2\cos\theta, \sqrt2\sin\theta, -1) \end{align} For each $\theta$, draw a line from $C_1(\theta)$ to $C_2(\theta + \tfrac{\pi}{2})$. This gives you the family of blue lines shown in the picture below. Similarly, you can get the red lines by joining $C_1(\theta)$ and $C_2(\theta - \tfrac{\pi}{2})$ for each theta:

To identify geodesics, we will use two facts that are fairly well known (they can be found in many textbooks):

Fact #1: Any straight line lying in a surface is a geodesic. This is because its arclength parameterization will have zero second derivative.

Fact #2: Any normal section of a surface is a geodesic. A normal section is a curve produced by slicing the surface with a plane that contains the surface normal at every point of the curve. The commonest example of a normal section is a section formed by a plane of symmetry. So, any intersection with a plane of symmetry is always a geodesic.

There are infinitely many geodesics passing through the point $(1,0,0)$. But, using our two facts, we can identify four of them that are fairly simple. They are the curves G1, G2, G3, G4 shown in the picture below:

1. G1: the circle $x^2+y^2 =1$ lying in the plane $z=0$. This is a geodesic by Fact #2, since the plane $z=0$ is a plane of symmetry. At each point along the curve G1, the curve's principal normal must be parallel to the surface normal at the point, by symmetry. If this geometric argument is not convincing, we can confirm by calculations. At any point $P=(x,y,0)$ on G1, the surface normal and the curve's principal normal are both in the direction $(x,y,0)$. This is illustrated in the picture below:

2. G2: the hyperbola $x^2 - z^2 = 1$ lying in the plane $y=0$. Again, this is a geodesic by Fact #2, since the plane $y=0$ is a plane of symmetry.

3. G3: the line through the points $(1,-1,1)$ and $(1, 1, -1)$. This is one of the blue lines mentioned in the discussion of ruled surfaces above. In fact its two defining points are $(1,-1,1) = C_1\big(-\tfrac{\pi}{4}\big)$ and $(1,1,-1) = C_2\big(\tfrac{\pi}{4}\big)$. It has parametric equation $$ G_3(t) = \big(x(t),y(t),z(t)\big) = (1,t,-t) $$ To check that $G_3$ lies on the surface, we observe that $$ x(t)^2 + y(t)^2 -z(t)^2 = 1 +t^2-t^2 = 1 \quad \text{for all } t $$ It's a geodesic by Fact #1.

4. G4: the line through the points $(1,-1,-1)$ and $(1, 1, 1)$. The reasoning is the same as for G3.