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 $(1, 0, 0)$.


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 $\gamma$.


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

  • 4
    $\begingroup$ Hint For two geodesics, consider the planes $\Pi$ of symmetry of the hyperboloid $H$ through $(1, 0, 0)$, and use symmetry and the uniqueness of geodesics to argue that the curves $\Pi \cap H$ must be geodesic. For the other two, one can use that the hyperboloid of one sheet is doubly ruled: en.wikipedia.org/wiki/Ruled_surface . $\endgroup$ Jan 5, 2016 at 18:48
  • 2
    $\begingroup$ @ Mary Star You might consider the fact that the hyperboloid of one sheet is a ruled surface (in fact doubly ruled) and that if a surface in $\mathbb{R}^{3}$ contains a Euclidean line, then that Euclidean is necessarily a geodesic of the surface. $\endgroup$
    – THW
    Jan 5, 2016 at 18:50
  • $\begingroup$ So, is one of the four geodesics a line? @THW $\endgroup$
    – Mary Star
    Jan 8, 2016 at 22:12
  • 1
    $\begingroup$ @MaryStar Sure, I've turned my comment into an answer and included some elaboration of "symmetry and the uniqueness of geodesics" (and also of my hint for the other two geodesics) in the spoiler box in my answer. $\endgroup$ Jan 10, 2016 at 11:12
  • 1
    $\begingroup$ @MaryStar -- in fact, two of the geodesics through $(1,0,0)$ are lines. $\endgroup$
    – bubba
    Jan 10, 2016 at 15:38

4 Answers 4


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:

enter image description here

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:

enter image description here

  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:

enter image description here

  1. 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.

  2. 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.

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

  • 2
    $\begingroup$ You say that four fairly simple geodesics passing through the point $(1,0,0)$. Why there are no other geodesics? $\endgroup$ Jan 10, 2016 at 16:38
  • 1
    $\begingroup$ It is possible to compute and plot $all$ geodesics through that point using geodesic polar coordinates. $\endgroup$
    – Narasimham
    Jan 10, 2016 at 17:53
  • 3
    $\begingroup$ @DiegoFonseca -- I didn't say there were only four. The question only asks us to find four, and these four are the easiest, I think. $\endgroup$
    – bubba
    Jan 10, 2016 at 23:51
  • 2
    $\begingroup$ @bubba You are right, my question was out of curiosity. $\endgroup$ Jan 11, 2016 at 0:03
  • 1
    $\begingroup$ @DiegoFonseca -- I see. As always, there are infinitely many geodesics through the given point. Apparently Narasimham can tell you how to plot them. But that's not the topic of the OP's question. $\endgroup$
    – bubba
    Jan 11, 2016 at 0:07


Note that our surface is a surface of revolution, putting us a general context, let $S$ be a surface of revolution with parametrization $X\left(u,v\right)=\left(f\left(u\right)\cos \left(v\right),f\left(u\right)\sin \left(v\right),g\left(u\right)\right)$.

Let $\gamma$ a curve in $S$, this is, $\gamma \left(t\right)=X\left(u\left(t\right),v\left(t\right)\right)$. It is easy to see that the covariant derivative can be expressed as: \begin{eqnarray*} \frac{D\gamma'}{dt} & = & \left(u''+\Gamma_{11}^{1}\left(u'\right)^{2}+2\Gamma_{12}^{1}u'v'+\Gamma_{22}^{1}\left(v'\right)^{2}\right)X_{u}\\ & & +\left(v''+\Gamma_{11}^{2}\left(u'\right)^{2}+2\Gamma_{12}^{2}u'v'+\Gamma_{22}^{2}\left(v'\right)^{2}\right)X_{v}. \end{eqnarray*}

Where $\Gamma ^{k} _{ij}$ for $i,j,k=1,2$ are the christoffel symbols of $S$.

So, for that $\gamma$ will be a geodesic we must have that $\frac{D\gamma '}{dt}=0$, then we have the system: $\tag 1 \begin{eqnarray*} u''+\Gamma_{11}^{1}\left(u'\right)^{2}+2\Gamma_{12}^{1}u'v'+\Gamma_{22}^{1}\left(v'\right)^{2} & = & 0\\ v''+\Gamma_{11}^{2}\left(u'\right)^{2}+2\Gamma_{12}^{2}u'v'+\Gamma_{22}^{2}\left(v'\right)^{2} & = & 0. \end{eqnarray*}$

For another hand, the Christoffel symbols of $S$ are:

\begin{eqnarray*} \Gamma_{11}^{1}=0,\quad & \Gamma_{11}^{2}=-\frac{ff'}{\left(f'\right)^{2}+\left(g'\right)^{2}},\quad & \Gamma_{12}^{1}=\frac{ff'}{f^{2}}, \end{eqnarray*}

\begin{eqnarray*} \Gamma_{12}^{2}=0,\quad & \Gamma_{22}^{1}=0,\quad & \Gamma_{22}^{2}=\frac{f'f''+g'g''}{\left(f^{'}\right)^{2}+\left(g'\right)^{2}}. \end{eqnarray*}

With the values above, system (1) becomes

$$ \tag 2 \begin{array}{rrr} u''+\frac{2ff'}{f^{2}}u'v' & = & 0 \\ v''-\frac{ff'}{\left(f'\right)^{2}+\left(g'\right)^{2}}\left(u'\right)^{2} +\frac{f'f''+g'g''}{\left(f^{'}\right)^{2}+\left(g'\right)^{2}}\left(v'\right)^{2}&= & 0 \\ \end{array}$$

In the case of the one-sheeted hyperboloid we have $f\left(u\right)=\sqrt{1+u^{2}}$ and $g\left(u\right)=u$. Then, the system (2) become

$$ \begin{array}{rrr} u''+\frac{2u u'v'}{u^{2}+1} & = & 0 \\ v''-\frac{u\left(u^{2}+1\right)}{u^{2}+u+1}\left(u'\right)^{2} +\frac{u}{\left(u^{2}+1\right)\left(u^2+u+1\right)}\left(v'\right)^{2}&= & 0 \\ \end{array}$$ The geodesics are found by solving the previous system.

  • $\begingroup$ I think your computation about christoffel symbols is wrong. Or your parametrization should exchange the position of u and v. $\endgroup$
    – cxh007
    Jun 5, 2021 at 12:35

Hint For two geodesics, consider the planes $\Pi$ of symmetry of the hyperboloid $H$ through $(1,0,0)$, and use symmetry and the uniqueness of geodesics to argue that the curves $\Pi \cap H$ must be geodesic. For the other two, one can use that the hyperboloid of one sheet is doubly ruled.

Additional hint For the first two, consider a geodesic $\gamma$ through $(1, 0, 0)$ tangent to $\Pi \cap H$ at that point. By symmetry, the reflection of $\gamma$ through $\Pi$, call it $\widetilde{\gamma}$, is geodesic and it has the same tangent vector at $(1, 0, 0)$ as $\gamma$. So, by uniqueness of geodesics, $\widetilde{\gamma} = \gamma$, and in particular $\gamma$ is fixed by reflection and so is contained inside $\Pi \cap H$. For the second two, since $H$ is doubly ruled there are two straight lines through $(1, 0, 0)$ contained in $H$. In particular the constant-speed parameterizations of these lines have zero acceleration, and hence they have zero normal acceleration (regarded as curves in $H$), so they are geodesics.

  • $\begingroup$ You're welcome, Mary, I hope you found it useful. $\endgroup$ Feb 3, 2016 at 10:35

There are infinitely many geodesics on it in each direction. The meridian, the circumference at neck ( minimum radius), two ruled straight line asymptotes are the 4 principal geodesics you refer to.

Their normal curvatures follow Euler's law

$$ k_n = k_1 \cos^2 \alpha + k_2 \sin ^2 \alpha \tag{1} $$

respectively for 180 degree rotation the four $k_n's $ are minimum,0,maximum,0.. which repeat as follows at $0, 30, 90, 150, 180 ...$ degrees for curvature ratio at a point in all directions of a tangent bundle

$$ \frac{k_1}{k_2} = - \frac{1}{3} \tag{2}$$

as shown in the graph including asymptotic directions $ \alpha = \tan^{-1}\sqrt{\frac{1}{3}} = 30^{\circ} $ for the 4 important geodesics :


Four Geodesics on 1Sht_Hyp

As it is a surface of revolution differential geometrical methods lead to the Clairaut's law.

$$ r \sin \alpha = C \tag{3} $$

After a study of second fundamental form of surface theory you appreciate that the above says the same thing for geodesic curvature (in tangential plane)

$$ k_g = 0. \tag{4} $$

For lines of (principal) curvature $ k_g=0, k_n =$ minimum or maximum.

For the slant geodesic and asymptotic lines occurring in between them we have $ (k_g=0, k_n = 0). $ Skew lines shown are Bamboo reeds.

The geodesics are the meridian, pair of straight lines, central latitude circle $$(x^2-z^2=1,y=0),(x \pm z=1,y=1),(x ^2+y^2=1).$$


For getting a $ r- \theta $ relation for any start angle, combine slope and Clairaut's law relations (3), a=1, in:

$$ r^2 - z^2 = a^2 ; \tan \phi = \sqrt { (r/a)^2 -1} ; dr/ \sin \phi = r d\theta \cot \psi; \tag{5}$$ and simplify.

$$ r= r_o \sin \alpha \tag{6} $$

for any geodesic start angle $\alpha$ chosen ( need not be among the four. )

$$ (dr/d \theta)^2 = r^2 ( r^2/r_o^2-1) ((r/a)^2-1)/(2(r/a)^2-1) \tag{7} $$

Elliptic integrals may be used for closed form but faster to numerically integrate and plot.

Nature of geodesics


from WolframAlpha

Geodesics on Hyperboloids

from me

It may be instructive here to mention three types of geodesic behaviour around a hyperbolic point, we can see it neatly in the easier_to_handle surfaces of revolution:

$ r_o < a $. As given already in the sketches the geodesic shoots through from one horn to the other.

$ r_o = a $. The geodesic goes round and round but never reaches $ r = a $ which is an asymptote.

$ r_o > a $. The geodesic U-turns ahead of $ r = r_o$ . In Filament winding practice it is called a turnaround. Google images by this name if wish intuition to match mathematical formulation.

The red wire shows behavior of returning geodesic ahead of the neck of a bamboo stool, a particularly good example of our surface with its straight ruled asymptotic ruled generators.


[ Please ignore this para for time being... A plane parallel to its axis and cutting exactly at circle of min radius produces the asymptotes. It may confuse in the begining that they are geodesics, asymptotes and rulings of the ruled surface all at same time].

  • $\begingroup$ How do we find that these are geodesics on the hyperboloid of one sheet? In my book I found a proposition, according to which any normal section of a surface is a geodesic. Did you use this property to find these geodesics? $\endgroup$
    – Mary Star
    Jan 9, 2016 at 22:18
  • 1
    $\begingroup$ Omg, which book? Either you reproduced it in a wrong context or the book is making a disastrous generalization.... To appreciate it, take a cone and cut it normally at any point P. The intersection conic curve (ellipse) is nowhere geodesic except at P, only locally. ( nowhere else $k_g =0 )$ !! $\endgroup$
    – Narasimham
    Jan 10, 2016 at 17:41
  • $\begingroup$ @Narasimham -- her book (Pressley, I think) is probably correct. It probably defines a "normal section" to be a curve produced by slicing with a plane that contains the surface normal at every point of the curve. $\endgroup$
    – bubba
    Jan 11, 2016 at 12:28
  • $\begingroup$ @bubba Question is whether the slicing plane is local or global all along the geodesic curving to contain the surface normal. .. not suitable as a comprehensive definition.I like to see a scan maybe something said before in context is missing. $\endgroup$
    – Narasimham
    Jan 11, 2016 at 14:47
  • $\begingroup$ and the geodesic is an envelope of such sections at each point ? ! $\endgroup$
    – Narasimham
    Jan 11, 2016 at 14:58

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