I'm trying to show that a meridian of a surface of revolution is a geodesic, except I cannot do it without solving a system of differential equations. And, how can we determine which circles of latitute are geodesics?


up vote 3 down vote accepted

In my text, it writes a surface of revolution as ${\bf x} = (r(t)cos(\theta),r(t)sin(\theta),z(t))$. Then, the meridians are the $t-curves$, and the circles of latitude are the $\theta - curves$. We get the $t-curves$ by fixing $\theta$, and vice versa.

We know that a curve ${\bf \gamma}$ is a geodesic if its second derivative is everywhere normal to the surface along ${\bf \gamma}$. Thus, what you should do is calculate ${\bf x_1}$ and ${\bf x_2}$, the partial derivatives of the surface with respect to $t$ and $\theta$, and then show that $<{\bf \gamma} '',{\bf x_i}>=0$. This implies that $\gamma ''$ is normal to the surface, since the normal vector ${\bf n} = {\bf x_1}\times{\bf x_2}$ , and thus ${\bf \gamma}$ is a geodesic.

For the $t-curves$, you just have to prove that $<{\bf \gamma} '',{\bf x_i}>=0$.

For the $\theta-curves$, you should calculate $<{\bf \gamma} '',{\bf x_i}>$, and determine what condition would make that innerproduct $0$.

I believe the condition should be that $r'=0$, meaning that the function $r(t)$ is in a local maximum or minimum. This makes sense, as we know that the equator of a sphere is a geodesic, and the equator can be viewed as a $\theta-curve$ for a surface of revolution of a half-circle.

You should be able to do it directly, writing up the integral for the path length and noting that the path is of minimum length. More precisely, assume you revolve the graph of $y=f(x)$ about the $x$ axis, and parametrize a path on the surface as $$(x,y,z)=(x,f(x)\cos\theta(x),f(x)\sin\theta(x))$$ for some function $\theta$. I expect that arclength $ds$ will be given by $$ds^2=\bigl(1+f'(x)^2+(f(x)\theta'(x))^2\bigr)dx^2$$ (thought I am too lazy to check that carefully), and this will minimize the arc length when $\theta'$ is identically zero.

I expect you can do a similar calculation in the transverse direction, and you should find that the circles of latitude corresponding to the critical points of $f$ are geodesics. To be precise, parametrize a curve using $\theta$ as a parameter: $$(x,y,z)=(x(\theta),f(x(\theta))\cos\theta,f(x(\theta))\sin\theta)$$ where $x$ is some function, and proceed by analogy to the above. You should get $$ds^2=\bigl(f(x(\theta))^2+(1+f'(x(\theta))^2)x'(\theta)^2\bigr)d\theta^2,$$ but this is admittedly trickier to analyze. You need to calculate the variational derivative in this case.

  • The hint I found for proceeding was incorporating the idea of the normal. And I seem to be confused about the next part of the question. – mary Oct 19 '12 at 9:44
  • I added a bit on the second part. Sorry, I don't have the time for more. If you're still having problems, perhaps someone else will step in with more details. – Harald Hanche-Olsen Oct 19 '12 at 9:54

First recall that a constant speed $\gamma$ in a surface $S$ is a geodesic iff, for any part $\gamma(t)=\sigma(u(t),v(t))$ in a patch of the surface $(\sigma,U)$, we have $\frac{d}{dt}(F_I{\dot{u} \choose \dot{v}})=(\dot{u} \quad \dot{v})(\frac{1}{2}{\frac{\partial}{\partial u} \choose \frac{\partial}{\partial v}}F_I){\dot{u} \choose \dot{v}}$, where $F_I$ is the First Fundamental Form of the surface, written as a matrix. These are usually called geodesics equations.

Let $\mu=(f(t),0,g(t))$ be a curve in the (x,z)-plane (we can assume it is of constant speed), the surface of revolution about the y-axis is $\sigma(u,v)=(f(u)\cos(v),f(u)\sin(v),g(u)$. In this case, the geodesics equations becomes: (1) $\ddot{u}=(ff')\dot{v}^{2}$, (2) $\frac{d}{dt}(f^{2}\dot{v})=0$ and (3) $\dot{u}^{2}+f^{2}\dot{v}^{2}=1$.

To see that a meridian is a geodesic, we show that the line of constant $v$ is a geodesic. Say $v=c$ for some constant $c$, then $\dot{v}=0$ solving (2); and $\dot{u}=\pm1$ solving (1) and $\ddot{u}=0$ solving (3).

Moreover, a parallel (line of constant $u$) is a geodesic iff $f'(u)=0$, that is iff $f(u)$ is a maximum or a minimum.

For surfaces of revolutions about other axes, the solution is similar.

  • We cant do the system of DE. Try invoking the definition of a normal vector or parametrization. – mary Oct 19 '12 at 19:03
  • Ok, i try to give a more geometrical argument. Remember that a curve is a geodesic if its geodesic curvature is zero. Let $\mu$ be the curve described above, and we rotate it by an arbitrary angle. Now consider the plane that contains such a curve, the binormal vector $b$ is orthogonal to this plane. Moreover, at any point of $\mu$, the binormal vector $b$ and the unit normal $t$ vector lie in the tangent plane of the surface (this plane is normal to some vector $N$). Therefore the principal normal vector $n=b \times t$, so n is parallel to N. This gives the geodesic curvature to be zero. – user39280 Oct 19 '12 at 20:10
  • Your welcome, I hope the argument was clear enough. – user39280 Oct 21 '12 at 8:30

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.