# Meridian curves of surfaces of revolution are geodesics

Could someone explain how to go about proving that the meridian curves on a surface of revolution are geodesics?

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I started an answer but...See do Carmo, page 256. This is Exercise 7.8 on pages 43-44 of John Thorpe, Elementary Topics in Differential Geometry. This is written up very well in numerous places. – Will Jagy Feb 6 '12 at 22:22
In P.255 and 256 of "Differential Geometry of curves and surfaces" of do Carmo, it talks about it. You can look at it to see what you couldn't understand, and then come back to share where you get stuck. – Paul Feb 6 '12 at 22:23
@Will: You beat me by 28 secs! – Paul Feb 6 '12 at 22:23
@Paul, right, I began an answer, then looked up longitude, realized that meridians refers to lines of constant longitude, and I was talking about constant latitude. I was not interested in re-typing, when this is dealt with so well in plenty of books. I really like the Thorpe book, my first DG course was from Thorpe with these as note handouts prior to publication. – Will Jagy Feb 6 '12 at 22:52
I just got hold of the do Carmo book. I followed it until the top of page 256 where it says "Since the first fundamental form along the meridian u=const, v=v(s) yields $((f')^2+(g')^2)(v')^2=1$...". I don't understand how they got this from the first fundamental form? – 09867 Feb 8 '12 at 14:56

Hint: At any point along a geodesic, the normal of the geodesic is parallel to the normal of the surface.

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Alternative hint: A meridian is is the set of fixed points for a particular isometric transformation of the surface. Isometries preserve geodesics. At every point on a smooth surface, there is exactly one geodesic in each direction.

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Remembering Clairaut formula for geodesics after seeing how it was obtained is also useful. The unit normal to surface and unit normal in Frenet frame (without surface) are identical vectors for a running geodesic.

$r_o= r*sin \psi$ ,$\psi=0$ is the basic meridian.

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