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A curve is a geodesic on a surface proof

JoeG

New member
Mar 5, 2019
3
Let $\gamma(u): I \to \mathbb{R}^3$be a unit speed curve and let $\vec{b}(u)$ be its binormal vector. Consider the surface $S$ given by the surface patch $\sigma(u,v) = \gamma(u)+v\vec{b}(u)$. Show that $\gamma$ is a geodesic of $S$.

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A curve on a surface is geodesic iff it's geodesic curvature is zero everywhere, so I understand that I've to show that $\displaystyle k_g = \ddot \gamma \cdot (n \times \dot \gamma) = 0 $ (where n is the unit normal to $\sigma$), which I've trouble calculating.

I've $\sigma_u =\dot \gamma +v b'(u)= \dot \gamma -v \tau \vec{n}$ and $\sigma_v = \vec{b}$. But it doesn't seem clear what $\sigma_u \times \sigma_v$ would be.

How do I show that $k_g = 0$? Thanks.
 
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GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
256
Hi JoeG ,

Let $\gamma(u): I \to \mathbb{R}^3$be a unit speed curve and let $\vec{b}(u)$ be its binormal vector. Consider the surface $S$ given by the surface patch $\sigma(u,v) = \gamma(u)+v\vec{b}(u)$. Show that $\gamma$ is a geodesic of $S$.

---

A curve on a surface is geodesic iff it's geodesic curvature is zero everywhere, so I understand that I've to show that $\displaystyle k_g = \ddot \gamma \cdot (n \times \dot \gamma) = 0 $ (where n is the unit normal to $\sigma$), which I've trouble calculating.

I've $\sigma_u =\dot \gamma +v b'(u)= \dot \gamma -v \tau \vec{n}$ and $\sigma_v = \vec{b}$. But it doesn't seem clear what $\sigma_u \times \sigma_v$ would be.

How do I show that $k_g = 0$? Thanks.
Let's use $n$ to denote the normal to the surface and $T, N$, and $b$ to denote the Frenet frame for $\gamma.$ Doing so we have $\sigma_{u} = \dot{\gamma}-v\tau N = T-v\tau N$. From here $n=\sigma_{u}\times\sigma_{v}=T\times b - v\tau N\times b.$ Using the Frenet frame $T\times b = c_{1}N$ and $N\times b = c_{2}T,$ for some constants $c_{1}$ and $c_{2}.$ Hence $n=c_{1}N-v \tau c_{2} T.$ Using this, see if you can show the desired result. Feel free to ask any follow up questions.
 

JoeG

New member
Mar 5, 2019
3
Hi JoeG ,



Let's use $n$ to denote the normal to the surface and $T, N$, and $b$ to denote the Frenet frame for $\gamma.$ Doing so we have $\sigma_{u} = \dot{\gamma}-v\tau N = T-v\tau N$. From here $n=\sigma_{u}\times\sigma_{v}=T\times b - v\tau N\times b.$ Using the Frenet frame $T\times b = c_{1}N$ and $N\times b = c_{2}T,$ for some constants $c_{1}$ and $c_{2}.$ Hence $n=c_{1}N-v \tau c_{2} T.$ Using this, see if you can show the desired result. Feel free to ask any follow up questions.
Hi GJA,

Thanks for the answer.


This is probably wrong, but here goes

$k_g = \ddot \gamma \cdot (n \times \dot \gamma) = \ddot \gamma \cdot ((c_{1}N-v \tau c_{2} T) \times \dot \gamma)$

$= \ddot \gamma \cdot ((v \tau c_{2} T \times \dot \gamma- c_{1} N \times \dot \gamma)$

but $T = \dot \gamma $ by definition so $T \times \dot \gamma =0 $

But also $\gamma$ is a unit speed curve so $N = \ddot \gamma $

So $k_g = N\cdot (-c_1 N \times T) = 0? $
 
Last edited:

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
256
Looks good to me. Only small comment I would make is that $\ddot{\gamma}=\kappa N$ where $\kappa$ is the curvature, but this doesn't change the fact that the result is zero.
 

JoeG

New member
Mar 5, 2019
3
Looks good to me. Only small comment I would make is that $\ddot{\gamma}=\kappa N$ where $\kappa$ is the curvature, but this doesn't change the fact that the result is zero.
Thanks again.

In the solution to this problem, it's written

$\sigma_u \times \sigma_v (u, 0) = -n, k_g = \ddot \gamma \cdot \dot g \times n = 0$.

What did they do in the first equality?

Not where $g$ comes from either.
 

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
256
They used the equation we had for $\sigma_{u}\times\sigma_{v} = T\times b - v\tau N$ and set $v=0$. From there they used the Frenet frame relations to write $T\times b = -N;$ i.e., $n(u,0) = -N(u).$