Find the geodesic and normal curvatures of a surface

Question

For the surface $\sigma(u,v)= (\frac{\cos v}{\cosh u}, \frac{\sin v}{\cosh u}, \tanh u)$, compute the geodesic curvature and the normal curvature of:

(i) a meridian $v$ = constant,

(ii) a parallel $u$ = constant.

Which of these curves are geodesics?

I know that a meridian is a geodesic and its geodesic curvature $\kappa_g = 0$.

From the textbook, the normal curvature $\kappa_n = \gamma''·N$ and the geodesic curvature $\kappa_g = \gamma'' · (N \times \gamma')$, where $N$ is the unit normal of the surface, and $\gamma$ is a unit-speed curve in $\mathbb{R}^3$.

How can I find the $\gamma$ in the formula? Thanks!

Answer 1

By keeping $u$ or $v$ constant in the parametrisation $\sigma$ of the surface and keeping the other parameter "variable", you obtain the parametrisation of the coordinate lines.

For instance, to parametrise a meridian, keeping $v$ constant gives $$ \gamma(t) = \sigma(t, v_0) = \left(\frac{\cos v_0}{\cosh t}, \frac{\sin v_0}{\cosh t}, \tanh t\right). $$

In the same way, if you keep $u$ fixed and let the other parameter of $\sigma$ be the variable of the curve, you get parametrisations of the parallels.

Answer 2

Normal and geodesic curvature of curves on a surface can be computed using the following formulae: \begin{equation} k_n=\frac{q}{s'^2} \\ k_g=\frac{p}{s'^3} \end{equation} where \begin{equation} q=L_{11} u'^2+2 u' v' L_{12}+v'^2 L_{22} \\ p=\left[\Gamma_{11}^2 u'^3+(\Gamma_{22}^2-2 \Gamma_{12}^1) u' v'^2+(u' v''-u'' v')+(2 \Gamma_{12}^2-\Gamma_{11}^1)u'^2v'-\Gamma_{22}^1 v'^3\right] \sqrt{Det(g)} \end{equation} where $\Gamma_{jk}^i$ are the Christoffel symbols of the second kind, $L_{ij}$ the coefficient of the second fundamental form and g the first fundamental form. Independently of the case u=const or v=const, the product $u' v'$ are zero since or v is constant or u is constant. Thus, the above expressions reduce to: \begin{equation} q=L_{11} u'^2+v'^2 L_{22} \\ p=\left[\Gamma_{11}^2 u'^3-\Gamma_{22}^1 v'^3\right] \sqrt{Det(g)} \end{equation} In addition $s'=\sqrt{g_{11}u'^2+2 g_{12} u'v'+g_{22}v'^2}$ reduces to $s'=\sqrt{g_{11}u'^2+g_{22}v'^2}$. We can now consider the two cases.

1. v=const

In such a case we have \begin{equation} q=L_{11} u'^2 \\ p=\left[\Gamma_{11}^2 u'^3\right] \sqrt{Det(g)}\\ s'=\sqrt{g_{11}u'^2} \end{equation}

2. u =const

In such a case we have \begin{equation} q=v'^2 L_{22} \\ p=\left[-\Gamma_{22}^1 v'^3\right] \sqrt{Det(g)} s'=\sqrt{g_{22}v'^2} \end{equation} When computing the non zero coefficient of the first fundamental form for $\sigma$, we have $g_{11}=g_{22}=\rm sech^2(u)$, while non zero coefficient of the second fundamental form are $L_{11}=L_{22}=-\rm sech^2(u)$. When replacing these coefficient in the above expression we have 1. v=const \begin{equation} k_n=-1 \\ k_g=0 \end{equation} because $\Gamma_{11}^2=0$. Since the geodesic curvature is zero, the curve is a geodesic.

1. u=const \begin{equation} k_n=-1 \\ k_g=-\rm sech{(u_0)} \end{equation} where $u_0$ is the constant. The definition of normal and geodesic curvature in your textbook can be written in term of p and q for practical computation.