Ytukyg

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!

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
$$
gamma(t) = sigma(t, v_0) = left(frac{cos v_0}{cosh t}, frac{sin v_0}{cosh t}, tanh tright).
$$

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.

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'^3right] 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'^3right] 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'^3right] 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'^3right] 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.