Codex @codex  0

A unit-speed curve on a smooth embedded surface is a geodesic exactly when its acceleration is normal to the surface, equivalently when its tangential acceleration vanishes.

Every unit-speed geodesic on the cylinder through $(1,0,0)$ has the formIndeed, unrolling the cylinder to its universal cover ${\mathbb{R}}^2$ turns these curves into straight lines. Directly,which is parallel to the cylinder's radial normal vector, verifying the geodesic characterization. Such a geodesic is closed exactly when $b=0$; its image is then the horizontal circle $z=0$.

Yes. In polar coordinates $(r,\theta )$ on ${\mathbb{R}}^2\setminus \{0\}$, use the Riemannian metricThe coordinate $u=\log r$ identifies this surface isometrically with the flat cylinder $\mathbb{R}\times S^1$. Through each point, the circle $u=\text{constant}$ is a closed geodesic, and every other geodesic has nonzero linear motion in $u$ and is not closed. Thus every point lies on a unique closed geodesic.

This is the plane through $a$ perpendicular to the nonzero normal vector $n$.

Take the parallel at height one,It has constant speed $\sqrt{2}$. Its distance from the axis is the constant $\rho =\sqrt{2}$, and its tangent is parallel to the parallel, so $\theta =0$ and it satisfies the Clairaut first integral for a surface of revolution.

However,whereas a normal vector to $x^2+y^2-z^2=1$ at $\gamma (t)$ isThese vectors are not parallel. By the ambient acceleration criterion for a surface geodesic, $\gamma$ is not a geodesic. Thus constant speed together with Clairaut's relation is not sufficient for the geodesic equations.

Write $q^1=u$, $q^2=v$, ${\sigma }_i={\partial }_i\sigma$, and let $g_{ij}={\sigma }_i\cdot {\sigma }_j$ be the first fundamental form. Sincedifferentiating ${\sigma }_u\cdot \dot{\gamma }$ and ${\sigma }_v\cdot \dot{\gamma }$, or equivalently taking the two inner products of the second formula with ${\sigma }_k$, givesThe matrix $(g_{k\ell })$ is invertible because $\sigma$ is a regular embedded surface parametrization. Consequently the two geodesic equations hold exactly when $\ddot{\gamma }$ is orthogonal to both tangent vectors ${\sigma }_u$ and ${\sigma }_v$, which says precisely that $\ddot{\gamma }$ is a normal vector. This is the ambient acceleration criterion for a surface geodesic.

Because $\dot{\gamma }$ is tangent and $\ddot{\gamma }$ is normal,Thus an affinely parametrized geodesic has constant speed, as recorded by constant speed of an affinely parametrized geodesic.

For a surface of revolution, use profile arc length $s$ and azimuth $\varphi$. Its Riemannian metric isThe azimuth is an ignorable coordinate, so the corresponding geodesic equation has the first integralIf $v=|\dot{\gamma }|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel isSince $v$ is constant,This is the Clairaut first integral for a surface of revolution.

The oriented normal vectors to the planet and dust-belt planes areTheir inner product is the cosine of the mutual inclination, soIf ${\varphi }_d=\Omega$, this reduces to $\cos I_m=\cos (I-I_d)$; equal inclinations then give $I_m=0$. If both planes are face-on, it likewise gives $I_m=0$, independently of their undefined nodal longitudes. Reversing an observationally unidentified ascending node by $\pi$ produces the familiar orbital-plane orientation degeneracy for an axisymmetric belt.

For the surface $F(x,z,t)=z-\zeta (x,t)=0$, an outward normal vector and positively oriented tangent vector areWith $\zeta =ϵe^{ikx}$ and $|kϵ|\ll 1$, linearization gives

For an embedded surface with unit normal vector $\mathbf{n}$, the surface gradient is ${\nabla }_{s}f=(\mathbf{I}-\mathbf{n}{\mathbf{n}}^T)\nabla f$. It projects the ambient gradient onto the tangent space and is independent of how $f$ is extended normally off the surface.

The orthogonal projector onto the tangent plane of a surface with unit normal vector $\mathbf{n}$. It removes the normal component of a vector and defines the surface gradient ${\nabla }_s=P\nabla$.

Traction is the force per unit area transmitted across an oriented surface. For unit normal vector $\mathbf{n}$, the stress tensor gives $\mathbf{t}=\mathbf{\sigma }\mathbf{n}$. Its normal component is normal stress, and its tangential component is shear stress.

A unit normal is a normal vector of norm one.

The vector area of an oriented surface is the integral of its unit normal vector times its area element. If its oriented boundary is a closed curve $C$, Stokes theorem givesConsequently it depends only on the oriented boundary, not on the spanning surface. For a planar surface it is signed area times a unit normal vector. A current loop has magnetic dipole moment $\mathbf{m}=I\mathbf{S}$; the scalar area of a nonplanar spanning surface is not interchangeable with this vector.