The derivative of paremeterized curve $r(t)$ gives the attached tangent vector at that point , if $\dot{r}(t)=\vec t$ is further differentiated , $r''(t)=\dot{\vec{t}}=\kappa n$. where n is the normal attached at that point but in the given illustration the $\dot{\vec{t}}$ is not in the direction of n . please help i am confused and quite naive in this topic. is the principal normal vector $\vec{p}$ different from the normal vector n? reference page 5 https://www.cmu.edu/biolphys/deserno/pdf/diff_geom.pdf


$\newcommand{\Vec}[1]{\mathbf{#1}}$In a word, "yes": There are two distinct "normal vectors" in your situation.

(Notation below altered slightly in edit to match the diagram below rather than the diagram in the question.)

  1. The unit normal field to the surface $S$, labeled "$\Vec{n}$" in the diagram below.

  2. The principal normal $\Vec{N}$ of the path $\Vec{r}$, which satisfies $\Vec{r}''(t) = \kappa \Vec{N}$.

If the image of a path $\Vec{r}$ lies on a surface $S$, the second derivative of $\Vec{r}$ may be decomposed into components parallel to and orthogonal to the surface normal. The magnitudes of these components are the normal curvature $\kappa_{n}$ and geodesic curvature $\kappa_{g}$, respectively.

If $\Vec{T}(t) = \dfrac{\Vec{r}'(t)}{\|\Vec{r}'(t)\|}$ is the unit tangent field of $\Vec{r}$, then $$ \Vec{r}''(t) = \kappa \Vec{N} = \kappa_{n} \Vec{n} + \kappa_{g} (\Vec{n} \times \Vec{T}). $$

The diagram below shows a latitude on a sphere, parametrized at unit speed. The principal normal $\Vec{N}$ lies in the plane of the latitude circle, and points directly at the center. The acceleration $\Vec{r}''(t)$, which is parallel to the principal normal (but neither parallel nor orthogonal to the unit surface normal $\Vec{n}$ in this instance), decomposes uniquely into a component $\kappa_{n} \Vec{n}$ normal to $S$ and a component $\kappa_{g} (\Vec{n} \times \Vec{T})$ tangent to $S$. The magnitudes of these components are the normal curvature $\kappa_{n}$ and geodesic curvature $\kappa_{g}$ of the latitude. Neither quantity is equal to the curvature $\kappa$ of the latitude. (In fact, the Pythagorean theorem gives $\kappa = \sqrt{\kappa_{n}^{2} + \kappa_{g}^{2}}$.)

The principal normal of a latitude line

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  • $\begingroup$ not very informative $\endgroup$ – Nebo Alex Mar 11 '16 at 7:56
  • $\begingroup$ @Boris: Given that I read your question to be "Is the principal normal different from the surface normal?", a "yes" answer together with contextual explanation related to the diagram seemed to be what you were seeking. If that's not the case, could you please be more specific about how this answer is "not very informative"? (If you're learning differential geometry on your own, not as part of a formal course, that information would be helpful.) Thanks. :) $\endgroup$ – Andrew D. Hwang Mar 11 '16 at 11:58
  • $\begingroup$ yes i am self learning differential geometry @andrew d. hawng . $\endgroup$ – Nebo Alex Mar 11 '16 at 14:13
  • $\begingroup$ if we attach a frenet frame to it along the curve then the normal will always be perpendicular to the tangent but principal normal is not normal to the frame normal $\endgroup$ – Nebo Alex Mar 11 '16 at 15:28
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    $\begingroup$ now it's more clear thank you very much $\endgroup$ – Nebo Alex Mar 11 '16 at 19:00

"The derivative of parameterized curve, r(t), gives a tangent vector"- there are many different tangent vectors, all pointing in the same direction with different lengths. If the parameter, t, is "arclength" the derivative with respect to t gives the unit tangent vector an in that case, the second derivative is the normal vector.

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  • $\begingroup$ is the principal normal vector p different from the normal vector n $\endgroup$ – Nebo Alex Mar 10 '16 at 16:16

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