Truly Bizarre - The unit tangent and unit normal vectors aren't orthogonal

Click For Summary
SUMMARY

The discussion centers on the non-orthogonality of unit tangent and unit normal vectors in differential geometry, specifically for the curve defined by the parametric equations α(t) = {Cos(t), 2Sin(t)}. The unit tangent vector is derived as T(t) = α'(t) / ||α'(t)||, while the unit normal vector is corrected to be T'(t) / ||T'(t)||, rather than α''(t) / ||α''(t)||. The Mathematica code provided illustrates the implementation of these concepts, highlighting the graphical representation of the vectors on the curve.

PREREQUISITES
  • Understanding of differential geometry concepts, particularly curves and vector fields.
  • Familiarity with Mathematica for implementing mathematical models and visualizations.
  • Knowledge of parametric equations and their derivatives.
  • Basic understanding of vector normalization and orthogonality.
NEXT STEPS
  • Explore the concept of frame fields in differential geometry.
  • Learn about the implications of curvature on unit tangent and normal vectors.
  • Investigate the use of Mathematica for visualizing vector fields and curves.
  • Study the mathematical principles behind differentiating vector functions.
USEFUL FOR

Mathematicians, physics students, and software developers interested in differential geometry and vector calculus, particularly those utilizing Mathematica for computational modeling.

jdinatale
Messages
153
Reaction score
0
OK, this looks like a differential geometry problem, which it is, but at the end of the day I am trying to figure out why the unit normal and unit tangent vectors to a curve aren't orthogonal, so even if you don't know about DG, please respond.

Joseph-2.png


Obviously the two choices for E_1 and E_2 are the unit normal and unit tangent vectors to the curve.

Using Mathematica...

alpha[t_] := {Cos[t], 2 Sin[t]};

alphaprime[t_] := {-Sin[t], 2 Cos[t]};

alphaprimeprime[t_] := {-Cos[t], -2 Sin[t]};

unittangentvector[t_] := alphaprime[t] / Norm[alphaprime[t]];

unitnormalvector[t_] := alphaprimeprime[t] / Norm[alphaprimeprime[t]];

Or by hand...


\alpha(t) = {Cos(t), 2Sin(t)}
\alpha'(t) = {-Sin(t), 2Cos(t)}
\alpha''(t) = {-Cos(t), -2Sin(t)}

However, graphically, the unit tangent and unit normal vectors are far from perpendicular on this curve!

642.png


Here is my mathematica code

VFieldOnCurve2D[dominterval_, CurveEq_, FrameField_, CodomainBox_,
Size_] :=
Module[{a2, b2, Content, IS, DomainPieces, DomainPiece1,
DomainPiece2, CodomainCenter, CodomainWidth, len, EE1, EE2,
ImagePieces, ImagePiece0, ImagePiece1, ImagePiece2},
{a2, b2} = dominterval;
IS = 300;
Content = Mapping12Content[dominterval, CurveEq];
DomainPiece1 = Content[[1]];
DomainPiece2[t_] := Points2D[{{0, t}}, .3];
DomainPieces[t_] :=
Show[DomainPiece1, DomainPiece2[t], ImageSize -> IS/4];

{CodomainCenter, CodomainWidth} = CodomainBox;
len = Length[FrameField];
If[len == 2, EE1 = FrameField[[1]];
EE2 = FrameField[[2]], {EE1} = FrameField];

ImagePiece0 = EmptySpace2DXCenter[CodomainCenter, CodomainWidth];
ImagePiece1 = Content[[2]];
ImagePiece2[t_] :=
If[len == 2, {Vec[CurveEq[t], EE1[t]], Vec[CurveEq[t], EE2[t]]},
Vec[CurveEq[t], EE1[t]] ];
ImagePieces[t_] :=
Show[ImagePiece0, ImagePiece1, ImagePiece2[t], ImageSize -> Size];

t0 = (a2 + b2)/2;
Manipulate[
Row[{DomainPieces[t], ImagePieces[t]}], {{t, t0, "t"}, a2, b2},
SaveDefinitions -> True]


]



alpha[t_] := {Cos[t], 2 Sin[t]};

alphaprime[t_] := {-Sin[t], 2 Cos[t]};

alphaprimeprime[t_] := {-Cos[t], -2 Sin[t]};

unittangentvector[t_] := alphaprime[t] / Norm[alphaprime[t]];

unitnormalvector[t_] := alphaprimeprime[t] / Norm[alphaprimeprime[t]];

E1[t_] = unitnormalvector[t];
E2[t_] = unittangentvector[t];


DomainInterval = {0, 2 \[Pi]};
initvalue = 0;
CodomainBox = {Origin2D, 2};
Size = 400;
VFieldOnCurve2D[DomainInterval, alpha, {E1, E2}, CodomainBox, Size]
 
Physics news on Phys.org
The unit normal isn't alphaprimeprime[t] / Norm[alphaprimeprime[t]]. It's the derivative of the UNIT tangent vector divided by the norm of the derivative of the UNIT tangent vector. alphaprime isn't UNIT.
 
Last edited:
Dick said:
The unit normal isn't alphaprimeprime[t] / Norm[alphaprimeprime[t]]. It's the derivative of the UNIT tangent vector divided by the norm of the derivative of the UNIT tangent vector. alphaprime isn't UNIT.

Thanks, and that would seem to do the trick, but mathematica is hating me right now. Any ideas what's going on? Here is the newly defined unit normal vector

alpha[t_] := {Cos[t], 2 Sin[t]};

alphaprime[t_] := {-Sin[t], 2 Cos[t]};

alphaprimeprime[t_] := {-Cos[t], -2 Sin[t]};

unittangentvector[t_] := alphaprime[t] / Norm[alphaprime[t]];

unitnormalvector[t_] :=
unittangentvector'[t] / Norm[unittangentvector'[t] ];

E1[t_] = unitnormalvector[t];
E2[t_] = unittangentvector[t];


failure.png



Implementation

VFieldOnCurve2D[dominterval_, CurveEq_, FrameField_, CodomainBox_,
Size_] :=
Module[{a2, b2, Content, IS, DomainPieces, DomainPiece1,
DomainPiece2, CodomainCenter, CodomainWidth, len, EE1, EE2,
ImagePieces, ImagePiece0, ImagePiece1, ImagePiece2},
{a2, b2} = dominterval;
IS = 300;
Content = Mapping12Content[dominterval, CurveEq];
DomainPiece1 = Content[[1]];
DomainPiece2[t_] := Points2D[{{0, t}}, .3];
DomainPieces[t_] :=
Show[DomainPiece1, DomainPiece2[t], ImageSize -> IS/4];

{CodomainCenter, CodomainWidth} = CodomainBox;
len = Length[FrameField];
If[len == 2, EE1 = FrameField[[1]];
EE2 = FrameField[[2]], {EE1} = FrameField];

ImagePiece0 = EmptySpace2DXCenter[CodomainCenter, CodomainWidth];
ImagePiece1 = Content[[2]];
ImagePiece2[t_] :=
If[len == 2, {Vec[CurveEq[t], EE1[t]], Vec[CurveEq[t], EE2[t]]},
Vec[CurveEq[t], EE1[t]] ];
ImagePieces[t_] :=
Show[ImagePiece0, ImagePiece1, ImagePiece2[t], ImageSize -> Size];

t0 = (a2 + b2)/2;
Manipulate[
Row[{DomainPieces[t], ImagePieces[t]}], {{t, t0, "t"}, a2, b2},
SaveDefinitions -> True]


]



alpha[t_] := {Cos[t], 2 Sin[t]};

alphaprime[t_] := {-Sin[t], 2 Cos[t]};

alphaprimeprime[t_] := {-Cos[t], -2 Sin[t]};

unittangentvector[t_] := alphaprime[t] / Norm[alphaprime[t]];

unitnormalvector[t_] :=
unittangentvector'[t] / Norm[unittangentvector'[t] ];

E1[t_] = unitnormalvector[t];
E2[t_] = unittangentvector[t];


DomainInterval = {0, 2 \[Pi]};
initvalue = 0;
CodomainBox = {Origin2D, 2};
Size = 400;
VFieldOnCurve2D[DomainInterval, alpha, {E1, E2}, CodomainBox, Size]
 
Not really, sorry. It looks right. But I haven't done Mathematica since other people stopped paying for it for me. It's really expensive and unfree. So I can't test that in any detail.
 
Dick said:
Not really, sorry. It looks right. But I haven't done Mathematica since other people stopped paying for it for me. It's really expensive and unfree. So I can't test that in any detail.

Well, at least theoretically, would you agree that my choice of E1 and E2 provide a frame field on the curve?
 
Sure. The unit tangent T points along your curve. T' must be perpendicular to that. Just differentiate T.T=1. It's got to work, right?