Local non-intrinsic property of surface








In differential geometry, the study of surfaces involves understanding the intrinsic and extrinsic properties that define a surface's shape, curvature, and topology. A surface in differential geometry is typically a two-dimensional manifold, meaning that locally around each point, it resembles a flat plane, but globally it may have more complex structure and curvature.

Local Non-Intrinsic Properties of Surface

Intrinsic property of a surface is an invariant, inherent or unchanging property of surface. Some example like Gaussian curvature, which remain unchanged in a surface

Next, non-intrinsic properties (or local non-intrinsic properties) of a surface are those characteristics that are not determined by the internal geometry of the surface (like distances and angles, area within the surface) but by how the surface interacts with the external space. Some example of non-intrinsic properties of a surface are (a) Normal Vector: the orientation of the normal vector is determined by the surface's embedding in space. (b) Curvature of sections: It is a measure of how the surface bends in the surrounding space. If a surface is bent in one direction more than another, the curvature of sections captures this asymmetry.
az = 1.00
el = 0.30
N
az = 1.00
el = 0.30
N
(a)Normal to the surface(b)Normal section to the surface



az = 1.00
el = 0.30
az = 1.00
el = 0.30
(a)Normal to the surface(b)Oblique section to the surface






Normal section of Surface

Let S:r=r(u,v) be a surface and P be a point on it, on which
Γ be a plane at P
Then intersection of SΓ is a curve C on the surface. This curve is called section of the surface.
Also, we knaow that
N is normal to the surface at P
If the plane Γ contains N , then C is called normal section to the surface,
otherwise C is called oblique section to the surface
az = 1.00
el = 0.30
N=n
az = 1.00
el = 0.30
N
n
(a)Normal to the surface(b)Oblique section to the surface



NOTE
  1. There are infinitely many normal sections on the surface.
  2. In a normal section, n=N



Normal Curvature and related Theorems

Let S:r=r(u,v) be a surface and C be a normal section on it. Then curvature of C is called normal curvature. The normal curvature of the surface is denoted by κn.
It is measure of how the surface bends in a particular direction.

Expression of Normal Curvature

Let S:r=r(u,v) be a surface and C be a normal section on it. Then curvature of C is
κn=r
orκnn=r [normal curvature is denoted by κn]
orκnN=r [for normal section N=N]
Operating dot product on both sides by N, we get
κn=N.r (A)
Next, equation of the surface is
r=r(u,v)
Differentiation of w. r. to. s, we get
r=r1duds+r2dvds
orr=r1u+r2v (i)
Again, differentiation of w. r. to. s, we get
r=dds(r1u+r2v)
or r=(r11u+r12v)u+(r12u+r22v)v+r1u+r2v
or r=r11u2+2r12uv+r22v2+r1u+r2v (ii)
By substitution of (ii) in (A), the expression of normal curvature is
κn=N.r
orκn=N.(r11u2+2r12uv+r22v2+r1u+r2v)
orκn=(Lu2+2Muv+Nv2)
orκn=L(duds)2+2Mdudsdvds+N(dvds)2
or κn=Ldu2+2Mdudv+Ndv2ds2
or κn=Ldu2+2Mdudv+Ndv2Edu2+2Fdudv+Gdv2
or κn=III
Thus is required expression of normal curvature.
Note
If first and second fundamental coefficients are proportional, the normal curvature has expression
κn=LE=MF=NG

Find normal curvature of a curve u=t2,v=t on r=(u,v,u2+v2)at t=1

Solution
Given surface is
r=(u,v,u2+v2) (i)
By a bit of calculation, we get
E=5,F=4,G=5 at t=1
L=25,M=0,N=23 at t=1
dudt=2,dvdt=1 at t=1
Thus, normal curvature at t=1 is
κn=Ldu2+2Mdudv+Ndv2Edu2+2Fdudv+Gdv2
or κn=L(dudt)2+2Mdudtdvdt+N(dvdt)2E(dudt)2+2Fdudtdvdt+G(dvdt)2
or κn=10123






Meusnier’s Theorem

Meusnier’s theorem shows a relation between curvature of normal and oblique sections, it was first announced by Jean Basptiste Meusnier in 1776. The essence of the theorem is that, all curves lying on a surface S having same tangent have the same normal curvature.

Theorem

If κ and κn are curvatures of oblique and normal sections and θ be angles between these sections then κn=κcosθ
az = 1.00
el = 0.30
N=n
t
az = 1.00
el = 0.30
N
n
t
(a)Normal to the surface(b)Oblique section to the surface
Proof
Let S:r=r(u,v) be a surface and N be the unit normal. Then, curvature of oblique section is
r=κn (i)
Taking dot product of both sides of (i) by N, we get
r.N=κn.N (A)
Next, curvature of normal section is
r=κn.N (ii)
Taking dot product of both sides of (ii) by N, we get
r.N=κn (B)
Also given that θ is the angle between these sections, then
cosθ=n.N (C)
By substituting (B) and (C) in (A) we get
κn=κcosθ
This completes the proof of the theorem.


az = 1.00
el = 0.30



Principal Sections

Let S:r=r(u,v) be a surface and P be a point on it. Then there are infinitely many normal sections at P. Now, two sections, which give maximum and minimum curvature (principal curvatures) at P, are called principal sections at P. These principal sections are always orthogonal.


Principal Directions

Let S:r=r(u,v) be a surface and P be a point on it. Then, tangents to principal sections are called principal directions.Since, there are two principal sections (in general), it is convenient to state that, there are two principal directions (in general) at every point on a surface and it will be always orthogonal.


Differential equation of Principal Section/direction

Let S:r=r(u,v) be a surface and κn be the principal curvature, then
κn=Ldu2+2Mdudv+Ndv2Edu2+2Fdudv+Gdv2
or(Ldu2+2Mdudv+Ndv2)κn(Edu2+2Fdudv+Gdv2)=0
Then, differentiating w r. to. du and dv separately, we get
(Ldu+Mdv)κn(Edu+Fdv)=0 (i)
(Mdu+Ndv)κn(Fdu+Gdv)=0 (ii)
Eliminating kn from (i) and (ii) we get
(Ldu+Mdv)(Mdu+Ndv)=(Edu+Fdv)(Fdu+Gdv)
or (EMFL)du2+(ENGL)dudv+(FNGM)dv2=0 (iii)
which is the required equation of principal sections/directions.
Note
The equation of principal section/directions can be written as
|EFGLMNdv2dudvdu2|=0



Find principal sections on hyperboloid 2z=7x2+6xyy2 at origin


Solution The position vector of the hyperboloid 2z=7x2+6xyy2 is
r=(x,y,12(7x2+6xyy2))
By computing the fundamental coefficients of the surface, we get
E=1,F=0,G=1,H=1,L=7,M=3,N=1
Now the equation of principal sections is
(EMFL)dx2+(ENGL)dxdy+(FNGM)dy2=0
or3dx28dxdy3dy2=0
ordx3dy=0,3dx+dy=0
orx3y=0,3x+y=0
This completes the solution.


Show that principal directions are always orthogonal

Proof
The equation of principal directions is
(EMFL)du2+(ENGL)dudv+(FNGM)dv2=0 (A)
The equation of double family of curves is
Pdu2+Qdudv+Rdv2=0 (B)
Comparing (A) and (B) we get
P=(EMFL),Q=(ENGL),R=(FNGM)
Now, condition of orthogonally for double family of curves is
ERQF+GP=0
Hence, we have
ERQF+GP=E(FNGM)F(ENGL)+G(EMFL)
orERQF+GP=0 TRUE
It shows that, principal directions are always orthogonal


Principal Curvature

Let S:r=r(u,v) be a surface and P be a point on it. Then there are infinitely many normal curvatures at P. Now, two curvatures, which are maximum and minimum at P, are called principal curvatures at P. These two curvatures are denoted by
κa and κb
The corresponding radii of principal curvatures are called principal radii and are denoted by
ρa and ρb
Since, Principal curvatures are the maximum and minimum (signed) curvatures of various normal slices, these maximum and minimum curvatures always occur at right angles to one another.


Differential equation of principal curvature

Let S:r=r(u,v) be a surface and κn be the principal curvature, then
κn=Ldu2+2Mdudv+Ndv2Edu2+2Fdudv+Gdv2
or(Ldu2+2Mdudv+Ndv2)κn(Edu2+2Fdudv+Gdv2)=0
Then, differentiating w r. to. duanddv separately, we get
(Ldu+Mdv)κn(Edu+Fdv)=0 and (Mdu+Ndv)κn(Fdu+Gdv)=0
or(LκnE)du+(MκnF)dv=0 and (MκnF)du+(NκnG)dv=0
Eliminating du and dv, we get
(MκnF)(LκnE)=(NκnG)(MκnF)
or(MκnF)(MκnF)=(LκnE)(NκnG)
orκn2(EGF2)κn(EN2FM+GL)+(LNM2)=0
This is the equation of principal curvatures
Solving this quadratic equation for κ , we get two principal curvatures (maximum or minimum). Thus, principal curvatures κa and κb are obtained by
κa+κb=EN2FM+GLEGF2and
κa.κb=LNM2EGF2
Note
  1. The sum of the principal curvatures i.e., κa+κb=EN2FM+GLEGF2 is called first curvature, it is denoted by J.
  2. The arithmetic mean of principal curvatures i.e., 12(κa+κb)=EN2FM+GL2(EGF2) is called mean curvature
    or mean normal curvature, it is denoted by μ.
  3. The product of principal curvatures i.e., κa×κb=LNM2EGF2 is called Gaussian curvature, it is denoted by K.
    It is also called specific curvature, second curvature or total curvature.
  4. Given a point P
    (a). if κa=κb=0 , we say P is planner point
    (b). if K=0,we say P is parabolic point
    (c). if K>0, we say P is elliptic point
    (d). if K<0, we say P is hyperbolic point



Show that principal curvature of hyperboloid 2z=7x2+6xyy2 at origin are 8 and -2.

Solution
The position vector of the hyperboloid 2z=7x2+6xyy2 is
r=(x,y,12(7x2+6xyy2))
By computing the fundamental coefficients of the surface, we get
E=1,F=0,G=1,H=1,L=7,M=3,N=1
Now the equation of principal curvature is
κn2(EGF2)κn(EN2FM+GL)+(LNM2)=0
orκn2(1)κn(6)+(16)=0
orκn25κn16=0
orκn=8 and κn=2
This completes the solution.


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