## Wednesday, October 26, 2022

### An Introduction To Differential Geometry - Conclusion

Of particular importance in differential geometry are special or minimal surfaces, the topic for this concluding entry on differential geometry.  All right conoids as well as helicoids,  fall under this category.  For example, the right circular helix in the above graphic which we also saw in the initial (April 8) post in this series.  Indeed, we will see that the classes of surfaces pertaining to right conoids and surfaces of revolution each  include exactly one type of minimal surface:  the right helicoid and catenoid respectively.

The catenoid, e.g.

is obtained by rotating the surface known as a catenary which in Cartesian coordinates has the form:  y = f(x) =  a cosh (x/a)  or more generally:

f(x) = a cosh (bx - c)

Below are shown two catenaries (y =  2 cosh(x/2)  and y=  2 cosh(x/4) done using the software program Graphmatic:

Now, we change to the more general form:  y = 2 cosh (x/2 - 1)

For which a = 2,  b = 1/2 and c = 1.  Note that a = 2 gives the distance of the lowest point to the x-axis.  This occurs also at x = 2, which is equal to c/ b

= 1/ (1/2)

When this catenary is rotated about the x-axis it generates the surface we call the catenoid. Three key points follow:

1) Catenoids are the only surfaces of revolution that are minimal surfaces,

2) Right helicoids are the only right conoids that are minimal surfaces

3) A catenoid when cut along a meridian can be isometrically mapped onto a suitably chosen right helicoid

Every surface of revolution whose axis is the x -axis of the Cartesian coordinate system can be shown to be represented in the form:

x3  =   g (r),  r    =  x12    +   x32

We can also choose a parametric representation in the form:

x ( u1 ,  ) =  (2 cos u1 , 2 sin u1 , h2)

Which will exhibit the form:

As a result:
g 11   =  (2 ) 2
g 12   =  0,    g 22   =  1  +  h'2

g  =  (2 ) 2   (1  +  h'2 )

Where a prime denotes the derivative with respect to 2

And the mn  (or    ab)  were defined in Part 6 in terms of the fundamental metric tensor. Furthermore, we can obtain from this basis the following:

b 11   =  2 h'/  Ö((1  +  h'2 ),   b 12   =  0,

b 22   = - h"/Ö((1  +  h'2 )

If  h2  is constant we obtain a plane parallel to the  x1 x2 -plane, e.g. see the bold circular plane in the 3 D depiction of the catenoid above. Also, if   h'     0, then the condition H (mean curvature) = 0 implies:

h" h' / h'2 (1  +  h'2 ) =   - 1/ 2

The mean curvature H   =  ½ ab g ab

is zero at every point of a minimal surface.  Further, for any Gaussian curvature K  = - h   we obtain H = 0.  However, for the case  H    0 we have:

K  + h / 2H  =  k u

So that at every point of the surface S the normal curvature ( ) is the same for all directions.

Suggested Problems:

1) (a) Sketch the catenary: y = 3 cosh (x/2 -  ¼)

(b) If u2  is the parallel  u2 =0.5  and h' =   c / Ö(22  -  c2

Compute the mean curvature H if:

H = -½ (h' /2(1 + h' 2)1/2 ] + [h' /2(1 + h' 2)3/2 ])

(c) If  K  = h = +1   for a positively curved minimal surface, find; k u

(2) Using a quantitative approach find a representation of the right helicoids which are isometric to a catenoid.

Hint: You may use:  (u1a cosh (x3 /a  ) as the representation of the meridian of the catenoid.