Men Posing In Ladies Panties

Supplement Supplement # 1 # 2

areas within the geometry of the plane-half


This section is a continuation of the discussion in the Supplement # 1 , where the notions of lengths at the Poincare upper-half are extended to the concept of surfaces in the same hyperbolic plane.

Having found a way to assign a number to the lengths measured at the upper-middle, the next step is to try to find a way to measure hyperbolic limited areas within a certain region of space. To do this, consider the following drawing in which have raised two perpendicular to the axis x with left perpendicular to coincide with what would be the core and a Cartesian plane, and three rectangular regions identified between these lines vertical (Region 1 color cyano, region 2 pink, and yellow region 3, the latter extending up to infinity):

If we were in Euclidean geometry, the region 1 area would undoubtedly have a 1 square meter (or 1 centimeter square, or whatever depending on the unit is used to measure distances), region 2 would also have an area equal to the area of \u200b\u200bregion 1, while the region 3, extending up indefinitely, would have an infinitely large area. But we are not already in the Euclidean plane, we are in the hyperbolic plane, where things work differently.

First get the Region 1 area located between the vertices A , B , C D and then get the area between points C , D, E and F , and finally see what we can do with the yellow region. But before you do that, we need to find a definition for our area hyperbolic. A natural extension to the definition of area in the Euclidean plane geometry within a region R :

inspired redefinition had taken place in the Poincaré half-plane to the hyperbolic length following would be a limited area in a region R :

Let us use this definition to calculate the areas of the three regions identified in the above figure. We begin with the calculation of the hyperbolic surface S _h for region 2:

then calculates the hyperbolic area of \u200b\u200bregion 1:

Finally, we carried out the calculation of the hyperbolic surface in Region 3:

The three regions shown in the example above are not each a rectangle the Poincare half-plane, because to have such a thing requires two pairs of parallel lines that determine a rectangle are straight hyperbolic, which must be perpendicular to vertical lines x or arcs of circles centered on axis x , and clearly the horizontal lines delimiting each region are not arcs. Here we have a real ring-half at the top, with a midline EF lines drawn between the two vertical hyperbolic:

Midline EF is perpendicular to both the hyperbolic straight AB as the hyperbolic straight CD to be perpendicular to the tangents through the points and E F . It is, in fact, common perpendicular to two parallel AB and CD , which can only be one between two parallel lines hyperbolic. Raising the curve AB up keeping the center of the circle that produces the x axis , be confused with a Euclidean line, and the internal angles at the corners and A B will be 90 degrees, will be at right angles. Be a quadrilateral in which AB base (looking inverted) internal angles are right, and on the summit of the interior angles will be acute. We have seen that ring before. It is the Saccheri quadrilateral . This was precisely what Saccheri unknowingly discovered by rejecting the parallel postulate of Euclid's famous fifth postulate.

Now for something about the area of a hyperbolic triangle in the upper-half plane.

We had seen earlier that in hyperbolic geometry the area within a triangle is measured not according to what measure the sides of a triangle, but based on what they measure their internal angles, based on what is missing from the sum of these angles to be 180 degrees, which we call defect of the triangle. Now see what is the triangle of maximum area that can draw in a hyperbolic geometry. We have already seen that in hyperbolic geometry the area of \u200b\u200ba triangle depends not on the lengths of the sides that define the triangle but the magnitudes of the internal angles the triangle, and we saw that the area was given by default δ triangle, defined as missing to the sum of internal angles of the hyperbolic triangle to give 180 degrees (measured in radians). Then the triangle with the largest possible area is the one with the highest possible defect, which occur when the sum of internal angles of the triangle is zero . This at first glance may seem to some as a bold statement, because if the internal angle sum becomes zero then rather than having a triangle containing the maximum amount we will have a triangle area that have disappeared from our sight. However, this is not true, and we must remember again that we must adjust our thinking to what happens in a radically different geometry Euclidean geometry. Below is a hyperbolic triangle ABC in the Poincaré half-plane bounded by straight lines k, l and m , whose three internal angles have a value of zero (the angle is measured by drawing tangents to hyperbolic straight lines at each point of contact):

We should be no doubt that this is a hyperbolic triangle valid. Its three sides are straight hyperbolic arcs are obtained by center on the axis x .

surprises do not end here. If we draw a hyperbolic triangle so that two sides are vertical lines on the Poincare half-plane (this can be obtained starting with a hyperbolic triangle "typical", greatly increasing the radii of two of its defining lines so such a confused ending with two vertical lines built in the flat-half), we have what is called an asymptotic triangle as shown below:

It turns out that this triangle has a finite area! Again, we must adapt our thinking to understand this. The vertex angle A be 90 degrees (pi / 2 radians), while the vertex angle B is zero degrees. And the third point, we can not see why it's up, high up in the "infinite" will have a value of zero degrees (that we can see an arc of a circle centered on the axis x passing through the A vertex and crossing the vertical line through point B , which will have the third side of a triangle "usual." As they increase the radius corresponding to the third side of the vertex angle forming the arc with the vertical line through point B will decrease until, in the vicinity of infinity, the angle is zero degrees, and we have just what showing the picture above. Since the area of \u200b\u200ba hyperbolic triangle is defined by the default δ triangle, the area of \u200b\u200bthis triangle is asymptotic:

δ = π - ( π / 2 + 0 + 0)

δ = asymptotic triangle area = π / 2

Then the asymptotic triangle shown above has a finite area π / 2 units hyperbolic.


Movement figures in the hyperbolic plane


In Euclidean geometry, it is assumed that the ideal bodies are perfectly rigid and that the space "plane" is homogeneous and isotropic (as preserving all its features) in any direction , why the movement of a geometric figure from one region to another leaves the figure intact, unchanged in size. However, this is not what happens at the Poincare upper-half, where as a figure recedes vertically from the boundary line x the figure keeps growing larger without limit, and where as a figure is approaching the boundary line vertically figure is becoming increasingly smaller.

A movement of a figure in the plane can be viewed in two ways completely equivalent (the two versions are known as alias and alibi ), or keeping all the fixed and moving the figure in a sense, or keeping Figure fixed and moving around the plane in the opposite direction. Suppose that all points of the upper-middle move in such a way that the hyperbolic length of any arc belonging to this flat-half ( half plane) is equal to the hyperbolic length of the bow in his new position. Such displacement of the points will be called hyperbolic motion . This concept is analogous to the concept of motion of the Euclidean plane, for example, the rotation of the Euclidean plane at an angle about any point of the plane. If the move turns hyperbolic figure in F F ', then the figures and F F' figures are called same hyperbolically. Let's look at the two simplest types of hyperbolic motion.

In the first type of movement, if it passes each point of the flat-top half over the same distance and in a direction parallel to the boundary line x , is that each figure is transformed into a hyperbolically like her. Does not vary as the magnitude or Euclidean distance of their points x . This type of movement we have shown in the figure below where the object is shifted to the right (or whatever it is, the upper-half plane completely shifted to the left):

So

Therefore, if the object receives a strictly horizontal translation movement, then under the relationship defined in the Supplement # 1 applied to each infinitesimal segment of the segments contained in the object will remain at the same distance and line x limit, then the integration will become trivial to maintain the same value for each infinitesimal segment of the object .

The second type of movement, the object is moved not only horizontally but also vertically, maintaining a relationship of similarity in the following figure:

transforms the segment in the segment PQ P'Q '. Denote as above and and and ' , Respectively, the distances of the points and Q Q 'to the boundary line x . Then by similar triangles OAQ ' and OBQ, we have:

OQ' / OQ = y '/ y

Furthermore, by similar triangles and OP'Q OPQ , we have:

PQ / OQ = P'Q '/ OQ'

P'Q '/ PQ = OQ' / OQ

P'Q '/ PQ = y' / and

P'Q '/ y' = PQ / y

It follows that during this transformation does not change hyperbolic length of any line when carried out in the manner shown above.


Beltrami-Klein model


geometry based on the Beltrami-Klein model (also known as the Klein model , the projective model, the Cayley-Klein model the disk Klein, etc.) geometry is obtained by deforming the Poincare hyperbolic disk so that the lines hyperbolic Poincare disk (arcs) end up as strings, in a manner as shown below:

Thus, the arc AB becomes AB string and the bow CD becomes the rope CD. As investment disk and Poincaré half-plane, the Klein model is also a model of hyperbolic space.

At first glance, the Klein model appears similar to the Poincaré disk, but it is not because the inhabitants of a Klein model are straight lines in his space as we see them from outside . Geodesics in the Klein model are Euclidean strings a unit disk. No But its angular measurements will be different from ours . A model of this type that the way to measure the internal angles is different from what we do is called non-conformal . A simplistic but valid way of seeing the Klein model is considering that, in some ways, the geodesics of the Poincaré disk, which are all arcs are "straightened" projecting into the ropes to touch the extreme points of geodesics.

Following is the comparison of several straight lines in the three models we have seen, the Klein model, the Poincaré disk and the upper-half plane:

Given a Poincaré disk containing geometric figures, these figures can be projected onto their equivalents in the Klein disk and vice versa, but this involves several steps that must be understood before it is carried out such transformation. To get an idea on how it is possible to carry out the transformation of this kind, there will be a sketch, but not accurate and oversimplified, demonstrate how such a thing is possible.

First of all, we can start by taking a disc of Klein, which is of unit radius, and start expanding the radius until it infinitely large, keeping at all times in the "center" of the disc. This has the effect of making the Klein disk which seem to us infinitely large plane no "edge" outer circle. This done, we can take a hemisphere of unit radius, resting his pole over the center of the "plane of Klein." Finally, we take the Poincare disk and align transveral area above the hemisphere, and as the Poincaré disk is also unitary, we can carry out a mathematically perfect alignment as shown below:

Disc # 1 shown above can be taken as the Poincaré disk, and disk # 2 can taken as the Klein disk when it begins to be "stretched" radially to infinity. This done, we can draw a line on the Euclidean "flat Klein", then we can go tracing rays from that line toward the center of symmetry of the hemisphere, marking the points of the rays that touch the spherical surface of the hemisphere:

now show that any line l as shown above to be projected onto the surface of a hemisphere H become an arc. Consider several points on the line l to the surface of the hemisphere. Projection rays of the line going to the center of symmetry O hemisphere H all part of a plane passing through the hemisphere. And as a plane through a sphere through its center of symmetry divides the area into two equal parts, then form a circle on the surface of the sphere, or in this case the hemisphere. Then the line k be an arc. Finally, we can project upwards to the top of the hemisphere, to Poincaré disk, each of the points that form the arc k. In the following figure, which represents a lateral cross section of the hemisphere, we see several points originally arrived projected disk # 2 to disk # 1 (the point has been highlighted P whose image ends up being the point P '):

Thus, by projecting each of the points as part of the arc k up on the disc that forms the top of the hemisphere will result in a line segment in the Poincaré disk that corresponds to the Euclidean line that was projected from the Klein model. Such transformations are what we call mathematics an isomorphism , each point on the disk # 1 is a point on the disk # 2, and vice versa, there is a correspondence of each one, one-one correspondence .

However, as already stated, what has been done above is an oversimplification of the matter in teaching. Understandably, not many mathematics texts, including undergraduate college, entering thoroughly into the details of the proof required to obtain images in the Poincaré hyperbolic disk the equivalent of a straight Euclidean Klein disk. However, the final mathematical result, the formulas needed to perform the transformations, it is easy to digest. The use of formulas to be presented here is based on the use of polar coordinates (r, θ):

instead of Cartesian coordinates (x , and ). With this, we can carry out the conversion point by point of a geometric figure from one disk to another using the following formulas for converting Klein-Poincaré:

while for the conversion we use the Poincaré-Klein following formulas: