|
Post by Admin on Feb 23, 2017 0:14:44 GMT
Gary
I don't understand what you are saying here. Do you really mean area of a pseudosphere?
Which vertex is transposed to the real line?
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Feb 23, 2017 1:10:22 GMT
Gary I don't understand what you are saying here. Do you really mean area of a pseudosphere? Which vertex is transposed to the real line? Vasco Vasco, That should have read "the area of a triangle on the pseudosphere". I am referring the the angles $\alpha$ and $\beta$ at the origin, but I see it as transposed onto the horizon rather than transposed to the real line. That is, it is on the real line, but it appears to be on the horizon. Gary
|
|
|
Post by Admin on Feb 23, 2017 8:15:44 GMT
Gary
I don't see this as Needham using the angles at the origin (which is not supposed to be part of the map), to calculate the area of the triangle on the pseudosphere. Without going into the theory of double integrals, we can see from the evaluation of the double integral on page 315 (derived from 32 on page 298), that we need to be able to express the $y$ coordinate of any point on the arc joining the vertices at $\alpha$ and $\beta$ as a function of $x$ measured along the horizon. We can use Euclidean geometry for this, which allows us to include the origin. It's then clear that $y=\sqrt{1-x^2}$. We also need to know the values of $x$ at the two vertices at $\alpha$ and $\beta$, and we can also calculate these by using Euclidean geometry.
Then we can use the integral to calculate the area of the triangle on the pseudosphere.
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Feb 24, 2017 1:48:50 GMT
Gary I don't see this as Needham using the angles at the origin (which is not supposed to be part of the map), to calculate the area of the triangle on the pseudosphere. Without going into the theory of double integrals, we can see from the evaluation of the double integral on page 315 (derived from 32 on page 298), that we need to be able to express the $y$ coordinate of any point on the arc joining the vertices at $\alpha$ and $\beta$ as a function of $x$ measured along the horizon. We can use Euclidean geometry for this, which allows us to include the origin. It's then clear that $y=\sqrt{1-x^2}$. We also need to know the values of $x$ at the two vertices at $\alpha$ and $\beta$, and we can also calculate these by using Euclidean geometry. Then we can use the integral to calculate the area of the triangle on the pseudosphere. Vasco Vasco, That all makes good sense to me. I wonder how this examination of h-lines pertains to the formulae for Euclidean and h-centres and Euclidean and h-radii on p. 309. h-circles are like Euclidean circles in that we draw them as semi-circles from Euclidean centers on the horizon. By the formula for h-circles, if the Euclidean centre is at x, so is the h-centre. But if the h-radius is $\rho$, the Euclidean radius is $ y sinh(\rho)$ = 0, because y = 0. If the Euclidean radius is a real number $a$, then $a = y sinh(\rho)$ = 0. So we can't get from the Euclidean to the h-radius this way. Perhaps we could say that if the Euclidean radius is $a$ and the Euclidean centre is $x$, the radius is $|ai|$ at $x + ai$. Then $a = ai sinh(\rho)$ and $\rho = arcsinh(a/ai) = arcsinh(-i)$ = -1.57i$, but it doesn't make much sense to have a constant h-radius for an h-line with a variable radius. Or does it? It appears that hyperbolic geometry that applies to h-circles does not necessarily apply to h-lines. Gary
|
|
|
Post by Admin on Feb 24, 2017 9:51:04 GMT
Gary
I assume that you mean "...on p. 309. h-lines are..." rather than h-circles.
An h-circle by definition must have its centre in the hyperbolic plane and so its h-centre cannot lie on the horizon. In figure 27 on page 308 if we imagine the h-centre moving down the vertical h-line and getting closer and closer to the horizon, the picture remains essentially the same with the h-circles becoming closer and closer together on the horizon side of the h-centre. The Euclidean centre gets closer and closer to its corresponding h-centre and all the h-circles collapse down towards a point on the horizon.
So an h-circle and an h-line are very different concepts - an h-line is not some special limiting case of an h-circle. If we let the radius of an h-circle go towards infinity, in the limit the h-circle becomes the horizon, not an h-line. This is in contrast to what happens in Euclidian geometry.
When talking about semicircular h-lines and vertical h-lines on page 302 Needham writes: "...if you were an inhabitant of the hyperbolic plane, there would be no way for you to distinguish between the semicircular h-lines and the vertical h-lines: every line is exactly like every other, it's just our map that makes them look different."
We could also say, (a sort of inverse of what Needham says): h-lines and h-circles (to someone in the hyperbolic plane), are completely different things, it's just our map that makes them look similar (i.e. like circles).
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Feb 24, 2017 17:23:04 GMT
Gary I assume that you mean "...on p. 309. h-lines are..." rather than h-circles. An h-circle by definition must have its centre in the hyperbolic plane and so its h-centre cannot lie on the horizon. In figure 27 on page 308 if we imagine the h-centre moving down the vertical h-line and getting closer and closer to the horizon, the picture remains essentially the same with the h-circles becoming closer and closer together on the horizon side of the h-centre. The Euclidean centre gets closer and closer to its corresponding h-centre and all the h-circles collapse down towards a point on the horizon. So an h-circle and an h-line are very different concepts - an h-line is not some special limiting case of an h-circle. If we let the radius of an h-circle go towards infinity, in the limit the h-circle becomes the horizon, not an h-line. This is in contrast to what happens in Euclidian geometry. When talking about semicircular h-lines and vertical h-lines on page 302 Needham writes: "...if you were an inhabitant of the hyperbolic plane, there would be no way for you to distinguish between the semicircular h-lines and the vertical h-lines: every line is exactly like every other, it's just our map that makes them look different." We could also say, (a sort of inverse of what Needham says): h-lines and h-circles (to someone in the hyperbolic plane), are completely different things, it's just our map that makes them look similar (i.e. like circles). Vasco Yes, I meant "...on p. 309. h-lines are..." Your discussion is very helpful. In section 3.4 Beltrami's hyperbolic plane. I find these points from p. 301, paragraph 3, relevant to this discussion. The boldface is my own emphasis: "the hyperbolic plane, namely, the entire shaded half-plane y > 0" "The points on the real axis are infinitely far from ordinary points and are not (strictly speaking) considered part of the hyperbolic plane. They are called ideal points, or points at infinity. The complete line y = 0 of points at infinity will be called the horizon." In my first reading, I missed the significance of this, lumping everything, including the horizon, into the UHP model of the h-plane. But perhaps it is misleading to draw radii and semi-circles from points on the horizon, as in [21], [22], [23], [24], [25], [26]. Drawing them in this way suggests that the point on the horizon is a part of the radius or h-line, when in fact it is infinitely far away. But what is the epistemological status of an ideal point on the horizon? To call it an ideal point seems to beg the question. An ideal point might be called an indexical point or a point of reference, which would suggest that it is useful in situating objects in the h-plane, but what quality of the h-plane gives indexicality to something outside itself, not part of the h-plane? Is it that we can only apprehend the complete h-plane by its connections and relationship to the Euclidean plane? Taking this argument to its conclusion, it appears that the hyperbolic plane is just a field of mathematics that represents negative curvature by reference to the Euclidean plane. If we allow the curvature to tend from negative to 0, the limit is the Euclidean plane. That is why Needham can say that Euclidean geometry is subsumed by (3-dimensional) hyperbolic geometry (p. 327). Since 3-dimensional h-geometry can be mapped to the UHP, the latter must also be subsumed. Conclusion: It would be useful to somewhere state: In the figures, ends and centres of h-lines are often drawn from points on the horizon as though there were a continuous connection, but in fact the ideal points in the horizon are not part of the UHP. They are used only to index the x-values of objects in the UHP, notably including, but not limited to, the ideal endpoints and centres of h-lines. Gary
|
|
|
Post by Admin on Feb 24, 2017 17:54:24 GMT
Gary
Just to try and answer one of your questions: These points are only infinitely far away to Ms Poincare in the h-plane. The diagrams you reference can be thought of as diagrams of the upper half-plane which we can treat as Euclidean if we don't adopt the Poincare metric and then we can include the horizon.
Vasco
|
|
Gary
GaryVasco
Posts: 3,352
|
Post by Gary on Feb 24, 2017 18:08:46 GMT
Vasco,
I agree. That seems to be the correct interpretation of the discussion on p. 300. I think the horizon is nothing but the real line as seen by Ms. P.
Gary
|
|