Lobatschewsky gave a method to calculate the curvature of space (see Bonola “Non-Euclidean Geometry” § 45)

But I don't understand his method. Can somebody explain?

I understand that the method now won't work anymore, Lobatschewsky assumes that light rays are straight lines while in modern physics light rays can be bent, but still ignoring this I don't understand it.

Why is there assuming hyperbolic geometry a maximum parallax while assuming euclidean geometry there is no maximum? Also I read somewhere that under elliptic geometry the parallax can be negative, why is that?

up vote 1 down vote accepted

Read carefully:

We could repeat the calculation for much smaller parallaxes, for example 0",1, and we would find $k$ to be greater than a million times the diameter of the earth's orbit.

Thus, if the Euclidean Geometry and the Fifth Postulate are to hold in actual space, $k$ must be infinitely great. That is to say, there must be stars whose parallaxes are indefinitely small.

So this is not so much about a maximal parallax but about a minimal one. The further a star is away, the smaller its parallax. In Euclidean geometry, the limit case would be a star at infinite distance, in which case the light rays to that ideal star would be parallel and the parallax would be zero.

In hyperbolic geometry, there are no parallels. Even for an ideal star, an infinite distance away, the angle at that star is zero but the angles at the base line still sum to less than $180°$, so you'd still see the star under different angles for different points on earth's orbit.

In elliptic geometry, things are reversed. Let's use earth's coordinate system (and thus a spherical model) for simplicity. Take one observation on the equator at the $0°$E meridian, and another at the $90°$E meridian. From both points, you'd see the north pole at a right angle to the meridian, i.e. zero parallax even though the north pole is a finite distance away from your observations. If a star is somewhere “behind” the north pole, then the parallax would indeed be negative.

If you have trouble imagining this, then try the following: you are sitting in a train which travels along the equator, and you are looking straight northwards, perpendicular to your railway tracks. Now for near objects, you'd expect that you'll have to swivel your telescope to the left as your train moves to the right. That's positive parallax. The further an object is away, the less you'll have to swivel. If the object in question is the north pole, you'll not have to move your telescope at all. That's zero parallax. And if the object is behind the north pole, then you'll have to swivel in the opposite direction. Perhaps try this with a globe in front of you.

Since your question started with maximal parallax, I guess you got confused by this section:

Denote by $2 p$ the maximum parallax of the star $A$.

I guess this simply means the maximum over time, i.e. you wait till the parallax reaches it's maximum and that's the number you use for your calculations. I guess that this takes care of the fact that while $AB$ is assumed to be perpendicular to $BC$, it need not be perpendicular to the plane of earth's orbit (the ecliptic). In any case, I'd consider this a minor point, far less important than the lower bound on parallaxes.

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