Maths - Stereographic 2D Derivation

Here we look at a two dimensional euclidean space embedded in a three dimensional projective space, we are using the stereographic model to do this projection. For a more general discussion of stereographic projection see page here.

hemisphere projection

In two dimensional projective space using stereographic model:

From the above we can see that, as we move away from the origin, lines that cross at 90° in euclidean space do not cross at 90° in projective space. So angles away from the origin are not conserved by the projection.

Derivation for translation between projective (stereographic) and euclidean spaces

As for the hemisphere model, since we are projecting using a straight line, there is a linear relationship between a point on the plane and its projection on the sphere:

x
y
z
= λ
u
v
1

where:

Dividing the top two rows by the bottom row gives the euclidean coodinates in terms of the projective coordinates:

u
v
=
x/z
y/z

To go the other way from the projective coordinates to the euclidean coodinates we need to use the information that the coordinates are constrained to a unit sphere centred at (0,0,1) so:

x² + y² + (z-1)² = 1

rearanging gives:

x²/z² + y²/z² + (z²-2z +1)/z² = 1/z²

u² + v² + 1 = 2z/z² = 2/z = 2/λ

λ = 2/(u² + v² + 1)

substituting this into the first vector equation gives:

x
y
z
= 2/(u² + v² + 1)
u
v
1

Alternative Coordinate System

So far I have chosen the origin of the x,y,z coordinate system as the projection point as this seems the simplest way to do it. However other sources (see Doran & Lasenby - Geometric Algebra for Physisits book on right of this page and also the Wikipedia page on stereographic projection) seem to use a coordinate system where the origin of the x,y,z coordinate system is at the centre of the sphere. So in order to relate to these other sources I will repeat the above with this coordinate system:

We will now have:

x
y
z-1
= λ
u
v
1

So

u
v
=
x/(z-1)
y/(z-1)

since the origin is on the plane being projected and

x² + y² + z² = 1

dividing both sides by (z-1)² gives:

x²/(z-1)² + y²/(z-1)² + (z²-1)/(z-1)² = 0

u² + v² + (z +1)/(z-1) = 0

u² + v² + (λ+2)/λ= 0

u² + v² + 1 +2/λ= 0

λ= -2/(u² + v² + 1)

substituting this into the first vector equation gives:

x
y
z-1
= -2/(u² + v² + 1)
u
v
1

adding 1 to both sides of row 3 gives:

x
y
z
= -2/(u² + v² + 1)
u
v
1-(u² + v² + 1)/2

which is:

x
y
z
= 1/(u² + v² + 1)
2u
2v
u² + v² - 1

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Correspondence about this page

Book Shop - Further reading.

Where I can, I have put links to Amazon for books that are relevant to the subject, click on the appropriate country flag to get more details of the book or to buy it from them.

flag flag flag flag flag flag Roger Penrose - The Road to Reality: Partly a 'popular science' book as it tries to minimise the number of equations (Not that I'm complaining much his book 'Spinors and space-time' went over my head in the first few pages) it still has lots of interesting results that its difficult to find elsewhere.

 

cover Spinors and Space-Time: Volume 1, Two-Spinor Calculus and Relativistic Fields (Cambridge Monographs on Mathematical Physics) by Roger Penrose and Wolfgang Rindler - This book is about the mathematics of special relativity, it very quickly goes over my head by I hope I will understand it one day.

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