Circumference and area of ellipses in non-Euclidean geometries Jan K¨arrman Department of Mathematics Uppsala University Sweden Abstract In this paper we derive formulas for circumference and area of an ellipse in hyperbolic geometry. Corresponding formulas for elliptic geometry are given without proof. We also find a relation connecting the formulas for circumference and area. 1 Preliminaries This is a term paper for the course Geometry D given at Uppsala University. Due to this nature of the document, derivation of the main formulas will be carried out in quite some detail. We are deriving the formulas under the simplified assumption that the curvature constant k =1. All real numbers in√ this paper will be non-negative; there will be several simpli- fications of the type x2 = x without further comments about the correctness of such equations. A well-known property of ellipses in Euclidean geometry is used to define a non- Euclidean ellipse: Definition 1. Given two points P and Q, and a positive real number l>d= PQ. The set of all points R such that PR + QR = l is called an ellipse with focal points P and Q, and eccentricity d/l. 2 Ellipses in hyperbolic geometry Given points P and Q, and a real number l, satisfying the conditions for an ellipse. α Let O be the midpoint of PQ.− Let−→ X be any point distinct from O, and let = QOX. Dedekind’s axiom gives that OX cuts the ellipse in exactly one point R=R(α). Let r r α s l−s = ( )=←−→OR and = QR, then PR = , see Figure 1. Choose one of the sides of the line PQ , and consider rays emanating from O on that side, with 0 α π/2. 1 R l-s X s r α P d/2O d/2 Q Figure 1 We have d d cosh s = cosh r cosh − sinh r sinh cos α 2 2 d d cosh (l − s)=coshr cosh − sinh r sinh cos (π − α) 2 2 d d = cosh r cosh +sinhr sinh cos α, 2 2 which gives d cosh (l − s)+coshs =2coshr cosh 2 d cosh (l − s) − cosh s =2sinhr sinh cos α. 2 Combining this with the trigonometrical identities l l − 2s cosh (l − s)+coshs =2cosh cosh 2 2 l l − 2s cosh (l − s) − cosh s =2sinh sinh , 2 2 we get d l − 2s cosh r cosh 2 cosh = l 2 cosh 2 l − s sinh r sinh d cos α 2 2 , sinh = l 2 sinh 2 from which follows l − 2s l − 2s 1=cosh2 − sinh2 2 2 2 l 2 2 d 2 l 2 2 d 2 sinh 2 cosh r cosh 2 − cosh 2 sinh r sinh 2 cos α = 2 l 2 l sinh 2 cosh 2 l l l d l d ⇒ sinh2 cosh2 =sinh2 (sinh2 r + 1) cosh2 − cosh2 sinh2 r sinh2 cos2α 2 2 2 2 2 2 2 sinh2 l (cosh2 l − cosh2 d ) ⇒ 2 r 2 2 2 . sinh = 2 l 2 d 2 l 2 d 2 (1) sinh 2 cosh 2 − cosh 2 sinh 2 cos α We define the major and minor axes of the ellipse in analogy with the Euclidean case: set a = r(0), b = r(π/2), then sinh2 l (cosh2 l − cosh2 d ) l 2 a 2 2 2 2 sinh = 2 l 2 d 2 l 2 d =sinh (cosh 2 − 1) cosh 2 − cosh 2 (cosh 2 − 1) 2 sinh2 l (cosh2 l − cosh2 d ) cosh2 l 2 b 2 b 2 2 2 2 . cosh =sinh +1= 2 l 2 d +1= 2 d sinh 2 cosh 2 cosh 2 So we have d cosh a l =2a, cosh = . 2 cosh b It is easy to see that a and b can be use as parameters instead of l and d. Then equation (1) becomes 2 a 2 a − cosh2 a sinh cosh cosh2 b sinh2 r = , 2 a cosh2 a − 2 a cosh2 a − 2α sinh cosh2 b cosh cosh2 b 1 cos cosh2 b which, after multiplying numerator and denominator with the factor sinh2 a cosh2 a , becomes 2 b − 2 b 2 b 2 r cosh 1 sinh sinh sinh = cosh2 a−cosh2 b = sinh2 a−sinh2 b = 2 2 (2) − 2α − 2α 1 − k1 cos α 1 sinh2 a cos 1 sinh2 a cos where sinh2 a − sinh2 b k1 = . sinh a π Note that b r(α) a, and that r(α) is strictly decreasing for 0 α 2 when b<a. Ri+1 δ i Ri ri+1 ri π/2n O Figure 2 We will first derive a formula for the circumference of an ellipse. Let n be a positive πi πi integer, ri = r( 2n ), Ri =R(2n ), and δi = RiRi+1, see Figure 2. Then πi π π 1 ri+1 = r + = ri + ri + O , 2n 2n 2n n2 3 which gives r r O 1 , sinh i+1 =sinh i + n and π cosh δi = cosh ri cosh ri+1 − sinh ri sinh ri+1 cos 2n 2 π = cosh ri cosh ri+1 − sinh ri sinh ri+1 1 − 2sin 4n 2 π =2sinhri sinh ri+1 sin +cosh(ri+1 − ri) 4n π 2 r 2 r − r O 1 =2sinh i sin n +cosh( i+1 i)+ n3 4 π 2 π 2 r r O 1 O 1 =2 n sinh i + cosh n i + n2 + n3 4 2 2 2 π 2 1 π 1 =2 sinh ri +1+ ri + O . 4n 2 2n n3 2 δi From cosh δi =2sinh +1,weget 2 2 2 δi π 2 2 1 sinh = sinh ri +(ri) + O 2 4n n3 δ π i 2 2 1 ⇒ sinh = sinh ri +(r ) + O 2 4n i n2 π 2 2 1 ⇒ δi = sinh ri +(r ) + O . 2n i n2 Now sum all δi,i=0, ...,n− 1, and pass to the limit as n →∞. This gives, by symmetry, one fourth of the circumference C of the ellipse. Hence π/2 C =4 sinh2 r(α)+r(α)2 dα. 0 From (2) we get d − 3 r α r α r α −k2 b α α − k2 2α 2 , cosh ( ) ( )=dα sinh ( )= 1 sinh sin cos 1 1 cos whence sinh2 r(α)+r(α)2 sinh2 b k4 sinh2 b sin2α cos2α = + 1 2 2 2 3 2 1 − k1 cos α 1 − k cos2α cosh r(α) 1 2 b k4 2 b − 2α 2α sinh 1 sinh 1 cos cos = 2 2 + 3 2 1 − k1 cos α − k2 2α sinh b 1 1 cos 1−k2 cos2α +1 1 2 2 2 2 2 2 4 2 2 2 sinh b 1 − k1 cos α cosh b − k1 cos α + k1 sinh b 1 − cos α cos α = 2 1 − k2 cos2α cosh2 b − k2 cos2α 1 1 2 2 2 2 2 cosh b − k1 cosh b +1− k1 cos α =sinh2 b , 2 2 2 2 2 2 1 − k1 cos α cosh b − k1 cos α 4 and hence π/2 2 2 2 2 2 cosh b − k1 cosh b +1− k1 cos α C =4sinhb dα. 2 2 2 2 2 0 1 − k1 cos α cosh b − k1 cos α This integral can be transformed into a more standardized form, by applying a change of variable α = arctan (u tan β), where the real number u will be chosen later in some suitable way. Then 1 1 cos2β 1 − sin2β cos2α = = = = . 1+tan2α 1+u2 tan2β cos2β + u2 sin2β 1 − (1 − u2)sin2β For convenience, we define 1 tanh2 a − tanh2 b cβ = ,m1 = . 1 − (1 − u2)sin2β tanh2 a We will also make use of the following equation 1 1 1 cosh2 a − cosh2 b m1 = − = tanh2 a cosh2 b cosh2 a sinh2 a cosh2 b sinh2 a − sinh2 b k2 = = 1 . sinh2 a cosh2 b cosh2 b To avoid getting huge expressions; we apply the change of variable to the different parts of the integral in turn. 2 2 2 2 2 i) cosh b − k1 cosh b +1− k1 cos α 2 2 2 2 2 1 − sin β = cosh b − k1 cosh b +1− k1 1 − (1 − u2)sin2β 2 2 2 2 2 2 2 = cβ cosh b 1 − (1 − u )sin β − k1(cosh b +1− k1)(1 − sin β) 2 2 2 2 2 2 2 2 2 = cβ (1 − k1)(cosh b − k1)+(k1(1 − k1) − (1 − k1 − u ) cosh b)sin β) Now choose u so that the term containing sin2β vanishes 2 2 2 2 4 sinh b 2 sinh b cosh b tanh b sinh b = cβ cosh b(1 − m1)=cβ = cβ . sinh2 a sinh2 a tanh2 a sinh2 a tanh2 a That is, u has been chosen such that 2 2 2 2 2 (1 − k1 − u ) cosh b = k1(1 − k1) 2 2 2 2 (1 − k1) cosh b − k1(1 − k1) 2 ⇒ u = = (1 − k )(1 − m1) 2 b 1 cosh sinh2 b tanh2 b sinh b tanh b = = . sinh2 a tanh2 a sinh a tanh a 5 2 2 2 2 1 − sin β 2 2 2 2 ii) 1 − k1 cos α =1− k1 = cβ 1 − k1 − (1 − k1 − u )sin β 1 − (1 − u2)sin2β 2 2 2 2 k1(1 − k1) 2 sinh b 2 = cβ 1 − k1 − sin β = cβ (1 − m1 sin β). cosh2 b sinh2 a 2 2 2 2 2 2 1 − sin β iii) cosh b − k1 cos α = cosh b − k1 1 − (1 − u2)sin2β 2 2 2 2 2 2 = cβ cosh b − k1 − ((1 − u ) cosh b − k1)sin β 2 2 2 2 2 2 2 2 = cβ cosh b − k1 − (k1(1 − k1)+k1 cosh b − k1)sin β 2 2 2 2 2 2 2 = cβ cosh b − k1 1 − k1 sin β = cβ cosh b(1 − m1)(1 − k1 sin β) 2 2 2 tanh b 2 2 sinh b 2 2 = cβ cosh b (1 − k1 sin β)=cβ (1 − k1 sin β).