Supplementary information to The Kepler problem from a differential geometry point of view

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1 Supplementary information to The Kepler problem from a differential geometry point of view Supplementary information to The Kepler problem from a differential geometry point of view January 9 07 Contents Dr. Thomas S. Ligon ORCID X Introduction... I. Differential Geometry... II. Symplectic Differential Geometry... III. The Significance of the Symplectic Form... IV. Reduction to a Smaller Dimension... V. The Kepler Problem and First Reduction... VI. Stepwise Reduction... 3 VII. Calculation of the Trajectories of H via a Single Reduction... 7 VIII. Discussion of the Symmetry and Trajectories of M... 3 IX. Stereographic Projection and global SO(4 (resp. SO(3 Symmetry... 5 X. Concluding Remarks... 79

2 Supplementary information to The Kepler problem from a differential geometry point of view Introduction This document consists of detailed calculations for the proofs in the main paper. I. Differential Geometry II. Symplectic Differential Geometry III. The Significance of the Symplectic Form IV. Reduction to a Smaller Dimension V. The Kepler Problem and First Reduction (V.5. Proof. p = ( dq = ( dq dq ] = dq dq = p p. p = dq = (dq dq = (p p. q = qq (by definition ( q = q q = (q q q = q ( q q = q q. q = q q = q q q = (q q p = p p = p = p p. (p p p = p p = p p p = H = p p qq p = p p. H = ( p p ( p p q q = q q. p p H = ( p p p p ] ( p p p p ] q H = ( ( ] p H = ( p ( p q. q ] p p ] p

3 Supplementary information to The Kepler problem from a differential geometry point of view 3 VI. Stepwise Reduction (VI.4. Proof. (i According to (II.4. we have e.g. n i= {L L } = ( L L L L q i p i p i q i = (0(q 3 (0(p 3 (p 3 (0 (0(0 (p (q (q (p = q p q p = L 3. (ii Because of the Jacobi identity (II.. we have {{L L } H} {{L H} L } {{H L } L } = 0. Since L and L are constants of the motion {L H} = {H L } = 0 and so {L 3 H} = {{L L } H} = 0. (VI.8. Proof. First remember for the -dimensional Model III: H(q p = p q L = q p q p. Then a few intermediate steps: L q = p L p = q L q = p L p = q H = ( = q i q i q ( (q q i = 3 H = ( p = p i = p i p i p i The main result: q i q 3. {L H} = L q H p L p H q L q H p L p H q = (p ( p (q ( q q 3 (p ( p (q ( q q 3 = 0. Now a few more intermediate results: A i = q q i q i ( (q q i (p 3 i p = q q i q 3 A i = q q i ( (q j q j (p 3 ip j = q iq j p ip j. q 3 (p j = q j A i p i = (q p q ip i q i (p i ] = q ip i q j p j q i p i q i (p i ] = q jp j. A i p j = (q jp i q i p j. The main result: p j. q 3

4 4 Supplementary information to The Kepler problem from a differential geometry point of view {A H} = A q H p q 3 (p = ( q A H A H A H p p q q p p q (q p ( q q 3 (q q p p q 3 (p (q p q p ( q q 3 = q 3 (q p q q p q q p q p q q p (p p p p = 0. {A H} = 0. {A H} = A q H A H A H A H p p q q p p q = ( q q p p q 3 (p (q p q p q 3 ( q ( q p q 3 (p q 3 (q p ( q = q 3 (q q p q p q q p q p q q p (p p p p = 0. {A H} = 0. da = A q dq A q dq A p dp A p dp = ( q p q 3 dq ( q q q 3 p p da = A q dq A q dq A p dp A p dp = ( q q p p dq q 3 ( q p q 3 dq ( q p dp ( q p q p dp. dq ( q p q p dp ( q p dp. (VI.0. Proof. First we ll solve one equation for x 3 and the other for x 4 and then substitute the expression for x 3 into the expression for x 4 resulting in x 4 in terms of x and x. One more substitution then gives us x 3 in terms of x and x. da (X = ( q da (X = 0 x 3 = p q 3 ( q p q p q 3 da (X = 0 x 4 = x ( q q x ( q q q 3 p p q 3 p p ( q q p p x q p q 3 ( q p q 3 x ( q p x 3 ( q p q p x 4 x ( q p q p x 4 ]. x ( q p q p x 3 ]. Now we substitute the expression for x 3 into the expression for x 4 and use the common denominator of q q p p. x 4 = ( q q p q p p q q p p q 3 x (q q q p p p q p ( q x 4 = ( q q p q p p q q p p q 3 p q 3 (q q q p p p q p ( q p q p x x 4 ( q q p q p p q q p q 3 p q 3 p q q p p q 3 q 3 q 3 x q q p p (q p q p ( q q q 3 p p x q p p x ( q q p q p p q q p q p p q q p q p p q q p p q 3 q 3 q 3 x

5 Supplementary information to The Kepler problem from a differential geometry point of view 5 ( q q p p q p q p 4q q p p q q p p Now we can multiply both sides by q q p p 0 = ( q q p q 3 p q p 3 q p p x q 3 ( q q p q q p q p p q p p x q 3 ( (q p q p x 4 Now we divide both sides by L = (q p q p 0 = ( q p q 3 x 4 = L ( q x ( q q p q 3 x 4 = γ 4 x γ 4 x. q 3 p p x ( q q x 4 and collect terms. x ( (q p q p x 4 q 3 p p x Now we can substitute this expression for x 4 into the original expression for x 3. x 3 = x 3 = x 3 = ( q p q p q 3 x ( q p q p q p L ( q q p ( q q q 3 p p x p q 3 q p q p q 3 (q p q p q p q q q 3 p p x L ( q ( q p q p q p L (q q p p q 3 p q 3 ] x (q p q p L (q q p p q 3 ] x q p q 3 p q q p q p p q p 3 q p q 3 L q 3 L ] x q q p p q q p q q p q p p q p p ] x q p q 3 L q 3 L x Now we need to multiply a few terms by L = (q p q p in order to get the common denominators of L q 3 and L. x 3 = x 3 = q q p q 3 p q 3 p q q p q p 3 q p p q p p q p 3 ] x q p L q 3 L q q p q q p q q p q q p q p p q p p q p p q p p ] x q p L q 3 L q q p q p p q p L q 3 L x 3 = L q q p p ] x q 3 L q x 3 = γ 3 x γ 3 x. ] x p q 3 q q p q p p q p L q 3 L ] x ] x (VI.. Proof. Let XY be vector fields on A (A 0. Then we have ω(x Y = x y 3 x y 4 x 3 y x 4 y ω(x Y = x (γ 3 y γ 3 y x (γ 4 y γ 4 y y (γ 3 x γ 3 x y (γ 4 x γ 4 x ω(x Y = (γ 3 γ 3 x y (γ 3 γ 4 x y (γ 4 γ 3 x y (γ 4 γ 4 x y ω(x Y = (γ 4 γ 3 (x y x y

6 6 Supplementary information to The Kepler problem from a differential geometry point of view ω(x Y = L ( q q 3 p (x y x y ω(x Y = ( H L (x y x y ω A (A 0 = ( H L dq dq. (VI.3. Proof. First some intermediate steps: L = (q p (q p = q p (q p First we show how the calculation would be done in 3 dimensions. A = q q p L q (p L = L (q p = L q A = q q (p L = q L q L = (q A q The main result: A = q q q (p L q (p L A = L q p L (p L A = ( p q L (since p L = 0 A = H L Now we repeat the calculations for dimensions. L = (q p q p L = q p q q p p q p L = q p q p q p q p q p q p q q p p L = (q q (p p (q p q p L = q p (q p. q A = q q (q p q p q A = q L L = (q A q. A = q q q (q p q p ] (q p p (q p p q p 4 A = (q p q A = L q q p (q p p L p q p 4

7 Supplementary information to The Kepler problem from a differential geometry point of view 7 A = H L H = ( A L = ( A (q A q. (VI.5. Proof. H = ( A (A q q = (A q q H q (q A q A q (q A q A q q = {q H} = ( L H H q because of (VI.. q = ( L (A q q (q A q A q q = {q H} = ( L H H q because of (VI.. q = ( L (A q q (q A q A q q = q q q q = ( L A sin φ ( q q A cos φ (sin φ = cos φ φ (sin φ = q A q A (q A q A q q A q A φ = d q dφ d q q L q. = q φ = q A sin φ A cos φ A d(cos φ = A cos φ ln q = ln( A cos φ const. q = const. A cos φ. = ( L q cos φ VII. Calculation of the Trajectories of H via a Single Reduction (VII.. Proof. A = q p L = q p (q p = q q q A = q A i = q i p(q pq p q p i (q pq i p q. q(p pp(q p q A = A A = ( q q p L ( q q p L A = q q (p L q q (p L (p L A = L (q p p L (p L q A = L p L 0 q = L ( p q

8 8 Supplementary information to The Kepler problem from a differential geometry point of view A = H L. L A = L ( q L A = 0. p L q L M = L A = 0. H For H < 0: but L q = q L = 0 and L (p L = p (L L = 0 M L = ( H A L = ( H A L = H A L M L = H M L = H. For H > 0: H L ( L = H M L = ( H A L = ( H A L = H A L M L = H M L = H. H L ( L = H {H L i } is already known from (VI.4. but here is a calculation: {H L i } = {H q j p k q k p j } {H L i } = q j {H p k } p k {H q j } q k {H p j } p j {H q k } {H L i } = q j ( q k q 3 p k ( p j q k ( q j q 3 p j ( p k (from the proof of (VI.8. {H L i } = 0. Now for some intermediate results: {H q i } = H p i = p i. {H p i } = H q i = q i ( q / ] = ( q 3/ q i {H p i } = q i q 3. {H q p} = {H q p} = q {H p} p {H q} {H q p} = q ( q q 3 p (p {H q p} = q q 3 p {H q p} = q p. {H } = {H q ( q / } {H } = q ( q 3/ q {H q}

9 Supplementary information to The Kepler problem from a differential geometry point of view 9 {H } = ( q q 3 q ( p {H } = q q p q 3. {H p p} = p {H p} {H p p} = p {H p p} = q p q 3. q q 3 Now for the main result: {H A i } = {H q i p i (q pq i p q {H A i } = q i {H } {H q q q i} p i q p {H q p} {H p i} q i {H A i } = q i ( q p q 3 q (p i p i ( {H A i } = q i ( q p q 3 ( p {H A i } = 0. {H M i } = 0. } p q q p ( q i i ( p i p q q 3 q i p p {H p p} {H q i} (q p q ( 3 p p (p i (VII.4. Proof. We begin by showing q = q cos θ by definition (VII.3. q = q sin θ = q A A. q = q (L A L A by definition (VII.3. q = L (q A L A q = L q q L A q = L A q = L (q p A q sin θ = q A A q sin θ = A q (p L q L p(q L L(q p q (p L ] q sin θ = A p(q L L(q p] q sin θ = L A (q p

10 0 Supplementary information to The Kepler problem from a differential geometry point of view q sin θ = ± L (q p A q = ± q sin θ. Main results: First we ll calculate q. q A = q q (p L q A = q L (q p q A = q L cos θ = q A q A by definition (VII.3. cos θ = L A A q A cos θ = L q q = L ( A cos θ. sin θ = q q = sin θ = q (L A q L A by definition (VII.3. sin θ = sin θ = q L A q L A q L A sin θ = L (q p q L A sin θ = L (q p A (q p = L (q A L (q q q q (p L L p(q LL(q p cos θ (( A L L A sin θ. ( A cos θ q = q cos θ q = by definition (VII.3. L cos θ ( A cos θ. q = q sin θ q = p = p A A L sin θ ( A cos θ. by definition (VII.3. p = A p q p (p L q

11 Supplementary information to The Kepler problem from a differential geometry point of view p = A q p p = p = p = q p A q L (p p q L A sin θ cos θ ( (( A A ( A cos θ L sin θ. L p = q ( L (q p cf. remark after (VII.3. p = ( A cos θ ( L L A sin θ L 4 ( A cos θ p = L ( A cos θ A sin θ p = L A cos θ A cos θ A sin θ p = L A cos θ A ]. p = p (L A L A by definition (VII.3. p = p = L A p (L q q L A L (q p q p = L A L p (L (p L p (p(l LL(L p p L q p = L p A q p = L cos θ ( A A L ( A cos θ L A cos θ p = ( A ( A cos θ L A L L A p = A cos θ L A A p = A cos θ]. L (VII.5. Proof. For q we have the Hamilton equation: dq dt = {q H} = H p = p dq dθ = p (θ(t dθ dx ( L d ( cos θ dθ A cos θ (dθ = ( sin θ. dt L d ( dθ cos θ A cos θ = sin θ ( A cos θ ( L sin θ A sin θcos θ A sin θcos θ ( ( A cos θ ( L ( sin θ ( A cos θ (dθ = ( sin θ dt L ( dθ = ( sin θ dt L

12 Supplementary information to The Kepler problem from a differential geometry point of view dθ dt = L 3 ( A cos θ. The other cases are analogous and produce the same result. We will calculate one of them. dq dt = {q H} = H = p p dq dθ = p (θ(t dθ dt ( L d ( sin θ dθ A cos θ (dθ = ( ( A cos θ. dt L L θ( A cos θsin θ( A sin θ cos ] dθ = ( A cos θ dt L L ( A cos θ. cos θ A cos θ A sin θ ] dθ = ( A cos θ. ( A cos θ dt L L cos θ A dθ ( A cos θ ] = dt L dθ dt = L 3 ( A cos θ. ( A cos θ (VII.7. Proof. cos u = A cos θ A cos θ by definition (VII.6. cos u A cos u cos θ = cos θ A cos θ ( A cos u = cos u A cos θ = A cos u A cos u. sin u = A sin θ A cos θ by definition (VII.6. sin u A sin u cos θ = A sin θ sin θ = sin θ = sin θ = sin u( A cos θ A sin u A A cos u ( A A cos u sin u A ( A cos u A A cos u A cos u sin θ = A sin u A cos u. ( A cos u( A cos u = A A cos u = A A cos u. (VII.8. Proof. cos θ = A cos u A cos u

13 Supplementary information to The Kepler problem from a differential geometry point of view 3 ( d(cos θ dt d(cos θ dt d(cos θ dt from (VII.7. d(cos θ dt = ( A sin uu ( A cos u. = sin θ dθ dt = L 3 sin θ ( A cos θ from (VII.5. = sin u L 3 ( A ( ( A A cos u A cos u from (VII.7. d(cos θ ( = dt ( = ( u = du = dt dt = t = L 3 ( A 3 L 3 ( A 3 ( A 5/ sin u. L 3 ( A cos u 3 ( A 3 L 3 A cos u ( A cos udu = L 3 (u A sin u. ( A 3 d(u A sin u VIII. Discussion of the Symmetry and Trajectories of M (VIII.. Proof. For convenience we will repeat a few things here: A = q A = q p(q pq p q p (q pq p q M = H A H = p q Then a few intermediate steps: {p i H} = H q i = q i ( q = ( q 3 (q i = q i q 3 {p i H} = H q i = q i q 3. {q i H} = H p i = p i ( p = p i = p i. Now for the main results. {q A } = ( q 3 A q A i= q i p i p i q i {q A } = A p =( (q p q 3 p 3.

14 4 Supplementary information to The Kepler problem from a differential geometry point of view {q A } = A p =( (q p q p. {q 3 A } = A p 3 =( (q 3p q p 3. 3 {p A } = ( p i= = A A p A q i p i p i q i q {p A } = q q ( q q 3 (p p 3 {p A }= q 3 (q q q 3 q (p p 3 {p A } = q 3 (q q 3 (p p 3. {p A } = A q = q ( q q 3 (p p {p A } = q 3 q q (p p. {p 3 A } = A q 3 = q ( q 3 q 3 (p p 3 {p 3 A } = q 3 q q 3 (p p 3. {q M } = {q H A } {q M } = ( H 3/ {q H }A H {q A } {q M } = H 3/ {q H } A H {q A } {q M } = H 3/ ±{q H} A ±H{q A } with sign for H > 0 {q M } = ± H 3/ p A H (q p q 3 p 3 with sign for H < 0. {q M } and {q 3 M } are analogous so we can present a shorter calculation. {q M } = {q H A } {q M } = H 3/ ±{q H} A ±H{q A } with sign for H > 0 {q M } = ± H 3/ p A H ( (q p q p with sign for H < 0. {q 3 M } = {q 3 H A } {q 3 M } = H 3/ ±{q 3 H} A ±H{q 3 A } with sign for H > 0 {q 3 M } = ± H 3/ p 3 A H ( (q 3p q p 3 with sign for H < 0. {p M } = {p H A }

15 Supplementary information to The Kepler problem from a differential geometry point of view 5 {p M } = ( H 3/ {p H }A H {p A } {p M } = H 3/ {p H } A H {p A } {p M } = H 3/ ±{p H} A ±H{p A } with sign for H > 0 {p M } = ± H 3/ {p H} A H{p A } with sign for H < 0 {p M } = ± H 3/ ( q q 3 A H ( q 3 (q q 3 (p p 3 with sign for H < 0. {p M } and {p 3 M } are analogous so we can present a shorter calculation. {p M } = {p H A } {p M } = ± H 3/ {p H} A H{p A } with sign for H < 0 {p M } = ± H 3/ ( q q 3 A H ( q 3 q q (p p with sign for H < 0. {p 3 M } = ± H 3/ {p 3 H} A H{p 3 A } with sign for H < 0 {p 3 M } = ± H 3/ ( q 3 q 3 A H ( q 3 q q 3 (p p 3 with sign for H < 0. (VIII.. Proof. We can introduce six new coordinates: q p ( L q H and A. These six functions are independent and three of them are constant namely L H and A. As a result we need to find the differential equations for q p and ( as functions of these six new q coordinates. A = q p (q pq p = q q A = q p (q p q 3 p 3 q (p p 3 q H = p = p p p 3 q q p p 3 = (H q p p (q p q p q 3 p 3 q (p p p 3 q A = q q p (q p q 3 p 3 q (H q q p p (q p q 3 p 3 = q (H q q p A q q p (q p q 3 p 3 = A q H q q q p q p q 3 p 3 = p A q H q q q p ].

16 6 Supplementary information to The Kepler problem from a differential geometry point of view q = {q M } = ± H 3/ p A H (q p q 3 p 3 q = ± H 3/ p A H p (A q H q q q p q = ± H 3/ p A H p (A q H q q p q q = ± H 3/ p (A Hq H p ( q q A Hq ]. p = {p M } = ± H 3/ ( q q 3 A H ( q 3 (q q 3 (p p 3 p = ± H 3/ ( q q 3 A H (q q q 3 ( p p p = ± H 3/ ( q q 3 A H (q q q 3 ( (H q p p = ± H 3/ ( q 3 (q A Hq H ( H p q q p = ± H 3/ ( q 3 ( A q Hq H ( H q (p ]. ( q = ( ( q 3 (q q q q q 3 q 3 ( q = ( q 3 (q {q M } q {q M } q 3 {q 3 M } A (q p q p q 3 p 3 ( q = H 3 ( q 3 H (q q p q q 3 p 3 q p q q p q 3 p q q 3 p 3 ( q = H 3 ( q 3 A q p A (q p q 3 p 3 H (q (q p q 3 p 3 (q q 3 p q (q p q 3 p 3 A q p H ( q = H 3 ( q 3 q p H q p ( A H q ( (A p Hq q q q p ( q = H 3 ( A ( Hq q 3 ( q p H q p ( A Hq ( p (A Hq q q p q ( = q H 3/ ( q 3 ( A Hq ( ( q A p q Hq ( Hp q ]. (VIII.3. Proof. {H L i } = 0 from (VI.. {L i L j } = L k from (VI.4. {H A i } = 0 from (VI.8..

17 Supplementary information to The Kepler problem from a differential geometry point of view 7 {L i A j } = {(q p i A j } = {q j p k q k p j A j } {L i A j } = q j {p k A j }p k {q j A j } q k {p j A j } p j {q k A j } {L i A j } = q j q 3 q jq k q j(p j p k p k ( (q ip i q k p k q k ( q 3 (q i q k q k ( (p i p k p j ( (q k p j q j p k {L i A j } = q 3 (q j q k q i q k q k 3 (q jp j p k q i p i p k q k p k q k p i q k p k q k p j q j p j p k {L i A j } = q 3 (q k q (q ip i p k q j p j p k q k p i q k p j {L i A j } = q k q (p k(q p q k p {L i A j } = A k. {L i A i } = {(q p i A i } = {q j p k q k p j A i } {L i A i } = q j {p k A i }p k {q j A i } q k {p j A i } p j {q k A i } {L i A i } = q j q 3 q iq k q j(p i p k p k ( (q jp i q i p j q k ( q 3 (q iq j q k ( (p i p j p j ( (q k p i q i p k {L i A i } = q 3 (q iq j q k q i q j q k (q jp i p k q j p i p k q i p jpk q k p i p j q k p i p j q i p j p k {L i A i } = 0. Now we need some more intermediate results. { q A i} = { q A i} = { q A i} = 3 ( (A q m q p i ( m= (A m p m q q i m 3 ( m= ( q 3 (q m (A p i 0] m 3 ( q m q 3 m= p m (A i ] { q A i}=( q 3 ( (q iq p q i p i q i p i q j p i q i q j p j q k p i q i q k p k { q A i}=( q 3 ( (q iq p q i p i q j p i q k p i q i q p q i p i { q A i}=( q 3 ( (q iq p q p i { A q i}=( (q iq p q 3 p i q. {q p A i } = 3 m= q m {p m A i } p m {q m A i } {q p A i } = q i q 3 (q j q k q i (p j p k q j q 3 (q iq j q j (p ip j q k q 3 (q iq k q k (p ip k

18 8 Supplementary information to The Kepler problem from a differential geometry point of view p i (q ip j q k p k p j (q jp i q i p j p k (q kp i q i p k {q p A i } = q 3 (q iq j q i q k q i q j q i q k ( q ip j q i p k q j p i p j q k p i p k q i p i p j q k p i p k q j p i p j q i p j q k p i p k q i p k {q p A i } = (q ip j q i p k q j p i p j q k p i p k {q p A i } = (q i p p i q p. { p A i } = p i {p i A i } p j {p j A i } p k {p k A i } { p A i } = q 3 p i(q j q k ] p i(p j p k ] q 3 p j(q i q j ] p j(p i p j ] q 3 p k(q i q k ] p k(p i p k { p A i } = q 3 (q j p i q k p i q i q j p j q i q k p k (p ip j p i p k p i p j p i p k { p A i } = q 3 (q i p i q j p i q k p i q i p i q i q j p j q i q k p k 0 { p A i } = q 3 (q i(q p p i q { p A i } = q i (q p p i. q 3 q Now back to the main results. {A i A j } = { q i p i (q pq i p q A j } {A i A j } = q {q i A j } q i { q A j} p i {(q p A j} q p {A i A j } = q ( q ip j q j p i ] q i ( q jq p q 3 q p q 3 q iq j (p ip j ] q i q j (q p q 3 {p i A j } q i { p A j } p {q i A j } p j q ] p i ( q j p p j q p] p j q ] p ( (q ip j q j p i = q 3 (q ip j q q j p i q q i q j q p q i p j q q i q j q p q i q j q p q i p j q (q jp i p p i p j q p p i p j q p q i p j p q j p i p {A i A j } = q 3 ((q ip j q j p i q ((q ip j q j p i p {A i A j } = q (L k (L k p {A i A j } = ( q p L k {A i A j } = H L k. For H < 0: {M i M j } = { H A i H A j} = H {A i A j } = L k.

19 Supplementary information to The Kepler problem from a differential geometry point of view 9 Now we ll prepare some more intermediate results. {q i q j } = {q i q j } = 0 {p i p j } = 0 {q i p i } = {q i p j } = 0 3 ( q i q j q i q j m= = 0 q m p m p m q m {q i L i } = {q i q j p k q k p j } = q j {q i p k } p k {q i q j } q k {q i p j } p j {q i q k } = q j (0 p k (0 q k (0 p j (0 = 0 {q i L i } = 0. {q i L j } = {q i q k p i q i p k } = q k {q i p i } p i {q i q k } q i {q i p k } p k {q i q i } {q i L j } = q k ( p i (0 q i (0 p k (0 = q k {q i L j } = q k. {p i L i } = 0. {p i L j } = {p i q k p i q i p k } = q k {p i p i } p i {p i q k } q i {p i p k } p k {p i q i } {p i L j } = q k (0 p i (0 q i (0 p k ( = p k {q i L j } = p k. {L i L j } = {q j p k q k p j L j } = q j {p k L j } p k {q j L j } q k {p j L j } p j {q k L j } {L i L j } = q j (p i p k (0 q k (0 p j (q i = q j p i q i p j = L k {L i L j } = L k. Now back to the main results. For H < 0: {( L±M i (L±M j }={L i ± M i L j ± M j } = {L 4 i L j } ± {L 4 i M j } ± {M 4 i L j } {M 4 i M j } {( L±M i (L±M j } = 4 L k ± 4 H {L i A j } ± 4 H {A i L j } 4 ( H {A i A j } {( L±M i (L±M j } = 4 L k ± 4 H A k ± 4 H A k 4 ( {( L±M i (L±M j } = L k ± M k. {( L±M i (L±M j } = (L±M k. H (H {( LM i (LM j }={L i M i L j M j } = {L 4 i L j } {L 4 i M j } {M 4 i L j } {M 4 i M j } {( LM i (LM j } = 4 L k 4 H {L i A j } 4 H {A i L j } 4 ( H {A i A j } {( LM i (LM j } = 4 L k 4 H A k 4 H A k 4 ( H (H L k L k

20 0 Supplementary information to The Kepler problem from a differential geometry point of view {( LM i (LM j } = 0. For H > 0: {M i M j } = ( H {A i A j } = ( H H L k = L k {M i M j } = L k. For H = 0: {A i A j } = H L k=0. (VIII.4. Proof. <<Citation>> (VIII.5. Proof. Because of (I.7. and (II.8. we need to calculate the Poisson brackets. for H 0: {L i M j } = H {L i A j } = H A k = M k (VIII.3. {L i M j } = M k. {L M } = H {L A } = 0 (VIII.3. {A M } = H {A A } = 0 {H M } = 0 (VII.. {L M L 3 M 3 M } = L {M M } M {L M } L 3 {M 3 M } M 3 {L 3 M } {L M L 3 M 3 M } = L {A H A } M (M 3 L {A 3 H 3 A } M 3 (M {L M L 3 M 3 M } = L {L M L 3 M 3 M } = 0. For H = 0: H H (L H 3 M M 3 L (L 3 H M M 3 = 0. {L A L 3 A 3 A } = L {A A } A {L A } L 3 {A 3 A } A 3 {L 3 A } {L A L 3 A 3 A } = L (0 A (A 3 L 3 (0 A 3 (A = 0 {L A L 3 A 3 A } = 0. For H < 0: {L M 3 M } = L {L M } M 3 {M 3 M } {L M 3 M } = L (M 3 M 3 (L = 0. {L M 3 M } = 0. {L 3 M M } = L 3 {L 3 M } M {M M } {L 3 M M } = L 3 (M M (L 3 = 0. {L 3 M M } = 0.

21 Supplementary information to The Kepler problem from a differential geometry point of view {M M 3 L L 3 M } = M {M 3 M } M 3 {M M } L {L 3 M } L 3 {L M } = {M M 3 L L 3 M } = M (L M 3 (L 3 L (M L 3 (M 3 = 0 {M M 3 L L 3 M } = 0. For H > 0: {L M 3 M } = L {L M } M 3 {M 3 M } {L M 3 M } = L (M 3 M 3 (L = 0. {L M 3 M } = 0. {L 3 M M } = L 3 {L 3 M } M {M M } {L 3 M M } = L 3 (M M (L 3 = 0. {L 3 M M } = 0. {M M 3 L L 3 M } = M {M 3 M } M 3 {M M } L {L 3 M } L 3 {L M } = {M M 3 L L 3 M } = M (L M 3 (L 3 L (M L 3 (M 3 = 0 {M M 3 L L 3 M } = 0. For H = 0: {A A } = 0. {A A } = H (L 3 = 0 (VIII.3.. {L A } = {q p 3 q 3 p A } = q {p 3 A } p 3 {q A } q 3 {p A } p {q 3 A } {L A } = q q 3 q q 3 (p p 3 ] p 3 ( (q p q p ] q 3 q 3 q q (p p ] p ( (q 3p q p 3 {L A } = ( q 3 q q q 3 q q q 3 ] ( q p p 3 q p p 3 q p p 3 q 3 p p q 3 p p q p p 3 ] = 0 {L A } = 0. {H A } = 0 (VI... (VIII.6. Proof. (i For H < 0: Because of (II.8. we have dl ds L = {L M } = M 3 M 3 = {M 3 M } = L (L im 3 = M 3 il = i(l im 3 L im 3 = C e i(sα L = C cos(s α M 3 = C sin(s α. L 3 = {L 3 M } = M M = {M M } = L 3 = L X M L = {L M } = M 3 and so

22 Supplementary information to The Kepler problem from a differential geometry point of view (L 3 im = M il 3 = i(l 3 im L 3 im = C e i(sα L 3 = C cos(s α M = C sin(s α. (L M = L M 3 L 3 M (L M = C cos(s α C sin(s α C cos(s α (C sin(s α (L M = (C sin (s α (C sin (s α sign disagrees with original paper (ii For H > 0: L = {L M } = M 3 M 3 = {M 3 M } = L (L M 3 = M 3 L = (L M 3 L M 3 = C 3 e (sα 3 (L M 3 = M 3 L = (L M 3 L M 3 = C 3 e (sα 3 L = C 3 ( e(sα 3 e (sα 3 ] = C 3 cosh(s α 3 M 3 = C 3 ( e(sα 3 e (sα 3 ] = C 3 sinh(s α 3 L 3 = {L 3 M } = M M = {M M } = L 3 (L 3 M = M L 3 = (L 3 M L 3 M = C 4 e (sα 4 (L 3 M = M L 3 = (L 3 M L 3 M = C 4 e (sα 4 L 3 = C 4 ( e(sα 4 e (sα 4 ] = C 4 cosh(s α 4 M = C 4 ( e(sα 4 e (sα 4 ] = C 4 sinh(s α 4 (L M = L M 3 L 3 M (L M = C 3 cosh(s α 3 (C 3 sinh(s α 3 C 4 cosh(s α 4 C 4 sinh(s α 4 (L M = (C 3 sinh(s α 3 (C 4 sinh(s α 4 - sign disagrees with original paper (iii Since A 3 = {A 3 A } = 0 A 3 is constant. L = {L A } = A 3 L = A 3 s C 5. L 3 = {L 3 A } = A L 3 = A s C 6. (L A = L A 3 L 3 A

23 Supplementary information to The Kepler problem from a differential geometry point of view 3 (L A = (A 3 s C 5 A 3 (A s C 6 A (L A = (A A 3 s A C 6 A 3 C 5. (VIII.7. Proof. For H < 0: d M ds d M ds d M ds d M ds d M ds = d M = M ds M { M M } = ( M {M M } {M M } {M 3 M } = ( M M {M M } M {M M } M 3 {M 3 M } = ( M M (L 3 M 3 (L = (L M. M For H > 0: d M ds d M ds d M ds d M ds d M ds = d M = M ds M { M M } = ( M {M M } {M M } {M 3 M } = ( M M {M M } M {M M } M 3 {M 3 M } = ( M M (L 3 M 3 (L = (L M. M For all H along A : d L = d L = ds L ds L { L A } d L ds = ( L {L A } {L A } {L 3 A } d L ds = ( L L {L A } L {L A } L 3 {L 3 A } d L ds = ( L L (A 3 L 3 (A d L = (L A ds L. For H 0 along M : d L = d L = ds L ds L { L M } d L ds = ( L H {L A } {L A } {L 3 A }

24 4 Supplementary information to The Kepler problem from a differential geometry point of view d L ds = H ((L A L d L = (L M ds L.. (VIII.8. Proof. Because of (VII.4. we can express q and p as functions of M L and θ. Because of (VIII.6. M and L can be expressed as functions of s. Then the differential equation for θ(s follows from (VIII... The proof is analogous to (VII.5.. Here we use the equation for p because it is the simplest. We sketch the proof for H 0: dp ds = d ds ( sinθ = L d L L ds With (VIII.7. we then have: ( dp ds sinθ L = ( L 3 (L M sinθ dθ cosθ. L ds On the other hand we have dp ds = d ds (p M M = dp ds M p dm M ds = cosθ dθ ds. p dm p M dm M M ds M ds and p {M M M } = ( ( (p L M = ± (A q ( with the upper sign for H < 0 q M and the lower sign for H > 0. This implies dp ds = dp M ± (A ds M q ( q M dp M = q M ( M q q p M ds M H q 3 H M q q 3 M p M (L M. From (VIII.. we get: p p dp ds = p M (L M q M q q M H q 3 H ( M M q q q q p p. Now we use (VII.4. and (VII.. and the expression q = q ( M M q ( (L M L M and arrive at ( dp ds = cosθ 3 M 4H M L 4] cos θ H M H M L 4] cos3 θ M H L 4] sinθcosθ (L M ] H L 3 M sinθcos θ (L M ] sinθ (L M ]. L 3 L 3 A comparison of (* and (** yields the result. (VIII.9. Proof. (i With the help of (VIII.6. we can see that the functions in (VIII.8. are continuous and bounded everywhere. Therefore the existence theorem of Peano (cf. Kamke 8 page 6 holds. (ii In (VII.4. q and p were expressed as functions of M L and θ. In (VIII.6. M and L were expressed as functions of s. Because of (VIII.8. and (VIII.9. (i the desired function θ(s exists. With that we have q and p as functions of the integral curve parameter s.

25 Supplementary information to The Kepler problem from a differential geometry point of view 5 IX. Stereographic Projection and global SO(4 (resp. SO(3 Symmetry (IX.. Proof. (i For H = H 0 < 0: ξ 0 = p H 0 p H 0 ξ 0 = p H 0 p H 0 p H 0 ξ k = H 0p k p H 0 p k = ξ k( p H 0 H 0 p k = ( ξ k H 0 (4H 0 ξ 0 p k = H 0 ξ k ξ 0. η k = ( p H 0 q k (q p p k q k = η k (q p p k] p H 0 η 0 q k = η k ( ( H 0 q k = η k ( η 0ξ k ξ 0 ] ( ξ 0 H 0 q k = ( η k (ξ 0 η 0ξ k ξ 0 ] ( ξ 0 H 0 = 4H 0 p H 0 ξ k H 0 (4H 0 q k = ( H 0 (η k ( ξ 0 η 0 ξ k. (ii For H = H 0 > 0: ξ k = H 0p k p H 0 p k = ξ k( p H 0 H 4H 0 0 p k = H 0 ξ k ξ 0. η k = ( p H 0 q k (q p p k q k = η k (q p p k] p H 0 η 0 ξ 0 ] ( ξ 0 q k = η k ( ( H H 0 0 ( ξ k ] ( ξ 0 ξ 0 q k = η k ( η 0ξ k ξ 0 ] ( ξ 0 H 0 q k = ( η k (ξ 0 η 0ξ k ξ 0 ] ( ξ 0 H 0 H 0 H 0

26 6 Supplementary information to The Kepler problem from a differential geometry point of view q k = ( H 0 (η k ( ξ 0 η 0 ξ k. (IX.3. Proof. First we ll get some intermediate results: Preliminary results for H = H 0 0: = p H 0 ξ 0 4H 0 p = H 0 4H 0 ξ 0 p = H 0 ( ξ 0 ξ 0 p = H 0 ( ξ 0 ξ 0. H 0 = p q q = p H 0 q = H 0 ( ξ 0 ξ 0 H 0 ( ξ 0 ξ 0 q = H 0 ξ 0. q = ( ξ 0 H 0. p H 0 = H 0(ξ 0 H 0 (ξ 0 ξ 0 p H 0 = 4H 0 ξ 0. p H 0 = H 0(ξ 0 H 0 (ξ 0 ξ 0 p H 0 = 4H 0 ξ 0 ξ 0. p H 0 p H 0 = (4H 0 ( ξ 0 ξ 0 ( 4H 0 ( ξ 0 p H 0 p H 0 = ξ 0. (i For H = H 0 < 0: η 0 = H 0 (q p (q p = ( η H 0 0. A = q p (q pq p q A = q ( p q (VIII.. p (q p A = ( η H ( ξ 0 η 0 ξ ] H 0 H 0 ξ 0 ( ξ 0 ] ( H 0 ( ξ ( η 0 ξ 0 ξ 0 H 0 0

27 Supplementary information to The Kepler problem from a differential geometry point of view 7 A = η ( ξ 0 η 0 ξ ] ( ξ 0 ξ 0 ξ η 0 ξ 0 A = η ( ξ 0 η 0 ξ ] ( ξ 0 ξ η 0 ξ 0 ξ 0 A = η ξ 0 ( ξ 0 η 0 ξ 0 ξ ξ η 0 ] ( ξ 0 A = η ξ 0 ( ξ 0 ξ η 0 ( ξ 0 ] ( ξ 0 A = ξ η 0 ξ 0 η. ξ 0 3 ξ k = k= ( p H 0 ( p H 0 ( H 0 p ξ 0 3 k= ξ k = ( ξ 0 4H 0 (4H 0 ( ξ 0 (8H ξ 0 (H 0 ( ξ 0 0 ξ 0 ξ 0 3 k= ξ k = ξ 0 ( ξ 0 ( ξ 0 ξ 0 3 k= ξ k =. ξ 0 η 0 3 k= ξ k η k = ( p H 0 ( H 0 H 0 p H 0 ( p p H 0 ( H 0 (q p ( H 0 (q p ( p H 0 p ξ 0 η 0 3 k= ξ k η k = ( H 0 ξ 0 ( H 0 ( η H 0 0 (q p ( η H 0 0 ( H 0 ( ( ξ 4H 0 0 ( η H 0 0(H 0 ( ξ 0 ξ 0 η 0 3 k= ξ k η k = η 0 ( ξ 0 η 0 ( ξ 0 ξ 0 η 0 3 k= ξ k η k = 0. η 0 3 k= η k = ( H 0 (q p ( p H 0 q ( p H 0 ( (q p ( (q p p ξ 0 η 0 3 k= η k = ( H 0 ( η H 0 0 (H 0 ( ( ξ 0 ξ 0 H 0 (H 0 ( ( ( η ξ 0 H 0 0 ( ( η H 0 0 (H 0 ( ξ 0 η 0 3 k= η k = η 0 ( η 0 η 0 (ξ 0 ξ 0 ξ 0 η 0 3 k= η k = η 0 3 k= η k =. ξ 0 (η 0 ξ 0 η 0 ξ 0 η 0 η 0 ξ 0 η 0 ξ 0

28 8 Supplementary information to The Kepler problem from a differential geometry point of view (ii For H = H 0 > 0: η 0 = H 0 (q p (q p = ( η H 0 0. A = q p (q pq p q from (VIII.. A = q ( p q p (q p A = ( η H ( ξ 0 η 0 ξ ] H 0 H 0 ξ 0 ( ξ 0 ] ( H 0 ( ξ ( η 0 ξ 0 ξ 0 H 0 0 A = η ( ξ 0 η 0 ξ ] ( ξ 0 ξ 0 ξ η 0 ξ 0 A = η ( ξ 0 η 0 ξ ] ( ξ 0 ξ η 0 ξ 0 ξ 0 A = η ξ 0 ( ξ 0 η 0 ξ 0 ξ ξ η 0 ] ( ξ 0 A = η ξ 0 ( ξ 0 ξ η 0 ( ξ 0 ] ( ξ 0 A = ξ 0 η ξ η 0. ξ 0 3 ξ k = k= ( p H 0 ( p H 0 ( H 0 p ξ 0 3 k= ξ k = ( ξ 0 4H 0 (4H 0 ( ξ 0 (8H ξ 0 (H 0 ( ξ 0 0 ξ 0 ξ 0 3 k= ξ k = ξ 0 ( ξ 0 ( ξ 0 ξ 0 3 k= ξ k =. ξ 0 η 0 3 k= ξ k η k = ( p H 0 ( H 0 H 0 p H 0 ( p ( H 0 p H 0 ((q p p ξ 0 η 0 3 k= ξ k η k = p H 0 ( H 0 ξ 0 ( H 0 (q p (q p ( η H 0 0 ( H 0 ( η H 0 0 ( H 0 ( ( ξ 4H 0 0 ( η H 0 0(H 0 ( ξ 0 ξ 0 η 0 3 k= ξ k η k = η 0 ( ξ 0 η 0 ( ξ 0 ξ 0 η 0 3 k= ξ k η k = 0. η 0 3 k= η k = ( p H 0 q ( H 0 (q p ξ 0

29 Supplementary information to The Kepler problem from a differential geometry point of view 9 ( p H 0 ( (q p ( (q p p η 0 3 k= η k = ( H 0 ( η H 0 0 (H 0 ( ( ξ 0 ξ 0 H 0 (H 0 ( ( ( η ξ 0 H 0 0 ( ( η H 0 0 (H 0 ( ξ 0 ξ 0 η 0 3 k= η k = η 0 ( η 0 η 0 (ξ 0 ξ 0 ξ 0 η 0 3 k= η k = η 0 3 k= η k =. ξ 0 (η 0 ξ 0 η 0 ξ 0 η 0 η 0 ξ 0 η 0 (IX.4. Proof. In each case the proof consists of a direct application of the definitions and a lengthy calculation. To reduce the volume we introduce a parameter σ=sign(h i.e. σ= if H > 0 and σ= if H < 0. Intermediate results for H 0: q = σ (σh 0 3/ ( (A (p (H 0 ( (q p q 3 p 3 from (VIII.. q = σ (σh 0 3/ because of (IX.. and (IX.3. ( σ(ξ 0η ξ η 0 (σ σh 0 ξ ξ 0 (H 0 ( ( σh 0 (η ξ 0 η ξ η 0 (σ σh 0 ξ ξ 0 (H 0 ( ( (η σh 3 ξ 0 η 3 ξ 3 η 0 (σ σh 0 ξ 3 0 ξ 0 ] q = ( H 0 ( (ξ 0η ξ η 0 (ξ (η ξ 0 η ξ η 0 (ξ (η 3 ξ 0 η 3 ξ 3 η 0 (ξ 3 ] ξ 0 q = ( ( ξ η0 ξ 0 ξ η ξ η ξ 0 ξ η ξ η0 ξ 3 η 3 ξ 0 ξ 3 η 3 ξ 3 η0 H 0 ξ 0 q = ( ( (ξ ξ ξ3 η0 ξ 0 (ξ η ξ η ξ 3 η 3 (ξ η ξ 3 η 3 H 0 ξ 0 q = ( ( (σ(ξ 0 η0 ξ 0 (σξ 0 η 0 (σξ 0 η 0 ξ η H 0 ξ 0 because of (IX.3. q = ( ( ση 0σξ 0 η0 σξ 0 η0 σξ 0 η 0 ξ η H 0 ξ 0 q = ( H 0 ( ση 0 (ξ 0 ξ η ξ 0 q = ( H 0 (ση 0 ξ η ξ 0 q = σ (σh 0 3/ ( (A (p (H 0 ( (q p q p

30 30 Supplementary information to The Kepler problem from a differential geometry point of view from (VIII.. q = σ (σh 0 3/ because of (IX.. and (IX.3. ( σ(ξ 0η ξ η 0 (σ σh 0 ξ ξ 0 (H 0 ( ( σh 0 (η ξ 0 η ξ η 0 (σ σh 0 ξ ξ 0 (H 0 ( ( ( (η σh ξ 0 η ξ η 0 (σ σh 0 ξ 0 ξ 0 ] q = ( H 0 ( (ξ 0η ξ η 0 (ξ (η ξ 0 η ξ η 0 (ξ ((η ξ 0 η ξ η 0 (ξ ] ξ 0 q = ( H 0 ( ξ 0ξ η ξ ξ η 0 ξ η ξ 0 ξ η ξ ξ η 0 ξ η ξ 0 ξ η ξ ξ η 0 ξ 0 q = ( H 0 ( ξ 0ξ η ξ η ξ 0 ξ η ξ η ξ 0 q 3 = σ (σh 0 3/ ( (A (p 3 (H 0 ( (q 3p q p 3 from (VIII.. q 3 = σ (σh 0 3/ because of (IX.. and (IX.3. ( σ(ξ 0η ξ η 0 (σ σh 0 ξ 3 ξ 0 (H 0 ( ( σh 0 (η 3 ξ 0 η 3 ξ 3 η 0 (σ σh 0 ξ ξ 0 (H 0 ( ( ( (η σh ξ 0 η ξ η 0 (σ σh 0 ξ 3 0 ξ 0 ] q 3 = ( H 0 ( (ξ 0η ξ η 0 (ξ 3 (η 3 ξ 0 η 3 ξ 3 η 0 (ξ ((η ξ 0 η ξ η 0 (ξ 3 ] ξ 0 q 3 = ( H 0 ( ξ 0ξ 3 η ξ ξ 3 η 0 ξ η 3 ξ 0 ξ η 3 ξ ξ 3 η 0 ξ 3 η ξ 0 ξ 3 η ξ ξ 3 η 0 ξ 0 q 3 = ( H 0 ( ξ 0ξ 3 η ξ η 3 ξ 0 ξ η 3 ξ 3 η ξ 0 p = σ (σh 0 3/ ( (A (q ( q 3 (H 0 ( q q ( q 3 from (VIII.. p = σ (σh 0 3/ because of (IX.. and (IX.3. (H 0 ( ( p p ] ( σ(ξ 0η ξ η 0 ( ( (η σh ξ 0 η ξ η 0 ( H ξ 0 (H 0 ( ( H 0 (H ξ 0 ( (η 0 σh ξ 0 η ξ η 0 ( H ξ 0 (H 0 ( (H 0 ( ξ 0 (H ξ 0 ( ( (σ σh 0 0 ξ ξ 0 ] (ξ 0 η ξ η 0 (η ξ 0 η ξ η 0 p = ( σ σh 0 (ξ 0 (( ξ 3 0 (η ξ 0 η ξ η 0 ( ξ 0 ( ξ 0 σξ ( ξ 0

31 Supplementary information to The Kepler problem from a differential geometry point of view 3 ξ η 0 η ξ 0 ξ η 0 η ξ η 0 ξ 0 η ξ 0 η ξ 0 ξ η 0 η p = ( σ σh 0 ξ 0 ξ 0 η ξ 0 η ξ η 0 (ξ 0 3 ξ 0 η ξ η 0 η ξ 0 ξ η 0 η ξ 0 ξ 0 ξ 0 ξ 0 ξ 3 0 σ(ξ ξ 0 ξ ] p = ( σ σh 0 (ξ 0 3 ξ η 0 η ξ 0 η ξ 0 ξ 0 η ξ 0 3 σξ σξ 0 ξ ]. p = σ (σh 0 3/ ( (A (q ( q 3 (H 0(q q ( q 3 (H 0 ( (p p from (VIII.. p = σ (σh 0 3/ because of (IX.. and (IX.3. ( σ(ξ 0η ξ η 0 ( ( (η σh ξ 0 η ξ η 0 ( H ξ 0 (H 0 ( (η σh ξ 0 η ξ η 0 ( (η 0 σh ξ 0 η ξ η 0 ( H ξ 0 (H 0 ( ( (σ σh 0 ξ ξ 0 (σ σh 0 ξ ξ 0 p = ( σ σh 0 (ξ 0 (ξ 0η ξ η 0 (η ξ 0 η ξ η 0 (η ξ 0 η ξ η 0 (η ξ 0 η ξ η 0 3 σξ ξ ( ξ 0 ξ η 0 η ξ 0 ξ η 0 η ξ ξ η 0 ξ 0 η η ξ 0 η η ξ 0 ξ η 0 η p = ( σ σh 0 (ξ 0 3 η η ξ 0 η η ξ η 0 η ξ 0 η η ξ 0 η η ξ 0 ξ η 0 η ξ η 0 η ξ 0 ξ η 0 η ξ ξ η 0 σ(ξ ξ ξ 0 ξ ξ p = ( σ σh 0 (ξ 0 3 η η ξ η 0 η ξ 0 η η σξ ξ σξ 0 ξ ξ ]. ] p 3 = σ (σh 0 3/ ( (A (q 3 ( q 3 (H 0(q q 3 ( q 3 (H 0 ( (p p 3 from (VIII.. p 3 = σ (σh 0 3/ because of (IX.. and (IX.3. ( σ(ξ 0η ξ η 0 ( ( (η σh 3 ξ 0 η 3 ξ 3 η 0 ( H ξ 0 (H 0 ( (η σh ξ 0 η ξ η 0 ( (η 0 σh 3 ξ 0 η 3 ξ 3 η 0 ( H ξ 0 (H 0 ( ( (σ σh 0 ξ ξ 0 (σ σh 0 ξ 3 ξ 0 p 3 = ( σ σh 0 (ξ 0 3 (σ(ξ 0 η ξ η 0 (η 3 ξ 0 η 3 ξ 3 η 0 (η ξ 0 η ξ η 0 (η 3 ξ 0 η 3 ξ 3 η 0 σξ ξ 3 ( ξ 0 ξ η 0 η 3 ξ 0 ξ η 0 η 3 ξ ξ 3 η 0 ξ 0 η η 3 ξ 0 η η 3 ξ 0 ξ 3 η 0 η p 3 = ( σ σh 0 (ξ 0 3 η η 3 ξ 0 η η 3 ξ η 0 η ξ 0 η η 3 ξ 0 η η 3 ξ 0 ξ 3 η 0 η ξ η 0 η 3 ξ 0 ξ η 0 η 3 ξ ξ 3 η 0 σ(ξ ξ 3 ξ 0 ξ ξ 3 p 3 = ( σ σh 0 (ξ 0 3 η η 3 ξ 3 η 0 η ξ 0 η η 3 σξ ξ 3 σξ 0 ξ ξ 3 ]. ]

32 3 Supplementary information to The Kepler problem from a differential geometry point of view A (q p (H (p p = σ (σh 0 3/ q 0 ( q (q p 3 (H 0 ( (p p p p from (VIII.. (p p = σ (σh 0 3/ because of (IX.. and (IX.3. p q 3 q ( σ(ξ 0η ξ η 0 ( H 0 3 ( η ξ 0 σh 0 0 (H 0 ( (η H ξ 0 η ξ η 0 ( H 0 3 ( η 0 ξ 0 H 0 0 (H 0 ( H 0 ξ 0 (σ σh 0 ξ ξ 0 (p p = H 0 ( ξ η 0 ξ 0 η 0 η η 0 η ξ 0 η 0 η ξ η 0 σ(ξ ξ 0 ξ (ξ 0 3 (p p = H 0 σ(ξ ξ 0 ξ η 0 η (ξ 0 3. ] (q p = σ (σh 0 3/ ( A q (H 0 ( (q p (H 0 ( (p (q p from (VIII.. (q p = σ (σh 0 3/ because of (IX.. and (IX.3. (q p = ( (q p = ( ( σ(ξ 0η ξ η 0 ( H 0 ξ 0 (H 0 ( ( σh 0 (η ξ 0 η ξ η 0 (H 0 ( ξ 0 ξ 0 (H 0 ( (σ σh 0 ξ ( η ξ 0 σh 0 0 ] σ (ξ η 0 ξ 0 η η ξ 0 η ξ 0 η ξ 0 η ξ η 0 ξ 0 ξ η 0 ξ η 0 σh 0 (ξ 0 σ (ξ 0η η ξ 0 η ξ 0 ξ η 0 ξ η 0 σh 0 (ξ 0. ( (A (p q = σ (σh 0 3/ p (H 0 ( (p (q p (H 0 ( (q p from (VIII.. (p q = σ (σh 0 3/ because of (IX.. and (IX.3. (p q = ( (p q = ( ( σ(ξ 0η ξ η 0 (H 0 ( ξ 0 ξ 0 (H 0 ( (σ σh 0 ξ ( η ξ 0 σh 0 0 (H 0 ( ( ((η σh ξ 0 η ξ η 0 (H 0 ( ξ 0 0 ξ 0 ] σ (ξ 0η ξ η 0 (ξ 0 ξ η 0 (η ξ 0 η ξ η 0 (ξ 0 ] σh 0 (ξ 0 σ σh 0 ((η ξ 0 η ξ η 0 (ξ 0 ξ η 0 (ξ 0

33 Supplementary information to The Kepler problem from a differential geometry point of view 33 (p q = ( (p q = ( σ σh 0 (η ξ 0 η ξ 0 η ξ 0 η ξ η 0 ξ 0 ξ η 0 ξ η 0 (ξ 0 σ σh 0 (η ξ 0 η ξ 0 η ξ η 0 ξ 0 ξ η 0 (ξ 0. Main results for H < 0: η 0 = ( σh 0 (q p p q from (IX.. η 0 = σ(ξ 0η η ξ 0 η ξ 0 ξ η 0 ξ η 0 η ξ 0 η ξ 0 η ξ η 0 ξ 0 ξ η 0 (ξ 0 from above η 0 = σ η ξ 0 = σ η ξ 0 η ξ 0 η ξ 0 = ση σ ξ 0η ξ 0 = ση ξ 0 η 0. η = ( σ ( p H 0 (q ( σ (p p (q ( σ (q p (p (p q (p (q p(p from (IX.. η = ( σ (4H 0 ( ξ 0 ( H 0 (ση 0 ξ η ξ 0 η = ( σ (H 0 σ(ξ ξ 0 ξ η 0 η (ξ 0 3 ( σh 0 (η ξ 0 η ξ η 0 ( σ ( σ ( σ ( σ (ξ 0η η ξ 0 η ξ 0 ξ η 0 ξ η 0 σh 0 (ξ 0 σh 0 (η ξ 0 η ξ 0 η ξ η 0 ξ 0 ξ η 0 (ξ 0 (σ σh 0 ξ ξ 0 (σ σh 0 ξ ξ 0 ( σ ( σh 0 η 0 ( σ σh 0 (ξ 0 3 ξ η 0 η ξ 0 η ξ 0 ξ 0 η ξ 0 3 σ(ξ ξ 0 ξ from above (IX.. (IX.3. and (IX.3. proof (ξ 0 3 σ (ση 0 ξ η ξ 0 ( ξ 0 (σξ σξ 0 ξ η 0 η (η ξ 0 η ξ η 0 σ(ξ 0 η η ξ 0 η ξ 0 ξ η 0 ξ η 0 ξ ( ξ 0 σ(η ξ 0 η ξ 0 η ξ η 0 ξ 0 ξ η 0 ξ ( ξ 0 (η 0 (ξ η 0 η ξ 0 η ξ 0 ξ 0 η ξ 3 0 σξ σξ 0 ξ ] η 0 ( ξ 0 σ(ξ η ( ξ 0 (σξ σξ 0 ξ η 0 η (η ξ 0 η ξ η 0 η = σ(η (ξ 0 3 ξ ( ξ 0 (η 0 (ξ η 0 η ξ 0 η ξ 0 ξ 0 η ξ 3 0 σξ σξ 0 ξ η 0 ξ 0 η 0 ξ 0 η 0 σξ η σξ 0 ξ η σξ η σξ 0 ξ η σξ η 0 σξ 0 ξ η η = (ξ 0 σξ 3 0 ξ η σξ 0 ξ η 0 η 0 η ξ 0 η 0 η ξ η 0 η σξ η σξ 0 ξ η ξ η 0 η ξ 0 η 0 η ξ 0 η 0 ξ 0 η 0 η 0 η ξ 3 0 η 0 σξ η 0 σξ 0 ξ η 0 η = η = (ξ 0 3 η 0 3ξ 0 η 0 3ξ 0 η 0 ξ 0 3 η 0 σξ η σξ 0 ξ η σξ 0 ξ η (ξ 0 3 η 0( 3ξ 0 3ξ 0 ξ 0 3 σξ η ( ξ 0 ξ 0

34 34 Supplementary information to The Kepler problem from a differential geometry point of view η = (ξ 0 3 η 0( ξ 0 3 σξ η ( ξ 0 η = η 0 σξ η ξ 0 = η 0 ξ η 0. η = ( σ ( p H 0 (q ( σ (p p (q ( σ (q p (p (p q (p (q p(p from (IX.. η = ( σ (4H 0 ( ξ 0 ( H 0 ( ξ 0ξ η ξ η ξ 0 ξ η ξ η ξ 0 η = η = η = η = η = ( σ (H 0 (σξ σξ 0 ξ η 0 η (ξ 0 3 ( σh 0 (η ξ 0 η ξ η 0 ( σ ( σ ( σ ( σ (ξ 0η η ξ 0 η ξ 0 ξ η 0 ξ η 0 σh 0 (ξ 0 σh 0 (η ξ 0 η ξ 0 η ξ η 0 ξ 0 ξ η 0 (ξ 0 (σ σh 0 ξ ξ 0 (σ σh 0 ξ ξ 0 ( σ ( σh 0 η 0 ( σ σh 0 (ξ 0 3 η η ξ η 0 η ξ 0 η η σξ ξ σξ 0 ξ ξ from above (IX.. (IX.3. and (IX.3. proof (ξ 0 3 (ξ 0 3 (ξ 0 3 σ(ξ 0 ξ η ξ η ξ 0 ξ η ξ η ( ξ 0 (σξ σξ 0 ξ η 0 η (η ξ 0 η ξ η 0 σ(ξ 0 η η ξ 0 η ξ 0 ξ η 0 ξ η 0 ξ ( ξ 0 σ(η ξ 0 η ξ 0 η ξ η 0 ξ 0 ξ η 0 ξ ( ξ 0 (η 0 (η η ξ η 0 η ξ 0 η η σξ ξ σξ 0 ξ ξ ] σ(ξ 0 ξ η ξ η ξ 0 ξ η ξ η ( ξ 0 (σξ σξ 0 ξ η 0 η (η ξ 0 η ξ η 0 σ(η ξ ( ξ 0 (η 0 (η η ξ η 0 η ξ 0 η η σξ ξ σξ 0 ξ ξ ] σξ 0 ξ η σξ η σξ 0 ξ η σξ η σξ 0 ξ η σξ 0 ξ η σξ 0 ξ η σξ 0 ξ η σξ η σξ 0 ξ η σξ ξ η 0 σξ 0 ξ η σξ 0 ξ η σξ 0 ξ ξ η 0 η 0 η η ξ 0 η 0 η η ξ η 0 η σξ η σξ 0 ξ η η 0 η η ξ η 0 η ξ 0 η 0 η η σξ ξ η 0 σξ 0 ξ ξ η 0 ] (ξ 0 3 σξ η σξ 0 ξ η σξ 0 ξ η (ξ 0 3 σξ η ( ξ 0 ξ 0 η = σξ η ξ 0 = ξ η 0. η 3 = ( σ ( p H 0 (q 3 ( σ (p p (q 3

35 Supplementary information to The Kepler problem from a differential geometry point of view 35 ( σ (q p (p 3 (p q (p 3 (q p(p 3 from (IX.. η 3 = ( σ (4H 0 ( ξ 0 ( H 0 ( ξ 0ξ 3 η ξ η 3 ξ 0 ξ η 3 ξ 3 η ξ 0 η 3 = η 3 = η 3 = η 3 = η 3 = ( σ (H 0 (σξ σξ 0 ξ η 0 η (ξ 0 3 ( σh 0 (η 3 ξ 0 η 3 ξ 3 η 0 ( σ ( σ ( σ ( σ (ξ 0η η ξ 0 η ξ 0 ξ η 0 ξ η 0 σh 0 (ξ 0 σh 0 (η ξ 0 η ξ 0 η ξ η 0 ξ 0 ξ η 0 (ξ 0 (σ σh 0 ξ 3 ξ 0 (σ σh 0 ξ 3 ξ 0 ( σ ( σh 0 η 0 ( σ σh 0 (ξ 0 3 η η 3 ξ 3 η 0 η ξ 0 η η 3 σξ ξ 3 σξ 0 ξ ξ 3 from above (IX.. (IX.3. and (IX.3. proof (ξ 0 3 (ξ 0 3 (ξ 0 3 σ(ξ 0 ξ 3 η ξ η 3 ξ 0 ξ η 3 ξ 3 η ( ξ 0 (σξ σξ 0 ξ η 0 η (η 3 ξ 0 η 3 ξ 3 η 0 σ(ξ 0 η η ξ 0 η ξ 0 ξ η 0 ξ η 0 ξ 3 ( ξ 0 σ(η ξ 0 η ξ 0 η ξ η 0 ξ 0 ξ η 0 ξ 3 ( ξ 0 (η 0 (η η 3 ξ 3 η 0 η ξ 0 η η 3 σξ ξ 3 σξ 0 ξ ξ 3 ] σ(ξ 0 ξ 3 η ξ η 3 ξ 0 ξ η 3 ξ 3 η ( ξ 0 (σξ σξ 0 ξ η 0 η (η 3 ξ 0 η 3 ξ 3 η 0 σ(η ξ 3 ( ξ 0 (η 0 (η η 3 ξ 3 η 0 η ξ 0 η η 3 σξ ξ 3 σξ 0 ξ ξ 3 ] σξ 0 ξ 3 η σξ η 3 σξ 0 ξ η 3 σξ 3 η σξ 0 ξ 3 η σξ 0 ξ η 3 σξ 0 ξ η 3 σξ 0 ξ 3 η σξ η 3 σξ 0 ξ η 3 σξ ξ 3 η 0 σξ 0 ξ η 3 σξ 0 ξ η 3 σξ 0 ξ ξ 3 η 0 η 0 η η 3 ξ 0 η 0 η η 3 ξ 3 η 0 η σξ 3 η σξ 0 ξ 3 η η 0 η η 3 ξ 3 η 0 η ξ 0 η 0 η η 3 σξ ξ 3 η 0 σξ 0 ξ ξ 3 η 0 ] (ξ 0 3 σξ 3η σξ 0 ξ 3 η σξ 0 ξ 3 η (ξ 0 3 σξ 3η ( ξ 0 ξ 0 η 3 = σξ 3η ξ 0 = ξ 3 η 0. ξ 0 = (p p (p p ( p H 0 = (p p (4H 0 p H 0 ( p H 0 ( p H 0 from (IX.. ξ 0 = (4H 0 (4H 0 (σξ σξ 0 ξ η 0 η (ξ 0 3 ( ξ 0 4H 0 from above and (IX.3. proof ξ 0 = (σξ σξ 0 ξ η 0 η (ξ 0

36 36 Supplementary information to The Kepler problem from a differential geometry point of view ξ 0 = σξ η 0 ( η ξ 0 = σξ ση 0 η 0. ξ = ( σh 0 ( p H 0 ( (p ( σh 0 (( p H 0 ( (p p (p from (IX.. ξ = ( σh 0 ( ξ 0 4H 0 (σ σh 0 ξ η 0 η ξ 0 η ξ 0 ξ 0 (ξ 0 3 η ξ 3 0 σξ σξ 0 ξ ] ξ = ξ = ξ = ξ = ( σh 0 ( ( ξ 0 4H 0 (H 0 (σξ σξ 0 ξ η 0 η (ξ 0 3 (σ σh 0 ξ ξ 0 from above and (IX.3. proof (ξ 0 ξ η 0 η ξ 0 η ξ 0 ξ 0 η ξ 0 3 σξ σξ 0 ξ (σξ σξ 0 ξ η 0 η ξ (ξ η 0 η ξ 0 η ξ 0 ξ 0 η ξ 3 0 σξ σξ 0 ξ (ξ 0 σξ σξ 0 ξ ξ η 0 η (ξ 0 (ξ 0η ξ 0 ξ 0 η ξ 0 3 (ξ 0 (η ξ 0 η ξ 0 ξ 0 ξ 0 ξ 0 3 ξ = η ξ 0 ξ 0 ξ 0 ξ = ξ 0 η ξ 0 = ξ 0 ση η 0. ξ = ( σh 0 ( p H 0 ( (p ( σh 0 (( p H 0 ( (p p (p from (IX.. ξ = ( σh 0 ( ξ 0 4H 0 (σ σh 0 (ξ 0 3 η η ξ η 0 η ξ 0 η η σξ ξ σξ 0 ξ ξ ] ξ = ξ = ξ = ( σh 0 ( ( ξ 0 4H 0 (H 0 (σξ σξ 0 ξ η 0 η (ξ 0 3 (σ σh 0 ξ ξ 0 from above and (IX.3. proof (ξ 0 η η ξ η 0 η ξ 0 η η σξ ξ σξ 0 ξ ξ (σξ σξ 0 ξ η 0 η ξ (ξ 0 (η η ξ η 0 η ξ 0 η η σξ ξ σξ 0 ξ ξ σξ ξ σξ 0 ξ ξ ξ η 0 η (ξ 0 (η η ξ 0 η η ξ = η η ξ 0 = η η 0. ξ 3 = ( H 0 ( p H 0 ( (p 3 ( H 0 (( p H 0 ( (p p (p 3 from (IX.. ξ 3 = ( H 0 ( ξ 0 4H 0 ( H 0 (ξ 0 3 η η 3 ξ 3 η 0 η ξ 0 η η 3 ξ ξ 3 ξ 0 ξ ξ 3 ]

37 Supplementary information to The Kepler problem from a differential geometry point of view 37 ξ 3 = ξ 3 = ξ 3 = ( H 0 ( ( ξ 0 4H 0 (H 0 (ξ ξ 0 ξ η 0 η (ξ 0 3 (σ σh 0 ξ 3 ξ 0 from above and (IX.3. proof (ξ 0 η η 3 ξ 3 η 0 η ξ 0 η η 3 ξ ξ 3 ξ 0 ξ ξ 3 (ξ ξ 0 ξ η 0 η ξ 3 (ξ 0 (η η 3 ξ 3 η 0 η ξ 0 η η 3 ξ ξ 3 ξ 0 ξ ξ 3 ξ ξ 3 ξ 0 ξ ξ 3 ξ 3 η 0 η (ξ 0 (η η 3 ξ 0 η η 3 ξ 3 = η η 3 ξ 0 = η 3 η 0. (IX.5. Proof. The formulas for ξ i η i are the solutions of the differential equations given in (IX.4. and the additional conditions on the integration constants follow from the additional conditions in (IX.3.. (i For H < 0: First we observe ξ 0 iξ i(η 0 iη ] = ξ 0 iξ iη 0 η ] = ξ η 0 η 0 iξ 0 η η 0 iη iξ 0 η 0 η 0 ξ η 0 = ξ 0 iξ iη 0 η ]( η 0. Then we check to see if the solution holds Be is e i(η 0α = Be is cos(η 0 α isin(η 0 α Be is e i(η 0α ] = ibe is e i(η 0α ibe is e i(η 0α η 0 = ibe is e i(η 0α ]( η 0. This gives us ξ 0 = (ξ 0 iξ (ξ 0 iξ ξ 0 = Beis cos(η 0 α B e is cos(η 0 α ]. ξ = (ξ i 0 iξ (ξ 0 iξ ξ = i Beis cos(η 0 α B e is cos(η 0 α ]. η 0 = (η 0 iη (η 0 iη η 0 = Beis sin(η 0 α B e is sin(η 0 α ]. η = (η i 0 iη (η 0 iη η = i Beis sin(η 0 α B e is sin(η 0 α ]. Now we observe (ξ iη = η η 0 iξ η 0 = i(ξ iη η 0

38 38 Supplementary information to The Kepler problem from a differential geometry point of view Then we check to see if the solution holds (ξ iη = B cos(η 0 α i sin (η 0 α ] (ξ iη = B e i(η 0α ] = ib e i(η 0α η 0 (ξ iη = i(ξ iη η 0. (ξ 3 iη 3 is analogous. ξ = B cos(η 0 α. ξ 3 = B 3 cos(η 0 α 3. η = B sin(η 0 α. η 3 = B 3 sin(η 0 α 3. = ξ 0 ξ ξ ξ 3 from (IX.3. = (ξ 0 iξ (ξ 0 iξ ξ ξ 3 = Be is cos(η 0 αb e is cos(η 0 α B cos (η 0 α B 3 cos (η 0 α 3. = BB cos(η 0 αcos(η 0 α B cos (η 0 α B 3 cos (η 0 α 3. = η 0 η η η 3 from (IX.3. = (η 0 iη (η 0 iη η η 3 = Be is sin(η 0 αb e is sin(η 0 α B sin (η 0 α B 3 sin (η 0 α 3. = BB sin(η 0 αsin(η 0 α B sin (η 0 α B 3 sin (η 0 α 3. = BB cos(η 0 αcos(η 0 α sin(η 0 αsin(η 0 α ] B cos (η 0 α sin (η 0 α ] B 3 cos (η 0 α 3 sin (η 0 α 3 = BB cos(η 0 α η 0 α B B 3 because cosxcosy sinxsiny = cos(x y and cos x sin x = = BB cos(η 0 α η 0 α B B 3 = B cosh(imα B B 3 because cos(ix = cosh(x for real x. 0 = ξ 0 η 0 ξ η ξ η ξ 3 η 3 0 = Beis cos(η 0 α B e is cos(η 0 α ] Beis sin(η 0 α B e is sin(η 0 α ] i Beis cos(η 0 α B e is cos(η 0 α ] i Beis sin(η 0 α B e is sin(η 0 α ] B cos(η 0 α B sin(η 0 α B 3 cos(η 0 α 3 B 3 sin(η 0 α 3 0 = 4 B e is sin(η 0 αcos(η 0 α BB sin(η 4 0 α cos(η 0 α

39 Supplementary information to The Kepler problem from a differential geometry point of view 39 4 BB sin(η 0 αcos(η 0 α 4 B e is sin(η 0 α cos(η 0 α 4 B e is sin(η 0 αcos(η 0 α 4 BB sin(η 0 α cos(η 0 α 4 BB sin(η 0 αcos(η 0 α 4 B e is sin(η 0 α cos(η 0 α B sin(η 0 α cos(η 0 α B 3 sin(η 0 α 3 cos(η 0 α 3 0 = BB sin(η 0 α cos(η 0 α BB sin(η 0 αcos(η 0 α B sin(η 0 α cos(η 0 α B 3 sin(η 0 α 3 cos(η 0 α 3 0 = B sin(η 0 Reα B sin(η 0 α B 3 sin(η 0 α 3 because sin(a = sin(acos(a and sin(a b = sin(acos(b cos(asin(b. Multiplying by and taking the square we get: ( : 0 = B 4 sin (η 0 Reα B 4 sin (η 0 α B 3 4 sin (η 0 α 3 B B sin(η 0 Reα sin(η 0 α B B 3 sin(η 0 Reα sin(η 0 α 3 B B 3 sin(η 0 α sin(η 0 α 3. Now we ll save this equation for the moment and create another equation for zero by subtracting the two expressions for one from each other: 0 = B cos(η 0 αcos(η 0 α sin(η 0 αsin(η 0 α ] B cos (η 0 α sin (η 0 α ] B 3 cos (η 0 α 3 sin (η 0 α 3 0 = B cos(η 0 Reα B cos(η 0 α B 3 cos(η 0 α 3 because cos(a b = cos(acos(b sin(asin(b and cos(a = cos (a sin (a. By taking the square we get: ( : 0 = B 4 cos (η 0 ReαB 4 cos (η 0 α B 3 4 cos (η 0 α 3 B B cos(η 0 Reα cos(η 0 α B B 3 cos(η 0 Reα cos(η 0 α 3 B B 3 cos(η 0 α cos(η 0 α 3. Now we can add the expressions (* and (** and use the identity sin (a cos (a = to get: 0 = B 4 B 4 B 3 4 B B sin(η 0 Reα sin(η 0 α cos(η 0 Reα cos(η 0 α ] B B 3 sin(η 0 Reα sin(η 0 α 3 cos(η 0 Reα cos(η 0 α 3 ] B B 3 sin(η 0 α sin(η 0 α 3 cos(η 0 α cos(η 0 α 3 ]. 0 = B 4 B 4 B 3 4 B B cos(reα α ] B B 3 cos(reα α 3 ] B B 3 cos(α α 3 because cos(a b = cos(acos(b sin(asin(b. A = ξ η 0 ξ 0 η from (IX.3. A = i Beis cos(η 0 α B e is cos(η 0 α ] Beis sin(η 0 α B e is sin(η 0 α ]

40 40 Supplementary information to The Kepler problem from a differential geometry point of view Beis cos(η 0 α B e is cos(η 0 α ] i Beis sin(η 0 α B e is sin(η 0 α A = 4i Beis cos(η 0 αb e is sin(η 0 α Be is sin(η 0 αb e is cos(η 0 α A = i BB sin(η 0 α η 0 α because sinxcosy cosxsiny = sin(x y A = i BB sin(iimα A = B sinh(imα because sin(ix = isinh(x for real x. (ii For H > 0: First we observe ξ 0 ξ η 0 η ] = ξ η 0 η 0 ξ 0 η η 0 η ξ 0 η 0 η 0 ξ η 0 ξ 0 ξ η 0 η ] = (ξ 0 ξ η 0 η ( η 0. Then we check to see if the solution holds ξ 0 ξ η 0 η ] = B 0 e s cosh(η 0 α 0 sinh (η 0 α 0 ξ 0 ξ η 0 η ] = B 0 e s e (η 0α 0 ] ξ 0 ξ η 0 η ] = B 0 e s e (η 0α 0 B 0 e s e (η 0α 0 η 0 ξ 0 ξ η 0 η ] = (ξ 0 ξ η 0 η ( η 0. Then we observe ξ 0 ξ (η 0 η ] = ξ η 0 η 0 ξ 0 η η 0 η ξ 0 η 0 η 0 ξ η 0 ξ 0 ξ (η 0 η ] = (ξ 0 ξ η 0 η ( η 0. Then we check to see if the solution holds ξ 0 ξ (η 0 η ] = B 0 e s cosh(η 0 α 0 sinh (η 0 α 0 ξ 0 ξ (η 0 η ] = B 0 e s e (η 0α 0 ] ξ 0 ξ (η 0 η ] = B 0 e s e (η 0α 0 B 0 e s e (η 0α 0 η 0 ξ 0 ξ (η 0 η ] = (ξ 0 ξ η 0 η ( η 0. Then we observe ξ 0 ξ η 0 η ] = ξ η 0 η 0 ξ 0 η η 0 η ξ 0 η 0 η 0 ξ η 0 ξ 0 ξ η 0 η ] = (ξ 0 ξ η 0 η ( η 0. Then we check to see if the solution holds ξ 0 ξ η 0 η ] = B e s cosh(η 0 α sinh (η 0 α ξ 0 ξ η 0 η ] = B e s e (η 0α ] ξ 0 ξ η 0 η ] = B e s e (η 0α B e s e (η 0α η 0 ξ 0 ξ η 0 η ] = (ξ 0 ξ η 0 η ( η 0. Then we observe ξ 0 ξ (η 0 η ] = ξ η 0 η 0 ξ 0 η η 0 η ξ 0 η 0 η 0 ξ η 0

41 Supplementary information to The Kepler problem from a differential geometry point of view 4 ξ 0 ξ (η 0 η ] = (ξ 0 ξ (η 0 η (η 0. Then we check to see if the solution holds ξ 0 ξ (η 0 η ] = B e s cosh(η 0 α sinh (η 0 α ξ 0 ξ (η 0 η ] = B e s e (η 0α ] ξ 0 ξ (η 0 η ] = B e s e (η 0α B e s e (η 0α η 0 ξ 0 ξ (η 0 η ] = (ξ 0 ξ (η 0 η (η 0. This gives us ξ 0 = (ξ 0 ξ (ξ 0 ξ ξ 0 = B 0e s cosh(η 0 α 0 B e s cosh(η 0 α ]. ξ = (ξ 0 ξ (ξ 0 ξ ξ = B 0e s cosh(η 0 α 0 B e s cosh(η 0 α ]. η 0 = (η 0 η (η 0 η η 0 = B 0e s sinh(η 0 α 0 B e s sinh(η 0 α ]. η = (η 0 η (η 0 η η = B 0e s sinh(η 0 α 0 B e s sinh(η 0 α ]. Then we observe ξ η ] = η η 0 ξ η 0 ξ η ] = (ξ η η 0. Then we check to see if the solution holds ξ η ] = B cosh(η 0 α sinh (η 0 α ] ξ η ] = B e (η 0α ] ξ η ] = B e (η 0α η 0 ξ η ] = (ξ η η 0. Then we observe ξ η ] = η η 0 ξ η 0 ξ η ] = (ξ η η 0. Then we check to see if the solution holds ξ η ] = B cosh(η 0 α sinh (η 0 α ] ξ η ] = B e (η 0α ] ξ η ] = B e (η 0α η 0 ξ η ] = (ξ η η 0. ξ 3 and η 3 are analogous. ξ = B cosh(η 0 α. ξ 3 = B 3 cosh(η 0 α 3.

42 4 Supplementary information to The Kepler problem from a differential geometry point of view η = B sinh(η 0 α. η 3 = B 3 sinh(η 0 α 3. = ξ 0 ξ ξ ξ 3 from (IX.3. = (ξ 0 ξ (ξ 0 ξ ξ ξ 3 = B 0 e s cosh(η 0 α 0 B e s cosh(η 0 α B cosh (η 0 α B 3 cosh (η 0 α 3 = B 0 B cosh(η 0 α 0 cosh(η 0 α B cosh (η 0 α B 3 cosh (η 0 α 3 = η 0 η η η 3 from (IX.3. = (η 0 η (η 0 η η η 3 = B 0 e s sinh(η 0 α 0 B e s sinh(η 0 α B sinh (η 0 α B 3 sinh (η 0 α 3 = B 0 B sinh(η 0 α 0 sinh(η 0 α B sinh (η 0 α B 3 sinh (η 0 α 3 = B 0 B cosh(η 0 α 0 cosh(η 0 α sinh(η 0 α 0 sinh(η 0 α ] B cosh (η 0 α sinh (η 0 α ] B 3 cosh (η 0 α 3 sinh (η 0 α 3 = B 0 B cosh (α 0 α B B 3 because coshxcoshy sinhxsinhy = cosh(x y and cosh x sinh x =. 0 = ξ 0 η 0 ξ η ξ η ξ 3 η 3 from (IX.3. 0 = B 0e s cosh(η 0 α 0 B e s cosh(η 0 α ] B 0e s sinh(η 0 α 0 B e s sinh(η 0 α ] B 0e s cosh(η 0 α 0 B e s cosh(η 0 α ] B 0e s sinh(η 0 α 0 B e s sinh(η 0 α ] B cosh(η 0 α B sinh(η 0 α B 3 cosh(η 0 α 3 B 3 sinh(η 0 α 3 0 = B 4 0 e s sinh(η 0 α 0 cosh(η 0 α 0 B 4 e s sinh(η 0 α cosh(η 0 α B 4 0B sinh(η 0 α cosh(η 0 α 0 sinh(η 0 α 0 cosh(η 0 α ] B sinh(η 0 α cosh(η 0 α B 3 sinh(η 0 α 3 cosh(η 0 α 3 B 4 0 e s sinh(η 0 α 0 cosh(η 0 α 0 B 4 e s sinh(η 0 α cosh(η 0 α

43 Supplementary information to The Kepler problem from a differential geometry point of view 43 4 B 0B sinh(η 0 α cosh(η 0 α 0 sinh(η 0 α 0 cosh(η 0 α 0 = B 0B sinh(η 0 α cosh(η 0 α 0 sinh(η 0 α 0 cosh(η 0 α ] 4 B sinh(η0 α cosh(η 0 α 4 B 3 sinh(η 0 α 3 cosh(η 0 α 3 0 = B 0B sinh(η 0 α 0 α B sinh(η 0 α B 3 sinh(η 0 α 3 because sinh(a b = sinh(acosh(b sinh(bcosh(aand sinh(a = sinh(acosh(a Multiplying by and taking the square we get: ( : 0 = B 0 B sinh (η 0 α 0 α B 4 sinh (η 0 α B 3 4 sinh (η 0 α 3 B 0 B B sinh(η 0 α 0 α sinh(η 0 α B 0 B B 3 sinh(η 0 α 0 α sinh(η 0 α 3 B B 3 sinh(η 0 α sinh(η 0 α 3 Now we ll save this equation for the moment and create another equation for zero by subtracting the two expressions for one from each other: 0 = B 0 B cosh(η 0 α 0 cosh(η 0 α sinh(η 0 α 0 sinh(η 0 α ] B cosh (η 0 α sinh (η 0 α ] B 3 cosh (η 0 α 3 sinh (η 0 α 3 0 = B 0 B cosh(η 0 α 0 α B cosh(η 0 α B cosh(η 0 α because cosh(a b = cosh(acosh(b sinh(asinh(b and cosh(a = cosh (a sinh (a. By taking the square we get: ( : 0 = B 0 B cosh (η 0 α 0 α B 4 cosh (η 0 α B 3 4 cosh (η 0 α 3 B 0 B B cosh(η 0 α 0 α cosh(η 0 α B 0 B B 3 cosh(η 0 α 0 α cosh(η 0 α 3 B B 3 cosh(η 0 α cosh(η 0 α 3. Now we can subtract the expression (* from (** to get: 0 = B 0 B sinh (η 0 α 0 α cosh (η 0 α 0 α ] B 4 sinh (η 0 α cosh (η 0 α ] B 3 4 sinh (η 0 α 3 cosh (η 0 α 3 ] B 0 B B sinh(η 0 α 0 α sinh(η 0 α ] B 0 B B cosh(η 0 α 0 α cosh(η 0 α ] B 0 B B 3 sinh(η 0 α 0 α sinh(η 0 α 3 ] B 0 B B 3 cosh(η 0 α 0 α cosh(η 0 α 3 ] B B 3 sinh(η 0 α sinh(η 0 α 3 cosh(η 0 α cosh(η 0 α 3 0 = B 0 B B 4 B 3 4 B 0 B B cosh(α 0 α α B 0 B B 3 cosh(α 0 α α 3 B B 3 cosh(α α 3 because cosh (a sinh (a = and cosh(a b = cosh(acosh(b sinh(asinh(b.

44 44 Supplementary information to The Kepler problem from a differential geometry point of view A = ξ 0 η ξ η 0 from (IX.3. A = B 0e s cosh(η 0 α 0 B e s cosh(η 0 α ] B 0e s sinh(η 0 α 0 B e s sinh(η 0 α ] B 0e s cosh(η 0 α 0 B e s cosh(η 0 α ] B 0e s sinh(η 0 α 0 B e s sinh(η 0 α A = B 4 0B cosh(η 0 α 0 e s sinh(η 0 α cosh(η 0 α e s sinh(η 0 α 0 A = sinh(α 0 α because sinhxcoshy sinhycoshx = sinh(x y. Remark. Because of ξ 0 and ξ 0 ξ we have ξ 0 ξ 0. Then (IX.5. (ii leads to B 0 0 and B 0. (IX.6. Proof. For H < 0 the equation is: η 0 = ( Be is sin(η 0 α B e is sin(η 0 α ]. For fixed s we will search for the intersection of the two functions that are on the left and the right of the equality sign. The left side is the straight line through zero with a slope of and the right side is a periodic bounded function of η 0. Therefore there is at least one intersection. The uniqueness stems from the fact that the slope of the function on the right side is always. If we denote this slope with S the we have: S = ( Be is cos(η 0 α B e is cos(η 0 α ] S = ( 4 B cos(η 0 αcos(η 0 α B cos(η 0 α cos(η 0 α B e is cos (η 0 α B e is cos ] (η 0 α S = ( 4(ξ 0 iξ (ξ 0 ξ (ξ 0 iξ (ξ 0 iξ ] = ξ 0. (ii For H > 0 the equation is: η 0 = ( B 0 e s sinh(η 0 α 0 B e s sinh(η 0 α ]. This time the right side is a function that behaves like e η 0 (resp. -e η 0 for η 0 0 (resp. η 0 0 since B 0 0 and B 0. As a result there is at least one intersection. The uniqueness stems from the fact that the slope of the function on the right side is always. If we denote this slope with S the we have: S = ( B 0 e s cosh(η 0 α 0 B e s cosh(η 0 α ]. S = ( 4 B 0 e s cosh (η 0 α 0 B e s cosh (η 0 α ] B 0 B cosh(η 0 α 0 cosh(η 0 α S = ( 4(ξ 0 ξ (ξ 0 ξ (ξ 0 ξ (ξ 0 ξ ] = ξ 0. In both cases the slope can only have a value of at isolated points since the function is analytical.

45 Supplementary information to The Kepler problem from a differential geometry point of view 45 (IX.8. Proof. For H < 0: 3 i=0 x i = ( q p cosψ H (q psinψ] 3 i=0 x i = q p 4 q p H q p (q p ( H (q p (q p q from (IX.7. 3 i=0 x i = H ( H q p ] cos ψ (q p (q p q p (q p ] cosψsinψ (q p p ] sin ψ ( H q p ( H q p ] cos ψ ( H (q p ( H (q p ] sin ψ from the definition of H 3 i=0 x i = cos ψ sin ψ 3 i=0 x i =. q p cosψ ( q q (q pp sinψ] 3 i=0 3 i=0 x i y i = ( q p H q p cosψ ( q x i y i = q p (q p H ( q p H q p (q p 3 i=0 x i y i = 3 i=0 x i y i = cosψ H (q pp q (q psinψ] sinψ] (q p (q p q p (q p H (q p q p H ( q q (q pp ] (q pcosψ H ( q p sinψ] ( q (q pp H q cosψsinψ ] cos ψ (q p (q p q p (q p ] sin ψ H ( q p H H (q p q p H ( q q (q pp ] cosψsinψ ( q p 4 q p (q p (q p p H q H ((q p q p cosψ ( q p sinψ ] cosψsinψ