To relate t to s , we identify Solving the quadratic equation for t versus s , we obtain the minimizing function t ∗ ( s ), which obviously obeys the boundary condition The symmetry s → 1 /s plays apparently an important role. It guarantees that the points −∞ , − 1, 0, +1, + ∞ are mapped onto themselves under the reparametrization s → t ∗ ( s ): t ∗ ( −∞ ) = −∞ , t ∗ ( − 1) = − 1, t ∗ (0) = 0, t ∗ (+1) = +1, t ∗ (+ ∞ ) = + ∞ . It is convenient to invert Eq. (28) using the Jacobi elliptic functions. Inverting Eq. (28a), we get The function sn has a nice trigonometric expansion in the parameter (nome) (Ref. [11], 8.146.1) giving This formula is applicable for − R/ 2 T < t < + R/ 2 T , when − √ µ < s < + √ µ . We can proceed in the same way with Eq. (28b), whose inverse is Using the addition formula ([11], 8.156.1) and the reduction of sn( x, 0) and cn( x, 0) to sin x and cos x , we obtain s ≈ √ µ cosh [ π 4 ( t − t − 1 + 2 T R )] . (34) s = √ µ sn ( K ( µ ) + i K ( µ ) ( t 2 − 1 2 t + T R ) , µ ) . (33) s ≈ √ µ sin ( πT R t (1 − t $^{2}$) ) . (32) exp ( − π K $_{(}$√$_{1}$ − ν $^{2}$) K ( ν ) ) ≈ ( µ/ 4) 2 ≪ 1 , (31) s = √ µ sn ( 2 K ( µ ) T R t 1 − t $_{2}$, µ ) , (30) − √ µ ≤ s ≤ √ µ or − √$_{T}$ 2 + R 2 T R ≤ t ≤ √$_{T}$ 2 + R 2 − T R . for t (0) = 0 . (29) R 2 K ( µ ) ∫ s √ µ d x √ x 2 − µ √ 1 − µx 2 − T 2 = R t 2 − 1 4 t (28b) for √ µ ≤ s ≤ 1 / √ µ or √$_{T}$ 2 + R 2 − T R ≤ t ≤ T $_{+}$√$_{T}$ 2 + R 2 R . R 2 K ( µ ) F ( s √ µ , µ ) = T t 1 − t 2 , (28a) for − √ µ ≤ s ≤ √ µ or − √$_{T}$ 2 + R 2 − T R ≤ t ≤ √$_{T}$ 2 + R 2 − T R 7