LENGTH OF SINE CURVE

Elliptic integrals

Takehito Yokoyama

Department of Physics, Tokyo Institute of Technology, 2-12-1 Ookayama, Meguro-ku, Tokyo 152-8551, Japan

(Dated: October 16, 2015)

LENGTH OF SINE CURVE

Consider a sine curve given by

y = b sin x

a . (1)

Its line element reads

ds ² = a ² + b ² a ²

1 − k ² sin ² x a

dx ² , k ² = b ²

a ² + b ² . (2)

Therefore, the length of the sine curve over a quarter period can be calculated as

√ a ² + b ² a

Z

a 0

r

1 − k ² sin ² x

a dx = p a ² + b ²

Z

0

q

1 − k ² sin ² ϕdϕ = p

a ² + b ² E(k) (3) where E(k) is the complete elliptic integral of the second kind.

MAGNETOSTATIC POTENTIAL BY A CHARGED CIRCLE

Let us consider the magnetostatic potential induced by charge uniformly distributed with unit line density on a circle with radius R. We consider the potential at a point P(a, b, c). The point on the circle can be written as Q(R cos θ, R sin θ, 0). Then, we have

PQ ² = p

a ² + b ² + R ²

+ c ² − 4R p

a ² + b ² cos ² θ − α

2 (4)

with tan α = _a ^b . Therefore, we can calculate the potential at P as U = 1

4πε Z 2π

0

Rdθ PQ = R

2πε Z π

0

dθ q √

a ² + b ² + R ²

+ c ² − 4R √

a ² + b ² cos ^{2 θ} ₂

= R πε

Z π/ 2 0

dϕ q √

a ² + b ² + R ²

+ c ² − 4R √

a ² + b ² sin ² ϕ

= R πε

1 q √

a ² + b ² + R ² + c ²

Z π/2 0

dϕ

p 1 − k ² sin ² ϕ = k 2πε

s

√ R

a ² + b ² K(k) (5)

where

k ² = 4R √ a ² + b ²

√ a ² + b ² + R ² + c ²

(6)

and K(k) is the complete elliptic integral of the first kind. In the limit of a, b → 0, we have U = _2ε ^√ _R ^R

+c

.

2 ANHARMONIC OSCILLATION

Here, let us consider the oscillation of a particle confined by anharmonic potentials. The law of energy conservation reads

E = 1 2 m

dx dt

²

+ U (x). (7)

We can factorize as

2

m (E − U (x)) = (x − a 1 )(a 2 − x)V (x). (8)

We assume that the oscillation occurs between a 1 and a 2 (a 1 < a 2 ). By integrating Eq.(7), we have t =

Z dx

p (x − a 1 )(a 2 − x)V (x) . (9) By setting x = ¹ ₂ (a 1 + a 2 ) + ¹ ₂ (a 1 − a 2 ) cos ϕ, we finally obtain the following expression of t:

t = Z dϕ

√ V . (10)

Quartic potential . As an example, let us consider a quartic potential of the form U (x) = α

2 x ² + β

4 x ⁴ (11)

with α, β > 0, which can be rewritten as 2

m (E − U (x)) = − β

2m (x ² − a ² )(b ² + x ² ) (12)

where E > 0 and

a ² = − α + p

α ² + 4βE

β , b ² = a ² + 2α

β . (13)

The period of the oscillation is thus calculated as T = 2

r 2m β

Z π 0

dϕ

p b ² + a ² cos ² ϕ = 4

s 2m

β(a ² + b ² ) K(k) = 4

s m

p α ² + 4βE K(k) (14)

where

k ² = a ²

a ² + b ² = βa ²

2(α + βa ² ) . (15)

In the limit of β → 0, we have T = 2π p _m

α .

Cubic potential. Next, let us consider a cubic potential of the form U (x) = α

2 x ² − γ

2 x ³ (16)

with α, γ > 0, which can be factorized as 2

m (E − U (x)) = γ

m (x − a 1 )(a 2 − x)(a 3 − x), a 1 < a 2 < a 3 (17) where E > 0, and thus V (x) = _m ^γ (a 3 − x). The period of the oscillation is given by

T = 2 r m

γ Z π

0

dϕ q

a 3 − ¹ 2 (a 1 + a 2 ) + ¹ ₂ (a 2 − a 1 ) cos ϕ

= 4

r m

γ(a 3 − a 1 ) K(k) (18) where k ² = ^a _a

^{− a}

− a

. In the limit of γ → 0, we have a 1 = − a 2 and γa 3 = α, and hence T = 2π p m

α .

3 ROTATION OF A RIGID BODY

Consider a free rotation of a rigid body described by the Euler’s equation of motion in a rotating frame: _dt ^d L+ω ×L = 0 where

ω =



 ω 1

ω 2

ω 3



 , L =



 Aω 1

Bω 2

Cω 3



 . (19)

Thus, we have

A d

dt ω 1 = (B − C)ω 2 ω 3 , B d

dt ω 2 = (C − A)ω 3 ω 1 , C d

dt ω 3 = (A − B)ω 1 ω 2 . (20) For C > B > A, we consider the solution of the form:

ω 1 = αcnλt, ω 2 = βsnλt, ω 3 = γdnλt. (21) Note ω 3 > 0. Inserting these ansatz into the equation of motion, we have

αβγ

λ = Aα ²

C − B = Bβ ²

C − A = k ² Cγ ²

B − A . (22)

The energy conservation law reads 2E = Aα ² + Cγ ² with the energy E. The magnitude of angular momentum is also conserved: L ² = A ² α ² + C ² γ ² . With these conditions, we finally obtain

α ² = 2EC − L ²

A(C − A) , β ² = 2EC − L ²

B(C − B ) , γ ² = − 2EA + L ²

C(C − A) , k ² = B − A C − B

2EC − L ²

− 2EA + L ² , λ ² = C − B

ABC ( − 2EA + L ² ). (23)