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A Isotropic two-dimensional Harmonic Oscillator
For Classical Mechanics A, Univ.Tokyo (2017)
Descriptions: The “Komaba” solution is presented for the two-dimensional harmonicoscillator in the polar coordinate representation, in contrast to the “Pasadena” solution
presented by Dr. S. Golwala (CalTech).
[June 16, 2017 ]1

Isotropic two-dimensional harmonic oscillator in the polar coordinates

The equation of motion for the isotropic two-dimensional harmonic oscillator is given asmd2r(t) = −mω2 r(t),
dt2

(1)

where the position vector r(t) is given as(r(t) = r(t)θ(t)
sin θ(t)cos)≡ r(t)er .

(2)

The unit vector er is often referred to as the radial basis vector in the two-dimensional poloar coordinates.
The other basis eθ is defined to be orthogonal to er . The usual definition reads()− sin θ(t)
eθ =.
(3)
cos θ(t)The left-hand side of the equation of motion, i.e., the second-order derivative of the position vector, isexpressed in the poloar coordinates as()
1 d ( 2 _)d2_2 e +r(t)=r−r
θr θ eθ .
r
dt2r dt(4)

The equation of motion thus gives the radial equation and the angular momentum conservationr − rθ_ 2 = −ω2 r,

2_mr θ = l(= const.)

(5)
(6)

Eliminating θ_ thanks to the angular momentum conservation, the radial equation becomesr = −ω2 r +2l2.
m 2 r3(7)

Energy integral

Multiplying r_ = dr
dt to the both sides in the above equation, we haveddt( 2)
r_ω 2 r2l2
= 0.
++222m2 r4(8)This equation corresponds to the energy conservation, and we haver_ 2 ω2 r2
l2= E(= const.)++
2
22m2 r21(9)3The potential energy

Regarding the third term in the energy conservation as the centrifugal energy, we can consider the sum ofthe second and the third term as the “potential energy”.
U(r) =

ω 2 r2l2+.
22m2 r2

The local minimum dU/dr = 0 occurs at√r0 =(10)

l,mω

(11)

lω.
m(12)

and the value of the local minimum is evaluated to beU(r0 ) =2
When E = lωm , the “kinetic energy” mr_ /2 must be zero due to the energy conservation. Then, the corresponding motion of the oscillator is circular with its radius being r0 . The rotational frequency is calculated
through the angular momentum conservation,

ω0 ≡ θ_ =4

l= ω.
mr20(13)

Scaling of dynamicl variables

Let us introduce the scaling of dynamical variables to simplify the equation of motion. First of all, theradial coordinate r is scaled to be√lρ,(14)r = r0 ρ =mωwhere ρ is a unitless dynamical variable to represent the radial coordinate. We may call r0 “the oscillatorlength”, as an analogy to the quantum theory of the harmonic oscillator.
Next, the natural time scale can be given in connection to the oscillator frequency ω. From the dimension
analysis, we may write
2πt = t0 τ =τ.
(15)ωWith these scaling, the energy conservation reads(orr0t0)212((dρdτ)2+ ω2 r20

1 dρ
2π dτ

ρ2
l2 1+= E.
22m2 r20 ρ2)2+ ρ2 +1= E,
ρ2(16)(17)

where the energy is scaled as
E = U(r0 )E.

5(18)Second integral

Now, we are going to attempt the second integral to obain r(t). Firstly, we rewrite the scaled energy
conservation equation to√dρ
1(19)= ±2π E − ρ2 − 2 .
dτρ2In the literature (including Goldstein), it is recommended to convert the time-derivative to θ-derivativewith help of the angular momentum conservation,
dr _l dr
dr=
θ=.
dtdθ
mr2 dθ(20)together with the variable conversion u = 1/r. Then, we can obtain the formal expression of the second
integral for the potential V = arn+1 ,∫udu√
θ = θ0 −.
(21)2mE
2ma −n−12u0−u−u22ll
According to Goldstein, we can solve this integral analytically for n = 1, −2, −3 in terms of the trigonometricexpressions, and for n = 5, 3, 0, −4, −5, −7 in terms of the (Jacobi’s) ellipitic functions. The ellipitic solutions
are out of our interest in the current consideration, so that we concentrate on the cases n = 1, −2, −3. The
harmonic oscillator corresponds to n = 1, while the gravity (the inverse-square law) and the centrifugal force
correspond to n = −2 and n = −3, respectively. The θ-derivative has an advantage for the negative-powercases, but the harmonic oscillator receives no benefit for this variable conversion, as demonstrated in thelecture (by coincidence ...). Therefoe, we directly carry out the time integral in the following.
For the harmonic oscillator, we have±2π(τ − τ0 ) =∫ρdρ√.
E − ρ2 − ρ−2ρ0(22)√To remove the square root, we introduce ρ = E cos φ. The left-hand side of the above equation then turnsto∫ρ∫φdρsin φdφ√√.
(23)=−2−22
1E −ρ −ρρ0
φ0sin φ − 2
2E cos φDue to an identity()221 − 4/E 2 − cos2 (2φ)
E 2 cos2 φ sin2 φ − 1sin (2φ) − 4/Esin φ −===,E 2 cos2 φ
E 2 cos2 φ
4 cos2 φ
4 cos2 φ21

the integral can be rewritten∫φ∫φ2 sin φ cos φdφ1sin(2φ)
√√dφ.
−= −√
(1 − 4/E 2 ) − cos2 (2φ)
(1 − 4/E 2 ) φ0 1 − cos2 (2φ)φ02(24)

(25)

1−4/E

√
This is the well-known integral to define the arccos function. By putting cos χ = cos(2φ)/ 1 − 4/E 2 , wehave the solution of the integral as (−χ + χ0 )/2. That is, the result of the second integral reduces to±4π(τ − τ0 ) = −χ + χ0 ,(26)

χ = cos (4πτ + δ0 ) .

(27)

orcos

Here, the phase δ0 is defined as the sum involving the constants τ0 and χ0 . Putting back the above result
to the expression with ρ, we have√cos(2φ) = 2 cos2 φ − 1 = 1 − 4/E 2 cos(4πτ + δ0 ),
(28)
or
ρ2 =

√
E 2 /4 − 1 cos(4πτ + δ0 ) + E/2

3(29)Finally, we can scale back to the original dynamical variables.
√r(t) = r0 A cos(2ωt + δ0 ) + B,

(30)

where constats A, B are given as
A=√(B − 1)(B + 1),

B = E/2,

E ≥ 2.

(31)

Let us call our solution the “Komaba solution”, in comparison to the “Pasadena solution” presented by Dr.
S. Golwala. It should be interesting to examine why the energy range is restricted to E ≥ 2, which is left tostudents for analysis.

6

Meaning of the doubling in rotational frequency

The above solution gives rise to the elliptic trajectory, which coincides with the trajectory produced by thegravity (the inverse-square law). However, the frequency in the Komaba solution is doubled in comparison
to the original frequency ω. The reason for this doubling should be understood through a physical consideration, which was actually discussed in the lecture. We will come back to this problem again in the gravity
section, where the elliptic trajectory is given in a diﬀerent mathematical expression,r(t) =a(1 − ϵ2 ),1 + ϵ cos(θ + δ)

(32)

where a = − GMm
2E is the semi-major axis and ϵ denotes the eccentricity√
ϵ=1+2l2 E.
m 3 G2 M 24(33)