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The Spherical Harmonic Oscillator as a basis Model for the solution of various Classical or Relativistic IVP problems in Astrophysics and Mechanics
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The Spherical Harmonic Oscillator
I.N. Galidakis,
Agricultural University of Athens,
Athens, Greece
July 2010
1
Contents
1 The Two-dimensional Harmonic Oscillator 3
2 The Three-dimensional Harmonic Oscillator 4
2.1 SphericalCoordinates ............................. 5
2.1.1 xy Plane ................................ 5
2.1.2 xz Plane................................. 7
2.1.3 yz Plane................................. 7
2.2 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.2.1 xy Plane ................................ 8
2.2.2 xz Plane................................. 8
2.2.3 yz Plane................................. 9
2.3 OnTheSphere ................................. 9
3 Signal 9
4 Non-commensurability 12
4.1 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
4.2 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
5 A Simple Time Machine 15
5.1 Mechanical ................................... 15
5.2 Optical ..................................... 16
5.3 TimeDilation.................................. 17
5.4 DopplerShift .................................. 18
6 Compressive and Decompressive Agents 18
6.1 Two-dimensional ................................ 19
6.2 The Eigenvalues of Cin R2.......................... 19
6.3 Three-dimensional ............................... 20
6.4 The Eigenvalues of Cin R3.......................... 20
6.5 The Iteration of C............................... 20
6.6 Applications of Ct............................... 21
6.6.1 Exponentially Collapsing Rotating Body . . . . . . . . . . . . . . . 21
6.6.2 Gravitationally Collapsing Rotating Body . . . . . . . . . . . . . . 23
6.6.3 Harmonically Pulsating Rotating Body . . . . . . . . . . . . . . . 25
7 Concluding Remarks 26
References 27
2
Figure 1: Harmonic Oscillator
Abstract
The Harmonic Oscillator is the simplest theoretical model which can describe all peri-
odic motions. Any (eventually) periodic signal can be modelled using such an oscillator.
Although Harmonic Oscillators are easily analyzed on the plane, the analysis of such oscil-
lators in three-dimensional space presents difficulties, because the motion in three-space
needs to be analyzed separately on all three planes xy,xz and yz. This thesis examines
and analyzes the Harmonic Oscillator in three-space and presents some applications in
Optics, Relativity and Physics.
1 The Two-dimensional Harmonic Oscillator
The Harmonic Oscillator is shown on Fig. 11. It is a rotating vector OA, commonly
called a phasor ([11]), whose tip completes a full rotation around Oin Tseconds, called
the period of the oscillator. If θis the rotation angle of a phasor which has angular
frequency ωwhich corresponds to a period T,θis given as a function of time in radians,
as:
θ(t, T ) = ωt =2πt
T(1.1)
The Cartesian coordinates OB and OC of the phasor, are given as functions of tas:
x(t, T ) = OB = cos(θ(t, T ))
y(t, T ) = OC = sin(θ(t, T )) (1.2)
Using Complex number notation on the Complex xy plane, a phasor with period T
and length ρRcan be represented as a function of t:
~v(t, T ) = ρeθ(t,T )i, 0 t < T (1.3)
If n, j N, 1 jn,~vj(t, Tj) are phasors with corresponding periods Tj, and cj
are constants, then any linear combination HS(t) = Pn
j=1 cj~vj(t, Tj) is again a phasor
according to vector addition, provided there exist mj, mkNsuch that:
mjωj=mkωk, 1 j, k n(1.4)
Because ω= 2π/T , (1.4) is equivalent to:
1Geometric figures were generated with [8] and [9].
3
Figure 2: Harmonic Oscillator in three dimensions
Tj
mj
=Tk
mk
, 1 j, k n(1.5)
Equation (1.4) then is also equivalent to:
Tj
Tk
=mj
mk
, 1 j, k n(1.6)
Periods Tj,Tksatisfying equation (1.6) are called commensurable.HS (t) is called a
Harmonic Phasor Sum and such sums always generate a periodic signal2. The period
of such a signal is given in [2]3and is T= LCM(Tj: 1 jn). The latter is defined
because the Tiare commensurable.
2 The Three-dimensional Harmonic Oscillator
A Harmonic Oscillator in three dimensions can be seen as a generalization of the two-
dimensional Harmonic Oscillator with two phasors, but instead of both phasors rotating
on the xy plane, one phasor rotates on the xy plane and the other rotates on the plane
which is always normal to the first phasor. Such an oscillator can be visualized as in Fig.
2.
A static instance of Fig. 2 suggests an immediate reduction to a Spherical Coordinate
system (ρ, θ, φ), with zenith direction zand azimuth axis xand with θ=DAB being
the inclination angle and φ=COA being the azimuth angle. Without loss of generality,
we can take ρ= 1, hence consider the phasor to be moving on the surface of the unit
sphere.
If Tφand Tθare the corresponding (commensurable) periods of the two phasors, the
phasor OB is given as a function of tas:
~v(t, T ) = (1, θ(t, Tθ), φ(t, Tφ)) = µ1,2πt
Tθ
,2πt
Tφ, 0 tT= LCM(Tθ, Tφ) (2.1)
2Phasor Sums with non-commensurable periods do not generate a periodic signal and hence are not
Harmonic.
3If the Tjare not integral, write Tj=aj/bj. Write M= LCM(bj: 1 jn) and nj=M aj/bj.
Then, the period will be given as T= LCM(nj: 1 jn)/M.
4
Figure 3: Cartesian coordinates of phasor ~v(t, T )
2.1 Spherical Coordinates
The Cartesian coordinates of ~v(t, T ) are given as functions of 0 tTby the transfor-
mations (Fig. 3):
~x(t, T ) = sin(θ(t, Tθ)) cos(φ(t, Tφ))
~y(t, T ) = sin(θ(t, Tθ)) sin(φ(t, Tφ))
~z(t, T ) = cos(θ(t, Tθ))
(2.2)
In order to understand the oscillation in three dimensions, one needs to understand the
oscillations the Harmonic Oscillator induces on all three planes, xy,xz and yz, separately.
Assume then that there exist m, n N, such that:
Tθ
m=Tφ
n(2.3)
2.1.1 xy Plane
Equation (2.3) implies θ= (n/m)φ. The oscillation on the Complex xy plane is described
by the complex number x(θ, φ) + y(θ, φ)i=x((n/m)φ, φ) + y((n/m)φ, φ)i, which is the
same curve as that described by the number x(nφ, mφ)+y(nφ, mφ)imodulo the argument
range, consequently both xand ycan be written as functions of φ, based on the equations
(2.2), as,
x(φ) = sin() cos()
y(φ) = sin() sin()(2.4)
After some elementary trigonometric transformations, equations (2.4) can be con-
verted to their equivalent exponential form, which give the projection of the phasor ~v(t, T )
on the plane xy as a function of φ, as:
~vxy (φ) = i
2³e(mn)φi e(m+n)φi´, 0 φ2πT (2.5)
Theorem 2.1 The curve of equation (2.5) is a folium with p-fold radial symmetry, where,
p=max (GCD(2, m),GCD(2, n)) n
GCD(m, n)
5
Figure 4: Folium ~vxy(φ), 0 φ2π T , for m= 3 and n= 10
Proof:
~vxy (φ) = i
2emφi ¡enφi enφi¢
Note that either p=n/ GCD(m, n) or p= 2n/ GCD(m, n), consequently we can write
2n/p =k. Then,
~vxy µφ+2π
p=i
2em(φ+2π
p)i³en(φ+2π
p)ien(φ+2π
p)i´
=i
2emφie2mπi
p³enφie2nπi
penφie2nπi
p´
=i
2emφie2mπi
p¡enφi(1)kenφi (1)k¢
=i(1)k
2e2mπi
p³e(mn)ie(m+n)i´
= (1)ke2mπi
p~vxy (φ)
The last equation implies that ~vxy(φ+ 2π/p) is exactly either a clockwise or a coun-
terclockwise rotation by an integer multiple of an angle of 2π/p radians and the theorem
follows. ¤
The folium for m= 3 and n= 10 is shown on Fig. 4. These folia are usually
called roses or rhodonea curves and are essentially the polar curve r= cos(), kQ
([16])(modulo a rotation). This curve is holomorphic4on the disk Dρ={|z|< ρ +²}.
When m= 1, this is also the shape traced by the Foucault Pendulum ([12]) as seen from
directly above (i.e. on the xy plane) because of precession5.
Let Ddenote the complex unit circle on the plane xy. From Theorem 2.1 follows
immediately that,
4For example, under an appropriate coordinate transformation which takes the arguments ρand φ
onto a standard polar coordinate system ρand θ, the curve can be seen to satisfy the polar form of the
Cauchy-Riemann equations.
5The folia of Theorem 2.1 also bare similar p-polygonal symmetry to the symmetry of the orbit
of the iterated exponential zz···
on the Complex plane ([3]), with the exception of that when either
m/ GCD(m, n) or n/ GCD(m, n) is divisible by 2, the period pis doubled. This symmetry is also similar
to the symmetry which is observed inside a reflecting unit circle, when an observer shines a light ray
inside it at an angle θrelative to the tangent at the point on the circle where the ray enters. In this case
one has p= 2θ.
6
Figure 5: ~vxz(φ), 0 φ2πT , for m= 3 and n= 10
Corollary 2.2 For some z0D,
{~vxy(φ) : 0 φ2πT } ∩ D={zC:zp=z0}
2.1.2 xz Plane
Similarly, the curve for the xz plane is given by the curve x(nφ, mφ) + z()i, and in
this case equations (2.2) give,
x(φ) = sin() cos()
z(φ) = cos()(2.6)
Converting the equations to their exponential form, we get the projection of the phasor
~v(t, T ) on the xz plane as a function of φ, as:
~vxz (φ) = i¡enφi +e(n+m)φi¢2¡emφi 1¢2
4e(n+m)φi , 0 φ2πT (2.7)
The trajectory ~vxz (φ) is shown on Fig. 5. Typically, it will be a projected Cl´elies
curve6and it will display symmetry only along one axis, since x(φ) = x(φ).
2.1.3 yz Plane
Similarly in this case as well, the curve for the yz plane is given by the curve y(nφ, mφ) +
z()i, and in this case equations (2.2) give,
y(φ) = sin() sin()
z(φ) = cos()(2.8)
Converting again the equations above to their exponential form, we get the projection
of the phasor ~v(t, T ) on the yz plane as a function of φ, as:
6This curve in general will not be holomorphic, as it may contain “cusps”, as, for example in the case
m= 1 and n= 2.
7
Figure 6: ~vyz (φ), 0 φ2π T , for m= 3 and n= 10
~vyz (φ) = ¡emφi +i¢2¡enφii+e(n+m)φi ¢2
4e(n+m)φi , 0 φ2πT (2.9)
The trajectory ~vyz (φ) is shown on Fig. 6. Again, in general, this curve will display
symmetry, but only along one axis, since y(φ) = y(φ). Typically, it will be the dual
Cl´elies curve7of the curve ~vxz (φ) as in this example for m= 3 and n= 10.
2.2 Cylindrical Coordinates
If we project the motion of the Oscillator to a cylinder tangent to the (unit) sphere on
which it moves, the Cartesian coordinates of ~v(t, T ) are given as functions of 0 tT
by the transformations (Fig. 3):
~x(t, T ) = cos(φ(t, Tφ))
~y(t, T ) = sin(φ(t, Tφ))
~z(t, T ) = cos(θ(t, Tθ))
(2.10)
Again in order to understand the oscillation in three dimensions on the cylinder, one
needs to understand the oscillations the Harmonic Oscillator induces on all three planes,
xy,xz and yz, separately. Assume then that there exist m, n N, such that equation
(2.3) is satisfied.
2.2.1 xy Plane
The oscillation on the Complex xy plane is described by the complex number x(φ) +
y(φ)i= cos(φ) + sin(φ)i, which is the equation of a circle and which is exactly the
equation of a two-dimensional Harmonic Oscillator on the xy plane.
2.2.2 xz Plane
Similarly, the curve for the xz plane is given by the curve x() + z()i, so in this
case equations (2.10) give,
7This curve in general will also not be holomorphic.
8
Figure 7: ~vxz(φ), 0 φ2πT , for m= 3 and n= 10
x(φ) = cos()
z(φ) = cos()(2.11)
The phasor trajectory of equations (2.11) is shown on Fig. 7. It is a Lissajous
or Bowditch curve([15])8, with an appropriate parameter determined by the ratio m/n.
Because in this case m/n Q, the curve is closed9.
2.2.3 yz Plane
The curve for the yz plane is given by the curve y()+ z()i, so in this case equations
(2.10) give,
y(φ) = sin()
z(φ) = cos()(2.12)
The phasor trajectory is shown on Fig. 8 and is again a Lissajous or Bowditch curve,
with an appropriate parameter determined by the ratio m/n. It is the Lissajous curve of
equations (2.11), modulo an argument shift, since sin(φ) = cos ¡π
2φ¢.
2.3 On The Sphere
The three dimensional curve traced on the surface of the sphere (Cl´elies)10 is shown on
Fig. 9.
3 Signal
The two-dimensional Harmonic Oscillator has a simple signal. If an observer is situated
at any point of the extended line OA, the oscillator has period Tand δis the Dirac delta,
the oscillator’s signal will be,
8These curves are holomorphic and are related to harmonic motion and the Harmonograph ([13]) and
hence to the trajectories of coupled pendulums.
9Closed in the sense of having finite length for 0 φ < .
10Investigated by Pappus of Alexandria.
9
Figure 8: ~vyz (φ), 0 φ2π T , for m= 3 and n= 10
Figure 9: ~v(t, T ) on the unit sphere, for m= 3, n= 10 and 0 tT
10
Figure 10: Fourier series F25(t) for signal sH O (t) with T= 1 and 0 t3T
sHO (t) = δ(tt0+kT ), t0
T=HOB
2π,kN(3.1)
Without loss of generality, we may take t0= 0 (observer along the positive x-axis),
and consider the signal only for 0 tT, in which case the Fourier coefficients of sHO (t)
will be given ([10]) in terms of the Heaviside step function H(t) ([14]) as11:
a0=1
TZT
0
δ(t)dt =H(T)1
T
an=1
TZT
2
T
2
δ(t) cos µ2nπt
Tdt =2H(T)1
T
bn=1
TZT
2
T
2
δ(t) sin µ2nπt
Tdt = 0
(3.2)
The signal sHO (t) can then be approximated by the Fourier series,
F(t) = a0
2+
X
n=1 ·ancos µ2πnt
T+bnsin µ2πnt
T¶¸ (3.3)
F25(t) is shown on Fig. 10. When n→ ∞, the representation becomes a Dirac comb
([18]), and will vanish everywhere except at tN, where the function will be undefined.
When 0 <t0
T=HOB
2π<1, the corresponding approximation will be that of Fig. 10
shifted right t0
Tunits.
The three-dimensional Harmonic Oscillator also has a simple signal. If the observer is
situated along the line O(ρ, θ(t, Tθ), φ(t, Tφ)), and Tθand Tφare commensurable, the
signal will be (Fig. 2):
sSH O(t) = δ(tt0+k T ), t0
T=COA
2π,kN,T= LCM(Tθ, Tφ) (3.4)
Again without loss of generality, we may take t0= 0 so that the observer may be
situated along the zenith direction z, in which case he will detect a signal once per
11Modulo a multiplicative constant.
11
Figure 11: ~vxy(φ), 0 φ20π, for m=πand n= 10
T= LCM(Tθ, Tφ). The Fourier coefficients will then be those of (3.2), but with the new
Tinstead. The corresponding Fourier series will be similar to (3.3) but the spikes of the
signal will be spaced Tunits apart.
The Inverse Fourier Transform of signal (3.4) will be:
F1{sSH O(t)}=Z
−∞
δ(t)e2πitξ = 1 (3.5)
Therefore the spectrum of the spherical Harmonic Oscillator will be unity ([10, p. 142])
and the spherical Harmonic Oscillator will broadcast at a frequency f(Hz), with,
f=1
T=1
LCM (Tθ, Tφ)(3.6)
4 Non-commensurability
When Tθ/TφR\Q, chaos enters the game, because the phasor’s motion is not harmonic.
4.1 Spherical Coordinates
With Spherical coordinates the projection of the phasor of equation (2.1) on the xy plane
will be a rose with an infinite number of petals. Further, as t→ ∞, the points on the
unit circle which are visited by the phasor will be dense12. The corresponding projections
of the phasor on the planes xy,xz and yz will also be dense on the unit disk as well.
Projections for m=πand n= 10 are shown on Figs. 11, 12 and 13.
4.2 Cylindrical Coordinates
On Cylindrical coordinates the corresponding Lissajous curves on the xz and yz planes
will not close and will be dense on their corresponding planes13 and are shown on Figs.
14 and 15.
12Exactly in a similar way the orbit of the iterated exponential zz···
becomes chaotic when a certain
parameter tof zis irrational. See footnote 5. Details on [3].
13See footnote 12, above.
12
Figure 12: ~vxz(φ), 0 φ20π, for m=πand n= 10
Figure 13: ~vyz (φ), 0 φ20π, for m=πand n= 10
13
Figure 14: ~vxz(φ), 0 φ20π, for m=πand n= 10
Figure 15: ~vyz (φ), 0 φ20π, for m=πand n= 10
14
Figure 16: Gyroscopic Mechanical Time Machine
5 A Simple Time Machine
When the Harmonic Oscillator of (1.3) oscillates with period T, it has angular frequency
ω= 2π/T . This gives rise to a linear velocity ~v, with k~vk=ωρ at the tip of the phasor.
Because ~v is a linear function of ρ, there exists ρfor which k~vk=c, with cbeing the
speed of light. Therefore the oscillating phasor creates a space-time singularity exactly
at the point E= (ρ0, θ), where ρ0is such that ωρ0=c. This singularity will be a
two-dimensional circle14 of radius ρ0. This principle can be used to create a simple time
machine, by employing, for example a rotating laser beam. The same principle can be
generalized in two dimensions two create a three-dimensional singularity. This can be
realized in at least two ways: Mechanically and optically.
5.1 Mechanical
A simple mechanical time machine can be realized as on Fig. 16. It consists of a gyroscopic
mechanism with two15 rotating rings16. The first ring is anchored against a non-moveable
base and the second is anchored either against the base or against the first one at an angle
of π/2 radians.
To create a singularity at E, the linear velocity there must be equal to at least c.
Since the linear velocity at Eis the vector sum of the linear velocities of the two rings,
we must have:
14For a relativistic explanation of what happens at Efor the Crab Nebula pulsar for example, see [6,
pp. 268-274].
15The analysis which follows easily generalizes to k > 2 rings. In this case, the anchors of each inward
ring should be spaced at angles α=π/k radians from the anchors of the previous ring. When k= 2,
α=π/2.
16Surprisingly this design is almost identical to the designs used in the films Contact (adapted from a
Carl Sagan novel) and X-Men ([23],[20]), although the radii of the mechanical rings used in these designs
were way too small and the number of rings (k) was not always equal to 2. Taking for example on average
ρ50m as in the film Contact with k= 4 rings and solving (5.1) below for T, we get T0.2·105s,
which implies at least 477464.8rps or 477.4kHz for each of the rings. That’s approximately 35.8 orders
faster than the fastest (13.3kHz) rotor (ultrasonic dental drill) we have ([22]).
15
EG +
EF =c
r°
°
°
EG°
°
°
2+°
°
°
EF °
°
°
2=c
q(ρωθ)2+ (ρωφ)2=c
ρqω2
θ+ω2
φ=c
ρ=c
qω2
θ+ω2
φ
=c
2πq1
T2
θ
+1
T2
φ
(5.1)
If the angular frequencies are commensurable, then θ=φand the last of (5.1)
gives the critical radius as a function of the angular frequency ωφas,
ρ=c
ωφq¡n
m¢2+ 1
=c
2π
Tφq¡n
m¢2+ 1
(5.2)
5.2 Optical
An optical time machine based on the above principle can also be realized again as a
gyroscopic mechanism, but this time instead of mechanical rings we can rotate laser
beams. To find out the details of this arrangement, we remember that any point on the
surface of a sphere can be described by its Euler Angles, (α, β, γ). Further, if a point A
is rotated to position Eby the Euler angles (α, β, γ ), it can be shown that there exist
angles (a, b, c) such that the following list of operations rotates point Aagain to E(see
for example [1, p. 161]):
1. Rotate by angle a around the z-axis.
2. Rotate by angle b around the new x-axis.
3. Rotate by angle c around the new z-axis.
There are only two axes involved in the overall rotation when using angles (a, b, c).
Therefore, in the simplest case we only need two rotating laser beams. Such a setup
however is equivalent to using one laser beam and two mirrors. A little judicious inspection
shows that we can use the mounting which is depicted on Fig. 17.
When using this setup then, the phasor which describes the laser beam rotational
singularity will be (by equations (2.1) and (5.1)),
(ρ0, θ(t), φ(t)) =
c
qω2
θ+ω2
φ
, ωθt, ωφt
=
c
2πq1
T2
θ
+1
T2
φ
,2πt
Tθ
,2πt
Tφ
(5.3)
When Tθand Tφare commensurable as θ=φ, the phasor is given as a function
of Tφ(by equation (5.2)) as:
(ρ0, θ(t), φ(t)) =
c
ωφq¡n
m¢2+ 1
,n
mωφt, ωφt
=
c
2π
Tφq¡n
m¢2+ 1
,2πnt
mTφ
,2πt
Tφ
(5.4)
16
Figure 17: Gyroscopic Optical Time Machine
5.3 Time Dilation
The time for an observer inside this setup, situated at a distance ρfrom the center of the
setup (Ion Fig. 16 or Bon Fig. 17), will be relativistically dilated and will be given as
t0=γt, where γis the Lorentz Factor,
γ=1
r1hk~vk
ci2(5.5)
Therefore,
t0=t
r1hk~vk
ci2(5.6)
Note that k~vk=ρqω2
θ+ω2
φ, so equation (5.6) becomes,
t0=t
s1·ρω2
θ+ω2
φ
c¸2=t
q1ρ2(ω2
θ+ω2
φ)
c2
(5.7)
When Tθand Tφare commensurable as θ=φ, equation (5.7) becomes,
t0=t
r1£ρωφ
c¤2h¡n
m¢2+ 1i
=t
r1h2πρ
Tφci2h¡n
m¢2+ 1i
(5.8)
As a simple example, if Tθ=Tφ= 60/16000 (266.6 Hz), then if the observer is situated
at ρ= 126km away from the center, he will experience a time dilation of γ10.2. This
means that the time for the observer is passing at a rate of 10.2 times faster than the
time for an external observer (at a distance greater than ρ0)17. The Lorentz Factor is
shown on Fig. 18.
17Surprisingly, this was also the case in the film Contact, where Dr. Arroway experiences 18 hours of
time to have been passed while inside the machine, relative to a few seconds for the external observers.
17
Figure 18: Lorentz Factor with 0 f110000/60, f2= 3f1and 0 ρ100km
5.4 Doppler Shift
The frequency of a source for an observer inside the optical setup, situated at a distance
ρfrom the center of the setup (Ion Fig. 16 or Bon Fig. 17), will be transversely
Doppler-shifted18 and will be given as fo=fs, where fsand foare the stationary and
observed frequencies and γis again the Lorentz Factor.
fo=fss1·k~vk
c¸2
(5.9)
When Tθand Tφare commensurable as θ=φ, in view of the equations of the
previous section, equation (5.9) becomes,
fo=fss1·2πρ
Tφc¸2·³n
m´2+ 1¸(5.10)
Again as a simple example, if Tθ=Tφ= 60/16000 (266.6 Hz), then if the observer is
situated at ρ= 10km away from the center and a green laser beam of wavelength 532nm
is being used with the optical time machine, using the equation f=c/λ the observer will
measure a wavelength of 533.6nm, hence the laser light will be red-shifted.
6 Compressive and Decompressive Agents
For a source Oof waves (EM or otherwise) of velocity ~w, which also rotates counter-
clockwise on the plane as a Harmonic Oscillator with period Tas on Fig. 19, the wave
propagation phasor at times tand t0=t+ ∆tis given as (see equation (1.3)),
~v(t, T ) = ρteθ(t,T )i
~v(t+ ∆t, T ) = ρt+∆teθ(t+∆t,T )i(6.1)
with,
ρt=k~wkt
ρt+∆t=k~wk(t+ ∆t)(6.2)
18To date, only one inertial experiment seems to have verified the redshift effect for a detector actually
aimed at 90 degrees to the object, [7].
18
Figure 19: Implied “decompression” of wave for a Harmonic Oscillator
This gives rise to an implied “decompression” of the phasor, as
OA +
AI =
OI. The
(implied) decompressive agent can be seen as
AI =
OI
OA. This can be seen also in
reverse with respect to time, implying a “compressive” agent
IA.
6.1 Two-dimensional
On the plane, the agent
IA (resp.
AI) can therefore be seen as an operator C:R2R2,
which acts as follows:
C[ρ, θ]=[ρ0, θ0] (6.3)
On the plane therefore, we can describe the operator Caccording to how it acts on ρ
and θ. Setting ρ0=λρρand θ0=λθθ, we find the general form of the operator as:
C[ρ, θ] = [λρρ, λθθ] (6.4)
Theorem 6.1 The operator Cdefined in (6.4) is linear.
Proof:
C[[ρ1, θ1]+[ρ2, θ2]] = C[ρ1+ρ2, θ1+θ2]
= [λρ(ρ1+ρ2), λθ(θ1+θ2)]
= [λρρ1, λθθ1]+[λρρ2, λθθ2]
=C[ρ1, θ1] + C[ρ1, θ1]
and,
C[aρ, aθ] = [ρρ, aλθθ]
=a[λρρ, λθθ]
=aC[ρ, θ]
and the Theorem follows. ¤
6.2 The Eigenvalues of Cin R2
The operator Chas eigenvalues λρand λθ. Depending on the values these eigenvalues
take, the action of Cis determined accordingly.
19
1. λρ= 1 and λθ= 1 + 2kπ/θ,kZ: Identity.
2. λρ= 1 and λθ1 + 2kπ/θ,kZ: Clockwise/counterclockwise rotation.
3. λρ1 and λθ= 1 + 2kπ/θ,kZ: Contraction/expansion.
4. λρ1 and λθ1 + 2kπ/θ,kZ: Contraction/expansion with rotation.
6.3 Three-dimensional
On the sphere, the agent
IA (resp.
AI) can be seen as an operator C:R3R3, which
acts as follows:
C[ρ, θ, φ] = [ρ0, θ0, φ0] (6.5)
In this case we can describe the operator Caccording to how it acts on ρ,θand φ.
Setting ρ0=λρρ,θ0=λθθand φ0=λφφ, we find the general form of the operator as:
C[ρ, θ, φ] = [λρρ, λθθ, λφφ] (6.6)
Theorem 6.2 The operator Cdefined in (6.6) is linear.
Proof: Similar to that of Theorem 2.1 ¤
6.4 The Eigenvalues of Cin R3
Similarly, the operator Chas eigenvalues λρ,λθand λφ. Depending on the values these
eigenvalues take, the action of Cis determined accordingly.
1. λρ= 1 and λθ= 1 + 2kπ/θ,λφ= 1 + 2kπ/φ,kZ: Identity.
2. λρ= 1 and λθ1 + 2kπ/θ,λφ1+2kπ/φ,kZ: Rotation19.
3. λρ1 and λθ= 1 + 2kπ/θ,λφ= 1 + 2kπ/φ,kZ: Contraction/expansion.
4. λρ1 and λθ1+2kπ/θ,λφ1 + 2kπ/φ,kZ: Contraction/expansion with
rotation.
6.5 The Iteration of C
We can now define the repeated operators in two and three-dimensional space.
Definition 6.3 Cn:N×R2R2, with
Cn[ρ, θ] = £λn
ρρ, λn
θθ¤
Definition 6.4 Cn:N×R3R3, with
Cn[ρ, θ, φ] = £λn
ρρ, λn
θθ, λn
φφ¤
Theorem 6.5 If λρ<1then the sequence Cn[ρ, θ](resp. Cn[ρ, θ, φ]), nN, converges
to the origin.
19In Spherical coordinates.
20
Proof: The origin [0,0] (resp. [0,0,0]) is a fixed point of the map C(resp. of Cin R3),
since C[0,0] = [λρ·0, λθ·0] = [0,0] (resp. C[0,0,0] = [λρ·0, λθ·0, λφ·0] = [0,0,0]). Further,
if λρ<1 then the sequence kCn[ρ, θ]k(resp. kCn[ρ, θ, φ]k) is monotone decreasing and
bounded below by 0 and the Theorem follows. ¤
Remark 1: Cases 3 and 4 of Section 6.4 are interesting in that they can be used to
model physical situations and are connected to the Mathematical Theory of Elasticity:
Case 3 with λρ<1 for example, corresponds to the operator C[ρ, θ, φ] = [λρρ, θ, φ],
which can be seen as uniform compression of a solid sphere. Any such compression is
(ultimately) characterized by the “crack” the compressive agent produces on the various
elastic media it is applied upon. This, has been analyzed completely in [5] for the “gen-
erally” isotropic or anisotropic disk under tension20. The implosion type nuclear weapon
on the other hand is another process which causes deformation and a “crack” (on a much
larger scale). This has been analyzed in ([4]).
Remark 2: Case 3 of Section 6.4 with λρ= 0 or the iteration Cn[ρ, θ, φ], nN, with
λρ<1 can be used to model a sudden or gradual partial or total collapse of a spherical
mass into a more compressed state or singularity and hence can be used as a model for
the more extreme phenomena in astrophysics, such as black holes or stars.
Remark 3: For case 3 of Section 6.4, the iteration Cn[ρ, θ, φ], nN, with λρ>1
can also be used to model the opposite process, the explosion of a spherical mass and
hence can be used as a model to describe the explosive details of any kind of explosive,
conventional or nuclear.
6.6 Applications of Ct
To aid all the above models, the operator Cncan be made continuous21 as a function
of t, so that Ct[ρ, θ, φ] = [fρ(ρ, t), fθ(θ, t), fφ(φ, t)], in which case the system behavior is
completely characterized by the initial value problem (IVP)22,
µ[Cn],·dnCt
dtn¸¶,nN(6.7)
The two-dimensional Harmonic Oscillator may be described for example by the system:
µ[ρ0, θ0],·
dt = 0,
dt =ωθ¸¶ (6.8)
The three-dimensional Harmonic Oscillator may be described by the system:
µ[ρ0, θ0, φ0],·
dt = 0,
dt =ωθ,
dt =ωφ,ωφ
ωθ
=m
n¸¶ (6.9)
6.6.1 Exponentially Collapsing Rotating Body
A rotationally23 collapsing spherical body with mass mdue to uniform compressive forces,
which conserves angular momentum, may be described as follows. Angular momentum
Lis preserved, so if Iis the body’s moment of inertia and ωis its angular frequency at
moment t, then,
ωI =L
dt =ω
I
dI
dt
(6.10)
20With the compressive agent being
IA in the direction ~n, giving component stresses Xnand Ynon
a linear element ∆sas in page 10, therein. This being the most general case, a model of any other case
can be constructed with a subset of [5].
21As continuous however, it may not necessarily be linear.
22Or, equivalently, the Equation of State (EOS).
23Around the zenith axis, so θis not needed, as
dt = 0.
21
Now if ciis the body’s inertial constant and ρits radius, then I=ci·m·(2ρ)2and
for a sphere ci= 2/5, so equation (6.10) gives24,
dt =2ω
ρ
dt (6.11)
Note that equation (6.11) may prohibit solutions with ρ= 0, since then the (momen-
tary) change in angular frequency ωmay become unbounded25. To be on the safe side, call
this radius ρ>0. If the collapse is λρ-exponential, it suffices to take ρ(t) = t
ρ+ρ.
The IVP then with the help of (6.10) becomes,
µ[ρ0, θ0, φ0, ρ1, θ1, φ1],·ρ(t) = t
ρ+ρ,
dt =ωφ,φ
dt =2ωφ
ρ
dt ¸¶ (6.12)
Solving the system ρ(0) = ρ0and ρ(1) = ρ1, gives,
A=λρ
λρ1(ρ0ρ1)
ρ=λρρ1ρ0
λρ1
(6.13)
Note, λρ=ρ0
ρ1and substituting this into the second of (6.13), above, forces ρ= 0
and A=ρ0. Therefore it suffices to consider ρ(t) = ρ0λt
ρand system (6.11) reduces to:
µ[ρ0, θ0, φ0, ρ1, θ1, φ1],·ρ(t) = ρ0λt
ρ,
dt =ωφ,φ
dt =2ωφln λρ¸¶ (6.14)
Solving the differential equation in system (6.14) for this ρ(t), we get:
ωφ(t) = 2t
ρ(6.15)
Note φ(t) = ωφ(t)t, therefore φ1=ωφ(1) and then φ1=2
ρ, from which
C=φ1
λ2
ρ
(6.16)
which substituted in (6.15) gives,
ωφ(t) = φ1λ2(t1)
ρ(6.17)
On the other hand, the linear velocity of any surface particle on this body at any time
cannot exceed the speed of light c, so we can get a relativistic time trel.
ωφ(t)ρ(t) = c
ρ0φ1λt2
ρ=c
trel =
ln hρ0φ1
ci
ln λρ
+ 2
(6.18)
The relativistic radius therefore will be ρrel =ρ(trel ), which evaluates as,
24Note that equation (6.10) is independent of the body’s mass and its inertial constant.
25This will depend on the asymptotic relationship between ωφand ρfor a body with non-zero angular
momentum.
22
ρrel =µρ0
λρ2φ1
c(6.19)
The relativistic angular frequency will be ωrel =ωφ(trel ), which evaluates as,
ωrel =µρ
ρ021
φ1
(6.20)
For the sun, for example k~v¯k ∼ 7.189 ·103km/h and ρ0=ρ¯6.955 ·105km.
This means that ω¯0.2871235721 ·105rad/sec26 and φ1=ω¯(1). If the sun was
to collapse under a force which has λρ= 2, it would collapse in trel 19.196secs to a
relativistic radius of ρrel 1.1574km, and the final object would have a relativistic angular
frequency ωrel 259202.6rad/sec, which corresponds to a period Trel = 2πrel
0.2424043917 ·104secs, or 41.253kHz. This happens because the norm of the linear
velocity becomes unbounded as a function of time t, being exponential.
6.6.2 Gravitationally Collapsing Rotating Body
A different model can be built if we consider the collapse to be gravitational. For a particle
at angle θfrom the (zenith) axis of rotation, taking into account centripetal forces because
of the object’s rotation ([19]) and the object’s gravitational constant g, the radius will be
given as:
ρ(t) = 1
2¡gω2ρ(t) sin(θ)¢+ρ0
ρ(t) = gt22ρ0
t2ω2sin(θ)2
(6.21)
Using equations (6.21), the system reduces to:
µ[ρ0, θ0, φ0, ρ1, θ1, φ1],·ρ=gt22ρ0
t2ω2sin(θ)2,
dt =ωφ, DE1¸¶ (6.22)
If φ
dt =ω0, then DE1 of (6.22) is:
ω0=4ωt ¡2g+ωsin(θ)t302ω2sin(θ)ρ02ωsin(θ)ρ00¢
(2 + t2ω2sin(θ))(t2g2ρ0)(6.23)
If C1and C2are constants, then the solution of differential equation (6.23) found by
Maple is given in terms of the “RootOf” function:
ω(t) = 1
RootOf [2z4+ (C1gt2C22ρ0)z3t2sin(θ)]2(6.24)
To avoid having to deal with all four solutions of (6.24), we can try to substitute some
values and see if we can solve (6.23) directly. Substituting for example ρ0=ρ¯,θ=π/4
and g¯= 9.81 ·27.94m/sec2, and solving the resultant differential equation we get the
implicit form:
26Which corresponds to T¯= 2π¯696564.1954πsec25.32 days.
23
Figure 20: Radius ρ¯/105and surface linear velocity ~v¯for a gravitationally collapsing
sun as functions of t
ln(ω(t)) 2 ln ³22 + t2ω(t)2´
+241860125 ·1017 ln ¡1370457t26955 ·109¢ω(t)4
(13704572 + 34775 ·108ω(t)2)2
1906305687 ·1010 ln ¡1370457t26955 ·109¢ω(t)22
(13704572 + 34775 ·108ω(t)2)2
+7512609555396 ln ¡1370457t26955 ·109¢
(13704572 + 34775 ·108ω(t)2)2=C
(6.25)
On equation (6.25) we substitute ω(1) = φ1and solve for C, giving C.
=44.3,
which we plug back to get the complete implicit solution. The resultant equation has the
solution,
ω(t) = 0.1736092821 ·1020
RootOf ht2z4e(886015077
107)22 + (1370457t2+ 6955 ·109)zi2(6.26)
Maple finds a series solution for equation (6.26), as:
ω(t).
= 0.2871234471 ·105+ 0.1131515507 ·1011t2
+ 0.3344265708 ·1018t4+ 0.8785490887 ·1025 t6
+ 0.2163554639 ·1031t8+ 0.5114361751 ·1038 t10 +O(t12)
(6.27)
With this ω(t) = ω¯(t) we can estimate ρ¯(t) and we can see the magnitude of the
surface linear velocity of the collapsing sun, as k~v¯(t)k=ω¯(t)ρ¯(t) . This is shown on
Fig. 20. Note that in this case gravity wins and the preservation of angular momentum
is not enough to prevent total collapse of the object. The sun would collapse in tc.
=
2252.7secs, roughly in 37.5 minutes into a singularity of radius ρ(tc) = 0, with an angular
frequency just before total collapse of ω(tc) giving a frequency of 0.9593555615 ·105Hz.
When ρ(tc) becomes zero, the angular velocity becomes zero of course. This would happen
for example if all nuclear reactions at the core suddenly stopped and did not resume under
any pressure.
24
Figure 21: Surface linear velocity ~v¯for a harmonically pulsating sun as a function of t
(in days)
6.6.3 Harmonically Pulsating Rotating Body
Another model which is of interest is that of a harmonically pulsating spherical body. If
its pulsation angular frequency is ωρand its pulsation width is A < ρ0, then without loss
of generality we may assume that its radial pulsation is given as,
ρ(t) = ρ0+Asin (ωρt) (6.28)
In this case, the system reduces to:
µ[ρ0, θ0, φ0, ρ1, θ1, φ1],·ρ0+Asin (ωρt),
dt =ωφ,φ
dt =2ωφ
ρ
dt ¸¶ (6.29)
System (6.29) has the solution,
ω(t) = 2φ1(ρ2
0+ 2ρ0Asin(ωρ) + A2sin(ωρ)2)
2ρ2
0+ 4ρ0Asin(ωρt) + A2sin(2ωρt)(6.30)
The numerator of expression (6.30) vanishes when,
ωρ= arctan Ãρ0
A,±pA2ρ2
0
A!(6.31)
The denominator vanishes when,
t=
arctan µρ0
A,±A2ρ2
0
A
ωρ
(6.32)
Because A < ρ0, expressions (6.31) and (6.32) are complex, therefore expression (6.30)
never vanishes. A graph of the surface linear velocity as a function of time, is shown on
Fig. 21.
To find the maximum pulsation width, set A=ρ0in equation (6.28), in which case
system (6.29) gives the solution,
25
ω(t) = 2φ1(2 + 2 sin(ωρ)cos(ωρ)2)
3 + 4 sin(ωρt)cos(2ωρt)(6.33)
The denominator of expression (6.33) vanishes when,
tk(ωρ) = 2kπ π
2
ωρ
(6.34)
Because the linear velocity becomes unbounded when the denominator of (6.33) van-
ishes, we get a relativistic time trel [0, t1(ωρ)], which after substitution into equation
(6.28) gives a relativistic radius ρrel =ρ(trel), as,
ρrel =ρ2
0φ1(2 sin(ωρ)2cos(ωρ)2)
c(6.35)
This means that the maximum width of any harmonically pulsating and rotating
spherical body will be:
A=ρ0ρrel (6.36)
For a harmonically pulsating sun with the sun’s rotating speed, ρrel evaluates to
4.63km.
7 Concluding Remarks
If on a body there is any compression or expansion or rotation along any axis, the model
can be sufficiently modelled using IVP (6.7), with parameters specifying the type of
movement.
The IVP (6.7) for the Hydrogen atom for example, is given by the Schr¨odinger Equa-
tion, [17], while IVP (6.7) for the Sun, is given by the Polytrope Model, [21]27 .
It seems likely then that any physical state therefore which moves in any of the above
modes can be modelled by IVP (6.7). One may characterize a mathematical model of
any such state as correct, when the corresponding IVP for this state takes on a form
which is solvable28. In this case, the IVP which results will have a stable solution as a
function of time tif the process is in equilibrium29 or an unstable or perhaps metastable
solution whenever the process is of a more extreme kind, such as an explosion, implosion
or reduction of any mass to near critical radius.
27Although the Polytrope Model is not a model of motion, it is an equilibrium model and as such it
describes perfectly well the forces involved in that equilibrium.
28Note that although IVP (6.7) may take on a more specific form, it may not necessarily be solvable
by analytical/algebraic methods. For example, the IVP which describes the state of any atom with more
than one electrons clearly exists, but it may not necessarily be solvable analytically. In either case the
harmonics of the corresponding atomic states are given by the element spectrum, which is exactly a
Fourier analysis of the atom’s signal under excitation.
29Such as Quasi-Hydrostatic Equilibrium in stars for example.
26
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