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## Curvilinear Motions in Newtonian Mechanics and Quantum Spin

### Contents

Centripetal and Centrifugal Forces
Orbital Motions
Rigid Body Dynamics
Differential Rotation
Quantum Mechanical Spins

### Centripetal and Centrifugal Forces

According to Newton's First Law, force-free object always moves in a straight line. In other word, some kind of force would be involved when the object's motion deviates from a straight line, i.e., in a curvilinear motion. The pulling is called centripetal force such as the gravitational attrcation Fp = -GmM/r2. The centrifugal force is in a direction pushing outward, this is the force resisting the change and in the form Ff = mv2/r (Figure 01a). These two forces cancel out exactly in circular motion as depicted in most of the popular images. In this special case the mathematics is especially simple with :
radial velocity vr = dr/dt = 0, r = constant, angular velocity v = d/dt = v/r = constant, and the angular momentum L = mvr = mvA/ = constant, where A = r2 is the area of the circle.

#### Figure 01a Curvilinear Motion [view large image]

The Kepler's three laws are very simple in this special case : 1. the circular trajectory is a special case of an ellipse, 2. Since both r and /t are constants, the area A sweeps out in interval t is always A = r2/2.
3. By equating Fp = Ff , it can be shown readily that the square of the period T2 = (42/GM) r3.

### Orbital Motions

While the angular momentum L = mvr remains to be a constant for orbital motion in general, the equation of motion in the radial direction becomes more complicated since the centripetal and centrifugal forces do not balance as shown below.

#### Figure 01b Kepler's Three Law [view large image]

This is the Kepler's 1st Law. The 2nd Law is expressed by dA/dt = L/2m = constant, where dA = (r2/2)d. While the 3rd Law is covered by T = ab/(dA/dt) giving T2 = (42/GM)a3.

#### Figure 02 Parabolic Motions [view large image]

Although there is no solution for three body problem in closed form, it turns out that there are points in a three body configuration with two large and a much smaller ones on which the forces are balanced. They can rotate together as a whole. The five such locations are called Lagrangian points L1, ... L5. Following is the mathematics to derive the location of the smaller object r at L1 with respect to M2; R is the distance between M1 and M2 (see Figure 03a). According to a theorem in two-body problem, the two body motion can be reduced to a one body equivalence with center of mass M = M1 + M2 and distance Rm = M1R/M (with origin at M2). Assuming circular motion, the equation for the balance of contripetal and contrifugal forces can be written as :

#### Figure 03a Lagrangian Points [view large image]

This is exactly the distance to L1 - the parking lot for many of the artificial satellites. However, it is known to be unstable, thruster burns manoeuvre is required to keep the object in place.
The equation for L2 is similar with (R-r) replaced by (R+r) and changing sign for the 2nd and 4th terms. The approximate distance r is the same as the one for L1. BTW, see "Cubic Formula" for cubic root calculation. The approximate formula for r is obtained by neglecting the r5, r4 powers and taking only the leading term in the cubic formula.

See "Synchronous Rotation" for another application of Newtonian mechanics to celestial bodies.

One more example of circular motion involves charged particle moving in magnetic field. It is part of the
"Lorentz Forces" :     F = e (v x B),
where v is the velocity and B the magnetic field. The motion is particularly simple when v is perpendicular to a constant B field giving F = Bev, which is balanced by the centrifugal force : Bev = mv2/r. Thus, the motion is circular with r = mv/eB = constant (Figure 03b). This is the principle in its application to cyclotron and synchrotron. The motion becomes a helix in case the velocity has a component along the z direction (see insert).

#### Figure 03b Lorentz Force [view large image]

Actually, at relativistic speed the circular motion of a charged particle turns into a spiral circling inward as it losses energy via synchrotron radiation unless to keep replenished such as in the case of synchrotron.

### Rigid Body Dynamics

The Newtonian's equations of motion cannot be solved in closed form for three particles and beyond. It seems that the case for many particles would be much worse. However, such system can be described with mathematical precision if the separation between the particles is constant, that is, it is a rigid body, which is a good approximation to many solid objects. It will be shown momentarily below that the dynamics can be reduced to two equations - one for the translational motion of the center of mass, and the other for the rotation around the center of mass (an additional vibrational mode has to be considered for microscopic oscillation of the individual constituents, see specific heat). As shown in Figure 04a, if the reference point for the particles in the solid is shifted to the center of mass such that i mir'i = 0, r'i = constant, and with the further restriction on the motion to be parallel to a plane, i.e., in 2-dimensional motion, then the equations of motion can be written down in terms of the force F and torque T summing over all the particles in the system :

#### Figure 04b Moment of Inertia

These formulas are applicable to 3-dimensional motion as well, and the rotational term can be neglected if the size of the solid is much smaller than the distance scale.
That's why a galaxy of size ~ 1022 cm can be treated as a point in the cosmological model at the scale of ~ 1028 cm. Another example is the planets with size no bigger than 1010 cm in orbit of about 1013 cm or further from the Sun. Actually, these examples are just one of the reductionist's views of the world, which maintains that details of lower levels can be neglacted in higher level (see "Effective Theories").

When a force is exerted at one point on a beam as shown in Figure 05a,b it will be transferred internally all along the beam. This is a special property of rigid body and is very useful in our daily life. The situation is drastically different when the beam comes to rest on two pivots as shown Figure 05,d. Since the beam is not moving, the total force and total moment (torque) must be zero. Thus, we have

#### Figure 06a Internal Shear Force [view large image]

P = F1 + F2, and F1d = F2(L-d) giving
F1 = [(L-d)/L]P, and F2 = (d/L)P.

The internal shear force Fs inside the beam is computed by the formula :

Fs = F1 + (all forces between A and x) Fi

with the convention that positive force points up, negative force down, and the force F2 at the other end doesn't count. Thus
 Fs = F1 from point A up to the load point, and on to point B with Fs = -F2, see

If we flip the diagram in Figure 05,d, i.e., F1 and F2 become the loads and P is the reaction force at the pivot (also see the wood gatherer ¾ö¤Ò in Figure 06c). The mathematical formulas remain unchanged but with the reverse direction of all the forces (including the shear force).

 Another example with continuous and constant load along the beam is shown in Figure Figure 06a where f is the load density in unit of N/m. The summation in the shear force computation is now replaced by the integral

The internal shear force is canceled out by the reaction of the material within the rigid body resulting in static equilibrium with no deformation.

The above illustrations assume that the beam is perfectly rigid. In the real world, all materials exhibit some degree of elastic property defined by the shear modulus S = Shear Stress / Strain = (P/A) / (L/L). The higher the number corresponds to stiffer material until it becomes infinity for a perfectly rigid body. Figure 05,e shows the deformation of a beam under a point load. The maximum amount of deflection is expressed by the formula max = (PL3) / (48EI), where E is the Young's modulus defined similarly to the shear modulus S except that the force is now applied perpendicular to the area; and I is the second moment of area in unit of (length)4. The material is said to be linear if it can return to its original shape (after running through damped harmonic oscillation, see also footnote) after the

#### Figure 06b Modulus [view large image]

removal of the load. All materials have this characteristic for very small displacement, but the modulus would vary with displacement at large displacement for non-linear materials. All materials will break with very large applying force or deformed to certain amount. For example, the steel rod will break at about 0.25 BPa (109N/m2).

Figure 06b lists the values of three different kinds of modulus for various materials. The Bulk modulus measures the substance's resistance to uniform compression. The other modulus are for linear deformation.

Figure 06c shows some of the indispensable working tools that make modern life much easier. The principle of lever comes from the balance of moments (torques) Faa = Fbb, or Fb = (a/b) Fa (see the illustrations in Figure 06c). Once such configuration has been arranged, then it takes very little effort to move the load at the other end. The factor (a/b) is the "mechanical advantage", which expresses the fact that the lower this number corresponds to lesser applying force (effort) to lift the object (resistance).

#### Figure 06c Mechanical Advantage

 Footnote : For a simply supported beam (as shown in Figure 05,e) released from linear compression, it will undergo damped harmonic motion with force P = kx. This force is derived from the elastic property of the material; in term of the deflection P = (48EI/L3)max. Upon identifing x = max, we obtain k = (48EI/L3). For a rectangular beam with thickness t, width b, I = (t3b)/12 (see illustration below ), then the fundamental frequency = (k/m)1/2 = (2t/L2)(E/)1/2, where is the density of the beam, and m = tbL.

 BTW, there are three types of supports as shown

### Differential Rotation

Another way to describe the many particles motion is to consider them collectively in the form of fluid motion - a velocity field depending on space and time, e.g., u(x,y,z,t). The mathematics is embodied in the Navier-Stokes Equations. For irrotational flow with , they assume the forms :

#### Figure 06d Differential Rotation

where u = velocity vector field, = thermodynamic internal energy, p = pressure, T = temperature, = density, = viscosity, KH = heat conduction coefficient, F = external force per unit mass = acceleration, , and . Further details can be found in "Fluid Dynamics and the Navier-Stokes Equations"

The differential rotation (Figure 06d) is one of the examples for its application. According to the mathematics of Spiral Flow in polar coordinates, the trajectory is in the form of an Archimedean spiral r = a + b (see insert in Figure 06d), while the solution of the velocity fields is :

ur2 = GM / r + 2G0r0 - r2u2,
r2u2 = r (GM + 2G 0r0 r) / (b2 + r2),
where M is the mass of the central black hole, while the mass on the disk within r is given by M' = 2 r dr, where = 0 (r0/r) is the surface density in unit of gm/cm2, 0 is its cutoff value at the edge of the galaxy r = r0, and the third term is related to the centrifugal force. The formula for (r u) shows a similar profile taking from the Milkway disk (Figure 06d). The observed curve takes into account of the dark matter in the halo. This kind of analysis is also applicable to the hurricane, and the drain in the sink although on a much smaller scale.

See "Formation of Spherical Body" for a hydrostatics application.

### Quantum Mechanical Spins

Apart from the orbital motion, the most prominent curvilinear motion in quantum mechanics is the spin associated with every elementary particles. The common feature of quantum spin is its discrete values (in unit of angular momentum), i.e., its value can only be a certain multiple of ~ 10-27 erg-sec. Other characteristics are summarized in the followings.

• Fermions and bosons - As indicated in Figure 07, elementary particles can carry spins in steps of (n/2), where n = 0, 1, 2, 4. Those have integer spin are bosons, while the one with 1/2 is fermion. Table 01 below lists some differences between these two kinds.

#### Figure 08 Elementary Particles

Table 01 Fermion vs Boson

The restriction on identical state occupation (for two fermions) is called "Exclusion Principle". However, such rule can be broken if two 1/2 spin particles combine to form a boson of spin 1. Figure 08 shows a list of the elementary particles with spin 1/2 or 1. The newly discovered Higgs boson has spin 0, while the graviton associated with gravity has spin 2.

• Angular Momentum Quantum Numbers - The 1/2 spin of the fermion has its origin in the derivation of the Dirac Equation, which involves the Pauli matrices :

#### Figure 09 Quantum Numbers S, L, and J [view large image]

The quantum mechanical orbital angular momentum follows the classical definition of L = r x p, with Lx = ypz - zpy, ...

#### Figure 11 Addition of Quantum Spins [view large image]

The tabulations in Figure 11 list the Clebsch-Gordan (CG) coefficients for j1 = 1 and j2 = 1/2 in forming the various j and m; while the specific choice of j = 3/2 and m = 1/2 is shown pictorially by vectors in nearly correct proportion. The j1 and j2 can be any orbital angular momentum or quantum spin. See more CG coefficients entries in "Table of Clebsch-Gordan coefficients".

• Angular Momentum Couplings - According to classical electromagnetism, curvilinear motion of charge e with angular momentum L would carry a magnetic dipole moment = eL/2me (Figure 12). In quantum mechanics, electron with S = /2 = L would have e = - (gs/2)e/2me 1 Bohn magneton = B. It is the crowning success of Quantum Field Theory to shown that the gyro-magnetic ratio e/L = gse/2me where gs -2 = 0.0023318416 is in agreement with experimental measurement up to such accuracy. Since particles with magnetic dipole would interact with each others, there are various kinds of couplings as summarized in the followings.
• Spin-orbit Coupling - A single electron moving around the nucleus would generate a magnet field B = (Ze0/8mer3)L, where 0 = 4x10-7 N/amp2 is the magnetic constant, Z is the number of protons in the nucleus, r is the averaged orbital radius. The interaction is -eB. For example, the 3p energy level for the outermost electron in sodium atom is split into two by S-L interaction generating the sodium doublet lines.

#### Figure 13 S-L Coupling [view large image]

Further splitting with various mj occurs in the presence an external magnetic field (Figure 13).

• J-J Coupling - For light atom with many coupling electrons, the individual spin and orbital angular momenta add together to form L = li, and S = si via electrostatic interaction. Then L and S combine to form J as described above. For heavy atoms, the nuclear charge Z becomes great enough to produce spin-orbit interaction (for each electron) comparable in magnitude to the electrostatic ones between li and between si. The S-L coupling scheme begins to break down. Each electron generates its own total angular momentum ji = li + si to alter the electronic configuration slightly resulting in various spectral splitings. This is the J-J coupling with J = ji.
• #### Figure 14 J-J Coupling [view large image]

BTW, the nuclear charge is involved in the orbital interaction because the nucleus becomes the merry-go-round object with respect to the electron's frame of reference (Figure 14).

• Spin-spin Interaction - This is the interaction between the electron and nucleus magnetic dipole moments e and I = gIII respectively where gI is the nuclear g-factor, I = Ze/2mN the nuclear magneton, and I the nuclear spin. The interaction Hamiltonian turns out to be a rather formidable looking formula, but luckily can be summarized into a much simplified form such as A(IJ) where A is the hyperfine structure constant to be determined by experiment. The net effect of this interaction is to introduce a very small splitting (hyperfine) in atomic energy levels. For example, the 1s state of hydrogen atom is separated by 6x10-6 ev according whether the two spins are parallel or anti-parallel. The 21 cm emission line from transition between these states has been used to produce the rotation curve of the Milkyway (Figures 15 and 06). Another important application is to make atomic clock by the hyperfine transition in the cesium atoms .
• #### Figure 15 Spin-spin Interaction

• Entanglement of Spins - The entanglement can occur naturally as shown in Figure 16 via the decay of a meson. On the other hand, two independent spins can be forced to merge and then separated as an entangled pair (Figure 17). Anyway, they are
described by the superposition of states, which are subjected to decoherence. Since it is known that the two electrons in the H2 molecule and the Helium atom are entangled (see insert in Figure 17), the coupling mechanism to create superposition state could be similar to the s-s coupling mentioned above.
The entanglement of photons involves another kind of mechanism. On rare occasions, two down-conversion photons in

#### Figure 17 Entanglement of Spins [view large image]

superposition state are emitted simultaneously to form an entangled pair.

Thus, there are many ways to create entangled pairs as long as the particles can be assembled into a superposition state initially.