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Orbital Motions

Rigid Body Dynamics

Differential Rotation

Quantum Mechanical Spins

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 F_{p} = -GmM/r^{2}. The centrifugal force is in a direction pushing outward, this is the force resisting the change and in the form F_{f} = mv^{2}/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 v _{r} = dr/dt = 0, r = constant, angular velocity v_{} = d/dt = v/r = constant, and the angular momentum L = mvr = mv_{}A/ = constant, where A = r^{2} is the area of the circle.
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## Figure 01a Curvilinear Motion |
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 = r^{2}/2. |

## Figure 01b Kepler's Three Law |
This is the Kepler's 1st Law. The 2nd Law is expressed by dA/dt = L/2m = constant, where dA = (r^{2}/2)d. While the 3rd Law is covered by T = ab/(dA/dt) giving T^{2} = (4^{2}/GM)a^{3}. |

## 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 M

## 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. |

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 = mv^{2}/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).
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## Figure 03b Lorentz Force |
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. |

## Figure 04a Rigid Body Dynamics |
## 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. |

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 05 Statics in Rigid Body [view large image] |
## Figure 06a Internal Shear Force [view large image] |
P = F_{1} + F_{2}, and F_{1}d = F_{2}(L-d) giving F _{1} = [(L-d)/L]P, and F_{2} = (d/L)P. |

The internal shear force F

F

with the convention that positive force points up, negative force down, and the force F

F_{s} = F_{1} from point A up to the load point, and on to point B with F_{s} = -F_{2}, see |

If we flip the diagram in Figure 05,d, i.e., F

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} = (PL^{3}) / (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 (10^{9}N/m^{2}). |

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) F _{a}a = F_{b}b, or F_{b} = (a/b) F_{a} (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/L^{3})_{max}. Upon identifing x = _{max}, we obtain k = (48EI/L^{3}). For a rectangular beam with thickness t, width b, I = (t^{3}b)/12 (see illustration below _{}), then the fundamental frequency = (k/m)^{1/2} = (2t/L^{2})(E/)^{1/2}, where is the density of the beam, and m = tbL. |

BTW, there are three types of supports as shown |

## Figure 06d Differential Rotation |
where u = velocity vector field, = thermodynamic internal energy, p = pressure, T = temperature, _{} = density, _{} = viscosity, K_{H} = heat conduction coefficient, F = external force per unit mass = acceleration, _{}, and _{}. Further details can be found in "Fluid Dynamics and the Navier-Stokes Equations" |

u

r

where M is the mass of the central black hole, while the mass on the disk within r is given by M' = 2

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

- 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
- 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 07 Quantum Mechanical Spins [view large image]

#### 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 L_{x}= yp_{z}- zp_{y}, ...

#### Figure 10 Legendre Polynomials [view large image]

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

The tabulations in Figure 11 list the Clebsch-Gordan (CG) coefficients for j_{1}= 1 and j_{2}= 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 j_{1}and j_{2}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/2m
_{e}(Figure 12). In quantum mechanics, electron with S = /2 = L would have_{e}= - (g_{s}/2)e/2m_{e}1 Bohn magneton =_{B}. It is the crowning success of Quantum Field Theory to shown that the gyro-magnetic ratio_{e}/L = g_{s}e/2m_{e}where g_{s}-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**= (Ze_{0}/8m_{e}r^{3})**L**, where_{0}= 4x10^{-7}N/amp^{2}is the magnetic constant, Z is the number of protons in the nucleus, r is the averaged orbital radius. The interaction is -_{e}**B**. 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. - J-J Coupling - For light atom with many coupling electrons, the individual spin and orbital angular momenta add together to form L =
_{}l_{i}, and S =_{}s_{i}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 l_{i}and between s_{i}. The S-L coupling scheme begins to break down. Each electron generates its own total angular momentum j_{i}= l_{i}+ s_{i}to alter the electronic configuration slightly resulting in various spectral splitings. This is the J-J coupling with J =_{}j_{i}. - Spin-spin Interaction - This is the interaction between the electron and nucleus magnetic dipole moments
_{e}and_{I}= g_{I}_{I}**I**respectively where g_{I}is the nuclear g-factor,_{I}= Ze/2m_{N}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(**I****J**) 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 . - 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

Thus, there are many ways to create entangled pairs as long as the particles can be assembled into a superposition state initially.described by the superposition of states, which are subjected to decoherence. Since it is known that the two electrons in the H _{2}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 16 Entanglement of Spins [view large image]

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

superposition state are emitted simultaneously to form an entangled pair.

#### Figure 12 Magnetic Dipole Moment [view large image]

#### Figure 13 S-L Coupling

[view large image]Further splitting with various m _{j}occurs in the presence an external magnetic field (Figure 13).

#### 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).

#### Figure 15 Spin-spin Interaction

_{} - Spin-orbit Coupling - A single electron moving around the nucleus would generate a magnet field