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User:Marshallsumter/Radiation astronomy1/Mathematics

Most of the mathematics needed to understand the information acquired through astronomical radiation observation comes from physics. But, there are special needs for situations that intertwine mathematics with phenomena that may not yet have sufficient physics to explain the observations. Both uses constitute radiation mathematics, or astronomical radiation mathematics, or a portion of mathematical radiation astronomy.

Astronomical radiation mathematics is the laboratory mathematics such as simulations that are generated to try to understand the observations of radiation astronomy.

The mathematics needed to understand radiation astronomy starts with arithmetic and often needs various topics in calculus and differential equations to produce likely models.

Notations
Notation: let the symbol $$ L_{\odot} $$ indicate the solar visible luminosity.

 * {| border="1" cellpadding="5" cellspacing="0" align="none"

! colspan="3" |Notational locations
 * Weight
 * Oversymbol
 * Exponent
 * Coefficient
 * Variable
 * Operation
 * Number
 * Range
 * Index
 * }
 * Range
 * Index
 * }
 * }

For each of the notational locations around the central Variable, conventions are often set by consensus as to use. For example, Exponent is often used as an exponent to a number or variable: 2-2 or x2.

In the Notations at the top of this section, Index is replaced by symbols for the Sun (⊙), Earth ($$R_\oplus$$), or can be for Jupiter (J) such as $$R_J$$.

A common Oversymbol is one for the average $$\overline{Variable}$$.

Operation may be replaced by a function, for example.

All notational locations could look something like

 * {| border="0" cellpadding="3" cellspacing="0" align="none"

where the center line means "a x Σ f(x)" for all added up values of f(x) when x = n from say 0 to infinity with each term in the sum before summation multiplied by bn, then divided by n for an average whenever n is finite.
 * bx
 * x = n
 * a
 * $$\sum$$
 * f(x)
 * n
 * }
 * f(x)
 * n
 * }
 * }
 * }
 * }
 * }

Astronomy
An initial use of mathematics in astronomy is counting entities, sources, or objects in the sky.

Objects may be counted during the daytime or night.

One use of mathematics is the calculation of distance to an object in the sky.

Most of the topics in mathematics that are applied to astronomy have been described in mathematical astronomy

Mathematics
Def. an "abstract representational system used in the study of numbers, shapes, structure and change and the relationships between these concepts" is called mathematics.

Theoretical radiation mathematics
Def. an action or process of throwing or sending out a traveling ray in a line, beam, or stream of small cross section is called radiation.

The term radiation is often used to refer to the ray itself.

Def. the "shooting forth of anything from a point or surface, like the diverging rays of light; as, the radiation of heat" is called radiation.

Dimensional analyses
Def. a single aspect of a given thing is called a dimension.

Usually, in astronomy, a number is associated with a dimension or aspect of an entity. For example, the Earth is 1.50 x 108 km on average from the Sun. Kilometer (km) is a dimension and 1.50 x 108 is a number.

Def. the study of the dimensions of quantities; used to obtain information about large complex systems, and as a means of checking equations is called dimensional analysis.

Numbers
Notation: let PeV denote 1015 eV.

Notation: let EeV denote 1018 eV.

Outside the nucleus, free neutrons are unstable and have a mean lifetime of 885.7±0.8 s (about 14 minutes, 46 seconds); therefore the half-life for this process (which differs from the mean lifetime by a factor of ln(2) = 0.693) is 613.9±0.8 s (about 10 minutes, 11 seconds).

202 stars complete the magnitude range 8.90 < V < 16.30.

Astronomical units
Def. "the distance traveled by light in one Julian year in a vacuum" is called the light-year (ly).

Arithmetics
Usually, pure arithmetic only involves numbers. But, when arithmetic is used in a science such as radiation astronomy, dimensional analysis is also applicable.

To build an observatory usually requires adding components together.

 * 1 dome + 1 telescope + 1 outbuilding + 1 control room + 1 laboratory + 1 observation room may = 1 observatory.

Yet,

 * 1 + 1 + 1 + 1 + 1 + 1 = 6 components in 1 simple observatory.

However, attempting to add 1 dome to 1 telescope may have little or no meaning. The operation of addition would be similar to the operation of construction.

If 1 G2V star is added to 1 M2V star the result may be a binary star. The operation of addition here usually requires an explanation (a theory).

Arithmetic can apply to tables of numbers by pairing the numbers together. A concept of relative intensity might cover the range from zero to 100, where the maximum observed intensity in dimensional units is eliminated by dividing each of the observed intensities by the maximum.

 * 55 photons, 22 photons, 11 photons divided by 55 photons yields 1, 0.4, 0.2, or when times 100: 100, 40, and 20. If each of these intensities (photons in this case) occurred over different types of radiation (X-ray, Gamma ray, and visual), then binary pairs can be created:

 * 1) 100 X-rays
 * 2) 40 gamma rays
 * 3) 20 visual rays, for example.

As each band may have an average wavelength, the pairs can become which can be ordered by wavelength and graphed to show a spectrum.
 * 1) 100 5 nm
 * 2) 40 0.5 pm
 * 3) 20 500 nm,

Arithmetic can also be performed at various notational locations.

 * 100 kgn0 coul → 99 kgp+1 coul + 1 kge-1 coul,

 * 100 kg → 99 kg + 1 kg, and
 * 0 coul of net charge → +1 coul + -1 coul of separated charges.

 * 4(X-rays of 1 nm) + 5(X-rays of 1 nm) = (4 + 5)(X-rays of 1 nm) = 9(X-rays of 1 nm),
 * 2x * 1y → (2x,1y), a binary pair, or
 * nΩ + 1 to ∞Ω → undefinedΩ.
 * Ψundefined + Ψundefined = Ψundefined.

But, the exponent can require a different operator of arithmetic.

 * e2 + e3 ≠ e5. Yet
 * e2 x e3 = e(2 + 3) = e5.

Here, the context determines the operation.

Free neutron decay
Free neutrons decay by emission of an electron and an electron antineutrino to become a proton, a process known as beta decay:

For the above relation
 * {| border="1" cellpadding="5" cellspacing="0" align="none"

! colspan="3" |Notational locations Starting with the left symbol, Weight is 1 (not mentioned), Oversymbol is not used, Exponent is replaced by Charge, the Coefficient is 1 (not mentioned), the Variable is a letter designation for the subatomic particle of interest (n for neutron), the Operation is actually a relation decays to (=>), Number is the atomic number Z = 0 for a neutron (not mentioned), the Range is not applicable, and no Index is being used. The neutron's decay products are a proton (p), electron (e), and a neutrino (ν), where Index is used to indicate that the neutrino is an electron neutrino and Oversymbol indicates it is actually an antineutrino. The Operation (+) is not mathematical addition, but indicates another decay product.
 * Weight
 * Oversymbol
 * Exponent
 * Coefficient
 * Variable
 * Operation
 * Number
 * Range
 * Index
 * }
 * Range
 * Index
 * }
 * }

Neutron ejection
Gamma radiation with an energy exceeding the neutron binding energy of a nucleus can eject a neutron. Two examples and their decay products:

 * 9Be + >1.7 Mev photon → 1 neutron + 2 4He

 * 2H (deuterium) + >2.26 MeV photon → 1 neutron + 1H

Algebras
Fundamentally, algebra uses letters to represent as yet unspecified numbers. The numbers may be integers, rational numbers, irrational numbers, or any real number or complex number. As an experimentalist, eventually you must find a way to change unspecified numbers into specified ones. But, as a theoretician, first you are free to leave the numbers in some algebraic form, then to have your theory tested by any experimentalist you need to relate the algebraic terms of your theory to real or complex numbers.

For radiation, there are spatial, temporal, energy, or wavelength choices, among others. Meteors penetrating some portion of a planet's atmosphere can be said to have energy, a spatial distribution, a temporal one, or be in a relationship to something else.

Binary pairs of the radiation, perhaps expressed as meteors/km2 at progressive time intervals: (meteors/km2,years). The km2 is a surface area through which some number of meteors are observed. One may state: each year over Madrid 200 meteors are observed, that would be an arithmetic fact of observation. Theoretically, this could be M km2 per year, where M indicates the air space over Madrid divided into squares so many kilometers on a side.

Your theory would relate M to your explanation for the meteors. Years may be expressed as Julian years, years AD, b2k, or another dimension.

Degrees of freedom
For other distributions, two-thirds of the average energy is often referred to as the temperature, since for a Maxwell-Boltzmann distribution with three degrees of freedom, $$\langle E \rangle = (3/2) \langle k_BT \rangle$$.

Metallicities
Blue main sequence stars that are metal poor ([Fe/H] ≤ -1.0) are most likely not analogous to blue stragglers.

The metallicity of the Sun is approximately 1.8 percent by mass. For other stars, the metallicity is often expressed as "[Fe/H]", which represents the logarithm of the ratio of a star's iron abundance compared to that of the Sun (iron is not the most abundant heavy element, but it is among the easiest to measure with spectral data in the visible spectrum). The formula for the logarithm is expressed thus:

$$ [\mathrm{Fe}/\mathrm{H}] = \log_{10}{\left(\frac{N_{\mathrm{Fe}}}{N_{\mathrm{H}}}\right)_{star}} - \log_{10}{\left(\frac{N_{\mathrm{Fe}}}{N_{\mathrm{H}}}\right)_{sun}} $$

where $$N_{\mathrm{Fe}}$$ and $$N_{\mathrm{H}}$$ are the number of iron and hydrogen atoms per unit of volume respectively. The unit often used for metallicity is the "dex" which is a (now-deprecated) contraction of decimal exponent. By this formulation, stars with a higher metallicity than the Sun have a positive logarithmic value, while those with a lower metallicity than the Sun have a negative value. The logarithm is based on powers of ten; stars with a value of +1 have ten times the metallicity of the Sun (101). Conversely, those with a value of -1 have one tenth (10 −1), while those with -2 have a hundredth (10−2), and so on. Young Population I stars have significantly higher iron-to-hydrogen ratios than older Population II stars. Primordial Population III stars are estimated to have a metallicity of less than −6.0, that is, less than a millionth of the abundance of iron which is found in the Sun.

Hydrogens
The energy differences between levels in the Bohr model, and hence the wavelengths of emitted/absorbed photons, is given by the Rydberg formula :

 * $$ {1 \over \lambda} = R \left( {1 \over (n^\prime)^2} - {1 \over n^2} \right) \qquad \left( R = 1.097373 \times 10^7 \ \mathrm{m}^{-1} \right)$$

where n is the initial energy level, n′ is the final energy level, and R is the Rydberg constant. Meaningful values are returned only when n is greater than n′ and the limit of one over infinity is taken to be zero.

Compositions
"Olivines are described by Mg2yFe2-2ySiO4, with y in[0, 1]." Substituting values for y from 0 to 1 produce ideal compositions from forsterite Mg2SiO4 to fayalite Fe2SiO4. "Amorphous olivine with y = 0.5 and crystalline olivine with y = 0.95 were taken into account for the olivine component." as best fits to observed data. "In the green, the polarization of the pure silicate composition qualitatively appears a better fit to the shape of the observed polarization curves". The silicates used to model the cometary coma dust are olivene (Mg-rich is green) and the pyroxene, enstatite.

Exponentials
When a gamma ray passes through matter, the probability for absorption is proportional to the thickness of the layer, the density of the material, and the absorption cross section of the material. The total absorption shows an exponential decrease of intensity with distance from the incident surface:

 * $$I(x) = I_0 \cdot e ^{-\mu x}$$

where μ = nσ is the absorption coefficient, measured in cm−1, n the number of atoms per cm3 in the material, σ the absorption cross section in cm2 and x the thickness of material in cm.

The time evolution of the number of emitted scintillation photons N in a single scintillation event can often be described by the linear superposition of one or two exponential decays. For two decays, we have the form:

N = A\exp\left(-\frac{t}{{\tau}_f}\right) + B\exp\left(-\frac{t}{{\tau}_s}\right) $$

where τf and τs are the fast (or prompt) and the slow (or delayed) decay constants. Many scintillators are characterized by 2 time components: one fast (or prompt), the other slow (or delayed). While the fast component usually dominates, the relative amplitude A and B of the two components depend on the scintillating material. Both of these components can also be a function the energy loss dE/dx.

In cases where this energy loss dependence is strong, the overall decay time constant varies with the type of incident particle. Such scintillators enable pulse shape discrimination, i.e., particle identification based on the decay characteristics of the PMT electric pulse. For instance, when BaF2 is used, gamma rays typically excite the fast component, while alpha particles excite the slow component: it is thus possible to identify them based on the decay time of the PMT signal.

The monochromatic flux density radiated by a greybody at frequency $$\nu$$ through solid angle $$\Omega$$ is given by $$F_{\nu} = B_{\nu}(T) Q_{\nu} \Omega $$ where $$B_{\nu}(T)$$ is the Planck function for a blackbody at temperature T and emissivity $$Q_{\nu}$$.

For a uniform medium of optical depth $$\tau_{\nu}$$ radiative transfer means that the radiation will be reduced by a factor $$e^{-\tau_{\nu}}$$. The optical depth is often approximated by the ratio of the emitting frequency to the frequency where $$\tau=1$$ all raised to an exponent β.

For cold dust clouds in the interstellar medium β is approximately two. Therefore Q becomes,

$$Q_{\nu}=1-e^{-\tau_{\nu}}=1-e^{-\tau_0 (\nu / \nu_{0})^{\beta}}$$. ($$\tau_0=1$$, $$\nu_0$$ is the frequency where $$\tau_0=1$$).

Lorentz factors
The Lorentz factor is defined as:

 * $$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - \beta^2}} = \frac{\mathrm{d}t}{\mathrm{d}\tau} $$

where:

 * v is the relative velocity between inertial reference frames,
 * β is the ratio of v to the speed of light c.
 * τ is the proper time for an observer (measuring time intervals in the observer's own frame),
 * c is the speed of light.

Dose equivalents
Def. the dose received in one hour at a distance of 1 cm from a point source of 1 mg of radium in a 0.5 mm thick platinum enclosure is called a sievert.

Equivalent dose to a tissue is found by multiplying the absorbed dose, in gray, by a weighting factor (WR). The relation between absorbed dose D and equivalent dose H is thus:

 * $$H = W_R \cdot D$$.

The weighting factor (sometimes referred to as a quality factor) is determined by the radiation type and energy range.

 * $$H_T = \sum_R W_R \cdot D_{T,R}\ ,$$

where
 * HT is the equivalent dose absorbed by tissue T
 * DT,R is the absorbed dose in tissue T by radiation type R
 * WR is the weighting factor defined by the following table

Thus for example, an absorbed dose of 1 Gy by alpha particles will lead to an equivalent dose of 20 Sv. The maximum weight of 30 is obtained for neutrons with L = 100 keV/μm.

Effective doses
The effective dose of radiation (E), absorbed by a person is obtained by averaging over all irradiated tissues with weighting factors adding up to 1:

 * $$E = \sum_T W_T \cdot H_T = \sum_T W_T \sum_R W_R \cdot D_{T,R}$$.

Electronic computers
A computer is a general purpose device that can be programmed to carry out a finite set of arithmetic or logical operations. Since a sequence of operations can be readily changed, the computer can solve more than one kind of problem.

Probabilities
Probability is a measure of the expectation that an event will occur or a statement is true. Probabilities are given a value between 0 (will not occur) and 1 (will occur). The higher the probability of an event, the more certain we are that the event will occur.

Def. a family of continuous probability distributions such that the probability density function is the Gaussian function
 * $$\varphi_{\mu,\sigma^2}(x) = \frac{1}{\sigma\sqrt{2\pi}} \,e^{ -\frac{(x- \mu)^2}{2\sigma^2}} = \frac{1}{\sigma} \varphi\left(\frac{x - \mu}{\sigma}\right),\quad x\in\mathbb{R}$$ is called a normal distribution.

Programmings
A computer program (also software, or just a program) is a sequence of instructions written to perform a specified task with a computer. A computer requires programs to function, typically executing the program's instructions in a central processor.

Computer programming (often shortened to programming or coding) is the process of designing, writing, testing, debugging, and maintaining the source code of computer programs.

Statistics
Statistics is the study of the collection, organization, analysis, interpretation, and presentation of data. It deals with all aspects of this, including the planning of data collection in terms of the design of surveys and experiments.

Geometries
The universe as often perceived may be described spatially, sometimes with plane geometry, other occasions with spherical geometry.

Trigonometries
Trigonometry studies triangles and the relationships between their sides and the angles between these sides. Trigonometry defines the trigonometric functions, which describe those relationships and have applicability to cyclical phenomena, such as waves.

Calculuses
Calculus [focuses] on limits, functions, derivatives, integrals, and infinite series.

Planetary sciences
"The nuclear processes that produce cosmogenic 36Cl in rocks are spallation, neutron capture, and muon capture. The first two processes dominate production on the land surface; muon production in Ca and K becomes more important with increasing depth (Rama and Honda, 1961)."

"The decay of radioactive U and Th also give rise to the production of 36Cl, via neutron capture (Bentley et al., 1986)."

"The production rate of cosmogenic 36Cl in bedrock and regolith exposed at Earth's surface is dependent on its calcium, potassium, and chloride content and can be expressed by the equation

 * $$P = \psi_{Ca}(C_{Ca}) + \psi_K(C_K) + \psi_n(\sigma_{35}N_{35}/\Sigma \sigma_iN_i),$$

where $$\psi_K$$ and $$\psi_{Ca}$$ are the total production rates (including production due to slow negative muons) of 36Cl due to potassium and calcium, respectively; $$C_K$$ and $$C_{Ca}$$ are the elemental concentrations of potassium and calcium, respectively; and $$\psi_n$$ is the thermal neutron capture rate, which is dependent on the fraction of neutrons stopped by 35Cl $$(\sigma_{35} N_{35}/\Sigma \sigma_i N_i),$$ as determined by the effective cross sections of 35Cl$$(\sigma_{35})$$ and all other absorbing elements $$(\Sigma \sigma)$$ and their respective abundances $$(N_{35}$$ and $$N_i)$$."

Theoretical radiation astromathematics
The overall theory of astronomy consists of three fundamental parts:
 * 1) the derivation of logical laws,
 * 2) the definitions of natural bodies (entities, sources, or objects), and
 * 3) the definition of the sky (and associated realms).

Here's a theoretical definition:

Def. the mathematics that describes the logical laws regarding radiation and astronomical entities, sources, and objects is called theoretical radiation astronomy mathematics, or theoretical radiation astromathematics.

Entities
"There is no correlation between n (the exponent for the change in particle diameter) and any of the other entities, which shows that n-values cannot be used as an index of maturation of a soil."

 * $$ C_D = S (D/D_0)^{-n},$$

where $$C_D$$ is change in particle diameter, $$S$$ is the constant of surface area rate of change, $$D$$ is the measured particle diameter in the soil, $$D_0$$ is the initial particle diameter for all samples, and $$n$$ is the exponent.

Peak wavelength for a Planckian radiator
Planck's equation describes the amount of spectral radiance at a certain wavelength radiated by a black body in thermal equilibrium.

In terms of wavelength (λ), Planck's equation is written as
 * $$B_\lambda(T) =\frac{2 hc^2}{\lambda^5}\frac{1}{ e^{\frac{hc}{\lambda k_\mathrm{B}T}} - 1}$$

where B is the spectral radiance, T is the absolute temperature of the black body, kB is the Boltzmann constant, h is the Planck constant, and c is the speed of light.

This form of the equation contains several constants that are usually not subject to variation with wavelength. These are h, c, and kB. They may be represented by simple coefficients: c1 = 2h c2 and c2 = h c/kB.

By setting the first partial derivative of Planck's equation in wavelength form equal to zero, iterative calculations may be used to find pairs of (λ,T) that to some significant digits represent the peak wavelength for a given temperature and vice versa.

 * $$\frac{\partial B}{\partial \lambda} = \frac{c1}{\lambda^6}\frac{1}{ e^{\frac{c2}{\lambda T}} - 1}[\frac{c2}{\lambda T}\frac{1}{ e^{\frac{c2}{\lambda T}} - 1}e^{\frac{c2}{\lambda T}} - 5] = 0.$$

Or,

 * $$\frac{c2}{\lambda T}\frac{1}{ e^{\frac{c2}{\lambda T}} - 1}e^{\frac{c2}{\lambda T}} - 5 = 0.$$

 * $$\frac{c2}{\lambda T}\frac{1}{ e^{\frac{c2}{\lambda T}} - 1}e^{\frac{c2}{\lambda T}} = 5.$$

Use c2 = 1.438833 cm K.

Electromagnetics
"[A] variety of geophysical and astrophysical phenomena can be explained by a net charge on the Sun of -1.5 x 1028 e.s.u." This figure was later reduced by a factor of five.

Electrostatics
Coulomb's law states that the electrostatic force $$F_q$$ experienced by a charge, $$0.2Q_{\odot}$$ at position $$r_q$$, in the vicinity of another charge, $$Q_{\odot}$$ at position $$r_Q$$, in vacuum is equal to:

 * $$F_q = {0.2Q^2_{\odot}\over4\pi\varepsilon_0}{1\over {r^2}},$$

where $$\varepsilon_0$$ is the electric constant or the permittivity of free space and $$r$$ is the distance between the two charges and the constant $$\varepsilon_0$$ is in SI units of C2 m−2 N−1, where C is Coulombs.

ε0 ≈ 8.854 x 10-12 C2 m−2 N−1, or

ε0 ≈ 8.854 x 10-12 C2 10-6 km−2 N−1, then

ε0 ≈ 8.854 x 10-18 C2 km−2 N−1.

1 e.s.u. ≈ 3.34 x 10-10 C.

 * $$F_q = \frac{0.2}{4\pi} \frac{[(-3 \times 10^{27} e.s.u.) (3.34 \times 10^{-10} C/e.s.u.)]^2}{(8.854 \times 10^{-18} C^2 km^{-2} N^{-1}) [(1.5 \times 10^8) 10^3 km]^2}, $$

 * $$F_q = \frac{0.2 (-3)^2 (3.34)^2}{4\pi (8.854) (1.5)^2} \frac{(10^{27} \times 10^{-10})^2}{10^{-18} (10^8 \times 10^3)^2} \frac{[(e.s.u.) (C/e.s.u.)]^2}{C^2 km^{-2} N^{-1} km^2}, $$

 * $$F_q = 8.02 \times 10^{28} N.$$

Each of the constants used are approximates so Fq ≈ 8.02 x 1028 N.

Magnetostatics
If the velocity noted on approach is the vector component pointed toward the Sun, then the angle between the velocity vector and the magnetic field of the Sun is about 180°.

 * $$1G(Gauss) = \frac{1T(Tesla)}{10,000} = \frac{1Ns}{10^4 C 10^{-3} km}.$$

 * $$F_{qB} = 0.2 (-3 \times 10^{27} e.s.u.) (3.34 \times 10^{-10} C/e.s.u.) (20 km s^{-1}) \frac{{-1} N s}{10^4 C km 10^{-3}}, $$

 * $$F_{qB} = 0.2 (-3) (3.34) (20) (-1) \frac{10^{27} \times 10^{-10}}{10^4 \times 10^{-3}} \frac{(e.s.u.) (C/e.s.u.) (km s^{-1}) N s}{C km}, $$

 * $$F_{qB} = 4.01 \times 10^{17} N. $$

Approximately, FqB ≈ 4.01 x 1017 N.

The total Lorentz force is predominantly the electrostatic force.

Emissions
"The prismoidal method provides a good approximation of the dust emission peak for cold sources [...], but probably overestimates the long-wavelength flux for warmer, flatter [spectral energy distribution] SED sources [...], resulting in a bias toward lower Tbol."

"In a binary formed via gravitational fragmentation, we would expect the separation to correspond to the local Jeans length (Jeans 1928):"

 * $$\lambda_J = ( { \pi c^2_s \over G\mu_p m_H n } )^{1/2},$$

"where cs is the local sound speed, and μp = 2.33 and n are the mean molecular weight and mean particle density, respectively. A Jeans length of 4200 AU would require a relatively high density (n ~ 6 x 105 cm−3, assuming cs = 0.2 km s−1). The mean density of the Per-Bolo 102 core, measured within an aperture of 104 AU, is 4 x 105 cm−3, close to the required value."

Alpha decay is characterized by the emission of an alpha particle, a 4He nucleus. The mode of this decay causes the parent nucleus to decrease by two protons and two neutrons. This type of decay follows the relation:

$${}_Z^A\!X\to {}_{Z-2}^{A-4}\!Y+ {}_4^2\alpha$$

Neutron activation is the process in which neutron radiation induces radioactivity in materials, and occurs when atomic nuclei capture free neutrons, becoming heavier and entering excited states. The excited nucleus often decays immediately by emitting particles such as neutrons, protons, or alpha particles. The neutron capture, even after any intermediate decay, often results in the formation of an unstable activation product. Such radioactive nuclei can exhibit half-lives ranging from small fractions of a second to many years.

Proton emission (also known as proton radioactivity) is a type of radioactive decay in which a proton is ejected from a nucleus. Proton emission can occur from high-lying excited states in a nucleus following a beta decay, in which case the process is known as beta-delayed proton emission, or can occur from the ground state (or a low-lying isomer) of very proton-rich nuclei, in which case the process is very similar to alpha decay.

Beta decay is characterized by the emission of a neutrino and a negatron which is equivalent to an electron. This process occurs when a nucleus has an excess of neutrons with respect to protons, as compared to the stable isobar. This type of transition converts a neutron into a proton; similarly, a positron is released when a proton is converted into a neutron. These decays follows the relation:

$${}_Z^A\!X\to {}_{Z+1}^A\!Y+ \bar{\nu} + \beta^-$$ $${}_Z^A\!X\to {}_{Z-1}^A\!Y+ \nu + \beta^+$$

Gamma ray emission is follows the previously discussed modes of decay when the decay leaves a daughter nucleus in an excited state. This nucleus is capable of further de-excitation to a lower energy state by the release of a photon. This decay follows the relation:

$${}^A\!X^* \to {}^A\!Y + \gamma $$

Generation of electromagnetic radiation can occur whenever charged particles pass within certain distances of each other without being in fixed orbits, the accelerations (or decelerations) may give off the radiation. This is partly illustrated by the diagram at right where an electron has its course altered by near passage by a positive particle. Bremsstrahlung radiation also occurs when two electrons or other similarly charged particles pass close enough to deflect, slow down, or speed up at least one of the particles.

Bremsstrahlung includes synchrotron and cyclotron radiation.

When high-energy radiation bombards materials, the excited atoms within emit characteristic "secondary" (or fluorescent) radiation.

Meteors
A particle on the exact design trajectory (or design orbit) of the accelerator only experiences dipole field components, while particles with transverse position deviation $$\scriptstyle x(s)$$ are re-focused to the design orbit. For preliminary calculations, neglecting all fields components higher than quadrupolar, an inhomogenic Hill differential equation

 * $$ \frac{d^2}{ds^2}\,x(s) + k(s)\,x(s) = \frac{1}{R} \, \frac{\Delta p}{p} $$

can be used as an approximation, with
 * a non-constant focusing force $$\scriptstyle k(s)$$, including strong focusing and weak focusing effects
 * the relative deviation from the design beam impulse $$\scriptstyle \Delta p / p$$
 * the trajectory curvature radius $$\scriptstyle R$$, and
 * the design path length $$\scriptstyle s$$,

thus identifying the system as a parametric oscillator. Beam parameters for the accelerator can then be calculated using Ray transfer matrix analysis; e.g., a quadrupolar field is analogous to a lens in geometrical optics, having similar properties regarding beam focusing (but obeying Earnshaw's theorem).

Cosmic rays
At right is an image indicating the range of cosmic-ray energies. The flux for the lowest energies (yellow zone) is mainly attributed to solar cosmic rays, intermediate energies (blue) to galactic cosmic rays, and highest energies (purple) to extragalactic cosmic rays.

There is "a correlation between the arrival directions of cosmic rays with energy above 6 x 1019 electron volts and the positions of active galactic nuclei (AGN) lying within ~75 megaparsecs."

The Oh-My-God particle was observed on the evening of 15 October 1991 over Dugway Proving Ground, Utah. Its observation was a shock to astrophysicists, who estimated its energy to be approximately $3 eV$ (50 joules)—in other words, a subatomic particle with kinetic energy equal to that of a baseball (142 g or 5 oz) traveling at 100 km/h (60 mph).

It was most probably a proton with a speed very close to the speed of light, so close, in fact, [(1 − $5$) × c], that in a year-long race between light and the cosmic ray, the ray would fall behind only 46 nanometers ($5$ light-years), or 0.15 femtoseconds ($1.5 s$).

Protons
For proton astronomy, “at the high end of the proton energy spectrum (above ≈ 1018 eV) [the Larmor radius] deflection becomes small enough that proton astronomy becomes possible.”

The Larmor radius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

“[F]or a particle of energy E in EeV and charge Z in a magnetic field B in µG [the Larmor radius (RL)] is roughly”

 * $$R_L = 1 kpc \frac{E}{Z B}$$

where
 * $$R_L \ $$ is the Larmor radius,
 * $$E\ $$ is the energy of the particle in EeV
 * $$Z \ $$ is the charge of the particle, and
 * $$B \ $$ is the constant magnetic field.

Electrons
"Various gaps and density minima have been observed in the Saturnian ring system. Attempts have been made to attribute these observations to gravitational resonances with the inner satellites, thus causing the removal of particles from the dark regions of the ring system (Alexander 1953, 1962)."

"According to an alternative theory (Alfvén and Arrhenius, 1976) the observed dark regions are the result of a 'cosmogonic shadow effect' produced by the 'two-thirds fall-down mechanism'. The basis of this mechanism is the following: plasma contained in the magnetic dipole field of a central (celestial) body is brought up to a state of partial corotation. In the corotating frame of reference, the plasma experiences an outward centrifugal force which drives a current of density J.

The "charge on a dust particle changes with latitude, i.e., [going] from the equator up to the '2/3' points. The electrostatic charging of dust particles in a plasma [where] charging processes such as photo-emission, field emission thermo-ionic emission, and secondary emission can be neglected [...] contributions to the total charging current come mainly from thermal fluxes of electrons and ions, i.e., $$I = I_i + I_e$$. A dust grain acquires its equilibrium charge -qe (q > 0) when"

 * $$I = I_i + I_e = 0.$$

"The electron and ion currents are given by (Mendis et al., 1984)"

 * $$I_e = -4\pi r^2_g n_e e {({kT \over 2 \pi m_e})^{1/2}} e^y,$$

 * $$I_i = -4\pi r^2_g n_p e {({kT \over 2 \pi m_i})^{1/2}} [1 - y],$$

"where y = -q(e2/kTC), C being the grain capacitance[, ] rg is the grain radius, $$n_e$$ the average electron density, and $$n_p$$ is the average ion (plasma) density."

Gamma rays
"Over the last few years, the cold dark matter cosmogony has become a fiducial model for the formation of structure. [...] The problem with detecting dark matter using annihilation radiation gamma rays has been that the expected signal is comparable to the background (Stecker 1988) and it would be difficult to separate a "cosmic-ray halo" from a dark halo."

"The flux of annihilation gamma rays is given by"

 * $$F = (\sigma V) X D^{-2} m^{-2}_{\lambda} \int_0^\infty \rho^2 r^2\, dr,$$

"where $$(\sigma V)$$ is the cross section, $$D$$ is the distance to the [dwarf spheroidals] dSph's, $$m_\lambda$$ is the [weakly interacting massive particles] WIMP's mass, $$X$$ is the average number of gamma rays in excess of 100 MeV per annihilation and depends weakly on $$m_\lambda$$ (Stecker 1988)."

In March 2010 it was announced that active galactic nuclei are not responsible for most gamma-ray background radiation. Though active galactic nuclei do produce some of the gamma-ray radiation detected here on Earth, less than 30% originates from these sources. The search now is to locate the sources for the remaining 70% or so of all gamma-rays detected. Possibilities include star forming galaxies, galactic mergers, and yet-to-be explained dark matter interactions.

Visuals
The visible spectrum is the portion of the electromagnetic spectrum that is visible to (can be detected by) the human eye. Electromagnetic radiation in this range of wavelengths is called visible light or simply light. A typical human eye will respond to wavelengths from about 390 to 750 nm. In terms of frequency, this corresponds to a band in the vicinity of 400–790 THz. A light-adapted eye generally has its maximum sensitivity at around 555 nm (540 THz), in the green region of the optical spectrum (see: luminosity function).

Infrareds
The Sakuma–Hattori equation is a mathematical model for predicting the amount of thermal radiation, radiometric flux or radiometric power emitted from a perfect blackbody or received by a thermal radiation detector.

In its general form it looks like:

 * $$S(T) = \frac{C}{\exp\left(\frac{c_2}{\lambda _x T}\right)-1}$$

where:

The Stokes parameters are a set of values that describe the polarization state of electromagnetic radiation.

The relationship of the Stokes parameters to intensity and polarization ellipse parameters is shown in the equations below and the figure at right.

 * $$ \begin{align}

S_0 &= I \\ S_1 &= p I \cos 2\psi \cos 2\chi\\ S_2 &= p I \sin 2\psi \cos 2\chi\\ S_3 &= p I \sin 2\chi. \end{align} $$

Here $$p I$$, $$2\psi$$ and $$2\chi$$ are the spherical coordinates of the three-dimensional vector of cartesian coordinates $$(S_1, S_2, S_3)$$. $$I$$ is the total intensity of the beam, and $$p$$ is the degree of polarization. The factor of two before $$\psi$$ represents the fact that any polarization ellipse is indistinguishable from one rotated by 180°, while the factor of two before $$\chi$$ indicates that an ellipse is indistinguishable from one with the semi-axis lengths swapped accompanied by a 90° rotation. The four Stokes parameters are sometimes denoted I, Q, U and V, respectively.

If given the Stokes parameters one can solve for the spherical coordinates with the following equations:

 * $$ \begin{align}

I &= S_0 \\ p &= \frac{\sqrt{S_1^2 + S_2^2 + S_3^2}}{S_0} \\ 2\psi &= \mathrm{atan} \frac{S_2}{S_1}\\ 2\chi &= \mathrm{atan} \frac{S_3}{\sqrt{S_1^2+S_2^2}}.\\ \end{align}$$

Superluminals
A relativistic jet coming out of the center of an active galactic nucleus is moving along AB with a velocity v. We are observing the jet from the point O. At time $$t_1$$ a light ray leaves the jet from point A and another ray leaves at time $$t_2$$ from point B. Observer at O receives the rays at time $$t_1^\prime$$ and $$t_2^\prime$$ respectively.

 * $$AB \ = \ v\delta t$$
 * $$AC \ = \ v\delta t \cos\theta$$
 * $$BC \ = \ v\delta t \sin \theta$$
 * $$t_2-t_1 \ = \ \delta t$$

 * $$t_1^\prime = t_1 + \frac{D_L+v\delta t \cos\theta}{c}$$
 * $$t_2^\prime =t_2+ \frac{D_L}{c}$$

 * $$\delta t^\prime = t_2^\prime - t_1^\prime = t_2 - t_1 - \frac{v\delta t \cos\theta}{c} = \delta t - \frac{v\delta t \cos\theta}{c} = \delta t (1-\beta \cos\theta)$$, where $$\beta=\frac{v}{c}$$

 * $$\delta t = \frac{\delta t^\prime}{1-\beta\cos\theta}$$

 * $$BC \ = \ D_L\sin\phi \approx \phi D_L = v\delta t \sin \theta \Rightarrow \phi D_L = v\sin\theta\frac{\delta t^\prime}{1-\beta\cos\theta}$$

Apparent transverse velocity along CB, $$v_T = \frac{\phi D_L}{\delta t^\prime}=\frac{v\sin\theta}{1-\beta\cos\theta}$$

 * $$\beta_T = \frac{v_T}{c} = \frac{\beta\sin\theta}{1-\beta\cos\theta}$$

 * $$\frac{\partial\beta_T}{\partial\theta} = \frac{\partial}{\partial\theta} \left[\frac{\beta\sin\theta}{1-\beta\cos\theta}\right] = \frac{\beta\cos\theta}{1-\beta\cos\theta} - \frac{(\beta\sin\theta)^2}{(1-\beta\cos\theta)^2} = 0$$

 * $$\Rightarrow \beta\cos\theta (1-\beta\cos\theta)^2 = (1 - \beta\cos\theta) (\beta\sin\theta)^2$$

 * $$\Rightarrow \beta\cos\theta (1-\beta\cos\theta) = (\beta\sin\theta)^2 \Rightarrow \beta\cos\theta - \beta^2\cos^2\theta = \beta^2sin^2\theta \Rightarrow \cos\theta_{max} = \beta$$

 * $$\Rightarrow \sin\theta_{max} = \sqrt{1-\cos^2\theta_{max}} = \sqrt{1-\beta^2} = \frac{1}{\gamma}$$, where $$\gamma=\frac{1}{\sqrt{1-\beta^2}}$$

 * $$\therefore \beta_T^{max} = \frac{\beta\sin\theta_{max}}{1-\beta\cos\theta_{max}} = \frac{\beta/ \gamma}{1-\beta^2} = \beta\gamma$$

If $$\gamma \gg 1$$ (i.e. when velocity of jet is close to the velocity of light) then $$\beta_T^{max} > 1$$ despite the fact that $$\beta < 1$$. And of course $$\beta_T > 1$$ means apparent transverse velocity along CB, the only velocity on sky that we can measure, is larger than the velocity of light in vacuum, i.e. the motion is apparently superluminal.

Plasma objects
When a beam or packet of ions hits the metal it gains a small net charge while the ions are neutralized. The metal can then be discharged to measure a small current equivalent to the number of impinging ions. Essentially the faraday cup is part of a circuit where ions are the charge carriers in vacuum and the faraday cup is the interface to the solid metal where electrons act as the charge carriers (as in most circuits). By measuring the electrical current (the number of electrons flowing through the circuit per second) in the metal part of the circuit the number of charges being carried by the ions in the vacuum part of the circuit can be determined. For a continuous beam of ions (each with a single charge)

 * $$\frac {N}{t} = \frac {I}{e}$$

where $$N$$ is the number of ions observed in a time $$t$$ (in seconds), $$I$$ is the measured current (in amperes) and $$e$$ is the elementary charge (about 1.60 × 10−19 C). Thus, a measured current of one nanoamp (10−9 A) corresponds to about 6 billion ions striking the faraday cup each second.

Gaseous objects
In astrophysics, particularly the study of accretion disks, the epicyclic frequency is the frequency at which a radially displaced fluid parcel will oscillate. It can be referred to as a "Rayleigh discriminant". When considering an astrophysical disc with differential rotation $$\Omega$$, the epicyclic frequency $$\kappa$$ is given by
 * $$\kappa^{2} \equiv \frac{2 \Omega}{R}\frac{d}{dR}(R^2 \Omega),$$

where R is the radial co-ordinate. This quantity can be used to examine the 'boundaries' of an accretion disc - when $$\kappa^{2}$$ becomes negative then small perturbations to the (assumed circular) orbit of a fluid parcel will become unstable, and the disc will develop an 'edge' at that point. For example, around a Schwarzchild black hole, the Innermost Stable Circular Orbit (ISCO) occurs at 3x the event horizon - at $$6GM/c^{2}$$. For a Keplerian disk, $$\kappa = \Omega$$.

Monatomic hydrogens
As temperature increases in an astronomical object composed of H2, the molecules begin to dissociate.

"At a temperature of 8000 K, hydrogen gas is 99.99 percent monatomic."

 * $$ \rho_H = \rho_{H_0} e^{E_T/{kT}},$$

where $$\rho_{H_0}$$ is an initial concentration [H] at low temperatures as partial particle density, $$E_T$$ is the dissociation energy 4.52 eV, k is Boltzmann's contant (8.6173324(78)×10−5 eV K-1), and T is temperature in K.

Using

 * $$ [H] = 70400 e^{-4.52/(0.00008617T)}$$

 * 1) what is the concentration of H ([H]) at T = 8000 K?
 * 2) what is [H] at T = 800 K?
 * 3) at what temperature is [H] = 1?
 * 4) what is [H] at T = 5778 K?

Lithiums
"Our calculations show that production of [lithium] in low-energy flares [by nucleosynthesis], taking place in the surfaces of T Tauri-like stars, is energetically possible and can give the observed excesses over the interstellar abundance." "[T]here is evidence of lithium production in some stars due to some undefined mechanism. The observations show that the Li abundance on some red giants ... and young stars exceeds the average abundance in the universe by 2 orders of magnitude". It is "suggested that Li produced in the helium envelopes of red giants comes to the surface of the stars as the result of a strong convection." For young stars, "the production of the light elements in nonthermal nuclear reactions seems the most appropriate mechanism that can be responsible for enrichment of stellar photospheres by Li." "At least 0.3 metric tons of excited Li and Be nuclei were produced during the solar flare of 1991 November 15. One can estimate the equilibrium concentration of 7Li nuclei in the solar atmosphere by assuming that they are produced only in solar flares and that a leakage of Li nuclei occurs with the solar wind."

Although 7Be is usually assumed to have been produced by the Big Bang nuclear fusion, excesses (100x) of the isotope on the leading edge of the Long Duration Exposure Facility (LDEF) relative to the trailing edge suggest that "most of the sun's fusion must occur near the surface rather than the core." The particular reaction

3He(α,γ)7Be

3He + 4He → 7Be + γ (429 keV)

and the associated reaction chains

7Be(e-,νe)7Li(p,α)α

and

7Be(p,γ)8B → 2α + e+ + νe

generate 14% and 0.1% of the α-particles, respectively, and 10.7% of the present-epoch luminosity of the Sun. Usually, the 7Be produced is assumed to be deep within the core of the Sun; however, such 7Be would not escape to reach the leading edge of the LDEF.

Materials
Astrognosy deals with the materials of celestial objects and their general exterior and interior constitution.

The theoretical constitution of the Earth is illustrated using the one-dimensional Preliminary Reference Earth Model (PREM) at right. The density in kg-m-3 of radial layers is plotted against radius in km.

Sun
As the Sun consists of a plasma and is not solid, it rotates faster at its equator than at its poles. This behavior is known as differential rotation, and is caused by convection in the Sun and the movement of mass, due to steep temperature gradients from the core outwards. This mass carries a portion of the Sun’s counter-clockwise angular momentum, as viewed from the ecliptic north pole, thus redistributing the angular velocity.

Solar rotation is able to vary with latitude because the Sun is composed of a gaseous plasma. The rate of rotation is observed to be fastest at the equator (latitude φ=0 deg), and to decrease as latitude increases. The differential rotation rate is usually described by the equation:

 * $$\omega=A+B\,\sin^2(\varphi)+C\,\sin^4(\varphi)$$

where ω is the angular velocity in degrees per day, φ is the solar latitude and A, B, and C are constants. The values of A, B, and C differ depending on the techniques used to make the measurement, as well as the time period studied. A current set of accepted average values is:

 * A= 14.713 deg/day (± 0.0491)

 * B= –2.396 deg/day (± 0.188)

 * C= –1.787 deg/day (± 0.253)

Limb brightness
Def. "the radial distance q from the Sun's center such that the following finite Fourier transform is zero:

 * $$F(G; q, a) = \int_{-1/2}^{+1/2} G(q + a \sin \pi s) \cos(2\pi s) ds,$$

where s is a dummy variable, G is the observed solar intensity as a function of the radius, and the parameter a determines the extent of the solar limb used" is called the solar edge.

"When F(G; q, a) = 0, the a dependence of q can be used to choose different points as the edge."

Coronal clouds
R⊙eq ≈ 6.955 x 105 km. The thickness of the photosphere is about 400 km. R⊙p ≈ 6.951 x 105 km.

 * $$V_{\odot p} = \frac{4\pi}{3} [R_{\odot eq}^3 - R_{\odot p}^3] km^3,$$

 * $$V_{\odot p} = \frac{4\pi}{3} [6.955^3 - 6.951^3] \times 10^{15} km^3,$$

 * $$V_{\odot p} = \frac{4\pi}{3} (0.580) \times 10^{15} km^3,$$

 * $$V_{\odot p} = 7.288 \times 10^{15} km^3.$$

Photospheres
The density of the Sun is about 2 x 10-4 kg m-3. Or,

 * $$\rho_{\odot p} = 2 \times 10^{-4} kg \cdot m^{-3},$$

 * $$\rho_{\odot p} = 2 \times 10^{-4} kg \cdot [10^{-3} km]^{-3},$$

 * $$\rho_{\odot p} = 2 \times 10^5 kg \cdot km^{-3}.$$

One mole of H2 (gas) has a mass of 2.016 x 10-3 kg. The molar density of the photosphere may be

 * $$\rho_{\odot p} = \frac{2 \times 10^5 kg}{2.016 \times 10^{-3} kg/mole} km^{-3},$$

 * $$\rho_{\odot p} = \frac{2}{2.016} \frac{10^5}{10^{-3}} \frac{kg}{kg/mole} km^{-3},$$

 * $$\rho_{\odot p} = 0.992 \times 10^8 moles \cdot km^{-3},$$

 * $$\rho_{\odot p} = 10^8 moles \cdot km^{-3}.$$

 * $$V_{\odot p} = 7.288 \times 10^{15} km^3.$$

 * $$H_{2 \odot p} = (10^8 moles \cdot km^{-3}) \cdot (7.288 \times 10^{15} km^3),$$

 * $$H_{2 \odot p} = 7.288 \times 10^{23} moles.$$

Photosphere volumes
For H2 (gas) the molar constant-volume heat capacity at 298 K is 20.18 J/(mol · K). At 2000 K it is about 25 J/(mol · K). Using a linear extrapolation,

 * $$C_{V,m} = (2.83 \times 10^{-3})T J/(mol \cdot K^2) + 19.3 J/(mol \cdot K),$$

for 5777 K, yields

 * $$C_{V,m} = (2.83 \times 10^{-3}) (5777) J/(mol \cdot K) + 19.3 J/(mol \cdot K),$$

 * $$C_{V,m} = 35.6 J/(mol \cdot K).$$

Before calculating the amount of energy or power necessary to heat the coronal clouds around the Sun, let's see if the influx of electrons from outside the heliosphere may be able to heat the surface of the photosphere (p) to 5777 K from 100 K.

 * $$\Delta Q = [35.6 J/(mol \cdot K)] \cdot (5777 - 100) K,$$

 * $$\Delta Q = (35.6) \cdot (5677) \frac{J}{mole \cdot K}{K},$$

 * $$\Delta Q = 2.02 \times 10^5 J/mole.$$

 * $$1 J = 6.24 \times 10^{18} eV.$$

 * $$\Delta Q = (2.02 \times 10^5 J/mole) \cdot (6.24 \times 10^{18} eV/J),$$

 * $$\Delta Q = (2.02) \cdot (6.24) \times 10^5 \times 10^{18} eV/mole,$$

 * $$\Delta Q = 12.6 \times 10^{23} eV/mole,$$

 * $$\Delta Q = 1.26 \times 10^{24} eV/mole.$$

Photosphere heating

 * $$\Delta Q = 1.26 \times 10^{24} eV/mole.$$

 * $$H_{2 \odot p} = 7.288 \times 10^{23} moles.$$

Voyager 1 is 17,932,000,000 km (119.9 AU) from the Sun at RA 17.163h Dec +12.44°, ecliptic latitude of 34.9°.

For this laboratory example, let the electron flux be 2 e- cm-2 s-1 diffusing into our solar system from elsewhere in the galaxy. Each of these electrons has an energy of 10 MeV.

 * $$\Phi_{e^-} = 2 e^- \cdot cm^{-2} \cdot s^{-1} \cdot \frac{(10^{-2} \cdot m \times 10^{-3} \cdot km/m)^{-2}}{cm^{-2}},$$

 * $$\Phi_{e^-} = 2 e^- \cdot cm^{-2} \cdot s^{-1} \cdot \frac{10^{10} \cdot km^{-2}}{cm^{-2}},$$

 * $$\Phi_{e^-} = 2 \times 10^{10} e^- \cdot km^{-2} \cdot s^{-1}.$$

If the electron flux measured by Voyager 1 is close to 2 e- cm-2 s-1 where each electron averages 10 MeV and these electrons are heading for the Sun, then each electron may strike the photosphere from anywhere in a sphere around the Sun.

To heat the photosphere to 5777 K takes

 * $$\Delta Q = (1.26 \times 10^{24} eV/mole) \cdot (7.288 \times 10^{23} moles),$$

 * $$\Delta Q = (1.26) \cdot (7.288) \times (10^{24} \times 10^{23}) \cdot eV,$$

 * $$\Delta Q = 9.18 \times 10^{47} eV.$$

The power (P) that may be deposited on the photospheric surface of the Sun is

 * $$P_{e^-} = 4 \pi R_{Voyager 1}^2 \cdot \Phi_{e^-} \cdot (10 MeV/e^-),$$

 * $$P_{e^-} = 4 \pi (1.7932 \times 10^{10} km)^2 \cdot (2 \times 10^{10} e^- \cdot km^{-2} \cdot s^{-1}) \cdot (10 MeV/e^-),$$

 * $$P_{e^-} = (4 \pi) \cdot (1.7932)^2 \cdot 2 \times (10^{20} \times 10^{10} \times 10^7) \cdot (km^2 \cdot e^- \cdot km^{-2} \cdot s^{-1} \cdot eV/e^-),$$

 * $$P_{e^-} = 80.8 \times 10^{37} \cdot eV \cdot s^{-1},$$

 * $$P_{e^-} = 8.08 \times 10^{38} eV \cdot s^{-1}.$$

The luminosity (in Watts, W) of the Sun is 3.846 x 1026 W. In eV/s this is

 * $$L_{\odot} = 3.846 \times 10^{26} W \cdot (10^7 erg/(s \cdot W)) \cdot 6.24 \times 10^{11} eV/erg,$$

 * $$L_{\odot} = (3.846) \cdot (6.24) \times (10^{26} \times 10^7 \times 10^{11}) \cdot (W \cdot erg/(s \cdot W)) \cdot eV/erg),$$

 * $$L_{\odot} = 24.0 \times 10^{44} \cdot eV \cdot s^{-1},$$

 * $$L_{\odot} = 2.40 \times 10^{45} \cdot eV \cdot s^{-1}.$$

If the energy of the incoming electrons is 700 MeV and the flux is 8.48 x 104 e- cm-2 s-1, then the power from the incoming electrons would be

 * $$P_{e^-} = (8.08 \times 10^{38} eV \cdot s^{-1}) \cdot (70) \cdot (8.48/2 \times 10^4),$$

 * $$P_{e^-} = (8.08) \cdot (70) \cdot (4.24) \times (10^4 \times 10^{38}) eV \cdot s^{-1},$$

 * $$P_{e^-} = 2400 \times 10^{42} eV \cdot s^{-1},$$

 * $$P_{e^-} = 2.40 \times 10^{45} eV \cdot s^{-1}.$$

Physics
Def. a "quantity that denotes the ability to do work and is measured in a unit dimensioned in mass × distance²/time² (ML²/T²) or the equivalent" is called energy.

Def. a "physical quantity that denotes ability to push, pull, twist or accelerate a body which is measured in a unit dimensioned in mass × distance/time² (ML/T²): SI: newton (N); CGS: dyne (dyn)" is called force.

In astronomy we estimate distances and times when and where possible to obtain forces and energy.

The key values to determine in both force and energy are (L/T²) and (L²/T²). Force (F) x distance (L) = energy (E), L/T² x L = L²/T². Force and energy are related to distance and time using proportionality constants.

 * {| style="padding:5px;"

 * colspan=2| Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them:
 * valign="top"|
 * $$F = G \frac{m_1 m_2}{r^2}\ $$,
 * valign="top"|
 * $$F = G \frac{m_1 m_2}{r^2}\ $$,

where:
 * F is the force between the masses,
 * G is the gravitational constant,
 * m1 is the first mass,
 * m2 is the second mass, and
 * r is the distance between the centers of the masses.
 * valign="top"|
 * }

In the International System of Units (SI) units, F is measured in newtons (N), m1 and m2 in kilograms (kg), r in meters (m), and the constant G is approximately equal to $6.674 N m^{2} kg^{−2}$.

Observationally, we may not know the origin of the force.

Coulomb's law states that the electrostatic force $$F$$ experienced by a charge, $$q$$ at position $$r_q$$, in the vicinity of another charge, $$Q$$ at position $$r_Q$$, in vacuum is equal to:

 * $$F = {qQ\over4\pi\varepsilon_0}{1\over {r^2}},$$

where $$\varepsilon_0$$ is the electric constant and $$r$$ is the distance between the two charges.

Coulomb's constant is

 * $$k_e = 1 / (4\pi\varepsilon_0\varepsilon_r),$$

where the constant $$\varepsilon_0$$ is in SI units of C2 m−2 N−1.

For reality, $$\varepsilon_r$$ is the relative (dimensionless) permittivity of the substance in which the charges may exist.

The energy $$E$$ for this system is

 * $$E = F \cdot D,$$

where $$D$$ is the displacement.

Unknown forces
Newton's second law of motion is that $$F = ma$$, where $$F$$ is the force applied, $$m$$ is the mass of the object receiving the force, and $$a$$ is the acceleration observed for the astronomical object. The newton is therefore:

 * $${1~{\rm N} = 1~{\rm kg} \frac {{\rm s}^2}}$$

where:
 * N: newton
 * kg: kilogram
 * m: metre
 * s: second.

In dimensional analysis:
 * $${\mathsf F} = \frac{\mathsf {ML}} {{\mathsf T}^2}$$

where
 * M: mass
 * L: length
 * T: time.

But, for a force of unknown type, mass or charge may be meaningless until proven applicable.

So that

 * $${\mathsf A} = {\mathsf {F/M}} = \frac{\mathsf L} {{\mathsf T}^2},$$

 * $${\mathsf A} = {\mathsf {F/Q}} = \frac{\mathsf L} {{\mathsf T}^2},$$

 * $${\mathsf P} = {\mathsf {E/M}} = \frac{{\mathsf L}^2} {{\mathsf T}^2},$$

and

 * $${\mathsf P} = {\mathsf {E/Q}} = \frac{{\mathsf L}^2} {{\mathsf T}^2},$$

where $${\mathsf P}$$ may be called an energy phantom, or astronomical energy phantom.

Technology
A cyclotron is a compact type of particle accelerator in which charged particles in a static magnetic field are travelling outwards from the center along a spiral path and get accelerated by radio frequency electromagnetic fields. Cyclotrons accelerate charged particle beams using a high frequency alternating voltage which is applied between two "D"-shaped electrodes (also called "dees"). An additional static magnetic field $$B$$ is applied in perpendicular direction to the electrode plane, enabling particles to re-encounter the accelerating voltage many times at the same phase. To achieve this, the voltage frequency must match the particle's cyclotron resonance frequency

 * $$f = \frac{q B}{2\pi m}$$,

with the relativistic mass m and its charge q. This frequency is given by equality of centripetal force and magnetic Lorentz force. The particles, injected near the center of the magnetic field, increase their kinetic energy only when recirculating through the gap between the electrodes; thus they travel outwards along a spiral path.

Cyclotron radiation is electromagnetic radiation emitted by moving charged particles deflected by a magnetic field. The Lorentz force on the particles acts perpendicular to both the magnetic field lines and the particles' motion through them, creating an acceleration of charged particles that causes them to emit radiation (and to spiral around the magnetic field lines). Cyclotron radiation is emitted by all charged particles travelling through magnetic fields, however, not just those in cyclotrons. Cyclotron radiation from plasma in the interstellar medium or around black holes and other astronomical phenomena is an important source of information about distant magnetic fields. The power (energy per unit time) of the emission of each electron can be calculated using:

$${-dE \over dt}={\sigma_t B^2 V^2 \over c \mu_o} $$

where E is energy, t is time, $$ \sigma_t $$ is the Thomson cross section (total, not differential), B is the magnetic field strength, V is the velocity perpendicular to the magnetic field, c is the speed of light and $$ \mu_o $$ is the permeability of free space.

A synchrotron is a particular type of cyclic particle accelerator originating from the cyclotron in which the guiding magnetic field (bending the particles into a closed path) is time-dependent, being synchronized to a particle beam of increasing kinetic energy. The synchrotron is one of the first accelerator concepts that enable the construction of large-scale facilities, since bending, beam focusing and acceleration can be separated into different components. Unlike in a cyclotron, synchrotrons are unable to accelerate particles from zero kinetic energy; one of the obvious reasons for this is that its closed particle path would be cut by a device that emits particles. Thus, schemes were developed to inject pre-accelerated particle beams into a synchrotron. The pre-acceleration can be realized by a chain of other accelerator structures like a linac, a microtron or another synchrotron; all of these in turn need to be fed by a particle source comprising a simple high voltage power supply, typically a Cockcroft-Walton generator.

Hypotheses

 * 1) Radiation mathematics accommodates the limitations of physics to the immensity of the known universe.


text:
User:Egm6322.s12.team1.zheng.zx/TP

= Modern Applications of Continued Fraction = Zexi Zheng

Abstract
In this paper, we describe several modern applications of continued fraction in some different fields. First, we start with the introduction of a motivation from which the continued fraction derived. After that, we recall the basic formulae, Euclid's algorithm and 3-term recurrence. Second, we discuss two basic applications of CF in mathematics. Third, we discuss its application to the Solar System. Then, the contribution of CF to modern calendar. Finally, we briefly describe the using of CF in Cryptography and RSA attack.

Motivation
How the ancient Greeks measured accurately length

Method of successive subtractions, using only unmarked measuring rod and compass

If we are given two segments AB and CD as shown above. Find the common denominator EF.

Firstly, we subtract the shorter segment CD from longer segment AB, as many times as possible, as b0 times, until the remaining lngth is shorter than CD.

So, for example, AB=135 and CD=24, then b0=5, and the remainder is r0=15/24.

Note that the numbers '135' and '24' are used here just to give a concrete idea for lengths AB and CD and how the process works. The method of course can be carried out without knowing the actual 'values' of AB and CD, since the goal is to measure AB relative to CD.

For these lengths AB and CD, the expected result is :

\displaystyle \overline{AB}= \{5,1,1,1,2\} \cdot \overline{CD} =5\frac{15}{24}\cdot \overline{CD} $$
 * }
 * }
 * }

The goal is to find the remainder 15/24 by the method of successive subtractions; to do so, one must successively find the quotients 5,1,1,1,2,i.e., the continued fraction{5,1,1,1,2}.

1. Subtract $$b_0=5$$ times the length CD from the length AB; let $$r_0 < \overline{CD}$$be the remainder.

2. Subtract $$b_1=1$$ times the length r0 from the length CD; let $$r_1 < r_0$$be the remainder.

3. Subtract $$b_2=1$$ times the length r1 from the length r0; let $$r_2 < r_1$$be the remainder.

4. Subtract $$b_3=1$$ times the length r2 from the length r1; let $$r_3 < r_2$$be the remainder.

5. Subtract $$b_4=2$$ times the length r3 from the length r2; the remainder is $$r_4=0$$.

See also Holme 2010, Geometry, p.46

Basic formulae
A finite continued fraction is an expression of the form
 * {| style="width:100%" border="0" align="left"

\displaystyle a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{ \ddots + \cfrac{1}{a_n} }}} $$
 * }
 * }
 * }

where $$a_0$$ is an integer(may be negtive), all other $$a_i$$ are positive integers, and $$n$$ is a non-negtive integer.

An infinite continued fraction can be written as
 * {| style="width:100%" border="0" align="left"

\displaystyle a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{a_4 + \ddots}}}} $$ with the same constraints on the $$a_i$$ as in the finite case.
 * }
 * }
 * }

Euclid's algorithm for continued fraction
The method of successive subtractions can be written in the Euclid's algorithm.

Let
 * {| style="width:100%" border="0" align="left"

\displaystyle \overline{AB}=:x_0>x_1:=\overline{CD} $$
 * }
 * }
 * }

So step 1 above can be written as
 * {| style="width:100%" border="0" align="left"

\displaystyle x_0=b_0x_1+x_2 $$
 * }
 * }
 * }

with the remainder satisfying the condition
 * {| style="width:100%" border="0" align="left"

\displaystyle \overline{CD}=:x_1>x_2:=r_0 $$
 * }
 * }
 * }

step 2 above can be written as
 * {| style="width:100%" border="0" align="left"

\displaystyle x_1=b_1x_2+x_3 $$
 * }
 * }
 * }

such that
 * {| style="width:100%" border="0" align="left"

\displaystyle x_2>x_3 $$
 * }
 * }
 * }

step 3 above can be written as
 * {| style="width:100%" border="0" align="left"

\displaystyle x_2=b_2x_3+x_4 $$
 * }
 * }
 * }

In general, we have
 * {| style="width:100%" border="0" align="left"

\displaystyle x_{k-1}>x_k $$
 * }
 * }
 * }

step k is then
 * {| style="width:100%" border="0" align="left"

\displaystyle x_{k-1}=b_{k-1}x_k+x_{k+1} $$
 * }
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\displaystyle \vdots $$
 * }
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\displaystyle x_{n-1}>x_n $$
 * }
 * }
 * }

step n is then:
 * {| style="width:100%" border="0" align="left"

\displaystyle x_{n-1}=b_{n-1}x_n+x_{n+1} $$
 * }
 * }
 * }

Rational numbers: If the process ends within a finite number of steps and the remainder is zero, then it's a rational number ,i.e.,$$x_{n+1}=0$$.The greatest common divisor of $$x_0$$and $$x_1$$is then $$x_n$$

Irrational numbers: If the process never ends and the remainder never reaches zero, it's an irrational number.

And for
 * {| style="width:100%" border="0" align="left"

\displaystyle x_{k-1}=b_{k-1}x_k+x_{k+1} $$
 * }
 * }
 * }

we may write as
 * {| style="width:100%" border="0" align="left"

\displaystyle \frac{x_{k-1}}{x_k}=b_{k-1}+\frac{1}{\displaystyle \frac{x_k}{x_{k+1}}}, \ for \ k=1,2,... $$
 * }
 * }
 * }

and we have the (simple) continued fraction:
 * {| style="width:100%" border="0" align="left"

\displaystyle \frac{x_0}{x_1}=b_0+\frac{1}{\displaystyle b_1+\frac{1}{\displaystyle b_2+\frac{1}{\displaystyle b_3+\frac{1}{\displaystyle b_4+\frac{1}{\displaystyle \frac{x_5}{x_6}}}}}} $$
 * }
 * }
 * }

3-term recurrence for general continued fraction

 * {| style="width:100%" border="0" align="left"

\displaystyle \frac{x_{k-1}}{x_k}=b_{k-1}+\frac{a_k}{\displaystyle \frac{x_k}{x_{k+1}}} \ for \ k=1,2,... $$
 * }
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\displaystyle \frac{x_0}{x_1}=b_0+\frac{a_1}{\displaystyle b_1+\frac{a_2}{\displaystyle b_2+\frac{a_3}{\displaystyle b_3+\frac{a_4}{\displaystyle b_4+\frac{a_5}{\displaystyle \frac{x_5}{x_6}}}}}} $$
 * }
 * }
 * }

And the corresponding 3-term recurrence relation as
 * {| style="width:100%" border="0" align="left"

\displaystyle x_{k-1}=b_{k-1}x_k+a_kx_{k+1} $$
 * }
 * }
 * }

Notations for continued fractions
The integers $$a_0\, a_1\, ...$$are called the quotients of the continued fraction. Then continued fraction can be abbreviated and written as
 * {| style="width:100%" border="0" align="left"

\displaystyle x=[a_0;a_1,a_2,a_3] $$
 * }
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 * }

or, in the notation of Pringsheim, as
 * {| style="width:100%" border="0" align="left"

\displaystyle x=a_0+\frac{1|}{|a_1}+\frac{1|}{|a_2}+\frac{1|}{|a_3} $$
 * }
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 * }

Here is another notation:
 * {| style="width:100%" border="0" align="left"

\displaystyle x=a_0+\frac{1}{a_1+}\frac{1}{a_2+}\frac{1}{a_3+} $$
 * }
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Sometimes we can use angle brackets, like this:
 * {| style="width:100%" border="0" align="left"

\displaystyle x=\langle a_0;a_1,a_2,a_3 \rangle $$
 * }
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 * }

Also one can define infinite simple continued fractions as limits:
 * {| style="width:100%" border="0" align="left"

\displaystyle [a_0;a_1,a_2,a_3,...]=\lim _{n \to \infty}[a_0;a_1,a_2,...,a_n] $$
 * }
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Solving quadratic equations, including the quadratic formula
We have several methods to solve a quadratic equation and the quadratic formula, such as the most 'normal' way: factoring and then graphing the equation. Or use a less common method like applying continued fractions. But it isn't used much often for it needs a lot of calculations. And it has a big limitation: sometimes only one of the two solutions can be obtained. Let's consider the following quadratic equation:
 * {| style="width:100%" border="0" align="left"

\displaystyle x^2-5x+6=0 $$ $$
 * $$\displaystyle (Equation\;2.1.1 )
 * }
 * }

If factoring the L.H.S, we find the roots are 2 and 3.

Now using continued fractions.

Both sides of equation 2.1.1 are divided by x, then rearranging:
 * {| style="width:100%" border="0" align="left"

\displaystyle x=5-\frac{6}{x} $$ $$
 * $$\displaystyle (Equation\;2.1.2 )
 * }
 * }

Next, express x as an infinite continued fraction.
 * {| style="width:100%" border="0" align="left"

\displaystyle x=5-\frac{6}{5-}+\frac{6}{x} $$ $$
 * $$\displaystyle (Equation\;2.1.3 )
 * }
 * }

Doing the same procedure again:
 * {| style="width:100%" border="0" align="left"

\displaystyle x=5-\frac{6}{5-}+\frac{6}{5-}+\frac{6}{x} $$ $$
 * $$\displaystyle (Equation\;2.1.4 )
 * }
 * }

If we set $$x=1$$ in the R.H.S of equation 2.1.2, we get
 * {| style="width:100%" border="0" align="left"

\displaystyle x=5-\frac{6}{1}=5-6=-1 $$ if we set $$x=1$$ to the R.H.S of equation 2.1.3, we get
 * }
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 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle x=5-\frac{6}{5-}+\frac{6}{1}=5-\frac{6}{\displaystyle 5-\frac{6}{1}}=5-\frac{6}{-1}=11 $$ If we continue the process, we can get an infinite sequence that approaches 3:
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle x=5-\frac{6}{5-}+\frac{6}{5-}\cdots \frac{6}{5-}+\frac{6}{1}=3 $$ That's one solution of equation 2.1.1, but we can't get the second solution i.e.$$x=2$$using this method. The method can also be generalized and applied to any quadratic equation.
 * }
 * }
 * }

We may verify the answer for the example above using the following Matlab program, at the same time, we should note that the initial number can be arbitrary, not necessarily be 1.

clear x=input('Please designate an initial number x: '); n=input('Please designate the times of process: '); s=x; for i=1:n s=5-6/s; end s

Express an irrational number as a continued fraction
We can use continued fractions to express a rational or an irrational number and to approximate the value. this kind of expression will let us easily study certain properties of a number that we are interested in. Let's look at an irrational number $$\,\sqrt 2 \,$$as an example.

We may find it's not easy to express this number as a continued fraction at first glance. So let's do some manipulations first.

We observe that $$\,\sqrt 2>1 \,$$, so we can decompose that into two parts:
 * {| style="width:100%" border="0" align="left"

\displaystyle \sqrt 2= 1+\frac{1}{x} $$ then solving for a real number:
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle x=\frac{1}{\sqrt 2-1} $$ We multiply the numerator and denominator by $$\,\sqrt 2 +1\,$$
 * }
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 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle x=\sqrt 2 +1 $$ Then, substituting back $$\,\sqrt 2\,$$
 * }
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 * {| style="width:100%" border="0" align="left"

\displaystyle x=1+\frac{1}{x}+1=2+\frac{1}{x} $$ Now we can do the substitution procedure and get the continued fraction representation:
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \sqrt 2=1+\frac{1}{\displaystyle 2+\frac{1}{\displaystyle 2+\frac{1}{\displaystyle 2+\frac{1}{\ddots}}}} $$ $$ Or
 * $$\displaystyle (Equation\;2.1.5 )
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \sqrt 2=[1;2,2,2,2,2,\cdots] $$ Other square roots or irrational numbers can also be expressed as a continued fraction similarly.
 * }
 * }
 * }

An application to the Solar System
Let's look at a kind of solar system: a set of cog wheels. A cog wheel has an integer number of teeth around its rim that mesh with the teeth on another cog. The ratio of the two number of teeth on each pair of cog wheels decide the speed ratio between them. For example, we have a pair of cogs, one has 40 teeth and the other has 10 teeth, the cog with 10 teeth will rotate 4 times faster than that with 40.

Solar systems are widely used, from cog wheels in a small watch to the gear box in a car. However, there exists some limitations on such systems. For example, the number of teeth is an integer, and the ratio between two integers is a rational fraction. So if we want the rotation ratio to be an irrational number such as $$\,\sqrt 2\,$$, we can't make that. But with the help of continued fraction, we may do some approximation. Sometimes if the number of teeth can be large, the result will be precise enough for engineering purpose.

For such case, if we want the ratio to be $$\,\sqrt 2 \,$$

First, let's write it as continued fraction: (see equation 2.1.5)
 * {| style="width:100%" border="0" align="left"

\displaystyle \sqrt 2=1+\frac{1}{\displaystyle 2+\frac{1}{\displaystyle 2+\frac{1}{\displaystyle 2+\frac{1}{\ddots}}}} $$ Defining:
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \begin{align} C_1 &:=1+\frac{1}{2}\\ C_2 &:=1+\frac{1}{2+}\frac{1}{2}\\ &\ \ \ \vdots\\ C_i &:=1+\underbrace{\frac{1}{2+}\frac{1}{2+}\cdots \frac{1}{2}}_{i\ terms} \end{align} $$ $$\,C_i\,$$as the $$i$$th convergence of $$\, \sqrt 2 \,$$. We see:
 * }
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 * {| style="width:100%" border="0" align="left"

\displaystyle C_0:=1 $$
 * }
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 * {| style="width:100%" border="0" align="left"

\displaystyle C_1:=1+\frac{1}{2}=\frac{3}{2} $$
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 * {| style="width:100%" border="0" align="left"

\displaystyle C_2:=1+\frac{1}{2+\frac{1}{2}}=1+\frac{2}{5}=\frac{7}{5} $$
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 * {| style="width:100%" border="0" align="left"

\displaystyle C_3:=1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}=1+\frac{1}{2+\frac{2}{5}}=1+\frac{5}{12}=\frac{17}{12} $$
 * }
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 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \begin{align} & \vdots \\ \lim_{i \to \infty}C_i &=\sqrt 2 \end{align} $$ We may use the follwoing Matlab program to verify the results and plot the convergence. clear
 * }
 * }
 * }

n=10;

f=sqrt(2); f=sym(f); f=char(f); fplot(f,[-2 2],'r','o'); legend('sqrt(2)'); hold on

disp(['C' char(48) '=']) disp(1)

f=1; f=sym(f); f=char(f); fplot(f,[-2 2]);

for x=1:n; s=0; for i=1:x; s=1/(2+s); end s=1+s; s=vpa(s); disp(['C' char(sym(x)) '=']) disp(s)

f=s; f=sym(f); f=char(f); fplot(f,[-2 2]);

end axis([-2 2 1.39 1.51])

So, the convergence says the more recurrent terms we apply, the more accurate answer we will get.

We can get a set of convergent fraction numbers after that:
 * {| style="width:100%" border="0" align="left"

\displaystyle \frac{3}{2},\,\frac{7}{5},\,\frac{17}{12},\,\frac{41}{29},\,\frac{99}{70},\cdots $$ For a pair of small cog wheels, we could have 5 teeth on one cog and 7 on the other, or 12 on one and 17 on the other; for a pair of larger cogs, we can apply 12 on one and 17 on another or if possible 99 and 70 teeth will give a closer approximation, then the error to $$\,\sqrt 2\,$$is only 0.007%.
 * }
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Astronomical year and modern calendar
A calendar year is the period between two dates that with the same name in a calendar. For the calendar we use today, the numbers of days are all integers, but for astronomical year there is no such integer number of days or lunar months, for a year is $$\,365.24219878\,$$ days and one lunar month is $$\,29.530589\,$$ days, so any calendar as long as it follows an astronomical year must have its own system of adjustment such as the leap year intercalculation. We have noted that scientific calculations nowadays often use a 365-day calendar.

Many kinds of calendars are still in use today all around the world: Chinese, Ethiopian, Hebrew, Hindu etc. At a first glance,these calendar systems seam so different and are just based on arbitrary choices. However almost all calendars are based on the motion of the astronomical system, i.e. the motion of the Sun and the Moon by observation.

About 3000 years ago the 365-day calendar was first introduced, which called Alexandrian calendar. It has a sixth epagomenal day every fourth year. Our calendar in use today is directly descended from the ancient Roman calendar and following is the story.

Around 46 B.C. Julius Caesar learned about the Alexandrian calendar from Egypt through his campaign. The Alexandrian calendar at that time had a 4-year leap year cycle which He found was much more precise than the one Roman in use with 365 days. Along with him after the campaign was the Alexandrian astronomer Sosigenes, he gave Julius some advice on how to improve the accruracy. Upon the advices he received Julius created the Julian calendar based on his reform. The mean length of a year for Julian calendar is 365.25 days, which is already very precise comparing with the number 365.24219878. And the Julian calendar accumulates only one day error in about a hundred years long.

Idea of modern calendar with continued fraction
To decide a rule of modern calendar, we need to think about the idea: to have a period spanning several years during which some of the years are leap year while others are not. And the mean length of one year should be very close to the astronomical year. The period length should be either very short or very easy to use. As the Julian 4-year leap year cycle and Gregorian 400-year leap year cycle are nice examples.

Suppose we have a cycle of $$\,q \,$$ years and during which there are $$\,p \,$$ leap years.

Then there will be $$\,365q+p \,$$days in all during this period.(the leap year here is the year with 366 days)

The mean length for a year should be
 * {| style="width:100%" border="0" align="left"

\displaystyle \frac{365q+p}{q}=365+\frac{p}{q} $$ days.
 * }
 * }
 * }

For we already know there are $$\, 365.24219878 \,$$days in each year. So let's define
 * {| style="width:100%" border="0" align="left"

\displaystyle \alpha=365.24219878-365=0.24219878 $$ Then we need to find a proper number$$\,q\,$$ to make $$\,\frac{p}{q}\,$$as close as possible to $$\,\alpha\,$$.
 * }
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Write $$\,\alpha\,$$as the form of continued fraction:
 * {| style="width:100%" border="0" align="left"

\displaystyle 0.24219878=\frac{1}{\displaystyle 4+\frac{1}{\displaystyle 7+\frac{1}{\displaystyle 1+\frac{1}{\displaystyle 3+\frac{1}{\displaystyle 5+\ddots}}}}} $$
 * }
 * }
 * }

Then we can write out a series of numbers of convergents.
 * {| style="width:100%" border="0" align="left"

\displaystyle C_1=\frac{p_1}{q_1}=\frac{1}{4} $$
 * }
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 * {| style="width:100%" border="0" align="left"

\displaystyle C_2=\frac{p_2}{q_2}=\frac{7}{29} $$
 * }
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 * {| style="width:100%" border="0" align="left"

\displaystyle C_3=\frac{p_3}{q_3}=\frac{8}{33} $$
 * }
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 * {| style="width:100%" border="0" align="left"

\displaystyle C_4=\frac{p_4}{q_4}=\frac{31}{128} $$
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 * {| style="width:100%" border="0" align="left"

\displaystyle C_5=\frac{p_5}{q_5}=\frac{163}{673} $$
 * }
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 * }

Then we may find$$\,C_1\,$$in the sequence corresponds to the Julian 4-year-cycle system during which cycle there exists one leap year. And the remaining values seem not so convenient to be used like $$\,7\,$$leap years in a $$\,29\,$$ -year cycle as it's hard to distribute the 7 leap years. If we want to have a cycle spanning several centuries long,i.e.
 * {| style="width:100%" border="0" align="left"

\displaystyle q=100n \ \ \ n=1,2,3,\cdots,9 $$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \alpha = \frac{p}{q}=\frac{p}{100n}=0.24219878 $$
 * }
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 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \frac{p}{n}=24.219878 $$ Then the problem turns to be finding p and n.
 * }
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 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle 24.219878=24+\frac{1}{\displaystyle 4+\frac{1}{\displaystyle 1+\frac{1}{\displaystyle 1+\frac{1}{\displaystyle 4+\ddots}}}} $$
 * }
 * }
 * }

The first 4 convergents are:
 * {| style="width:100%" border="0" align="left"

\displaystyle C_1=\frac{p_1}{n_1}=\frac{97}{4} $$
 * }
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 * {| style="width:100%" border="0" align="left"

\displaystyle C_2=\frac{p_2}{n_2}=\frac{121}{5} $$
 * }
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 * {| style="width:100%" border="0" align="left"

\displaystyle C_3=\frac{p_3}{n_3}=\frac{218}{9} $$
 * }
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 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle C_4=\frac{p_4}{n_4}=\frac{993}{41} $$ For we wish $$\ n\ $$ to be within [1,9]
 * }
 * }
 * }

We have three candidates 4,5 and 9.

And we may find $$\ n_1=4\ $$corresponds to $$\ q_1=100n_1=400\ $$, that's our Gregorian calendar.

For Gregorian calendar, there are 97 leap years in 400-year cycle. Then how do we assign these 97 years?

We define the year divisible by 4 except 100th, 200th and 300th be the leap year, there will be 100-3=97, just meet with the demands.

For the other two candidates, if we choose 5 as 500 years cycle.

We need 121 leap years in each cycle, we may still choose 4 year as a period, then
 * {| style="width:100%" border="0" align="left"

\displaystyle \frac{500}{4}=125 $$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle 125-121=4 $$ We will have 4 more years if we just follow the only rule. So let's make some constraints: The year divisible by 100 with the exception of years divisible by 500 will not be accounted as leap years. Then all demands are met. This system is just as simple and as convenient as the Gregorian calendar and can provide a much better accuracy. As the last candidate 9, although it seems more accurate, the span is too long--900 years, and the distribution of 218 leap years is also a big problem.
 * }
 * }
 * }

Lastly, let's consider the situation if our calendar was based on a 500 year cycle. The problem may turn out to be the following expansion.
 * {| style="width:100%" border="0" align="left"

\displaystyle 500\times 0.24219878=121+\frac{1}{\displaystyle 10+\frac{1}{\displaystyle 16+\frac{1}{\displaystyle 3+\frac{1}{\displaystyle 2+\frac{1}{\displaystyle 2+\ddots}}}}} $$ The convergents will be:
 * }
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 * {| style="width:100%" border="0" align="left"

\displaystyle [121,\ \frac{1211}{10},\ \frac{19497}{161},\ \frac{59702}{493},\ \frac{138901}{1147},\ \frac{337504}{2787},\cdots] $$ Let's see the second convergent,1211/10. It says 1211 leap years in a cycle of 5,000 years. The 5,000 year-cycle calendar will cause an accumulation of 1 day error in a 1 million years. So we can designate Feburary of year 5,000 to have 30 days. This system was proposed by Bernard Rasof.
 * }
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 * }

About Public-Key Cryptography and RSA
Public-key cryptography refers to a cryptographic system that requires two different keys to fully operate, one key is in charge of locking and encrypting the plaintext, and the other is in charge of unlocking and decrypting the cyphered text. The two keys will function seperately and neither of them can do both. After that, one of the two keys is published to public and the other should be kept secret. The cryptographic approach will use asymmetric key algorithms such as RSA.

RSA is an algorithm for public key cryptography, which is based on the assumption that it's hard to do the factoring of large integers. RSA stands for three names: Ron Rivest, Adi Shamir and Leonard Adleman, who first described it in 1978. The idea of RSA is that firstly a user creates a number which is the product of two large prime numbers. Then the product along with the auxiliary value will be published to public as their public key. The two prime numbers should be kept secret, anyone can use the public key to encrypt a plaintext, but only with the currently published methods. Breaking RSA encryption is like solving an open question, but it's as hard as factoring a large number.

Wiener's attack on a typical RSA with small secret exponent
The security of RSA is based on the difficulty of finding the prime factors of large integers.

Let's define the modulus of a RSA cryptosystem as $$\,\,n\,\,$$, and it's the product of two large primes $$\,\,p\,\,$$ and $$\,\,q\,\,$$ so
 * {| style="width:100%" border="0" align="left"

\displaystyle n=p\times q $$ $$ The public exponent $$e$$ and the secret exponent $$d$$ are related by
 * $$\displaystyle (Equation\;2.4.1 )
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle e\times d\equiv 1,\ \ \ (mod \,\,\phi(n)) $$ $$ Where
 * $$\displaystyle (Equation\;2.4.2 )
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \phi (n)=(p-1)(q-1)=n-p-q+1 $$ $$ In a typical RSA cryptosystem p and q have approximately the same number of bits, and e$$\displaystyle (Equation\;2.4.3 )
 * }
 * }

In 1990, Wiener developped a polynomial time algorithm for breaking through a typical(p and q are of the same size and e < n) RSA cryptosystem if the secret exponent d has at most 1/4 number of bits as the modulus n. i.e.
 * {| style="width:100%" border="0" align="left"

\displaystyle p\,<\,q\,<\,2p $$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle e\,<\,n $$
 * }
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 * {| style="width:100%" border="0" align="left"

\displaystyle d\,<\,\frac{1}{3}\sqrt[4]{n} $$ then d is the denominator of a convergent of the expansion of e/n in continued fraction.
 * }
 * }
 * }

''' For here I don't understand ! ''' There is an integer k such that
 * {| style="width:100%" border="0" align="left"

\displaystyle ed-k\phi(n)=0 $$ Now
 * }
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 * {| style="width:100%" border="0" align="left"

\displaystyle \phi(n)\approx n $$ implies
 * }
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 * {| style="width:100%" border="0" align="left"

\displaystyle \frac{k}{d}\approx \frac{e}{n} $$ More precisely, we have
 * }
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 * {| style="width:100%" border="0" align="left"

\displaystyle n-3\sqrt{n} < \phi(n) < n $$ and
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \left | \frac{k}{d}-\frac{e}{n}\right | < \frac{3k}{d\sqrt{n}} < \frac{1}{2d^2} $$ By Legendre's theorem, $$\,\frac{k}{d}\,$$ is a convergent of continued fraction expansion of $$\,\frac{e}{n}\,$$.
 * }
 * }
 * }

For the denominators grow exponentially, thus the total number of convergents of $$\,\frac{e}{n}\,$$ is of order O(log n). If we can test the convergent in polynomial time, then there will be a polynomial algorithm applicable to determine d.

Let's look at the following method to test convergents.

Let $$\frac{a}{b}$$ be a convergent of $$\frac{e}{n}$$, if it's the correct guess for $$\frac{k}{d}$$ , then $$\phi(n)$$ can be computed from
 * {| style="width:100%" border="0" align="left"

\displaystyle \phi(n)= (p-1)(q-1)=\frac{be-1}{a} $$ Now
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \frac{pq-(p-1)(q-1)+1}{2}=\frac{p+q}{2} $$ $$ and
 * $$\displaystyle (Equation\;2.4.4 )
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \left ( \frac{p+q}{2}\right ) ^2 -pq=\left ( \frac{q-p}{2}\right ) ^ 2 $$ $$ We can compute $$\frac{p+q}{2}$$ and $$\frac{q-p}{2}$$ from equation 2.4.4 and equation 2.4.5.
 * $$\displaystyle (Equation\;2.4.5 )
 * }
 * }

Factoring n with the help of continued fraction expansion of $$\,\frac{e}{n}\,$$
For example, let
 * {| style="width:100%" border="0" align="left"

\displaystyle n=7978886869909 $$
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle e=3594320245477 $$ and assume
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle d < 561 $$ Then, continued fraction expansion of $$\,\,\frac{e}{n}\,\,$$ is
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \frac{e}{n}=\frac{3594320245477}{7978886869909}=\frac{1}{\displaystyle 2+\frac{1}{\displaystyle 4+\frac{1}{\displaystyle 1+\frac{1}{\displaystyle 1+\frac{1}{\displaystyle 4+\frac{1}{\displaystyle 1+\frac{1}{\displaystyle 2+\frac{1}{\displaystyle 31+\frac{1}{\displaystyle 21+\ddots}}}}}}}}} $$ the convergents are:
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle 0,\frac{1}{2},\frac{4}{9},\frac{5}{11},\frac{9}{20},\frac{41}{91},\frac{50}{111},\frac{141}{313},\frac{4421}{9814},\cdots $$ Applying test $$(2^e)^d\equiv 2\,\,(mod\,\,n)$$, we obtain $$\,\,d=313\,\,$$.
 * }
 * }
 * }

Then from the convergents result, we may find 141/313 with the denominator 313. So a/b=141/313, with
 * {| style="width:100%" border="0" align="left"

\displaystyle \begin{align} a &= 141 \\ b &= 313 \end{align} $$ After that, we can get
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \phi(n)= (p-1)(q-1)=\frac{be-1}{a}=\frac{313\times 3594320245477-1}{141}=7978881112300 $$ Then, using equation 2.4.4:
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \frac{p+q}{2}=\frac{pq-(p-1)(q-1)+1}{2}=\frac{n-\phi(n)+1}{2}=\frac{7978886869909-7978881112300+1}{2}=2878805 $$ And equation 2.4.5:
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \frac{q-p}{2}=\sqrt{\left ( \frac{p+q}{2}\right )^2 -pq}=\sqrt{8287518228025-7978886869909}=555546 $$ This yields
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle \begin{align} p &= 2878805-555546=2323259 \\ q &= 2878805+555546=3434351 \end{align} $$ So, the factorization of n:
 * }
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\displaystyle n=p\times q=2323259\times 3434351=7978886869909 $$
 * }
 * }
 * }

Summary
The paper consists of two parts. The first part introduced the definition and notation of continued fraction started from a motivation. The second part is the core part. We discussed the applications of CF in three different areas. Firstly, we discussed its two basic applications in mathematics. Secondly, we discussed its application to the Solar System. Thirdly, we talked about the relationship between CF and modern calendar. Lastly, we briefly described the using of CF in Cryptography and RSA attack.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Egm6322.s12.team1.zheng.zx/TP


text:
User:Egm4313.s12.team8.tclamb/R6-1

=R6.6 - Verify Particular Solution to ODE=

Problem Statement
Verify the following:
 * $$6 e^{-2x} ( n \cos{3x} - m \sin{3x} )$$

As the left hand side to the following ODE:
 * $$y_p'' + 4 y_p' + 13y_p = 2 e^{-2x} \cos{3x}$$

Where:
 * $$y_p(x) = x e^{-2x} ( m \cos{3x} + n \sin{3x} )$$

Then, solve for m and n.

Solution
First, we find and simplify each term of the left hand side:
 * $$\begin{array}{rl}

y_p &= x e^{-2x} ( m \cos{3x} + n \sin{3x} ) \\ &= e^{-2x} ( m x \cos{3x} + n x \sin{3x} ) \\ 13 y_p &= e^{-2x} ( 13 m x \cos{3x} + 13 n x \sin{3x} ) \\ &\\ y_p' &= e^{-2 x} x (3 n \cos{3 x}-3 m \sin{3 x})+e^{-2 x} (m \cos{3 x}+n \sin{3 x})-2 e^{-2 x} x (m \cos{3 x}+n \sin{3 x}) \\ &= e^{-2 x} ((-2 m x+m+3 n x) \cos{3 x}+ (-3 m x-2 n x+n)\sin{3 x}) \\ 4 y_p' &= e^{-2 x} ((-8 m x+4m+12 n x) \cos{3 x}+ (-12 m x-8 n x+4n)\sin{3 x}) \\ &\\ y_p'' &= e^{-2 x} (3 n \cos{3 x}-3 m \sin{3 x}) \\ &\;\;+\left(e^{-2 x}-2 e^{-2 x} x\right) (3 n \cos  (3 x)-3 m \sin{3 x})\\ &\;\;+e^{-2 x} x (-9 m \cos{3 x}-9 n \sin{3 x})\\ &\;\;-2 e^{-2 x} (m  \cos{3 x}+n \sin{3 x})\\ &\;\;+4 e^{-2 x} x (m \cos{3 x}+n \sin{3 x})\\ &\;\;-2 e^{-2 x} (x (3 n \cos{3 x}-3 m \sin{3 x})+m \cos{3 x}+n \sin{3 x}) \\ &= e^{-2 x} ( (-5 x m - 4 m - 12 x n + 6 n) \cos{3 x} + (12 m x - 6 m - 5 x n - 4 n) \sin{3 x}) \end{array}$$

Then, we combine the three terms and simplify:
 * $$\begin{array}{rl}

y_p'' + 4 y_p' + 13 y_p &= e^{-2 x} ( (-5 x m - 4 m - 12 x n + 6 n) \cos{3 x} + (12 m x - 6 m - 5 x n - 4 n) \sin{3 x}) \\ &\;\;+e^{-2 x} ((-8 m x+4m+12 n x) \cos{3 x}+ (-12 m x-8 n x+4n)\sin{3 x}) + e^{-2x} ( 13 m x \cos{3x} + 13 n x \sin{3x} ) \\ &= 6 e^{-2x} ( n \cos{3x} - m \sin{3x} )\end{array}$$

This verifies the given left hand side. To solve for m and n, we set this equal to the right hand side of the ODE:
 * $$\begin{array}{rcl}

6 e^{-2x} ( n \cos{3x} - m \sin{3x} ) &=& 2 e^{-2x} \cos{3x} \\ 6 ( n \cos{3x} - m \sin{3x} ) &=& 2 \cos{3x} \\ n \cos{3x} - m \sin{3x} &=& \frac13 \cos{3x} \end{array}$$

Therefore:
 * $$\begin{array}{rl}

n &= \frac13 \\ m &= 0 \\ y_p(x) &= \frac13 x e^{-2x} \sin{3x} \end{array}$$

title: User:Egm4313.s12.team8.tclamb/R6-1


text: File:Egm6322.s12.team1.zheng.zx.motivation.JPG


text:
User:Gjaksec/ENES100/CC123DTutorial

Steps
1) Draw the sleeve (the scuba material part which will vacuum seal to your forearm). Select the "cone object," and create it with its center   point at the origin of the left vertical plain.The cone's radius is 6cm, its angle is -30cm, and its length is 12cm.

This is the real version of the Core Control System: http://www.instructables.com/id/CoreControl-DIY/

title: User:Gjaksec/ENES100/CC123DTutorial


title: File:Report6.5n=9.png


text:
User:Collingwood

Collingwood was globally locked as a result of a WikiMedia Foundation office finding that the user was related to Poetlister, who is globally banned. The finding was based on secret evidence, not disclosed. Collingwood was not blocked on Wikiversity, and no disruption on Wikiversity was alleged -- other than possibly block evasion, because an openly used Poetlister account, Poetlister1, was blocked here (based on the global ban discussion, not for disruption here), but Collingwood cannot log into the account due to the global lock.

If anyone wishes to contact Collingwood, the email interface should work. --Abd (discuss • contribs) 02:24, 3 March 2014 (UTC)

/Todo/

title: User:Collingwood


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text:
User:Egm4313.s12.team8.lucas/R6.7

=R6.7=

Problem Statement
Find the seperated ODEs for the heat equation

$$\ \frac{\partial u}{\partial t}=\kappa \frac{\partial ^2 u}{\partial x^2}$$

Solution
Assume $$\ u(x,t)=F(x)\cdot G(t)$$

where:

$$\ F(x)=f(x)$$, a function of only x

$$\ G(t)=f(t)$$, a function of only t

Finding first and second partial derivatives of u with restpect to x gives:

$$\ \frac{\partial u}{\partial x}=F'(x) \cdot G(t)$$

$$\ \frac{\partial^2 u}{\partial x^2}=F''(x) \cdot G(t)$$

Finding first and second partial derivates of u with respect to t gives:

$$\ \frac{\partial u}{\partial t}=F(x) \cdot \dot{G}(t)$$

$$\ \frac{\partial^2 u}{\partial ^2t}=F(x) \cdot \ddot{G}(t)$$

Plugging these into the heat equation gives:

$$\ \frac{\partial u}{\partial t}=\kappa \frac{\partial ^2u}{\partial x^2}  =>  F(x) \cdot \dot{G}(t)=\kappa F''(x) \cdot G(t)$$

Combining like terms gives:

$$\ \frac{\dot G(t)}{\kappa G(t)}=\frac{F''(x)}{F(x)}=c$$

where c is a constant

$$\ \frac{\dot G(t)}{\kappa G(t)}=c   => \dot G(t)=c \kappa G(t)  =>  \dot G(t)-c \kappa G(t)=0$$

$$\ \frac{F(x)}{F(x)}=c    => F(x)=cF(x)     =>    F''(x)-cF(x)=0$$

$$\ \dot G(t)-c \kappa G(t)=0$$

$$\ F''(x)-cF(x)=0$$

title: User:Egm4313.s12.team8.lucas/R6.7


text: File:F x.png

title: File:F x.png


text: Portal:Modern Hebrew/Department News

title: Portal:Modern Hebrew/Department News


text:
Template:Modern Hebrew/Department News

April 10, 2012 April 9, 2012 Go To Archives
 * Created the page Where to start?.
 * Creation of the Modern Hebrew Department.

title: Template:Modern Hebrew/Department News


text: Portal:Hebrew/Related News

title: Portal:Hebrew/Related News


text:
Portal:Modern Hebrew/Beginners

{| cellpadding="15" cellspacing="5" style="width: 100%; background-color: inherit; margin-left: auto; margin-right: auto"
 * style="width: 15%; background-color: Honeydew; border: 1px solid Gray; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; height: 60px;" colspan="1" |[[Image:Flag of Israel.svg|center|200px]]

 * style="width: 100%; background-color: #ffa; border: 1px solid #777777; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" colspan="1"|

Where to start?
Are you feeling a bit overwhelmed? We can help! The Modern Hebrew Department is overwhelming because it has so much to offer, which is great! But sometimes having so many resources makes it difficult figuring out where to begin. We have created this page to give you an understanding of how the Modern Hebrew Department works and what it has to offer you. If you are still confused after this guide we have included people who you can contact for further help. Our community is warm and welcoming and we look forward to getting to know you. {| cellpadding="10" cellspacing="5" style="width: 100%; background-color: inherit; margin-left: auto; margin-right: auto" |
 * }

 * style="width: 60%; background-color: LightYellow; border: 1px solid Gray; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" rowspan="2" |
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Title Left

 * Text left

 * style="width: 50%; background-color: Seashell; border: 1px solid Gray; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px;" |

Helpful Links
Text top right
 * style="width: 40%; background-color: cornsilk; border: 1px solid Gray; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; height: 200px;" |
 * style="width: 40%; background-color: cornsilk; border: 1px solid Gray; vertical-align: top; -moz-border-radius-topleft: 8px; -moz-border-radius-bottomleft: 8px; -moz-border-radius-topright: 8px; -moz-border-radius-bottomright: 8px; height: 200px;" |

Staff Members You Can Contact
The following is a list of staff members who are extremely active in the Modern Hebrew Department. If you have any questions or need help with something ask one of them and they will get back to you promptly.
 * Readyokaygo

Also, it never hurts to check out this page's (or any page's) Talk Page as people often participate in conversation on various issues there.
 * }

title: Portal:Modern Hebrew/Beginners


text: Portal talk:Modern Hebrew/Beginners

title: Portal talk:Modern Hebrew/Beginners


text:
Template:Modern Hebrew/Intro

The Modern Hebrew Department offers free and open-source modern Hebrew language courses and resources for all levels of learners. We are passionate about learning and education and believe everyone should have free and quality access to materials that further their interests. Learning Modern Hebrew gives allows you to communicate with over ten million people in every corner of the world. It allows you to order a coffee and shop in Tel Aviv or ask for directions to the Western Wall in the holy city of Jerusalem. Modern Hebrew is the only "dead" language to be revived.

title: Template:Modern Hebrew/Intro


text:
Controversies in Science/What killed the dinosaurs/A critique of Mass extinction of birds at the cretaceous-palogene (k-pg) boundary

(Review Paper) Cited in

Points Made

 * The extinction event that they suggest took place around this time period involves the very least one asteroid impact and likely an increase in volcanic activity. This would create a large amount of dust and ash which would linger in the atmosphere reducing the amount of sunlight able to reach the surface of the earth.

 * Molecular studies cannot clearly identify which species of birds were present at the end of the Cretaceous period which is why these authors relied on fossil records to execute this study.

 * The authors focus on 17 species of birds which were shown to have existed in the Cretaceous and none of them made the transition into the Paleogene era. Therefore, the authors believe that there was an extinction event (K-PG) near the end of the Cretaceous.

 * An interesting point raised by the others was the large diversification of birds towards the end of the Cretaceous but a lack of the very small and vary large birds that are present today..

Methods
The authors studies the fossils of birds found in three different locations.
 * Hell Creek Formation (Montana, North & South Dakota)
 * Lance Formation (Wyoming)
 * Frechman Formation (Saskatchewan)

These fossils were most likely deposited at these locations in the later part of the Cretaceous period, about 300,000 years before the Paleogene period.

Researchers examined the fossils of the birds found in the regions described above and focused their researched on the coracoid bone to prevent overlapping of birds into the same categories. They chose this bone because it is quite commonly preserved and also has very little variation within the same species. This allowed them to determine if any of the species that they have identified are also present after the K-PG extinction event.

Since this data is focused on the western part of North America they researchers cannot conclusively infer that the findings will hold true all over the world but, although centered in North America, this was likely a global event.

Results
The studies found that the body size of birds varied greatly prior to the end of the Cretaceous period, though they didn't appear to represent the largest volant birds - larger species were found in the Mesozoic era. Larger birds (>10kg) and smaller birds (<200g) do not appear to be present immediately after the cretaceous period. There have been no recovered fossils of these species from the Paleogene era, but this only accounts for fossils found in North America. The study focused on 17 specific species of birds that were shown to exist in the Cretaceous period but didn't exist in the Paleogene period. These results indicate that an event occurred (at least one asteroid, if not more) that would lead to a mass extinction of these species during the Cretaceous era.


text:
123D Basics tutorial

Overview
This tutorial will introduce the basics of the "123D" 3D drawing software. You will be learning the fundamentals of 123D as you create a 3D model of a house.

Lets get started
1. The first step that should be taken whenever you are going to use 123D is to CHOOSE THE UNITS OF MEASUREMENT. Look down at the bottom   right corner of the screen, there you should see a bar with the centimeters abbreviation(cm). The units will start in centimeters by default. Click on the "cm" and a list of options will appear. For this tutorial, keep the units in "cm".

2. The next step is to begin the drawing. On the top of the screen you will see a bar with multiple symbols. The two you will be using through this tutorial are the two on the left, The pencil symbol (labeled "sketch") and the box symbol (labeled primitives).

3. To begin the drawing, click on the Box symbol and choose the Box option. When you do this, three grid axis will appear. Drag the mouse over the bottom axis. This will allow you to create on that axis. Next, you could either click on the grid you wish the first corner to be, or just type in the coordinates. Whichever you choose, start the box at (6,6), and make the box 8cm by 8cm. I would recommend typing the coordinates. Now that the base is established, the program will ask you to "extrude" the box. This just means adding the third dimension to create the box. Extrude the Box 8cm.

11. After creating this house, you know how to build just about any basic polygonal object.

title: 123D Basics tutorial


text:
Template:Modern Hebrew/Things You Can Do

The following is a list of tasks that need to be worked on or completed in the Hebrew division and subsequent departments:
 * Organize Category:Hebrew by making sure all pages are properly categorized and structured accordingly.
 * Ensure the templates on the main page are consistently updated.
 * Make sure no red links exist on this page.