text:
x-ray diffraction, Bragg's law and Laue equation

1. x-ray diffraction and Bragg's law

1.1 The history

1.1.1 X-ray
In November 1895, Wilhelm Röntgen discovered X-rays while working at the University of Wurzburg, Germany. Röntgen was investigating cathode rays in different types of evacuated glass tubes and trying to determine their range in air. He noticed that while the rays were being produced, a screen coated in fluorescent barium platinocyanide would glow. He was intrigued because the screen was too far from the tube to be affected by the cathode rays.

He assumed unkown rays, X-rays, were being emitted from the walls of the tube while the cathode ray tube was running. To his amazement, Röntgen found that the rays could pass straight through his hand and cast shadows of his bones on the fluorescent screen. He spent several weeks privately investigating the rays before publishing his results at the end of the year.

1.1.2 Bragg's Law
Bragg's law is the result of experiments derived by physicist Sir William Lawrence Bragg in 1912 and first presented on the same year to the Cambridge Philosophical Society. [1] William Lawrence Bragg and his father, Sir William Henry Bragg, were awarded the Nobel Prize in physics in 1915 for their work in determining crystal structures beginning with NaCl, ZnS, and diamond.

After Wilhelm Roentgen discovered X rays in 1895, William Henry Bragg，pioneered the determination of crystal structure by X-RAY diffraction methods, began a lifelong investigation of the nature of radiation, principally X rays but also alpha and beta particles and gamma rays. After the discovery of the diffraction of X rays by crystals in 1912, Bragg and his son, William L., derived Bragg's law, which relates the wavelength of X rays to the glancing angle of reflection. In 1913 the elder Bragg built the first X-ray spectrometer, which he initially used to study X-ray spectral distributions. Within several years they were able to use this instrument and Bragg's law to derive the structure of crystals and show the exact positions of atoms. Subsequently, they demonstrated that the properties and behavior of a large variety of substances can be related to the position of their constituent atoms.

William Lawrence Bragg went on to become director of the Cavendish Laboratory in Cambridge England. It was at this lab, while he as director, in the early 1950's that J.D. Watson and F.H. C. Crick, using the X-ray diffraction techniques that Bragg pioneered, deduced the double helical structure of Deoxyribonucleic acid. (DNA)

Biotechnology has many forms today: Crime detection, forensic science, recombinant DNA to strengthen plants and animals as well as fetal analysis to detect birth defects just to name a few. None of these would be possible if Watson and Crick had not deduced the structure of DNA using X-ray diffraction. Just as the engineer who is building a tall building must understand steel and cement, a scientist must understand the double helix before she can build a molecule.[2]

1.1.3 Nobel Prize for Bragg diffraction
Although simple, Bragg's law confirmed the existence of real particles at the atomic scale, as well as providing a powerful new tool for studying crystals in the form of X-ray and neutron diffraction.

[3]

In 1915, William Henry Bragg and William Lawrence Bragg were awarded the Nobel Prize for their contributions to crystal structure analysis. They were the first and (so far) the only father-son team to have jointly won the prize. Other father/son laureates include Niels and Aage Bohr, Manne and Kai Siegbahn, J. J. Thomson and George Thomson, Hans von Euler-Chelpin and Ulf von Euler, and Arthur and Roger Kornberg, who were all awarded the prize for separate contributions.

W. L. Bragg was 25 years old at the time, making him the youngest Nobel laureate to date.

W.H Bragg (1862-1942) and W.L. Bragg (1890-1971) on Swedish postage stamp.

1.2 The principle of Bragg’s Law and X-ray diffraction

1.2.1 The mathmetical form of Bragg's Law

(1)
\begin{align} n\lambda=2d\cdot\sin\theta \end{align}

where
• n is an integer determined by the order given,
• λ is the wavelength of x-rays, and moving electrons, protons and neutrons,
• d is the spacing between the planes in the atomic lattice, and
• θ is the angle between the incident ray and the scattering planes.

1.2.2 The derivation of Bragg's Law

Bragg diffraction occurs when electromagnetic radiation or subatomic particle waves with wavelength comparable to atomic spacings, are incident upon a crystalline sample, scattered by the atoms in the system and undergo constructive interference in accordance to Bragg's law. [2]

Consider conditions necessary to make the phases of the beams coincide when the incident angle equals and reflecting angle. The rays of the incident beam are always in phase and parallel up to the point at which the top beam strikes the top layer at atom z. The second beam continues to the next layer where it is scattered by atom B. The second beam must travel the extra distance AB + BC if the two beams are to continue traveling adjacent and parallel. This extra distance must be an integral (n) multiple of the wavelength ($\lambda$) for the phases of the two beams to be the same:

(2)
\begin{align} n\lambda=AB+BC \end{align}

Recognizing d as the hypotenuse of the right triangle Abz, we can use trigonometry to relate d and $\theta$ to the distance (AB + BC). The distance AB is opposite $\theta$ so,

(3)
\begin{align} AB=d\sin\theta \end{align}

Because AB = BC eq. (2) becomes,

(4)
\begin{align} n\lambda=2AB \end{align}

Substituting eq. (3) in eq. (4) we have,

(5)
\begin{align} n\lambda=2d\cdot\sin\theta \end{align}

and Bragg's Law has been derived.

Note that if only two rows of atoms are involved, the transition from constructive to destructive interference as $\theta$ changes is gradual. However, if interference from many rows occurs, then the constructive interference peaks become very sharp with mostly destructive interference in between. This sharpening of the peaks as the number of rows increases is very similar to the sharpening of the diffraction peaks from a diffraction grating as the number of slits increases.

1.2.3 Discussion

According to the figure on the right side above, we know the relation:

(6)
\begin{align} \overrightarrow{k}=\overrightarrow{k^{'}}+\overrightarrow{g} \end{align}

so

(7)
\begin{align} |\overrightarrow{g}|=2|\overrightarrow{k}|\sin\theta \end{align}

To ensure Bragg's Law, $\overrightarrow{g}$ should be set to equal to $\overrightarrow{G}$, where

(8)
\begin{align} |\overrightarrow{G}|=2\pi/d_{h,k,l} \end{align}

Compare Eq. (7) and Eq. (8), we have

(9)
\begin{align} \lambda=2/d_{h,k,l}\sin\theta \end{align}

So, only if

(10)
\begin{align} \overrightarrow{g}=\overrightarrow{G} \end{align}

the Bragg's Law can be satisfied.[9]

The definition of reciprocal lattice can be seen as follows:

In crystallography, the reciprocal lattice of a Bravais lattice is the set of all vectors $\overrightarrow{k}$ such that:

(11)
\begin{align} \exp({i}\overrightarrow{k}\cdot\overrightarrow{r})=1 \end{align}

This reciprocal lattice is itself a Bravais lattice, and the reciprocal of the reciprocal lattice is the original real lattice.

1.3 The application of Bragg's Law—Bragg diffraction

Bragg diffraction (also referred to as the Bragg formulation of X-ray diffraction) was first proposed by William Lawrence Bragg and William Henry Bragg in 1913 in response to their discovery that crystalline solids produced surprising patterns of reflected X-rays (in contrast to that of, say, a liquid). They found that in these crystals, for certain specific wavelengths and incident angles, intense peaks of reflected radiation (known as Bragg peaks) were produced. W. L. Bragg explained this result by modeling the crystal as a set of discrete parallel planes separated by a constant parameter d. It was proposed that the incident X-ray radiation would produce a Bragg peak if their reflections off the various planes interfered constructively, just as what we showed above.

The concept of Bragg diffraction applies equally to neutron diffraction and electron diffraction processes.

1.3.1 X-ray diffraction

A diffraction pattern is obtained by measuring the intensity of scattered waves as a function of scattering angle. Very strong intensities known as Bragg peaks are obtained in the diffraction pattern when scattered waves satisfy the Bragg's Law.

Following Bragg's law, each dot (or reflection) in the diffraction pattern above forms from the constructive interference of X-rays passing through a crystal. The data can be used to determine the crystal's atomic structure.[4]

The figures above show experimental x-ray diffraction patterns of cubic SiC using synchrotron radiation. We can see Bragg peaks clearly from the figures.[5]

1.3.2 Neutron diffraction

This is the neutron diffraction pattern of the Li0.86(Ni0.80Fe0.20)1.14O2 phase ( = 1.5939 Å). The difference between the observed and calculated profiles is given. The vertical bars show the Bragg reflections.[6]

1.3.3 Electron diffraction

Electron diffraction provides a basis for studying the structure of crystals and of identifying materials. Metals tend to give very strong electron diffraction patterns, whereas biological specimens generally diffract quite weakly.

The figure above is electron diffraction patterns from selected small areas. The micrograph shows a field of crystalline particles outlined by a large selection aperture (6 µm at the specimen). The corresponding 'spotty ring' type of diffraction pattern (upper left) indicates that the crystals are oriented in a number of different directions. Using a smaller selecting aperture (2.3 µm at the specimen) positioned as indicated, the pattern is formed from the chosen crystal alone, giving the single-crystal pattern seen at lower left. [7]

The patterns and explanation below are similar to the figure above, but the field of view is less than 1 µm across and hence too small for conventional s.a.d. A single particle has been selected for diffraction analysis by focusing the illumination into a spot covering the particle. The diffraction spots are discs rather than points because the electron beam is no longer parallel at the specimen, but a cone; each diffracted beam is therefore also a cone, which becomes a disc in the film plane. Discs can overlap and obscure others; hence, analysis of smaller crystals from µ-diffraction patterns is less precise than conventional s.a.d. [7]

The figure below shows the scattering intensity as a function of the scattering angle of a gaseous sample. The gaseous sample presents all possible orientations of atom-atom separations to the electron beam and the resulting diffracting pattern is like an X-ray powder photograph. The intensity of the diffracted beam decreases steadily with increasing scattering angle and the overall pattern consists of a series of concentric undulations on a background.[8]

1.4 Aknowlegements
I would like to thank Dharma Maddala for precious information, and ching-chang chung, liang Dong, Satyesh Yadav for many useful advice. Also thank Prof. Rampi to give me a chance to build the webpage with what we learned in class MSE 5317.

1.5 References
[1] W.L. Bragg, "The Diffraction of Short Electromagnetic Waves by a Crystal", Proceedings of the Cambridge Philosophical Society, 17 (1913), 43–57.
[2]http://en.wikipedia.org/wiki/Bragg%27s_law
[3]http://nobelprize.org/nobel_prizes/physics/laureates/1915/index.html
[4]The figure is from work of Jeff Dahl, and the figure is from http://www.goiit.com/posts/list/community-shelf-diffraction-83055.htm
[5]This part is from Paul J. Schields, Center for High Pressure Research, Department of Earth & Space Sciences, State University of New York at Stony Brook, Stony Brook, NY 11794-2100.
[6]G. Prado, E. Suard, L. Fournes and C. Delmasb, "Cationic distribution in the Li1-z(Ni1-yFey)1+zO2 electrode materials", J. of Mater. Chem. 10 (2000) 2553
[7]"The principles and practice of electron microscopy", Ian M. Watt, 1997, Cambridge University Press
[8]http://perso.fundp.ac.be/%7Ejwouters/DRX/diffraction.html

2. Laue Equation and its relation with Bragg's Law

2.1 The history

2.1.1 Biography of Laue

Max Theodor Felix von Laue (9 October 1879 – 24 April 1960) was a German physicist who won the Nobel Prize in Physics in 1914 for his discovery of the diffraction of X-rays by crystals.[1]

From 1909 to 1912, Laue was a Privatdozent at the Institute for Theoretical Physics, under Arnold Sommerfeld, at Ludwig Maximilians University of Munich(LMU). During the 1911 Christmas recess and in January 1912, Paul Peter Ewald was finishing the writing of his doctoral thesis under Sommerfeld. It was on a walk through the Englischer Garten in Munich in January, that Ewald told Laue about his thesis topic. The wavelengths of concern to Ewald were in the visible region of the spectrum and hence much larger than the spacing between the resonators in Ewald’s crystal model. Laue seemed distracted and wanted to know what would be the effect if much smaller wavelengths were considered. In June, Sommerfeld reported to the Physikalische Gesellschaft of Göttingen on the successful diffraction of x-rays by Laue, Paul Knipping and Walter Friedrich at LMU, for which Laue would be awarded the Nobel Prize in Physics, in 1914. While at Munich, he wrote the first volume of his book on relativity during the period 1910 to 1911.

2.1.2 Nobel Prize Awarded

# The Nobel Prize in Physics 1914:"for his discovery of the diffraction of X-rays by crystals"

2.1.3 Bragg's Law

After the discovery of the diffraction of X rays by crystals in 1912, Bragg and his son, William L., derived Bragg's law, which relates the wavelength of X rays to the glancing angle of reflection. In 1913 the elder Bragg built the first X-ray spectrometer, which he initially used to study X-ray spectral distributions. Within several years they were able to use this instrument and Bragg's law to derive the structure of crystals and show the exact positions of atoms. Subsequently, they demonstrated that the properties and behavior of a large variety of substances can be related to the position of their constituent atoms.

William Lawrence Bragg went on to become director of the Cavendish Laboratory in Cambridge England. It was at this lab, while he as director, in the early 1950's that J.D. Watson and F.H. C. Crick, using the X-ray diffraction techniques that Bragg pioneered, deduced the double helical structure of Deoxyribonucleic acid. (DNA)

Biotechnology has many forms today: Crime detection, forensic science, recombinant DNA to strengthen plants and animals as well as fetal analysis to detect birth defects just to name a few. None of these would be possible if Watson and Crick had not deduced the structure of DNA using X-ray diffraction. Just as the engineer who is building a tall building must understand steel and cement, a scientist must understand the double helix before she can build a molecule.[2]

2.2 The principle of Laue Equation

2.2.1 Reciprocal space

All the reciprocal lattice points can be described by a linear combination of multiples of two basis vectors: $\vec{b_{1}},\vec{b_{2}},\vec{b_{3}}$. Its definition is equivalent to the following relations:

(12)
\begin{align} \vec{b_{1}}=\frac{\vec{a_{2}}\times\vec{a_{3}}}{\vec{a_{1}}\cdot(\vec{a_{2}}\times\vec{a_{3}})} \end{align}
(13)
\begin{align} \vec{b_{2}}=\frac{\vec{a_{3}}\times\vec{a_{1}}}{\vec{a_{1}}\cdot(\vec{a_{2}}\times\vec{a_{3}})} \end{align}
(14)
\begin{align} \vec{b_{3}}=\frac{\vec{a_{1}}\times\vec{a_{2}}}{\vec{a_{1}}\cdot(\vec{a_{2}}\times\vec{a_{3}})} \end{align}

An interesting property of the reciprocal lattice vector is that the relations are still valid if we replace lattice vectors by reciprocal lattice vectors and inversely reciprocal lattice vectors by lattice vectors. If V* is the volume of the reciprocal unit cell, the reader can show that the expression VV* = 1 is also valid.

We can now define a reciprocal lattice vector $\vec{h}$ by a linear combination of integer multiples of the three reciprocal lattice vectors:

(15)
\begin{align} \vec{h}=h\vec{b_{1}}+k\vec{b_{2}}+l\vec{b_{3}} \end{align}

We shall see later that the reciprocal vectors play an important role in diffraction. Before, lets us relate the reciprocal lattice vectors to some macroscopic properties of crystals.

2.2.2 Miller indices

By construction, every reciprocal lattice vector is normal to a series of lattice planes. We know also that the faces of a crystal are parallel to lattice planes and therefore normal to the reciprocal vectors. We can therefore uniquely characterise each face of a crystal by the integer components of a reciprocal vector with one important subtlety. A face normal to vector $\vec{h}$ cannot be distinguished from a normal to vector $n\vec{h}$. Therefore, we shall characterise each face of a crystal by three integer components but without any common denominator. The triplet (h,k,l) satisfying this condition is called the Miller indices of the face. With this specification, we realise that Miller indices are only a subset of the possible reciprocal lattice vectors.

The figure above is an abstraction of a crystalline structure where only the nodes of the lattice are given. We have also reported a series of lattice planes with distances $d_{i}$ and the corresponding normal $\vec{h_{i}}$ to the series of planes. Without further demonstration we give here an important relation between the distances d and the reciprocal lattice vector $\vec{h}$ normal to the series of equidistant planes :

(16)
\begin{align} d=\frac{1}{|\vec{h}|} \end{align}

2.2.3 Laue Equation in one dimension

Now that we have introduced the definition of the reciprocal lattice vectors, we are ready to express the physical law governing the diffraction of crystalline material. [3]

Assume a row of scatterers separated by constant repeat, a. Radiation of wavelength $\lambda$ is incident on this row at an angle $\alpha_{0}$. Examine the the scatter from this row at an angle $\alpha_{n}$. The path difference of rays scattering at points A and D is just AB-CD. If the incoming rays are in phase, the path difference must be some integral multiple of the wavelength for constructive interference to occur. This leads to the first Laue equation:

(17)
\begin{align} (AB-CD)=a(\cos(\alpha_{n}-\cos\alpha_{0}))=n_{x}\lambda \end{align}

This result is valid for any scattered ray that makes an angle $\alpha_{n}$ with the unit cell axis. Thus the Laue condition is consistent with a cone of scattered rays centered about the a axis.

This equation can be restated in vector terms. The repeat distance a, becomes a unit cell vector $\vec{a}$. Call a unit vector parallel to the incoming rays, $\vec{S_{0}}$, and one parallel to the scattered rays, $\vec{S}$. There are then some simple vector dot products:

(18)
\begin{align} \vec{a}\cdot\vec{S}=a\cos\alpha_{n} \end{align}
(19)
\begin{align} \vec{a}\cdot\vec{S_{0}}=a\cos\alpha_{0} \end{align}
(20)
\begin{align} a\cos(\alpha_{n}-\cos\alpha_{0})=\vec{a}\cdot(\vec{S}-\vec{S_{0}})=n_{x}\lambda \end{align}

2.2.4 Laue Equation in three dimension

For a crystal with cell parameters $\vec{a},\vec{b},\vec{c}$, we have three Laue equations:

(21)
\begin{align} a\cos(\alpha_{n}-\cos\alpha_{0})=h\lambda \end{align}
(22)
\begin{align} b\cos(\beta_{n}-\cos\beta_{0})=k\lambda \end{align}
(23)
\begin{align} c\cos(\gamma_{n}-\cos\gamma_{0})=l\lambda \end{align}

where $\cos\alpha_{0}$, $\cos\beta_{0}$, $\cos\gamma_{0}$ are the direction cosines of the incident ray, and $\cos\alpha_{n}$, $\cos\beta_{n}$, $\cos\gamma_{n}$ are the direction cosines of the reflected ray in the crystal axis. So we must also satisfy:

(24)
\begin{align} \cos^{2}\alpha_{0}+\cos^{2}\beta_{0}+\cos^{2}\gamma_{0}=1 \end{align}
(25)
\begin{align} \cos^{2}\alpha_{n}+\cos^{2}\beta_{n}+\cos^{2}\gamma_{n}=1 \end{align}

Using that the angle between the incident and reflected ray is $2\theta$

(26)
\begin{align} \cos2\theta=\cos\alpha_{n}\alpha_{0}+\beta_{n}\beta_{0}+\gamma_{n}\gamma_{0} \end{align}

Take $\vec{k_{i}}$ to be the wavevector for the incoming (incident) beam and $\vec{k_{o}}$ to be the wavevector for the outgoing (diffracted) beam. $\vec{k_{o}}-\vec{k_{i}}=\Delta\vec{k}$ is the scattering vector and measures the change between the two wavevectors. Take $\vec{a_{1}},\vec{a_{2}},\vec{a_{3}}$ to be the primitive vectors of the crystal lattice. The three Laue conditions for the scattering vector, or the Laue equations, for integer values of a reflection's reciprocal lattice indices (h,k,l) can be also written as follows:

(27)
\begin{align} \vec{a_{1}}\cdot\Delta\vec{k}=2\pi h \end{align}
(28)
\begin{align} \vec{a_{2}}\cdot\Delta\vec{k}=2\pi k \end{align}
(29)
\begin{align} \vec{a_{3}}\cdot\Delta\vec{k}=2\pi l \end{align}

These conditions say that the scattering vector must be oriented in a specific direction in relation to the primitive vectors of the crystal lattice.

In crystallography, the Laue equations give three conditions for incident waves to be diffracted by a crystal lattice. They can be reduced to the Bragg law as discussed in the following section.[1]

2.3 The relation of Laue Equation to Bragg's Law

If $\vec{G}=h\vec{b_{1}}+k\vec{b_{2}}+l\vec{b_{3}}$ is the reciprocal lattice vector, we know:

(30)
\begin{align} \vec{G}\cdot(h\vec{a_{1}}+k\vec{a_{2}}+l\vec{a_{3}})=2\pi(h+k+l) \end{align}

The Laue equations specify:

(31)
\begin{align} \Delta\vec{k}\cdot(h\vec{a_{1}}+k\vec{a_{2}}+l\vec{a_{3}})=2\pi(h+k+l) \end{align}

where

(32)
\begin{align} \vec{k_{o}}-\vec{k_{i}}=\vec{G} \end{align}

As a result, the diffraction condition can be derived as follows:

(33)
\begin{align} \vec{k_{o}}-\vec{k_{i}}=\vec{G} \end{align}
(34)
\begin{align} (\vec{k_{i}}+\vec{G})^{2}=\vec{k_{o}}^{2} \end{align}
(35)
\begin{align} |\vec{k_{i}}|^{2}+2\vec{k_{i}}\cdot\vec{G}+|\vec{G}|^{2}=|\vec{k_{o}}|^{2} \end{align}

Considerring elastic scattering, we have:

(36)
\begin{align} |\vec{k_{i}}|^{2}=|\vec{k_{o}}|^{2} \end{align}

Finally, we have:

(37)
\begin{align} 2\vec{k_{i}}\cdot\vec{G}=|\vec{G}|^{2} \end{align}

So the diffraction condition $2\vec{k_{i}}\cdot\vec{G}=|\vec{G}|^{2}$ reduces to the Bragg's law $2d\sin\theta=n\lambda$.[2]

2.4 References

[1]http://en.wikipedia.org/wiki/Max_von_Laue
[2]Jialan's another webpage:http://electrons.wikidot.com/x-ray-diffraction-and-bragg-s-law
[3]http://capsicum.me.utexas.edu/ChE386K/html/laue_equations.htm
[4]Kittel, C. (1976). Introduction to Solid State Physics, New York: John Wiley & Sons. ISBN 0-471-49024-5
[5]W.L. Bragg, "The Diffraction of Short Electromagnetic Waves by a Crystal", Proceedings of the Cambridge Philosophical Society, 17 (1913), 43–57.


text:
# opposite of squares

the square fills approximately 0.6366 of its circumscribed circle. A Study of Definition", Information Age Publishing, 2008, p. 59, Chakerian, G.D. "A Distorted View of Geometry." For example, say expansive Jupiter was in analytical, micromanaging Virgo—forming a square to love-planet Venus in adventurous, caution-to … In the opposite corners of a square there are two identical ions. The meaning of the square is about the physical world. Squaring the circle, proposed by ancient geometers, is the problem of constructing a square with the same area as a given circle, by using only a finite number of steps with compass and straightedge. Possible antonyms could include circular, since a square has no curved sides and a circle has no straight sides. The properties of rectangle are somewhat similar to a square, but the difference between the two is, a rectangle has only its opposite sides equal. If f(x)=-x^2 (the opposite of x squared), what is the domain/range? Every acute triangle has three inscribed squares (squares in its interior such that all four of a square's vertices lie on a side of the triangle, so two of them lie on the same side and hence one side of the square coincides with part of a side of the triangle). . Squaring is the same as raising a number to the power of two. Find the coordinates of the remaining angular points. Squaring a number means multiplying the figure by itself. The opposite of squaring and cubing are called square root and cube root. A square is a special case of many lower symmetry quadrilaterals: These 6 symmetries express 8 distinct symmetries on a square. What are synonyms for square? However, Aristotle did not draw any diagram. This page was last edited on 27 November 2020, at 15:27. It has the same vertex arrangement as the square, and is vertex-transitive. A crossed square is sometimes likened to a bow tie or butterfly. Find more ways to say square dance, along with related words, antonyms and example phrases at Thesaurus.com, the world's most trusted free thesaurus. It can also be defined as a rectangle in which two adjacent sides have equal length. The opposite of finding the square root of a number is squaring the number. Rather, squares in hyperbolic geometry have angles of less than right angles. The diagonals of a square bisect its angles. The opposite of squaring and cubing are called square root and cube root. Synonyms for square in Free Thesaurus. Each has a charge of + e. The length of one side of he square is L. What is the net electric potential caused by the two positive ions at both of the empty corners of the square? As you can see, a diagonal of a square divides it into two right triangles,BCD and DAB. http://mathispower4u.com. verb adapt, regulate. four-sided. Its thought to cause internal friction in the native’s life and thus prompts some action to reduce … Antonyms for square Antonyms for (adj) square. The three special parallelograms — rhombus, rectangle, and square — are so-called because they’re special cases of the parallelogram. In hyperbolic geometry, squares with right angles do not exist. The area enclosed by a square is S × S, or S 2. A square has equal sides (marked "s") and every angle is a right angle (90°) Also opposite sides are parallel. For example, the square root of four is two, and two squared is four. Square antonyms. If You Know the Length of One Side Find the length of one side of the square. “Unfortunately, they don't clearly distinguish between the two, and in the process, Opposite of to form with right angles and straight lines, or flat surfaces, Opposite of plural for a person considered to be old-fashioned or boringly conventional in attitude or behavior, “He is always open to new ideas, and it is unsurprising to find out he is a. Zalman Usiskin and Jennifer Griffin, "The Classification of Quadrilaterals. 168 synonyms for square: town square, close, quad, market square, quadrangle, village square, conservative, dinosaur, traditionalist, die-hard, stick-in-the-mud. Only the g4 subgroup has no degrees of freedom, but can seen as a square with directed edges. This was done several centuries later by Apuleius and Boethius. The doctrine of the square of opposition originated with Aristotle inthe fourth century BC and has occurred in logic texts ever since.Although severely criticized in recent decades, it is still regularlyreferred to. The origin of the square can be traced back to Aristotle making the distinction between two oppositions: contradiction and contrariety. Anonymous. It’s really very simple…. old-fashioned, conventional. In non-Euclidean geometry, squares are more generally polygons with 4 equal sides and equal angles. This will probably be … The diagonals have the following properties: The two diagonals are congruent (same length). adj. Squares: to be in agreement on every point. 25 is an example of a perfect square. ℓ A square is a special case of a rhombus (equal sides, opposite equal angles), a kite (two pairs of adjacent equal sides), a trapezoid (one pair of opposite sides parallel), a parallelogram (all opposite sides parallel), a quadrilateral or tetragon (four-sided polygon), and a rectangle (opposite sides equal, right-angles), and therefore has all the properties of all these shapes, namely: The three-level hierarchy you see with in the above quadrilateral family tree works just like A dog is a […] Definition: having four equal sides and four right angles or forming a right angle. Remember, the square of a number is that number times itself. The square is in two families of polytopes in two dimensions: The square is a highly symmetric object. Domain: Set of all real numbers. Square antonyms. The diagram for the traditional square of opposition is: The theses embodied in this diagram I call ‘SQUARE’. g2 defines the geometry of a parallelogram. In 1882, the task was proven to be impossible as a consequence of the Lindemann–Weierstrass theorem, which proves that pi (π) is a transcendental number rather than an algebraic irrational number; that is, it is not the root of any polynomial with rational coefficients. The circumradius of this square (the radius of a circle drawn through the square's vertices) is half the square's diagonal, and is equal to In Geometry, a square is a two-dimensional plane figure with four equal sides and all the four angles are equal to 90 degrees. -5² = -25 NOT -5² ≠ 25 Many people will feel certain that the answer here is 25. (In addition, the square is a special case or type of both the rectangle and the rhombus.) Antonyms for squares include beats, betters, exceeds, passes, smashes, tops, surpasses, bests, eclipses and overtakes. So if n^2=p, then I n I=I p^1/2 I. The perimeter, or distance around a square is S+S+S+S, or 4S. "Regular polytope distances". A square also fits the definition of a rectangle (all … the little squares in each corner mean "right angle" A square has equal sides (marked "s") and every angle is a right angle (90°) Also opposite sides are parallel. The opposite of the adjective square would be "not square." {\displaystyle {\sqrt {2}}.} The diagonal of the square is the hypotenuseof these triangles.We can use Pythagoras' Theoremto find the length of the diagonal if we know the side length of the square. SEE DEFINITION OF square. What are the characteristics of squares, rhombi, kites, and trapezoids? The square has Dih4 symmetry, order 8. Squares happen when two planets are 90 degrees apart. Antonyms & Near Antonyms for square. You can think of it as the "root" of the square or the number that was used to make the square. http://mathispower4u.com. verb correspond, agree. The coordinates for the vertices of a square with vertical and horizontal sides, centered at the origin and with side length 2 are (±1, ±1), while the interior of this square consists of all points (xi, yi) with −1 < xi < 1 and −1 < yi < 1. Square, Point on the Inscribed Circle, Tangency Points. Antonyms for square include nonsquare, parallel, aligned, collateral, alongside, equidistant, lateral, side-by-side, in the same direction and running alongside. the crossed rectangle is related, as a faceting of the rectangle, both special cases of crossed quadrilaterals.[13]. There are four lines of, A rectangle with two adjacent equal sides, A quadrilateral with four equal sides and four, A parallelogram with one right angle and two adjacent equal sides. The semi-square is considered an important yet hard minor aspect. The square of opposition is a chart that was introduced within classical (categorical) logic to represent the logical relationships holding between certain propositions in virtue of their form. They share the same astrological “quality” (cardinal, mutable or fixed) , but nothing else. 2 It has two lines of reflectional symmetry and rotational symmetry of order 2 (through 180°). Dually, a square is the quadrilateral containing the largest area within a given perimeter. Larger spherical squares have larger angles. This led to the use of the term square to mean raising to the second power. Therefore, a rectangle is called a square only if all its four sides are of equal length. ◻ By doing so, I will have a plus or minus case. A square has two diagonals. O2ke/L O ke/2L ke/L O sqrt(2) ke/L In spherical geometry, a square is a polygon whose edges are great circle arcs of equal distance, which meet at equal angles. There are 2 dihedral subgroups: Dih2, Dih1, and 3 cyclic subgroups: Z4, Z2, and Z1. It has half the symmetry of the square, Dih2, order 4. In terms of the inradius r, the area of the square is. There are more rules we can use with indices. However, the square symbolizes the opposite. Squares, cubes and higher powers are shown as small digits called indices. Squaring, which we learned about in a previous lesson , has an inverse too, called "finding the square root." John H. Conway, Heidi Burgiel, Chaim Goodman-Strauss, (2008) The Symmetries of Things, "List of Geometry and Trigonometry Symbols", "Properties of equidiagonal quadrilaterals", "Quadrilaterals - Square, Rectangle, Rhombus, Trapezoid, Parallelogram", "Geometry classes, Problem 331. 7 in. The square had a focus of the points of the compass to the pagans. Perfect squares are the squares of integer values: >>> >>> Top antonyms for square (opposite of square) are hypocritical, outrageous and round. The interior of a crossed square can have a polygon density of ±1 in each triangle, dependent upon the winding orientation as clockwise or counterclockwise. The K4 complete graph is often drawn as a square with all 6 possible edges connected, hence appearing as a square with both diagonals drawn. The square root is just the opposite of the square. Top antonyms for square (opposite of square) are hypocritical, outrageous and round. Main entry: square. since the area of the circle is Click hereto get an answer to your question  A(3,4) and C(1, - 1) are the two opposite angular points of a square ABCD. Squaring a number means multiplying the figure by itself. In other words, x 2 + 14x + 49 could be factored as. Usage: a square peg in a round hole; a square corner A square has a larger area than any other quadrilateral with the same perimeter. ‘Every S is P’ and ‘No S is P’ are contraries. Ex 7.4, 4 The two opposite vertices of a square are (–1, 2) and (3, 2). The square is the area-maximizing rectangle. This graph also represents an orthographic projection of the 4 vertices and 6 edges of the regular 3-simplex (tetrahedron). The opposite of finding the square root of a number is squaring the number. In simple arithmatic, they might simply mean the negative of the number, so the answer would be - √3. "Opposite" is a term lacking adequate mathematical definition. You can think of it as the "root" of the square or the number that was used to make the square. Unlike the square of plane geometry, the angles of such a square are larger than a right angle. Also opposite sides are parallel and of equal length. Therefore, a rectangle is called a square only if all its four sides are of equal length. Each subgroup symmetry allows one or more degrees of freedom for irregular quadrilaterals. Click hereto get an answer to your question  Let the opposite angular points of a square be (3,4) and (1, - 1) . The fraction of the triangle's area that is filled by the square is no more than 1/2. This equation means "x2 or y2, whichever is larger, equals 1." An obtuse triangle has only one inscribed square, with a side coinciding with part of the triangle's longest side. The circle will represent things that are spiritual and sacred in nature. As to the opposite of A square, it is a negative square, or a square moving backwards through time. The area can also be calculated using the diagonal d according to, In terms of the circumradius R, the area of a square is. If a circle is circumscribed around a square, the area of the circle is, If a circle is inscribed in the square, the area of the circle is. [7] Indeed, if A and P are the area and perimeter enclosed by a quadrilateral, then the following isoperimetric inequality holds: with equality if and only if the quadrilateral is a square. This article is about the polygon. honest, genuine. is. 1 decade ago. The distance from one corner of a square to the opposite corner is sometimes called the diagonal. John Conway labels these by a letter and group order.[12]. In term logic, the square of opposition is a diagram representing the relations between the four basic categorical propositions. A crossed square is a faceting of the square, a self-intersecting polygon created by removing two opposite edges of a square and reconnecting by its two diagonals. The properties of rectangle are somewhat similar to a square, but the difference between the two is, a rectangle has only its opposite sides equal. can also be used to describe the boundary of a square with center coordinates (a, b), and a horizontal or vertical radius of r. The following animations show how to construct a square using a compass and straightedge. Park, Poo-Sung. {\displaystyle \ell } Opposite of to make level or equal in amount, degree, or status, “One benefit of this approach is that players always have a reason to keep playing and the only goal is to, Opposite of to be compatible or consistent with, “She complained that a standard witness oath would, Opposite of to adapt something to be consistent with, “The warlords and diplomats are still wedded to the old false theories, therefore we shall, Opposite of to deal with, or reach an agreement about, something. A square is a rectangle with four equal sides. , Squaring is the same as raising a number to the power of two. The circle symbolizes the sacred earth. Larger hyperbolic squares have smaller angles. This video explains how to simplify the opposite of square roots of perfect square variable expressions. Antonyms for square. Square Meaning Differences between Square and Circle. {\displaystyle \pi R^{2},} In Geometry, a square is a two-dimensional plane figure with four equal sides and all the four angles are equal to 90 degrees. Although relatively simple and straightforward to deal with, squares have several interesting and notable properties. The reciprocal is well defined and is 1/√3. The square is the n=2 case of the families of n-. In geometry, a square is a regular quadrilateral, which means that it has four equal sides and four equal angles (90-degree angles, or 100-gradian angles or right angles). The square of opposition is a group of theses embodied in a diagram.The … verb pay off, satisfy. r8 is full symmetry of the square, and a1 is no symmetry. ‘No S is P’ and ‘Some S is P’ are contradictories. The presence of a squared term inside the radical symbol tells me that I will apply the square root operation on both sides of the equation to find the inverse. This is a situation where I will make a decision on which one to pick as the correct inverse function. All sides of a square have the same length. The Square. Ch. They are: SQUARE ‘Every S is P’ and ‘Some S is not P’ are contradictories. I know that you can substitute anything for x, right, so all real numbers? If a figure is both a rectangle (right angles) and a rhombus (equal edge lengths), then it is a square. These two forms are duals of each other, and have half the symmetry order of the square. Elearning, Online math tutor, LMS", http://forumgeom.fau.edu/FG2016volume16/FG201627.pdf, "Cyclic Averages of Regular Polygons and Platonic Solids", Animated course (Construction, Circumference, Area), Animated applet illustrating the area of a square, https://en.wikipedia.org/w/index.php?title=Square&oldid=990968392, Wikipedia pages semi-protected against vandalism, Creative Commons Attribution-ShareAlike License, A quadrilateral where the diagonals are equal, and are the perpendicular bisectors of each other (i.e., a rhombus with equal diagonals), A convex quadrilateral with successive sides. The opposite of The square equals the respective square root. 2 d2 is the symmetry of an isosceles trapezoid, and p2 is the symmetry of a kite. Find the coordinates of the other two vertices. Favorite Answer. Sign for Square Root The sign for square root looks like this: Some examples of square roots: Finding the Square Root There really isn't a good way to find a square root other than using your calculator. Squares: Both opposite sides are congruent and parallel, both opposite angles are congruent, has 4 sides Rhombi: Both opposite sides are congruent and parallel, both opposite angles are congruent, diagonals are perpendicular, has 4 sides Kites: two pairs of consecutive congruent sides, has a perpendicular bisector, has 4 sides Trapezoids: one pair of opposite sides parallel, has 4 sides, legs are congruent … Although relatively simple and straightforward to deal with, squares have several interesting and notable properties. adj. Subtraction is the opposite of addition, division is the inverse of multiplication, and so on. A square with vertices ABCD would be denoted In a right triangle two of the squares coincide and have a vertex at the triangle's right angle, so a right triangle has only two distinct inscribed squares. This video explains how to simplify the opposite of square roots of perfect square variable expressions. In the figure above, click 'reset'. In this example, n, the square root, is 5. adj. The opposite of square (fair, … It appears as two 45-45-90 triangle with a common vertex, but the geometric intersection is not considered a vertex. 7. In the figure above, click 'show both diagonals', then drag the orange dot at any vertex of the square and convince yourself this is so. The point of this entry is to trace its history from thevantage point of the early twenty-first century, along with closelyrelated doctrines bearing on empty terms. R Find more opposite words at wordhippo.com! Math teacher Master Degree. In classical times, the second power was described in terms of the area of a square, as in the above formula. For example, the square root of four is two, and two squared is four. {\displaystyle \square } For other uses, see. It is usually denoted by putting a two in superscript after the number. It indicates a challenging or stimulating energy, like a square but less intense and more internal. Reciprocal of n = 1/n For example, -9 squared equals 81, but the square root of 81 can be 9 or -9. The fundamental definition of a square is as follows: A square is both a rectangle and a rhombus and inherits the properties of both (except with both sides equal to each other). noun municipal park. Because it is a regular polygon, a square is the quadrilateral of least perimeter enclosing a given area. x 2 + 14x + 49 = (x + 7) 2. The equation, specifies the boundary of this square. ‘Some S is P’ and ‘Some S is not P’ are subcontraries. Each one is a line segment drawn between the opposite vertices (corners) of the square. The square root is just the opposite of the square. A square and a crossed square have the following properties in common: It exists in the vertex figure of a uniform star polyhedra, the tetrahemihexahedron. Synonyms: accords, agrees, answers… Antonyms: differs (from), disagrees (with)… Find the right word. The fundamental definition of a square is as follows: A square is both a rectangle and a rhombus and inherits the properties of both (except with both sides equal to each other). Then the circumcircle has the equation. Another word for square dance. noun person who is old-fashioned, conventional. [1][2], A convex quadrilateral is a square if and only if it is any one of the following:[3][4], A square is a special case of a rhombus (equal sides, opposite equal angles), a kite (two pairs of adjacent equal sides), a trapezoid (one pair of opposite sides parallel), a parallelogram (all opposite sides parallel), a quadrilateral or tetragon (four-sided polygon), and a rectangle (opposite sides equal, right-angles), and therefore has all the properties of all these shapes, namely:[6], The perimeter of a square whose four sides have length A square is a rectangle with four equal sides. d4 is the symmetry of a rectangle, and p4 is the symmetry of a rhombus. It is usually denoted by putting a two in superscript after the number. College, SAT Prep. The square is the area-maximizing rectangle. The square root, then, is the number n, which when multiplied by itself yields the square, x. Find more opposite words at wordhippo.com! Note: If the coefficient of x had been any number but 14, this would not have been a perfect square trinomial. ABCD. This is possible as 4 = 22, a power of two. bohemian, boho, counterculturist, deviant, enfant terrible, free spirit, heretic, iconoclast, All four angles of a square are equal (each being 360°/4 = 90°, a right angle). π Squares, cubes and higher powers are shown as small digits called indices. Case or type of both the rectangle, both special cases of crossed quadrilaterals. [ 13.. Physical world what are the characteristics of squares, cubes and higher powers are shown small! ◻ { \displaystyle { \sqrt { 2 } }. two adjacent have. Sacred in nature see, a square are equal to 90 degrees.. In term logic, the angles of a number is squaring the number was... Learned about in a round hole ; a square to mean raising to the second.... Second power was described in terms of the square is S × S, or 2... Likened to a bow tie or butterfly case of Many lower symmetry quadrilaterals these! Exceeds, passes, smashes, tops, surpasses, bests, eclipses overtakes. Hard minor aspect an inverse too, called  finding the square root ''!, smashes, tops, surpasses, bests, eclipses and overtakes is squaring the.! ( each being 360°/4 = 90°, a right angle right triangles, BCD DAB. Curved sides and a circle has no straight sides and all the four basic propositions... Cases of crossed quadrilaterals. [ 13 ] one to pick as the  ''... 81, but the square. what are the characteristics of squares, cubes and powers... Of finding the square or the number they might simply mean the negative of the square root, the. Also opposite sides are of equal distance, which we learned about in round. A special case or type of both the rectangle, and so on opposite! ( in addition, the square root, is 5 and trapezoids it into two right triangles BCD. Every point spherical geometry, a right angle the term square to the pagans between... Vertices of a square corner all sides of a square only if all four. Not square. multiplying the figure by itself ” ( cardinal, mutable or ). An isosceles trapezoid, and Z1 or -9 triangle 's area that is by. To be in agreement on Every point 7 ) 2 when multiplied by itself whose. All sides of a rectangle is related, as a faceting of the and... Group of theses embodied in a round hole ; a square are equal ( each being 360°/4 90°. Right triangles, BCD and DAB coinciding with part of the 4 and! ’ and ‘ no S is not P ’ are contraries line segment drawn between opposite! And ‘ no S is P ’ are contradictories we learned about a. This video explains how to simplify the opposite of squaring and cubing are called square.... R, the square. these 6 symmetries express 8 distinct symmetries on a square is in two families polytopes. An orthographic projection of the compass to the opposite of square roots of perfect square expressions., agrees, answers… antonyms: differs ( from ), but nothing else this would not been. Was last edited on 27 November 2020, at 15:27 inverse of multiplication, and so on express! Forms are duals of each other, and 3 cyclic subgroups: Z4, Z2, and?. A situation where I will make a decision on which one to pick as the square of plane,., like a square there are two identical ions longest side lesson, has an too! Square ( fair, … the semi-square is considered an important yet hard aspect. Is sometimes likened to a bow tie or butterfly is filled by the or. Right angles or forming a right angle a two in superscript after the number n, the square opposite of squares. Have the following properties: the two diagonals are congruent ( same length ) equals.... { \displaystyle \square } ABCD symmetry quadrilaterals: these 6 symmetries express 8 distinct on... ( tetrahedron ) the meaning of the square root of opposite of squares square are (,! Other, and trapezoids ( fair, … the semi-square is considered an important yet hard minor.... Vertices ABCD would be  not square. possible antonyms could include circular, since a square have the properties. The square. times, the square root is just the opposite of squaring and cubing called! Vertex, but nothing else has an inverse too, called  finding the square, Dih2, 4... Triangles, BCD and DAB, whichever is larger, equals 1. edited on 27 November 2020 at! A regular polygon, a rectangle in which two adjacent sides have equal length x )! No more than 1/2 ) … Find the right word Many lower symmetry:... Quadrilateral of least perimeter enclosing a given perimeter include beats, betters, exceeds, passes smashes. You can think of it as the  root '' of the square of opposition a. The correct inverse function or stimulating energy, like a square only if its! These 6 symmetries express 8 distinct symmetries on a square is S × S or! N I=I p^1/2 I least perimeter enclosing a given perimeter, equals 1. an important yet hard aspect! By the square of a square only if all its four sides are of equal.! Answer would be - √3 perfect square variable expressions minor aspect, since square! Power of two, … the semi-square is considered an important yet hard aspect! Each one is a special case of Many lower symmetry quadrilaterals: these 6 symmetries 8... Done several centuries later by Apuleius and Boethius square also fits the definition of a number means multiplying the by... Of an isosceles trapezoid, and Z1 has half the symmetry order of the points opposite of squares the compass the! 4 the two diagonals are congruent ( same length of quadrilaterals. [ 13 ] situation! Are more rules we can use with indices n=2 case of the square of plane geometry squares! Be factored as and of equal length — rhombus, rectangle, and trapezoids graph also represents an projection... Which two adjacent sides have equal opposite of squares means multiplying the figure by itself and the rhombus. addition, square. They might simply mean the negative of the inradius r, the square. vertices 6... Meaning of the compass to the power of two and cubing are called square root, 5. People will feel certain that the answer would be - √3, betters, exceeds, opposite of squares, smashes tops! Into two right triangles, BCD and DAB number, so all real?. Regular polygon, a square peg in a diagram.The … squares: to be in agreement on point! Duals of each other, and trapezoids they are: square ‘ Every S is P ’ and Some. Geometric intersection is not P ’ are contradictories a perfect square trinomial,. Congruent ( same length what is the same as raising a number is squaring number. Squaring is the number square corner all sides of a number means multiplying figure. Each subgroup symmetry allows one or more degrees of freedom for irregular quadrilaterals. [ 12 ] be... With, squares in hyperbolic geometry, squares with right angles 's area is. }. diagonal of a square moving backwards through time squares are more we. Square root of 81 can be 9 or -9 would be  not square. in... Representing the relations between the four basic categorical propositions than right angles or forming a right angle.. If all its four sides are of equal length adequate mathematical definition if the coefficient of x squared,. Sides have equal length squaring the number that was used to make the square is no more 1/2. Four is two, and p4 is the opposite of square ) hypocritical! Largest area within a given area or butterfly ( corners ) of 4! Only one inscribed square, it is usually denoted by putting a two superscript. Is squaring the number that was used to make the square can be 9 or -9 but the,. Tops, surpasses, bests, eclipses and overtakes as small digits called indices,... = 22, a square divides it into two right triangles, BCD DAB..., n, which we learned about in a round hole ; a square with directed edges quadrilaterals. 12. Relations between the opposite of x squared ), disagrees ( with ) … Find the right.! Square ‘ Every S is not P ’ are contradictories case or type of both the rectangle, square... Is considered an important yet hard minor aspect tetrahedron ) \sqrt { }... Called a square has no curved sides and equal angles identical ions parallelograms rhombus! Sides are parallel and of equal length, … the opposite of square are! Less intense and more internal \square } ABCD hyperbolic geometry, a power of two square, a. Right, so all real numbers be denoted ◻ { \displaystyle { \sqrt { }. Of equal length }. with indices graph also represents an orthographic projection of the families of polytopes in families. Into two right triangles, BCD and DAB: if the coefficient of x squared ) but. Example, the second power term logic, the square of plane geometry, a square the!: Dih2, Dih1, and two squared is four \sqrt { 2 } } }. Accords, agrees, answers… antonyms: differs ( from ), but can seen as rectangle.

format: plain


text:
# An ansatz approach to the zeta function

Published:

The evaluation of Riemann’s zeta function

$\zeta(s) = \sum_{k \geq 1} \frac{1}{k^s}$

at even integers is well-known, going back to the venerable Leonhard Euler and his answer to the Basel problem, namely

$\zeta(2) = \frac{\pi^2}{6}.$

The Basel problem stumped even the great Bernoulli’s, and Euler’s solution required the ingenuity and insight found throughout his work. Nearly three-hundred years later, is this easier to answer? I claim that not only is it easy, but that it is entirely routine to both guess the answer and to prove it, provided that you allow some basic Fourier analysis. Let’s see how.

Given a suitably nice function $f$ on $[0, 1]$, we define its Fourier transform by

$\hat{f}(t) = \int_0^1 f(x) e^{-2\pi i t x}\ dx.$

These are the coefficients of the Fourier series

$\sum_k \hat{f}(k) e^{2\pi i k}.$

There is a deep theory about what functions are suitably nice and what properties the Fourier transform satisfies, but for now we are only interested in one fact:

The family $\{e^{2\pi i n x}\}_n = \{e(n)\}_n$ is an orthonormal basis for the square-integrable functions on $[0, 1]$ equipped with the usual integral inner product

$(f, g) = \int_0^1 f \overline{g}.$

In particular,

$||f||_2 = \sum_k |(f, e(n))|^2 = \sum_k |\hat{f}(k)|^2.$

In one direction, this tells us that sums whose terms are Fourier coefficients can be evaluated as an integral. This is exactly what we will do.

First, we need a humanly-proved lemma (though in principle a computer could likely figure this out).

Lemma. The Fourier transform of $f_n(x) = x^n$ at integers satisfies $\hat{f_n}(0) = (n + 1)^{-1}$, and $\hat{f}_n(k)$ is a polynomial in $(2\pi i k)^{-1}$ of degree $n$ for all nonzero integers $k$.

Proof. Apply integration by parts to prove that the sequence $\hat{f}_n(k)$ satisfies

$\hat{f}_n(k) = n(2\pi i k)^{-1} \hat{f}_{n - 1}(k) - (2\pi i k)^{-1}.$

The claim follows immediately by induction once we note that $\hat{f}_0(k) = 0$. $\blacksquare$

At this point we are in possession of a very powerful ansatz. The Fourier transform of polynomials at integers gives us reciprocals of powers of integers! There must be a connection with the zeta function. In particular, we want to find a polynomial

$f(x) = \sum_{k = 0}^n a_k x^k$

such that $\hat{f}(k)$ is some multiple of $k^{-n}$ times a factor independent of $k$. Given the lemma above, we know that $\hat{f}(k)$, for nonzero $k$, is a polynomial in $(2\pi i)^{-1}$, meaning that we can look at their coefficients and require ones smaller than $(2\pi i)^{-n}$ to vanish. We can stipulate that $a_n = -1$ and that $\hat{f}(0) = 0$, which gives us a total of $n + 1$ linear equations in $n + 1$ unknowns, which we can (probably) solve!

So, in light of this, it suffices to write a program that will equate these coefficients and solve the resulting linear equations. This will then, a posteriori, provide evaluations of $\zeta(n)$ for positive, even integers $n$. (Only the evens since, of course, we must take squares.)

# The program

Here is one such program that will do the job.

import sympy as sp
from sympy.abc import x, t
from sympy import I, pi

def zeta_ansatz(n):
xs = sp.symbols("a:{}".format(n + 1))
f = sum(xs[k] * x**k for k in range(n + 1))

k = sp.symbols("k", integer=True, zero=False)
ft = sp.integrate(f * sp.exp(-2 * pi * I * k * x), (x, 0, 1))
ft = ft.simplify().expand()
# Replace 2 pi I with a dummy variable 1 / t to grab coefficients.
ft = ft.subs(k, 1).subs(I, 1).subs(pi, 1 / t)

coeffs = ft.collect(t, evaluate=False)
zero_eqns = [coeff for key, coeff in coeffs.items() if key != t**n]
eqns = zero_eqns + [xs[-1] + 1, sum(x / (k + 1) for k, x in enumerate(xs))]
soln = sp.solve(eqns, xs)

f = sum(soln[xs[k]] * x**k for k in range(n + 1))

coeff = sp.factorial(n) / (2 * pi)**n

return f, sp.integrate(abs(f)**2, (x, 0, 1)) / coeff**2 / 2

An example:

In [1]: time [zeta_ansatz(k) for k in range(1, 5)]
CPU times: user 1.85 s, sys: 19 µs, total: 1.85 s
Wall time: 1.85 s
Out[1]:
⎡⎛          2⎞  ⎛               4⎞  ⎛          2        6⎞  ⎛                         8 ⎞⎤
⎢⎜         π ⎟  ⎜   2       1  π ⎟  ⎜   3   3⋅x    x   π ⎟  ⎜   4      3    2   1    π  ⎟⎥
⎢⎜1/2 - x, ──⎟, ⎜- x  + x - ─, ──⎟, ⎜- x  + ──── - ─, ───⎟, ⎜- x  + 2⋅x  - x  + ──, ────⎟⎥
⎣⎝         6 ⎠  ⎝           6  90⎠  ⎝        2     2  945⎠  ⎝                   30  9450⎠⎦

So there, we have answered the Basel problem plus evaluated the next three terms of the sequence $\zeta(2n)$, all in under two seconds and without any foreknowledge of the answer. Not too shabby, eh, Euler?

# Connections to known results

These results are obviously not new. There are known closed-form evaluations of $\zeta(2n)$ for all positive integers $n$. A cursory glance suggests that the polynomials we get are exactly the negatives of the Bernoulli polynomials $B_n(x)$, and mentioned in that article is that the Fourier transform of $B_n(x)$ is

$\hat{B}_n(x) = -\frac{n!}{(2\pi i)^n} \sum_{k \neq 0} \frac{e^{2\pi i k x}}{k^n},$

which matches exactly what we have said here.

Does this idea uniquely define the Bernoulli polynomials? The Fourier transform is invertible, so saying “let $B_n(x)$ be the preimage of such and such function under the Fourier transform on $[0, 1]$” is a fine definition. It is surprising that such a thing is a polynomial, but nevertheless true, and lucky for us that it was. Conversely, I am pretty sure that polynomials will only ever give you linear combinations of the $\zeta$ function evaluated at even integers, so we have completely exhausted the usefulness of polynomials coupled with Fourier transforms.


text:
# The banana problem

There is a SQUARE field and 44 BANANA trees are planted across the square field. Find the peremeter of that square field, if the distance between 2 BANANA tree is 2 metre.

Note by Jeet Dubey
7 years ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

352 m is right i suppose considering question is right ! 44 banana across the field (assuming the line of 44 bananas parallel to one of the side) having 2 metre between each tree so length of one side comes to 88m . Since perimeter is asked we have to take addition four time because it is square, so 88*4=352.

- 6 years, 11 months ago

4 Banana trees are planted ONE at each corner of a square. and 10 trees are planted at each side between two corners of a side so Perimeter of the square field=4+(10*4)=44 meter

- 6 years, 11 months ago

88

- 4 years, 9 months ago

- 7 years ago

44 is not a square number so the field can't be square

- 7 years ago

I agree that this question doesn't make much sense, and that 144 (likely a typo) would be a much better interpretation.

However, you're making the assumption that the banana trees must be planted parallel to the sides of the field, which is not necessary. Hence, an interpretation of this problem (which likely isn't the one originally intended) could be to find the minimum area of a field that contains exactly 44 banana trees.

Staff - 7 years ago

If Jeet D intended to type 144,then the answer would be the same as Luca B's answer.Otherwise it should be $6 \times 2m \times 4=48m$ because 49 is the smallest square number larger than 44 and $\sqrt{49}=7$

- 7 years ago

Not quite. This is actually much more interesting. You are still making the same assumption that the banana trees must be planted parallel to the sides of the field.

For example, if we only wanted 2 trees, then your approach will require a square field of side length 2, which has 4 banana trees on the corners. Whereas, if you take the 'diamond', you can see that a square field of side length $\sqrt{2}$ is sufficient, with 2 banana trees on the opposite corners. Can you find an argument to show that this is the minimum length possible?

What happens in the case for 3 banana trees? Can we also do it with a square field of side length $\sqrt{2}$? If not, what is the minimum length?

What happens in the case for 4 banana trees? Is the minimum length 2? Can we not do any better? Why, or why not?

What happens in the case for 44 banana trees?

Staff - 7 years ago

Double interpretation with same result LOL

- 7 years ago

Perimeter is 88 m. 12 banana trees per side, 22 metre per side.

- 7 years ago

Why are there 12 banana trees per side?

Staff - 7 years ago

I thought the 44 trees are in the perimeter...12 per side with 4 in commons in the corners. This is my only way to interpret the problem so that it's logical...

- 7 years ago

That's certainly a valid interpretation, given the ambiguity in the question. Thanks for explaining!

Staff - 7 years ago

No problem :)

- 7 years ago


text:
SCIENTIA SINICA Informationis, Volume 49, Issue 4: 464-485(2019) https://doi.org/10.1360/N112018-00196

## A watermarking algorithm for image content authentication in double-compression environment

• AcceptedDec 17, 2018
• PublishedApr 11, 2019
Share
Rating

### Abstract

Malicious tampering of digital image content may lead to serious consequences. As image resolution increases and its circulations on the network grows rapidly, images undergo double compression in the following environment: compression (JPEG/JPEG2000) for first release, decoding, application processing (conventional signal processing/security attacks/malicious content tampering), and compression (JPEG2000/JPEG) for rerelease before detection. Effectively determining whether the image content has been tampered and to locate the tampering position in a double-compression environment is an urgent problem. In this paper, a novel watermark expression method based on rotating vector and its modulation algorithm is proposed. Based on this method, a semi-fragile watermarking scheme is established for the image content authentication. The stability of watermarked data is analyzed theoretically, providing the watermark scheme with a theoretical basis. The authentication schemes with feature extraction and reconstruction, watermark embedding and extraction, and tamper detection and localization are mainly elaborated. The robustness of watermark, its security, and related detection performance are analyzed. Theoretical analysis and experiments show that the scheme features a good watermark transparency, good robustness, and stable distribution for different attacks and can effectively distinguish malicious tampering from content-preserving operators and locate tampering area under double-compression environment. In addition, the proposed scheme can resist watermark attacks, collage attacks, and forge attacks. Compared with related schemes, the proposed scheme has superior comprehensive performance and is suitable for content authentication in double-compression environment, which expands the application scope of watermark-based content authentication.

### References

[1] Lin C Y, Chang S F. Semi-fragile watermarking for authenticating JPEG visual content. In: Proceedings of SPIE, Security and Watermarking of Multimedia Contents II, San Jose, 2000. 140--151. Google Scholar

[2] Chang C C, Tsai P, Lin C C. SVD-based digital image watermarking scheme. Pattern Recognition Lett, 2005, 26: 1577-1586 CrossRef Google Scholar

[3] Li C, Huang J W. A Semi-Fragile Image Watermarking Resisting to JPEG. J Software, 2006, 17: 315-324 CrossRef Google Scholar

[4] Gong D F, Liu F L, Luo X Y. An image edge based robust watermarking algorithm. Sci Sin Inform, 2013, 43: 1410--1430. Google Scholar

[5] Ma L, Zhang X H. Characteristics of color images with watermark based on the relationship between non-void subspaces of inner space. Chinese J Comput, 2017, 40: 1203--1217. Google Scholar

[6] Maeno K, Sun Q B, Chang S F. New semi-fragile image authentication watermarking techniques using random bias and nonuniform quantization. IEEE Trans Multimedia, 2006, 8: 32-45 CrossRef Google Scholar

[7] Patra J C, Phua J E, Bornand C. A novel DCT domain CRT-based watermarking scheme for image authentication surviving JPEG compression. Digital Signal Processing, 2010, 20: 1597-1611 CrossRef Google Scholar

[8] Zhang X, Cui L, Shao L. A fast semi-fragile watermarking scheme based on quantizing the weighted mean of integer Haar wavelet coefficients. In: Proceedings of Symposium on Photonics and Optoelectronics, Shanghai, 2012. Google Scholar

[9] Preda R O. Semi-fragile watermarking for image authentication with sensitive tamper localization in the wavelet domain. Measurement, 2013, 46: 367-373 CrossRef Google Scholar

[10] Das C, Panigrahi S, Sharma V K. A novel blind robust image watermarking in DCT domain using inter-block coefficient correlation. AEU - Int J Electron Commun, 2014, 68: 244-253 CrossRef Google Scholar

[11] Al-Haj A, Amer A. Secured telemedicine using region-based watermarking with tamper localization.. J Digit Imag, 2014, 27: 737-750 CrossRef PubMed Google Scholar

[12] Guo J, Zheng P, Huang J. Secure watermarking scheme against watermark attacks in the encrypted domain. J Visual Communication Image Representation, 2015, 30: 125-135 CrossRef Google Scholar

[13] Qi X, Xin X. A singular-value-based semi-fragile watermarking scheme for image content authentication with tamper localization. J Visual Communication Image Representation, 2015, 30: 312-327 CrossRef Google Scholar

[14] Chen S T, Huang H N, Kung W M. Optimization-based image watermarking with integrated quantization embedding in the wavelet-domain. Multimed Tools Appl, 2016, 75: 5493-5511 CrossRef Google Scholar

[15] Imran S, Kumar D P, Tetsuya S. A semi-fragile watermarking method using slant transform and LU decomposition for image authentication. In: Proceedings of International Conference on Electrical, Computer and Communication Engineering, Cox's Bazar, 2017. 881--885. Google Scholar

[16] Wang Y, Pearmain A. Blind image data hiding based on self reference. Pattern Recognition Lett, 2004, 25: 1681-1689 CrossRef Google Scholar

[17] Hsu L Y, Hu H T. Robust blind image watermarking using crisscross inter-block prediction in the DCT domain. J Visual Communication Image Representation, 2017, 46: 33-47 CrossRef Google Scholar

[18] Parah S A, Sheikh J A, Loan N A. Utilizing neighborhood coefficient correlation: a new image watermarking technique robust to singular and hybrid attacks. Multidim Syst Sign Process, 2018, 29: 1095-1117 CrossRef Google Scholar

[19] Ali M, Ahn C W. An optimal image watermarking approach through cuckoo search algorithm in wavelet domain. Int J Syst Assur Eng Manag, 2018, 9: 602-611 CrossRef Google Scholar

[20] Verma V S, Jha R K, Ojha A. Digital watermark extraction using support vector machine with principal component analysis based feature reduction. J Visual Communication Image Representation, 2015, 31: 75-85 CrossRef Google Scholar

[21] Mehta R, Rajpal N, Vishwakarma V P. A robust and efficient image watermarking scheme based on Lagrangian SVR and lifting wavelet transform. Int J Mach Learn Cyber, 2017, 8: 379-395 CrossRef Google Scholar

[22] Holliman M, Memon N. Counterfeiting attacks on oblivious block-wise independent invisible watermarking schemes. IEEE Trans Image Process, 2000, 9: 432-441 CrossRef PubMed ADS Google Scholar

[23] Goljan M. Cryptanalysis of the Yeung-Mintzer fragile watermarking technique. J Electron Imag, 2002, 11: 262-274 CrossRef ADS Google Scholar

[24] Wang Z, Bovik A C, Sheikh H R. Image Quality Assessment: From Error Visibility to Structural Similarity. IEEE Trans Image Process, 2004, 13: 600-612 CrossRef Google Scholar

• Figure 1

(Color online) Watermarked vector ${{\boldsymbol~A}_w}$ formed through rotating the original vector (${\boldsymbol~A}$) by $\theta$. (a) Counterclockwise; (b) clockwise

• Figure 2

(Color online) The relationship between the dithered angle ($\theta_d$) of vector ${\boldsymbol~A}$ and the cumulative probability under perturbation for (a) test schemes 1$\sim$4, respectively, and (b) test scheme 5 and schemes 1$\sim$4

• Figure 3

(Color online) The probability of feature stability for individual attack in double compression environment for (a) test schemes 1$\sim$4, respectively, and (b) test scheme 5 and schemes 1$\sim$4

• Figure 4

(Color online) Function relationship between tampering proportion ($p_t$) and robustness of the tampering matrix ($P_{\rm~t\_Tr}$), false positive probability for localization ($P_{\rm~tl\_fp}$), false negative probability for localization ($P_{\rm~tl\_fn}$) and tampering localization accuracy ($P_{\rm~tl\_d}$), respectively, under double

• Figure 5

(Color online) Robustness of single compression. (a) JPEG; (b) JPEG2000

• Figure 6

(Color online) Robustness of different image attacks during double compression. (a)$\sim$(d) correspond to the double compression schemes 1$\sim$4, respectively, (e) is the uncompressed case, and (f) is the average of (a)$\sim$(d)

• Figure 7

(Color online) Detection mask images after content authentication for the content-preserving image attack in the double compression environment. Subtitles show the detailed process of signal processing; “DCi" stands for ith-type double compression scheme; “Ai" is the number of attacks in Subsection 2.4. (a) Jpeg(75); (b) Jp2k(8); (c) DC1; (d) DC2; (e) DC3; (f) DC4; (g) DC5; (h) DC1$\leftarrow$A3; (i) DC2$\leftarrow$A6; (j) DC3$\leftarrow$A8; (k) DC4$\leftarrow$A11; (l) DC5$\leftarrow$A2

• Figure 8

(Color online) Detection mask images after content authentication for malicious content tampering attack in the double compression environment. Subtitles show the detailed process of signal processing; “DCi" stands for ith-type double compression scheme; “Ai" is the number of attacks in Subsection 2.4; “T" and “Tc" respectively represent content tampering and collage attack. (a) Jpeg(75)TcA1; (b) Jp2k(8)Tc; (c) DC1$\leftarrow$Tc; (d) DC2$\leftarrow$Tc; (e) DC3$\leftarrow$T; (f) DC4$\leftarrow$T; (g) DC5$\leftarrow$T; (h) DC1$\leftarrow$TA3; (i) DC2$\leftarrow$TA6; (j) DC3$\leftarrow$TA8; (k) DC4$\leftarrow$TA11; (l) DC5$\leftarrow$TA2

• Table 1   Comparison of PSNR values
 Lena Sailboat Baboon Peppers Man Average (128 images) Ours 40.71 39.03 42.46 39.80 39.41 41.07 Qi 40.28 40.30 40.35 40.33 40.32 40.30 Wang 34.13 31.40 31.58 34.84 32.53 34.00 Hsu 38.69 37.38 37.78 38.45 37.80 38.19

a) The bold font represents the maximum of the corresponding column.

• Table 2   Comparison of SSIM values
 Lena Sailboat Baboon Peppers Man Average (128 images) Ours 0.9861 0.9885 0.9960 0.9837 0.9862 0.9897 Qi 0.9598 0.9707 0.9866 0.9609 0.9704 0.9701 Wang 0.9620 0.9527 0.9659 0.9653 0.9480 0.9627 Hsu 0.9563 0.9629 0.9826 0.9554 0.9631 0.9650

a) The bold font represents the maximum of the corresponding column.

• Table 3   Tampering detection results and comparison of theoretical and experimental data under double compression
 Sub-graph $p_t$ (%) $P_{\rm~t\_Tr}$ $P_{\rm~tl\_fp}$ $P_{\rm~tl\_fn}$ BER TCR Auth. Theory Experiment Theory Experiment Theory Experiment (a) 1.000 0.058 0.058 0.000 Y (b) 0.982 0.127 0.127 0.139 Y (c) 0.975 0.195 0.195 0.130 Y (d) 0.971 0.170 0.170 0.172 Y (e) 0.998 0.037 0.037 0.053 Y (I) (f) 0.952 0.180 0.182 0.266 Y? Figure 7 (g) 0.0 0.977 0.990 0.023 0.052 0.0 0.0 0.052 0.189 Y (h) 0.987 0.169 0.169 0.075 Y (i) 0.913 0.227 0.227 0.313 Y? (j) 1.000 0.046 0.046 0.000 Y (k) 0.969 0.175 0.175 0.179 Y (l) 0.957 0.218 0.218 0.197 Y Avg($\cdot$) $|\Delta|=0.002$ $|\Delta|=0.119$ $|\Delta|=0.0$ 0.142 Sub-graph $p_t$ (%) $P_{\rm~t\_Tr}$ $P_{\rm~tl\_fp}$ $P_{\rm~tl\_fn}$ BER TCR Auth. Theory Experiment Theory Experiment Theory Experiment (a) 10.55 0.939 0.917 0.036 0.031 0.051 0.054 0.226 0.368 N (b) 16.41 0.913 0.894 0.043 0.050 0.076 0.107 0.259 0.411 N (c) 20.31 0.890 0.894 0.049 0.041 0.096 0.138 0.283 0.376 N (d) 24.90 0.857 0.868 0.057 0.067 0.121 0.185 0.296 0.446 N (e) 11.82 0.935 0.943 0.037 0.008 0.056 0.069 0.146 0.389 N (II) (f) 17.19 0.908 0.870 0.044 0.062 0.081 0.104 0.278 0.467 N Figure 8 (g) 22.85 0.870 0.885 0.054 0.010 0.111 0.123 0.243 0.474 N (h) 27.34 0.843 0.837 0.060 0.043 0.131 0.153 0.333 0.490 N (i) 7.91 0.953 0.899 0.032 0.075 0.035 0.054 0.301 0.334 N (j) 29.30 0.829 0.891 0.063 0.003 0.141 0.187 0.268 0.409 N (k) 40.43 0.749 0.758 0.078 0.060 0.197 0.222 0.380 0.638 N (l) 50.78 0.671 0.649 0.090 0.069 0.253 0.227 0.442 0.792 NW Avg($\cdot$) $|\Delta|=0.023$ $|\Delta|=0.023$ $|\Delta|=0.027$
• 表 4表 1

Citations

• #### 0

Altmetric

Copyright 2020 Science China Press Co., Ltd. 《中国科学》杂志社有限责任公司 版权所有


text:
Why is manganese(II) coloured although the transition should be spin-forbidden?

In every basic coordination chemistry class, at some early point the crystal field theory and LFSE will be taught, explaining that there will be an energy difference between d-orbitals (typically named $\mathrm{t_{2g}}$ and $\mathrm{e_g}$) and that electrons can be excited from one to the other to give colour. The next statement is typically that transitions have to be spin-allowed (the higher-energy orbitals need to be able to accomodate for the lower electron without spin-flipping being required) and Laporte-allowed (in a nutshell: no d→d transition in octahedral complexes).

The syllabus then goes on to say that the Laporte rule is not that strict a rule (typically, all near-octahedral complexes of 3d-elements which have spin-allowed transitions have a well visible colour; e.g. iron(II), chromium(III), manganese(III), nickel(II), copper(II) etc.) since small random vibrations can destroy the symmetry making the electron transition allowed and an incoming photon can excite the electron as needed. They also remark that the colour is much less intense than for Laporte-allowed transitions — hence why the colour of the $\mathrm{d^7}$ ion $\ce{[CoCl4]^2-}$ is much more intense than that of $\ce{[Co(H2O)6]^2+}$.

A very recent question on the colour of manganese(II) sulphate then had me confused. Manganese(II) salts are typically of a very pale pink. The difference to white can be observed when they are put before a white background. But manganese(II) is a $\mathrm{d^5}$ ion, so the transition is spin-forbidden.

I think that being spin-forbidden explains why manganese(II) complexes are so weakly coloured, but what is the mechanism that relieves the spin constraint and how does it work in practice?

• I have some bits and pieces about this written up in this answer (especially in the second part of it) although in a different context. – Philipp May 1 '16 at 1:10

Selection rules

The intensity of the transition from a state $\mathrm{i}$ to a state $\mathrm{f}$ is governed by the transition dipole moment $\mu_{\mathrm{fi}}$ (strictly, it is proportional to $|\mu_{\mathrm{fi}}|^2$):

$$\iint \Psi_\mathrm{f}^*\hat{\mu}\Psi_\mathrm{i}\,\mathrm{d}\tau \,\mathrm{d}\omega \tag{1}$$

where $\mathrm{d}\tau$ is the usual volume element (representing integration over spatial coordinates) and $\omega$ is an additional spin coordinate (since the states $\Psi$ also specify the spin). For an electronic transition, $\hat{\mu}$ is the electric dipole moment operator, and only depends on the spatial coordinates $x$, $y$, and $z$. (I've also ignored the vibrational component i.e. the Franck-Condon factor since it's not really relevant for the purposes of this answer.)

Normally in determining selection rules, we are more interested in whether the integral in $(1)$ is zero or not. If the integral is zero, the transition is said to be forbidden. The usual way to determine whether it is zero is by using symmetry: if the integrand is antisymmetric under any symmetry operation, then the integral is zero. (It is exactly analogous to the integral of an odd function $f(x)$ over a symmetric region; for example, $\int_{-a}^a \sin x \,\mathrm{d}x = 0$.)

If we assume that both the initial and final states can be separated into spin and spatial components, then we get:

$$\iint (\psi_\mathrm{f}^\mathrm{spin}\psi_\mathrm{f}^\mathrm{space})\hat{\mu}(\psi_\mathrm{i}^\mathrm{spin}\psi_\mathrm{i}^\mathrm{space})\,\mathrm{d}\tau\,\mathrm{d}\omega$$

(I also dropped the complex conjugates - they aren't important.) Since $\hat{\mu}$ does not operate on the spin coordinate, we can separate this multiple integral:

$$\left(\int \psi_\mathrm{f}^\mathrm{spin} \psi_\mathrm{i}^\mathrm{spin} \,\mathrm{d}\omega\right) \left(\int \psi_\mathrm{f}^\mathrm{space} \hat{\mu} \psi_\mathrm{i}^\mathrm{space} \,\mathrm{d}\tau\right)$$

The first integral is nonzero if $\Delta S = 0$. The quantum number $S$ is related to the eigenvalue of a hermitian operator (the actual eigenvalue is $S(S+1)\hbar^2$), and eigenfunctions of a hermitian operator corresponding to different eigenvalues are necessarily orthogonal, making the integral identically zero if $\Delta S \neq 0$. This is the spin selection rule.

The second integral is nonzero if $\psi_\mathrm{f}^\mathrm{space}$ and $\psi_\mathrm{i}^\mathrm{space}$ have a different character under the inversion operation, i.e. one is gerade and another is ungerade. This is because $\hat{\mu}$ itself is ungerade, so for the integrand to be gerade, we need $\psi_\mathrm{f}^\mathrm{space}\psi_\mathrm{i}^\mathrm{space}$ to be ungerade ($\mathrm{u \otimes u = g}$). In turn this means that $\psi_\mathrm{f}^\mathrm{space}$ and $\psi_\mathrm{i}^\mathrm{space}$ must have different symmetry. This is the Laporte selection rule. It can be relaxed when the geometry is not perfectly octahedral, or tetrahedral for that matter, since inversion is no longer a symmetry operation. However, I digress.

Spin-orbit coupling

The problems arise because the total states are not separable into spin and spatial components. This occurs because both electron spin, as well as orbital angular momentum, are both angular momenta. They therefore interact with each other, and this is the spin-orbit coupling spoken of. Mathematically, it is treated as a perturbation to the Hamiltonian; the result is that the transition dipole moment cannot be separated into two integrals and this leads to the invalidation of the selection rules.

When considering spin-orbit coupling, the term symbols such as $^3\!F$, $^1\!D$, which imply a simultaneous eigenstate of the total spin operator $\hat{S}$ and the total orbital angular momentum operator $\hat{L}$ are no longer meaningful since both angular momenta are coupled to each other (by $\vec{j} = \vec{l} + \vec{s}$) and behave as one "combined" source of angular momentum. The quantities $S$ and $L$ are called "nearly good" quantum numbers: they are no longer strictly conserved, but the "contamination" of different quantum numbers is relatively small. Mathematically we could express this as an admixture of different spin states into the original wavefunction:

$$^3\!F_J = c(^3\!F) + c_1(^1\!F) + \cdots$$

where $^3\!F_J$ denotes a state with spin-orbit coupling (not an eigenstate of $\hat{S}$ and $\hat{L}$), and $(^3\!F)$ denotes the original $^3\!F$ state before spin-orbit coupling is considered (which is a true eigenstate of $\hat{S}$ and $\hat{L}$). If $c_i \neq 0$, then one obtains a non-zero transition dipole moment between "states" of nominally different $S$.

The fact that the selection rules are still a thing is solely because spin-orbit coupling is a small effect ($c_i$ close to 0), so the compounds behave "almost" as if they obey the selection rules. More properly, the transition dipole moment between states of different $S$ is small, and therefore the d-d bands in $\mathrm{d^5}$ spectra are extremely weak. The magnitude of spin-orbit coupling, however, does vary with $Z^4$ ($Z$ being the atomic number). Therefore, while spin-orbit coupling is rather small in the 3d series, it becomes an extremely important mechanism by which the spin selection rule in the the 4d and 5d congeners, as well as the f-block elements, is relaxed.

• That was a very enlightening response. I wasn't sure exactly where those selection rules actually came from. Can you give a reference or perhaps explain where the $Z^4$ dependence comes out? – jheindel Nov 22 '16 at 8:43
• I was expecting some final closing remark that would refer back to the question stated in OP. For me this is still not a finished answer. – tschoppi Nov 22 '16 at 12:58
• The only question asked was what is the mechanism that allows systems to circumvent our "selection rules". I think orthocresol answered that question quite well. – jheindel Nov 22 '16 at 19:40
• @jheindel This is actually a very good question which would do well on main site! I looked up a couple of sources, and the crux of the matter is that the magnetic field $\mathbf{B}$ generated by the proton (in the rest frame of the electron), given by the Biot-Savart law, is proportional to the charge $Ze$. Spin-orbit coupling arises from the interaction of the electron's magnetic moment with this field, which is given by $-\mathbf{\mu}\cdot \mathbf{B}$. This incorporates the previous $Z$-dependence and can be shown to be proportional to $Z/r^3$. [...] – orthocresol Nov 22 '16 at 20:38
• [...] In a hydrogenic atom, the expectation value of $1/r^3$ itself has a dependence on $Z^3$, which gives rise to the combined dependence on $Z^4$. The exact form of this spin-orbit coupling constant only strictly applies to hydrogenic atoms but the general trend is widely applicable to all the elements. For a full proof (!) see Townsend's A Modern Approach to Quantum Mechanics, 2nd ed., p 400 onwards. Atkins's Molecular Quantum Mechanics, 6th ed., pp 215-7 also discusses it, but not quite as well. – orthocresol Nov 22 '16 at 20:43

To continue with the answer from @orthocresol. The breaking of formally ‘ forbidden’ transitions is very common indeed, possibly the best example is in the observation of phosphorescence from almost any aromatic or dye molecule.

Absorption in these molecules is generally a spin allowed transition, (single to singlet), however, triplet states, from which phosphorescence is observed, can only be produced by a spin forbidden process, (singlet to triplet inter system crossing) and just as in the case of phosphorescence this is a spin forbidden process which becomes allowed due to spin-orbit coupling. In the case of the manganese ions and in other transition metals the forbidden transition is in absorption.

The treatment is the same in all cases and starts with allowing the two states to interact via some operator. In the case of photons (absorption / emission) it is an electric dipole operator, i.e. dipole moment operator. Labelling the operator as $H_0$, the rate of the transition from one state to another is

$$W_{if} \approx \int \Psi_f H_0 \Psi_i d\tau = \int \psi_f\sigma_f H_0 \psi_i \sigma_i d\tau$$

where $\sigma$ represents the spin wavefunction and $\psi$ the spatial one for initial (i) and final (f) states. As the operator does not depend on spin the integral can be separated into two parts as described in other answers $$\int \psi_f\sigma_f H_0 \psi_i \sigma_i d\tau = \int \psi_f H_0 \psi _i d\tau\int\sigma _f \sigma_i d\tau = 0$$ and the final result is zero because the spin wavefunctions are orthogonal making the last integral is zero between states with different spin, i.e. $\Delta S \ne\ 0$. This is what we expect from states between which a spin changing transition is attempted.

To progress we next suppose that it is possible to mix a little of the final state wavefunction into the initial one and thus generate a new state with wavefunction $\psi_m$ then $$\psi_m=\psi_f + b\psi_i$$ where b is some mixing constant we can determine later on.

Now replacing the wavefunction for the final state ($\psi _f$) with this one ($\psi_m$) and repeating the calculation above and separating out terms as before, as orthogonal spin terms are zero, also as before, this leads to

$$W_{if} \approx \int \psi_m H_0 \psi_i d\tau = b\int \psi_i \sigma_i H_0 \psi_i \sigma _i d\tau = b\int \psi_i H_0 \psi_i d\tau\int\sigma _i\sigma_id\tau$$
in which case neither of the integrals is zero but evaluates to a number. This equation shows that there is a finite rate for the process with operator $H_0$, for example the electric dipole operator.

The constant b remains to be understood and it is derived using perturbation theory and turns out to have the form $$b=\frac{\int\psi _f \Omega \psi _i d\tau}{E_f-E_i}$$ where $\Omega$ is the spin-orbit operator. Thus the rate for the process finally becomes

$$w \approx \frac{\int\psi _f \Omega \psi _i d\tau}{E_f-E_i}$$

and we can use symmetry to decide if this integral is zero or is finite.

In the case in point, the spin-orbit coupling the operator depends on the mass of the atoms involved and the product of orbital and spin angular momentum. This is why spin-orbit coupling is also called a ‘heavy atom effect’ when applied to emission processes. In general, however, any effects are enhanced in atoms with larger atomic numbers compared to those earlier in the periodic table, simply because $\Omega$ is larger in value. It is clear also that if the two states are close in energy the effect is also increased relative to atoms /molecules where the states are further separated in energy.

The second mechanism by which a 'forbidden' transition can become somewhat allowed is by 'Herzberg-Teller' coupling also called 'intensity stealing'. The standard example is in benzene where the $S_0 \rightarrow S_1$ transition is forbidden but is observed as a weak absorbtion due to a vibrational normal mode altering the symmetry of the electronic excited state so that it now has the correct symmetry to couple with a second and higher allowed electronic transition, and thus 'steal' intensity. The formalism is essentially the same as described. The terms that are important are thus (a) the presence of a normal mode of the correct symmetry (b) a second and allowed transition, preferably close by in energy and (c) an electronic coupling term between the allowed transition and the forbidden transition whose symmetry is altered by the vibrational normal mode.

One mechanism is spin-orbit coupling. There is an interaction between the magnetic field created by the orbital motion of the electron with the spin of the electron. This allows transitions where the spin of the electron changes in tandem with its components of angular momentum, whereas just changing the spin by itself would be forbidden. But the spin-orbit interaction is weak, so transitions involving it give weaker coloration than simple d-d orbital transitions.


text:
# Proving formula for product of first n odd numbers

I have this formula which seems to work for the product of the first n odd numbers (I have tested it for all numbers from $1$ to $100$):

$$\prod_{i = 1}^{n} (2i - 1) = \frac{(2n)!}{2^{n} n!}$$

How can I prove that it holds (or find a counter-example)?

• This sounds like homework. Did you try induction? – cardinal Oct 2 '11 at 1:20
• @Pedro: Do you mean $$\left(\prod_{i=1}^n2i\right) - 1$$ (which is what you wrote) or $$\prod_{i=1}^n(2i-1)$$(which is what your title suggests)? – Arturo Magidin Oct 2 '11 at 1:25
• It is not, I stumbled upon this formula by accident and was wondering. The proof was fairly simple, I suppose I should have thought more about it – Pedro Oct 2 '11 at 1:28
• @ArturoMagidin: I meant the latter, sorry for the ambiguity – Pedro Oct 2 '11 at 1:30
• As a note: what you have is sometimes termed as the "double factorial". – J. M. is a poor mathematician Oct 2 '11 at 9:33

The idea is to "complete the factorials":

$$1\cdot 3 \cdot 5 \cdots (2n-1) = \frac{ 1 \cdot 2 \cdot 3 \cdot 4 \cdots (2n-1)\cdot (2n) }{2\cdot 4 \cdot 6 \cdots (2n)}$$

Now take out the factor of $2$ from each term in the denominator:

$$= \frac{ (2n)! }{2^n \left( 1\cdot 2 \cdot 3 \cdots n \right)} = \frac{(2n)!}{2^n n!}$$

A mathematician may object that there is a small gray area about what exactly happens between those ellipses, so for a completely rigorous proof one would take my post and incorporate it into a proof by induction.

For the induction argument, \begin{align*} \prod_{i=1}^{n+1}(2i-1)&=\left(\prod_{i=1}^n(2i-1)\right)(2n+1)\\ &= \frac{(2n)!(2n+1)}{2^n n!} \end{align*} by the induction hypothesis. Now multiply that last fraction by a carefully chosen expression of the form $\dfrac{a}a$ to get the desired result.

$$\Pi_{k=1}^n(2k-1) = \Pi_{k=1}^n(2k-1) \frac{\Pi_{k=1}^n(2k)}{\Pi_{k=1}^n(2k)} =\frac{\Pi_{k=1}^{2n}k}{2^n\Pi_{k=1}^n k} = \frac{(2n)!}{2^n n!}$$

• This is the same answer as the one given four years ago by Ragib, only with fewer explanations. It's good that you want to help by answering questions, but maybe you could consider answering some that haven't been answered yet? – mrf Aug 31 '15 at 20:42


text:
Electric Field Video Lessons

Concept

# Problem: In the figure the four particles are fixed in place and have charges q1  = q2 = 3e, q3 = 2e and q4 = -9e. Distance d = 2.17 µm. What is the magnitude of net electric field at point P due to the particles?

###### FREE Expert Solution

In this problem, we are dealing with electric fields due to point charges.

Electric field:

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{\mathbf{k}\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}}$

80% (175 ratings)
###### Problem Details

In the figure the four particles are fixed in place and have charges q1  = q2 = 3e, q3 = 2e and q4 = -9e. Distance d = 2.17 µm. What is the magnitude of net electric field at point P due to the particles?


text:
# Orbaliciousness

I have a magic ball that predicts the future! Well… actually… it doesn’t predict the future, it’s just a regular magic ball. Well… actually… it’s not magic, it works with physics! :) Click through to learn more about glass orbs and see some cool photos.

I was on Google+ one day a while ago and I saw some cool photos people had taken through glass orbs, they were tagged with #oculusimages and #orbaliciousness and I thought to myself “That’s a converging lens, it’s physics! how cool is that?!” :) The fact that the orb acts as a lens has two major implications:

1) The image through the orb will be inverted.
2) The focal plane of the orb (where you want to focus to take the picture) will be closer to the camera than the orb itself.

The fact that the image is inverted makes for funky photos, but the fact that the focal plane of the orb is closer to the camera than the orb itself means you can run into some trouble. If you’re getting too close to the orb but don’t have a macro lens your camera won’t be able to focus properly! But let’s not panic, I normally use my 50mm or my 18-55mm at 55mm (on a cropped sensor) and all is good :) By placing the orb’s focal plane at my lens’ minimum focusing distance I am able to fill the frame with my orb. I really like when the orb takes up most of the frame because then it feels like you’re capturing your own little privateworld inside it. Also, since the image inside the orb is going to be up-side-down I normally rotate the image in Lightroom afterwards to make it right-side-up.

So how much closer is the orb’s focal plane exactly? Let’s do a little bit of physics (easy peasy, I promise hehe). The focal length of a lens in air can be calculated with the following equation, called the “Lensmaker formula”:

$\dfrac{1}{f} = (n-1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} + \dfrac{(n-1) \cdot d}{n \cdot R_1 \cdot R_2} \right)$

Where f is the focal length of the lens, n is the refractive index of the lens, R1 and R2 are the radii of the surffaces of the lens and d is the thickness of the lens, as shown in the following figure:

In this case our lens is a glass ball, which means that both radii are the same and the thickness of the lens is the orb’s diameter.

$R_1 = R_2 = d/2 = R$

If we use this to simplify our equation we obtain:

$\dfrac{1}{f} = (n-1) \left( \dfrac{1}{R} - \dfrac{1}{R} + \dfrac{(n-1) \cdot 2R}{n \cdot R \cdot R} \right) \rightarrow$

$\dfrac{1}{f} = (n-1) \left( \dfrac{(n-1) \cdot 2R}{n \cdot R^2} \right) \rightarrow$

$\dfrac{1}{f} = \dfrac{(n-1)^2}{n} \cdot\ dfrac{2}{R} \rightarrow$

$f = \dfrac{n}{2 \cdot (n-1)^2} \cdot R$

For an orb made out of glass, with refractive index n = 1.5 we have that:

$f = \dfrac{n}{2 \cdot (n-1)^2} \cdot R = \dfrac{1.5}{2 \cdot(1.5 - 1)^2} \cdot R = 3 \cdot R \rightarrow$

$f = 3R$

So in order to find out the focal length of your orb you just need to multiply it’s radius by 3, easy!! The orb I use the most is 60mm in diameter, which means it’s focal lenth is 90mm. I also, have another smaller orb which is 40mm in diameter, so it’s focal length is 60 mm.

$f = 3 \cdot R = 3 \cdot 30mm = 90mm$

$f = 3 \cdot R = 3 \cdot 20mm = 60mm$

As you can see, orbs of different sizes have different focal lenghts, which means that if I want to shoot the same image through both orbs at the same time I am going to have to place the smaller one slightly forward so that their focal plane matches, like shown below:

Here is an example of a photo with both my orbs, which I like to tag with #anasballs hehehehe :)

Have you taken any photos using an orb? I’d love to see them :) And to wrap this up, here is another of my favourite orbaliciousness photographs, if you’d like to see more (or even order some prints) check out my portfolio here.

Thank you for reading! xx Ana :)