text:
# 10. Miscelleaneous problems

### 1007. Minimum Domino Rotations For Equal Row

In a row of dominoes, tops[i] and bottoms[i] represent the top and bottom halves of the ith domino. (A domino is a tile with two numbers from 1 to 6 - one on each half of the tile.)

We may rotate the ith domino, so that tops[i] and bottoms[i] swap values.

Return the minimum number of rotations so that all the values in tops are the same, or all the values in bottoms are the same.

If it cannot be done, return -1.

Example:

Input: tops = [2,1,2,4,2,2], bottoms = [5,2,6,2,3,2] > Output: 2 Explanation: The first figure represents the dominoes as given by tops and bottoms: before we do any rotations.
If we rotate the second and fourth dominoes, we can make every value in the top row equal to 2, as indicated by the second figure.

code

from collections import Counter

class Solution:
def minDominoRotations(self, tops: List[int], bottoms: List[int]) -> int:

def minRotationWithRef(arr: List[int], swap:List[int]) -> int:
"""Returns a minimum rotations to normalize arr"""

arrCount = Counter(tops)
arrCount = dict(sorted(arrCount.items(), key=lambda x:x[1], reverse=True))

minTurns = float("inf")

for ref, v in arrCount.items():

turns = 0

for i in range(len(arr)):

if arr[i] == ref:
continue

if arr[i] != ref and swap[i] != ref:
break

turns += 1

if i == len(arr)-1:
minTurns = min(minTurns, turns)

return minTurns

minRotateTops = minRotationWithRef(tops, bottoms)
minRotateBottom = minRotationWithRef(bottoms, tops)

ans = min(minRotateTops, minRotateBottom)

if ans == float("inf"):
return -1

return ans

format: plain


text:
# Sodium azide to Nitrogen ratio in airbags

I've been given this question and I don't know how to solve it.

Airbags are designed to protect the occupants of a car during a collision. They use the decomposition of sodium azide, $\ce{NaN3}$. Three chemical reactions occur rapidly after impact and produce nitrogen gas ($\ce{N2}$) to fill the airbag.

The equations summarise the overall process that takes place.

1. $\ce{2 NaN3 -> 2 Na + 3 N_2}$
2. $\ce{10 Na + 2 KNO3 -> K2O + 5 Na2O + N2}$
3. $\ce{K2O + Na2O + 2 SiO2 -> K2O3Si + Na2O3Si}$

What is the ratio between the number of sodium azide ($\ce{NaN3}$) molecules used and the number of nitrogen ($\ce{N2}$) molecules produced in the overall process?

I initially just used the first equation to get a ratio of 2:3, but obviously that's not right. I have no clue where to start. How do you do this? (How this can be considered a Year 10 question baffles me)

• It baffles me also. You need also to consider the second equation, which produces N2 gas as well. The first equation generates Na, and this is consumed in the second equation. – long Apr 25 '16 at 4:20

Assuming you have an excess of $$\ce{KNO3}$$, the ratio of $$\ce{NaN3}$$ consumed to $$\ce{N2}$$ produced would be 5:8.
Let's renormalize the first two equations, dividing the first through by 2, and the second through by 10. By doing this we can see, directly, that in the overall reaction, 3/2 + 1/10 = 8/5 = 1.6 molecules of $$\ce{N_2}$$ are produced for every molecule of $$\ce{NaN3}$$ consumed. If you prefer an integer ratio, it's 1:(8/5) = 5:8.
1. $$\ce{NaN3 -> Na + 3/2 N_2}$$
2. $$\ce{Na + 1/5 KNO3 -> 1/10K2O + 1/2 Na2O + 1/10 N2}$$
The initial ratio 2:3 is correct. $$\ce{2NaN3 = 2 Na}$$ molecules as well as $$\ce{6 N}$$ molecules. The products of the reaction, $$\ce{2Na}$$ is equal to two $$\ce{Na}$$ molecules and $$\ce{3N2}$$ is equal to $$\ce{6 N}$$ molecules. So as you can see the equation is perfectly balanced. Hope this helps.


text:
# Could the Sun's bending of light be measured on photographic plates before Einstein's prediction?

We are searching data for your request:

Forums and discussions:
Manuals and reference books:
Data from registers:
Wait the end of the search in all databases.
Upon completion, a link will appear to access the found materials.

It was predicted from Newtonian physics already in the 18th century that gravity should bend light, but not by as much as general relativity predicts. This was first confirmed during a solar eclipse in 1919. Surely, solar eclipses were imaged long before that. Didn't anyone notice that stars near the Sun were displaced? Hadn't anyone thought of testing the newtonian prediction of bent light?

I suggest reading the paper on the 1919 expedition to get a clearer picture of why they did it at that time and why it hadn't been done before. From reading chapter 2 I think the main reason was the astrophotography equipment required for the experiment and the alignment of bright enough stars close to the sun to observe the effect.

Of course before Einsteins prediction in 1911 no one had any specific reason to observe the stars close to a solar eclipse before and during totality, it might seem obvious now but that's hindsight for you.

## Einstein, Eddington and the 1919 eclipse

Peter Coles is Professor of Theoretical Physics at Maynooth University in Ireland.

You can also search for this author in PubMed Google Scholar

No Shadow of a Doubt: The 1919 Eclipse That Confirmed Einstein’s Theory of Relativity Daniel Kennefick Princeton University Press (2019)

Gravity’s Century: From Einstein’s Eclipse to Images of Black Holes Ron Cowen Harvard University Press (2019)

Einstein’s War: How Relativity Triumphed Amid the Vicious Nationalism of World War I Matthew Stanley Dutton (2019)

In 1916, Albert Einstein published his general theory of relativity in full mathematical detail. That opened the window on a radically new framework for physics, abolishing established notions of space and time and replacing Newton’s formulation of the laws of gravity. Einstein’s revolution was to change the course of science but in the years immediately after publication, there was no definitive observational evidence that his theory was correct.

Enter Arthur Stanley Eddington. An astronomer interested in Einstein’s theory because of its wide-ranging implications for astrophysics and cosmology, Eddington took on the task of proving it. By harnessing a total solar eclipse, he argued that the deflection, or bending, of light by the Sun’s gravity could be measured. This was a critical test, because Einstein’s theory predicted a deflection precisely twice the value obtained using Isaac Newton’s law of universal gravitation. The needed eclipse came 100 years ago, in 1919. Eddington is now forever associated with two expeditions to view it: from Sobral in northern Brazil, and the island of Príncipe off the coast of West Africa. Those momentous ventures form the kernel of three books commemorating the centenary: No Shadow of a Doubt by physicist Daniel Kennefick, Gravity’s Century by science journalist Ron Cowen, and science historian Matthew Stanley’s Einstein’s War.

Einstein’s theory, eight years in the making, sprang from insights he had developed after he published his theory of special relativity in 1905. One of the effects predicted by the new theory was that light rays passing close to a massive body, such as a star, should be bent by its gravitational field. This effect had been predicted qualitatively using Newton’s theory of gravity. Tantalizingly, Newton himself had written in his 1704 opus Opticks: “Do not Bodies act upon Light at a distance, and by their action bend its Rays…?” But there is no evidence that he calculated the magnitude of the effect (the first full calculation was published by German mathematician Johann Georg von Soldner, in 1804).

Newton’s theory of gravity did not, of course, formulate gravity as a consequence of curved space. That was Einstein’s innovation. And when he calculated the effect, he confirmed that light is deflected (as in the Newtonian theory), but through curved space. It is this curvature that doubles the deflection.

## Staring at the Sun

General relativity abandoned Newton’s idea that gravity is a force pulling objects together. It reimagined gravity as a warping of time and space — a distortion in the fabric of the universe. According to the mathematics of relativity, light traveling through this distortion will change its path, accommodating the universe’s warps and wefts. The more massive an object, the bigger the distortion, and the more its gravity can bend light.

During the decade Einstein spent developing his theory, he realized that the sun was massive enough to make this effect noticeable. As the sun moves in the sky toward a background star, he said, it should bend the star’s light. The star will appear to move.

Of course, testing this prediction wasn’t easy because stars aren’t visible during the day they’re washed out by the sun’s light. And at night, when stars do appear, the sun isn’t around to bend their light. Only when the sun is out but its light blocked could Einstein’s work be checked. That’s why, while working out the kinks in his theory in 1911, he asked astronomers to start looking to the heavens during eclipses.

The first to answer that call was the young German astronomer Erwin Finlay-Freundlich, a tragic figure who dedicated much of his life to proving Einstein’s light-bending right and never succeeded. He started by analyzing photographs of historical eclipses, but the stars weren’t clear enough to test Einstein’s idea. So Freundlich raised money — nearly borrowing a sum from Einstein himself — for a voyage to present-day Ukraine, where an eclipse was expected in 1914. His team arrived, armed with telescopes and cameras and glass photographic plates, just as war broke out. Russian soldiers captured Freundlich and confiscated his instruments.

Freundlich wasn’t the only one watching in 1914. Astronomers from California’s Lick Observatory also tried to photograph the eclipse from near Kiev, but they fared little better. Though spared imprisonment because they were Americans, the scientists were thwarted by nature: Clouds obscured their view.

These failures actually may have been a stroke of luck for Einstein. While reviewing his calculations, he found errors. His predictions about how much the stars should move were off. Einstein always was a better physicist than mathematician (though the popular story that he failed math in high school is not true).

By the time Einstein had corrected his mistakes and published the completed theory of general relativity, the Great War was in full swing. And after the war, Germany was in shambles, too wrecked to mount expeditions to the distant parts of the world where an eclipse in 1919 would be visible.

## This Is How, 100 Years Ago, A Solar Eclipse Proved Einstein Right And Newton Wrong

Not only is the Sun's corona visible during a total solar eclipse, but so are, under the right . [+] conditions, stars located a great distance away. With the right observations, one can test the validity of Einstein's General Relativity against the predictions of Newtonian gravity. The total solar eclipse of May 29, 1919, was now a full 100 years ago, and marks perhaps the greatest advance in humanity's scientific history.

Miloslav Druckmuller (Brno U. of Tech.), Peter Aniol, and Vojtech Rusin

On May 29, 1919, the world changed forever. For hundreds of years, Isaac Newton's theory of gravity — the law of universal gravitation — had gone unchallenged, as its predictions matched every observation or measurement that had ever been made. But a mismatch between Newton's predictions for the orbit of Mercury and what astronomers saw surfaced in the mid-19th century, and scientists struggled to explain it.

Perhaps we needed to modify the laws of gravity, after all. Evidence mounted when special relativity came out, demonstrating that there was no such thing as absolute distance. Newton's theory predicted an instantaneous force, again violating relativity. In 1915, Albert Einstein put forth a new alternative theory of gravity: General Relativity. The way to test it against Newton's theory was to wait for a total solar eclipse. 100 years ago today, Einstein was proven right. Here's how.

An event like a total solar eclipse can provide a unique test of Einstein's relativity, as the light . [+] paths of distant astronomical objects will be deflected as they pass near the Sun, but will still be visible to skywatchers on Earth because of the darkened skies as the Sun is blocked out. This method was employed on May 29, 1919, to provide the first confirmation of Einstein's General Relativity.

NASA's scientific visualization studio

Today, Albert Einstein's General theory of Relativity is arguable the most successful theory of all-time. It explains everything from GPS signals to gravitational redshift, from gravitational lensing to merging black holes, and from the timing of pulsars to the orbit of Mercury. The predictions of General Relativity have never once failed.

When this theory was first introduced in 1915, it was attempting to replace Newton's gravitation. Although it could reproduce the earlier Newtonian successes and explain the orbit of Mercury (where Newton could not), the most critical test would come in the form of a new prediction that severely differed from the predictions of the universal law of gravitation. A total solar eclipse would provide a unique and straightforward opportunity.

The curvature of space, as induced by the planets and Sun in our Solar System, must be taken into . [+] account for any observations that a spacecraft or other observatory would make. General Relativity's effects, even the subtle ones, cannot be ignored in applications ranging from space exploration to GPS satellites to a light signal passing near the Sun.

NASA/JPL-Caltech, for the Cassini mission

In Newton's gravity, anything with mass attracts anything else with mass. Even though light is massless, it has an energy, and therefore you can assign an effective mass to it through Einstein's E = mc 2 . (You find that m = E/c 2 .) If you allow a photon to pass near a large mass, you can use this effective mass to predict how much the starlight should bend by, and you get a specific value. Near the limb of the Sun, it's just under 1" (arc-second), or 1/3600th of 1°.

But in Einstein's General Relativity, both space and time are distorted by the presence of mass, whereas in Newton's gravity, only an object's motion through space is affected by the gravitational force. This means that Einstein's theory predicts an extra factor of 2 (actually slightly more, especially as you get closer to the mass in question) over Newton's, or a deflection near the Sun of closer to 2".

An illustration of gravitational lensing showcases how background galaxies — or any light path — is . [+] distorted by the presence of an intervening mass, but it also shows how space itself is bent and distorted by the presence of the foreground mass itself. Before Einstein put forth his theory of General Relativity, he understood that this bending must occur, even though many remained skeptical until (and even after) the solar eclipse of 1919 confirmed his predictions. There is a significant difference between Einstein's and Newton's predictions for the amount of bending that should occur, due to the fact that space and time are both affected by mass in General Relativity.

The history of how Einstein's General Relativity came to be is fascinating, because it's only the fact that Newton's gravitation eventually had problems that motivated Einstein to formulate his new concept.

Newtonian gravity, put forth in 1687, is an extraordinarily simple law: put any masses anywhere in the Universe, a fixed distance apart, and you immediately know the gravitational force between them. This explained everything from the terrestrial motion of cannonballs to the celestial motion of comets, planets, and stars. After 200 years, it had passed every single test that was thrown its way. But one pesky observation threatened to derail everything: the detailed motion of the innermost planet in our Solar System.

After discovering Neptune by examining the orbital anomalies of Uranus, scientist Urbain Le Verrier . [+] turned his attention to the orbital anomalies of Mercury. He proposed an interior planet, Vulcan, as an explanation. Although Vulcan did not exist, it was Le Verrier's calculations that helped lead Einstein to the eventual solution: General Relativity.

Wikimedia Commons user Reyk

Every planet moves in an ellipse around the Sun. However, this ellipse isn't static, returning to the same fixed point in space with every orbit, but rather, it precesses. Precession is like watching that ellipse rotate in space over time, albeit very slowly. Mercury had been observed with incredible precision since Tycho Brahe in the late 1500s, so with 300 years of data, our measurements were extraordinary.

According to Newton's theory, its orbit should have precessed by 5,557"-per-century, due to the precession of Earth's equinoxes and the gravitational effects of all the planets on Mercury's orbit. But observationally, we observed 5,600"-per-century instead. That difference, of 43"-per-century (or just 0.00012°-per-year), had no explanation in Newton's framework. Either there was an extra planet interior to Mercury (which observations ruled out), or something was wrong with our old theory of gravity.

According to two different gravitational theories, when the effects of other planets and the Earth's . [+] motion are subtracted, Newton's predictions are for a red (closed) ellipse, running counter to Einstein's predictions of a blue (precessing) ellipse for Mercury's orbit.

Wikimedia Commons user KSmrq

But Einstein's new theory could explain the mismatch. He spent years developing the framework for General Relativity, where gravitation wasn't caused by masses attracting other masses, but rather by matter and energy curving the very fabric of space, which all objects then move through. When gravitational fields are weak, Newton's law is a very good approximation to what Einstein's theory laid down.

Close to very large masses or at high speeds, however, Einstein's predictions differed from Newton's, predicting exactly that 43"-per-century difference. But the bar to overthrow a scientific theory is higher than that. To supersede the old theory, a new one must do the following:

1. Reproduce all the successes that the old theory enjoyed (otherwise, the old theory is still superior in some way),
2. Succeed in the regime where the old theory could not (otherwise, your new theory doesn't fix the problem with the old one),
3. And to make a new prediction that you can go out and test, distinguishing between the old-and-new ideas (otherwise, you don't have any scientifically predictive power).

That last piece is where the solar eclipse comes in.

During a total eclipse, stars would appear to be in a different position than their actual . [+] locations, due to the bending of light from an intervening mass: the Sun. The magnitude of the deflection would be determined by the strength of the gravitational effects at the locations in space which the light rays passed through.

E. Siegel / Beyond the Galaxy

When the stars appear in the night sky, the starlight travels to our eyes from a different location in the galaxy, many light years away. If Newton was correct, that light should either travel in a completely straight line, undeflected by any masses it passes near (since light is massless), or that it should bend due to the gravitational effects of mass-energy equivalence. (After all, if E = mc 2 , then perhaps you can treat light as have an effective mass of m = E/c 2 .)

But Einstein's theory, particularly if light passes very close by a large mass, offers a prediction different from both of these numbers. That extra factor of 2 (or, rather, 2 and an extra few parts-per-million) is a unique and very specific prediction from Einstein's theory, and one that could be tested by making two observations at different times of the year.

While one could argue that Newtonian gravity either predicted no deflection or deflection of a . [+] specific amount due to the force law and E=mc^2, Einstein's predictions were definitive and different from them both.

NASA / Cosmic Times / Goddard Space Flight Center, Jim Lochner and Barbara Mattson

The largest mass we have close by Earth is the Sun, which normally renders starlight invisible during the day. As starlight passes near the edge of the Sun, according to Einstein, it should travel along that curved space, causing the light-path to appear bent. During a total solar eclipse, however, the Moon passes in front of the Sun, blocking its light and causing the sky to become as dark as night, enabling the stars to be seen during the daytime.

If you previously measured those stellar positions to an accurate enough precision, you could see whether they've shifted or not — and by how much — due to the presence of that large, nearby mass. If you could detect a deflected position at the sub-arc-second level, you could know definitively whether Newton's, Einstein's, or neither prediction was correct.

An early photographic plate of stars (circled) identified during a solar eclipse all the way back in . [+] 1900. While it's remarkable that not only the Sun's corona but also stars can be identified, the precision of the stellar positions is insufficient to test the predictions of General Relativity.

Chabot Space & Science Center

Photographic plates of the Sun during a total solar eclipse had revealed not only details in the Sun's corona before, but the presence and positions of stars during the daytime. However, none of the pre-existing photographs were high enough in quality to determine the deflected positions of nearby stars to the necessary accuracies the deflection of starlight is a very small effect requiring very precise measurements to detect!

After Einstein set forth his general theory of relativity in 1915, there were a few chances to test it: 1916, which World War I interfered with, 1918, where attempted observations were defeated by clouds, and 1919, which is where the first successful test took place. Arthur Eddington masterminded an expedition that involved two teams, one in Brazil and one in Africa, to photograph and measure these stellar positions during one of the 20th century's longest total eclipses: nearly 7 minutes in duration.

Actual negative and positive photographic plates from the 1919 Eddington Expedition, showing (with . [+] lines) the positions of the identified stars that would be used for measuring the light deflection due to the Sun's presence. This was the first direct, experimental confirmation of Einstein's General Relativity.

The results of those observations was compelling and profound: Einstein's theory was right, while Newton's broke down in the face of the bending of starlight by the Sun. Although the data and analysis was controversial, as many accused (and some still accuse) Arthur Eddington of "cooking the books" to get a result that confirmed Einstein's predictions, subsequent eclipses have shown definitively that General Relativity works where Newton's gravity does not.

In addition, careful reanalysis of Eddington's work shows that it was, in fact, good enough to confirm the predictions of General Relativity. The features in newspapers around the world trumpeted this tremendous success, and even a century later, some of the world's best science writers are still publishing wonderful books on this remarkable achievement.

A headline from the New York Times (L) and the Illustrated London News (R), show not only a . [+] difference in the quality and depth of reporting, but also in the level of excitement expressed by journalists in two different countries at this incredible scientific breakthrough. Light, indeed, was found to be bent in the proximity of mass, by the amount predicted by Einstein.

New York Times, 10 November 1919 (L) Illustrated London News, 22 November 1919 (R)

Today, May 29, 2019, marks the 100th anniversary of the day, the event, and the expedition that validated Einstein's General Relativity as humanity's leading theory of how gravitation works. Newton's laws are still incredibly useful, but only as an approximation to Einstein's theory with a limited range of validity.

General Relativity, meanwhile, has gone on to successfully predict everything from frame-dragging to gravitational waves, and still has yet to encounter an observation that conflicts with its predictions. Today marks a full century of General Relativity's demonstrated validity, with not even a hint of how it might someday break down. Although we certainly don't know everything about the Universe, including what a quantum theory of gravity might actually be like, today is a day for celebrating what we do know. 100 years after our first critical test, our best theory of gravity still shows no signs of slowing down.

## How to Watch a Solar Eclipse

What you need to know about eclipses, how to be safe during an eclipse and some fun experiments you can try during this rare event.

So the universe was still up for grabs in March 1919, when Eddington and his colleagues set sail for Africa to observe the next eclipse. Astronomically, the prospects were as good as they could get. During the eclipse, the sun would pass before a big cluster of stars known as the Hyades, so there ought to be plenty of bright lights to see yanked askew.

Eddington was the right man for the job. A math prodigy and professor at Cambridge, he had been an early convert to Einstein’s new theory, and an enthusiastic expositor to his colleagues and countrymen.

A story went that he was once complimented on being one of only three people in the world who understood the theory. Admonished for false modesty when he didn’t respond, Eddington replied that, on the contrary, he was trying to think of who the third person was.

General relativity was so obviously true, he said later, that if it had been up to him he wouldn’t have bothered trying to prove it.

But it wasn’t up to him, due to a quirk of history. Eddington was also a Quaker and so had refused to be drafted into the army. His boss, Frank Dyson, the Astronomer Royal of Britain, saved Eddington from jail by promising that he would undertake an important scientific task, namely the expedition to test the Einstein theory.

Eddington also hoped to help reunite European science, which had been badly splintered by the war, Germans having been essentially disinvited from conferences. Now, an Englishman was setting off to prove the theory of a German, Einstein.

According to Einstein’s final version of the theory, completed in 1915, as their light rays curved around the sun during an eclipse, stars just grazing the sun should appear deflected from their normal positions by an angle of about 1.75 second of arc, about a thousandth of the width of a full moon.

According to old-fashioned Newtonian gravity, starlight would be deflected by only half that amount, 0.86 second, as it passed the sun during an eclipse.

A second of arc is about the size of a star as it appears to the eye under the best and calmest of conditions from a mountaintop observatory. But atmospheric turbulence and optical exigencies often smudge the stars into bigger blurs.

So Eddington’s job, as he saw it, was to ascertain whether a bunch of blurs had been nudged off their centers by as much as Einstein had predicted, or half that amount — or none at all. It was Newton versus Einstein.

And what if Eddington measured twice the Einstein deflection?, Dyson was asked by Edwin Cottingham, one of the astronomers on the expedition. “Then Eddington will go mad and you will come home alone,” Dyson answered.

To improve the chances of success, two teams were sent: Eddington and Cottingham to the island of Principe, off the coast of Africa, and Charles Davidson and Andrew Crommelin to Sobral, a city in Brazil. The fail-safe strategy almost didn’t work.

## Today in science: This solar eclipse proved Einstein right

Einstein’s triumph. This early photograph shows the total solar eclipse on May 29, 1919. See the tick marks around stars near the eclipsed sun? It was the precise measurement of the positions of these stars that proved the sun bends starlight, in accordance with Einstein’s theory of general relativity. Image via Wikimedia Commons.

May 29, 1919, is the date of a solar eclipse that caused a revolution in science. The eclipse is famous for testing Albert Einstein’s theory of general relativity. Einstein was relatively unknown at the time. He had proposed general relativity in 1915, and scientists had been intrigued by the entirely new way of thinking about gravity – for example, the idea that mass causes space to curve – but no one had experimentally proven the theory to be correct. Then, on May 29, 1919, an expedition of English scientists – led by Sir Arthur Eddington – traveled to the island of Príncipe off the west coast of Africa to observe a total solar eclipse. If the theory were right, the light from stars should be bent by the gravity of the sun and appear displaced. An eclipse, where the moon blocks the sunlight enough for stars to be seen near the sun, was the perfect opportunity to test this.

The scientists’ measurements during the eclipse showed that, astoundingly, Einstein’s predictions were indeed correct. The locations of the now visible stars appeared displaced, due to the fact that their light had to travel to us on the curved space around the sun caused by its gravity, as described by Einstein.

### From anonymity to stardom via a solar eclipse

Later that year – on November 6, 1919, in London – England’s Astronomer Royal, Frank Dyson, who had organized the expedition, presented the results at a joint meeting of the Royal Astronomical Society and the Royal Society. Dyson said “there can be no doubt” that measurements made during the May 29, 1919, solar eclipse “confirm Einstein’s prediction.”

As part of the celebration of the 100th anniversary of this legendary solar eclipse, Caltech physicist Sean Carroll explained to NBCNews in 2019:

General relativity was the poster child for being a crazy, new, hard-to-understand theory, with dramatic implications for the nature of reality. And yet you could see [the results] you could photograph it. So people got caught up in that excitement.

And so Albert Einstein was catapulted to rock star fame, a status in popular culture he has retained ever since.

During a solar eclipse, stars normally not visible in the glaring sunlight appear on the side of the sun and are displaced from the location they’d normally be in. Why? Because – just as Einstein’s theory said it should – light bends in the presence of mass, in this case the mass of a star, our sun. Rather than traveling in a straight path, the light of distant stars is forced to travel a curved path along the curved space near the sun. Note that the bending of starlight is exaggerated in this image. In reality, the stars are displaced by up to 1.75 arcseconds (about 0.0005 degrees). Image via GSFC/ NASA/ DiscoverMagazine.com.

### A new perspective on gravity and the universe

Einstein’s general theory of relativity underlies our most basic modern cosmology, our way of looking at the universe as a whole. Before Einstein, scientists relied on Isaac Newton’s theory of gravity. Newton’s way of looking at gravity is still valid and is still taught to physics students. But while Newton’s formulation of gravity is more of a special case under specific conditions, Einstein’s theory is a refinement of scientists’ understanding of gravity that covers the big picture … and what a mind-blowing big picture! Einstein proposed that mass causes space to curve. So, for example, although there appears to be a “force” (as described by Newton) that causes our Earth to be pulled towards the sun by gravity, that force can “simply” be described as Earth traveling in curved space around the sun, according to Einstein.

Einstein’s general theory of relativity not only explains the motion of Earth and the other planets in our solar system. In our modern cosmology, it also describes extreme examples of curved space, such as around black holes. And it helps to describe the history and expansion of the universe as a whole.

### The solar eclipse was the first proof of many

In the century and a bit since the 1919 total solar eclipse, Einstein’s relativity theory has been proven again and again, in many different ways. You might have seen the recent first-ever photo of a black hole?. It also proved, once again, that Einstein was right.

This image captured people’s imaginations when it was released in 2019: the first-ever actual image of a giant black hole, in the center of galaxy M87. It also proves Einstein’s theory, which predicted the observations from M87 with unerring accuracy. Image via Event Horizon Telescope Collaboration.

### Now and then

The Royal Astronomical Society (RAS) described modern-day practical applications of Einstein’s theory:

The theory fundamentally changed our understanding of physics and astronomy, and underpins critical modern technologies such as the satellite-based Global Positioning System (GPS).

The theory of relativity is essential for the correct operation of GPS systems, which in turn are relied on in many common applications including vehicle satellite navigation (SatNav) systems, weather forecasting, and disaster relief and emergency services. However, the world had to wait decades before the applications of such a blue skies result could be realized.

Back in the day of the 1919 eclipse, Sir Arthur Eddington attended a dinner of the same organization – RAS – shortly after the successful expedition. He then showed his humorous side by reciting a verse he had written on the feat:

Oh leave the wise our measures to collate
One thing at least is certain, light has weight
One thing is certain and the rest debate
Light rays, when near the sun, do not go straight.

Sir Arthur Eddington led the expedition that provided the first proof of Einstein’s theory of general relativity. Image via Wikimedia Commons. Albert Einstein in 1912.

Bottom line: The solar eclipse of May 29, 1919, was the day astronomer Sir Arthur Eddington verified Einstein’s general theory of relativity, by observing how stars near the sun were displaced from their normal positions. This apparent change in position happens because, according to Einstein’s theory, the path of light is bent by gravity when it travels close to a massive object like our sun.

## The bending of starlight is twice the Newtonian prediction

I was wrong about there being no bending of light. It can also happen in a prism:

Doubling the Deflection
.
Intuitively, Einstein’s 1911 prediction was only half of the correct value because he did not account for the cumulative effect of spatial curvature over a sequence of small regions of spacetime, within each of which the principle of equivalence applies. This can be understood from the figure below, which depicts a ray of light passing through a sequence of “Einsteinian elevators” near the Sun.

Yes, it is true that assuming there could be a "uniform gravitational field" globally was a problem with Einstein's 1911 calculation.

I think it's worth expanding some on why it is a problem. Take a step back and consider the flat spacetime of SR in a global inertial frame (which of course always exists in flat spacetime). It is obvious that in this global frame, there is no bending of light. In order to have bending of light in flat spacetime, you must be at rest in a uniformly accelerated frame (at least if we are talking about vacuum--with a material medium present you can, of course, have light bending in an inertial frame in flat spacetime, as @PeroK's example of a prism shows). But there is no global uniformly accelerated frame in flat spacetime. (Of course there isn't one in any curved spacetime either, but in 1911 Einstein hadn't quite gotten to curved spacetime conceptually.) We know that now because we know about Rindler coordinates (which were not discovered until decades later, IIRC), and we know that they only cover a portion of flat spacetime, not all of it.

So in extending his 1911 calculation globally, Einstein was in fact doing invalid math he just didn't realize it.

## Sun and Moon Mark Holborn

Struck out price AUD$120.00 Price AUD$72 Struck out price CAD$105.00 Price CAD$63 Struck out price &euro69.95 Price &euro41.97 Struck out price £59.95 Price £35.97 Struck out price T79.95 Price T47.97 Struck out price USD$79.95 Price USD$47.97

Gift options available at checkout

A unique pictorial history of astronomical exploration from the earliest prehistoric observatories to the latest satellite images

With 280 spectacular images and an inspiring story imparting the excitement of discovery, Sun and Moon marks the anniversary of the first moon landing by Apollo 11 in July 1969, and the 40th anniversary of NASA's geological survey of the moon, with its extraordinary cartography.

It illustrates how the development of photography and cartography - the means of documenting other worlds - is linked indelibly to the charting of the heavens, from the first image on a glass plate to the Hubble Space Telescope.

Sun and Moon is the gift of the season for anyone who has ever gazed at the stars or looked through a telescope.

Specifications:

• Format: Hardback
• Size: 308 x 292 mm (12 1/8 x 11 1/8 in)
• Pages: 376 pp
• Illustrations: 200 illustrations
• ISBN: 9780714876566

Mark Holborn is an editor and book designer who has worked internationally with such diverse artists as Lucian Freud, Richard Avedon, Irving Penn, Issey Miyake, William Eggleston and James Turrell. He is also a writer and curator with a particular interest in Japan, about which he has written books that include The Ocean in the Sand and Beyond Japan. In affiliation with the Beijing studio of the artist and designer Tim Yip, he has created a number of exhibitions in China. He is currently working on a series of historical surveys in Saudi Arabia and the Middle East with the photographer Humberto da Silver.

"Sun and Moon amply demonstrates that the efforts of scientists and explorers to comprehend our wider environment arises from an appreciation of beauty as well as the pursuit of reason-from hearts as well as minds."&mdashTimothy Ferris, bestselling author and filmmaker (hailed by the Washington Post as "the best science writer of his generation.")

". A sublime history of man and space rendered in gorgeous detail."&mdashOuter Places

"Sun and Moon is a thrilling catalog of important cosmic discoveries."&mdashCool Hunting

". A deeply researched, richly photographed history of modern space exploration. Sun and Moon captures our progress and our enduring fascination with space."&mdashWorth.com

"Sun and Moon celebrates our fascination with the skies above."&mdashWerd.com

". A quintessential coffee-table book by Mark Holborn that traces centuries of beyond-Earth imagery and imagination."&mdashGeekwire

". is a thrilling catalog of important cosmic discoveries. [it] gives new (and old) insight to curious readers."&mdashCool Hunting

"Holborn marries history, sociology, art and science to unpick the ways that space has defined our existence here on Earth… This all-encompassing view of our relationship with space is made all the more enjoyable by the rich visual material that has been amassed…the true gems are densely packed photographs that capture the wonders of our stars, made possible by rapidly changing technologies. To be able to comprehend the surface of the moon, or the wonders of the Eagle Nebula and the Whirlpool Galaxy, is something truly heavenly."&mdashElephant magazine

"Taken as a whole, the book shows how dreary life on Earth would be without the beauty of the sun by day and the splendor of the moon at night."&mdashAir and Space/Smithsonian Magazine

"Capturing a sense of the infinite unknown that enraptures dedicated stargazers, Mark Holborn's Sun and Moon: A Story of Astronomy, Photography, and Mapping is an extensive-and stunning-visual history of space exploration. Elegantly designed, with nearly 300 images, Sun and Moon is an altogether grand retrospective of humankind's attempts to make sense of the mysteries of space."&mdashBookPage

"In the year of the 50th anniversary of the moon landing, a design editor's approach to the cosmos, tracing the side-by-side developments of artistic and scientific attempts to explain it."&mdashThe New York Times Online

"In the year of the 50th anniversary of the moon landing, a design editor's approach to the cosmos, tracing the side-by-side developments of artistic and scientific attempts to explain it."&mdashThe New York Times

## 100 years on: the pictures that changed our view of the universe

Total solar eclipse, 29 May 1919. Glass positive photograph of the corona, taken at Sobral in Brazil, with a telescope of 4in in aperture and 19ft focal length. Photograph: Science & Society Picture Library/SSPL via Getty Images

Total solar eclipse, 29 May 1919. Glass positive photograph of the corona, taken at Sobral in Brazil, with a telescope of 4in in aperture and 19ft focal length. Photograph: Science & Society Picture Library/SSPL via Getty Images

Arthur Eddington’s photographs of the 1919 solar eclipse proved Einstein right and ushered in a century where gravity was king

Last modified on Tue 14 May 2019 11.49 BST

A hundred years ago this month, the British astronomer Arthur Eddington arrived at the remote west African island of Príncipe. He was there to witness and record one of the most spectacular events to occur in our heavens: a total solar eclipse that would pass over the little equatorial island on 29 May 1919.

Observing such events is a straightforward business today, but a century ago the world was still recovering from the first world war. Scientific resources were meagre, photographic technology was relatively primitive, and the hot steamy weather would have made it difficult to focus instruments. For good measure, there was always a threat that clouds would blot out the eclipse.

These risks doubtless caused worries but they were well worth facing, Eddington reckoned, for he believed his observations could prove, or disprove, the most revolutionary scientific idea to have been put forward in modern science: Albert Einstein’s theory of general relativity.

In his 1915 theory, Einstein argued that gravity was not a force that acted at a distance between objects, as Isaac Newton had stated. Instead, he maintained, it was the result of an object’s mass causing space to curve. From this perspective, a body in orbit around the Sun is actually going in a straight line but through space that has been bent by the mass of the Sun. Even a beam of light would bend as it passed along this section of curved space.

“Einstein used existing astronomical observations to support his theory – for example, known anomalies in the orbit of Mercury round the Sun,” says Carolin Crawford, of the Institute of Astronomy, Cambridge. “But these were post hoc rationalisations. What was needed was a specific testable prediction to show his theory was right. The May 1919 eclipse provided that opportunity.”

During a total solar eclipse, the disc of the Moon passes in front of the Sun. This blots out its blindingly bright rays and allows astronomers to study the relatively dim light of background stars. By comparing existing photographs of a particular cluster of stars with images of them taken during an eclipse, it should be possible to discover whether the latter have shifted position because space is being bent by the Sun as it passes in front of them.

And that is what Eddington set out to prove – along with a second group of British astronomers who were sent to Sobral, in northern Brazil, which also lay under the eclipse path. Both expeditions were organised by the British astronomer royal Frank Watson Dyson and both would study stars of the Hyades cluster in the constellation Taurus – in front of which the eclipsed Sun would pass. If the apparent positions of these stars moved compared with standard, night-time photographs of the region, this would indicate that the Sun’s mass was causing space to curve.

There was one slight complication. “Newtonian physics also forecast that star positions could be shifted during an eclipse – but less so,” says Crawford. “Einstein’s theory predicted a greater deflection.”

Eddington therefore faced a daunting double problem. Could he detect any star-position changes caused by the Sun? More to the point, could he do it accurately enough to determine whether they were occurring in response to the physics of Newton or the science of Einstein? According to the former, star images would be deflected by about 0.8 of an arcsecond while Einstein said they would be moved by about 1.8 arcseconds. Given that an arcsecond is 1/3,600th of a degree, such extremely small differences would be very hard to detect.

Worse, the conditions on Príncipe turned out to be much grimmer than Eddington and his assistant Edwin Cottingham had anticipated. The pair had to work under mosquito nets and chase off monkeys which continually tried to steal bits of their equipment. “Then on the morning of 29 May, after travelling thousands of miles to view the eclipse, Eddington and Cottingham awoke to a rainstorm,” writes the author Ron Cowen in his recent history of the event: Gravity’s Century: From Einstein’s Eclipse to Images of Black Holes.

The clouds slowly cleared but there were still intermittent patches of mist in the sky when the eclipse was well under way. Eddington worked feverishly and managed to take 16 photographic plates. Later he discovered that only two of these contained enough stars to tell whether their light might have been bent or not. Finally, steamboat schedules forced them to leave Príncipe within days – without having started their measurements of the stars on the plates.

Then there were the observations taken in Sobral, in Brazil, by a second group of astronomers who were led by Andrew Crommelin and Charles Rundle Davidson of the Royal Greenwich Observatory. They experienced better observing conditions. However, to their horror they subsequently found that all 19 images they had taken with their main telescope were out of focus, the Sun’s heat having caused its mirror to expand unequally and produce blurred images. Fortunately, eight other plates taken using a smaller backup telescope produced perfect results.

The scientists gathered in August and began measuring the positions of stars on their plates. They obtained two sets of results. From the Sobral images, they found a light-bending figure of 1.98 arcseconds. From the Príncipe plates, they got a value of about 1.6.

These findings, although based on limited data, were completely in line with Einstein’s theory. It was a sensational result, one that made the event “probably the most important eclipse in history,” says the US physicist Daniel Kennefick in his book No Shadow of a Doubt: The 1919 Eclipse That Confirmed Einstein’s Theory of Relativity.

Armed with their data and analysis, the team – including Dyson, Eddington and Crommelin – stood before a packed meeting in the Royal Society in London on the evening of 6 November and outlined their results. Their audience was stunned. Two hundred years of Newtonian physics had been overturned. “This is the most important result obtained in connection with the theory of gravitation since Newton’s day,” announced the Royal Society’s president, the Nobel laureate JJ Thomson.

The world’s journalists were equally impressed. “Revolution in science: new theory of the universe: Newtonian ideas overthrown,” ran headlines in the Times the next day, while the New York Times plumped for: “Lights all askew in the heavens: Einstein’s theory triumphs.” For his part, Einstein was transformed into a global celebrity.

This was the beginning of gravity’s century – 100 years in which its warping effects on space have come to dominate astronomy. Subsequent solar eclipses also generated results entirely consistent with Einstein’s theory, while later photographs taken by the Hubble telescope have revealed even more spectacular distortions of space produced by powerful gravitational fields. In some images, starlight around massive clusters of galaxies is twisted into long stretched stripes, so mighty is the curving of space around these vast star collections.

## This Is How, 100 Years Ago, A Solar Eclipse Proved Einstein Right And Newton Wrong

On May 29, 1919, the world changed forever. For hundreds of years, Isaac Newton’s theory of gravity — the law of universal gravitation — had gone unchallenged, as its predictions matched every observation or measurement that had ever been made. But a mismatch between Newton’s predictions for the orbit of Mercury and what astronomers saw surfaced in the mid-19th century, and scientists struggled to explain it.

Perhaps we needed to modify the laws of gravity, after all. Evidence mounted when special relativity came out, demonstrating that there was no such thing as absolute distance. Newton’s theory predicted an instantaneous force, again violating relativity. In 1915, Albert Einstein put forth a new alternative theory of gravity: General Relativity. The way to test it against Newton’s theory was to wait for a total solar eclipse. 100 years ago today, Einstein was proven right. Here’s how.

Today, Albert Einstein’s General theory of Relativity is arguable the most successful theory of all-time. It explains everything from GPS signals to gravitational redshift, from gravitational lensing to merging black holes, and from the timing of pulsars to the orbit of Mercury. The predictions of General Relativity have never once failed.

When this theory was first introduced in 1915, it was attempting to replace Newton’s gravitation. Although it could reproduce the earlier Newtonian successes and explain the orbit of Mercury (where Newton could not), the most critical test would come in the form of a new prediction that severely differed from the predictions of the universal law of gravitation. A total solar eclipse would provide a unique and straightforward opportunity.

In Newton’s gravity, anything with mass attracts anything else with mass. Even though light is massless, it has an energy, and therefore you can assign an effective mass to it through Einstein’s E = mc². (You find that m = E/c².) If you allow a photon to pass near a large mass, you can use this effective mass to predict how much the starlight should bend by, and you get a specific value. Near the limb of the Sun, it’s just under 1" (arc-second), or 1/3600th of 1°.

But in Einstein’s General Relativity, both space and time are distorted by the presence of mass, whereas in Newton’s gravity, only an object’s motion through space is affected by the gravitational force. This means that Einstein’s theory predicts an extra factor of 2 (actually slightly more, especially as you get closer to the mass in question) over Newton’s, or a deflection near the Sun of closer to 2".

The history of how Einstein’s General Relativity came to be is fascinating, because it’s only the fact that Newton’s gravitation eventually had problems that motivated Einstein to formulate his new concept.

Newtonian gravity, put forth in 1687, is an extraordinarily simple law: put any masses anywhere in the Universe, a fixed distance apart, and you immediately know the gravitational force between them. This explained everything from the terrestrial motion of cannonballs to the celestial motion of comets, planets, and stars. After 200 years, it had passed every single test that was thrown its way. But one pesky observation threatened to derail everything: the detailed motion of the innermost planet in our Solar System.

Every planet moves in an ellipse around the Sun. However, this ellipse isn’t static, returning to the same fixed point in space with every orbit, but rather, it precesses. Precession is like watching that ellipse rotate in space over time, albeit very slowly. Mercury had been observed with incredible precision since Tycho Brahe in the late 1500s, so with 300 years of data, our measurements were extraordinary.

According to Newton’s theory, its orbit should have precessed by 5,557"-per-century, due to the precession of Earth’s equinoxes and the gravitational effects of all the planets on Mercury’s orbit. But observationally, we observed 5,600"-per-century instead. That difference, of 43"-per-century (or just 0.00012°-per-year), had no explanation in Newton’s framework. Either there was an extra planet interior to Mercury (which observations ruled out), or something was wrong with our old theory of gravity.

But Einstein’s new theory could explain the mismatch. He spent years developing the framework for General Relativity, where gravitation wasn’t caused by masses attracting other masses, but rather by matter and energy curving the very fabric of space, which all objects then move through. When gravitational fields are weak, Newton’s law is a very good approximation to what Einstein’s theory laid down.

Close to very large masses or at high speeds, however, Einstein’s predictions differed from Newton’s, predicting exactly that 43"-per-century difference. But the bar to overthrow a scientific theory is higher than that. To supersede the old theory, a new one must do the following:

1. Reproduce all the successes that the old theory enjoyed (otherwise, the old theory is still superior in some way),
2. Succeed in the regime where the old theory could not (otherwise, your new theory doesn’t fix the problem with the old one),
3. And to make a new prediction that you can go out and test, distinguishing between the old-and-new ideas (otherwise, you don’t have any scientifically predictive power).

That last piece is where the solar eclipse comes in.

When the stars appear in the night sky, the starlight travels to our eyes from a different location in the galaxy, many light years away. If Newton was correct, that light should either travel in a completely straight line, undeflected by any masses it passes near (since light is massless), or that it should bend due to the gravitational effects of mass-energy equivalence. (After all, if E = mc², then perhaps you can treat light as have an effective mass of m = E/c².)

But Einstein’s theory, particularly if light passes very close by a large mass, offers a prediction different from both of these numbers. That extra factor of 2 (or, rather, 2 and an extra few parts-per-million) is a unique and very specific prediction from Einstein’s theory, and one that could be tested by making two observations at different times of the year.

The largest mass we have close by Earth is the Sun, which normally renders starlight invisible during the day. As starlight passes near the edge of the Sun, according to Einstein, it should travel along that curved space, causing the light-path to appear bent. During a total solar eclipse, however, the Moon passes in front of the Sun, blocking its light and causing the sky to become as dark as night, enabling the stars to be seen during the daytime.

If you previously measured those stellar positions to an accurate enough precision, you could see whether they’ve shifted or not — and by how much — due to the presence of that large, nearby mass. If you could detect a deflected position at the sub-arc-second level, you could know definitively whether Newton’s, Einstein’s, or neither prediction was correct.

Photographic plates of the Sun during a total solar eclipse had revealed not only details in the Sun’s corona before, but the presence and positions of stars during the daytime. However, none of the pre-existing photographs were high enough in quality to determine the deflected positions of nearby stars to the necessary accuracies the deflection of starlight is a very small effect requiring very precise measurements to detect!

After Einstein set forth his general theory of relativity in 1915, there were a few chances to test it: 1916, which World War I interfered with, 1918, where attempted observations were defeated by clouds, and 1919, which is where the first successful test took place. Arthur Eddington masterminded an expedition that involved two teams, one in Brazil and one in Africa, to photograph and measure these stellar positions during one of the 20th century’s longest total eclipses: nearly 7 minutes in duration.

The results of those observations was compelling and profound: Einstein’s theory was right, while Newton’s broke down in the face of the bending of starlight by the Sun. Although the data and analysis was controversial, as many accused (and some still accuse) Arthur Eddington of “cooking the books” to get a result that confirmed Einstein’s predictions, subsequent eclipses have shown definitively that General Relativity works where Newton’s gravity does not.

In addition, careful reanalysis of Eddington’s work shows that it was, in fact, good enough to confirm the predictions of General Relativity. The features in newspapers around the world trumpeted this tremendous success, and even a century later, some of the world’s best science writers are still publishing wonderful books on this remarkable achievement.

Today, May 29, 2019, marks the 100th anniversary of the day, the event, and the expedition that validated Einstein’s General Relativity as humanity’s leading theory of how gravitation works. Newton’s laws are still incredibly useful, but only as an approximation to Einstein’s theory with a limited range of validity.

General Relativity, meanwhile, has gone on to successfully predict everything from frame-dragging to gravitational waves, and still has yet to encounter an observation that conflicts with its predictions. Today marks a full century of General Relativity’s demonstrated validity, with not even a hint of how it might someday break down. Although we certainly don’t know everything about the Universe, including what a quantum theory of gravity might actually be like, today is a day for celebrating what we do know. 100 years after our first critical test, our best theory of gravity still shows no signs of slowing down.

1. Verddun

It is remarkable, very much the helpful information

2. Shakakree

Willingly I accept. In my opinion, it is actual, I will take part in discussion. I know, that together we can come to a right answer.

3. Davison

I'm ready to re-read the article again. Good material and written simply! That's what you need.

4. Jesper


text:
MATH 422 Introduction to Geometric Topology, 2018 Spring

Instructor: Professor Jae Choon Cha

Office hour: by an appointment
Instructor’s web page: http://gt.postech.ac.kr/~jccha/

Classroom and hour

Monday and Wednesday 14:00-15:15, Science Building II 105

See POVIS LMS.

Final Project

The final project is writing a short article on the notion of the linking number.

Due date is June 12, by 11:59pm. Hand in your homework at the MA422 collection box on the first floor of Math Building.

Homework Problems and announcements

Homework problems will be posted on this webpage, and will be collected in class on the due date. Turn it in BEFORE the lecture starts.  $\def\C{\mathbb{C}} \def\R{\mathbb{R}} \def\Z{\mathbb{Z}} \def\Q{\mathbb{Q}} \def\Ker{\operatorname{Ker}} \def\rel{\text{ rel }} \def\sm{\smallsetminus}$

Problem Set #6 (due 06/01)

Hand in your homework at the MA422 collection box on the first floor of Math Building, by 1:00pm on June 1st.

1. For two smooth manifolds $M\subset \R^k$ and $N\subset \R^\ell$, show that $M\times N \subset \R^{k+\ell}$ is a smooth manifold and $T(M\times N)_{(x,y)} = TM_x \times TM_y$ in $\R^{k+\ell}$.
2. For a smooth map $f\colon M\to N$, the graph of $f$ is defined to be $G(f)=\{(x,f(x))\in M\times N \mid x\in M\}$.  Show that $G(f)$ is a smooth manifold, and that the tangent space $TG(f)_{(x,f(x))}\subset TM_x \times TN_{f(x)}$ equals the graph of the linear map $df_x$. (It’s a good idea to think about the case of $f\colon \R\to \R$ as an example)
3. For a smooth manifold $M\subset \R^k$, define the tangent bundle of $M$ by $TM := \{(x, v)\subset \R^k\times \R^k\mid x\in M,\ v\in TM_x\}$ and the normal bundle of $M$ by $\nu M:=\{(x, v)\subset \R^k\times \R^k\mid x\in M,\ v\in (TM_x)^\perp\}$.  Here, $(TM_x)^\perp:=\{v\in \R^k \mid v \perp TM_x\}$ designates the orthogonal complement. Show that $TM$ and $\nu M$ are smooth manifolds.
4. Show that the following surfaces in $\R^3$ are smooth manifolds of dimension 2.  (Assume $a,b,c>0$.)
1. Ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$
2. Hyperboloid: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{z^2}{c^2}\pm 1$
3. Hyperbolic paraboloid: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=\frac{z}{c}$
5. Suppose $A(t)=[a_{ij}(t)]$ is an $n\times m$ matrix whose entries $a_{ij}(t)$ are real-valued continuous maps defined on an open subset $U\subset\R$. Suppose $A(t_0)$ has rank $n$. Show that there is an open neighborhood of $t_0$ on which $A(t)$ has rank $n$.
6. Suppose $f\colon M\to N$ is a smooth map.
1. Suppose $x\in M$ is a regular point for $f$. Show that there is an open neighborhood of $x$ in $M$ which consists of regular points for $f$.
2. Suppose $M$ is compact, and $y\in N$ is a regular value for $f$. Show that there is an open neighborhood of $y$ in $N$ which consists of regular values for $f$.
7. Suppose $M$ is a smooth manifold of dimension $n$ with boundary. Recall the definition of $\partial M$: for $x\in M$, $x\in \partial M$ if and only if there is a parametrization $g\colon U\cap \R^n_+ \to M$ such that $x\in g(U\cap \partial \R^n_+)$. Prove that $\partial M$ is well-defined, independent of the choice of a parametrization.
8. Suppose $M$, $N$ and $P$ are compact connected smooth manifolds of the same dimension, and $f\colon M\to N$, $g\colon N\to P$ are smooth maps. Show that $\deg gf = \deg g\cdot \deg f$.

Problem Set #5 (due 05/14)

1. Draw a sequence of knot diagrams, in such a way that the knot given by each diagram is obviously seen to be equivalent to the next one, to show that the following knot is trivial.
2. Let $K$ be the figure eight knot. Recall that we have computed a Wirtinger presentation for $K$. See Lecture Note #11. Use the Wirtinger presentation to show that there is an epimorphism of $G=\pi_1(\R^3\sm K)$ onto the dihedral group $D_{10}$. Conclude that the figure eight knot $K$ is nontrivial. (Hint: try the map $x\mapsto (1\,2)(3\,4)$, $w\mapsto (2\,3)(4\,5)$, where $x$ and $w$ are the generators used in Lecture Note #11.)
3. Recall from Lecture Note #12 that we computed the Alexander polynomial of the figure eight knot.  In the computation, we claimed that the vectors $y_1$ and $y_3$ form a basis for the vector space $G’/G^{\prime\prime}\otimes \Q$.
1. Express each of $t^{-1}y_2$, $y_2$, $ty_2$, $t^2y_2$ as a linear combination of $y_1$ and $y_3$.
2. Why are $y_1$ and $y_3$ linearly independent? Provide a detailed proof.
4. Compute a Wirtinger presentation of the following knot $K$. (Show explicitly your choice of generators in the diagram!)
5. Compute the Alexander polynomial of the above knot $K$. Show that $K$ is not equivalent to any one of the trivial knot, trefoil knot and figure eight knot.
6. Recall from Lecture Note #11 that certain conjugates of the four Wirtinger relators of the figure eight knot diagram satisfy the relation (P).  For the trefoil knot diagram in Lecture Note #12, find conjugates of the three Wirtinger relators which satisfies an analogous relation. Explain how you found them, using a picture of some loops on the knot diagram.  Do the same for the knot in the above #4.
7. This is a continuation of #6. Generalize the relation (P) to a theorem for an arbitrary knot diagram.  Give a proof.
8. An $m$-component link in $\R^3$ is defined to be the union of $m$ disjoint simple closed curves embedded in $\R^3$. For example, a knot is a 1-component link.
1. Let $L_1$ and $L_2$ be the 2-component links in $\R^3$ illustrated below.
$L_1$ =          $L_2$ =
Find a presentation of $\pi_1(\R^3-L_1)$ and $\pi_1(\R^3-L_2)$, generalizing the Wirtinger presentation of knots.
2. Compute the abelianization of $\pi_1(\R^3-L_1)$ and $\pi_1(\R^3-L_2)$.
3. In general, for an $m$-component of $L$ in $\R^3$, can you guess what the abelianization of $\pi_1(\R^3-L)$ must be? State and prove.
9. Suppose $U$ is the trivial knot in $S^3$.
1. Show that the set of free homotopy classes of maps $S^1\to \R^3\sm U$ is in 1-1 correspondence with $\Z$.
2. Using (a), prove that there is no homeomorphism $h\colon \R^3\to \R^3$ such that $h(L_1)=L_2$, where $L_1$ and $L_2$ are the links in #8. In other words, show that the links $L_1$ and $L_2$ are not equivalent.

Problem Set #4 (due 04/25)

1. Show that $G :=\langle x, y \mid xyx^{-1}y^{-1}, x^2 y \rangle$ is isomorphic to $\Z$.  Find a word in $x,y$ which is a generator of the group $G$.
2. For two given maps $i\colon C \to A$ and $j\colon C\to B$ of topological spaces $A$, $B$, $C$, define $X=A\cup B/\sim$ where $a\sim b$ ($a\in A$, $b\in B$) if and only if $a=i(c)$ and $b=j(c)$ for some $c\in C$.  $X$ is a topological space endowed with the quotient topology. Show that $X$ endowed with the naturally defined maps $f\colon A\to A\cup B \to X$ and $g\colon B\to A\cup B \to X$ is a pushout in the following sense: if $\alpha\colon A\to Y$ and $\beta\colon B \to Y$ are maps satisfying $\alpha i = \beta j$, then there is a unique map $\phi\colon X \to Y$ satisfying $\phi f=\alpha$ and $\phi g=\beta$.
3. Give detailed proofs of the following facts. In what follows, all graphs are assumed to be connected and with finitely many vertices and edges.
1. A tree is contractible. (Here a tree is a graph with no cycle; use induction on the number of edges.)
2. For a graph $X$, $\pi_1(X)$ is a free group with finitely many generators.  (Take a maximal tree of $X$, and use induction on the number of edges not contained in the maximal tree.)
4. Let $X$ be the graph with one vertex $*$ and $n$ edges $x_1,\ldots,x_n$ attached to $*$.  Each $x_i$ can be regarded as a loop.
1. Show that $\pi_1(X,*)$ is isomorphic to the free group generated by (the classes of) the loops $x_1,\ldots, x_n$.
2. Given a word $w=x_{i_1}^{\epsilon_1}\cdots x_{i_r}^{\epsilon_r}$, let $f_w\colon (S^1,1) \to (X,*)$ be the loop representing $w$. Define $Y_w$ to be the space $(D^2 \cup X)\mathbin{/}z\sim f_w(x)$ for $z\in S^1 \subset D^2$.  Compute a presentation for $\pi_1(Y_w,*)$.
3. Show the following by generalizing (b): for a group $G=\langle x_1,\ldots,x_n\mid r_1,\ldots,r_m\rangle$, there is a path connected space $K$ such that $\pi_1(K)\cong G$.
5. Let $h\colon S^1\times S^1 \to S^1\times S^1$ be the map defined by $h(z, w)=(z^aw^b, z^cw^d)$ where $a,b,c,d\in \Z$. (As usual, regard $S^1$ as the space of unit length complex numbers.)
1. Compute the map $h_*\colon \pi_1(S^1\times S^1) \to \pi_1(S^1\times S^1)$, as a $2\times 2$ matrix over $\Z$ representing a linear map $\Z^2 \to \Z^2$,  identifying $\pi_1(S^1\times S^1)$ with $\Z^2$.
2. Show that $h$ is a homeomorphism if and only if $h_*$, as a matrix, has determinant $\pm 1$.
3. Let $X=(D^2 \times S^1) \cup (D^2\times S^1)/(z,w)\sim h(z,w)$ (here $(z,w)\in S^1\times S^1\subset D^2\times S^1$).  Show that $X$ is a 3-dimensional compact manifold if $h_*$ has determinant $\pm1$. Compute $\pi_1(X)$.
6. Let $K$ be a bounded subset in $\R^n$ with $n\ge 3$. Viewing $K\subset \R^n \subset S^n=\R^n\cup\{\infty\}$, Show that $\pi_1(\R^n-K)$ is isomorphic to $\pi_1(S^n-K)$.
7. Define the $n$-dimensional real projective space by $\R P^n = S^n/x\sim -x$.
1. Show that the quotient map $S^n \to \R P^n$ is a covering map.
2. Show that $\pi_1(\R P^n)\cong \Z_2$ for $n\ge 2$.
8. Let $SO(n)$ be the space of $n\times n$ matrices $A$ satisfying $A^T A=I$ and $\det(A)=1$. We regard $SO(n)$ as a subspace of $\R^{n^2}$, by viewing each entry of a matrix as a coordinate. $SO(n)$ acts on $\R^n$ by multiplication on the left, viewing $x\in \R^n$ as a column vector.
1. Show that any $A\in SO(3)$ has 1 as an eigenvalue. Using this, show that $A$ is a rotation around a line passing through the origin.
2. Given a unit vector $x$ in $\R^3$ and $\theta\in [0,\pi]$, associate a rotation $A\in SO(3)$ with axis $x$ and angle $\theta$ along the right hand orientation.  Prove that the correspondence $A \to (\theta/\pi)x$ induces a homeomorphism between $SO(3)$ and $D^3/x\sim -x \; (x\in S^2)$.
3. Using #7 and (b), compute $\pi_1(SO(3))$.
4. Give an alternative computation of $\pi_1(SO(3))$, by applying the Seifert-van Kampen theorem directly to the quotient space $D^3/x\sim -x \; (x\in S^2)$.  Here, use $U=q(\operatorname{int} D^3)$ and $V=q(D^3\sm \{0\})$, where $q\colon D^3 \to D^3/x\sim -x \; (x\in S^2)$ is the quotient map.
5. Find a loop $s\mapsto A_s\in SO(3)$ ($0\le s \le 1$) representing a nontrivial element in $\pi_1(SO(3))$. Describe your matrix $A_s$ explicitly.

Problem Set #3 (due 04/11)

1. We say that a covering $p\colon E\to B$ with $E$ connected is regular if $p_*(\pi_1(E,e_0))$ is a normal subgroup of $\pi_1(B,b_0)$ for every $b_0\in B$ and $e_0\in p^{-1}(b_0)$.  Otherwise, we say that $p\colon E\to B$ is irregular. Prove that the following are equivalent. [Typo fixed on 04/09]
1. $p\colon E\to B$ is regular.
2. For a loop $\alpha$ in $B$, either all the lifts of $\alpha$ are loops, or all the lifts of $\alpha$ are not loops.
2. Explain why every connected covering of $S^1$ is regular. Find an example of an irregular cover of the graph with one vertex and two edges, with a proof.
3. Suppose $B$ is connected, locally path connected and semi-locally simply connected. Prove the following: for every surjective homomorphism $\pi_1(B,b_0)\to G$ onto a group $G$, there is a regular covering $p\colon E\to B$ such that $E$ is connected and the covering transformation group is isomorphic to $G$. [Typo fixed on 04/09]
4. Prove that a path connected space $X$ is simply connected if and only if for every pair of paths $\alpha$, $\beta\colon I\to X$ such that $\alpha(0)=\beta(0)$ and $\alpha(1)=\beta(1)$, $\alpha$ and $\beta$ are homotopic rel $\{0,1\}$.
5. Recall, from the proof of the realization part of the classification of coverings, that we defined the subsets of the form $\langle\alpha,V\rangle$. (See Lecture Note #6.) Give a fully detailed proof that they form a base for a topology.
6. Consider the complex exponential map $\exp\colon \C \to C\sm \{0\}$ defined by $\exp(z)=e^z$. As usual, for a domain $D\subset \C\sm\{0\}$, we say that a continuous map $L\colon D\to \C$ is a logarithm if $\exp(L(z))=z$ for every $z$. Give a topological proof that there does not exist a logarithm on $\C\sm \{0\}$, using the lifting criterion.  In addition, for a given domain $D\subset \C\sm \{0\}$, find a necessary and sufficient condition for a logarithm $L\colon D\to \C$ to exist, in terms of the fundamental groups.
7. Suppose a group $G$ acts on a connected space $X$, on the left: $G\times X \to X$. (Recall our convention that maps are assumed to be continuous unless; here $G$ has discrete topology.) Let $B$ be the orbit space with the quotient topology, that is, $B=X/\sim$ where $x\sim y$ if $y=gx$ for some $g\in G$. Suppose, for each $x\in X$, there is an open neighborhood $U$ such that $gU\cap U=\emptyset$ for every $g\in G$ which is not the identity.  Prove the following.
1. The quotient map $p\colon X\to B$ is a covering map.
2. $p$ is regular.
3. The covering transformation group is isomorphic to $G$.
4. If $X$ is locally path connected and simply connected, then $\pi_1(B,*)\cong G$.

Problem Set #2 (due 03/28)

1. The $n$-dimensional real projective space is a quotient space of $S^n$ which is defined by $\R P^n = S^n/x\sim -x$.
1. Show that the quotient map $S^n \to \R P^n$ is a covering map.
2. It is known that $S^n$ is simply connected, that is, $\pi_1(S^n)$ is trivial for any choice of a basepoint. (We will prove this later!) Using this, show that $\pi_1(\R P^n)$ is isomorphic to the cyclic group of order 2 for $n\ge 2$.
2. Suppose $p\colon E\to B$ and $f\colon X \to B$.  Define $f^*(E)=\{(e,x)\in E\times X \mid p(e)=f(x)\}$.
1. Show that the projections of $E\times X$ onto $E$ and $X$ give rise to maps $q\colon f^*(E) \to X$ and $\tilde f\colon f^*(E)\to E$ making the following diagram commute:
$$\begin{array}{c@{}c@{}c} f^*(E) & \longrightarrow & E \\ \downarrow & & \downarrow\\ X & \longrightarrow & B \end{array}$$
2. Show that if $p$ is a covering map, then $f^*(E) \to X$ is a covering map.
3. Show that  $p^{-1}(f(x))$ is in 1-1 correspondence with $q^{-1}(x)$.
3. Suppose $p\colon E\to B$ is a covering map and $b_0\in B$, $e_0\in E$ are basepoints such that $p(e_0)=b_0$.  For two loops $\alpha$, $\beta$ in $B$ based at $b_0$, let $\tilde\alpha$, $\tilde\beta$ be lifts in $E$ starting from $e_0$. Show that $\tilde\alpha(1)=\tilde\beta(1)$ if $[\alpha]=[\beta]$ in $\pi_1(B,b_0)$.
4. Let $f\colon E\to B$ be the covering described at the end of Lecture Note #4. See the following diagram.
1. Show that $[x],\, [y]\in \pi_1(B,*)$ are nontrivial elements, by applying the above #3.
2. Show $[x]\ne [y]$ in $\pi_1(B,*)$. Show $[x]\ne [x]^2$ in $\pi_1(B,*)$.
3. Explain why it is not possible to prove that $[x]^2$ is nontrivial in $\pi_1(B,*)$ by applying the above #3 to this covering. Find the maximal number of elements in $\pi_1(B,*)$ which can be seen to be mutually different by applying #3 to this covering. Explain why.
4. Show that $[x]^2$ is nontrivial in $\pi_1(B,*)$. Try to find two different proofs, one by applying the above #3 to this covering, and another by using an appropriate map $B \to S^1$.
5. Let $B$ be the base space of the covering in the above #4.
1. Show that $[x][y]\ne [y][x]$ in  $\pi_1(B,*)$, using lifts in a covering map.
2. Find two maps $f,g\colon S^1\to B$ with $f(1)=*=g(1)$ such that $f$ and $g$ are freely homotopic but represent different elements in $\pi_1(B,*)$.

Problem Set #1 (due 03/12)

1. Give a detailed proof that homotopy is an equivalence relation.
2. Show that a homotopy inverse of a homotopy equivalence $f\colon X\to Y$ is unique up to homotopy, that is, two homotopy inverses of $f$ are homotopic.
3. Prove that the union of the $x$-axis, $y$-axis and $z$-axis in $\R^3$ is contractible.
4. Show that if $X$ and $Y$ are homotopy equivalent, then $X\times Z$ and $Y\times Z$ are homotopy equivalent.
5. Prove that each of the following pairs are homotopy equivalent.
1. $D^n=\{x\in \R^n \mid |x|\le 1\}$ and $I^n=I\times\cdots\times I$.
2. $S^n$ and $\R^{n+1}-\{0\}$.
3. $S^1$ and $\R^3-\{z$-axis$\}$.
6. Suppose $X$ is a space, and $\alpha\colon I\to X$ be a path. Let $Y$ be the quotient space $X\cup (I\times I)/\alpha(s)\sim(s,0)$ where $0\le s\le 1$. That is, for two points $p$, $q$ in the disjoint union $X\cup (I\times I)$, $p\sim q$ if and only if either (i) $p=q$, or (ii) $\{p,q\}=\{(s,0),\alpha(s)\}$ for some $s\in I$, or (iii) $p=(s,0)$ and $q=(t,0)$ for some $s,t\in I$ such that $\alpha(s)=\alpha(t)$. Show that $X$ and $Y$ are homotopy equivalent. (Hint: try to prove that the inclusion $X\hookrightarrow Y$ is a homotopy equivalence.  Here a homotopy on $I\times I$ which sends $I\times I$ onto $I\times [0,1-t]$ at time $t$ may be useful.)
7. Suppose $X$ is a path connected space and $x_0,x_1\in X$.  Show that if $\pi_1(X,x_0)$ is abelian, then for any two paths $\alpha$ and $\beta$ in $X$ from $x_0$ to $x_1$, the isomorphisms $\alpha_*$, $\beta_*\colon \pi_1(X,x_1)\to \pi_1(X,x_0)$ are equal.
8. Suppose $\alpha_1$, $\ldots$, $\alpha_n\colon I\to X$ are paths satisfying $\alpha_i(1)=\alpha_{i+1}(0)$ for $i=1,\ldots,n-1$. Define $\beta\colon I\to X$ by $\beta(s)=\alpha_i(n\cdot s-i+1)$ for $\frac{i-1}{n} \le s \le \frac{i}{n}$, $i=1,\ldots,n$. Show that $\beta\simeq (\cdots(\alpha_1*\alpha_2)*\alpha_3 *\cdots )*\alpha_n \rel \{0,1\}$.


text:
## Stream: new members

### Topic: fundamental theorem of algebra

#### Mii Nguyen (Apr 05 2019 at 10:31):

I try to formal the Fundamental theorem of algebra. Here it's my work:

theorem fundamental_theorem_of_algebra: ∀(p: polynomial ℂ) (degree p ≥ 1), ∃ (z: ℂ), is_root p z := sorry

I would like to declare non-constant polynomial p by using degree p >= 1. But lean demand the condition

⊢ has_le (polynomial ℂ)

that I don't understand. Could somebody help me explain it?

However, I try another formal

theorem fundamental_theorem_of_algebra_2: ∀(p: polynomial ℂ), ∀ (c:ℂ) ( p ≠ C c), ∃ (z: ℂ), is_root p z := sorry

and it work, but I wonder if there are a better method to formal a non-constant polynomial?

#### Kenny Lau (Apr 05 2019 at 10:43):

(H : degree p ≥ 1)

#### Johan Commelin (Apr 05 2019 at 10:56):

@Mii Nguyen Are you aware of https://github.com/leanprover-community/mathlib/pull/851?

#### Mii Nguyen (Apr 05 2019 at 11:02):

@Johan Commelin I didn't update that

#### Johan Commelin (Apr 05 2019 at 11:02):

No, it's just that I wondered if you realised that someone else had formalised FTA as well.

#### Mii Nguyen (Apr 05 2019 at 11:06):

@Johan Commelin No, I didn't realised it. I study this version of FTA now for understand. Thanks for your info.

Last updated: May 17 2021 at 21:12 UTC


text:
# Baron Ferejohn (1989) - Bargaining In Legislatures

Article
Has bibtex key
Has article title Bargaining In Legislatures
Has author Baron Ferejohn
Has year 1989
In journal
In volume
In number
Has pages
Has publisher

Handwritten Notes from Class. Also: Summary downloaded from Muriel Niederle's homepage here.

## Reference(s)

Baron, D. and J. Ferejohn (1989), Bargaining in Legislatures, American Political Science Review 83 (December), 1181. pdf

## Abstract

Bargaining in legislatures is conducted according to formal rules specifying who may make proposals and how they will be decided. Legislative oucomes depend on those rules and on the structure of the legislature. Althrought the social choice literature provides theories about voting equilibria, it does not endogenize the formation of the agenda on which the voting is based and rarely takes into account the institutional structure found in legislatures. In our theory members of the legislature act noncooperatively in choosing strategies to serve their own districts, explicitly taking into account the strategies that members adopt in response to the sequential nature of proposal making and voting. The model permits the characterization of a legislative equilibrium reflecting the structure of the legislature and also allows consideration of the choice of elements of that structure in a context in which the standard, institution-free model of social choice yields no equilibrium.

## Key Concepts

Amendment rules:

• Open: An amendment may be added and voted on (essentially instead of the motion), or the motion can be moved (to an immediate vote).
• Closed: Motion voted on immediately against the status quo. There are no amendments.

## The Model

The legislature consists of:

• $n\,$ members - each represents a district. $n\,$ is assumed odd.
• a recognition rule that determines who may make a proposal: this is random and exogenous
• an amendment rule: Open or Closed
• a voting rule: simple majority voting is used

There is one (non-negative) unit of benefits to be split among the districts. The members are risk neutral and their utility depends only on the benefits to their district. The game is one of perfect information. Members can not make binding commitments - their strategies must be self-enforcing at all points. There for the solution concept is SPNE.

The model assumes:

• $p_i\,$ is the probability that a member is recognized.
• $x^i = (x_1^i, \ldots, x_n^i)\,$ is a proposal for the distribution such that $\sum_j x_j^i \le 1\,$
• The status quo is no allocation and under a closed rule members vote against the status quo.
• For open rules an amendment is a new proposal from a different member ($j \ne i\,$) who is recognised with probability $\frac{p_j}{\sum_{k \ne i} p_k}\,$. Only one amendment can be made, or the motion can be passed forward. If an amendment is made the vote is between the original motion and the amended motion. If the vote is for the amendment it becomes the motion on the floor, otherwise the motion on the floor persists into the next period. The game ends when a vote on the motion on the floor is passed.
• The discount factor is common: $\delta \le 1\,$.
• Members have utility: $u^j(x^k,t) = \delta^t x_j^k\,$
• A pure strategy at time $\tau\,$ is: $s_{\tau}^i: H_{\tau} \to [yes,no]\,$, where $H_{\tau}\,$ is the history
• Mixed strategies $\sigma_{\tau}^i\,$ is a probability distribution over $s_{\tau}^i\,$
• $v_i(t,g)\,$ is the value of sub-game $g\,$, and $\delta v_i(t,g)\,$ is the continuation value if the legislature moves to sub-game $g\,$. $v_i\,$ is the ex-ante value at the beginning of the game.
• Voting occurs sequentially and openly, allowing the elimination of weakly dominated strategies.

### Closed Rule - Two Sessions

Suppose equal probabilities of recognition, and that members get zero if nothing is past at the end of the final session. The tie-break rules are:

1. A member votes for a bill if indifferent between the its distribution and the continuation value.
2. A member whose vote will not be decisive votes for a bill iff its distribution is at least as great as the continuation value.

The game is solved by backwards induction.

An strategy is a SPNE iff:

• If recognized in period 1 propose:
• Give $\frac{\delta}{n}\,$ to any $\frac{(n-1)}{2}\,$ other members
• Keep $1-\frac{\delta (n-1)}{2n}\,$ for himself
• If recognized in period 2 propose:
• Keep everything
• Vote for:
• Any first period proposal that gives at least $\frac{\delta}{n}\,$
• Vote for any second period proposal

The proof is straight forward:

$v_i(2,g) = 0\,$

As the game ends after period two, the continuation value is zero.

$v_i(1,g) = \frac{1}{n}\,$

As each member has equal probability of being recognized in the next period, is risk neutral, and can assign all benefits to themselves.

Therefore vote yes iff offered at least $\frac{\delta}{n}\,$. And the minimal majority needed to pass the vote is:

$\frac{(n-1)}{2}\,$.

Therefore keep:

$1-\frac{\delta (n-1)}{2n}\,$

In a three member legislature this is:

$1-\frac{\delta (1)}{3}\,$

As $n \to \infty\,$ this becomes:

$1-\frac{\delta (1)}{2}\,$

Therefore the payoffs to the first proposer are:

$u \in \left[1-\frac{\delta (1)}{2}, 1-\frac{\delta (1)}{3} \right]\,$

Key notes:

1. The distribution reflects the majority rule used
2. Recognition in the first period, in conjuntion with the closed rule, gets the member the largest share. This is agenda power. The proposed gets at least half the benefits.
3. The initial offer is accepted and the legislature adjourns after 1 period. This results from impatience.
4. If the members have different probabilities of recognition their continuation value in period 1 is equal to their probability of recognition.
5. A recognised member will choose to share with the members with the lowest continuation values, and the member with the highest probability will have the lowest ex-ante value for the game.
6. The SPNE doesn't say who should be choosen to share with, and can randomize. Randomization provides a stationary symmetric solution.

### Closed Rule - Infinite Sessions

#### No equilibrium restriction

From proposition 2 in the paper, if:

$1 \gt \delta \gt \frac{(n+2)}{2(n-1)} \mbox{ and } n \ge 5\,$

then: Any distribution of benefits ($x\,$) may be supported.

To support an arbitrary distribution $x \in X\,$ then:

1. A member proposes $x\,$ when recognized, everyone is to vote for $x\,$
2. If a majority rejects $x\,$, then the next member proposes $x\,$
3. If a member is recognized and proposes $y \ne x\,$ then
1. A majority $M(y)\,$ is to reject $y\,$
2. The next member proposes $z(y)\,$ such that for the deviator $z_j(y) = 0\,$ and everyone in $M(y)\,$ is to vote for $z(y)\,$ over $y\,$
3. If the next member doesn't propose $z(y)\,$ repeat the above stage to punish that member.

However, this is unsatisfactory as it requires a complete history at all points (which is unrealistic if $\delta\,$ is a reelection probability and new members can't know the history), and if a member were indifferent between enforcing and not, it is only weakly credible.

#### With equilibrium restriction: Exactly the same as Closed Rule +Finite Horizon

To restrict the equilibrium space the paper considers Stationary Equilibrium.

Two sub-games are structurally equivalent iff:

1. The agenda is identical
2. Set members who may be recognized (at the next node) are identical
3. The strategy sets of the members are identical

An equilibrium is stationary if the continuation values for each structurally equivalent subgame are the same. This necessarily has that strategies are stationary - members take the same actions in structurally equivalent subgames. Not that stationary strategies are history independent.

In the case of equal probabilities, majority rule and infitite session, proposition 3 in the paper states that for all $\delta \in [0,1]\,$ a stationary SPNE in pure strategies exists iff:

• A recognised member proposes to give $\frac{\delta}{n}\,$ to $\frac{(n-1)}{2}\,$ randomly choosen other members, and to keep $1-\frac{\delta (n-1)}{2n}\,$ for himself
• Each member votes for any proposal that gives him at least $\frac{\delta}{n}\,$
• The first vote recieves a majority, so the legislature completes in one session

This is the same as closed-rule, finite session.

Note that we have minimum winning coalitions, a value of the game $v_{i}=1/n$ proposal power equal to $x_{i}^{i}\geq 1/2 \geq \delta/n$. We have fairness because the value of the game is equal across all players.

Comparitive statics around proposal power: $x_{i}^{i}=1-\frac{\delta(n-1)}{2n}$. Note that patience reduces proposal power: $\frac{\partial x_{i}^{i}}{\partial\delta}=-\frac{n-1}{2n}\lt 0$. Larger legislatures reduce proposal power:$\frac{\partial x_{i}^{i}}{\partial n}=-\frac{\delta}{2}+\frac{\delta(n-1)}{2n^{2}}=\frac{-\delta(n^{2}-n-1}{2n^{2}}\lt 0$.

Difference in equality between proposer and coalition members=$\Delta=1-\frac{\delta(n-1)}{2n}-\frac{\delta}{n}$. $\frac{\partial \Delta}{\partial n}=\frac{\delta}{n}\gt 0$. Coalition members also worse off as legislature size gets bigger.

### Open Rule: Infinite session

Editor: What about open rule finite session? Doesn't seem to be in the paper anywhere.

Open rule: Member is recognized and makes a proposal. Another member is recognized and can either move the previous question to a vote agains the status quo (which is zero for everyone), or offer an alterative proposal to be voted against the previous proposal. If first proposal ends: Game over. If amendment wins -- another recognition round to offer proposal.

The equilibrium strategy is as follows:

• If recognized, keep $\hat{y}^{a}$ for yourself and distribute the remainder to $m(\delta,n)$ other members, where $1-n\geq m \geq (n-1)$. $1\gt m/2\geq 1/2$.
• If recognized as an amender who is part of the aforementioned group of $m$: Move to a vote.
• If recognized as an amender who is NOT part of the aforementioned group of $m$: Make a proposal to keep $\hat{y}^{a}$ for yourself and distribute the remainder to $m(\delta,n)$ other members -- including all those who are not in the first proposer's majority. Recognizer is never included because he is too expensive to pay off.
• If you're a voter: Same rules as above. Vote for whichever pays you higher, and for the amendement if you're indifferent.

In equilibrium, the game procedes sequentially until nature randomly selects an amender who is in the coalition of the original proposer, when the proposal is approved.

#### Proof

Let $y^{a}$ be what proposer (j) keeps for self m members have continuation values offered to $V_{j}^{m}(y^{j})$ be the value of the game to $j$ when $y^{j}$ is on the floor.

Note: Stationarity implies that $y^{j\ast}$ will be the same in all recognized rounds. Note: N members, member 1 is recognized, member j is excluded member recognized, member

A: $\frac{1-\hat{y}^{a}}{m}\geq \delta V_{i}(y^{i})$. This means that the amendment $\max V_{i}^{m\ast}(y_{i}).$

#### Substantive conclusions

• Possibility of delay.
• Size of winning coalition can be larger than the minimum winning coalition.
• $\frac{\partial m}{\partial \delta}\lt 0$, and (separately) $\frac{\partial m}{\partial n}\lt 0$ (Rui says this latter one needs to be checked).


text:
Calculates the log-likelihood of the mixture model based on estimates for theta and phi.

get.llk.mixmod(phi, theta, ndata.types, dat, nobs, nmaxclust)

## Arguments

phi A list of proportion estimates that characterize distributions (bins) for each data stream and possible behavioral state. numeric. A vector of values that sum to one. numeric. The number of data streams being analyzed. A data frame containing only columns of the discretized data streams for all observations. numeric. The total number of rows in the dataset. numeric. A single number indicating the maximum number of clusters to test.

## Value

A numeric value of the log-likelihood based upon the current values for phi and theta.


text:
# undefined control sequence \abx@aux@sortscheme

I am trying to start a basic tex document with bibtex citations as follows:

\documentclass{article}

\usepackage[backend=bibtex]{biblatex}
\bibliography{biblo}

\begin{document}
\printbibliography
\end{document}

However, when I run pdflatex mytexfile.tex I receive the error that \abx@aux@sortscheme is undefined. How can I resolve this?

• The syntax should be \addbibresource{biblo.bib}. Sep 9, 2016 at 22:01
• I think this error occurs after having changed some options to biblatex or removing the inclusion itself, then recompiling --> there is some code left in .aux file. Try to remove the aux-file
– user31729
Sep 9, 2016 at 22:04
• @Bernard: Yes, but \bibliography{biblo} still works (however not recommended)
– user31729
Sep 9, 2016 at 22:04
• @Christian Hupfer: That Biblatex is really a good lad… Sep 9, 2016 at 22:06
• @Bernard: On page 71 of current biblatex.pdf: \bibliography ... The legacy command... and as a marginnote: Deprecated (see at the bottom of the page)
– user31729
Sep 9, 2016 at 22:08

The reported error \abx@aux@sortscheme being undefined is a consequence of changing options to biblatex or having removed the \usepackage[...]{biblatex} entry followed by recompilation.
@aux indicates that this setting is written to the .aux file, after changing or removing this entry is still left in the .aux file and causes the error message during the next run.
Solution: Delete the .aux file and recompile.
A similar error occurs when changing language options from babel.
Side note: \bibliography is a deprecated command in conjunction with biblatex -- use \addbibresource instead.


text:
Question 6

# Find the mean proportional between 2 and 98?

Solution

Let,

The mean proportional of 2 and 98 is the number ‘n’ in the equation,

$$\frac{2}{n}=\frac{n}{98}$$

Solving for n by cross-multiplication , get that

or, $$\left(n\right)\left(n\right)=\left(98\right)\left(2\right)$$

or, $$n^2=\left(7\right)\left(7\right)\left(2\right)\left(2\right)$$

or, $$n^2=7^2\times 2^2$$

or, $$\sqrt{n^2}=\sqrt{(7^2\times 2^2)}$$   (by taking square root on the both side)

or, $$n=\left(7\right)\times\left(2\right)$$

or, $$n=14$$

So, the mean proportion of 2 and 98 is 14 .

Create a FREE account and get:

• Download RRB Study Material PDF
• 45+ RRB previous papers with solutions PDF
• 300+ Online RRB Tests for Free

Comments

OR

Boost your Prep!


text:
# How to understand the formation of covalent bond from the viewpoint of wavefunction interaction?

The formation of the covalent bond is usually understood from the electron sharing, like explained by the following figure:

However, this simple physical picture seemingly hinders many things. In quantum mechanics or chemistry, the electron is represented fully by the wavefunction.

Therefore, how can I understand the formation of the covalent bond from electron/wavefunction interaction?

I have gained some initial ideas when two wavefunctions are touched, such as:

• Deformation;
• Electrostatic attraction;
• Exchange;
• Polarization;
• Charge transfer.

These processes/concepts play an important role when two wavefunctions interacted. (I'm sorry I can't find concrete references for these concepts or processes.) In these processes, the attraction interaction (bonding) will compete with the repulsion interaction (antibonding) to find the ground state.

I post this question to look a full understanding of the formation of the covalent bond from electron wavefunction interaction.

• When you calculate the full system, let said, two atoms, you will have only one wavefunction for the whole system, not two "interacting" wavefunctions (one for each atom). By the way, the wavefunctions do not interact one with other. The bond is "defined" starting from the electronic density that is calculated from the probability density using the wavefunction.
– Camps
Oct 28 '20 at 13:43
• Basically, when two atoms are very far apart, the wavefunction of the total system can be taken as a product of two wavefunctions to a high degree of accuracy. In other words, the system hamiltonian is the sum of the hamiltonians for each atom, with no coupling terms. So, what you're asking is basically how good the atom-only basis is at representing the case when there is a strong interaction between two atoms. The answer is that it will be very bad. You can attempt to say what physical effects lead to this breakdown, as you've suggested, but this is never unique and always contentious. Oct 28 '20 at 18:10
• @jheindel you should submit that as an answer Oct 29 '20 at 16:10
• @Camps Great! you should submit that as an answer Oct 29 '20 at 16:11
• This is a good question, and probably the best result will be to get multiple answers explaining different ways of thinking about it. Oct 29 '20 at 16:12

There is more than one way to answer this question well. I'll give one answer here.

Let's treat the two atoms (that will form a covalent bond) as two localized election states (orbitals). The detailed shape isn't important. The Hamiltonian of this system could be written:

$$H = V (n_{\uparrow,1} n_{\downarrow,1}+ n_{\uparrow,2} n_{\downarrow,2}) - t \sum \limits_{s=\pm1/2} (c^\dagger_{s,2} c_{s,1} + c^\dagger_{s,1} c_{s,2})$$

Here $$n_{s,i}$$ is the total number of electrons on site $$i$$ (0,1, or 2) and the $$c^\dagger_{s,1} c_{s,2}$$ terms 'hop' an electron with spin $$s$$ from site 1 to 2 (and vice versa).

Basically here you have $$V$$ representing an electron-electron repulsion (they don't want to be on the same site) and $$t$$ representing a tunneling between the two sites. This tunneling is roughly proportional to the wavefunction overlap between the site 1 and 2:

$$t \approx \int d^3r \psi_1 \psi_2$$

Now, when the two atoms are far away, that overlap will be small and $$t\to 0$$. In this limit we just have two isolated atoms. If we put two electrons in the system, the ground state would have one on each site with any combination of spins.

As you bring the two atoms closer together, the tunneling becomes finite and it breaks the degeneracy, lowering the energy of a pair of opposite aligned spins because they can tunnel back and forth and pick up a $$-t^2/V$$ decrease in energy (from second order degenerate perturbation theory.

This example is essentially the Hubbard model.

• Very nice ideas. When the competition between kinetic energy and interaction energy is minimized, the covalent bond is formed?
– Jack
Oct 30 '20 at 0:57


text:
Pinned toot

I wrote a little post about the war of attrition (game theory) with two players. It's intended mostly for a graduate audience:
mattwthomas.com/blog/war-of-at

All of this was in other places, but it took me a while to understand this fully and felt it was good to share it all in one place since it would have saved me a lot of time.

Pinned toot

I just finished a post about QF (quadratic finance) which some are pushing as the "mathematically optimal" way to fund open source.

mattwthomas.com/blog/fund-open

I hope to clear up some of the confusion that I've heard about it and point out some problems with the way that people outside of Economics are interpreting the original paper.

Ladybugs are the lion of the garden. Look at these fine bois feasting on scale insects (or aphids)

I created a python package to generate approximate solutions to Volterra and Fredholm integral equations of the first kind:

pypi.org/project/inteq/

I don't think there is another package that implements these routines. The Fredholm one isn't that great, but it's a difficult problem since uniqueness isn't guaranteed.

A funny music video: youtu.be/92-eHSMdBG0

Does anyone know best practices for warning that data is for academic use only in a software package?

I am including a dataset with such a restriction in an R package.

The source code for R is pretty nice and easy to follow.

(timelapse)

Announcing Meme It So! It's a Star Trek The Next Generation Meme and GIF Generator, inspired by Frinkiac.

It has every line of dialogue spoken on STTNG tied to screencaptures. Which makes it good for looking up lines as well.

memeit.so/

I wrote up another set of notes about game theory papers that I'm working through.

mattwthomas.com/blog/covert-di

This one is about putting handicaps in contests to make them more competitive. I'm particularly interested in anonymous/covert handicaps. A handicap like this is fair in the sense that everyone has the same rules. However, it can be designed to internationally advantage a particular player.

Travel agents

If you advertise that you can vest your retirement account in "angel investments", then you are deeply illegitimate.

Test

$$∑_{i=1}^{n} \frac{1}{n} \to \log(n)$$

$$\tilde{g}_{-i}(s) = \frac{c'_{i}(s)}{v_{i}(s; s)} - \int_0^s \frac{v'_{i}(s; y)}{v_{i}(s; s)} \tilde{g}_{-i}(y) dy$$

I decided to make the text really big for desktop users on my site:
mattwthomas.com/

Does anyone like or dislike? I basically set the root font size to 20 if your screen is more than 800px wide.

For anyone who visits CBC's website, they now have a lighter version that is much, much smaller: CBC Lite

cbc.ca/lite

I wrote a little post about the war of attrition (game theory) with two players. It's intended mostly for a graduate audience:
mattwthomas.com/blog/war-of-at

All of this was in other places, but it took me a while to understand this fully and felt it was good to share it all in one place since it would have saved me a lot of time.

coqui.ai/

Ex mozilla speech devs to develop open speech service!

Got a new printer. I guess some things never change.

@floppy @verbumfeit @pcastela we are, just like wikipedia, operating not-for-profit. Our mission is to ensure freedom, build a free and safe home for Free and Open Source projects.

Death to endnotes, footnotes are the only acceptable way to insert parenthetical remarks.

Unless you're Ada Lovelace, in that case write your frickin magnum opus in the endnotes laying the foundation for all modern mechanical computation, that's fine.

But for the rest of us, please don't make us save our spot, flip to a page to read something, and then flip back.

Opinion: all modern languages should have builtin support for *basic* YAML syntax

Show older

The social network of the future: No ads, no corporate surveillance, ethical design, and decentralization! Own your data with Mastodon!