text:
# How do astronomer measures the size of any celestial objects?

What techniques and tools are available to the astronomers to measure the size of any celestial objects such as, stars or perhaps black holes that doesn't emit light nor reflects starlight?

If a celestial body is larger than the resolving power of a telescope, its size can be measured directly. This is the case for most galaxies, molecular clouds in the Milky Way and nearby galaxies, and even for a few nearby stars. EDIT: See discussion by Rob Jeffries on how these measurements are carried out for stars using interferometry.

For more distant stars, we can rely on our understanding of stellar evolution, which tells us pretty accurately the radius, once we know its spectrum (EDIT: or just a assume blackbody radiation and use the formula given by Rob). If the star is a member of a binary system whose orbit we observe roughy edge-on, we can measure how the luminosity declines as one star occults the other, and calculate the radius. This can also be used to measure the sizes of exoplanet. And for stars, the same technique is even possible using our own Moon as occulter. See a description here.

Black holes (BH) that don't emit light, cannot be measured (at least not until we are able to detect gravitational waves), but often BHs are surrounded by a disk of accreting gas, which is heated to million of degrees by friction as it spirals down the drain. Measuring the temperature of this gas tells us the mass of the BH, which is directly proportional to its radius ($R_{mathrm{BH}} simeq 3 M/M_odot$ km).

A nice technique for measuring the size of the quite small region of gas clouds around a supermassive BHs, even though they are billions of lightyears away, is called reverberation mapping. Here, some of the light emitted from the BH's accretion disk travels directly in our direction, while some of if travels in other directions, illuminating the clouds around it. When we measure the light from those clouds, the signal looks like the directly observed signal, but with a delay $t$ corresponding to extra length of the path that the light has taken. Since we know the speed of light $c$, we can calculate the extra distance as $d = ct$, i.e. the size of the system.

The main tool to measure the diameter of a star is interferometry combined with a parallax-based distance measurement - a brief review by Kervella (2008) might be useful. The principles behind interferometry are described here.

Interferometry involves measuring the light from a star using two (or more) telescopes that are separated by some distance. Together, the signals from these telescopes can be combined to give an angular resolution that can be (in the best circumstances) equivalent to a telescope with a diameter equal to the telescope separation. These measurements give the angular size of the star, which must then be multiplied by their distances to get a physical diameter.

One of the most successful experiments is the Chara array, which has yielded diameters for many nearby stars. Precisions can be as good as a few percent, but more usually 10% and of order 100 (predominantly nearby) stars have had their radii measured in this way.

A second main direct technique is to use eclipsing binary systems. The measured light curve can be used in an almost model-independent way to estimate the radii of the two stars involved. Of course most eclipsing binaries are close pairs with short orbital periods and with orbital inclinations that allow us to see the eclipse. They are therefore highly prized objects. Radii can be measured with precisions of 1%. A reasonably complete catalogue of the $sim 100$ known eclipsing binaries with precise radii can be found here.

Another technique is lunar occultation. The passage of a star behind the limb of the moon results in a changing diffraction pattern that can be used to estimate the angular size of the star. Again a distance is required to convert this into an actual diameter.

More distant stars are inaccessible - their angular diameters are simply too small. At the moment only indirect estimates of their radii are possible. For example, if we were to assume that a star radiates as a blackbody, then its luminosity ($L$), radius ($R$) and temperature ($T$) are related by Stefan's law. $$L = 4pi R^2 sigma T^4,$$ where $sigma$ is the Stefan constant. If the star has a measured flux at the Earth and we know how far away it is, then $L$ can be estimated. If we take a spectrum and estimate its temperature, then the equation above can be rearranged to give the radius in terms of the measured luminosity and temperature. Real stars are more complicated than blackbodies, but the principle is the same.

Neither of the above techniques can work for black holes, and the sizes (event horizon or Schwarzschild radius) of black holes have not yet been directly measured. The physics of a black hole is relatively simple(!) and so there is a direct relationship between their Schwarzschild radii and their masses (modified somewhat by rotation). Basically it is 3 km multiplied by the mass in solar units. Therefore a measurement of the black hole mass gives its "radius". The masses of black holes are measured by looking at the motions of stars and gas around them and applying our knowledge of how gravity works.

## How do astronomer measures the size of any celestial objects? - Astronomy

I am starting out in amateur astronomy, but would like to get more out of the exercise of star-gazing than just - well, star-gazing. I would like to observe (not just read) how far the celestial objects are to us, how fast they are moving, and what gases they are composed of. To this end, a telescope may be a good beginning point, but certainly not the final stop - it would be great to have access to a spectrometer and an UV ray-detecting telescope. Are these hopelessly outside the price range of a conventional amateur?

Hi. It seems that you want to see how easy it is to do science with amateur astronomy equipment, and how much would it cost. The answer is that yes, amateur astronomers can do some significant science, not only redo old experiments but also contribute to the field. To do any UV astronomy is pretty impractical, but there are things you can do with other equipment, for different price ranges.

1) A very simple spectroscope: you can get pretty cheaply (a couple hundred $or less) a small prism spectroscope, but what you can see is mainly limited by the size of your telescope. In fact, that is true for almost anything. But with this spectrometer (you could even make it yourself) on a moderately sized telescope (say 8-12" diameter) you can see spectral lines in some of the brighter stars. You cannot do too much that is really scientific, since all you see are the lines on a piece of paper. For details, look in the back of a Sky & Telescope magazine. 2) A CCD. For about$500 you can get a very decent CCD camera (an electronic camera) for use on almost any telescope. Things people do are: search for supernovae, search for asteroids/comets, or monitor variable stars. All three areas have had very significant contributions made by amateur astronomers, and they continue to do so. You can find more information on the Sky & Telescope site (www.skyandtelescope.com), or search for the AAVSO (American Association of Variable Star Observers). A CCD will also allow you to do other things like look at Cepheid variables, track the motion of the moons of Jupiter/Saturn, and similar things that have been done but are nonetheless interesting.

3) A telescope. It is really essential to everything you do. The bigger, the better, but that isn't the most important thing. For info on telescope selection you can also look at the S&T site, but one of the more important things are good sky conditions and a motorized mount (essential for CCD work).

Another option is radio astronomy. This really requires some technical background because it is much more do-it-yourself than optical astronomy, but there is a lot of room for science. Look for either the Society of Amateur Radio Astronomers (SARA) or the amateur radio astronomy mailing list (ara). You can do things like monitor the rotation of the galaxy, listen to noise storms on Jupiter, monitor solar flares, observe meteors, or even search for extra-terrestrial intelligence. You can find info about all of this on the web.

This page was last updated July 18, 2015.

### About the Author

#### Dave Kornreich

Dave was the founder of Ask an Astronomer. He got his PhD from Cornell in 2001 and is now an assistant professor in the Department of Physics and Physical Science at Humboldt State University in California. There he runs his own version of Ask the Astronomer. He also helps us out with the odd cosmology question.

## Elementary School Science

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This lesson contains affiliate links to products I have used and personally recommend. At no cost to you, I make a commission for purchases made through the links or advertisements. These commissions help to pay for the costs of the site and enable it to remain free for anyone who wants to use it.

The students will learn about astronomy.

The students will be able to define the terms: astronomy, astronomer, and telescope.

The students will be able to name famous astronomers and explain their contributions to astronomy.

The students will be able to describe the five fields of astronomy.

Questions that encompasses the objective:

Think about the Solar System. How did we learn that there are planets, stars, and other objects in space?

Prepare the Learner: Activating Prior Knowledge.

How will students prior knowledge be activated?

Warm up by asking students:

What do you know about astronomy?

Can you name any famous astronomers?

Common Core State Standards:

Materials and Free Resources to Download for this Lesson:

Page 9: History of Astronomy

Pages 10-12: Famous Astronomers

Pages 13-14: Fields of Astronomy

&ldquoA Day in the Life of&hellip&rdquo activity materials:

&ldquoFamous Astronomers&rdquo &ndash Here are some ideas

Arno Penzias and Robert Wilson

Giovanni Domenico Cassini

**The teacher can either provide fact sheets for the students or allow the students to do research on their own to gather information about their assigned astronomer**

###### Space Lesson Plans: How Big Is The Solar System?

Input:
What is the most important content in this lesson?
To reach this lesson&rsquos objective, students need to understand:

The field of astronomy and its importance in science.

The definitions to the terms: astronomy, astronomer, and telescope.

The names of famous astronomers and their contributions to astronomy.

The five fields of astronomy.

How will the learning of this content be facilitated?

For the first few minutes, the students will fill out the &ldquoWhat I Know About Astronomers&rdquo worksheet.

The teacher will begin the class by handing out the &ldquoGuess the Astronomer&rdquo worksheet. On the worksheet is a box with the names of the three famous astronomers: Galileo Galilei, Nicolaus Copernicus, and Isaac Newton. There will be nine questions (three for each of the astronomers). The teacher should allow the students about 10 minutes to read through each question and write down their guess of which astronomer refers to each question. The teacher should tell the students that each astronomer will have three statements on the worksheet. At the end of 10 minutes, the teacher should review the worksheet with the students.

Galileo [G] Newton [N] Copernicus [C]

Believed the Earth was not the center of the Earth, but did not believe that the moon caused tides. (G)

First scientist to prove that the Solar System revolved around the Earth. (G)

Invented and improved the telescope. (G)

Defined the 3 laws of motion and universal gravitation. (N)

Invented the reflecting telescope in 1668. (N)

Explained that gravity helps the planet orbit around the Sun. (N)

Stated that the Sun is in the center and that the planets orbit around it (theory called &ldquoHeliocentrism&rdquo) (C)

Not only was he an astronomer, but a physician, scholar, economist, translator, mathematician, artist, and diplomat. (C)

Many people believe he started modern astronomy. (C)

After the worksheet is reviewed, the teacher will begin presenting the information on the Astronomers. If it is possible, project each page of the teacher&rsquos copy of the &ldquoAll About the Solar System&rdquo Science Journal worksheet onto the board using a projector or put into a PowerPoint document and project. The teacher&rsquos copy of the journal has certain words/phrases that are bolded red and highlighted. It is important the teacher explain to the students those words/phrases are to be highlighted in their (students) journal. For this lesson, the teacher should review these pages:

Page 9: History of Astronomy

**This page reviews what astronomy is and its history**

Pages 10-12: Famous Astronomers

**These pages will review the three famous astronomers from the beginning of class: Galileo Galilei, Nicolaus Copernicus, and Isaac Newton**

Pages 13-14: Fields of Astronomy

**This page explains the different fields of astronomy: Observational Theoretical Solar Planetary Stellar**

After the information has been presented, the students will break into pairs. On the board, the teacher should write the names of famous astronomers (See activity section). Each pair will pick an astronomer. Together, the students will create a poster board depicting what life was/is like for that astronomer. (Refer to materials section as to how information/research should be obtained). Allow the students about 15-20 minutes to complete the poster board.

**Each student/pair can either present their &ldquoPlanet&rdquo or a gallery walk can be set up. For the gallery walk, each pair will place their planet poster on desks throughout the room. The students will walk around and look at their peers&rsquo poster**

After the presentation or gallery walk is over, the students should reconvene and the teacher should discuss the activities. The teacher should review the vocabulary words with the students.

The students will then fill out the &ldquoWhat I Learned About Astronomers&rdquo worksheet.

They will then share with the class, or with a partner, some things they wrote down in their &ldquoWhat I Learned About Astronomers&rdquo page.

**The students should keep both of the "What I Know" and "What I Learned" sheets in a folder to put them all together into a book when the unit is finished.**

## How do astronomers determine masses of planets and other celestial objects? It's not hard to scale their volume, but it's not enough to find mass out. Do they define prevailing materials and substances and speculate on the density to calculate mass using volume or what?

Usually masses are determined by observing the orbits of things around them, in the case of planets it's usually the planets' moons that are used. A lot of stars are in binary systems, and the binary orbits were used to understand the relationship between brightness and mass, which can be used to determine the mass of solo stars. For more distant things like galaxies, we can see how much the light of even farther objects gets deflected by gravity, and use that to calculate the mass.

And what do you do with planets that have no moons like Mercury or Venus?

For clarity, the planet-moons part would only apply in the Solar System. For exoplanets, it's usually done with the same technique as binary stars (using the star-planet orbit). There's some cases where the planet mass is estimated from theoretical models (but those are very uncertain) and through microlensing (which involves the bending of light by massive objects).

/u/iorgfeflkd answered your question well, but I wanted to add that for objects outside our solar system, it's usually easier to directly measure the mass than the radius/volume. For almost every binary star system, the stars are just points of light to us, so directly measuring the radius is impossible - but watching their movement and change in velocity can be done, which allows us to work out the mass.

Typically we measure the mass of stars indirectly though, because we have a really good handle on stellar evolution, and because stars are fairly simple, so it's possible to give the mass of a star based on unresolved stuff like its colour & brightness.

If the binary system is an eclipsing binary system, that is the stars pass in front of one another as they orbit, then it is possible to get a measurement of their radii relative to the size of their orbit (specifically the semi-major axis). To measure the size of the orbit you need a number of radial velocity (velocity along observer's line of sight) measurements, but fairly routine for eclipsing binaries. Eclipsing binaries are one of the rare examples where it is possible to measure both the mass and radius of the stars involved.

Although we do have a fairly good idea of how a star evolves, there are still lots of uncertainties present in the models used for stellar evolution. For example, there are uncertainties associated with how convection is modelled, there are examples where the models cannot reproduce the radii and temperature of stars in observed binary systems, at least for low mass stars (say <2 solar masses). There is currently a lot of research going into trying to improve these models because for single stars, the models are often the only way to obtain a mass, something that is important if you are trying to understand the environment around a planet-hosting star.

## Astronomers measure new distances to nearby stars

Astronomers from the U.S. Naval Observatory (USNO), in collaboration with others from the REsearch Consortium On Nearby Stars (RECONS), have determined new distances to a group of faint young stars located within 25 parsecs (pc) of the sun. These measurements, based on parallax observations obtained over periods ranging from nine to twelve years, include new measures of the star known as TRAPPIST-1, which has been recently identified as having a system of up to seven Earth-sized planets orbiting around it.

The paper describing the measurements, whose lead author is Dr. Jennifer Bartlett of the USNO, has been published in the Astronomical Journal.

Measuring the distances to nearby stars is accomplished by a technique called "trigonometric parallax," in which the tiny apparent annual shift of a star's position is related to the diameter of the Earth's orbit around the sun. By measuring the tiny angle produced by this motion, astronomers can use trigonometry to determine the distance to the star.

The distances to these stars are measured in parsecs, short for "parallax second." One parsec is the distance at which a star would show an annual parallax shift of one second of arc on the plane of the sky it is equivalent to 3.26 light-years.

Utilizing the 0.9-meter telescope at the Cerro Tololo Inter-American

Observatory (CTIO) in Chile, Dr. Bartlett and her colleagues measured the parallaxes of 32 stellar systems, most of which are very cool faint red dwarf stars. Of these systems, 17 have never had previous parallax measurements, and out of those, 14 have been found to lie less than 25 pc from the Earth. One of these new, nearby stars, 2MASS 2351-2537AB, also shows evidence of actually being two new nearby stars, i.e., a binary.

In addition to these newly-measured star systems, the astronomers also obtained new parallax measurements for 15 other, previously-known nearby stars. Among these is the star known as "TRAPPIST-1," which has been recently shown to host a system of at least seven planets. Using over 12 years of observations made with the 0.9-meter telescope at CTIO, Dr. Bartlett presents a new parallax of 79.29 +/- 0.96 mas, yielding a distance of 12.61 pc, about 4 percent more distant than previous measurements.

The team also measured the brightness of each of these star systems in three regions of the optical and near infrared electromagnetic spectrum and analyzed how the brightness of each system varied over the course of their parallax observations. Although the team caught TRAPPIST-1 flaring during July 2009, they found the overall variability of this star to be low. In other words, its brightness varied enough to be detectable but not enough to be considered significant.

For many of these star systems, the team obtained optical spectra using the 1.5 m CTIO telescope. This study is the first to identify LP 991-84 as a M 4.5 V type star, confirming its cool, dim, red nature. As a graduate student, Dr. Bartlett discovered that this star is within 10 pc of the sun. This study measured its parallax more precisely to be 115.90 +/- 1.33 mas, or 8.63 pc.

From their assessment of the positions, motions, variability, and spectra of these 32 star systems, the astronomers concluded that 13 of them are young, probably less than 120 million years old. TRAPPIST-1 and 2MASS 2351-2537AB, however, do not appear to be particularly youthful.

On the other hand, LP 991-84 may be 1 billion years old, or more.

"I am amazed at what we can find in our backyard—well, if your backyard is 25 pc deep," Dr. Bartlett said. "We are still looking and identifying stars within the sun's immediate vicinity, its neighborhood so to speak. What will we find next?"

## What is Extragalactic Astronomy? (with pictures)

The dawn of extragalactic astronomy was in 1917, when American astronomer Heber Curtis observed a stellar nova within M31, the formal name for what was then called the Great Andromeda Nebula. At the time, spiral nebulae such as Andromeda were thought to lie within our own galaxy, with a size only several times larger than that of our solar system and a distance less than 50,000 light-years. They thought the Milky Way Galaxy represented the entire universe.

After observing the nova in M31, Curtis searched the photographic record, noticing 11 additional novae in the region. If M31 was just a stellar nebulae, why were there so many novae within it, and why were these characteristically fainter than other novae? Reasoning from the observation that these novae were about 10 magnitudes fainter than novae known to occur in our own galaxy, Curtis declared that the Great Andromeda Nebula was in fact an "island universe," distinct from the Milky Way and located 500,000 light-years away. Astronomers did not accept his hypothesis at first, and a scientific debate began.

In 1920, Harlow Shapley, another American astronomer, challenged Curtis to a Great Debate on important astronomical issues of the time, including whether spiral nebulae like Andromeda were really outside our own galaxy. Many fellow astronomers followed the debate, but the final results were inconclusive. It was not until 1925, when Edwin Hubble (after whom the Hubble Space Telescope is named) published observations from the 100-inch Hooker telescope, then the largest in the world, that he had discovered Cepheid variable stars in the Andromeda nebulae and used them to measure its distance, found to be an enormous 2.5 million light-years. The era of extragalactic astronomy had begun, and the Andromeda Nebula was renamed the Andromeda Galaxy.

For the past 80 years, extragalactic astronomy has been an active area of research. By measuring the relative speed of galaxies using their optical signature, it was found that all galaxies are moving away from each other and the entire universe is expanding. In 1998 observations of Type Ia supernova even suggested that the expansion is accelerating. Cosmologists now think it is likely that the universe will end in a "Heat Death" where accelerating expansion causes all matter to disperse and freeze.

An important episode in extragalactic astronomy is the discovery and investigation of quasars, QUasi-stellAR radio sources. These bright point sources were known to be very luminous and very remote, among the most distant objects known, with some as far away as 13 billion light-years. Although quasars were first observed in the 1950s, it wasn't until the 1970s that a scientific consensus began to emerge on the nature of quasars: they were active galactic nuclei, consisting of supermassive black holes sucking in several solar masses worth of material per century and releasing tremendous amounts of radiation in the process. Formal models have been built to describe this, and one of the greatest mysteries in extragalactic astronomy was solved.

Today, millions of galaxies have been photographed and classified by scientists, sometimes even using the help of the public (as in GalaxyZoo). Galaxies are either spiral or elliptical. It is estimated that there exist about a hundred billion galaxies in the observable universe. Interestingly, this is about the same as the number of neurons in a human brain.

Michael is a longtime contributor who specializes in topics relating to paleontology, physics, biology, astronomy, chemistry, and futurism. In addition to being an avid blogger, Michael is particularly passionate about stem cell research, regenerative medicine, and life extension therapies. He has also worked for the Methuselah Foundation, the Singularity Institute for Artificial Intelligence, and the Lifeboat Foundation.

Michael is a longtime contributor who specializes in topics relating to paleontology, physics, biology, astronomy, chemistry, and futurism. In addition to being an avid blogger, Michael is particularly passionate about stem cell research, regenerative medicine, and life extension therapies. He has also worked for the Methuselah Foundation, the Singularity Institute for Artificial Intelligence, and the Lifeboat Foundation.

## Frequently bought together

### Review

". has infectious enthusiasm that makes people want to buy a telescope

### From the Back Cover

The night sky is alive with many wonders??distant planets, vast star clusters, glowing nebulae, and expansive galaxies, all waiting to be explored. Let respected astronomy writer Philip Harrington introduce you to the universe in Star Watch, a complete beginner?s guide to locating, observing, and understanding these celestial objects. You?ll start by identifying the surface features of the Moon, the banded cloud tops of Jupiter, the stunning rings of Saturn, and other members of our solar system. Then you?ll venture out beyond our solar system, where you?ll learn tips and tricks for finding outstanding deep-sky objects from stars to galaxies, including the entire Messier catalog??a primary goal of every serious beginner.

Star Watch features a detailed physical description of each target, including size, distance, and structure, as well as concise directions for locating the objects, handy finder charts, hints on the best times to view each object, and descriptions of what you?ll really see through a small telescope or binoculars and with the naked eye.

Star Watch will transport you to the farthest depths of space??and return you as a well-traveled, experienced stargazer.

## An astronomer's guide to stargazing from your backyard

With the proper equipment, you can enjoy the beauty of the night sky.

This is a challenging time for families. Schools across the US are struggling to provide a meaningful online experience. The coronavirus pandemic has cut off or restricted many entertainment options. As an astronomer, I believe a great way for families to fill the void and have a meaningful science experience in the time of COVID-19 is to turn their attention to the stars they can see right outside their homes.

The night sky is, and always has been, safe and free.

Here are five ways you can get started.

### Naked eye

You can see a lot with just your eyes. But the night sky is a strange landscape for most people. Just as you would when traveling somewhere unfamiliar, you’ll need a map. Sky and Telescope – an astronomy news publication – has a guide to get started, with printable sky maps for any month. The objects of the night sky migrate through a complete cycle over the course of a year.

As the Earth moves around the sun, different stars and constellations come into view, so you can enjoy new sights all through the year. You may find it more convenient to have a planisphere, or sky wheel, a rotating plastic disk that shows the night sky for any date and time. They can be bought for US$10-$20 online.

Spotting planets are trickier since they move among the stars, but there are interactive maps online that show them in the night sky for any time at any location. Your view of the night sky depends on your latitude, so it varies with your specific location. Even easier, there are smartphone apps that take all the work – and maybe also the fun – out of navigating the night sky.

Hold your phone up and the apps identify stars and overlay the constellation shapes. Some respond to voice commands and add detailed information on celestial objects or show the International Space Station as it whizzes overhead.

### Binoculars

You probably have a pair of binoculars somewhere in your house. If you don’t or you want an upgrade, new binoculars cost anywhere from $35 to over$500, and the lower end of the range is just fine for stargazing.

Perhaps you use them at concerts or for bird-watching. Well, they’re also perfect for stargazing. There are two numbers on binoculars. They represent the magnification and the lens diameter, so 7 x 50 will magnify an image by a factor of 7 using 50-millimeter lenses. At a dark location, your naked eye will see about 3000 stars. With binoculars, this number goes up to 100,000.

The moon is spectacular through binoculars. It’s fun for kids to track a cycle of the moon phases over a month, and then do an activity that shows why the moon has phases.

### Small telescopes

If you want to make a bigger commitment to exploring the night sky, consider getting a small telescope. Peering through a telescope opens up a world of star clusters, galaxies, and nebulae. You can see Saturn’s rings and the moons of Jupiter.

It’s a good idea to read a guide before you take the plunge. Small telescopes range from a hundred to several thousand dollars, but you can get a good starter one for as little as $200. These basic telescopes usually have a viewfinder attached to help locate objects, but you’ll need sky maps to get the most out of them. If you can shell out at least$400, you can acquire what are called “GoTo” telescopes that have motors and are computer-controlled, where the telescope does the work of finding the deep sky objects. You just have to type the name in or choose from a list. Now you’ll be ready to learn tricks of the trade, like using a red LED flashlight to preserve your night vision and looking slightly to the side, which lets you see deeper because the cells near the edge of your retina are more sensitive to low light levels.

### Online resources

The internet is a great resource for backyard astronomy. You might want to start with a Crash Course in naked-eye astronomy. In addition to reviews of apps and binoculars and telescopes, there are tutorials on how to use your new telescope.

You’ll also want to check out Sky and Telescope’s weekly “Sky at a Glance.” BBC’s Sky at Night magazine has a longer monthly summary of what you can observe.

And if you just want to be inspired by the visual splendor of a dark night sky, there are a number of time-lapse videos you can enjoy.

### Dark skies

Hopefully, you’re excited about backyard astronomy, but what you can actually see will depend on where you live.

For tens of thousands of years, the night sky was a familiar friend to our ancestors, and they used it to navigate, tell time, and project their myths into the stars and constellations. But the glories of the night sky have been steadily eroded by industrial activity and artificial lights.

You can see recent measures of night sky brightness in a zoomable map of the U.S., where it’s clearly harder to find a dark sky in the eastern half of the country.

To measure how sky brightness affects what you can see, amateur astronomers use something called the Bortle scale, where 9 is an inner-city and 1 is a pristine wilderness. Light pollution is the effect of artificial lighting on the night sky, and you can also see how it affects the familiar Big Dipper and Orion constellations.

National parks are great places to enjoy the sky because they try to protect it from artificial lights, and in normal times many national parks offer astronomy programs. Many are also beginning to resume those programs with COVID-19 restrictions in place.

If you want to contribute to the effort to raise awareness of light pollution by monitoring the sky brightness where you live, the U.S. national observatories run a project called “Globe at Night.”

Anyone can collect data and help a research project by doing citizen science, which is when non-scientists gather data and contribute to a collective research effort. You can become a citizen scientist and submit your own observations from a computer or smartphone.

This article was originally published on The Conversation by Chris Impey at the University of Arizona. Read the original article here.


text:
# Lattice Erasure Codes of Low Rank with Noise Margins

We consider the following generalization of an (n,k) MDS code for application to an erasure channel with additive noise. Like an MDS code, our code is required to be decodable from any k received symbols, in the absence of noise. In addition, we require that the noise margin for every allowable erasure pattern be as large as possible and that the code satisfy a power constraint. In this paper we derive performance bounds and present a few designs for low rank lattice codes for an additive noise channel with erasures.

## Authors

• 4 publications
• ### On additive MDS codes over small fields

Let C be a (n,q^2k,n-k+1)_q^2 additive MDS code which is linear over 𝔽_q...
12/11/2020 ∙ by Simeon Ball, et al. ∙ 0

• ### Twisted Reed-Solomon Codes With One-dimensional Hull

The hull of a linear code is defined to be the intersection of the code ...
09/16/2020 ∙ by Yansheng Wu, et al. ∙ 0

• ### Partial MDS Codes with Local Regeneration

Partial MDS (PMDS) and sector-disk (SD) codes are classes of erasure cod...
01/14/2020 ∙ by Lukas Holzbaur, et al. ∙ 0

• ### Tensor Representation of Rank-Metric Codes

We present the theory of rank-metric codes with respect to the 3-tensors...
04/10/2019 ∙ by Eimear Byrne, et al. ∙ 0

• ### Infinite families of near MDS codes holding t-designs

An [n, k, n-k+1] linear code is called an MDS code. An [n, k, n-k] linea...
10/18/2019 ∙ by Cunsheng Ding, et al. ∙ 0

• ### Algebraic lattice codes for linear fading channels

In the decades following Shannon's work, the quest to design codes for t...
12/19/2017 ∙ by Laura Luzzi, et al. ∙ 0

• ### Study of Distributed Robust Beamforming with Low-Rank and Cross-Correlation Techniques

In this work, we present a novel robust distributed beamforming (RDB) ap...
11/27/2019 ∙ by H. Ruan, et al. ∙ 0

##### This week in AI

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## I Introduction

We consider low rank lattice codes for transmission over a noisy erasure channel as illustrated in Fig. 1. In this figure

information symbols are mapped by an encoder/modulator to a vector

, where is a rank- lattice in . The output of the additive noise channel is , where is a noise vector independent of and with independent components. Components of are then erased by an erasure network, whose outputs are obtained by retaining only those symbols of indexed by subsets in a given sub-collection of subsets; thus coincides with is the positions identified by . As an example, with , and ,

. A decoder estimates the source symbols based on

with a probability of error denoted

. The objective is to minimize for each by designing a single codebook which satisfies a power constraint , where

denotes expectation with respect to a uniform distribution on the codebook. Here we consider as our sub-collection

, all -subsets of . Our paper is organized as follows. Prior work and lattice background is in Sec. II. Two performance bounds are presented in Sec. III, constructions for codes in dimension are presented and compared to the derived bounds in Sec. IV. A summary is in Sec. LABEL:sec:summary.

We use the acronym w.l.o.g to mean ‘without loss of generality’.

## Ii Prior Work and Review of Lattice Terminology

The problem considered here may be viewed as a code design problem for a special case of the compound channel, see e.g.  [blackwell1959capacity], [csiszar1991capacity]. This work was motivated by a study on cross layer coding that appeared in [courtade2011optimal]. For prior contributions on the Gaussian erasure channel, please refer to [ozccelikkale2014unitary] and the references therein.We now develop notation and some basic definitions for low rank lattices in . Let be a collection of orthonormal column vectors in and let denote the associated orthonormal matrix. Let denote a generator matrix of full rank for a lattice . We will refer to as the mother lattice. Let

 Λ=ΦΛV:=ΦVZk. (1)

is a rank- lattice in . Let denote the Gram matrix of ( is the transpose operator).

The determinant of a lattice is defined in terms of the determinant of its Gram matrix by Let denote the radius of the largest inscribed sphere in a Voronoi cell of . The packing density of is defined in terms of the volume of a unit-radius Euclidean ball in by

 Δk(Λ)=Vkρk/√detΛ. (2)

We denote by , the largest packing density that can be acheived by any lattice in . The problem of finding lattices that maximize the packing density is a classical problem in number theory and geometry, with several excellent references [SPLAG][gruber1987geometry].

The following definitions are from [gruber1987geometry]. A body captures the notion of a solid subset of , specifically, is a body if it has nonempty interior and is contained in the closure of its interior [gruber1987geometry]. A body is said to be centrally symmetric if , where . A closed body with the property that for any , the point for every is called a star body. While convex bodies are star bodies, the converse is not true. A simple example, and one directly relevant to us is the star body formed by the union of centrally symmetric ellipsoids in . A lattice is said to be admissible for , or -admissible, if no non-zero point in lies in . The greatest lower bound of over all admissible lattices is called the lattice constant of , denoted (which is set to is there are no admissible lattices). A -admissible lattice with is said to be a critical lattice for .

A lattice is said to be a packing lattice for a body if the sets and are disjoint for all non-zero . It is known, Thm.1, Ch. 3, Sec 20 [gruber1987geometry], that is a packing lattice for centrally symmetric, convex body if and only if it is admissible for . Thus, for a convex body, the problem of finding a packing lattice for is equivalent to that of finding an admissible lattice for . The connection between packings and admissibility for non-convex bodies is messier. The distinction arises because for a centrally symmetric body , is a lattice packing of if and only if it is admissible for , where denotes the set sum or Minkowski sum. If the centrally symmetric body is also convex, then and thus packing problems and admissibility problems are closely related. On the other hand, if is centrally symmetric but non-convex, in order to solve a packing problem for one must solve an admissibility problem for , and this set may not be as easily described as .

In our application, we need to index body by subset in a given sub-collection of subsets and our problem is one of packing , which is non-convex. While admissibility for non convex centrally symmetric body says nothing in general about packings for , it turns out that our design problem is equivalent to finding a critical lattice for because the decoder knows . Thus it is possible to draw on the theory of admissible lattices for star bodies. This theory provides several key ingredients to help find a solution to this problem. Most notably, in the chapter on Mahler’s compactness theorem [cassels2012introduction], Theorem VII states that every critical lattice for a bounded star body has linearly independent points on the boundary of .

## Iii Bounds

Let . For , let be the lattice obtained be retaining only those coordinates that are in or equivalently is the projection of into the subspace where is the th unit vector in . For any k-subset , we denote by the submatrix obtained by extracting from the rows identified by . The generator matrix for is and its Gram matrix .

Define the (packing volume) contraction ratio

 βS=(ρ(ΛS)/ρ(ΛV))k (3)

let and let .

### Iii-a Determinant Upper Bound

We will use symbols ,

to denote the arithmetic mean and geometric mean, respectively, of the real numbers

over some index set . When a -dim mother lattice is set in using a basis , the projections on the subsets , cannot all be simultaneously good. There are two important factors that measure the ‘goodness’ of the projections—the packing density and the scale of the lattices . The following theorem develops one of two bounds presented in this paper.

###### Theorem 1.

(Determinant Bound) Given a mother lattice and orthonormal basis , let and be respectively, the geometric mean of the volume contraction ratios and packing densities of the child lattices , taken over all -subsets of . Then

 (β#Δ(ΛV))2≤(Δ#)2(nk). (4)

Equality holds if and only if all child lattices have equal determinants.

###### Proof.

The packing densities of the mother lattice and child lattice are related by the following identity

 Δ2(ΛV)β2S=Δ2(ΛS)detΛSdetΛV. (5)

Compute the geometric mean of both sides over the collection of -subsets to get

 Δ2(ΛV)(β#)2=(Δ#)2(∏SdetΛSdetΛV)1(nk). (6)

From the arithmetic-geometric mean inequality it follows that

 Δ2(ΛV)(β#)2≤(Δ#)21(nk)(∑SdetΛSdetΛV) (7)

and equality holds if and only if is a constant with respect to . However

 ∑SdetΛS = ∑Sdet(G(ΛS)) (8) = ∑Sdet((ΦSV)t)det(ΦSV) (a)= det((ΦV)t(ΦV)) = detΛV,

where in (a) we have used the Cauchy-Binet formula, see e.g. [horn2012matrix]. The remainder of the proof follows directly. ∎

The following corollary is immediate.

###### Corollary 1.

Given mother lattice and orthonormal basis , let be the minimum volume contraction ratio of the child lattices , taken over all -subsets of , and the geometric mean of the packing densities. Then

 (βminΔ(ΛV))2≤(Δ#)2(nk)≤Δk(opt)2(nk). (9)

Equality holds in the left inequality iff all the contraction ratios are equal and all the child lattices have equal determinants. Equality holds in the right inequality iff all child lattices achieve the optimal packing density in dimension .

### Iii-B Trace Upper Bound

###### Theorem 2.

(Trace Bound) For an code, the compaction ratio is bounded as

 β2/kmin≤¯¯¯¯¯¯¯¯¯¯β2/kS≤kn. (10)

Equality holds if the shortest vector of each lattice is the image of the shortest vector in .

###### Proof.

Upon summing over all -subsets we obtain

 ∑SG(ΛS) = ∑SVtΦtSΦSV (11) = Vt(∑SΦtSΦS)V = Vt(n−1k−1)ΦtΦV = (n−1k−1)VtV.

By definition the smallest packing radius of any child lattice satisfies

 ρmin2≤ρ(ΛS)2≤(1/2)utG(ΛS)u (12)

for any non-zero and any -subset . Upon averaging over subsets we obtain the upper bound

 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ρ(ΛS)2 ≤ 12(nk)∑SutG(ΛS)u (13) = 12(nk)ut∑SG(ΛS)u = (n−1k−1)2(nk)utG(ΛV)u.

Equality holds if is the shortest vector in for all . Thus

 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ρ(ΛS)2≤knρ2(ΛV) (14)

and (10) follows immediately. ∎

## Iv Analysis of Some (4,k) Codes

We construct for for various values of and various mother lattices . Numerical results for the , are presented in Fig. 2, in which is plotted as a function of the packing density of the mother lattice. We have plotted the determinant bound using both the optimal and the cubic lattice for the child lattices. We have also plotted the trace bound.

Observe that in the case there is a significant gap between the best possible construction and the upper bounds. In the case performance close to the determinant bound is achieved by setting the mother lattice to be the cubic lattice. Also with the cubic lattice as the mother lattice, since the trace bound is lower than the determinant bound, this is proof that it is impossible to simultaneously achieve the packing density of when the mother lattice is the cubic lattice.

Geometrically, the ability to decode correctly, post-erasure, with iid Gaussian noise is determined by the largest noise sphere which can be packed by the projected (or child) lattice in each of the subspaces . When each noise sphere is projected back onto the subspace spanned by the columns of , a noise sphere is transformed into an ellipsoid. To see this consider the noise sphere in subspace . Setting , this leads to the noise ellipsoid . The packing of the noise ellipsoids , (recall that is the half the length of the shortest non-zero vector in ), by the mother lattice is shown in the six panels in Fig. 3(a), one for each subspace. Fig. 3(b) shows the star body . This illustrates the star body, which is the union of the six ellipses, two of which are circles. Also shown are points of the lattice , which in this case is an admissible lattice for this star body and illustrates also the interpretation as a code design problem for the compound channel. Observe that is simultaneously good as a packing for all six erasure configurations, i.e. for each of the bodies . Also, is simultaneously critical for five of the six bodies (notice that one circle does not touch any of the lattice points).

### Iv-a (4,1)

Let , . We obtain . The trace and determinant upper bounds yield . Hence this construction is optimal.

### Iv-B (4,2)

With , and

 Φt=(aaa0a−a0a)

we obtain six child lattices with Gram matrices

 (2a2002a2), (a2a2a22a2), (2a2a2a2a2), (a2−a2−a22a2), (2a2−a2−a2a2), (a200a2). (15)

All six child lattices are similar to . The first one has shortest vector of square length and all the others have square length . This code achieves , .

The trace upper bound is , regardless of the mother lattice, while the determinant upper bound depends on the mother lattice, and is and for (hexagonal lattice) and , resp. Thus the determinant bound is tighter than the trace bound but greater than . This construction does not meet the trace bound or the determinant bound with equality. However with , the following theorem shows this to be the best possible. A computer-based search has also failed to reveal any improvements in .

###### Theorem 3.

Let be a matrix with orthonormal columns, so that . Let . Let where is a 2-subset of . Let be the length of the shortest non-zero vector in . Then for at least one of the six 2-subsets of .

###### Proof.

For this proof we will write

 Φt=(x1x2x3x4y1y2y3y4). (16)

Our proof is by contradiction. Suppose that the shortest non-zero vector in all has square length . Now , in at most one position , else the shortest would be smaller than . The same is true for , and . Hence there exists one position , w.l.o.g. , such that , and . It follows that (i) and must be of opposite sign and (ii) . (i) is true because if , are of the same sign and then or . Assuming , we have which contradicts the hypothesis that . (ii) is true for if not or , and by the same argument as in (i) cannot exceed .

Since , it follows that for positions . In at least one of these positions, say , and . Now and must be of the same sign and , by a proof similar to that used before. Thus for , and . Again for , both and cannot hold for the same , hence, either and or and . We assume the first case. The proof for the other case is similar.

We have already proved that and are of the same sign. Assume they are both positive (if not reverse signs of all elements of ). Further, assume that and consider positions (if , then the same proof applies but for positions ). We now break up the proof into two cases:

Case 1: ( and of the same sign): Either (a) or (b) . If (a) then and . Thus , hence .

If (b) then let (). Thus and it follows that . Since it follows that and that and thus . Thus . But then and it follows that .

format: plain


text:
# egb(...,Algorithm=>...) -- algorithm choice for egb

## Description

Buchberger: This is a top level implementation of the equivariant Buchberger algorithm.

Incremental: This strategy uses Macaulay2's built in Gröbner basis algorithm gb. A Gröbner basis is computed for each truncated ideal. If no new elements are discovered up to Inc-action are discovered between the n truncation and the 2n-1 truncation for some n larger than the width of the generators, then the result is returned.

Signature: This is an implementation of an equivariant variant of the Gao-Volny-Wang signature based Gröbner basis algorithm. Experimental!

 i1 : R = buildERing({symbol x}, {1}, QQ, 2); i2 : egb({x_0+x_1}, Algorithm=>Buchberger) o2 = {x } 0 o2 : List i3 : use R; i4 : egb({x_0+x_1}, Algorithm=>Incremental) o4 = {x } 0 o4 : List i5 : use R; i6 : egb({x_0+x_1}, Algorithm=>Signature) -- TOTAL covered pairs = -6 o6 = {x + x , 2x } 1 0 0 o6 : List

## Further information

• Default value: Buchberger
• Function: egb -- computes equivariant Gröbner bases
• Option key: Algorithm -- an optional argument

## Functions with optional argument named Algorithm :

• egb(...,Algorithm=>...) -- algorithm choice for egb
• "gb(...,Algorithm=>...)" -- see gb -- compute a Gröbner basis
• "syz(...,Algorithm=>...)" -- see syz(Matrix) -- compute the syzygy matrix


text:
# A Retrospective on the 2014 NeurIPS Experiment

Computer Lab Seminar Series

## NeurIPS in Numbers

• To review papers we had:
• 1474 active reviewers (1133 in 2013)
• 92 area chairs (67 in 2013)
• 2 program chairs

## NeurIPS in Numbers

• In 2014 NeurIPS had:
• 1678 submissions
• 414 accepted papers
• 20 oral presentations
• 62 spotlight presentations
• 331 poster presentations
• 19 papers rejected without review

## The NeurIPS Experiment

• How consistent was the process of peer review?
• What would happen if you independently reran it?

## The NeurIPS Experiment

• We selected ~10% of NeurIPS papers to be reviewed twice, independently.
• 170 papers were reviewed by two separate committees.
• Each committee was 1/2 the size of the full committee.
• Reviewers allocated at random
• Area Chairs allocated to ensure distribution of expertise

## Timeline for NeurIPS

• Submission deadline 6th June
1. three weeks for paper bidding and allocation
2. three weeks for review
3. two weeks for discussion and adding/augmenting reviews/reviewers
4. one week for author rebuttal
5. two weeks for discussion
6. one week for teleconferences and final decisons
7. one week cooling off
• Decisions sent 9th September

## Quantitative Evaluation

• 10: Top 5% of accepted NIPS papers, a seminal paper for the ages.

I will consider not reviewing for NIPS again if this is rejected.

• 9: Top 15% of accepted NIPS papers, an excellent paper, a strong accept.

I will fight for acceptance.

• 8: Top 50% of accepted NIPS papers, a very good paper, a clear accept.

I vote and argue for acceptance.

## Quantitative Evaluation

• 7: Good paper, accept.

I vote for acceptance, although would not be upset if it were rejected.

• 6: Marginally above the acceptance threshold.

I tend to vote for accepting it, but leaving it out of the program would be no great loss.

## Quantitative Evaluation

• 5: Marginally below the acceptance threshold.

I tend to vote for rejecting it, but having it in the program would not be that bad.

• 4: An OK paper, but not good enough. A rejection.

I vote for rejecting it, although would not be upset if it were accepted.

## Quantitative Evaluation

• 3: A clear rejection.

I vote and argue for rejection.

• 2: A strong rejection. I’m surprised it was submitted to this conference.

I will fight for rejection.

• 1: Trivial or wrong or known. I’m surprised anybody wrote such a paper.

I will consider not reviewing for NIPS again if this is accepted.

• Quality

• Clarity

• Originality

• Significance

## NeurIPS Experiment Results

Table: Table showing the results from the two committees as a confusion matrix. Four papers were rejected or withdrawn without review.

 Committee 1 Accept Reject Committee 2 Accept 22 22 Reject 21 101

## A Random Committee @ 25%

Table: Table shows the expected values for the confusion matrix if the committee was making decisions totally at random.

 Committee 1 Accept Reject Committee 2 Accept 10.4 (1 in 16) 31.1 (3 in 16) Reject 31.1 (3 in 16) 93.4 (9 in 16)

## NeurIPS Experiment Results

 Committee 1 Accept Reject Committee 2 Accept 22 22 Reject 21 101

## A Random Committee @ 25%

 Committee 1 Accept Reject Committee 2 Accept 10 31 Reject 31 93

## Reviewer Calibration Model

$y_{i,j} = f_i + b_j + \epsilon_{i, j}$

$f_i \sim \mathcal{N}\left(0,\alpha_f\right)\quad b_j \sim \mathcal{N}\left(0,\alpha_b\right)\quad \epsilon_{i,j} \sim \mathcal{N}\left(0,\sigma^2\right)$

## Fitting the Model

• Sum of Gaussian random variables is Gaussian.
• Model is a joint Gaussian over the data.
• Fit by maximum likelihood three parameters, $\alpha_f$, $\alpha_b$, $\sigma_2$

## NeurIPS 2014 Parameters

$\alpha_f = 1.28$ $\alpha_b = 0.24$ $\sigma^2 = 1.27$

## Conference Simulation

• Calibration model suggests score is roughly 50% subjective, 50% objective.
• Duplicate experiment backs this up with roughly 50% correlation.

## Experiment

• Simulate conference scores which are 50% subjective/objective.
• Study statistics of conference.

## Conclusion

• Inconsistent errors are better than consistent errors
• NeurIPS and Impractical Knives


text:
# Sum up different arrays into a new array

I have a question regarding sums in arrays.

So I have the following array:

list=RandomReal[{0,1},{5,2}]

(*{{0.693551,0.447185},{0.274842,0.637526},{0.745271,0.0288363},{0.894933,0.937219},{0.605447,0.0337067}}*)

And from that I want to have the splitting for each pair like that

Subsets[Range@Length@list, {2}]
{{1,2},{1,3},{1,4},{1,5},{2,3},{2,4},{2,5},{3,4},{3,5},{4,5}}

Lets say every array in the list is a pair of x and y coordinates. Now I compute the distance between the different points using:

dist = EuclideanDistance @@@ Subsets[list, {2}] .

But now I want to have the sum of all the distances where particle 1 is occuring, so these are the first 5 entries in the array, for particle 2 its the first and the 5, 6 and 7th etc.

In the end I want to have a list containing 5 arrays with the sum for each particle.

So can anyone help me please?

You can use DistanceMatrix and use Total with the upper part of the matrix:

rowtotaldistances = Total[UpperTriangularize[DistanceMatrix[#],1],{2}]&

m = Partition[Range[0,10],2];

TeXForm @ MatrixForm @ m

$$\left( \begin{array}{cc} 0 & 1 \\ 2 & 3 \\ 4 & 5 \\ 6 & 7 \\ 8 & 9 \\ \end{array} \right)$$

 TeXForm @ MatrixForm[UpperTriangularize[DistanceMatrix[m],1]]

$$\left( \begin{array}{ccccc} 0 & 2 \sqrt{2} & 4 \sqrt{2} & 6 \sqrt{2} & 8 \sqrt{2} \\ 0 & 0 & 2 \sqrt{2} & 4 \sqrt{2} & 6 \sqrt{2} \\ 0 & 0 & 0 & 2 \sqrt{2} & 4 \sqrt{2} \\ 0 & 0 & 0 & 0 & 2 \sqrt{2} \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right)$$

rowtotaldistances @ m // TeXForm

$$\left\{20 \sqrt{2},12 \sqrt{2},6 \sqrt{2},2 \sqrt{2},0\right\}$$

Note: Using {2} as the second argument of Total gives the row totals:

mat = Array[Subscript[a, ##] &, {5, 5}];
TeXForm@MatrixForm[mat]

$$\left( \begin{array}{ccccc} a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} & a_{1,5} \\ a_{2,1} & a_{2,2} & a_{2,3} & a_{2,4} & a_{2,5} \\ a_{3,1} & a_{3,2} & a_{3,3} & a_{3,4} & a_{3,5} \\ a_{4,1} & a_{4,2} & a_{4,3} & a_{4,4} & a_{4,5} \\ a_{5,1} & a_{5,2} & a_{5,3} & a_{5,4} & a_{5,5} \\ \end{array} \right)$$

TeXForm @ MatrixForm[Total[m, {2}]]

$$\left( \begin{array}{c} a_{1,1}+a_{1,2}+a_{1,3}+a_{1,4}+a_{1,5} \\ a_{2,1}+a_{2,2}+a_{2,3}+a_{2,4}+a_{2,5} \\ a_{3,1}+a_{3,2}+a_{3,3}+a_{3,4}+a_{3,5} \\ a_{4,1}+a_{4,2}+a_{4,3}+a_{4,4}+a_{4,5} \\ a_{5,1}+a_{5,2}+a_{5,3}+a_{5,4}+a_{5,5} \\ \end{array} \right)$$

For column totals leave the second argument out (or use 1 as the second argument):

TeXForm @ MatrixForm[Total[mat]]

$$\left( \begin{array}{c} a_{1,1}+a_{2,1}+a_{3,1}+a_{4,1}+a_{5,1} \\ a_{1,2}+a_{2,2}+a_{3,2}+a_{4,2}+a_{5,2} \\ a_{1,3}+a_{2,3}+a_{3,3}+a_{4,3}+a_{5,3} \\ a_{1,4}+a_{2,4}+a_{3,4}+a_{4,4}+a_{5,4} \\ a_{1,5}+a_{2,5}+a_{3,5}+a_{4,5}+a_{5,5} \\ \end{array} \right)$$

• Okay, thanks but if I do that now with rowtotaldistances = Total[Most @ UpperTriangularize[DistanceMatrix[list],1],{2}] I just get 4 entries, but I have 5 particles – Armani42 Jun 21 '19 at 13:07
• @Armani42, removed Most to get a 5X5 distance matrix. – kglr Jun 21 '19 at 13:11

Thanks, but now I have another problem, now I have

ener[r2_] = 4* (1/r2^12 - 1/r2^6);

And I want to make now

dist = DistanceMatrix[list, DistanceFunction->(#1-#2 &)];

Which produces me 2x2 vectors in a matrix. But now I want to do that

e = Apply[ener @ {##}&, dist, {2}];

But here is the problem, that some entries in the vectors are 0 and dividing by zero is not possible.

So how can I avoid that maybe by an if statement or something like that?

• @kglr do you have an idea for that problem? :) – Armani42 Jun 21 '19 at 13:48


text:
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# Normalized frequency (fiber optics)

In an optical fiber, the normalized frequency, V (also called the V number), is given by

${\displaystyle V={2\pi a \over \lambda }{\sqrt {{n_{1}}^{2}-{n_{2}}^{2}}}\quad ={2\pi a \over \lambda }\mathrm {NA} ,}$

where a is the core radius, λ is the wavelength in vacuum, n1 is the maximum refractive index of the core, n2 is the refractive index of the homogeneous cladding, and applying the usual definition of the numerical aperture NA.

In multimode operation of an optical fiber having a power-law refractive index profile, the approximate number of bound modes (the mode volume), is given by

${\displaystyle {V^{2} \over 2}\left({g \over g+2}\right)\quad ,}$

where g is the profile parameter, and V is the normalized frequency, which must be greater than 5 for the approximation to be valid.

For a step index fiber, the mode volume is given by 4V2/π2. For single-mode operation is required that V < 2.4048, which is the first root of the Bessel function J0.

• 1/3
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• V-Number or Normalized Frequency of Optical Fiber
• Comsol Step by Step: Refraction, Total Internal Reflection
• Attenuation in Optical Fibres

## References

•  This article incorporates public domain material from the General Services Administration document: "Federal Standard 1037C". (in support of MIL-STD-188)
This page was last edited on 30 September 2020, at 20:33
Basis of this page is in Wikipedia. Text is available under the CC BY-SA 3.0 Unported License. Non-text media are available under their specified licenses. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc. WIKI 2 is an independent company and has no affiliation with Wikimedia Foundation.


text:
1
JEE Main 2020 (Online) 4th September Evening Slot
+4
-1
A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at t = 0, then the time at which the energy stored in the inductor reaches $$\left( {{1 \over n}} \right)$$ times of its maximum value, is :
A
$${L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n + 1}}} \right)$$
B
$${L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n - 1}}} \right)$$
C
$${L \over R}\ln \left( {{{\sqrt n + 1} \over {\sqrt n - 1}}} \right)$$
D
$${L \over R}\ln \left( {{{\sqrt n - 1} \over {\sqrt n }}} \right)$$
2
JEE Main 2020 (Online) 4th September Morning Slot
+4
-1
Choose the correct option relating wave lengths of different parts of electromagnetic wave spectrum:
A
$$\lambda$$radio waves > $$\lambda$$micro waves > $$\lambda$$visible > $$\lambda$$x-rays
B
$$\lambda$$visible > $$\lambda$$x-rays > $$\lambda$$radio waves > $$\lambda$$micro waves
C
$$\lambda$$visible < $$\lambda$$micro waves < $$\lambda$$radio waves < $$\lambda$$x-rays
D
$$\lambda$$x-rays < $$\lambda$$micro waves < $$\lambda$$radio waves < $$\lambda$$visible
3
JEE Main 2020 (Online) 3rd September Evening Slot
+4
-1
The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is
$$\overrightarrow E = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$$.
The magnetic field $$\overrightarrow B$$ , at the moment t = 0 is :
A
$$\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat j$$
B
$$\overrightarrow B = {{{E_0}} \over {\sqrt {{\mu _0}{ \in _0}} }}\cos \left( {kx} \right)\widehat k$$
C
$$\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat k$$
D
$$\overrightarrow B = {E_0}\sqrt {{\mu _0}{ \in _0}} \cos \left( {kx} \right)\widehat j$$
4
JEE Main 2020 (Online) 3rd September Morning Slot
+4
-1
The magnetic field of a plane electromagnetic wave is
$$\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i$$ T
where c = 3 $$\times$$ 108 ms–1 is the speed of light. The corresponding electric field is :
A
$$\overrightarrow E = - {10^{ - 6}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m
B
$$\overrightarrow E = - 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m
C
$$\overrightarrow E = 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$ V/m
D
$$\overrightarrow E = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k$$
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Organic Chemistry
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text:
## Cryptology ePrint Archive: Report 2014/405

Indistinguishability Obfuscation versus Multi-Bit Point Obfuscation with Auxiliary Input

Christina Brzuska and Arno Mittelbach

Abstract: In a recent celebrated breakthrough, Garg et al. (FOCS 2013) gave the first candidate for so-called indistinguishability obfuscation (iO) thereby reviving the interest in obfuscation for a general purpose. Since then, iO has been used to advance numerous sub-areas of cryptography. While indistinguishability obfuscation is a general purpose obfuscation scheme, several obfuscators for specific functionalities have been considered. In particular, special attention has been given to the obfuscation of so-called point functions that return zero everywhere, except for a single point $x$. A strong variant is point obfuscation with auxiliary input (AIPO), which allows an adversary to learn some non-trivial auxiliary information about the obfuscated point $x$ (Goldwasser, Tauman-Kalai; FOCS, 2005).

Multi-bit point functions are a strengthening of point functions, where on $x$, the point function returns a string $m$ instead of $1$. Multi-bit point functions with auxiliary input (MB-AIPO) have been constructed from composable AIPO by Canetti and Dakdouk (Eurocrypt 2008) and have been used by Matsuda and Hanaoka (TCC 2014) to construct CCA-secure public-key encryption schemes and by Bitansky and Paneth (TCC 2012) to construct three-round weak zero-knowledge protocols for NP.

In this paper we present both positive and negative results. We show that if indistinguishability obfuscation exists, then MB-AIPO does not. Towards this goal, we build on techniques by Brzuska, Farshim and Mittelbach (Crypto 2014) who use indistinguishability obfuscation as a mean to attack a large class of assumptions from the Universal Computational Extractor framework (Bellare, Hoang and Keelveedhi; Crypto 2013). On the positive side we introduce a weak version of MB-AIPO which we deem to be outside the reach of our impossibility result. We build this weak version of MB-AIPO based on iO and AIPO and prove that it suffices to construct a public-key encryption scheme that is secure even if the adversary can learn an arbitrary leakage function of the secret key, as long as the secret key remains computationally hidden. Thereby, we strengthen a result by Canetti et al. (TCC 2010) that showed a similar connection in the symmetric-key setting.

Category / Keywords: foundations / indistinguishability obfuscation, differing-inputs obfuscation, point function obfuscation, multi-bit point function obfuscation, auxiliary input obfuscation, leakage resilient PKE

Original Publication (with minor differences): IACR-ASIACRYPT-2014

Date: received 31 May 2014, last revised 14 Nov 2014

Contact author: arno mittelbach at cased de

Available format(s): PDF | BibTeX Citation

Note: improved presentation

Short URL: ia.cr/2014/405

[ Cryptology ePrint archive ]


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# Tag Info

0

The method you are using here is Euler-Cromer. It only really works for sytems with two variables (although I guess it could be modified for systems with any even number of variables - e.g. find $x, y, z$ then $vx, vy, vz$). There is a more accurate method, which is more general, but more complicated to program. This is the 4th order Runge Kutta (RK) ...

0

I just discovered: opticalraytracer From the manual: OpticalRayTracer is a free (GPL) cross-platform application that analyzes systems of lenses and mirrors. It uses optical principles and a virtual optical bench to predict the behavior of many kinds of ordinary and exotic lens types as well as flat and curved mirrors. OpticalRayTracer ...

1

I suspect your teacher glanced at it very quickly, and didn't realize you were using timesteps. When you work with timesteps you are doing a discrete approximation of the differential equations which describe the relationships, namely: $\frac{dv}{dt} = a$ and $\frac{dx}{dt} = v$ which in the discrete form, and arranged to match your code, become: $\Delta ... 0 Your solution is correct at the level of your course. We don't see what else is in your code, so there might be another problem. I can only guess what he means by "acceleration / 2". Perhaps he didn't read your program carefully enough; he might have been expecting a solution involving$1/2 a t^2$, and when he didn't see it, moved on. Your solution ... 0$s = v \cdot t$is only true if$v$stays constant. If$v$changes with time, the proper relation is $$s(t) = \int_{t_0}^{t} v(t') \mathrm dt'$$ For$v = a \cdot t$with a constant acceleration$a$this becomes $$s = \frac{1}{2} a t^2$$ So instead of calculating the position from the velocity you could calculate it from acceleration directly. 0 WSxM is free and does this sort of statistics using the "flood" tool. 1 If you want the conceptual answer of what the equations would be for the exact answer, rather than practical numerical rules that approximate the results of the integral, the basic answer is that you first break down each force into components along each coordinate axis ($x$,$y$and$z\$ if you're using Cartesian coordinates), then you add the components of ...

Top 50 recent answers are included


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# Why did Standard Model never sense a requirement to include gravitational quantum? [closed]

Standard Model is advanced (lorentz invariant) version of Quantum physics. It tried to include everything which came in the way while understanding quantum world. It even didn't bother to include even Higgs Boson which was hypothetical at that time. Did they never find gravitation in the way of other quantum interaction.

Note: I know, there were many unsuccessful attempts to add gravitation with SM to make Theory of Everything. My question: Why didn't Standard Model keep gravitation as raw ingredients (with unresolved relationship with others)?

-

## closed as primarily opinion-based by Dilaton, Dan, Waffle's Crazy Peanut, user1504, Manishearth♦Jul 11 '13 at 4:20

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise.If this question can be reworded to fit the rules in the help center, please edit the question.

Firstly, the concept of the Higgs actually came into play when the SM was being integrated--it fixed a problem with the electroweak interaction. Secondly, the SM wasn't "planned"--it sprouted bit by bit from the unification of forces. Thirdly, it's not as simple to "add" gravitation as a raw ingredient--there's a lot of mathematics behind it, and anyway, like I said, the model was built when the theory behind each force was being unified. – Manishearth Jul 6 '12 at 6:17
Also, keeping gravity with an unresolved relationship with others makes no sense--these models attempt to explain the working of the forces mathematically. "Keeping gravity with an unresolved relationship" is exactly the same as "assuming that some day, gravity will be added". I see no difference, can you clarify on that? – Manishearth Jul 6 '12 at 6:19
I disagree with the first sentence of the last paragraph. It is too early to decide if the attempts for a TOE are successful or not; and they are still work in progress ... – Dilaton Jul 6 '12 at 18:27
The standard model is just defined as the low energy theory excluding gravity. It isn't that people willfully excluded gravity, that's just not what the theory is describing. It is therefore incomplete in this way, but it is complete in every other way (so far). The low energy theory is an important thing to understand, so people weren't wasting time making the standard model. Also, the Higgs was a part since the beginning in 1967. What's the question exactly? – Ron Maimon Jul 6 '12 at 18:35
@SachinShekhar: Huh? What does that mean? The gravitational field is not a ghost. It has positive propagators in the physical part. This is also a completely different question. I am voting to close. – Ron Maimon Jul 7 '12 at 5:03

Read this link to get a framework of where the SM stands as far as interactions go. The SM is a mathematical shorthand of our data for the microcosm of quarks and leptons.

Look at table 1 and you will see that at the level of quarks and leptons the gravitational interaction is so weak that it is completely irrelevant and certainly its effect on the values used in the standard model cannot be measured with our present experimental accuracies.

-

In Quantum Electrodynamics, things are simple, because the photons are uncharged, so they themselves do not interact through Electromagnetdism, . Of course, there are still divergencies, and you still do need to renormalise, but things are very simple, compared to...

Quantum Chromodynamics,. Things now get much more complicated. Well, the Lagrangian Density takes almost the same form as in Quantum Electrodynamics, but compute anything, is a horror. Why? Gluons themselves have colour charge. SSo the additional gluon potentials, which ignoring constants, is $A^\mu=\nabla^\mu-\partial^\mu$ in the contra - variant form, have colour charges themselves, and interact through the strong force themselves. And thus, there is a lot of problems with the divergencies, but still, it is renormalisable.

Since everything was renormalisable, strong coupling, weak coupling, and they could make a TOEEG (theory of everything except gravity), called the standard model, they thought the same elegance could be extended to gravity, to general relativity. I mean, the standard model does incorporate special relativity, so why not general?

The most obvious way to do so, was to introduce a gravitational quantum, the graviton. But sadly, it was done in a very naive way. The gravitons themselves contributed to the gravitational field, just like in Quantum Chromodynamics, and whatever it is, the end result diverged. But unlike in Quantum Chromodynamics, for Quantum Gravity, it was non-renormalisable,.

So, a less naive theory of gravitons is required, and such a theory is string theory.

So, thus, it was not about the standard model not sensing a requirement to include gravity, it couldn't. One needs string theory for that; to make everything known to man, and everything that is true, to come out naturally.

There are, of course, other TOQGs (theories of quantum gravity), but the problem is that most of them are lorentz asymmetric (cf. Lubos Motl's criticism to Loop Quantum Gravity), they don't allow for other forces (interactions) other than gravity, they are mathematically inconsistent, etc. Or they just lead to string theory (e.g. supergravity, kaluza - klein theory).

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# Requirements space

Originally coined in linear programming, this represents the set of non-negative linear combinations of the columns of the matrix $LaTeX: \textstyle (A): \left \{Ax: x \ge 0 \right \}.$ For a linear program in standard form, the requirements space is the set of feasible right-hand sides, and it is a convex, polyhedral cone.


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# Make a ring larger but same thickness

Is it possible to make a ring larger, but maintain the same thickness?

I only selected the vertices of the ring, the ring is part of a larger object and I don't want to separate it.

If I simply scale it with S, it becomes thicker as well.

## 1 Answer

Try scaling in the y axis how much you want, then scale the same in the x axis. (Example with things you type: S,Y 90, S,X 90)


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# Concepts through Pictures

Who discovered $E=\Delta mc^2$? Even Einstein admitted that Poincaré derived the relation in 1900 by analyzing the recoil of an artillery canon---five years before he did.

Is $S\propto A\propto M^2$ the entropy of a black hole where $A$ is the area of the event horizon and $M$ is its mass, implying that $T\propto M^{-1}$ be its temperature? There is no physical system, or any final state of one, that would admit such a thermodynamic description for it would violate the same law it claims to have derived---the second law.

Are there gravitational waves, and if there are, why should they propagate at the speed of light? The general theory evolved as a generalization of the Minkowski metric which claims to describe inertial systems. The generalization would constitute systems which accelerate, but all forms of acceleration are not equivalent. Consequently, there is no reason why gravitational waves should propagate at the speed of light, and that far from gravitating bodies the metric should reduce to the Minkowski metric.

Are the curvilinear triangles that make up the tessellation different in size? No, Poincaré discovered a concept of length under which all the cells of the tessellation are of equal size, and so introduced hyperbolic geometry into mainstream mathematics---and even relativity!

Has anyone seen bodies contract in the direction of motion? The contraction hypothesis was used to explain the lack of time differences in the two journeys, parallel and perpendicular to the direction of the motion of the aether, in the Michelson-Morley experiment based on an incorrect emission theory.

Does the Sagnac effect contradict relativity? The Sagnac effect was used as an experimental proof that there is a time difference for light to cover a circular distance in the direction of rotation of the disc and in the opposite direction---giving rise to an interference pattern and a shift in the fringes.

Can the acceleration of a rocket ship nullify the force of gravity? Gamow's sketch of Einstein's gendanken experiment, showing the equivalence between acceleration and gravity, cannot be realized because gravitational and centrifugal forces are different---in the optical analogy, gravitation would influence the index of refraction whereas the centrifugal potential is built into the metric.

The "twin paradox" is used to show that a clock in motion goes slower than a stationary clock so that an astronaut who makes a round trip in space will have aged less than his twin brother who stayed at home. The clocks are in uniform motion so that either clock can be taken to be the "moving" one. The gedanken experiment assumes that acceleration and deceleration have no effect, but without them the astronaut could not be propelled into space and complete a round trip.


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0 energy points

# IIT JEE integral limit

Video transcript
The value of the limit as x approaches 0 of 1 over x to the third times the definite integral from 0 to x of t natural log of 1 plus t, all of that over t to the fourth plus 4dt is. Now, when you see an integral like this, you're first temptation is to try to solve the integral, try to find the antiderivative of this expression here. And you might try to experiment maybe with inverse trig functions, maybe arctangent looks tempting over here. But you have this natural log. Maybe you can do it with integration by parts. But very quickly, or maybe not so very quickly, you'll realize that this is a nearly impossible integral to simply solve using analytical methods. But, lucky for us, this problem isn't just about solving this integral. It's actually about evaluating this a limit. And if we rearrange this expression a little bit, it might become a little obvious to you how we can approach this problem. So this is the limit as x approaches 0 of. Let me write this in the numerator, because this is could be multiplying by 1 over x to the third is the same thing as dividing by x to the third. So this is the integral from 0 to x of t natural log of 1 plus t, all of that over t to the fourth plus 4dt. And all of that over x to the third. And now, this is probably ringing bells in some of your brains. And what we see here is we don't know how to evaluate this integral over here. But what does this integral approach as x approaches 0? What is the limit of this numerator expression as it approaches 0? So let me just write it. The limit as it approaches 0, so the limit as x approaches 0 of 0 to x, all of that stuff inside dt is going to equal the integral from 0. It's going to approach the integral from 0 to 0 of that stuff dt. And a definite integral from 0 to 0, we're not moving at all along the horizontal or the t axis. And so, this is just going to be 0. And if we do the same thing in the denominator, that's much simpler. The limit as x approaches 0 of x to the third, that's clearly also going to be 0. So we took the limits of the numerator and denominator and we got 0/0. That's the indeterminate form. And so now the bells should be ringing very loudly in your brain and screaming L'Hopital's rule. And I've done several videos on this in the calculus playlist. But this just tells us if we evaluate the limit of the numerator as it approaches 0 and the limit of the denominator approaches 0, and we get an indeterminate form, then this limit is going to be the exact same thing, assuming the limit exist, of the limit as x approaches 0 of the derivative of this expression over the derivative of this expression. Now let's do the denominator first, since that's easier. What's the derivative of x to the third? Well, that's just 3x squared. And what's the derivative of the numerator? And just as a reminder, and I'm just going to break out the fundamental theorem of calculus. The derivative with respect to x from some constant to x of f of t dt is just equal to f of x. You're undoing the integration here. You just get f of x, it's just the function with respect to x. So the derivative of this with respect to x is just going to be this function with respect to x instead of t. So it's going to be x natural log-- that was an ugly natural log-- of 1 plus x, all of that over x to the fourth plus 4. Now, when we try to evaluate the limit here, what happens? If we do the numerator, if we do the same thing we did up here, this numerator still evaluates to 0. You get 0 times natural log of 1. That's 0 as well. So that's all 0. The denominator still evaluates to 0. So we have indeterminate form again. So we might be tempted to just go straight, do L'Hopital's rule again, take the derivative of the numerator and then take the derivative of the denominator. And that would work. Or maybe you'll see that this is a pretty hairy thing to take a derivative of. And maybe we might have to take a derivative again after that. So this could get very, very, very hairy. So, instead of just straight up just taking a derivative of the numerator and derivative of the denominator again, let's see if we can simplify this a little bit. And we can. So this is the same thing. This is equal to the limit as x approaches 0. I'm just going to put the x to the fourth plus 4 in the denominator. So it's x natural log of 1 plus x, all of that over this 3x squared. And if I'm dividing by something and dividing by something else, that's just like dividing by that first thing. So this is x to the fourth plus 4. I just put this in the denominator, and we can simplify this. We can divide the numerator and denominator by x. And now, let's try to evaluate this limit. As x approaches 0 in the numerator, we get natural log of 1. So that's 0. So this is approaching 0 over. And if we do it in the denominator, we have 3 times 0 times 4. So that's going to be 0. Once again, indeterminate form. So now let's apply L'Hopital's rule again. But we can apply it on a simpler expression, on natural log of 1 plus x over 3x times x to the fourth plus 4. So let's do that. So this is going to be equal to, assuming the limit exists, this is equal to the limit as x approaches 0 of the derivative of the numerator. The derivative of natural log of 1 plus x, that's just 1 over 1 plus x over the derivative of the denominator. We just have to do a little product rule here. Derivative of the first function is derivative of 3x is just 3 times the second function, which is x to the fourth plus 4 plus the first function, 3x, it's just the product rule, times the derivative of the second function. The derivative of this is 4x to the third. Now, when we approach 0, what happens to this value? What is the numerator and the denominator approach? The numerator approaches 1 over 1 plus 0, or just 1. The denominator approaches 3 times 0 plus 4 plus this is going to be 0, plus 0. So this is equal to 1/12. And we're done. This crazy looking limit with this integral that we couldn't solve is just equal to 1/12. So thank goodness we had L'Hopital's rule so we didn't have to actually take the antiderivative of this expression.


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# Vacuum airship

Francesco Lana de Terzi's flying boat concept c.1670

A vacuum airship, also known as a vacuum balloon, is a hypothetical airship that is evacuated rather than filled with a lighter-than-air gas such as hydrogen or helium. First proposed by Italian jesuit priest Francesco Lana de Terzi in 1670,[1] the vacuum balloon would be the ultimate expression of displacement lift power.

## History

From 1886 to 1900 Arthur De Bausset attempted in vain to raise funds to construct his "vacuum-tube" airship design, but despite early support in the United States Congress, the general public was skeptical. Illinois historian Howard Scamehorn reported that Octave Chanute and Albert Francis Zahm "publicly denounced and mathematically proved the fallacy of the vacuum principle", however the author does not give his source.[2] De Bausset published a book on his design[3] and offered \$150,000 stock in the Transcontinental Aerial Navigation Company of Chicago.[4][5] His patent application was eventually denied on the basis that it was "wholly theoretical, everything being based upon calculation and nothing upon trial or demonstration."[6]

In 1921, Lavanda Armstrong discloses a composite wall structure with a vacuum chamber "surrounded by a second envelop constructed so as to hold air under pressure, the walls of the envelope being spaced from one another and tied together", including a honeycomb-like cellular structure, however leaving some uncertainty how to achieve adequate buoyancy given "walls may be made as thick and strong as desired".[7]

In 1983, David Noel discussed the use of geodesic sphere covered with plastic film and "a double balloon containing pressurized air between the skins, and a vacuum in the centre".[8]

In 1982-1985 Emmanuel Bliamptis elaborated on energy sources and use of "inflatable strut rings".[9]

In 2004-2007 Akhmeteli and Gavrilin address choice of materials ("beryllium, boron carbide ceramic, and diamond-like carbon" or aluminum) in honeycomb double layer craft to address buckling issues.[10]

## Principle

An airship operates on the principle of buoyancy, according to Archimedes' principle. In an airship, air is the fluid in contrast to a traditional ship where water is the fluid.

The density of air at standard temperature and pressure is 1.28 g/l, so 1 liter of displaced air has sufficient buoyant force to lift 1.28 g. Airships use a bag to displace a large volume of air; the bag is usually filled with a lightweight gas such as helium or hydrogen. The total lift generated by an airship is equal to the weight of the air it displaces, minus the weight of the materials used in its construction including the gas used to fill the bag.

Vacuum airships would replace the helium gas with a near-vacuum environment and would theoretically be able to provide the full lift potential of displaced air, so every liter of vacuum could lift 1.28 g. Using the molar volume, the mass of 1 liter of helium (at 1 atmospheres of pressure) is found to be 0.178 g. If helium is used instead of vacuum, the lifting power of every liter is reduced by 0.178 g, so the effective lift is reduced by 14%. A 1 liter volume of hydrogen has a mass of 0.090 g.

The main problem with the concept of vacuum airships however is that with a near-vacuum inside the airbag, the atmospheric pressure would exert enormous forces on the airbag, causing it to collapse if not supported. Though it is possible to reinforce the airbag with an internal structure, it is theorized that any structure strong enough to withstand the forces would invariably weigh the vacuum airship down and exceed the total lift capacity of the airship, preventing flight.[citation needed]

## Material constraints

From the analysis by Akhmeteli and Gavrilin:[10]

The total force on a hemi-spherical shell of radius ${\displaystyle R}$ by an external pressure ${\displaystyle P}$ is ${\displaystyle \pi R^{2}P}$. Since the force on each hemisphere has to balance along the equator the compressive stress will be

${\displaystyle \sigma =\pi R^{2}P/2\pi Rh=RP/2h}$

where ${\displaystyle h}$ is the shell thickness.

Neutral buoyancy occurs when the shell has the same mass as the displaced air, which occurs when ${\displaystyle h/R=\rho _{a}/(3\rho _{s})}$, where ${\displaystyle \rho _{a}}$ is the air density and ${\displaystyle \rho _{s}}$ is the shell density, assumed to be homogeneous. Combining with the stress equation gives

${\displaystyle \sigma =(3/2)(\rho _{s}/\rho _{a})P}$.

For aluminum and terrestrial conditions Akhmeteli and Gavrilin estimate the stress as ${\displaystyle 3.2\cdot 10^{8}}$ Pa, of the same order of magnitude as the compressive strength of aluminum alloys. Akhmeteli and Gavrilin note, however, that this disregards buckling, and using R. Zoelli's formula for the critical buckling pressure of a sphere

${\displaystyle P_{cr}={\frac {2Eh^{2}}{\sqrt {3(1-\mu ^{2})}}}{\frac {1}{R^{2}}}}$

where ${\displaystyle E}$ is the modulus of elasticity and ${\displaystyle \mu }$ is the Poisson ratio of the shell. Substituting the earlier expression gives a necessary condition for a feasible vacuum balloon shell:

${\displaystyle E/\rho _{s}^{2}={\frac {9P_{cr}{\sqrt {3(1-\mu ^{2})}}}{2\rho _{a}^{2}}}}$

The requirement is about ${\displaystyle 4.5\cdot 10^{5}kg^{-1}m^{5}s^{-2}}$.

Akhmeteli and Gavrilin assert that this cannot even be achieved using diamond (${\displaystyle E/\rho _{s}^{2}\approx 1\cdot 10^{5}}$), and propose that dropping the assumption that the shell is a homogeneous material may allow lighter and stiffer structures (e.g. a honeycomb structure).[10]

## Atmospheric constraints

A vacuum airship should at least float (Archimedes law) and resist external pressure (strength law, depending on design, like the above R. Zoelli's formula for sphere). These two conditions may be rewritten as an inequality where a complex of several physical constants related to the material of the airship is to be lesser than a complex of atmospheric parameters. Thus, for a sphere (hollow sphere and, to a lesser extent, cylinder are practically the only designs for which a strength law is known) it is ${\displaystyle k_{\rm {L}}<{\sqrt {1-{\frac {P_{\rm {int}}}{P}}}}\cdot L_{\rm {a}}}$, where ${\displaystyle P_{\rm {int}}}$ is pressure within the sphere, while ${\displaystyle k_{\rm {L}}}$ («Lana coefficient») and ${\displaystyle L_{\rm {a}}}$ («Lana atmospheric ratio») are:[11]

${\displaystyle k_{\rm {L}}=2.79\cdot {\frac {\rho _{s}}{\rho _{\rm {atm}}}}\cdot {\sqrt {\frac {P_{\rm {atm}}}{E}}}\cdot (1-\mu ^{2})^{0.25}}$ (or, when ${\displaystyle \mu }$ is unknown, ${\displaystyle k_{\rm {L}}\approx 2.71\cdot {\frac {\rho _{s}}{\rho _{\rm {atm}}}}\cdot {\sqrt {\frac {P_{\rm {atm}}}{E}}}}$ with an error of order of 3% or less);
${\displaystyle L_{\rm {a}}={\frac {\rho _{a}}{\rho _{\rm {atm}}}}\cdot {\sqrt {\frac {P_{\rm {atm}}}{P}}}}$ (or, when ${\displaystyle \rho _{a}}$ is unknown, ${\displaystyle L_{\rm {a}}=10\cdot {\sqrt {\frac {P_{\rm {atm}}}{P}}}\cdot {\frac {M_{a}}{T_{a}}}}$),

where ${\displaystyle P_{\rm {atm}}=101325}$ ${\displaystyle Pa}$ and ${\displaystyle \rho _{\rm {atm}}=1.22}$ ${\displaystyle kg/m^{3}}$ are pressure and density of standard Earth atmosphere at sea level, ${\displaystyle M_{a}}$ and ${\displaystyle T_{a}}$ are molar mass (kg/kmol) and temperature (K) of atmosphere at floating area. Of all known planets and moons of the Sun system only the Venusian atmosphere has ${\displaystyle L_{\rm {a}}}$ big enough to surpass ${\displaystyle k_{\rm {L}}}$ for such materials as some composites (below altitude of ca. 15 km) and graphene (below altitude of ca. 40 km). Both materials may survive in the Venusian atmosphere. The equation for ${\displaystyle L_{\rm {a}}}$ shows that exoplanets with dense, cold and high-molecular (${\displaystyle CO_{2}}$, ${\displaystyle O_{2}}$, ${\displaystyle N_{2}}$ type) atmospheres may be suitable for vacuum airships, but it is a rare type of atmosphere.

## In Fiction

In Passarola Rising, novelist Azhar Abidi imagines what might have happened had Bartolomeu de Gusmão built and flown a vacuum airship.

Spherical vacuum body airships using the Magnus effect and made of carbyne or similar superhard carbon are glimpsed in Neal Stephenson's novel The Diamond Age.

## References

1. ^ "Francesco Lana-Terzi, S.J. (1631-1687); The Father of Aeronautics". Retrieved 13 November 2009.
2. ^ Scamehorn, Howard Lee (2000). Balloons to Jets: A Century of Aeronautics in Illinois, 1855-1955. SIU Press. pp. 13–14. ISBN 978-0-8093-2336-4.
3. ^ De Bausset, Arthur (1887). Aerial Navigation. Chicago: Fergus Printing Co. Retrieved 2010-12-01.
4. ^ "Aerial Navigation" (PDF). New York Times. February 14, 1887. Retrieved 2010-12-01.
5. ^ "To Navigate the Air" (PDF). New York Times. February 19, 1887. Retrieved 2010-12-01.
6. ^ Mitchell (Commissioner) (1891). Decisions of the Commissioner of Patents for the Year 1890. US Government Printing Office. p. 46. 50 O. G., 1766
7. ^ US patent 1390745, Lavanda M Armstrong, "Aircraft of the lighter-than-air type", published Sep 13, 1921, assigned to Lavanda M Armstrong
8. ^ David Noel (1983). "Lighter than Air Craft Using Vacuum" (PDF). Correspondence, Speculations in Science and Technology. 6 (3): 262–266.
9. ^ US patent 4534525, Emmanuel Bliamptis, "Evacuated balloon for solar energy collection", published Aug 13, 1985, assigned to Emmanuel Bliamptis
10. ^ a b c US application 2007001053, AM Akhmeteli, AV Gavrilin, "US Patent Application 11/517915. Layered shell vacuum balloons", published Feb 23, 2006, assigned to Andrey M Akhmeteli and Andrey V Gavrilin
11. ^ E. Shikhovtsev (2016). "Is FLanar Possible?". Retrieved 2016-06-19.


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# Quod Erat Demonstrandum

## 2009/04/20

### Just reply to my so-called ETV

Filed under: Fun,Junior Form Mathematics — johnmayhk @ 12:45 上午

Here is just a reply to my primary school mathematics ETV (magic square).

When I was a primary school boy, my mathematics teacher taught us how to construct magic square, today, I still remember what she had taught me. For your reference, here is an easy way to construct magic square of odd order:

http://en.wikipedia.org/wiki/Magic_square#A_method_for_constructing_a_magic_square_of_odd_order

Students, try to construct magic squares of order 5 and 7 by using the method mentioned above.

(Sudden question: will the magic squares of order 5 and 7 etc. have the same property as described in the video?)

To see (not prove) the reason why there is a property as described in the video, let’s look at the magic square of order 3 again and do some junior form mathematics.

– – – – – – M1

It is actually ‘equivalent’ to the following magic square:

– – – – – – M2

M2 is formed by subtracting 5 from each entry in M1.

Look at M2, it may be easier to see the sums of entries of the same row, same column and same diagonal are zero. That is actually the property of magic square. Right?

For simplicity, let $a = 100, b = 10$.

Look at M1, the number formed from the first row, i.e. 816, can be written as $8a + b + 6$.

To have the linkage with M2, I write

$816^2$
$= (8a + b + 6)^2$
$= (3a - 4b + 1 + 555)^2$
$= (3a - 4b + 1)^2 + c(3a - 4b + 1) + d$ – – – – – – (*)

(where $c = 2 \times 555 = 1110, d = 555^2 = 308025$)

See, 3, -4, 1 are numbers in the first row in M2.

Look at M1 again, the number formed from the third row, i.e. 492, and the ‘reverse’ of this number is 294.

In fact, 294 is quite ‘related’ to the number 816, consider

$294^2$
$= (2a + 9b + 4)^2$
$= (-3a + 4b - 1 + 555)^2$
$= (-3a + 4b - 1)^2 + c(-3a + 4b - 1) + d$
$= (3a - 4b + 1)^2 - c(3a - 4b + 1) + d$

Compare with (*), their middle terms are different from a negative sign only.

This is not by accident, it is all about the construction of magic square.

The ‘proof’ is done by the ‘zero sum’ in M2. What? Okay, let me write the whole story.

$816^2 + 357^2 + 492^2$
$= (3a - 4b + 1)^2 + c(3a - 4b + 1) + d$
$+ (-2a + 2)^2 + c(-2a + 2) + d$
$+ (-a + 4b -3)^2 + c(-a + 4b -3) + d$
$= (3a - 4b + 1)^2 + (-2a + 2)^2 + (-a + 4b -3)^2 + 3d$ – – – – – – (#)

The construction of magic square ensures the terms involving $c$ being cancelled.

Now, the ‘reversed sum’

$618^2 + 753^2 + 294^2$
$= 294^2 + 753^2 + 618^2$
$= (3a - 4b + 1)^2 - c(3a - 4b + 1) + d$
$+ (-2a + 2)^2 - c(-2a + 2) + d$
$+ (-a + 4b -3)^2 - c(-a + 4b -3) + d$
$= (3a - 4b + 1)^2 + (-2a + 2)^2 + (-a + 4b -3)^2 + 3d$

Also, the terms involving $c$ are cancelled, and the result is just the same as (#) and that is why

$816^2 + 357^2 + 492^2 = 618^2 + 753^2 + 294^2$

Instead of considering rows, as shown in the video, the property remains when we consider numbers formed from columns. And the so-called proof is exactly the same as before.

Urm, we may ask naturally that will the following be true for some positive integer $n$ (other than $n = 1, 2$)?

$816^n + 357^n + 492^n = 618^n + 753^n + 294^n$

Students, explore it if you are interested.


text:
# homomorphisms and vector spaces

Let $$A=\{f\in \operatorname{Hom}(V,V) \mid g \circ f = 0\}$$. Find a basis of $$A$$. Here $$g$$ is a homomorphism of the vector space $$V$$ with the basis $$\{e_1,e_2,e_3,e_4\}$$ (canonical vectors) such that \begin{align} g(e_1+e_2)&=-e_1\\ g(e_1-e_2)&=2e_2\\ g(e_1+e_3)&=e_1+e_4\\ g(e_1-e_4)&=e_2+e_4; \end{align}

Could you find some clever way to get the solution?

• See edits to learn how to properly format a question. – John Hughes Mar 16 at 23:26
• What's the language used? – psidaga Mar 16 at 23:35
• Do you mean $g(e_1 + e_2)=...$, etc. ? – zugzug Mar 17 at 0:24
• It's called MathJax, and it's a flavor of LaTeX. Here's a quick tutorial -- the first few paragraphs give you 90% of what you need: math.meta.stackexchange.com/questions/5020/… – John Hughes Mar 17 at 0:41
• Thank you so much. I was using the representative matrix of g with the canonical basis... The system was really annoying – psidaga Mar 17 at 6:52

The range of $$g$$ is the span of $$e_1,e_2,e_4$$, hence its kernel is one dimensional. Find a nonzero element $$v$$ of the kernel, that will generate it.
Now, $$g\circ f\, (x)=0$$ iff $$f(x)\in\ker g$$, so the range (=column space) of $$f$$ must be contained in $${\rm span}(v)$$, and you can obtain a basis $$(f_i)$$ of $$A$$ by putting $$v$$ in the $$i$$th column and $$0$$ otherwise in (the standard matrix of) $$f_i$$.
Hint: the matrix for $$g$$ rel the bases $$\{e_1+e_2,e_1-e_2,e_1+e_3,e_1-e_4\}$$ and the standard basis is: $$\begin{pmatrix}-1&0&1&0\\0&2&0&1\\0&0&0&0\\0&0&1&1\end{pmatrix}$$.
So, the matrix of $$f$$, when multiplied by this matrix on the left, is zero, for any $$f$$ in $$A$$.
This gives a homogeneous linear system of $$16$$ equations in $$16$$ unknowns. So you can form a $$16×16$$ matrix and then row-reduce, to solve.