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Saturday, July 21, 2012

I am in a nothing-to-do month. I am really glad I was selected to problem set SRM 550 in Topcoder , else this would not have been a very productive month.

When you got nothing to do, you try to remember things that you were not able to do back when you were very busy. To me, my ever growing issue was that my ubuntu version in my main computer was ages behind. Until this week, I was using 10.10.

My history with ubuntu begins in 2005!, I installed a 5.something version. In retrospect it was not that good. With time ubuntu has been improving in somethings and getting worse in others.

The problem with keeping an old ubuntu version is that, eventually, you cannot upgrade packages anymore. Your repositories are no longer supported... And these two years were full of exciting things like 600 new firefox versions and the shift from OpenOffice to LibreOffice. So, I figured I needed to upgrade...

The plan

I am the proud owner of a 500GB external hard drive that connects to USB. This is one of the most useful things ever invented. So, I just made a backup of my ubuntu partition and saved it in the hard drive.

THREE HOURS I had to wait copying it. But it has so many benefits. I would not lose any data, and configuration for things that are not system-critial can be restored from my own home, if I just now where to find them and where to place them...

Install

A clean install is so much easier than upgrading. You can even test the ubuntu version from the install CD. A great sign was that things like 3D acceleration , sound, resolution and printer were working completely out of the box in this CD.

Something ominous was the new interface. The new ubuntu version has this thing called unity . Which was casually something they started using just AFTER version 10.10. So I avoided 2 whole years of development towards this unity thing. The change is VERY drastic from the GNOME thing I was used to (which was the default since before I started using ubuntu).

Oops

So, I start installing, it offers me to download updates for packages while installing. That is a novelty and sounds useful (In the past, after installing you would be greeted with the request to download tons of package updates so your install wasn't finished). So I click yes.

Then I find a cute surprise. Apparently the new installer can actually install ubuntu without deleting your data files. Perhaps the whole "copy partition to external hard drive" thing was not needed. I did not use the option though.

Format partition, install it. Then it downloads the packages. And it takes 2 hours to download everything. The good thing about the live CD installer is that you can actually use your computer and browse the web WHILE installing the Operating system. After the 2 hours, it begins installing the new packages and... THE IDIOT CRASHED!. Yes, installer crashed, I panicked, if clean install is messed up, then I would have a messed up install and the same kind of problems I had for upgrading so many times.

I restart, and somehow ubuntu boots just fine without complaining of everything. All hardware seems to be working. But apparently, updates weren't installed. So I have to... download again. But this time I also download all the things I remember are useful: jEdit, valgrind, wine (from its ppa), firefox 14 (from its ppa), jdk, g++, avidemux... I tried to remember everything that to me is vital. New download took ages, but thankfully this time everything installed fine. I found my own game, Xye in the repositories, great.

Unity... sucks

I began to notice this when playing with the live CD. Then I tried to use Unity for a while. Casually, by the time I finished setting the basics up, it was Thursday and I received the shock surprise that I was assigned to work in setting up SRM 550. Suddenly, I needed to use my computer for real, programming work. Yet, my programming environment was not setup, and I had to improvise with Unity.

After about 3 hours of attempting to work using unity. I really got sick of it. Sorry, but this vertical apple clone thing is GREAT at getting in the way. I can't tell if icons in it are launchers or already opened windows or both. When I have multiple windows of the same app, I need to browse whole menus to get the correct ones. Switch desktops? It takes seconds to play animations and double, triple click to enter the desktop.

You cannot do the minimum configuration. Want to change the interface's colors? NO. Want to move the launcher bar horizontally? No.

Maybe users new to computers like this stuff, as they are used to taking ages to do the most basic things. But I do not.

What I liked was the menu that appears when you click the main button. It automatically searches for applications or files or both. And remembers the ones you use the most. But I was fed up with just about everything else. I looked for alternatives.

GNOME 3, classic GNOME, classic GNOME without compiz...

So, the good thing is that I can actually install different interfaces and keep the operating system. Can you say something like this about Winods, OX/S, iHOS or ArDoird, or whatever they are called?

I did it: installed GNOME 3. Turns out GNOME 3 is also very different to the GNOME I was used to. When I tried GNOME 3. It was even MORE obscure than unity. Or perhaps I was just tired of trying funny things that I was not used to?. I really could not get why it shows the name of the active app at the top. Complete mystery.

The good thing is that it allows you to use the CLASSIC gnome too!. So I switched to that option. That was closer to what I use, but still not much. I could not configure the GNOME panels like in the past.

Google around, and it turns out that you cannot anymore right click to edit the panel, so I used the complicated keyboard combination and edited everything.

But even the classical gnome had some difference. There were definitely less options in the panel configuration, and I could not change the interface colors (seems new GTK+ is lacking in this department). Also, those flashy desktop effects and animations that were so annoying in unity are still around. I have been here before, It is because of something called compiz. I tried to like compiz in the past, but it makes everything slower and is too annoying at times. Specially when switching between Desktops. (Sorry, but I do not need animations there, I just want to switch in 0 seconds).

The last tweak was, pick the "Classic GNOME without Effects" option. But now, you do not have cute shadows and you cannot use transparency! Do not worry, you can now Enable metacity compositing.. This does the good things that compiz does. Without doing the childish, annoying things it does.

Some interesting results

• Sun Java does not exist. And Oracle Java cannot be installed from the repositories anymore because those Oracle guys are jerks that want to ruin open source completely. But it turns out OpenJDK from the repositories does all that I need it to do ... I can compile Java stuff, run Java stuff and the Topcoder Arena and MPSQAS work well!
• GCC 4.0 was missing from the repositories for ages. So I used GCC 4.2 to test Topcoder stuff. Now GCC 4.2 is also missing. GCC 4.3 is missing too! So I now have to use GCC 4.4. Which is very different. Topcoder really need to upgrade their GCC version. 4.0 is too antique (and bugged too).
• Classic/fallback gnome is much faster than the stuff I used in 10.10. For some reason. My environment is flying!
• Nautilus removed the emblems feature, now I have a lot of folders that I cannot differentiate from each other.
• LibreOffice is much faster and better than OpenOffice.
• My iPod (Won it at a Google code jam), still cannot be mounted. This is a bug that I started to experience at version 10.10, and thought it was because of the upgrade.
• The lack of customization options are KILLING me.
• I restored EVERYTHING from my old firefox by just replazing the .mozilla folder in the new install with the .mozilla folder from the old one. I mean everything. Semantic bar, addons, addons's settings...(/home/yourusename/.mozilla
• I restored EVERYTHING from my topcoder arena by doing the same with a file called /home/yourusername/ContestApplet.conf.
• The new version of GRUB loads much faster. If you install a program called grub-customizer, you can configure it and it is much easier than before.
• Ubuntu 12.04 boots VERY quickly.
• Say what you will about Ubuntu, and in fact its interface changes are terrible. But hardware support has really improved since the old years. Applications are very competitive now. And ubuntu specifically is great at one thing: packages. There are many things in the repositories, and then they invented ppas. It is very easy to install things on ubuntu. I got so used to the repositories that I cannot try another distribution. I can always change the interface and tweak it, but the packaging is something internal to the Linux distribution.

SRM 550: I regret nothing!

I wrote SRM 550. The editorial is coming soon-ish. I will use this entry not so much to explain the problems (will do in the editorial). But to comment about the match.

I was assigned this match on Thursday. I had something that looked like a match but not a very complete proposal. Was looking more to first confirm something in div1 hard difficulty before getting a match. But I think the admins had difficulty finding something better, and I was the one with the closest thing to a div1 hard (we really thought the problem that was ultimately used as a 850 pointer was not hard enough.

Those two days of development were intense. We replaced many problems with other problems. Most of the difficulty in the 850 pointer comes from nice modifications from rng_58 and mystic_tc.

This 850 pointer turned out to be the big screw up of the match. Just as the contest started, lyrically was already reporting bugs with the statement. Later Petr found out we did not have a constraint that specified that only a,b and c are allowed in the input. And ir5 got confused about something - apparently, the statement was never clear enough to say that each letter is changed one at a time.

Ultimately, it turns out that the 850 points problem was actually normal div1 hard difficulty and not deserving of the low score value. What happened? Maybe the statement was too confusing? Seems not, because we did not receive many complaints about this. Apparently, we tweaked the original version of the problem (only moves a->b, b->a , max cost <= 3, length <= 33) so much that we turned the problem that seemed like a div1 600 at best into a legit div1 hard.

But besides div1 850, I regret nothing!.

A. No sorry, can't do. This is part of what I do for a living. And it is always fun. The only viable way in which I can stop writing problems is if admins stop accepting them.

If you are interested in traveling to the past and stopping me from writing this match, a good fix would have been to give the admins an alternative to my problem set so that they did not have to use mine for this SRM in the rush of two days missing.

*Actual quote.

Q. Example #5 in div1 300/div2 550 is wrong/ do not get why it is wrong / etc

A. We got this a lot during the match. No, it was not wrong.

We decided to include this example and many others to make sure the success rate stayed high in this problem. So, yes, you should be happy this case was in the examples...

To be honest, I intended this problem as an easy problem for TCO rounds like round 3 or maybe the semifinals. The original version had movements < 1000000000. I had to use this problem for a normal SRM unfortunately, because the previous problem idea I had available fro this slot turned out to be a very tricky problem. That is right, we used this problem because it was the least tricky problem available for div1 easy.

My post there explains it.

Q. Div1 500 is too mathematical!

A. I call humbug!. When I thought of this problem it was because I had an assignment that said "Design a cell automaton to generate Sierpinski's triangle". It turns out this task was rather standard and everybody knows how to do this triangle from top to bottom (Using Pascal's triangle modulo 2). Yet I somehow thought of a solution for the assignment that does it diagonally.

I thought I could capitalize this strange idea - This fractal triangle is a very easy fractal, and I really like this fractal. So I thought of this problem that can be solved by two steps: a) Find out the pattern is a fractal. And b) Use simple recursion to generate the fractal for each requested cell.

So, to me, this has always been a problem that is meant to be solved through programming (recursion). All those guys that used a formula related to combinations and modulo 2, well, you used that formula because YOU are Mathematically-inclined, not because the problem is Mathematically-inclined :). I am more of a programming-inclined guy, so I was not aware of those formulas until the admins told me :)

Monday, July 09, 2012

SRM 549: The hell is an apex?

SRM 549... I had a bad day with a very slow 250 and a 600 that I solved but could not debug fast enough. I think the problem set itself was fine and interesting, albeit perhaps imbalanced (which happens 90% of the time). But the statements were just not clearly explained. Specially, the lack of images or better explanations in 250.

Div1 250: The one with cones

So, what is an apex? Those of us who did not learn basic geometry in English probably have no idea. I had to google apex, it turns out it is just the top vertex of the cone. So in fact, the distance from a cone's base to the apex is giberish for the word "height". It was still pretty hard to understand what the condition for a top cone to match a bottom cone.

Anyway, we have a list of top cones, each with radius and heights, there is a correct way (which is impossible to understand) which defines a condition in which a top cone can be used in combination with a bottom cone to make a wizard hat. What is the maximum number of hats you can make?

The problem is just a maximum bipartite matching problem. There is probably a way to solve the problem without max flow or the Hungarian algorithm. But that requires you to actually understand the rule that determines whether you can match a top cone with a bottom one. But I solved it with max flow. Basically, pasting max flow and preparing the bipartite matching was the easiest part that took me less than a minute. The problem was to then just find what the condition was...

At the end, I guessed it. If you are going to place a cone on top of another, then you can think of the cones as straight triangles. Then when placing a cone on top of another, you are interested in the height at which the width of the top cone intersects the lower cone. It is much, much, much easier with a image:

The trick is to find the line equation that would give us the y-position at which the bottom cone distance from axis to hypotenuse is equal to the top cone's width. Then you can find the position at which the apex (top vertex) of the top cone will end. Compare this position with the bottom cone's height, it must be higher. (Also compare the widths before doing all of this, the top cone's width must be smaller.

bool canDo(int topHeight, int topRadius, int bottomHeight, int bottomRadius) {     if (topRadius < bottomRadius) {         // The equation is:         // y = bottomHeight - (x/bottomRadius) * bottomHeight         // then:         // bottomHeight - (x/bottomRadius) * bottomHeight + topHeight         //  >         // bottomHeight         //         // or:         return ( topRadius*bottomHeight < topHeight*bottomRadius );     }     return false; }

Once I guessed this rule, I implemented it and it passed the examples. I submitted it and it turns out to be fine. Sorry but, I would have rather not spent 95% of the time trying to guess what the problem statement wanted. It is worse because at the end few people solved 600 so the difference between a good rank and a bad one was mostly defined by speed in this problem.

Div1 600: The one with hats in grid

This one had the clarity issue, albeit not as badly as the 250.

You are given a board (at most 13x13) which contains hats (at most 13) in some of the cells. There is a list of coins (less than or equal to the number of hats), that the wizard placed in such a way that each hat has a coin. You can make at most numGuesses guesses which involve picking one hat, one at a time. After you pick a hat but before you get to see its contents, the wizard is allowed to "reshuffle" the coins in the hats. This was the first point of confusion, after asking the admins, it turns out that the "reshuffling" is not random, instead, the wizard will place each coin in whatever hat he wants. However, the reshuffling will not change the contents of hats you have already picked before.

So in fact, whenever you pick a hat, the wizard can do whatever he likes and thus it does not really matter what was inside the hat before you picked it, the wizard will just give you the coin he likes the most or no coin at all. The game would be pointless if it was not for the following condition: At any point in time, for each row, the sum between the number of coins and hats in the row is even. And for each column, the sum between the number of coins and hats is even too.

It was such a coincidence (albeit ultimately not a very happy one, because I still could not solve the problem). That I recently remembered a problem of mine called RobotKing, which was used in a netEase tournament. The problems are similar in the game theory aspect and in having to encode the state in base 3...

Base 3, so imagine we determine the current state of the game as a list of values, one for each hat in the board (there are at most 13 of them). Each value is 0 if we still have not picked the hat. 1 if we picked the hat and it turned out to have a coin and 2 if we picked the hat and it turned not to have a coin. This state is enough to represent the game. Let us say there are correctGuesses 1s in the state, it means that you have already gotten correctGuesses coins. If the wizard decides to give you a coin, there is no reason for the wizard to give you a coin with value higher than the smallest value available. So you can assume that the correctGuesses smallest coins have already been given to you. Let usedGuesses be the number of values in the state that are not 0 (the number of known hats), then you also know you have made only usedGuesses in total.

Thus we have a function f(state), where the state is such an array of values (0,1 or 3) for each of the hats. That returns the number of hats you will get if you and the wizard play optimally - make the best choices possible. The state is currently an array, but you can also represent it as a single number mask. Considering it as a base 3 number, you can extract values (0,1 or 2) from it. So, if the number of used guesses is equal to the guess limit, we got a base case, we cannot win any more coins. Else we can pick one of the hats as a guess. So, let us try each possible choice. This will allow us to implement the recurrence with dynamic programming or memoization.

Let us say we decide to uncover the i-th hat (the value of the hat must be 0, unknown). Then the wizard may have two possible choices. a) To give you no coins at all. b) To give you the smallest coin available. The wizard will pick the decision that will ultimately give you the least number of coins.

The problem is that there are times at which placing a coin in a certain hat is not possible, or not placing a coin at all in the hat is not possible (To follow the condition regarding parity of rows and columns). I basically spent 20 minutes thinking of a way to do this. At the end, I thought it was very silly I did not think of this before. The constraints are still slow - 13. So in fact, we can do this with another dynamic programming function. couldHappen(mask), the mask is in the same format as the state in the other function. It returns if a possible setting (with some hats being unknown, other hats having coins and some definitely not having coins) is possible. The base case is when there all coin positions are known, we calculate the parities and if the parities are correct, then it is possible, else it is not.

The recursion involves, when not all coin positions are known, to decide to place a coin in one of the unknown positions. If any of these possible moves leads us to a correct setting, then the original setting is also correct.

.
struct MagicalHats {     int pow3[14];      int r, c;     int numGuesses;     int mem[1594323];     char could[1594323];     vector<int> coins;     int n;     int x[13];     int y[13];     vector<int> rowpar, colpar;                   int couldHappen(int mask)     {         // Is the state given by the mask possible?         char & res = could[mask];         if (res == -1) {             res = 0;             int known = 0;             {                 vector<int> rp = rowpar;                 vector<int> cp = colpar;                 for (int i=0; i<n; i++) {                     int ch = (mask / pow3[i]) % 3;                     // ch == 0: unknown contents                     // ch == 1: coin                     // ch == 2: no coin.                     if ( ch == 0) {                     } else if (ch == 1) {                         known++;                         rp[ x[i] ] ^= 1;                         cp[ y[i] ] ^= 1;                     }                 }                 // Base case, all coin positions are known,                 // verify the parities are correct.                 if (known == coins.size() ) {                      res = 1;                     for (int i=0; i<r; i++) {                         if (rp[i] == 1) {                             res = 0;                         }                     }                     for (int i=0; i<c; i++) {                         if (cp[i] == 1) {                             res = 0;                         }                     }                     return res;                 }             }             if ( known < coins.size() ) {                 // decide to place a coin in an unknown place                 res = 0;                 for (int i=0; i<n; i++) {                     if ( (mask / pow3[i]) % 3 == 0 ) {                         // this updates the mask, the place is                         // now known to have a coin:                         int nmask = mask + pow3[i];                         if (couldHappen(nmask)) {                             res = 1;                         }                     }                 }             }         }         return res;     }          // What is the number of coins the player wins after the      // state that is given by the mask?     int rec(int mask)     {         int & res = mem[mask];         if (res == -1) {             res = 0; // maximize this!             int usedGuesses = 0;             int correctGuesses = 0;             for (int i=0; i<n; i++) {                  int ch = (mask / pow3[i])%3;                 if ( ch != 0) {                     usedGuesses ++;                 }                 if ( ch == 1) {                     correctGuesses++;                 }             }             if (usedGuesses == numGuesses) {  //no more guesses                 res = 0;             } else {                 // guess something...                 for (int i=0; i<n; i++) {                                        // what happens if I pick hat i?                     if ( (mask / pow3[i])%3 == 0) {                         // The wizard actually wants to minimize the result                         int wiz = 130000;                                                  // What happens if the wizard places a coin here?                         int nmask = mask + pow3[i];                                              if ( couldHappen(nmask)) {                             // coins[correctGuesses] is the smallest coin available                             wiz = std::min(wiz, coins[correctGuesses] + rec(nmask));                         }                         // can the wizard place nothing here?                         nmask = nmask + pow3[i];                         if ( couldHappen(nmask)) {                             wiz = std::min(wiz, rec(nmask));                         }                         // the maximum is what we want...                         res = std::max(res, wiz);                     }                 }             }         }         return res;     }      int findMaximumReward(vector <string> board, vector <int> coins, int numGuesses)     {         sort(coins.begin(), coins.end());         this->coins = coins;         this->numGuesses = numGuesses;          //init powers of 3:                 pow3[0] = 1;         for (int i=1; i<=13; i++) {             pow3[i] = 3*pow3[i-1];         }         //init parities and positions:                 r = board.size(); c = board[0].size();         rowpar.resize(r);         colpar.resize(c);          n = 0;         for (int i=0; i<r; i++) {             for (int j=0; j<c; j++) {                 rowpar[i] ^= ( board[i][j] == 'H');                 colpar[j] ^= ( board[i][j] == 'H');                 if (board[i][j] == 'H') {                     y[n] = j;                     x[n++] = i;                 }             }         }         memset(could, -1, sizeof(could));         if (! couldHappen(0) ) {             return -1;         }         memset(mem, -1, sizeof(mem));         return rec(0);                       } };

I was not so lucky during the contest. Ultimately, the one bug that prevented me from submitting was : I typed int & res = mask; instead of int & res = mem[mask] ;

Outcome

Barely nothing in rating increase.

Again, I think the problems were good and interesting, but the clarity issues sort of broke the match for me. We don't want topcoder to become codeforces. Well, at least I don't...


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# Daniel Kokotajlo comments on Utility Maximization = Description Length Minimization

• Thanks for the reply, but I might need you to explain/dumb-down a bit more.

--I get how if the variables which describe the world can only take a finite combination of values, then the problem goes away. But this isn’t good enough because e.g. “number of paperclips” seems like something that can be arbitrarily big. Even if we suppose they can’t get infinitely big (though why suppose that?) we face problems, see below.

--What does it mean in this context to construct everything as limits from finite sets? Specifically, consider someone who is a classical hedonistic utilitarian. It seems that their utility is unbounded above and below, i.e. for any setting of the variables, there is a setting which is a zillion times better and a setting which is a zillion times worse. So how can we interpret them as minimizing the bits needed to describe the variable-settings according to some model M2? For any M2 there will be at least one minimum-bit variable-setting, which contradicts what we said earlier about every variable-setting having something which is worse and something which is better.

• I’ll answer the second question, and hopefully the first will be answered in the process.

First, note that , so arbitrarily large negative utilities aren’t a problem—they get exponentiated, and yield probabilities arbitrarily close to 0. The problem is arbitrarily large positive utilities. In fact, they don’t even need to be arbitrarily large, they just need to have an infinite exponential sum; e.g. if is for any whole number of paperclips , then to normalize the probability distribution we need to divide by . The solution to this is to just leave the distribution unnormalized. That’s what “improper distribution” means: it’s a distribution which can’t be normalized, because it sums to .

The main question here seems to be “ok, but what does an improper distribution mean in terms of bits needed to encode X?”. Basically, we need infinitely many bits in order to encode X, using this distribution. But it’s “not the same infinity” for each X-value—not in the sense of “set of reals is bigger than the set of integers”, but in the sense of “we constructed these infinities from a limit so one can be subtracted from the other”. Every X value requires infinitely many bits, but one X-value may require 2 bits more than another, or 3 bits less than another, in such a way that all these comparisons are consistent. By leaving the distribution unnormalized, we’re effectively picking a “reference point” for our infinity, and then keeping track of how many more or fewer bits each X-value needs, compared to the reference point.

In the case of the paperclip example, we could have a sequence of utilities which each assigns utility to any number of paperclips X < (i.e. 1 util per clip, up to clips), and then we take the limit . Then our unnormalized distribution is , and the normalizing constant is , which grows like as . The number of bits required to encode a particular value is

Key thing to notice: the first term, , is the part which goes to with , and it does not depend on . So, we can take that term to be our “reference point”, and measure the number of bits required for any particular relative to that reference point. That’s exactly what we’re implicitly doing if we don’t normalize the distribution: ignoring normalization, we compute the number of bits required to encode X as

… which is exactly the “adjustment” from our reference point.

(Side note: this is exactly how information theory handles continuous distributions. An infinite number of bits is required to encode a real number, so we pull out a term which diverges in the limit , and we measure everything relative to that. Equivalently, we measure the number of bits required to encode up to precision , and as long as the distribution is smooth and is small, the number of bits required to encode the rest of using the distribution won’t depend on the value of .)

Does this make sense? Should I give a different example/use more English?

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# Relative Pose Estimation of Calibrated Cameras with Known $\mathrm{SE}(3)$ Invariants

15 Jul 2020Bo LiEvgeniy MartyushevGim Hee Lee

The $\mathrm{SE}(3)$ invariants of a pose include its rotation angle and screw translation. In this paper, we present a complete comprehensive study of the relative pose estimation problem for a calibrated camera constrained by known $\mathrm{SE}(3)$ invariant, which involves 5 minimal problems in total... (read more)

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• . , , () : ed., . .
• # SkyrmionsIn Chiral Magnets

In the original sense of the word, a Skyrmion is a topological soliton solution known to occur in a non-linear field theory for interacting pions originally conceived by the nuclear physicist Tony Skyrme [1]. In a more permissive interpretation of the word Skyrmions, as mathematical objects, have found versatile application in a variety of different areas in physics. In this chapter we briefly review the historic origin of the Skyrmion and define the generalised concept that is nowadays understood as a Skyrmion. We outline previous applications in different fields of physics and then forward to 2009 to give a concise account of the discovery of the Skyrmion lattice phase in the chiral magnet MnSi.

### What is a Skyrmion?

In 1961 before the advent of quantum chromodynamics (QCD) the nuclear physicist T.H.R. Skyrme conjectured that the interior of a nucleus is dominated by a medium formed from three pion fields [1]. He introduced the Skyrme model, a non-linear sigma model, with the intention to describe baryons as the quantised soliton solutions of a field theory which involves only bosonic degrees of freedom. The model is understood as an intermediate between the traditional models which represent the nucleons as point particles interacting through a potential, and a complete description based on quarks and gluons [2]. The pion fields $\mathbf{\pi} = (\pi_1,\pi_2,\pi_3)$ are combined into a $SU(2)$-valued field $$U(\mathbf x) = \sqrt{1 - \mathbf{\pi}(\mathbf x)\cdot\mathbf{\pi}(\mathbf x)}~\mathbb{1} + i \mathbf{\pi}(\mathbf x) \cdot \mathbf{\sigma}~,$$ where $\mathbf{\sigma}$ is the vector of Pauli matrices and we have suppressed a possible time dependence of the fields. For static fields the energy in the Skyrme model is given by $$E = \int d^3\mathbf r \left( -\frac{1}{2} \text{Tr}(R_i R_i) - \frac{1}{16} \text{Tr}([R_i,R_j][R_i,R_j]) \right)~,$$ where we have introduced an associated current $R_i=(\partial_i U)U^\dagger$ and $[\cdot,\cdot]$ denotes the commutator. The vacuum is represented by $U(\mathbf x)=\mathbb{1}$. For the energy to be finite, $U$ must approach a constant at infinity [3]. The energy is invariant under translations and rotations in $\mathbb{R}^3$ and also under the transformation $U \to A U A^\dagger$ with $A \in SU(2)$, one may thus choose $U(\mathbf r \to \infty)=\mathbb{1}$. Effectively, due to this boundary condition space is then topologically (but not metrically) compactified to $\mathbb{S}^3$, and since the group manifold of $SU(2)$ is also $\mathbb{S}^3$, $U$ defines a map from $\mathbb{S}^3$ to $\mathbb{S}^3$. The structure of the homotopically distinct maps $U$ is given by the third homotopy group $\pi_3(\mathbb{S}^3)$ which happens to be isomorphic to $\mathbb{Z}$. The space of all maps $U: \mathbb{S}^3 \to \mathbb{S}^3$ decomposes into distinct subsets characterised by an integer-valued topological charge $B=\int d^3\mathbf r~\mathcal{B}$ with the topological charge density $\mathcal{B}$ given by $$\mathcal{B} = -\frac{1}{24 \pi^2} \epsilon_{ijk} \text{Tr}(R_i R_j R_k)~.$$ The minimal energy solutions for each $B$ are called Skyrmions and their energy is identified with their mass and $B$ with the Baryon number of the nucleus. The $B=1$ Skyrmion has the spherically symmetric hedgehog form [3] $$U(\mathbf x) = \exp(i f(r) \hat{\mathbf{r}} \cdot \mathbf{\sigma})~,$$ where $f(r)$ is a radial profile function obeying an ordinary differential equation with the boundary conditions $f(0)=\pi$ and $f(r\to\infty)=0$.

Skyrmions in their original sense are therefore smooth, topologically stable extremal field configurations which are trivial at spatial infinity and have a finite energy. They are defined by surjective mappings into the order parameter space $\mathbb{S}^3$ and characterised by a non-trivial topological charge $B$. Since the $n$th homotopy group of $\mathbb{S}^n$ is isomorphic to $\mathbb{Z}$ for any $n\ge1$, a more permissive definition of Skyrmion is given by

A Skyrmion is a smooth field configuration defined by a topologically non-trivial, surjective mapping from a base manifold $\mathcal{M}$ into the order parameter space $\mathcal{T}\simeq\mathbb{S}^n$, trivial on the surface of $\mathcal{M}$ and characterised by a finite integer-valued topological charge.

Fig. 1 shows the construction recipe for $\mathcal{M} = \mathbb{R}^2$ and $\mathcal{T}=\mathbb{S}^2$. We start with the identity map, $\hat{\Omega}(\hat{\mathbf x})=\hat{\mathbf x}$, with $\hat{\mathbf x} \in \mathbb{S}^2$ which can be visualised as a hedgehog configuration (c.f. Fig. 1, top left). The stereographic projection $\mathcal{P}$ maps the sphere onto the two-dimensional plane, $\mathcal{P}(\hat{\mathbf x})\in \mathbb{R}^2$ and thus a Skyrmion configuration is given by the mapping $\hat{M}: \mathbb{R}^2 \to \mathbb{S}^2$, $\hat{M}: \mathbf r \mapsto \hat{\Omega} \circ \mathcal{P}^{-1}(\mathbf r)$. The corresponding topological charge $W$ is given by $\int d^2\mathbf r~\mathcal{W}$ where we have defined the topological charge density $\mathcal{W}$ as $$\mathcal{W} = \frac{1}{8\pi} \hat{M} \cdot (\partial_x \hat{M} \times \partial_y \hat{M})~.$$ $W$ counts the number of times the mapping $\hat{M}$ sweeps out the target manifold $\mathbb{S}^2$. From the construction recipe above it is obvious that for the Skyrmion (Fig. 1, lower left) $W=-1$.

The color code has been chose such that arrows pointing to the north pole are plotted in red and those to the south pole as blue and the equator in green.

A chiral, non-inversion symmetric Skyrmion can be constructed if the hedgehog is additionally combed' by performing a $\pi/2$ rotation $\mathcal{R}$ about the $\hat{z}$-axis in order parameter space (Fig.1, top right), $\hat{N}: \mathbf r \mapsto \mathcal{R} \circ \hat{\Omega} \circ \mathcal{P}^{-1}(\mathbf r)$. $\mathcal{R}$ is a linear map on $\mathcal{S}^2$ and therefore $W=-1$ for this configuration as well. These later, chiral Skyrmions (Fig. 1, lower right) will be the main focus of this thesis. In contrast to a Skyrmion, a vortex does not sweep out the whole sphere. For example a vortex configuration is given by the map $V: \mathbb{R}^2 \to \mathbb{S}^2$, $V: \mathbf r \mapsto \hat e_\phi(\mathbf r)$, where $\hat{e}_\phi = (-\sin(\phi),\cos(\phi),0)^T$ in polar coordinates $(r,\phi)$. The vortex only sweeps out the equator, is singular at $\mathbf r=0$ and has a non-trivial winding for $\mathbf r \to \infty$.

### Skyrmions in other areas of physics

Within this generalised understanding Skyrmions have found versatile application in many different fields of physics. Here we only mention a few. In 1985 Klebanov proposed the possibility of a Skyrmion crystal [4]. A phenomenological application of this kind of a solution could be a neutron crystal, which may exist under high pressure inside neutron stars [5]. The theory might resolve puzzles concerning discrepancies about the maximum mass of stable neutron stars between observations and predictions by more traditional equation of state descriptions [6].

Liquid crystals are states of matter which show characteristics of those of a conventional liquid and those of a solid crystal. Many interesting ordering phenomena have been reported in these systems where the local order parameter is describe by a director field ( a field of headless vectors) rather than a vector field [7]. Among these the blue phases which have a regular three-dimensional cubic structure of defects with lattice periods of several hundred nanometers are particularly interesting. Here so-called $2\pi$ disclinations are singular line defects where the $2\pi$ indicates that the director rotates a full $360^\circ$ as the singular line is encircled. These singular defect configurations are unstable towards a non-singular configuration that differs from its original one only in the immediate neighborhood of the formerly singular line. For the $2\pi$ disclinations these non-singular configurations are given by Skyrmion configurations of directors in $n=2$. Recently it was shown theoretically [8], with the aid of numerical methods, that a highly chiral nematic liquid crystal can accommodate a quasi-two-dimensional Skyrmion lattice as a thermodynamically stable state, when it is confined to a thin film between two parallel surfaces.

Skyrmions were predicted to occur in quantum Hall systems close to the Landau level filling fraction $\nu =1$ for sufficiently small Zeeman splitting $g \mu_B B$ (compared to the the cyclotron gap $\omega_c = e B / mc$) [9]. The incompressible ground state of a two-dimensional electron gas at this filling fraction is ferromagnetic. For sufficiently small $g < g_c$ the charged excitations of the system were argued to be Skyrmions where their winding number is related to the charge $\nu e$ of the Skyrmion. The equivalence of physical charge and topological charge in the system is a consquence of the quantum Hall effect and is responsible for the dominating role of Skyrmions in determining many physical properties [10]. Brey and collaborators proposed that ground state close to $\nu=1$ is a crystal of charged Skyrmions [11]. Nuclear magnetic resonance measurements in $\mathrm{GaAs}$ provided only indirect evidence [12,13].

Topologically, skyrmions are equivalent to certain magnetic bubbles (cylindrical domains) in ferromagnetic thin films, which were extensively explored in the 1970s for data storage applications [14]. In ferromagnets where long-range order is frustrated due to long-range dipole-dipole interactions a wealth of different magnetic patterns can be seen, such as domain walls, vortices and periodic stripes. In Ref. [15] Lorentz transmission electron microscopy (LTEM) was used to show that a magnetic field applied perpendicular to a thin film of hexaferrite turns the periodic stripe domain state into a periodic, hexagonal lattice of chiral Skyrmion bubbles (c.f. Fig 1, lower right). In contrast to other materials however where the inversion symmetry of the atomic unit cell is broken, in hexaferrite the helicity of the Skyrmion is not fixed by crystal structure, but represent a $\mathbb{Z}_2$ degree of freedom and a random distribution of different helicities in the lattice can be observed. Here even helicity reversals within a single Skyrmion where observed. Note that the helicity is independent of the winding number which can be seen from the fact that one may smoothly deform helicities into one another.

Bogdanov and collaborators studied the mean-field theory of easy-axis ferromagnets with chiral spin-orbit interactions. They argued that in certain parameter regimes a mixed state with a finite density of Skyrmions much like the vortex lattice in type II superconductors becomes the thermodynamically stable phase [16]. Although the stability analysis was carried out in the circular unit cell approximation the Skyrmion lattice was predicted to be hexagonal [16]. Here the presence of easy-axis anisotropy turned out to be a necessary ingredient for the stabilisation of the mixed phase within the mean-field treatment. Also they assumed the magnetization vector to be homogeneous along the z-axis [17].

### Discovery of the Skyrmion lattice in MnSi

In 2009 Mühlbauer et al. reported the discovery of a Skyrmion lattice phase in the chiral magnet $\mathrm{MnSi}$ by a small angle neutron scattering study (SANS). Although in this section we concentrate on $\mathrm{MnSi}$ as the first chiral magnet the spontaneous formation of this phase of whirling magnetisation has been observed in, the Skyrmion lattice phase has since then been discovered in many other compounds as well. In 2010 the same group discovered a Skyrmion lattice phase in the doped semiconductor $\mathrm{Fe_{1-x}Co_x Si}$ [18,19]. The Skyrmion lattice phase in this material was also later confirmed by real-space images using Lorentz transmission electron microscopy (Lorentz TEM) [19]. Since then the Skyrmion lattice has been observed in a variety of different materials both as a bulk phase as well as in thin films. Here we only want to mention that the electronic properties of this set of compounds is very diverse: Among these are metals, insulators, semi-conductors and also a multi-ferroic material. This show that the Skyrmion lattice is not a peculiarity of $\mathrm{MnSi}$ but rather a general phenomenon in this class of materials.

The unifying property for all of these materials is that they crystallise in the so-called B20 structure. The symmetry transformations are described by the space group $P2_13$ with a cubic Bravais lattice [20]. With only 12 symmetry operations this space group is among the smallest compatible with the cubic lattice crystal system. The point symmetry at the component sites is $C_3$, the cyclic group of 3-fold $2\pi/3$ rotations about an appropriate $[111]$ axis. The nonsymmorphic group $P2_13$ contains in addition 3 screw rotations which involve 2-fold rotations about one of the three $[100]$ axis followed by an appropriate non-primitive translation $(0,\frac{1}{2},\frac{1}{2})$. Most notably the list of symmetry transformations does not include the inversion. The lack of inversion symmetry has profound consequences for the Ginzburg-Landau free energy description of these materials and for the symmetry constraints on the magnetic configuration that the materials can show. Materials with non-inversion symmetric atomic unit-cells can support non-inversion symmetric magnetic structures. There are other mechanisms by which Skyrmion lattice phase can be stabilised. We will return to this point at the end of this chapter. Although we concentrate prodominantly on $\mathrm{MnSi}$ in this chapter, the magnetic phases of $\mathrm{MnSi}$ are generic for chiral magnets. Particularly the phase diagram Fig. 2 can be seen as a generic phase diagram for $B20$ compounds that order helimagnetically.

The primitive cell of manganese silicide ($\mathrm{MnSi}$) contains four pairs of the 2 component formula units $\mathrm{Mn}$ and $\mathrm{Si}$ located at $(u,u,u)$, $(\frac{1}{2}+u,\frac{1}{2}-u,\bar u)$, $(\frac{1}{2}-u,\bar u,\frac{1}{2}+u)$, and $(\bar u,\frac{1}{2}+u,\frac{1}{2}-u)$ with $u_{\rm Mn}=0.138$ and $u_{\rm Si}=0.845$. $\mathrm{MnSi}$ is an itinerant ferromagnet with a fluctuating magnetic moment of 0.4 $\mu_{\rm B}$ and a saturated moment of 2.2 $\mu_{\rm B}$ per manganese atom. Before the discovery of the skyrmion lattice phase, it already attracted attention due to a high pressure anomaly: Although described very well by Fermi-liquid theory at ambient pressure, $\mathrm{MnSi}$ shows a non-Fermi liquid phase above a critical pressure of $p_c \sim 14.6~\rm kbar$ with the temperature dependence of the resistivity given by $T^{3/2}$ [22,23]. In addition in the pressure region $12~\mathrm{kbar}-20~\mathrm{kbar}$ a state of partial magnetic order was encountered in neutron scattering experiments [24].

At ambient pressure and zero applied magnetic field, $\mathrm{MnSi}$ develops magnetic order below a transition temperature $T_c = 29.5 K$ that is the result of three hierarchical energy scales. The strongest scale is the ferromagnetic exchange favoring a uniform spin polarisation (spin alignment). The lack of inversion symmetry of the cubic B20 crystal structure results in chiral spin-orbit interactions, which may be described by the rotationally invariant Dzyaloshinsky Moriya (DM) interaction favoring canted spin configurations [25,26]. The DM interaction originates from relativistic effects, i.e. spin orbit coupling $\lambda_{\rm SO} \sim 10^{-2}$, and is the lowest order chiral spin-orbit interaction [27,28,29]. In addition there are very weak crystalline field interactions which break the rotational symmetry and align the ordering wave vector of the magnetic structures along the $[111]$ axes [29].

### Magnetic phases of MnSi

The magnetic phase diagram of MnSi, Fig. 2, shows four distinct magnetic phases: a helical phase, a conical phase, a field-polarized phase and the previously mentioned skyrmion lattice phase (for historical reasons referred to as the `A-phase'' in the diagram). In the following we briefly describe the magnetic order in each of these.

#### Helical phase

Cooling the system at zero or only small applied magnetic field below the critical temperature $T_c \sim 29~\mathrm{K}$ a phase transition to the helical phase is encountered. In this phase the magnetization winds around an axis parallel to the spiral propagation vector $\mathbf q$ as shown in Fig. 4 with the local magnetic moment $\mathbf M$ perpendicular to $\mathbf q$. The period of the helix, $\lambda_{\rm h} = 2 \pi / |\mathbf q|$ is controlled by the competition of the ferromagnetic exchange with the chiral spin-orbit coupling. The weakness of the spin-orbit interaction leads to a wavelength $\lambda_{\rm h} \sim 190~\overset{\circ}{A}$ which is large as compared with the lattice constant, $a \sim 4.56~\overset{\circ}{A}$. This large separation of length scales results in an efficient decoupling of the magnetic and atomic structures. The direction of propagation $\hat q = \mathbf q / |\mathbf q|$ is determined by tiny crystal field anisotropies. Therefore, the alignment of the helical spin spiral along the cubic space diagonal $[111]$ is weak and is only fourth power in the small spin-orbit coupling, $\lambda_{\rm SO}^4$. The decoupling from the underlying atomic structure results in an extremely coherent helical phase with a huge correlation length of $10^4~\overset{\circ}{A}$ as reported in this neutron scattering study [30]. While the paramagnetic to helical transition is expected to be second order on a mean-field level, interactions between the helimagnetic fluctuations were theoretically predicted to give rise to important corrections. Indeed it was recently shown that a Brazovskii-type scenario is realized where an abundance of strongly interacting fluctuation distributed uniformly over a sphere in momentum space drives the transition first order [31].

#### Conical phase

Setting out in the helical phase one finds a phase change upon increasing the applied magnetic field above $B_{c_1} \sim 0.1~\mathrm{T}$. The stronger magnetic field allows for a net reduction in free energy by building up a uniform magnetic moment in the direction of the applied field. While for high magnetic fields above $B_{c_2} \sim 0.6~\mathrm{T}$ the DM interaction can be completely neglected and the magnetic configuration completely polarizes, there is an intermediate field range where the magnetization winds both around a spiral propagation vector $\mathbf q$ parallel to $\mathbf B$ and in addition possesses a uniform magnetization in the direction of $\mathbf B$ as the the magnetisation vectors tilt towards $\hat q = \hat B$.

The phase is referred to as the concial phase and is despicted in Fig. 5. On general grounds a crossover between the helical and the conical phase is expected where the ordering wave vector $\mathbf q$ rotates continously from the helical $[111]$ direction towards the direction of the applied field. If applied along special high symmetry axis one may encounter a second order phase transition however. The angle between the propagation vector $\mathbf q$ and the local magnetization $\mathbf M$ is smooth function of the applied magnetic field $\mathbf B$ and decreases to zero for $\mathbf B > \mathbf B_{c_2}$.

#### Skyrmion lattice phase

A first order phase transition separates a tiny pocket in the magnetic phase diagram close to $T_c$ at finite magnetic field from the surrounding conical phase. This region, termed for historical reasons "A-phase", has a hexagonal lattice of anti-skyrmions perpendicular to the applied magnetic field as its ground state. An illustration of the skyrmion lattice is despicted in Fig. 6. The configuration possesses a translational invariance along the direction of the applied magnetic field. The magnetisation configuration should therefore be imagined as an ordered arrangement of whirling tubes similar to the flux lattice in a type II superconductor. Fig. 6 shows only a single layer. The magnetic configuration can be approximated by a superposition of three helices with their propagation vectors lying in a plane perpendicular to the applied magnetic field and relative angles of $120^\circ$ plus a uniform magnetic moment antiparallel to the applied field. The relative phases are aligned such that the magnetization in the center of the skyrmion points antiparallel to $\mathbf{B}$. The lattice constant is therefore given by $2 \lambda_{\rm h}/\sqrt{3}$. The large lattice constant ensures an efficient decoupling of the magnetic structure from the underlying atomic lattice and allows for the orientation towards the applied field. The orientation of the hexagonal lattice within the plane however is determined by crystal field anisotropies. For a magnetic field in the $[001]$ direction, for instance, one of the three $\mathbf q$ vectors pins weakly in the $[110]$ direction of the atomic crystal. The building blocks of the lattice are referred to as \textit{anti}-skyrmions as their their winding number per magnetic unit cell $$W =\frac{1}{4 \pi} \int_{\rm UC} \hat M \left( \partial_x \hat M \times \partial_y \hat M \right) \label{eq:2WindingNumberDensity}$$ is quantized to -1. Here $\hat M = \mathbf M / |\mathbf M|$ and the integration is taken over the two-dimensional magnetic unit cell, which contains exactly one "knot".

The experimental technique used by Mühlbauer et al. was small angle neutron scattering (SANS). Neutron scattering is an ideal tool for the study of magnetic order in bulk phases as neutrons predominantly scatter from the magnetic structure in a solid-state system due to their magnetic moment. The lack of an electric charge allows them to penetrate deep into the system under investigation. The neutrons scatter elastically due to the interaction of their spin with the nuclei and unpaired electrons of the magnetic atoms in the sample and the scattered neutrons are recorded by detectors placed behind the sample. The Fourier modes in the magnetic order are recorded as Bragg peaks in reciprocal space. The Skyrmion lattice can be approximated by three helices with their ordering wavevectors in a plane normal to the applied magnetic field and relative angles of $120^\circ$. In a typical neutron scattering experiment the incoming neutron beam is perpendicular to the applied magnetic field. In such a setup not all of the 6 reflection spots (two per helix at $+\mathbf q$ and $-\mathbf q$) can be seen simultaneously. The setup chosen by Mühlbauer et al. aligned the incoming beam parallel to the applied field. This setup is much more advantageous and allows to record all 6 spots at the same time, c.f. Fig. 7.
Other experimental techniques were also able to prove the existence of the Skyrmion lattice. In recent years powerful real-space imaging techniques have been modified and applied to chiral magnetic systems which allow for a direct visualization of the spatial magnetization configuration. The advantage of such methods is that not only a single spin texture, but also the crystallization and melting process during phase conversions can be observed. Fig. 8 shows images of the Skyrmion lattice phase in a thin film of $\mathrm{Fe_{0.5}Co_{0.5}Si}$ recorded by Lorentz transmission elctron microscopy (LTEM). LTEM is a modification of traditional electron microscopy in which the Lorentz forces between the electrons in a beam and the sample are utilised to generate images which allow for the real-space observation of the magnetic structure of materials. The drawback of LTEM is that samples have to be electron transparent and therefore the technique can only be applied to thin films. Also LTEM images only the in-plane component of the magnetisation. Real-space images of the surface of bulk materials can be recorded using the magnetic field microscopy (MFM). MFM images forces between the surface of a sample and the magnetic stray field of a cantilever tip coated with a ferromagnetic film. The total force acting on the cantilever is inferred from small changes in its resonance frequency. It is complementary to LTEM in the sense that it is only sensitive to the out-of-plane component of the magnetisation. Fig. 9 shows MFM images of Skyrmions from the surface of bulk $\mathrm{Fe_{0.5}Co_{0.5}Si}$. Red (blue) color indicates an out-of-plane component of the magnetisation that is anti-parallel (parallel) to the line of sight. For more information about real-space imaging techniques. % - mention susceptibilty Other physical quantities also show signatures in the Skyrmion lattice phase. For instance the magnetic AC suscpetibility $\chi$ shows a sudden drop to a lower when entering the Skyrmion phase from the conical phase by increasing the applied field. It then rises exceeding the value in the conical phase before entering the conical phase once again for higher magnetic fields [32]. A more dramatic effect can be seen in measurements of the Hall effect in $\mathrm{MnSi}$. Here due to the unique topology of the Skyrmion lattice an additional top hat contribution to the Hall signal can be seen in the Skyrmion lattice phase [33]. % alle Zitate unterbringen Ever since the original discovery of Skyrmions in chiral magnets in 2009, many exciting developments have deepened our unstanding of these fascinating structures. Here we mention only a few. Neubauer et al. showed that the topological properties of the Skyrmion lattice lead to additional contribution to the Hall signal, called the topological Hall effect [34]. Everschor et al. analyzed the spin-transfer effects resulting from an electric current driven through a Skyrmion lattice, and, in particular, focussed on the current-induced rotation of the magnetic texture by an angle in such a setup [35]. Schulz and collaborators have shown that the forces acting on conduction electrons moving through a Skyrmion lattice can be accounted for by the introduction of emergent (fictious) electromagnetic fields. This offered fundamental insights into the connection between the emergent and real electrodynamics of skyrmions in chiral magnets [36]. Iwasaki et al. showed in a numerical study that a single skyrmion can be created by an electric current in a simple constricted geometry comprising a plate-shaped specimen of suitable size and geometry [37]. In experimental realisation of Skyrmion creation however with a different mechanism was reported by Romming and collaborators in 2013. They showed that on an ultrathin magnetic film in which individual skyrmions can be written and deleted in a controlled fashion with local spin-polarized currents from a scanning tunneling microscope [38]. There have been many more interesting and noteworthy publications which we cannot mention here and without question there will be many more.


text:
E15-13 (Stock Split and Stock Dividend) The common stock of Warner Inc. is currently selling at $110 per share. The directors wish to reduce the share price and increase share volume prior to a new issue. The per share par value is$10; book value is $70 per share. Five million shares are issued and outstanding. Instructions Prepare the necessary journal entries assuming the following. (a) The board votes a 2-for-1 stock split. (b) The board votes a 100% stock dividend. (c) Briefly discuss the accounting and securities market differences between these two methods of increasing the number of shares outstanding. (Kieso 882) ## Answers (2) • (a) No entry—simply a memorandum note indicating the number of shares has increased to 18 million and par value has been reduced from$10 to $5 per share.  (b) Retained Earnings ($10 X 9,000,000)................ 90,000,000 Common Stock Dividend Distributable....... 90,000,000 Common Stock Dividend Distributable............ 90,000,000 Common Stock............................................. 90,000,000

(c)      Stock dividends and splits serve the same function with regard to the securities markets. Both techniques allow the board of directors to increase the quantity of shares and reduce share prices into a desired “trading range.”

For accounting purposes the 20%–25% rule reasonably views large stock dividends as substantive stock splits. In this case, it is necessary to capitalize par value with a stock dividend because the number of shares is increased and the par value remains the same. Earnings are capitalized for purely procedural reasons

• (a) No entry—simply a memorandum note indicating the number of shares has increased to 18 million and par value has been reduced from $10 to$5 per share.

(b)
Retained Earnings (\$10 X 9,000,000)................
90,000,000

Common Stock Dividend Distributable.......

90,000,000

Common Stock Dividend Distributable............
90,000,000

Common Stock.............................................

90,000,000

(c) Stock dividends and splits serve the same function with regard to the securities markets. Both techniques allow the board of directors to increase the quantity of shares and reduce share prices into a desired “trading range.”

For accounting purposes the 20%–25% rule reasonably views large stock dividends as substantive stock splits. In this case, it is necessary to capitalize par value with a stock dividend because the number of shares is increased and the par value remains the same. Earnings are capitalized for purely procedural reasons
Get homework help


text:
# set / class

#### highmath

##### Member
What the differences between set and class?

#### castor28

##### Well-known member
MHB Math Scholar
What the differences between set and class?
Hi highmath ,

Each set can be viewed as a class, but the converse is not true. You can have sets of sets, but not necessarily sets of classes.

A class is basically any collection you can think of; this was the approach used by Cantor in his initial theory (the naive set theory). However, it soon became apparent that attempting to build a set theory on that basis led to contradictions; one of these contradictions is Russel's paradox.

To solve that problem, modern set theory uses an axiomatic method: there is a set of axioms (the standard system is called ZFC) that define a limited set of operations that can be used to construct sets. The collection of objects that satisfy a property is a class, but, unless you can construct it using the axioms of the theory, you cannot assume that it is a set; in particular, you cannot use it as a member of a set.

A class that is not a set is called a proper class. For example, the class of all sets is a proper class: there is no set of all sets.


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# Surface integral

The surface integral or surface integral is a generalization of the concept of integral on flat or curved surfaces . The integration area is therefore not a one-dimensional interval , but a two-dimensional set in three-dimensional space . For a more general representation in with see: Integration on Manifolds . ${\ displaystyle {\ mathcal {F}}}$ ${\ displaystyle \ mathbb {R} ^ {3}}$${\ displaystyle \ mathbb {R} ^ {n}}$${\ displaystyle n \ geq 2}$

A general distinction is made between a scalar and a vector surface integral, depending on the shape of the integrand and the so-called surface element . They are

${\ displaystyle \ iint _ {\ mathcal {F}} f \, \ mathrm {d} \ sigma}$with scalar function and scalar surface element as well as${\ displaystyle f}$${\ displaystyle \ mathrm {d} \ sigma}$
${\ displaystyle \ iint _ {\ mathcal {F}} {\ vec {v}} \ cdot \ mathrm {d} {\ vec {\ sigma}}}$with vector-valued function and vectorial surface element .${\ displaystyle {\ vec {v}}}$${\ displaystyle \ mathrm {d} {\ vec {\ sigma}}}$

## terms and definitions

When integrating over surfaces, parameterizations of the surface take the place of the integration variable and surface elements take the place of the infinitesimal (infinitely small) interval width . ${\ displaystyle \ mathrm {d} x}$

### Parameterization

As a two-dimensional set, a surface can be represented ( parameterized ) as a function of two variables . If a set, the edge of which does not contain double points, is continuously differentiable , not infinitely long and, furthermore, is a mapping of into , then one says, is the parameterization of the area if is At this point it should be pointed out that a large part of the difficulties in dealing with surface integrals is related to the parameterization. It is not clear a priori that different parameterizations produce the same value for the integral. A change of coordinates for surface integrals is not trivial and is therefore a motivation for the use of differential forms . ${\ displaystyle B \ subset \ mathbb {R} ^ {2}}$${\ displaystyle \ varphi}$${\ displaystyle B}$${\ displaystyle \ mathbb {R} ^ {3}}$${\ displaystyle \ varphi}$${\ displaystyle {\ mathcal {F}}}$${\ displaystyle {\ mathcal {F}} = \ varphi (B)}$

In general, an area can be represented with two parameters and in the following form: ${\ displaystyle \ mathbb {R} ^ {3}}$${\ displaystyle u}$${\ displaystyle v}$

${\ displaystyle \ varphi \ colon B \ to \ mathbb {R} ^ {3}, \ quad \ left (u, v \ right) \ mapsto {\ vec {\ varphi}} \ left (u, v \ right) = \ left ({\ begin {matrix} x \ left (u, v \ right) \\ y \ left (u, v \ right) \\ z \ left (u, v \ right) \\\ end {matrix }} \ right)}$

On the surface of the form of curves or the coordinate lines . These cover the area with a coordinate network, with two coordinate lines running through each point. Thus every point on the surface has unique coordinates . ${\ displaystyle {\ vec {\ varphi}} \ left (u, v \ right)}$ ${\ displaystyle u = \ mathrm {const}}$${\ displaystyle v = \ mathrm {const}}$ ${\ displaystyle \ left (u_ {0}, v_ {0} \ right)}$

#### Example 1: Representation of parameters

The surface of a sphere with a radius can be parameterized as follows: is the rectangle and ${\ displaystyle R}$${\ displaystyle B}$${\ displaystyle [0, \ pi] \ times [0.2 \ pi]}$

${\ displaystyle {\ vec {\ varphi}} (u, v) = {\ begin {pmatrix} R \ sin (u) \ cos (v) \\ R \ sin (u) \ sin (v) \\ R \ cos (u) \ end {pmatrix}}}$.

This parameterization fulfills the spherical equation (see also spherical coordinates ). is the polar angle (usually or ) and the azimuth angle (usually or ). ${\ displaystyle x ^ {2} + y ^ {2} + z ^ {2} = R ^ {2}}$${\ displaystyle u}$${\ displaystyle \ vartheta \,}$${\ displaystyle \ theta \,}$${\ displaystyle v}$${\ displaystyle \ varphi \,}$${\ displaystyle \ phi \,}$

#### Example 2: Explicit representation

If a function and the area are given in the form , then and are the two parameters; the parameterization of the area looks like this: ${\ displaystyle f \ colon B \ to \ mathbb {R}, \ \ left (x, y \ right) \ mapsto f \ left (x, y \ right)}$${\ displaystyle z = f (x, y)}$${\ displaystyle x}$${\ displaystyle y}$

${\ displaystyle {\ vec {\ varphi}} \ left (x, y \ right) = \ left ({\ begin {matrix} x \\ y \\ f \ left (x, y \ right) \\\ end {matrix}} \ right)}$

### Surface element

If in the one-dimensional case this represents the width of an infinitely small interval, then in the two-dimensional case it is obvious to replace it with the area of an infinitely small area . With the parameterization described in the previous section, two tangents can be placed at each point on the surface (see also: Curvilinear coordinates ): One is the tangent that arises if one leaves constant and varies minimally, and one with exchanged variables. That means two tangents to the two coordinate lines in the point under consideration . These tangents can be expressed by two infinitesimal tangent vectors (be the parameterized form of the surface): ${\ displaystyle \ mathrm {d} x}$${\ displaystyle \ mathrm {d} \ sigma}$${\ displaystyle v}$${\ displaystyle u}$${\ displaystyle \ left (u_ {0}, v_ {0} \ right)}$${\ displaystyle {\ vec {\ varphi}} \ left (u, v \ right)}$

${\ displaystyle \ left. {\ frac {\ partial {\ vec {\ varphi}}} {\ partial u}} \ right | _ {u_ {0}, v_ {0}} \ mathrm {d} u}$   and   ${\ displaystyle \ left. {\ frac {\ partial {\ vec {\ varphi}}} {\ partial v}} \ right | _ {u_ {0}, v_ {0}} \ mathrm {d} v}$

In the following, the compact notation is used for the partial derivatives:

${\ displaystyle {\ vec {\ varphi}} _ {u} = {\ frac {\ partial {\ vec {\ varphi}}} {\ partial u}}}$   and   ${\ displaystyle {\ vec {\ varphi}} _ {v} = {\ frac {\ partial {\ vec {\ varphi}}} {\ partial v}}}$

If these tangents are not parallel at any point on the surface, one speaks of a regular parameterization . The cross product of the tangent vectors is then a vector whose length is not equal to zero.

${\ displaystyle \ left | \ left | {\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} \ right | \ right | \ neq 0}$

The two tangent vectors lie in the tangent plane of the surface at the point under consideration. The area of the parallelogram spanned by the two tangent vectors now corresponds to the amount of their cross product .

If the surface is now parameterized regularly, one defines: ${\ displaystyle {\ vec {\ varphi}} (u, v)}$

• Scalar surface element
${\ displaystyle \ mathrm {d} \ sigma = \ left | \ left | {\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} \ right | \ right | \ mathrm {d} u \, \ mathrm {d} v}$
• Vectorial surface element
${\ displaystyle \ mathrm {d} {\ vec {\ sigma}} = {\ hat {n}} \ \ mathrm {d} \ sigma = {\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} \ \ mathrm {d} u \, \ mathrm {d} v}$     with the unit normal vector of the surface element     ${\ displaystyle {\ hat {n}} = {\ frac {{\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v}} {\ left | \ left | { \ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} \ right | \ right |}}}$

According to the properties of the cross product, the vectorial surface element is perpendicular to the surface; its amount corresponds precisely to the size of the infinitesimal surface piece.

In the form presented above, the vectorial surface element is not well-defined , since its direction depends on whether one calculates or . The two possibilities are antiparallel to each other. If you look at closed surfaces, you usually agree that the vectorial surface element pointing outwards is to be used. ${\ displaystyle {\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v}}$${\ displaystyle {\ vec {\ varphi}} _ {v} \ times {\ vec {\ varphi}} _ {u} = - \ left ({\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} \ right)}$

#### Example 1: Representation of parameters

As shown above, the surface of the sphere with radius R can be parameterized by the polar angle and the azimuth angle . The area element results from the following calculation: ${\ displaystyle u}$${\ displaystyle v}$

{\ displaystyle {\ begin {aligned} & {\ vec {\ varphi}} = R \ left ({\ begin {matrix} \ sin u \ \ cos v \\\ sin u \ \ sin v \\\ cos u \\\ end {matrix}} \ right), \ quad {\ vec {\ varphi}} _ {u} = R \ left ({\ begin {matrix} \ cos u \ \ cos v \\\ cos u \ \ sin v \\ - \ sin u \\\ end {matrix}} \ right), \ quad {\ vec {\ varphi}} _ {v} = R \ left ({\ begin {matrix} - \ sin u \ \ sin v \\\ sin u \ \ cos v \\ 0 \\\ end {matrix}} \ right), \\ & \ pm \ left ({\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} \ right) = \ pm R ^ {2} \ sin u \ left ({\ begin {matrix} \ sin u \ \ cos v \\\ sin u \ \ sin v \\\ cos u \\\ end {matrix}} \ right), \ quad \ left | \ left | \ pm \ left ({\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} \ right) \ right | \ right | = R ^ {2} \ sin u, \\ & {\ hat {n}} = \ pm \ left ({\ begin {matrix} \ sin u \ \ cos v \\\ sin u \ \ sin v \\\ cos u \\\ end {matrix}} \ right), \ quad \ mathrm {d} {\ vec {\ sigma}} = {\ hat {n}} \, \ mathrm {d} \ sigma = {\ hat {n}} \ R ^ {2} \ sin u \ \ mathrm {d} u \, \ mathrm {d} v \\\ end { aligned}}}

Two solutions are possible for the normal vector ( ), depending on the order of and in the cross product. Typically, the positive solution is chosen here, in which points away from the convex spherical surface (so-called "outer normal"). ${\ displaystyle \ pm}$${\ displaystyle {\ vec {\ varphi}} _ {u}}$${\ displaystyle {\ vec {\ varphi}} _ {v}}$${\ displaystyle {\ hat {n}}}$

#### Example 2: Explicit representation

If the surface is in the form specified, pressing the surface element by the differentials of the coordinates , from. ${\ displaystyle z = f (x, y)}$${\ displaystyle x}$${\ displaystyle y}$

${\ displaystyle {\ vec {\ varphi}} = \ left ({\ begin {matrix} x \\ y \\ f (x, y) \ end {matrix}} \ right), \ quad {\ vec {\ varphi}} _ {x} = \ left ({\ begin {matrix} 1 \\ 0 \\ f_ {x} \ end {matrix}} \ right), \ quad {\ vec {\ varphi}} _ {y } = \ left ({\ begin {matrix} 0 \\ 1 \\ f_ {y} \ end {matrix}} \ right)}$
${\ displaystyle \ pm \ left ({\ vec {\ varphi}} _ {x} \ times {\ vec {\ varphi}} _ {y} \ right) = \ pm \ left ({\ begin {matrix} - f_ {x} \\ - f_ {y} \\ 1 \ end {matrix}} \ right), \ quad \ left | \ left | \ pm \ left ({\ vec {\ varphi}} _ {x} \ times {\ vec {\ varphi}} _ {y} \ right) \ right | \ right | = {\ sqrt {f_ {x} ^ {2} + f_ {y} ^ {2} +1}} \, \ quad {\ hat {n}} = \ pm {\ frac {1} {\ sqrt {f_ {x} ^ {2} + f_ {y} ^ {2} +1}}} \ left ({\ begin {matrix} -f_ {x} \\ - f_ {y} \\ 1 \ end {matrix}} \ right)}$

Thus the surface element and the vectorial surface element are the same:

${\ displaystyle \ mathrm {d} \ sigma = {\ sqrt {f_ {x} ^ {2} + f_ {y} ^ {2} +1}} \ \ mathrm {d} x \, \ mathrm {d} y}$
${\ displaystyle \ mathrm {d} {\ vec {\ sigma}} = {\ hat {n}} \ {\ sqrt {f_ {x} ^ {2} + f_ {y} ^ {2} +1}} \ \ mathrm {d} x \, \ mathrm {d} y = {\ begin {pmatrix} -f_ {x} \\ - f_ {y} \\ 1 \ end {pmatrix}} \ \ mathrm {d} x \, \ mathrm {d} y}$

### Projection onto a surface with a known surface element

In the following, we assume that a surface with its surface element and associated normal vector is known. E.g. ${\ displaystyle A}$${\ displaystyle \ mathrm {d} A}$${\ displaystyle {\ hat {n}} _ {A}}$

${\ displaystyle \ mathrm {d} A = \ mathrm {d} x \ mathrm {d} y}$   and   ${\ displaystyle {\ hat {n}} _ {A} = {\ hat {e}} _ {z} = \ left ({\ begin {matrix} 0 \\ 0 \\ 1 \\\ end {matrix} } \ right)}$
• Lateral surface of a circular cylinder with radius :${\ displaystyle \ rho}$
${\ displaystyle \ mathrm {d} A = \ rho \ mathrm {d} \ varphi \ mathrm {d} z}$   and   ${\ displaystyle {\ hat {n}} _ {A} = {\ hat {e}} _ {\ rho} = \ left ({\ begin {matrix} \ cos \ varphi \\\ sin \ varphi \\ 0 \\\ end {matrix}} \ right)}$
• Spherical surface with radius :${\ displaystyle r}$
${\ displaystyle \ mathrm {d} A = r ^ {2} \ sin \ vartheta \ mathrm {d} \ vartheta \ mathrm {d} \ varphi}$   and   ${\ displaystyle {\ hat {n}} _ {A} = {\ hat {e}} _ {r} = \ left ({\ begin {matrix} \ sin \ vartheta \ \ cos \ varphi \\\ sin \ vartheta \ \ sin \ varphi \\\ cos \ vartheta \\\ end {matrix}} \ right)}$

The surface element is to be determined for another surface with a normal vector . The area is roughly given by and thus the normal vector is equal . ${\ displaystyle {\ mathcal {F}}}$${\ displaystyle {\ hat {n}} _ {\ mathcal {F}}}$${\ displaystyle \ mathrm {d} \ sigma}$${\ displaystyle g (x, y, z) = 0}$${\ displaystyle {\ hat {n}} _ {\ mathcal {F}} = \ nabla g / || \ nabla g ||}$

We now project along to . Then the surface elements can be linked using for : ${\ displaystyle {\ mathcal {F}}}$${\ displaystyle {\ hat {n}} _ {A}}$${\ displaystyle A}$${\ displaystyle \ mathrm {d} A = \ mathrm {d} {\ vec {A}} \ cdot {\ hat {n}} _ {A} = | \ mathrm {d} {\ vec {\ sigma}} \ cdot {\ hat {n}} _ {A} | = \ mathrm {d} \ sigma \, | {\ hat {n}} _ {\ mathcal {F}} \ cdot {\ hat {n}} _ {A} |}$${\ displaystyle {\ hat {n}} _ {\ mathcal {F}} \ cdot {\ hat {n}} _ {A} \ neq 0}$

${\ displaystyle \ mathrm {d} \ sigma = {\ frac {\ mathrm {d} A} {| {\ hat {n}} _ {\ mathcal {F}} \ cdot {\ hat {n}} _ { A} |}} = {\ frac {|| \ nabla g || \, \ mathrm {d} A} {| \ nabla g \ cdot {\ hat {n}} _ {A} |}}}$

Each line along the normal vectors may only intersect the surface once. Otherwise you have to divide the area into smaller areas , the projection of which is then clear, or choose a different base area . ${\ displaystyle {\ hat {n}} _ {A}}$${\ displaystyle {\ mathcal {F}}}$${\ displaystyle {\ mathcal {F}}}$${\ displaystyle {\ mathcal {F}} _ {1}, \, {\ mathcal {F}} _ {2}, \, \ dotsc}$${\ displaystyle A}$

The vectorial surface element is:

${\ displaystyle \ mathrm {d} {\ vec {\ sigma}} = {\ hat {n}} _ {\ mathcal {F}} {\ frac {\ mathrm {d} A} {| {\ hat {n }} _ {\ mathcal {F}} \ cdot {\ hat {n}} _ {A} |}} = {\ frac {\ nabla g} {|| \ nabla g ||}} {\ frac {| | \ nabla g || \, \ mathrm {d} A} {| \ nabla g \ cdot {\ hat {n}} _ {A} |}} = {\ frac {\ nabla g \, \ mathrm {d } A} {| \ nabla g \ cdot {\ hat {n}} _ {A} |}}}$

#### example 1

Given a surface of the form , then applies and thus: ${\ displaystyle {\ mathcal {F}}}$${\ displaystyle z = f (x, y)}$${\ displaystyle g (x, y, z) = zf (x, y)}$

${\ displaystyle \ nabla g = {\ begin {pmatrix} -f_ {x} \\ - f_ {y} \\ 1 \ end {pmatrix}} \, \ quad || \ nabla g || = {\ sqrt { f_ {x} ^ {2} + f_ {y} ^ {2} +1}} \, \ quad {\ hat {n}} _ {\ mathcal {F}} = {\ frac {\ nabla g} { || \ nabla g ||}} = {\ frac {1} {\ sqrt {f_ {x} ^ {2} + f_ {y} ^ {2} +1}}} {\ begin {pmatrix} -f_ {x} \\ - f_ {y} \\ 1 \ end {pmatrix}}}$

This area is now projected into the plane with and ; is there ${\ displaystyle xy}$${\ displaystyle \ mathrm {d} A = \ mathrm {d} x \ mathrm {d} y}$${\ displaystyle {\ hat {n}} _ {A} = {\ hat {e}} _ {z}}$

${\ displaystyle \ mathrm {d} \ sigma = {\ frac {|| \ nabla g || \, \ mathrm {d} x \ mathrm {d} y} {| \ nabla g \ cdot {\ hat {e} } _ {z} |}} = {\ frac {{\ sqrt {f_ {x} ^ {2} + f_ {y} ^ {2} +1}} \, \ mathrm {d} x \ mathrm {i.e. } y} {| {\ hat {e}} _ {z} \ cdot {\ hat {e}} _ {z} |}} = {\ sqrt {f_ {x} ^ {2} + f_ {y} ^ {2} +1}} \, \ mathrm {d} x \ mathrm {d} y}$
${\ displaystyle \ mathrm {d} {\ vec {\ sigma}} = {\ frac {\ nabla g \, \ mathrm {d} x \ mathrm {d} y} {| \ nabla g \ cdot {\ hat { e}} _ {z} |}} = {\ begin {pmatrix} -f_ {x} \\ - f_ {y} \\ 1 \ end {pmatrix}} \ mathrm {d} x \ mathrm {d} y }$

#### Example 2

We are looking for the surface element of a solid of revolution around the axis with , that is . ${\ displaystyle z}$${\ displaystyle \ rho = f (z)}$${\ displaystyle g (\ rho, \ varphi, z) = \ rho -f (z)}$

${\ displaystyle \ nabla g = {\ hat {e}} _ {\ rho} -f_ {z} {\ hat {e}} _ {z} \, \ quad || \ nabla g || = {\ sqrt {1 + f_ {z} ^ {2}}} \, \ quad {\ hat {n}} _ {\ mathcal {F}} = {\ frac {\ nabla g} {|| \ nabla g ||} } = {\ frac {{\ hat {e}} _ {\ rho} -f_ {z} {\ hat {e}} _ {z}} {\ sqrt {1 + f_ {z} ^ {2}} }}}$

By projecting onto the lateral surface of a circular cylinder with a radius , the surface element is obtained: ${\ displaystyle \ rho = f (z)}$

${\ displaystyle \ mathrm {d} \ sigma = {\ frac {|| \ nabla g || \, \ rho \, \ mathrm {d} \ varphi \ mathrm {d} z} {| \ nabla g \ cdot { \ hat {e}} _ {z} |}} = {\ frac {{\ sqrt {1 + f_ {z} ^ {2}}} \, f (z) \, \ mathrm {d} \ varphi \ mathrm {d} z} {| ({\ hat {e}} _ {\ rho} -f_ {z} {\ hat {e}} _ {z}) \ cdot {\ hat {e}} _ {\ rho} |}} = {\ sqrt {1 + f_ {z} ^ {2}}} \, f (z) \, \ mathrm {d} \ varphi \ mathrm {d} z}$
${\ displaystyle \ mathrm {d} {\ vec {\ sigma}} = ({\ hat {e}} _ {\ rho} -f_ {z} {\ hat {e}} _ {z}) \, f (z) \, \ mathrm {d} \ varphi \ mathrm {d} z}$

## The integrals

With the parameterizations and the surface elements you can now define the surface integrals. These multi-dimensional integrals are Lebesgue integrals , but can be calculated as multiple Riemann integrals in most applications .

### The scalar surface integral

The scalar surface integral of a scalar function over a surface with regular parameterization with is defined as ${\ displaystyle f \ colon \ mathbb {R} ^ {3} \ rightarrow \ mathbb {R}}$${\ displaystyle {\ mathcal {F}}}$${\ displaystyle \ varphi \ colon B \ rightarrow \ mathbb {R} ^ {3}}$${\ displaystyle B \ subset \ mathbb {R} ^ {2}}$

${\ displaystyle \ iint _ {\ mathcal {F}} f ({\ vec {x}}) \, \ mathrm {d} \ sigma = \ iint _ {B} f \ left ({\ vec {\ varphi} } (u, v) \ right) \, || {\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi}} _ {v} || \, \, \ mathrm {d} (u, v)}$

For example , if one sets , the scalar surface integral is simply the area of the surface. ${\ displaystyle f ({\ vec {x}}) = 1}$

### The vectorial surface integral

The vector surface integral of a vector-valued function over a surface with regular parameterization with is defined as ${\ displaystyle f \ colon \ mathbb {R} ^ {3} \ rightarrow \ mathbb {R} ^ {3}}$${\ displaystyle {\ mathcal {F}}}$${\ displaystyle \ varphi \ colon B \ rightarrow \ mathbb {R} ^ {3}}$${\ displaystyle B \ subset \ mathbb {R} ^ {2}}$

${\ displaystyle \ iint _ {\ mathcal {F}} {\ vec {f}} ({\ vec {x}}) \ cdot \ mathrm {d} {\ vec {\ sigma}} = \ iint _ {B. } {\ vec {f}} \ left ({\ vec {\ varphi}} (u, v) \ right) \ cdot ({\ vec {\ varphi}} _ {u} \ times {\ vec {\ varphi }} _ {v}) \, \, \ mathrm {d} (u, v) =: \ Phi _ {\ mathcal {F}} ({\ vec {f}})}$.

A clear presentation of this integral takes place via the flow of a vector field through the surface : The size indicates the contribution to the total flow made by the infinitesimally small surface vector ; namely, how much of flows through the patch . The flux is maximum when the vector field is parallel to the surface normal , and zero when it is perpendicular to , i.e. tangential to the surface - then it "flows" along the surface, but not through it. ${\ displaystyle \ Phi}$ ${\ displaystyle {\ vec {f}}}$${\ displaystyle {\ mathcal {F}}}$${\ displaystyle {\ vec {f}} \ cdot \ mathrm {d} {\ vec {\ sigma}}}$${\ displaystyle \ Phi _ {\ mathcal {F}} ({\ vec {f}})}$${\ displaystyle \ mathrm {d} {\ vec {\ sigma}} = {\ hat {n}} \, \ mathrm {d} \ sigma}$${\ displaystyle {\ vec {f}}}$${\ displaystyle \ mathrm {d} {\ sigma}}$${\ displaystyle {\ vec {f}}}$${\ displaystyle {\ hat {n}}}$${\ displaystyle {\ vec {f}}}$${\ displaystyle {\ hat {n}}}$${\ displaystyle {\ vec {f}}}$

## literature

• G. Bärwolff: Higher mathematics for natural scientists and engineers . 2nd Edition. Spectrum Academic Publishing House, 2006, ISBN 978-3-8274-1688-9
• KF Riley, MP Hobson: Mathematical Methods for Physics and Engineering . 3. Edition. Cambridge University Press, 2006, ISBN 978-0-521-67971-8


text:
# Which are the factors that affect the permeability of soil? (i) Particle size (ii) Void ratio (iii) Degree of saturation (iv) Adsorbed water

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closed

Which are the factors that affect the permeability of soil?

(i) Particle size

(ii) Void ratio

(iii) Degree of saturation

1. (i) and (ii)
2. (i), (ii) and (iii)
3. (ii) and (iv)
4. (i), (ii), (iii) and (iv)

by (117k points)
selected

Correct Answer - Option 4 : (i), (ii), (iii) and (iv)

Explanation

Permeability depends on both soil properties and Fluid properties

The kozney – Carman equation is quite useful, it reflects the effect of factors that affect permeability

$k = \frac{{{\gamma _w}}}{\mu } \times \frac{{{e^3}}}{{1 + e}} \times D_{10}^2$

Where,

D10 is effective grain size

e is the void ratio

μ is viscosity , γ w is unit weight of water

a) Grain size: The coefficient of permeability includes D102, where D10 is a measure of grain size

If the void ratio is the same then permeability is more in coarse soil than in the fine soil

b) Void ratio: From the equation, it is clear that k ∝ e2

If the particle size is the same, then loose soils are more permeable than dense soils

c) Degree of saturation: Permeability is directly proportional to the degree of saturation

d) Particle shape:

It is expressed in terms of specific surface area and permeability relates to specific surface area as ‘k ∝ 1/S

Adsorbed water is the water layer formed around the soil particle especially in the case of fine-grained soils. This reduces the size of the void space by about 10%. Hence, permeability reduces.