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# The interferometer

67 interferometers are optical instruments used for measuring flatness and determining the length of slip gauges by direct reference to the wavelength of light the interferometer incorporates the extension of the application of the optical flat it overcomes some of [. Lloyd's mirror and stellar interferometry the michelson interferometer other interferometers and interference phenomena with interference in scattered light and mach-zehnder, sagnac, jamin, fabry-perot, series and speckle interferometers fourier optics. Phasecam dynamic twyman-green laser interferometers: high resolution measurement despite vibration, 5000x faster than phase shifting laser interferometers. Norland products has over 20 years of interferometric measuring experience and has produced the first low cost, automated, non-contact interferometer for fiber optics call (609)395-1966 for pricing. Bringing together what is ligo and what is an interferometer content to explain ligo's ifo design. Hammar directed one leg of his interferometer through a heavy metal pipe plugged with lead if aether were dragged by mass, it was theorized that the mass of the sealed metal pipe would have been enough to cause a visible effect. An ftir is typically based on the michelson interferometer experimental setup an example is shown in figure 1the interferometer consists of a beam splitter, a fixed mirror, and a mirror that translates back and forth, very preciselythe beam splitter is made of a special material that transmits half of the radiation striking it and reflects.

Laser interferometers from zygo were first introduced commercially in 1972 today, zygo continues to maintain the leadership role in surface form metrology using laser interferometry. 2 interferometry: principles and applications of two-beam and multiple-beam interferometry by prof hiroshi komatsu institute for materials research. Scientific 1 012-02675b interferometer equipment scientific model os-8501 interferometer the pasco scientific model os-8501 michelson interferometer is a precision instru. Interferometer, an instrument that uses the interference patterns formed by waves (usually light, radio, or sound waves) to measure certain characteristics of the waves themselves or of materials that reflect, refract, or transmit the waves. Interferometry page for astrophysics option of nsw hsc physics course. The interferometer system is a high-precision, movable-mirror interferometer that can be used to perform michelson, fabry-perot & twyman-green interferometry.

4d technology manufactures laser interferometers, surface roughness profilers and surface defect gauges, for measuring precision surfaces and optics. Optical interferometer: optical interferometer,, instrument for making precise measurements for beams of light of such factors as length, surface irregularities, and index of refraction it divides a beam of light into a number of beams that travel unequal paths and whose intensities, when reunited, add or subtract. Scanning fabry-perot interferometers the core of thorlabs' scanning fabry-perot interferometers is an optical cavity formed by two, nearly identical, spherical mirrors separated by their common radius of curvature as shown in figure 1 this configuration is known as a mode-degenerate, confocal cavity design, which is generally referred to as a. Chapter 1 the michelson interferometer 11 prelab in this lab you have to nd the position of mirror m1 (see fig 11) such that the optical path length from the beam splitter to m1 and back is identical to.

A clear and easy-to-understand introduction to the science of interferometers, what they are, and how they work. Interferometers i the largest fully steerable dishes have diameters $d \approx 100$ m the angular resolution of a diffraction-limited telescope is $\theta \approx \lambda/d$ radians, so impossibly large diameters are needed to achieve sub-arcsecond resolution at radio wavelengths. With the interferometer which he invented, michelson found no evidence of the ether, to his and everyone else's surprise michelson's terse description of the experiment: the interpretation of these results is that there is no displacement of the interference bands. Terahertz interferometer system radiabeam technologies has developed an inexpensive, flexible michelson interferometer for far infrared to mm-wave radiation.

## The interferometer

Most precise ruler ever constructed this animation illustrates how the twin observatories of ligo work one observatory is in hanford, washington, the other.

• I the michelson interferometer discussion the michelson interferometer is often introduced in the context of special relativity where its historic role in providing evidence against the existence of an absolute rest frame defined by the electromagnetic ether is emphasized.
• Sumix interferometers are designed for inspection of fiber optic ferrules, patch cords, pigtails and bare fiber in the process of polishing and assembly.
• Using michelson interferometer, one can determine the wavelength of light from a monochromatic sourceit can also be used to calculate the refractive index of a thin film.
• Interferometers 21 3 figure 2 the michelson interferometer three times, while the other traverses it only once accordingly a compensating plate of the same thickness as the beam splitter is introduced in the second beam to equalize the.
• In optics, a fabry-pérot interferometer (fpi) or etalon is typically made of a transparent plate with two reflecting surfaces, or two parallel highly reflecting mirrors (precisely, the former is an etalon and the latter is an interferometer, but the terminology is often used inconsistently)its transmission spectrum as a function of.

1 the michelson interferometer invented by aa michelson, also famous for measuring the speed of light the original purpose of an interferometer. Fringe contrast adjust the interferometer alignment to maximize the fringe contrast in the optical signal then, use electronic filtering and amplification to maximize the signal for analog-to-digital conversion high dc offset leads. Experiment3 1 the$michelson$interferometer$and$the$he4ne$laser$physics2150$experiment$no$3$university$of$colorado$ $introduction$ themichelsoninterferometerisoneexampleofanopticalinterferometerthe. Experimental study of the role of trap symmetry 2 1 introduction atom interferometers [1, 2] have demonstrated excellent performance in measuring. Interferometer definition, optics a device that separates a beam of light into two ray beams, usually by means of reflection, and that brings the rays together to produce interference, used to measure wavelength, index of refraction, and astronomical distances see more. Basic schematic of ligo's interferometers with an incoming gravitational wave depicted as arriving from directly above the detector (image: ligo) what is an interferometer interferometers are investigative tools used in many fields of science and engineering.

The interferometer
Rated 4/5 based on 11 review

2018.


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# 29 Embedding images

There are a few ways to embed images in a plotly graph. Perhaps the easiest is to point the image source to a hyperlink containing a bitmap file (e.g., png, jpeg, tiff, etc).

plot_ly() %>%
layout(
images = list(
x = 0, y = 1,
sizex = 0.2, sizey = 0.1,
xref = "paper", yref = "paper",
xanchor = "left", yanchor = "bottom"
),
margin = list(t = 50)
)

The approach in Figure 29.1 has a downside though – if that hyperlink breaks, then so does your plot. It’d be better to download the file to your machine and use the dataURI() function from the base64enc package (or similar) to embed the image as a data URI (Urbanek 2015).

plot_ly() %>%
layout(
images = list(
source = base64enc::dataURI(file = "images/rstudio.png"),
x = 0, y = 1,
sizex = 0.2, sizey = 0.1,
xref = "paper", yref = "paper",
xanchor = "left", yanchor = "bottom"
),
margin = list(t = 50)
)

Another approach is to convert a raster object into a data URI, which the raster2uri() function in plotly is designed to do. R actually ships with native support for raster objects and many image processing R packages either build on this data structure or provide a utility to convert to a raster object (perhaps via as.raster()). For example, the readPNG() function from the png package reads image data in an R array, which can be converted to a raster object (Urbanek 2013).

pen <- png::readPNG("images/penguin.png")

plot_ly() %>%
layout(
images = list(
source = raster2uri(as.raster(pen)),
x = 2, y = 2,
sizex = 2, sizey = 1,
xref = "x", yref = "y",
xanchor = "left", yanchor = "bottom",
sizing = "stretch"
)
)

Moreover, thanks to the R package magick it’s possible to rasterize non-raster file formats (e.g., pdf, svg, etc) directly in R, so if you wanted, you could also embed non-bitmap images by using image_read() and image_convert() to generate a raster object (Ooms 2019).

format: plain


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# How do you find the domain and range of f(x)=2/(3/(x-6))-7?

Dec 28, 2017

Starting with a domain and range of the set of all real numbers, exclude domain values for which division by zero may occur, and exclude range values that would result from the forbidden domains, if no other domain maps to said range values. This yields:

Domain: $\left\{x \in \mathbb{R} | x \ne 6\right\}$

Range: $\left\{y \in \mathbb{R} | y \ne - 7\right\}$

#### Explanation:

Our function is

$f \left(x\right) = \frac{2}{\frac{3}{x - 6}} - 7$.

Since there are no radicals (where the imaginary number may reside), we could safely assume that the domain and range is the set of all real numbers...

$\left\{x | x \in \mathbb{R}\right\}$, $\left\{y | y \in \mathbb{R}\right\}$

... But we need to remember the wrath of division by zero.

To prevent any divisions by zero, we need to make sure that the denominator cannot equal zero. In this function, there is a fraction

$\frac{2}{\frac{3}{x - 6}}$

where the denominator

$\frac{3}{x - 6}$

cannot equal zero:

$\frac{3}{x - 6} \ne 0$

However, inside this denominator, there is also a denominator

$x - 6$

which also cannot equal zero:

$x - 6 \ne 0$.

To figure out what the forbidden numbers are, we could solve for $x$ as if it's an equal sign. In this case, add both sides by $6$:

$x - 6 + 6 \ne 0 + 6$

$x \ne 6$

But how about that bigger denominator?

$\frac{3}{x - 6} \ne 0$

Well, we could first try to multiply by the smaller denominator...?

$\frac{3}{x - 6} \cdot \left(x - 6\right) \ne 0 \cdot \left(x - 6\right)$

But wait! We need to make sure $\left(x - 6\right)$ isn't zero! That means assuming $x - 6 \ne 0$ and therefore $x \ne 6$ as we have solved earlier. Simplfying out:

$3 \ne 0$

Well, of course $3 \ne 0$, so that's pretty much all the forbidden numbers. It's just $x \ne 6$. So, the domain should be all real numbers except $6$:

Domain: $\left\{x \in \mathbb{R} | x \ne 6\right\}$

And the range? Is it all real numbers too? Take a moment to think about it. If $6$ wasn't a forbidden number, what would its output be? What if we try to simplify the function?

$f \left(x\right) = \frac{2}{\frac{3}{x - 6}} - 7$

Hmm... we could start by rewriting the fraction as plain division:

${f}_{\text{modified}} \left(x\right) = 2 \div \left(\frac{3}{x - 6}\right) - 7$

Division should "flip" the numerator and denominator (due to being the multiplicative inverse):

${f}_{\text{modified}} \left(x\right) = 2 \cdot \left(\frac{x - 6}{3}\right) - 7$

We could multiply the $2$ over:

${f}_{\text{modified}} \left(x\right) = \frac{2 \left(x - 6\right)}{3} - 7$

${f}_{\text{modified}} \left(x\right) = \frac{2 x - 12}{3} - 7$

And "split" the $3$:

${f}_{\text{modified}} \left(x\right) = \frac{2 x}{3} - \frac{12}{3} - 7$

Leaving us to simplify:

${f}_{\text{modified}} \left(x\right) = \frac{2}{3} x - 4 - 7$

${f}_{\text{modified}} \left(x\right) = \frac{2}{3} x - 11$

Nice, this is a linear function! Since its graph is a line, each range should also be associated to one and only one domain. With no divisions by $x$, we could input $6$ in:

${f}_{\text{modified}} \left(x\right) = \frac{2}{3} \left(6\right) - 11$

And solve:

${f}_{\text{modified}} \left(x\right) = \frac{2 \cdot 6}{3} - 11$

${f}_{\text{modified}} \left(x\right) = \frac{12}{3} - 11$

${f}_{\text{modified}} \left(x\right) = 4 - 11$

${f}_{\text{modified}} \left(6\right) = - 7$

Ah! So that value, $- 7$, could not possibly be in the original function's range, due to it being a linear function. Therefore,

Domain: $\left\{x \in \mathbb{R} | x \ne 6\right\}$

Range: $\left\{y \in \mathbb{R} | y \ne - 7\right\}$


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# What is y_c when finding y_R?

1. Sep 30, 2013

### aaronfue

1. The problem statement, all variables and given/known data

During the calculation of yR in fluid mechanics problem, hydrostatic force on a plane surface

2. Relevant equations

yR = yc+$\frac{I_xc}{y_c×A}$

3. The attempt at a solution

I can see how it is calculated but I've read through the chapter and I don't see how it is described. The textbook is Fundamentals of Fluid Mechanics (6th Ed.) by Munson.

#### Attached Files:

• ###### fluids example.JPG
File size:
20.6 KB
Views:
51
2. Sep 30, 2013

### SteamKing

Staff Emeritus
It looks like yc is the location of the centroid of the gate, while yR is the location of the resultant of the pressure acting on the gate, or the center of pressure.


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My bibliography  Save this paper

# Supplement to Martingale properties of self-enforcing debt''

## Author

Listed:
• Bidian, Florin
• Bejan, Camelia

## Abstract

We present some complementary results to Bidian and Bejan (2012). Part 1 provides necessary and sufficient transversality conditions for an agent's optimization problem. They are extensions to stochastic environments of the conditions given by Kocherlakota (1992), or alternatively, extensions to nonzero debt constraints of the corresponding conditions in Forno2003}. Part \ref{ap:B} presents an elementary proof of the characterization of NTT debt limits (Theorem 3.5 in the main paper) for the case when debt constraints bind in bounded time, that requires no martingale techniques or boundedness assumptions on the discounted debt limits. Part \ref{ap:C} complements results in Section 5.1 (in the main paper), showing that all the equilibria that can sustain bubbles under an interdiction to trade can be achieved from fixed, zero initial wealth for the agents. Thus endogeneity of debt limits causes multiplicity of not only asset prices (through bubbles), but also of real equilibrium allocations.

## Suggested Citation

• Bidian, Florin & Bejan, Camelia, 2011. "Supplement to Martingale properties of self-enforcing debt''," MPRA Paper 36610, University Library of Munich, Germany, revised 12 Feb 2012.
• Handle: RePEc:pra:mprapa:36610
as

File URL: https://mpra.ub.uni-muenchen.de/36610/1/MPRA_paper_36610.pdf
File Function: original version

## References listed on IDEAS

as
1. Kocherlakota, Narayana R., 1992. "Bubbles and constraints on debt accumulation," Journal of Economic Theory, Elsevier, vol. 57(1), pages 245-256.
2. Florin Bidian & Camelia Bejan, 2015. "Martingale properties of self-enforcing debt," Economic Theory, Springer;Society for the Advancement of Economic Theory (SAET), vol. 60(1), pages 35-57, September.
Full references (including those not matched with items on IDEAS)

## Citations

Citations are extracted by the CitEc Project, subscribe to its RSS feed for this item.
as

Cited by:

1. Gaetano Bloise & Pietro Reichlin & Mario Tirelli, 2013. "Fragility of Competitive Equilibrium with Risk of Default," Review of Economic Dynamics, Elsevier for the Society for Economic Dynamics, vol. 16(2), pages 271-295, April.
2. Bejan, Camelia & Bidian, Florin, 2014. "Bubbles and trading in incomplete markets," Journal of Mathematical Economics, Elsevier, vol. 53(C), pages 137-144.
3. Werner, Jan, 2014. "Rational asset pricing bubbles and debt constraints," Journal of Mathematical Economics, Elsevier, vol. 53(C), pages 145-152.

### Keywords

rational bubbles; transversality conditions; endogenous debt limits; not-too-tight constraints; self-enforcing debt; limited enforcement;

### JEL classification:

• E0 - Macroeconomics and Monetary Economics - - General
• G0 - Financial Economics - - General

### NEP fields

This paper has been announced in the following NEP Reports:

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text:
QUESTION

# Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3,508, what was the amount invested in scheme B?

Hint: To solve this question, we will let the amount invested in scheme B be x and then apply a formula of simple interest to find the amount.

Now, let the amount invested in B be Rs. x. Then, the amount invested in a is equal to Rs. (13,900 – x). The rate of interest per annum for A and B is 14% and 11% respectively. Now, simple interest is calculated by the formula,

Simple interest = $\dfrac{{{\text{P}} \times {\text{R}} \times {\text{T}}}}{{100}}$, where P is the principal amount invested, R is the rate of interest and T is the number of years.

According to question, S.I1 + S.I2 = 3508, where S.I1 is the simple interest earned by scheme A and S.I2 is the amount earned by B. So, applying the formula, we get

$\dfrac{{(13900{\text{ - x)(2)(14)}}}}{{100}}$ + $\dfrac{{({\text{x)(2)(11)}}}}{{100}}$ = 3508

14(13900 – x) + 11x = 175400

194600 – 14x + 11x = 175400

3x = 19200

x = 6400

So, amount invested in scheme B = Rs. x = Rs. 6400

Note: When we come up with such types of questions, we will first let the value be found as a variable. Then we will use the formula to find simple interest. Then, we will apply the values given in question in the formula. After it, we will solve the formed equation to find the value of the variable which is our required answer.


text:
# Uniqueness of solutions of Young differential equations

Consider the following one dimensional Young differential equation: $$Y_t=\int_0^t Y_s dX_s,\quad t\in[0,1].$$

Here the driving process $$X$$ is a bounded functions $$[0,1]\to\mathbb{R}$$, which is $$\beta$$-Holder with $$\beta<1/2$$.

If $$Y$$ is an $$\alpha$$-Holder function, $$\alpha>1-\beta$$, then this equation is well defined (because the integral then becomes just the Young integral).

Question: how to prove that the only solution to this equation in the class of $$\alpha$$-Holder functions is $$Y\equiv0$$?

Warning: note that $$\beta<1/2$$! If $$\beta>1/2$$, then this result is standard, but what to do if $$\beta<1/2$$?

Failed solution attempt

Denote by $$[X]_{\beta,[0,T]}$$, $$[Y]_{\alpha,[0,T]}$$ the corresponding Holder norms of $$X$$ and $$Y$$ on the interval $$[0,T]$$, respectively. Then the standard inequality for the Young integral implies $$|Y_t-Y_s-Y_s(X_t-X_s)|\le C[X]_{\beta,[0,T]}[Y]_{\alpha,[0,T]} (t-s)^{\alpha+\beta},\quad s,t\in[0,T].$$ This in turn leads $$|Y_t-Y_s|\le C[X]_{\beta,[0,T]}[Y]_{\alpha,[0,T]} (t-s)^{\alpha+\beta}+|Y_s|\,|X_t-X_s|,$$ and thus $$[Y]_{\beta,[0,T]}\le C[X]_{\beta,[0,T]}[Y]_{\alpha,[0,T]} T^{\alpha}+[X]_{\beta,[0,T]}\sup_{r\in[0,T]}|Y_r|.$$

However, because $$\beta<\alpha$$, the last inequality gives us nothing (we are estimating a smaller norm by a larger norm). The iteration over $$T$$ also seems hopeless. So what to do?

• Just notice that your first inequality and the condition $Y\in Hol_\alpha$ forces $X\in Hol_\alpha$ as long as $Y$ is separated from $0$. Jul 26, 2020 at 18:52
• @fedja I agree and I also thought about using this fact, but unfortunately I don't see how it helps. It might be that $X$ is alpha Holder on any interval $[\epsilon,1]$ (with its Holder norm going to infinity as $\epsilon$ approaches $0$) and still not alpha Holder on $[0,1]$. Think about $\sqrt x$. Its Lipschitz on any interval $[\epsilon,1]$, but not Lipschitz on $[0,1]$. Maybe I am missing something here, but the proof definitely does not follow immediately from your comment. Do you have a formal proof or could you please expand your ideas?
– Oleg
Jul 26, 2020 at 21:55
• "Maybe I am missing something here, but the proof definitely does not follow immediately from your comment." It does. Let $t>0$. Choose small $\varepsilon>0$ and consider the last moment $t_1\in[0,t]$ when $|Y_t|\le\varepsilon$. Then $|Y_s|\ge\varepsilon$ on $[t_1,t]$, so $X\in Hol_\alpha([t_1,t])$ and then $Y_s=Y_{t_1}\exp(X_s-X_{t_1})$ for $s\in[t_1,t]$. In particular $|Y_t|\le\varepsilon\exp(2\|X\|_{C([0,t])})$. But $\varepsilon>0$ is arbitrarily small! Jul 27, 2020 at 12:50
• @fedja thanks! that's indeed a great solution. If you would formally post is as a solution, I would be very happy to award you the promised bounty.
– Oleg
Jul 28, 2020 at 16:00
• I was hoping though that one can solve this problem by iterating the inequalities I wrote. My motivation is that i am stuck with a similar problem, but for PDEs. I want to show that dY/dt=\Delta Y+YdX has a unique zero solution (where now Y=Y(x,t)) under the same regularity assumptions. In this case one cannot write the solutions explicitly, but one can get very similar bounds to the ones presented in my question.
– Oleg
Jul 28, 2020 at 16:11

For such small $$\beta$$, we need to use Rough path theory to make sense of the integral and so below I go over that. (Indeed for $$\beta\in (\frac{1}{2},1]$$, there is an ODE theory for Young integrals eg see theorem 2.5 in "Stochastic differential equations driven by fractional Brownian motions")

In "Rough Path Theory" by Andrew L. Allan (available free online), I found in Proposition 6.12 a solution in the spirit of what mentioned in the comments.

In the uniqueness, they do go over an estimates argument in terms of Holder norms that might be useful in your SPDE setting.

In your case instead of $$1$$, you will have $$0$$ and so the same argument will apply. Meaning that for general $$y_{0}\in \mathbb{R}$$ the solution to

$$Y_{t}=y_{0}+\int_{0}^{t}Y_{s}d\mathbf{X}_{s}$$

can be similarly obtained by Ito's formula for rough paths (prop 6.11)

$$Y_{t}=y_{0}exp\left(X_{t}- \frac{1}{2}[\mathbf{X}]_{t}\right).$$

If you want me to add more details on the argument, please let me know.

For smaller $$\beta$$, the situation might break down. In here fractional Brownian Motion driven stochastic integrals, Hairer mentions some other approaches that might even apply to your SPDE setting (eg. regularity structures).


text:
Jannis Teunissen

# Differences

This shows you the differences between two versions of the page.

 — blog:computational_scientist [2019/05/03 21:07] (current) Line 1: Line 1: + ==== Becoming a computational scientist ==== + + //Note: the text below is copied from Appendix A of my [[publications:thesis|PhD + thesis]] from 2015.// + + I have now been working on computational science and computational physics + problems for more than six years. What has surprised me, in hindsight, is the + great number of things that one has to be familiar with in order to be + productive. The reason is probably the relatively large amount of DIY (do it + yourself) in computing, compared to other disciplines. Below I will try to + summarize what skills I have found to be generally important. + + === Selecting problems === + + For doing research, the most important skill is perhaps the ability to pick the + right problems. + This is a rather difficult skill to master, and I am confident that I have not + done so yet. + Still, there are a couple of simple questions that I find useful for selecting + problems: + + * Are you interested in the problem? + * Are others interested in the problem? + * Do you expect to learn something useful when studying the problem? + * Does the problem seem feasible to you? If this is not clear, how much time do you approximately have to invest to answer this? + * Suppose that everything works out: you solve the problem. What would that mean to you? And what could you do next? + * How long do you give yourself? And suppose that you are unable to solve the full problem, is there then an intermediate result that could be of value? + * How hard will it be to write about the results? For example, for certain types of results a carefully written introduction, motivation, discussion or analysis might be required. + + Another question that becomes more relevant towards the end of a PhD is whether + it is possible to obtain (future) grants or funding for a topic. + + ===Theoretical skills=== + + Below, I briefly discuss some of the theoretical topics that I believe to be + important for a computational scientist. + The most important topic is missing however, namely knowledge of the domain that + you are working in. + Such knowledge will help in selecting the right problems and in making the right + approximations. + + ====Applied mathematics==== + + These are some of the topics in applied mathematics that I think are important + for a computational scientist: + + * Linear algebra: many problems can be written as a system of linear equations. + * Calculus: ordinary differential equations and Taylor series are important for many numerical methods. Also helps for knowing what can be calculated analytically or for being able to construct reference solutions. + * Statistics: Monte Carlo methods are quite common; to work with them, at least a basic understanding of statistics is required. The same goes for problems that are probabilistic in nature or contain data with noise. + + ====Computer science==== + + When we want to solve a problem on a computer, we have to select the appropriate + //algorithm//. Algorithms can be classified by their difficulty' or + computational cost, which is the main topic of //computational complexity + theory//. Knowing and understanding the computational cost of algorithms is not + only important for efficiently solving a problem, but also for predicting what + problems are feasible. For example, if you recognize that you are trying to + solve an [[https://en.wikipedia.org/wiki/NP-hard|NP-hard]] problem, then you + immediately know that you are limited to small problem sizes. With parallel + computing, it is usually possible to go to larger problem sizes. To what extent + this is the case depends on how well the algorithmic components can be + parallelized, i.e., on the amount of local computation versus global + communication. + + The practical cost of algorithms also has to do with the device that performs + the algorithmic steps or computations. Modern processors operate in a rather + complicated way, but knowledge of the cost of typical operations is important + when you have to develop an efficient numerical method. The hardware in a + processor also determines what integer and floating point numbers you can use. + Understanding floating point arithmetic and its subtleties can save you a lot of + time debugging weird' behavior. + + ====Computational science==== + + Although there are many types of computations, most of them can be categorized + into just a few categories: + + * Solving linear systems of equations, i.e., solve $A x = b$ for a given matrix $A$ and vector $b$. Surprisingly many problems can be transformed into such a linear system. + * Optimization, for example: find the shortest path between $N$ cities, find the ground state energy of a quantum system or find the minimum of a function. + * Ordinary and partial differential equations. Many (physical) systems can be described by such equations. Different types of partial differential equations require quite different solution strategies. + + A computational scientist should probably be familiar with the basic methods for + solving problems from these categories, so that one is able to find and select + the best methods when the need arises. + To prevent reinventing the wheel, some knowledge of the available libraries and + codes is valuable. + + ===Practical skills=== + + The best strategy for solving a problem depends on what tools are already + available. If sufficiently many other people have worked on a (similar) problem, + software might be available that you can directly use. Take for example CFD + (computational fluid dynamics), for which there are many different simulation + tools. Selecting the right one then becomes one of the most important aspects of + solving your problem. + + The other extreme would be that no existing software exists for your problem, so + that you have to develop everything yourself. There are of course also many + cases in between, for example when existing tools have to be modified to suit + your needs. This means that it is often necessary to write computer code. Below, + some of the practical aspects of writing your own code and reusing others' code + are discussed. + + ====Computer basics==== + + For computing, the ''*nix'' operating systems appear to be most popular. + Being familiar with a variant of e.g., GNU/Linux, BSD or OS X is therefore quite + helpful -- this allows you to quickly use the code and tools that others have + written. + + Good command of a text editor such as ''vim'' or ''emacs'', or a suitable IDE + (integrated development environment) will speed up your code and text editing. + This might also reduce the risk of developing RSI (repetitive strain injury), + because most editors can be operated without a mouse ((In my experience, the + combination of stress and mouse usage is most likely to cause physical + discomfort.)). There are many useful tools included in a ''*nix'' system, but + ''ssh'' gets a special mention, because it allows you to work on remote systems. + + There exist a number of software suites for doing numerical or symbolic + computations. Commercial packages are for example Matlab and Mathematica, + whereas [[https://www.gnu.org/software/octave/|Octave]] or + [[http://www.sagemath.org/|SageMath]] are examples of free software + alternatives. The many built-in functions can help you to quickly develop a + computational method. Even if you eventually have to implement this solver in a + different environment, it can be helpful to start from a simple + proof-of-concept. The generality of such suites is also their drawback: + typically they will not be as efficient as a special purpose solution. + + ====Programming==== + + When you develop a method from scratch, you can use your preferred programming + language -- this is of course not possible when you have to modify an existing + method. The traditional languages for computing are C and Fortran. Especially C + is quite `low-level', so that experience with C will be useful for understanding + how a computer and other languages work. Fortran was specifically designed for + numerical computing, which can make code development more convenient. Another + popular //compiled// language is C++, which allows for many programming + styles. This flexibility can be good for the expert but is sometimes hard for + the beginner. Performance wise, there are no major differences between these + languages as long as you know what you are doing. + + For certain tasks, scripting or interpreted languages such as Python can be more + convenient. + Such languages can for example be used to glue together other programs, to + process data or to visualize results. + Python can also be used for computations, although the numerical work is then + typically performed by routines written in C or Fortran, which are made + available by Python modules such as ''numpy''. + + Numerical code is no different from other code: many things can go wrong. + Sometimes a program simply does not compile or run, but at other times it might + not be clear whether there is a //bug// or whether there is a failure for + another reason. Code often depends on (particular versions of) libraries, which + is a source of compilation errors; understanding how code is compiled will help + in figuring out what is required. Another example are the + //Makefiles//((Makefiles contain rules that describe how a collection of source + files should be compiled. Another common build system is ''CMake''.)) included + with numerical software: they might not work on your machine, in which case you + need to know how to modify them. As most programs contain bugs, basic debugging + skills are very valuable. The larger a project grows, the more important these + skills become. + + Being familiar with a version control system such as git has various benefits: + you can keep tracks of your changes, get the latest version of a code, + collaborate with others etcetera. Perhaps even more important is being able to + visualize your results. There exist many tools for this, examples of popular + open source packages are gnuplot, [[https://visit.llnl.gov/|Visit]] and + [[https://www.paraview.org/|Paraview]].


text:
# What is the compressibility of this simple “book”?

Compressibility is defined as

$$C=\frac{2^{HN}}{2^{H_{max}N}}$$

The book is made up of a simple alphabet of only {a,b,c,d} which occur with probabilities $$P(a)=0.2, P(b)=0.4, P(c)=0.1, P(d)=0.3$$

In class we were given the example of calculating the entropy of a simple string of coin tosses, where $$N$$ is the number of coin tosses.

So, for example, we could say that $$P(heads)=0.1$$ and $$P(tails)=0.9$$. The entropy of each coin toss is therefore $$H(p=0.9)=0.469$$ If we toss the coin 4 times then we end up with a compressibility of: $$C=\frac{2^{HN}}{2^{H_{max}N}}=\frac{2^{0.469\cdot4}}{2^{1\cdot 4}}=22.3\%$$ How do we extend this to the book case?

The entropy of the letter occurance is $$H(p=0.2,p=0.4,p=0.1,p=0.3)=1.85$$

and the maximum entropy is just 2 bits (i.e. we require 2 bits to descibe four characters)

Since this is all the information given in the question, I presume that we just set $$N=4$$ and plug in the values for $$H$$ and $$H_{max}$$ to get the entropy. However, I am struggeling to see why. Why is $$N$$ not the number of letters that make up this book?

• Can you add in the question an accessible reference where your compressibility is introduced? – John L. Jan 14 '19 at 5:43


text:
Talk:List of moments of inertia

WikiProject Physics (Rated List-class, Mid-importance)
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Polygon

"Thin, solid, polygon shaped plate with vertices $\vec{P}_{1}$, $\vec{P}_{2}$, $\vec{P}_{3}$, ..., $\vec{P}_{N}$ and mass $m$."

Does this work for ALL polygons? Convex/Non-convex etc. I think this should be made clear. --Smremde 10:16, 21 June 2007 (UTC)

If you consider a thin, flat sheet of material then it should work, since finding the moment of inertia is easily done if you know the surface area of the material. The formula given should have considered the formula for the surface area of an n-sided polygon. Anyway, if n → infinity the moment of inertia should tend towards that of the this flat circular disc. AquaDTRS (talk) 16:39, 22 March 2008 (UTC)

I'm going to fix some mistakes in the statement of the conditions for this one. It's only valid for rotation about an axis perpendicular to the plane, since the moment of inertia would otherwise have to depend on the orientation of the axis, and the expression given is independent of that orientation. It's also not necessary for it to be a regular polygon, since the expression is just an area-weighted sum of the r^2's of the centers of mass of the triangles. It is necessary for it to be a plane polygon, but it's not necessary for it to be thin.--207.233.88.250 (talk) 00:16, 12 December 2008 (UTC)

Do there really need to be norms around the vector products? In 2d they are scalars anyway and I believe the norms (getting absolute value functions) to be wrong. —Preceding unsigned comment added by 131.246.120.30 (talk) 14:28, 16 April 2010 (UTC)

Although the shapes are 2D, the formula holds for shapes embedded in a 3D space. So the vectors might have nonzero x, y, and z components. I prefer making it explicit that we mean the scalar magnitudes of the cross products. CosineKitty (talk) 19:59, 16 April 2010 (UTC)

Shouldn't the summation be over N triangles, with P(N+1)=P(1)? Or have I missed some 'degree of freedom' type thing? 86.4.250.43 (talk) 00:25, 9 December 2013 (UTC)

I have corrected the summation to go from 1 to N. I suspect the restriction on the polygon being star-shaped (wrt the origin) can be relaxed by taking care with the placement of the modulus signs, but I don't have a reference. Also added an entry for a triangle rotating about one vertex, from which the polygon formula is an easy corollary. A picture would be nice but I don't know how to do this. David Brightly 86.4.250.43 (talk) 22:54, 11 December 2013 (UTC)

?

I know it's generally obvious from the picture, but isn't it technically necessary to specify the axis of rotation as well as the body under consideration? Hammerite 00:51, 23 October 2005 (UTC)

duplicate item in area moments of inertia

There are two rows called "an axis collinear with the base", and somewhat worryingly they have different formulae. Could somebody check and delete/clarify one or the other? Bathterror 09:56, 8 March 2006 (UTC)

The two rows referred to a rectangular area and a triangular area. Should be a little clearer now. Hemmingsen 14:16, 22 April 2006 (UTC)

Thick cylinder with open ends mistake in formula?

The formula I of z axis has the multiplier 1/2 but shouldnt it be 1/8 ?

No, the first formula is OK. $\scriptstyle{{1\over 2}m(R_{ext}^2+R_{int}^2)}$ LPFR 12:39, 4 September 2006 (UTC)

This page does NOT display correctly on several on several of my machines (Mac OSX, Linux, and Windows). WHY are these colons in front of the math directive needed and why is it responsible to put in formating that causes parse errors to the casual observer of these pages?

It displays fine on mine (Windows XP. IE and Netscape). LPFR 12:39, 4 September 2006 (UTC)
If the colons your are talking about are " \,[/itex] ", this is a TeX/LaTeX directive that adds a small space in mathmode. It cannot create a parse error in any decent browser. I think that you have a problem with your machines or your browsers. I think that this page is OK and well written. But if you think that there are things that can be ameliorated, just do it. LPFR 12:58, 4 September 2006 (UTC)

No, the first formula with R1 + R2 is not OK. If the two radii are equal the wall thickness is zero, so the mass in such a case must be zero, and the inertia is zero. Do the math: for a material of density ρ, the triple integral from height -h/2 to +h/2, radius ri to ro, and direction 0 to 2π for the cylinder yields a moment of inertia of 2πρh(ro-ri). — Preceding unsigned comment added by JeffCalTech (talkcontribs) 13:13, 23 March 2014 (UTC)

Correction on the previous comment: The formula is unfortunately easily misunderstood and hence miseading in that most people think from the vantage point of making an object of a particular shape out of a particular material. They would like to vary the dimensions and see what happens to the inertia and do not expect the mass to remain constant as if it were being compressed. This confusion would be eliminated if the mass were replaced with the density yielding a moment of inertia of 2πρh(ro-ri). — Preceding unsigned comment added by JeffCalTech (talkcontribs) 13:34, 23 March 2014 (UTC)

Splitting the article

Would anyone have a problem if I split this into two seperate articles? I can't see any point in having these two distinct properties in the one article. I think it adds to the common confusion between these two quantities, and the article is a bit long also. Brendanfox 02:02, 25 April 2006 (UTC)

That sounds like a good idea to me. Hemmingsen 18:23, 25 April 2006 (UTC)
Well, this should be long enough time for people to voice their objections. Going ahead with the split... (to List of area moments of inertia) Hemmingsen 15:03, 30 December 2006 (UTC)

Formula for a thick cylinder

January 2010 (UTC) The correct formula is $\scriptstyle{{1\over 2}m(R_{ext}^2+R_{int}^2)}$.

Someone using IP 216.148.248.31 (CERFN California Education and Research Federation Network) wrongly "corrected" the equation changing the + sign for a minus sign. I suggest to this user to verify this formula in a physics book. If you did find this formula with the minus sign, verify in others books and correct the wrong formula.

As you do not seem to be able to calculate the inertia moment by yourself, I can demonstrate why the sign must be a + sign. If $\scriptstyle{R_{int}}$ increases, but maintaining the total mass m unchanged, this means that you are "sending" some of the mass from a small radius to a larger one. As the moment of inertia is proportional to the squared distance to axis of rotation, the moment of inertia should increase. That is, if $\scriptstyle{R_{int}}$ increases the moment of inertia increases. Then, the sign must be a plus sign. LPFR 07:59, 7 September 2006 (UTC)

The mistake is understandable - usually one would expect the mass to decrease if one made the hole larger, and if the hole was the same diameter as the outer then the mass and MoI would be zero! If the mass stays constant then the density must increase, but the positive sign is then correct. If the density stays constant (which is the practical case of most interest) then the formula becomes (I think, if my late-night integration is correct) (pi/2)*density*length*(R2^4 - R1^4). If you substitute mass/volume for density in this you get back to the correct formula given above. 217.147.104.220 10:51, 20 December 2006 (UTC)

The mass of the cylinder could be kept constant without increasing the density of the material, by increasing the length of the cylinder.Gregorydavid 11:23, 20 December 2006 (UTC)

The formula Iz = (pi/2)*density*length*(R2^4 - R1^4) couldn't possibly be correct, since R1 = R2 gives Iz = 0, if I'm not somehow mistaken. /Andreas —Preceding unsigned comment added by 129.16.53.239 (talkcontribs) 13:45, 5 April 2008 (UTC)

That formula differs from the other formula also discussed above by assuming a fixed density instead of a fixed total mass. With a fixed, finite density and R1 = R2 the mass will be zero because the volume will be zero, and a body with no mass will have no moment of inertia. The formula looks right to me. Hemmingsen 16:17, 5 April 2008 (UTC)
Ok, I see. However, setting that expression equal to $\frac{1}{2} m\left({r_1}^2 + {r_2}^2\right)$ seems incorrect and misleading. I think some clarification is in order as to what assumptions correspond to each expression for I_z. /Andreas —Preceding unsigned comment added by 193.11.234.85 (talk) 19:00, 5 April 2008 (UTC)
That does seem like a good point. The two sets of assumptions are equivalent in all cases except for the one where R1 = R2 and calculating the density as mass/volume would be a division by zero, but that special case does seem to be a case people are interested in. I will attempt to rewrite that part of the list slightly; let me know if you think it is insufficient or have a better idea. Hemmingsen 19:40, 5 April 2008 (UTC)
That seems to me like a good solution! /Andreas

I agree. The equation shows itself incorrect when you realize that as the interior radius approaches the exterior radius, this formula would give you a moment of inertia of 0 while the correct value, m*r^2 comes about when you use the formula with the plus sign.
Dexter411 18:03, 20 December 2006 (UTC)

I think the confusion above from the minus sign could be avoided if a extra column was made for the inertia as a function of the density, not the mass (would make easier reference as well). —Preceding unsigned comment added by 129.230.248.1 (talk) 13:03, 23 July 2008 (UTC)

The formula is unfortunately easily misunderstood and hence miseading in that most people think from the vantage point of making an object of a particular shape out of a particular material. They would like to vary the dimensions and see what happens to the inertia and do not expect the mass to remain constant as if it were being compressed. This confusion would be eliminated if the mass were replaced with the density yielding a moment of inertia of 2πρh(ro-ri). — Preceding unsigned comment added by JeffCalTech (talkcontribs) 13:27, 23 March 2014 (UTC)

Formula for Sheet rotated about one end or through center

I came here looking for this formula, it is in any intro physics text, and is rather important. Someone should add this. I don't have time now. - Anonymous

I checked the foumula by subtracting the moment of inertia of a solid cylinder of r1 from a solid cylinder of r2. I did not get the same value as presented by the foumula 12.4.26.248 (talk) 15:06, 2 November 2009 (UTC) R Friedman

oblate spheroid error

The oblate spheroid entry looks odd ... shouldn't it have different moments of inertia for rotation about the different axes? Also, the 2/3rds suggests a hollow rather than solid object - if this is the intention, the description should specify.

I think the formula for the ellipsoid is correct on both counts, assuming the formula for a solid ball is correct. The formula works for any values a, b, c of semiaxes, assuming that you are rotating around a. If you are actually rotating around b or c, you could do one of two things: change the formula to use the other two letters (other than the one you are rotating about, that is), or you could just change the variables you are using so that a becomes the one you are rotating about. If you let a=b=c, you will see that this is the same formula as for the solid ball, so this is a solid ellipsoid. I will change it to reflect that. (I'm not sure where the 2/3 is that you are talking about, though.) CosineKitty (talk) 15:40, 14 March 2010 (UTC)

Flat disc formula

-I'm not really sure where to put this, so I'll add it here, because this section's pointing out an error as well. I believe that the entry for a flat disc rotated about the x- and y-axes is incorrect. It lists the moment of inertia as $MR^2 / 3$, when it really should be $MR^2 / 4$ —Preceding unsigned comment added by 129.21.69.255 (talk) 04:25, 14 March 2010 (UTC)

[Note: I created a separate section for this issue. I will respond soon.] CosineKitty (talk) 15:31, 14 March 2010 (UTC)
Yes, you were right, and I corrected it. I had to re-derive the formula from scratch to be sure. I am also working on citing some references for this article so people can look up the formulas and confirm them. CosineKitty (talk) 16:22, 14 March 2010 (UTC)
Follow-up: I confirmed several of these formulas from my college physics book (Serway) and cited it in the article. CosineKitty (talk) 18:59, 14 March 2010 (UTC)

MOI of various triangles perhaps?

Would it be possible to source any formulas for the MOI of a triangle object or frame?

I know there is the general vector form for any polygon, but triangles are relatively common objects to come across (at least in text book material) and hence I think it is necessary to have it. I suppose a formula would make sense for a right-angle triangle, equilateral triangle and isosceles triangle.

Anyone know how difficult it may be to obtain the formulas? (My text book doesn't indicate any formulas)

Discussion much appreciated! (Eug.galeotti (talk) 07:02, 28 June 2010 (UTC))

Explanation of product of vector terms

The equation for the calculation of the moment of inertia for a general polygon consists of terms such as P^2; ie the square of a vector. I'm familiar with the dot and cross products of vectors but what does the square of a vector mean? —Preceding unsigned comment added by 81.187.174.42 (talk) 19:12, 4 August 2010 (UTC)

Merge List of moment of inertia tensors

The following discussion is closed. Please do not modify it. Subsequent comments should be made in a new section. A summary of the conclusions reached follows.
The result of the discussion was no consensus.

I suggest to merge List of moment of inertia tensors into the current article; since all tensors in the principal axes have only three non-zero components, and this information is already listed in this article.  // stpasha »  16:37, 17 August 2010 (UTC)

If the resulting article is not too long, I would think this a good suggestion. Baccyak4H (Yak!) 17:44, 17 August 2010 (UTC)
I don't think this is a good suggestions. Tensors are too confusing.Eregli bob (talk) 15:43, 13 October 2010 (UTC)
The article should only be merged if the information about the tensors don't overtake the simple information given in this article. Bande-Ali (talk) 02:02, 1 May 2011 (UTC)

Perhaps they tensors could be added as optional additional info, that way someone who is only looking for the basic formula will not be confused by a scary looking matrix — Preceding unsigned comment added by 147.4.36.65 (talk) 15:54, 30 April 2012 (UTC)

The above discussion is preserved as an archive. Please do not modify it. Subsequent comments should be made in a new section.

Right circular cone inertia incorrect?

The inertia is not around the center of mass but around the apex I believe. This should be stated in the notes. 132.239.186.132 (talk) 00:37, 28 January 2011 (UTC)

Error in cylinder formula

As far as I can tell the formula for a solid cylinder in Ix and Iy has an error. The displayed formula Ix = Iy = (1/12)*m*(3r^2 + 4h^2) should not have a 4*h I think. It should read Ix = Iy = (1/12)*m*(3r^2 + h^2) anyone agree? — Preceding unsigned comment added by 62.189.28.130 (talk) 15:03, 27 July 2011 (UTC)

I found the original error when I was working through the integrals for the moment of inertia. I then verified this with wolfram alpha. The h^2 should be 4h^2. http://www.wolframalpha.com/input/?i=moment+of+inertia+of+a+cylinder. If my math is wrong and wolfram alpha is wrong, then my update is wrong. Otherwise this newer version with 4h^2 is the right one. Hanavi (talk) 07:51, 29 July 2011 (UTC)
Also, someone should check the thick walled cylinder equation. I think it has the same error, but I never did the integral for that one. I am betting that it should also have the 4h^2 Hanavi (talk) 07:57, 29 July 2011 (UTC)
Compare with a rod, and see http://scienceworld.wolfram.com/physics/MomentofInertiaCylinder.html: it looks like it was correct without the 4, and http://www.wolframalpha.com/input/?i=moment+of+inertia+of+a+cylinder is wrong.--Patrick (talk) 08:24, 29 July 2011 (UTC)
Ok, I reworked the integral and found the mistake. I had mistakenly used the limits of integration from 0 to h and it should have been from -h/2 to h/2. The way I had done it would give the moments of inertia around the end of the cylinder (as can be seen by comparing the rod formulae). The original and now current versions are correct. Hanavi (talk) 21:31, 29 July 2011 (UTC)
I would also recomend updating the entry on the solid cylinder to include the moment of inertia about the z axis. MSWurmstein (talk) 00:59, 12 April 2012 (UTC)

Error in spherical shell

Article says I about axis of spherical shell is 3/5 MR^2.

I'm no physics PhD -- just a high school teacher -- but when I do out the integrals I get 2/3 not 3/5.

My text book agrees with me. (Halliday, Resnick, Walker, Fundamentals of Physics 9th ed, p255)

Is this just a typo? and if so -- I'm not a regular wikipedian -- can I just correct this?

Dstras (talk) 00:48, 3 February 2013 (UTC)david strasburger strasburger@nobles.edu

Solids of revolution

I found this paper describing how to get the mass moment of inertia for solid of revolution about the revolving axis as well as the perpendicular axis.

Maybe these formulas need to be included in the list. This theory would be applicable when one is looking for the moment of inertia of tapered rolling element, or a parabolic shape.

iou (talk) 17:06, 16 August 2013 (UTC)

Words Clipped for exported PDF

The styling of the webpage requires a certain width for each of the cells. When sent to the pdf, the size of the margins clips off a number of words from the Comments and Description column. — Preceding unsigned comment added by 132.194.3.169 (talk) 17:23, 9 November 2013 (UTC)

Error in torus description

The part of article about torus below:

Torus of tube radius a, cross-sectional radius b and mass m. About the VERTICAL axis: \left(a^2 + \frac{3}{4}b^2\right)m [4]

has an error: The sited source [4] is correct, but has a different picture with different orientation. Axis that is vertical on the source is HORIZONTAL on the wikipicture, therefore the leftmost description about the axis used "VERTICAL" is wrong. The left most formula is correct for an axis going through the torus, and that is horizontal axis. Some native english speaker should correct that in unambigious way.86.50.116.35 (talk) 15:03, 10 June 2014 (UTC)


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# Find the magnitude of a force

[SOLVED] find the magnitude of a force

## Homework Statement

A window washer pushes his scrub brush up a vertical window at constant speed by applying a force $$\vec{F}$$ as shown in the figure. The brush weighs 12.1 N and the coefficient of kinetic friction is 0.110.

## Homework Equations

1. $$\Sigma$$$$\vec{F}$$ = m$$\vec{a}$$

2. $$\Sigma$$$$\vec{Fx}$$ = max

3. $$\Sigma$$$$\vec{Fy}$$ = may

4. fk = $$\mu$$kn

5. $$\vec{F}$$ = $$\sqrt{Fx^2+Fy^2}$$

## The Attempt at a Solution

$$\Sigma$$Fx = -Fcos53.1 + fk = 0

$$\Sigma$$Fy = N + Fsin53.1 - w = 0

Fcos53.1 = fk

--> Fcos53.1 = $$\mu$$kn

N = w-Fsin53.1

---> Fcos53.1 = $$\mu$$k(w-Fsin53.1)

--> Fcos53.1 = $$\mu$$kw-$$\mu$$kFsin53.1

--> Fcos53.1 + $$\mu$$kFsin53.1 = $$\mu$$kw

--> F(cos53.1 + $$\mu$$ksin53.1) = $$\mu$$kw

--> F = $$\mu$$kw / (cos53.1 + $$\mu$$ksin53.1)

--> N = Fcos53.1/$$\mu$$k

my answer was F = 1.9N, which was wrong and I was so sure I did everything right.

if anybody can offer any help it would be greatly appreciated.

Related Introductory Physics Homework Help News on Phys.org
Tom Mattson
Staff Emeritus
Gold Member
The frictional force points in the y direction opposite the motion of the scrubber head, and the normal force points in the x direction, normal to the wall (that's why it's called the normal force!).

You've got them backwards.

The frictional force points in the y direction opposite the motion of the scrubber head, and the normal force points in the x direction, normal to the wall (that's why it's called the normal force!).

You've got them backwards.
I just realized that now, it seems the equations will turn out to be

mg + un = Fsintheta

with n = Fcostheta

eventually F = mg/(sintheta-ucostheta)


text:
# normal (ordinal) function

###### Definition.

A function $F\colon\operatorname{\mathbf{On}}\to\operatorname{\mathbf{On}}$ is continuous if and only if for each $u\subset\operatorname{\mathbf{On}}$ such that $u\neq\emptyset$ it holds that $F(\sup(u))=\sup\{F(\alpha)|\alpha\in u\}$.

###### Definition.

A function $F\colon\operatorname{\mathbf{On}}\to\operatorname{\mathbf{On}}$ is order preserving if and only if for each $\alpha,\beta\in\operatorname{\mathbf{On}}$ such that $\alpha<\beta$ it follows that $F(\alpha).

###### Definition.

A function $F\colon\operatorname{\mathbf{On}}\to\operatorname{\mathbf{On}}$ is a normal function if and only if $F$ is continuous and order preserving.

Title normal (ordinal) function NormalordinalFunction 2013-03-22 15:33:10 2013-03-22 15:33:10 florisje (7763) florisje (7763) 7 florisje (7763) Definition msc 03E10 continuous (for ordinal functions) order preserving (for ordinal functions) normality normal function


text:
Classes

Change the way you learn with Maqsad's classes. Local examples, engaging animations, and instant video solutions keep you on your toes and make learning fun like never before!

Class 9Class 10First YearSecond Year
##### (iv) If x is no larger than 10 then \ldots \ldots . (a) x \geq 8 (b) x \leq 10 (c) x<10 (d) x>10

Example 3Solve the double inequality -2<\frac{1-2 x}{3}<1 where x \in \mathrm{R} .

Example 2Solve \frac{1}{2} x-\frac{2}{3} \leq x+\frac{1}{3} where x \in \mathrm{R} .

1. Solve the following equations.(x) \frac{2}{3 x+6}=\frac{1}{6}-\frac{1}{2 x+4} x \neq-2

1. Solve the following equations.(vi) \frac{x}{3 x-6}=2-\frac{2 x}{x-2} x \neq 2

3. Answer the following short questions.(iv) Seven times the sum of an integer and 12 is at least 50 and at most 60 . Write and solve the inequality that expresses this relationship.

2. Solve for x (vi) \frac{1}{2}|x+3|+21=9


text:
# How to multiply using big $O$ notation

Suppose I have the following expression:

$$f(n) = \frac{\bigg(1+O(1/n)\bigg)}{\bigg(1+O(1/n)\bigg)\bigg(1+O(1/n)\bigg)}.$$

Is there a way to manipulate $$f$$ (i.e. multiply by terms involving $$n$$) in order to get an expression that looks like $$(1+O(1))$$?

If $$a_n,b_n,c_n\in O(1/n)$$, then \begin{align} \frac{1+a_n}{(1+b_n)(1+c_n)}-1 &=\frac{1+a_n-(1+b_n)(1+c_n)}{(1+b_n)(1+c_n)}\\ &=\frac{a_n-b_n-c_n-b_nc_n}{(1+b_n)(1+c_n)}\\ &\leq\frac{|a_n|+|b_n|+|c_n|+|b_n||c_n|}{|1+b_n||1+c_n|}\\ &=O(1/n) \end{align} hence $$f(n)=\frac{1+O(1/n)}{(1+O(1/n))(1+O(1/n))}=1+O(1/n)$$ consequently, $$f(n)=1+O(1)$$ as well.


text:
# C3.4 Algebraic Geometry (2016-2017)

## Primary tabs

We have updated our Undergraduate exams guidance in preparation for the Trinity Term examinations.

Please see our new webpages dedicated to TT exams.

2016-2017
Lecturer(s):
Prof. Alexander Ritter
General Prerequisites:

Part A Rings and Modules. B3.3 Algebraic Curves useful but not essential.

Course Term:
Michaelmas
Course Lecture Information:

16 lectures.

Course Weight:
1.00 unit(s)
Course Level:
M

### Assessment type:

Course Overview:

Algebraic geometry is the study of algebraic varieties: an algebraic variety is roughly speaking, a locus defined by polynomial equations. One of the advantages of algebraic geometry is that it is purely algebraically defined and applied to any field, including fields of finite characteristic. It is geometry based on algebra rather than calculus, but over the real or complex numbers it provides a rich source of examples and inspiration to other areas of geometry.

Course Synopsis:

Affine algebraic varieties, the Zariski topology, morphisms of affine varieties. Irreducible varieties.

Projective space. Projective varieties, affine cones over projective varieties. The Zariski topology on projective varieties. The projective closure of affine variety. Morphisms of projective varieties. Projective equivalence.

Veronese morphism: definition, examples. Veronese morphisms are isomorphisms onto their image; statement, and proof in simple cases. Subvarieties of Veronese varieties. Segre maps and products of varieties. Categorical products: the image of the Segre map gives the categorical product.

Coordinate rings. Hilbert's Nullstellensatz. Correspondence between affine varieties (and morphisms between them) and finitely generated reduced $k$-algebras (and morphisms between them). Graded rings and homogeneous ideals. Homogeneous coordinate rings.

Categorical quotients of affine varieties by certain group actions. The maximal spectrum.

Primary decomposition of ideals.

Discrete invariants of projective varieties: degree, dimension, Hilbert function. Statement of theorem defining Hilbert polynomial.

Quasi-projective varieties, and morphisms between them. The Zariski topology has a basis of affine open subsets. Rings of regular functions on open subsets and points of quasi-projective varieties. The ring of regular functions on an affine variety is the coordinate ring. Localisation and relationship with rings of regular functions.

Tangent space and smooth points. The singular locus is a closed subvariety. Algebraic re-formulation of the tangent space. Differentiable maps between tangent spaces.

Function fields of irreducible quasi-projective varieties. Rational maps between irreducible varieties, and composition of rational maps. Birational equivalence. Correspondence between dominant rational maps and homomorphisms of function fields. Blow-ups: of affine space at a point, of subvarieties of affine space, and of general quasi-projective varieties along general subvarieties. Statement of Hironaka's Desingularisation Theorem. Every irreducible variety is birational to a hypersurface. Re-formulation of dimension. Smooth points are a dense open subset.


text:
One of the reasons why Bitcoin keeps attracting so much attention is its exceptional "math". Satoshi Nakamoto has managed to create a system that can function with a complete absence of trust between its participants. All interactions are based on strict mathematics, no human factor involved. That was the revolutionary nature of the idea, and not the peer-to-peer networking, as many people may think. Therefore, we’ve decided to devote the first part of the article to the mathematical basics of Bitcoin.

Below we’ll explain the most basic things like elliptic curves, ECC (elliptic-curve cryptography), private/public keys and so on. Whenever possible, we will illustrate with examples of code, mostly in Python 2.7.

## Introduction

Cryptography is the fundamental part of Bitcoin; nothing would work without it. And that's why we will start here.

The so-called elliptic-curve cryptography is used in Bitcoin. It’s based on a special function—an elliptic curve (not to be confused with an ellipse). We’ll describe what this function is like and why it’s so remarkable.

## Elliptic curve

According to Wikipedia, an elliptic curve over field $K$ is a non-singular cubic curve on the projective plane over $\hat{K}$ (the algebraic closure of field $K$) given by a third-degree equation with coefficients from the field $K$ and a "point at infinity."

In lay terms, an elliptic curve may seem to be a rather simple function. As a rule, it can be written in the so-called Weierstrass’s form: $y^2 = x^3 + ax + b$

Depending on the values of parameters $a$ and $b$, the graph of this function may look different:

import numpy as np
import matplotlib.pyplot as plt

def main():
a = -1
b = 1

y, x = np.ogrid[-5:5:100j, -5:5:100j]
plt.contour(x.ravel(), y.ravel(), pow(y, 2) - pow(x, 3) - x * a - b, [0])
plt.grid()
plt.show()

if __name__ == '__main__':
main()

According to Wikipedia, this function was first seen in the works of Diophantus. Later, in the 17th century, Newton became interested in it. His research, that we’re going to discuss, has largely led to formulas for the addition of points on an elliptic curve. From this point on, we shall consider an elliptic curve $\alpha$.

Let there be two points $P, Q \in \alpha$. Their sum is point $R \in \alpha$ that is defined in the simplest case as follows: draw a line through $P$ and $Q$ —it will intersect the $\alpha$ curve  at a single point, let’s call it $R$. By changing the $y$ coordinate of point $-R$ to the opposite in sign, we obtain point $R$ that we will call the sum of $P$ and $Q$, that is, $P + Q = R$.

It’s important to note that we're introducing this addition operation. If you add points in the usual sense, that is, by adding the appropriate coordinates, you will get a completely different point $R'(x_1 + x_2, y_1 + y_2)$, which most likely has nothing to do with $R$ or $-R$ and doesn’t even lie on the $\alpha$ curve.

Careful readers have already asked themselves what will happen if we draw, for example, a straight line through two points with coordinates of form $P(a, b)$ and $Q(a, -b)$. That is, the straight line going through them will be parallel to the y-axis (number three in the picture below).

It is easy to see that in this case there is no third intersection with the $\alpha$ curve we called $-R$. To avoid this event, we introduce the so-called point of infinity usually denoted as $O$ or simply $0$ as in the picture. We will say that if there’s no intersection, $P + Q = O$.

Of particular interest to us is the case when we want to add a point to itself (number two in the picture above, point $Q$). In this case, simply draw a tangent to the $Q$ point and reflect the resulting intersection point with respect to $y$.

Now, with a subtle movement of the hand, we can introduce an operation of multiplying the point by the $\mathbb{N}$ number.  As a result, we obtain a new point $K = G * k$, that is, $K = G + G + ... + G, k$ times. This picture should make it all clear:

## Elliptic curve over a finite field

Absolutely the same curve is used as in the ECC, but it’s considered over some finite field $F_p = \mathbb{Z}/\mathbb{Z}_p = \{0, 1, ..., p - 1\}$, $p$ is a prime number. That is,

$y^2\ mod\ p = x^3 + ax + b\ (mod\ p)$

All the above properties (addition, multiplication, a point at infinity) remain valid for this function. However, if we try to render this function, it will only vaguely (at best) resemble the usual elliptic curve. As for the concept of the "tangent to a function at a point," it loses its meaning, but there’s nothing to worry about. Here is an example of the function $y^2 = x^3 + 7$ for $p = 17$:

As for $p = 59$, it’s almost a random set of points. The only thing that still resembles the origin of this graph is the symmetry regarding the $X$ axis.

P. S. In case you want to know how to compute the coordinates of  point $R(x_3, y_3)$ in the instance of a curve over a finite field knowing coordinates of $P(x_1, y_1)$ and $Q(x_2, y_2)$, take a look at “An Introduction to Bitcoin, Elliptic Curves and the Mathematics of ECDSA” by N. Mistry. Everything is described there in detail; it’s understandable if you know 8th grade level Math.

P.P.S. In case our examples did not satisfy your inquisitive mind, here’s a website for drawing curves of all kinds, so have some fun.

## SECP256k1

Getting back to Bitcoin, the SECP256k1 curve is used in it. It has the form of $y^2 = x^3 + 7$ and is considered over field $F_p$, where $p$ is a huge prime number, namely $2^{256} - 2^{32} - 2^9 - 2^8 - 2^7 - 2^6 - 2^4 - 1$.

The so-called base point, aka generator, is defined for SECP256k1. It’s just a point, usually denoted as $G$, lying on this curve. It’s necessary for creating a public key that will be described below.

Here’s a simple example: using Python, let’s check whether point $G(x, y)$ belongs to the SECP256k1 curve.

>>> p = 115792089237316195423570985008687907853269984665640564039457584007908834671663
>>> x = 55066263022277343669578718895168534326250603453777594175500187360389116729240
>>> y = 32670510020758816978083085130507043184471273380659243275938904335757337482424
>>> (x ** 3 + 7) % p == y**2 % p
True

## Digital signature

A digital signature is a mathematical scheme for demonstrating the authenticity of digital messages or documents. A valid digital signature gives a recipient reason to believe that the message was created by a known sender (authentication), that the sender cannot deny having sent the message (non-repudiation), and that the message was not altered in transit (integrity) — Wikipedia

The general idea is this: Alice wants to transfer 1 BTC to Bob. To do this, she creates a message like:

{
"from" : 1FXySbm7jpJfHEJRjSNPPUqnpRTcSuS8aN, // Alice's address
"to" : 1Eqm3z1yu6D4Y1c1LXKqReqo1gvZNrmfvN, // Bob's address
"amount" : 1 // Send 1 BTC
}

Then Alice takes her private key (for now you can assume that it’s a number known only to Alice), a hash of the message, and a function of form $sign\_text(private\_key, text)$. At the output, she obtains the signature of her message. In the case of ECDSA, it will be a pair of integers, while the signature may look different for other algorithms. After that, she sends out the original message, the signature and her public key to all participants of the network

As a result, every Bob can optionally take this trio, the function of form $validate\_signature(public\_key, signature, text)$ and check whether the owner of the private key has signed this message. And if everyone inside the network knows that the $public\_key$ belongs to Alice, you can understand whether she has sent this money or someone else is trying to do it on her behalf.

Moreover, suppose that there is a person who got in between Alice and the rest of the network. Let’s say he has intercepted Alice's message and changed something in it, literally 1 bit out of a billion. But even in this case the signature verification for validity $validate\_signature(public\_key, signature, text')$ will show that the message has been changed.

This is a very important feature for Bitcoin since the network is distributed. We cannot know in advance who gets our transaction with the requirement to transfer 1000 BTC. But he cannot change it (for example, specify his address as recipient) since the transaction has been signed by your private key and other participants of the network will immediately understand that there is something wrong.

ATTENTION! In reality, the process is quite different from the one described above. We just described in lay terms what a digital signature is and why we need it. The real algorithm is described in the part about Transaction.

## Private key

Private key is quite a general term and different types of private keys can be used in various electronic signature algorithms.

As you may have noticed, Bitcoin uses the ECDSA algorithm. In its case, a private key is a natural 256 bit number, that is, an integer from $1$ to $2^{256}$. Technically, even number 123456 will be a valid private key, but you will soon learn that your coins only "belong" to you until the attacker has your private key, and values like 123456 are very easy to traverse.

It is important to note that it’s impossible to traverse all the keys due to the fact that $2^{256}$ is an enormously large number.

But let’s try to imagine it: according to this article, there are a bit less than $10^{22}$ grains of sand on the Earth. Let's use the fact that $2^{10} \approx 10^3$, that is $10^{22} \approx 10^{80}$ grains of sand. And we have in the total of $2^{256}$ addresses, about $2^{80^3}$.

Hence we can take all the sand on the Earth and turn every grain of sand into a new Earth.  Then, we will turn every grain of sand on each planet in the resulting amount of planets into a new Earth. The total number of grains of sand will still be much less than the number of possible private keys.

For the same reason, most Bitcoin clients simply take 256 random bits when creating a private key—the probability of collision is extremely low. The following are three examples:

## Python

>>> import random
>>> private_key = ''.join(['%x' % random.randrange(16) for x in range(0, 64)])
>>> private_key
'9ceb87fc34ec40408fd8ab3fa81a93f7b4ebd40bba7811ebef7cbc80252a9815'
>>> # or
>>> import os
>>> private_key = os.urandom(32).encode('hex')
>>> private_key
'0a56184c7a383d8bcce0c78e6e7a4b4b161b2f80a126caa48bde823a4625521f'

## Python, ECDSA

>>> import binascii
>>> import ecdsa # sudo pip install ecdsa
>>> private_key = ecdsa.SigningKey.generate(curve=ecdsa.SECP256k1)
>>> binascii.hexlify(private_key.to_string()).decode('ascii').upper()
u'CE47C04A097522D33B4B003B25DD7E8D7945EA52FA8931FD9AA55B315A39DC62'

## Bitcoin-cli

$bitcoin-cli getnewaddress 14RVpC4su4PzSafjCKVWP2YBHv3f6zNf6U$ bitcoin-cli dumpprivkey 14RVpC4su4PzSafjCKVWP2YBHv3f6zNf6U
L3SPdkFWMnFyDGyV3vkCjroGi4zfD59Wsc5CHdB1LirjN6s2vii9

## Public key

Let $k$ be our private key and $G$ is the base point. Then the public key is $K = G * k$. So the public key is actually a point lying on the SECP256k1 curve.

There are two important things here. First, it's easy to see that the operation of obtaining a public key is uniquely defined. That is, there is always only one public key for a particular private key. Second, the reverse operation is computationally difficult. Normally we can obtain a private key from the public one only by a complete traversal of the first one.

Below you will find out that there’s exactly the same connection between the public key and the address, but it’s about the irreversibility of hash functions.

Examples of obtaining a Public Key:

### Python, ECDSA

>>> import binascii
>>> import ecdsa
>>> private_key = ecdsa.SigningKey.generate(curve=ecdsa.SECP256k1)
>>> public_key = private_key.get_verifying_key()
>>> binascii.hexlify(public_key.to_string()).decode('ascii').upper()
u'D5C08F1BFC9C26A5D18FE9254E7923DEBBD34AFB92AC23ABFC6388D2659446C1F04CCDEBB677EAABFED9294663EE79D71B57CA6A6B76BC47E6F8670FE759D746'

### C++, libbitcoin

#include <bitcoin/bitcoin.hpp>
#include <iostream>

int main() {
// Private key
bc::ec_secret secret = bc::decode_hash(
"038109007313a5807b2eccc082c8c3fbb988a973cacf1a7df9ce725c31b14776");
// Get public key
bc::ec_point public_key = bc::secret_to_public_key(secret);
std::cout << "Public key: " << bc::encode_hex(public_key) << std::endl;
}

For compilation and startup, we use (after previously installing libbitcoin):

$g++ -o public_key <filename>$$(pkg-config --cflags --libs libbitcoin)$ ./public_key

You can see that public key formats in the first and second examples are different (at least in length). We'll discuss this in more detail below.

### Base58Check encoding

We will see this encoding throughout the book. That's why we’d better understand how it works and why we need it.

The idea is to write the sequence of bytes as short as possible in a readable format and, at the same time, make the probability of possible misprints even less. You understand that safety cannot be superfluous with Bitcoin. Use one wrong character and money will go to an address, the keys to which will most likely never be found. Here’s a comment about encoding in base58:

/**
* Why base-58 instead of standard base-64 encoding?
* - Don't want 0OIl characters that look the same in some fonts and
*      could be used to create visually identical looking data.
* - A string with non-alphanumeric characters is not as easily accepted as input.
* - E-mail usually won't line-break if there's no punctuation to break at.
* - Double-clicking selects the whole string as one word if it's all alphanumeric.
*/

The easiest way to implement shortness of the record is by using the fairly common Base64 encoding. That is, using the 64 base numeral system, where digits 0,1, ..., 9, and letters a-z and A-Z  are used for the record. This gives 62 characters and the remaining two may be anything, depending on the implementation.

The first difference of Base58Check is that the characters 0, O, l, I have been removed in case someone decides to confuse them. Thus, there are 58 characters, you can check it yourself:

b58 = '123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz'

def base58encode(n):
result = ''
while n > 0:
result = b58[n%58] + result
n /= 58
return result

# print "Base58 encode for '123123':", base58encode(123123)
# # Base58 encode for '123123': dbp

The second difference is the check itself. checksum is added again at the end of the string—the first four bytes of SHA256(SHA256(str)). Well, we also need to add as many ones as the number of leading zeros there used to be before encoding in base58 but that’s easy.

import hashlib

def base58encode(n):
b58 = '123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz'
result = ''
while n > 0:
result = b58[n % 58] + result
n /= 58
return result

# Will be used to decode raw bytes and then encode them to the base58
def base256decode(s):
result = 0
for c in s:
result = result * 256 + ord(c)
return result

count = 0
for c in s:
if c == '\0':
count += 1
else:
break
return count

checksum = hashlib.sha256(hashlib.sha256(s).digest()).digest()[0:4]
result = s + checksum
return '1' * countLeadingZeroes(result) + base58encode(base256decode(result))

### Private key formats

The most obvious way to store a private key is to write 256 bits in the form of a heap of zeros and ones. But any technically literate person understands that it’s going to be much easier to provide the same sequence in the form of 32 bytes where each byte corresponds to two characters in the hexadecimal notation. Reminding you that  numbers 0,1,...,9 and letters A,B,C,D,E,F are used in this case. We’ve used this format in the examples above. It’s sometimes even separated by spaces for clarity.

E9 87 3D 79 C6 D8 7D C0 FB 6A 57 78 63 33 89 F4 45 32 13 30 3D A6 1F 20 BD 67 FC 23 3A A3 32 62

Another more progressive format is WIF (Wallet Import Format). It’s quite simply built:

1. Take a private key, say 0C28FCA386C7A227600B2FE50B7CAE11EC86D3BF1FBE471BE89827E19D72AA1D
2. Write it in Base58Check with the prefix 0x80. That’s all.
private_key = '0a56184c7a383d8bcce0c78e6e7a4b4b161b2f80a126caa48bde823a4625521f'

def privateKeyToWif(key_hex):
return base58CheckEncode(0x80, key_hex.decode('hex'))

# print "Private key in WIF format:", privateKeyToWif(private_key)
# # Private key in WIF format: 5HtqcFguVHA22E3bcjJR2p4HHMEGnEXxVL5hnxmPQvRedSQSuT4

### Public key formats

First recall that a public key is just a point on the SECP256k1 straight line. The first and most common variant of writing it is the uncompressed format, 32 bytes for X and Y coordinates. To avoid confusion, the 0x04 prefix is used for a total of 65 bytes.

import ecdsa

private_key = '0a56184c7a383d8bcce0c78e6e7a4b4b161b2f80a126caa48bde823a4625521f'

def privateKeyToPublicKey(s):
sk = ecdsa.SigningKey.from_string(s.decode('hex'), curve=ecdsa.SECP256k1)
vk = sk.verifying_key
return ('\04' + sk.verifying_key.to_string()).encode('hex')

uncompressed_public_key = privateKeyToPublicKey(private_key)

# print "Uncompressed public key: {}, size: {}".format(uncompressed_public_key, len(uncompressed_public_key) / 2)
# # Uncompressed public key: 045fbbe96332b2fc2bcc1b6a267678785401ee3b75674e061ca3616bbb66777b4f946bdd2a6a8ce419eacc5d05718bd718dc8d90c497cee74f5994681af0a1f842, size: 65

However, as you might have guessed from the name of it, it’s not the best way for storing a public key.

You will be surprised but the second format is called compressed. Its idea is as follows: a public key is a point on a curve, that is, a pair of numbers satisfying the equation $y^2\ mod\ p = x^2 + ax + b\ (mod\ p)$. This means you can write only the X coordinate and we’ll simply solve the equation if we need the Y coordinate. By doing so, we reduce the size of the public key by almost 50%!

There’s just one problem—if a point lies on a curve, there exist two solutions to solve the equation for its X coordinate (look at the graphs above in case you are in doubt). Usually we would simply save the sign for the Y coordinate, but when it comes to a function over a finite field, we should use the following feature: if there are two solutions of the equation for the X coordinate, one of the points will have an even Y coordinate and the second will have an odd one (but then again, you can make sure yourself).

Prefix 0x02 is used in the first case, and 0x03 in the second one. Here is an illustration of the process:

As has already been mentioned, the address is obtained from a public key in a unique way. Moreover, it is impossible to reverse the operation, since cryptographically persistent hash functions are used—RIPEMD160 and SHA256. Here is the algorithm for converting a public key into the address:

1. Take a private key, for example 45b0c38fa54766354cf3409d38b873255dfa9ed3407a542ba48eb9cab9dfca67
2. Obtain the public key from it in the uncompressed format, in our case it is 04162ebcd38c90b56fbdb4b0390695afb471c944a6003cb334bbf030a89c42b584f089012beb4842483692bdff9fcab8676fed42c47bffb081001209079bbcb8db.
3. Calculate RIPEMD160(SHA256(public_key)), and obtain 5879DB1D96FC29B2A6BDC593E67EDD2C5876F64C
4. Convert in Base58Check with prefix 0x0017JdJpDyu3tB5GD3jwZP784W5KbRdfb84X. This is the address.
def pubKeyToAddr(s):
ripemd160 = hashlib.new('ripemd160')
ripemd160.update(hashlib.sha256(s.decode('hex')).digest())
return base58CheckEncode(0, ripemd160.digest())


text:
# Magnetic orders proximal to the Kitaev limit in frustrated triangular systems: Application to Ba 3 IrTi 2 O 9

@article{Catuneanu2015MagneticOP,
title={Magnetic orders proximal to the Kitaev limit in frustrated triangular systems: Application to Ba 3 IrTi 2 O 9},
author={Andrei Catuneanu and Jeffrey G. Rau and Heung-Sik Kim and Hae Young Kee},
journal={Physical Review B},
year={2015},
volume={92},
pages={165108}
}
• Published 22 July 2015
• Materials Science
• Physical Review B
Frustrated transition-metal compounds in which spin-orbit coupling (SOC) and electron correlation work together have attracted much attention recently. In the case of $5d$ transition metals, where SOC is large, ${j}_{\text{eff}}=1/2$ bands near the Fermi level are thought to encompass the essential physics of the material, potentially leading to a concrete realization of exotic magnetic phases such as the Kitaev spin liquid. Here, we derive a spin model on a triangular lattice based on ${j}_… 31 Citations ## Figures from this paper Kitaev materials • Materials Science, Physics Physics Reports • 2022 Comprehensive study of the global phase diagram of the J−K−Γ model on a triangular lattice • Materials Science, Mathematics • 2021 The celebrated Kitaev honeycomb model provides an analytically tractable example with an exact quantum spin liquid ground state. While in real materials, other types of interactions besides the Kitaev-like honeycomb magnets: Global phase behavior and emergent effective models • Physics Physical Review B • 2019 Compounds of transition metal ions with strong spin-orbit coupling recently attracted attention due to the possibility to host frustrated bond-dependent anisotropic magnetic interactions. In general, Structural, electronic, and magnetic properties of nearly ideal Jeff=12 iridium halides • Physics, Materials Science Physical Review Materials • 2020 Heavy transition metal magnets with$J_{\rm eff}=$1/2 electronic ground states have attracted recent interest due to their penchant for hosting new classes of quantum spin liquids and Z2 vortices in the ground states of classical Kitaev-Heisenberg models • Physics • 2019 The classical nearest neighbor Kitaev-Heisenberg model on the triangular lattice is known to host$\mathbb{Z}_2$spin-vortices forming a crystalline superstructure in the ground state. The Persistent low-temperature spin dynamics in the mixed-valence iridate Ba 3 InIr 2 O 9 • Physics • 2017 Using thermodynamic measurements, neutron diffraction, nuclear magnetic resonance, and muon spin relaxation, we establish putative quantum spin-liquid behavior in Ba$_3$InIr$_2$O$_9$, where unpaired Unconventional magnetism in the 4d4 -based S=1 honeycomb system Ag3LiRu2O6 • Physics Physical Review B • 2019 We have investigated the thermodynamic and local magnetic properties of the Mott insulating system Ag$_{3}$LiRu$_{2}$O$_{6}$containing Ru$^{4+}$(4$d^{4}$) for novel magnetism. The material PhD Thesis: Searching for an emergent SU(4) symmetry in real materials The enhancement of the spin-space symmetry from the usual$\mathrm{SU}(2)$to$\mathrm{SU}(N)$with$N>2\$ is promising for finding nontrivial quantum spin liquids, but the realization of
Magnon damping in the zigzag phase of the Kitaev-Heisenberg- Γ model on a honeycomb lattice
• Physics
Physical Review B
• 2020
Author(s): Smit, RL; Keupert, S; Tsyplyatyev, O; Maksimov, PA; Chernyshev, AL; Kopietz, P | Abstract: We calculate magnon dispersions and damping in the Kitaev-Heisenberg model with an off-diagonal
Quantum-classical crossover in the spin- 12 Heisenberg-Kitaev kagome magnet
• Physics
• 2020
The spin-1/2 Heisenberg kagome antiferromagnet is one of the paradigmatic playgrounds for frustrated quantum magnetism, with an extensive number of competing resonating valence bond (RVB) states


text:
Lemma 10.60.14. Let $(R, \mathfrak m)$ be a Noetherian local ring. Suppose $x_1, \ldots , x_ d \in \mathfrak m$ generate an ideal of definition and $d = \dim (R)$. Then $\dim (R/(x_1, \ldots , x_ i)) = d - i$ for all $i = 1, \ldots , d$.

Proof. Follows either from the proof of Proposition 10.60.9, or by using induction on $d$ and Lemma 10.60.13. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
Kinematics HMWK help

1. Sep 18, 2006

Kinematics HMWK help!!!

hi, im in gr 12 physics and i need help in a kinematics question about Projectile motion... here is the question:

"a medievil prince is trapped in a castle wraps a message around a rock and throws it from the top of the castlewall with an initial velocity of 12 m/s [42 degrees above the horizontal]. The rock lands just on the far side of the casltes moat, at a level of 9.5 m below the initial level. Determine the (a) time of flight (b) width of moat and (c) velocity at impact"

i really cant figure this out and i need help ASAP

2. Sep 18, 2006

mbrmbrg

Hey, welcome aboard!
What work have you done on this problem?
What do you know? What are you looking for? Any ideas of what equation(s) you'll need to use?

3. Sep 18, 2006

ok well what i have so far is my vertical and horizontal velocities, which are 8.0 m/s and 8.9 m/s respectively....and i dunno where to go from here... i tried using d=v1t + 1/2 at*2 but when i put it in the quadratic formula i cant get it to work so i dunno where to go from here

4. Sep 18, 2006

mbrmbrg

OK, so it sounds like you're solving for x and y components separately, which is very good.

You know more than that: assign a value for the initial x position (I'd use 0m) and y position (which you could set at zero, but since the rock falls below the throw line, you'd have a lot of negatives flying around. I don't like negatives, so I'd set the initial y position as the height of the wall--which is given). You also know that there is no acceleration in the x direction, and the acceleration in the y direction is acceleration due to gravity.
I suggest starting by solving for time, using an equation involving y-components only.

5. Sep 18, 2006

ok ... now i have figured out the time ... it is 2.4 s ... how would i figure out the length of the moat? would i have to use the x components now? and if i do how would i do that?

6. Sep 18, 2006

mbrmbrg

yes, yes!

I'd use $x=x_{0}+v_{0x}t+\frac{1}{2}at^2$

Remember that x acceleration = 0.

7. Sep 18, 2006

ok i got the length of the moat (22 m) !!!! thanx for ur help... but i have one more question that i hope u can answer... how would i figure out the velocity at impact?

8. Sep 18, 2006

nvm... i got it ... i just used $V_2^2=V_1^2+2ad$... and i got the answer :) thnx again for ur help

9. Dec 6, 2006

pack60

hey im new here and need some help with a few basic questions

10. Dec 6, 2006

pack60

is anyone online???

11. Dec 6, 2006

pack60

can anyone in here help me with some questions

12. Dec 6, 2006

mbrmbrg

it might help if you posted your question...


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Math Help - integration by substitution

1. integration by substitution

$\int\frac{1}{e^x+1}dx ($let $u=e^x+1)$

$\int\frac{1}{e^x+1}dx=\int\frac{1}{u}\frac{1}{u-1}du$

$=\int\frac{1}{u^2-u}du$

$=\int\frac{1}{(u-\frac{1}{2})^2}-\frac{1}{4}du$

$=ln(\frac{u-1}{u})+c$

$=ln(\frac{e^x}{e^x+1})+c$

2. Re: integration by substitution

You can wright the numerator as:
$\int \frac{1}{e^x+1}dx= \int \frac{(e^x+1)-(e^x)}{e^x+1}dx=\int \frac{e^x+1}{e^x+1}dx - \int \frac{e^xdx}{e^x+1}$
Now use the substitution $t=e^x+1$ for the second integral.

3. Re: integration by substitution

Originally Posted by Punch
$\int\frac{1}{e^x+1}dx ($let $u=e^x+1)$

$\int\frac{1}{e^x+1}dx=\int\frac{1}{u}\frac{1}{u-1}du$

$=\int\frac{1}{u^2-u}du$

$=\int\frac{1}{(u-\frac{1}{2})^2}-\frac{1}{4}du$

$=ln(\frac{u-1}{u})+c$

$=ln(\frac{e^x}{e^x+1})+c$
This is the same mistake you made in the last question. If there is a u^2 term in the denominator, the integral is NOT a logarithm unless there is a u term in the numerator.

I would start by multiplying top and bottom by e^x...

4. Re: integration by substitution

Here's another way

$\int \dfrac{1}{e^x+1}dx = \int \dfrac{e^{-x}}{1 + e^{-x}}dx$

and let $u = 1 + e^{-x}$


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# lambda

lambda — An anonymous function.

a = lambda x,y : x+y
print(a(5, 6))
# Python code to illustrate cube of a number
# showing difference between def() and lambda().
def cube(y):
return y*y*y;

g = lambda x: x*x*x
print(g(7))

print(cube(5))
• inline function.
• Can have any number of arguments but only one expression, which is evaluated and returned.
• syntactically restricted to a single expression.
• Semantically, they are just syntactic sugar for a normal function definition.

lambda

## lambda — Structure map

Clickable & Draggable!


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# Maxwell's equations from continuum limit

In appendix A.6 of Schroeder's Thermal Physics, he mentions (in regards to classical fields):

The usual approach is to first pretend that the continuous object is really a bunch of point particles connected together by little springs, and eventually take the limit where the number of particles goes to infinity and the space between them goes to zero. The result is generally some kind of partial differential equation (for instance, the linear wave equation or Maxwell's equations) that governs the motion as a function of place and time.

I take the part in parentheses to mean that there's some way to derive Maxwell's equations from the continuum limit of some sort of system of springs. I'm familiar with deriving the wave equation from a system of springs, but have never heard of obtaining Maxwell's equations and can't seem to find anything that mention it other than this somewhat vague reference. I'm not even sure what the "springs" would be for this case unless we're talking about oscillations of the electromagnetic field. in which case the derivation would seem kind of circular to me. Is this some remnant of aether theory or something? An explanation of the procedure would be appreciated.

• I am not completely sure but here it goes: While not exactly springs, I think the author might be alluding to the derivation of macroscopic Maxwell equations from microscopic ones, by, for instance, considering a spacial averaging over bounded and free charges/currents, which will, at first approximation, include vibrational effects of atoms/molecules around their center of mass. Take a look of section 6.6 in Jackson's book, Classical Electrodynamics, 3ed, and the references suggested in there. – secavara Mar 9 '18 at 1:37
• Apparently, this is how Maxwell derived his equations. – flippiefanus Mar 9 '18 at 4:22
• It is interesting to note that quatum mechanical averaging is quite different (the opposite) from the "classical" one. – Vladimir Kalitvianski Sep 30 '19 at 15:15

I think you're interpreting him a little too literally. His meaning can be taken more literally if you use his first example, a wave equation for sound waves in a solid. I don't think he literally means that you can get Maxwell's equations by coupling together massive particles with springs -- when you do that, you get sound waves, not light waves. There are fundamental reasons why you can't hope to get a description of the electromagnetic field by literally using this method:

(1) The speed at which wave disturbances propagate will be less than $$c$$. (This holds if the matter that the springs and masses are made of are any reasonable form of matter -- specifically, one that obeys the dominant energy condition.)

(2) You're going to have a hard time getting rid of the longitudinal vibrations and getting the transverse ones. The prohibition on longitudinal vibrations really only makes sense relativistically, for a wave propagating at $$c$$.

(3) You will essentially be constructing an aether theory, which will not be consistent with Maxwell's equations (or at least, not with how we interpret them nowadays).

The discrete degrees of freedom that he has in mind for the EM field are probably more like the normal modes of a cavity, or something of a similar nature in QED.

The derivation of wave equation that you alluded in your question can be adapted to derive Maxwell equations. Perhaps, the best exposition of this technique is given by J.D. Jackson in section 12.7 of his 'Classical Electrodynamics' (3rd edition). It is so lucid that I reproduce it as is below.

The Lagrangian approach to continuous fields closely parallels the techniques used for discrete point particles. The finite number of coordinates $$q_i(t)$$ and $$\dot{q}_i(t)$$, $$i = 1, 2, \ldots n$$, are replaced by an infinite number of degrees of freedom. Each point in space time $$x^\alpha$$ corresponds to a finite number of values of the discrete index $$i$$. The generalized coordinate $$q_i$$ is replaced by a continuous field $$\phi_k(x)$$, with a discrete index ($$k = 1, 2, \ldots, n$$) and a continuous index $$x^\alpha$$. The generalized velocity $$\dot{q}_i$$ is replaced by the $$4$$-vector gradient $$\partial^\beta\phi_k$$. The Euler-Lagrange equations follow from the stationary property of the action integral with respect to variations $$\delta\phi_k$$ and $$\delta(\partial^\beta\phi_k)$$ around the physical values. We thus have the following correspondences: $$i \rightarrow x^\alpha, k$$, $$q_i \rightarrow \phi_k(x)$$, $$\dot{q}_i \rightarrow \partial^\alpha\phi_k(x)$$,
$$$$L = \sum_i L_i(q_i, \dot{q}_i) \rightarrow \int\mathcal{L}(\phi_k, \partial^\alpha\phi_k)d^x$$$$ and $$$$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}_i}\right) = \frac{\partial L}{\partial q_i} \rightarrow \partial^\beta\frac{\partial\mathcal{L}}{\partial(\partial^\beta\phi_k)} = \frac{\partial\mathcal{L}}{\partial\phi_k},$$$$ where $$\mathcal{L}$$ is a Lagrangian density, corresponding to a definite point in space-time and equivalent to the individual terms in a discrete particle Lagrangian. For the electromagnetic field, the "coordinates" and "velocities" are $$A^\alpha$$ and $$\partial^\beta A^\alpha$$.

I think this paragraph beautifully summarizes how one can go from masses-with-springs to field-equations and how to use the latter to get a Lagrangian description of the electromagnetic fields.

Jackson further writes

In analogy with the situation with discrete particles, we expect the free-field Lagrangian at least to be quadratic in the velocities, that is, $$\partial^\beta A^\alpha$$ or $$F^{\alpha\beta}$$, the electromagnetic field tensor.

Lastly, he also argues for a term $$J_\alpha A^\alpha$$ to account for the interaction of charged particles with electromagnetic fields. You may want to go over the rest of the section for his derivation of the inhomogeneous Maxwell equations. (The homogeneous ones follow from the definition of $$F^{\alpha\beta}$$.)

Before closing let me add that I could not find a mass-with-springs analogy in volume 1 of Maxwell's treatise on electricity and magnetism. In fact, his derivation (article 76) of Gauss law is similar to what we find in modern textbooks, except that he calls $$R$$ the magnitude of the electric field.

It's actually not springs but rather writing a Lagrangian and then deriving it. If I understand properly it just refers to the ability to write a Lagrangian density $$\mathcal{L} = \mathcal{L}_{\mathrm{field}} + \mathcal{L}_{\mathrm{int}} = - \frac{1}{4 \mu_0} F^{\alpha \beta} F_{\alpha \beta} - A_{\alpha} J^{\alpha}$$ for classical electrodynamics. Euler-Lagrange equations for the above equations will give the Maxwell's equations. The fact that one can consider the components of the electromagnetic potential $$A^{\lambda}$$ as field quantities which are a function of space and time. Just like in the case of a system of springs as one approaches the "limit where the number of particles goes to infinity and the space between them goes to zero". Under such an approximation one can replace the displacement of the individual particles $$\delta _i$$ by a function of space$$\delta (x)$$.


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# Proving or disproving expression with implies operators

I'm having a hard time reducing expressions involving "implies" operators. I did some reading about the actual meaning of the "implies" operator and browse for other Q&A on this website; however, I don't know how to interpret expressions. For example, I have:

$$(C1) A \implies B$$ $$(C2) C \implies B$$

And I as a definition given in the context I know that: $$C \implies A$$

Now, I'm being asked to prove or disprove: $$a) C1 \implies C2$$ and $$b) C2 \implies C1$$

Here is where my problem/question, for a) is it valid to do the following?

$$(A \implies B) \implies (C \implies B)$$

and then (somehow):

$$A \implies C$$

Based on my definition given a above, I'm thinking I could "disprove" a) and "prove" b) but I'm not sure if logic has any foundation.

Any help would be appreciated

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Are you allowed to use truth tables, or do you actually have to derive "$C_1 \implies C_2$"? It seems like the only way you could disprove such a statement is to find a case where $C_1$ is true but $C_2$ is false. –  you Apr 8 '12 at 23:13
I think I'm allowed to; however, I don't see a straight forward way of doing for my particular scenario. I'll keep it in mind, thanks! –  Ulises Apr 9 '12 at 18:32

You know that $C\implies A$. If you also knew $C1$, you’d have $$C\implies A\quad\text{ and }\quad A\implies B\;,$$ from which you may certainly infer $C\implies B$, which is $C2$. Thus, given that $C\implies A$, you must have that $C1\implies C2$. We’ve just proved (a).

As for (b), does knowing that $C\implies A$ and $C\implies B$ guarantee that $A\implies B$? What if $A,B$, and $C$ are:

$\qquad C:\quad n$ is a multiple of $6$,

$\qquad A:\quad n$ is a multiple of $2$, and

$\qquad B:\quad n$ is a multiple of $3$?

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Does this mean that the "implies" operator is transitive? in other words if we have have: $$C \implies A and A \implies B$$ Can we infer the following? $$C \implies B$$ –  Ulises Apr 9 '12 at 14:19
yes, implication is transitive. –  Ronald Apr 9 '12 at 16:39
Thank you guys, I guess this means $$(A \implies B) \implies (C \implies B)$$ cannot be reduced to: $$A \implies C$$ –  Ulises Apr 9 '12 at 18:29