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# Why are more massive white dwarfs smaller?

I understand that a white dwarf is supported by the Pauli Exclusion Principle and that the larger the gravitational force against them, the closer the electrons must pack.

But I have two queries: One is that for more massive white dwarf stars, there would be more particles present in the star's core, which should take up more space. Is there any way in seeing why this isn't a significant effect?

Also, we haven't studied quantum mechanics and the Pauli Exclusion Principle in detail so it seems weird to me that electrons can just pack closer. This is my understanding, the Pauli Exclusion Principle means no two electrons with the same spin can share the same quantum state. For this effect to produce a force, wouldn't it mean that the electrons were already the closest they could be, as any closer they would be in the same quantum state. How does the Pauli Exclusion Principle work to counteract gravity here?

As for why a more massive white dwarf becomes smaller, this is just a consequence of the particular equation of state. If the pressure is proportional to density to the power of $$\alpha$$ with $$4/3 <\alpha < 2$$, then the equilibrium radius is a decreasing function of mass. Essentially, the requirement of an increased pressure gradient to support the increased weight can only be satisfied by a decrease in radius, because the central pressure does not increase enough with density. As a consequence of being supported by electron degeneracy pressure, $$\alpha$$ varies between about 5/3 in a low-mass, low density white dwarf to nearly 4/3 in a more massive, high density white dwarf.


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## Real dice.

Suppose you want to draw randomly a number between 0 and 1, with multiple throws of a six sided dice, what would you write down on its face?

If you want to form numbers between 0 (inclusive) and 1 (exclusive), you could assign the faces a subset of the usual digits (0 to 9), and throw the dice multiple time, each time giving you a new decimal. So throwing 3, 4, 7, 8, (say) would give you 0.3478. But how do we chose what to write on the faces?

Intuitively, we would think that spacing the digits as evenly as possible is the way to go, something like 1, 3, 5, 7, 8, 9. But is our intuition correct? There are $\displaystyle\binom{10}{6}=210$ subsets of six digits, so there’s not that many to check, so we can use brute force enumeration. But how do we measure the evenness of the spread of generated numbers? We simply measure the (square of the) distance between all possible numbers generated.

The test therefore relies on three steps. 1) generate all possible subsets, 2) generate all possible numbers we can represent with the digits from each of the subsets, and 3) measure distance.

The first step, generate all possible subsets, can be done using the “shuttle algorithm” that takes a permutation of (not necessarily distinct symbols) and yield the next one in lexicographical order. We’ve already met this algorithm in a previous entry. We start with the initial permutation 0,0,0,0,1,1,1,1,1,1. The next will be 0,0,0,1,0,1,1,1,1,1. Eventually, we reach the last one, 1,1,1,1,1,1,0,0,0,0. We use the permutation as a bit-vector to select the subset of digits: 0,1,1,0,0,0,1,1,1,1 selects 1,2,6,7,8,9 from 0,1,2,…,9.

The second step is done with a base six counter, that will count, on a fix length, from 0 to n-digit overflow. This counters goes 0, 1, 2, 4, 5, 10, 11, 12, 13, 15, 25,… A presentation layer converts the number represented by the counter to the actual symbols from the subset. That re-coded number is converted to a float (or an integer).

The third, and last, step, just measures the spacing between successive numbers. This means that a number that goes from xxxx15 to xxxx20 will incur a larger penalty than a number that goes from yyyy18 to yyyy20.

*
* *

Running the program will let us discover the best subset of digits to write on the dice’s faces:

```024689	425723808
135789	520243666
146789	733009458
246789	827529316
256789	1040295108
345678	1339694288
356789	1347580758
456789	2080397992
```

So for this almost kind-of ten-sided dice on a six-side dice, the best combination is 0, 2, 4, 6, 8, 9. It’s not the only one. The program only selects a new combination if it has a strictly better score than the previous one. But we also find 0,1,2,3,5,7,9, 0,2,3,5,7,9, 0,2,4,5,7,9, and 0,2,4,6,7,9 with the same discrepancy, or, alternatively, with the same evenness.

format: plain


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# Reproducing kernel Hilbert sapce

I encountered the following claim (verbatim):

Theorem Let $V$ be a subspace of $L^2(\mathbb{R})$ and $\{e_n\}$ be a orthonormal basis of $V$. The $V$ is a reproducing kernel Hilbert space with kernel

$$K(x,y) = \sum_n e_n(x)e_n(y).$$

For any function $f \in V$,

$$f(y) = \int f(x) K(x,y) dy.$$

Questions:

• I am under the impression that a RKHS is a Hilbert space of functions, where pointwise evaluation is a bounded functional. $L^2$ is not even a space of functions, how does the claim make sense (e.g. take an ONB for the entire $L^2$)?
• Also, are there no convergence requirements on the reproducing kernel $K$, or is it purely formal?
-


text:
Transverse, Longitudinal, and Surface Waves

Note by , created almost 6 years ago

Contains how waves move and wave formulas.

1887
7
0
 Created by Selam H almost 6 years ago
Physical Science 2nd Quarter Assessment
Properties & States of Matter Vocab.
Classification of Matter
Life of a Star
#2 Information security and data protection
Energy, Mass, & Conversions
Rocks and Minerals Quiz
Kinetic & Potential Energy
Plate Tectonics

Page 1

↑F

G↓

A = Wavelength- distance from any part of the wave to the same point on the next cycle of the waveB = Frequency (also the Direction of Wave)- the number of wave cycles per unit (unit=hertz)C = Period- time it takes for one wave cycle to completeD = Amplitude- maximum height to the rest positionE = Direction of EnergyF = Trough- the lowest point of a waveG = Crest- the highest point of a waveH = Equilibrium (also the Rest Point)- part where the energy is the same throughout the wave

← H (the line)

Movement of a Transverse Wave:

Movement of Longitudinal Waves:

A

B

E

D

||||||C

Movement of Surface Waves:

A = Rarefaction- the part of a wave that stretches out the most B = Compression- the part of a wave that stretches out the most C = Rest Position- area of rest in a waveD = Direction of EnergyE = Direction of Wave

A= Crest- highest point of surface waveB= Wavelength- distance from one point of a surface wave to another which is the same part but of a the next cycleC= Rest PositionD= Trough- the lowest point of a surface waveE= Wave Height- distance from crest to trough

A

B

C                  D

E

WAVE FORMULAS

Speed= $Wavelength·Frequency$

Speed= $\frac{Wavelength}{Period}$

Frequency=$\frac{Speed}{Wavelength}$

Wavelength= $Speed·Frequency$

$\lambda=Wavelength$

$v=Velocity$(Speed)

$T=Period$

Transverse Waves

Longitudinal Waves

Surface Waves

Formulas


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# Calculate: y=(2)/((12x^4-7x^3)^5)

## Expression: $y=\frac{ 2 }{ {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} }$

Take the derivative of both sides

$y '=\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \frac{ 2 }{ {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} } \right)$

Use differentiation rule $\frac{ \mathrm{d} }{ \mathrm{d}x} \left( \frac{ a }{ f } \right)=-a \times \frac{ \frac{ \mathrm{d} }{ \mathrm{d}x} \left( f \right) }{ {f}^{2} }$

$y '=-2 \times \frac{ \frac{ \mathrm{d} }{ \mathrm{d}x} \left( {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} \right) }{ {\left( {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} \right)}^{2} }$

Use the chain rule $\frac{ \mathrm{d} }{ \mathrm{d}x} \left( f\left( g \right) \right)=\frac{ \mathrm{d} }{ \mathrm{d}g} \left( f\left( g \right) \right) \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( g \right)$, where $g=12{x}^{4}-7{x}^{3}$, to find the derivative

$y '=-2 \times \frac{ \frac{ \mathrm{d} }{ \mathrm{d}g} \left( {g}^{5} \right) \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( 12{x}^{4}-7{x}^{3} \right) }{ {\left( {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} \right)}^{2} }$

Find the derivative

$y '=-2 \times \frac{ 5{g}^{4} \times \frac{ \mathrm{d} }{ \mathrm{d}x} \left( 12{x}^{4}-7{x}^{3} \right) }{ {\left( {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} \right)}^{2} }$

Find the derivative of the sum or difference

$y '=-2 \times \frac{ 5{g}^{4} \times \left( 12 \times 4{x}^{3}-7 \times 3{x}^{2} \right) }{ {\left( {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} \right)}^{2} }$

Substitute back $g=12{x}^{4}-7{x}^{3}$

$y '=-2 \times \frac{ 5{\left( 12{x}^{4}-7{x}^{3} \right)}^{4} \times \left( 12 \times 4{x}^{3}-7 \times 3{x}^{2} \right) }{ {\left( {\left( 12{x}^{4}-7{x}^{3} \right)}^{5} \right)}^{2} }$

Simplify the expression

$y '=\frac{ 210{x}^{2}-480{x}^{3} }{ {\left( 12{x}^{4}-7{x}^{3} \right)}^{6} }$

Random Posts
Random Articles


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# Relativity question

1. Jan 20, 2006

### Tony11235

Melissa is on a flatcar moving east at 0.6c relative to Bob. She holds a flashbulb in each hand and causes them to flash simultaneously. Melissa's hands are 2.0 m apart and her arms are oriented in an east-west direction. According to Bob, which flashbulb flashes first and by how much?

For some reason I cannot get myself started on this problem. Any hints or help?

2. Jan 20, 2006

### Gamma

I have forgotten most of relativity that I learnt. But it is tempting to give it a try. So if I am wrong I hope some one will correct me here.

The time difference between the two light pulse must be caused by the separation of the light sources i.e 2m. However, Bob sees this length contracted due to length contraction

So for Bob, separation between the light sources = L' = L/gamma = 2/gamma

L' = 1.6 m ( gamma = 1.25)

So the time difference between the pulses = L' / c = 5.3 ns.

3. Jan 20, 2006

### Tom Mattson

Staff Emeritus
Gamma: Please don't give offer complete solutions to homework problems. We want students to do most of the work, and we help.

That said, I don't agree with your solution. In order for the 2.0m to be considered a length in Bob's frame he would have to measure the distance between Melissa's hands simultaneously. But if he uses the flashes to measure the distance he does not make the measurements simultaneously, because in his frame the flashes are not simultaneous.

Tony: Let Bob's frame be $S$ and let Melissa's frame be $S^{\prime}$. Let Event 1 be "East light flashes." and Event 2 be "West light flashes." Then begin as follows.

1.) Write down spacetime coordinates for Events 1 and 2 in $S^{\prime}$.
2.) Compute spacetme intervals from the coordinates found in 1.).
3.) Use the Lorentz transformation to find the time between the flashes in $S$.

4. Jan 20, 2006


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During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

# Difference between revisions of "1993 AJHSME Problems/Problem 25"

## Problem

A checkerboard consists of one-inch squares. A square card, $1.5$ inches on a side, is placed on the board so that it covers part or all of the area of each of $n$ squares. The maximum possible value of $n$ is

$\text{(A)}\ 4\text{ or }5 \qquad \text{(B)}\ 6\text{ or }7\qquad \text{(C)}\ 8\text{ or }9 \qquad \text{(D)}\ 10\text{ or }11 \qquad \text{(E)}\ 12\text{ or more}$

## Solution

Using Pythagorean Theorem, the diagonal of the square $\sqrt{(1.5)^2+(1.5)^2}=\sqrt{4.5}>2$. Because this is longer than $2$( length of the sides of two adjacent squares), the card can be placed like so, covering $12$ squares. $\rightarrow \boxed{\text{(E)}\ 12\ \text{or more}}$.

$[asy] for (int a = -2; a <= 2; ++a) { draw((-2,a)--(2,a)); draw((a,-2)--(a,2)); } pair A,B,C,D; A=(0,sqrt(2.25)); B=(sqrt(2.25),0); C=(0,-sqrt(2.25)); D=(-sqrt(2.25),0); draw(A--B--C--D--cycle); fill(A--B--C--D--cycle,lightgray); [/asy]$

 1993 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast Problem 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions


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## What you need to know

Things to remember:

• We can use estimation to simplify fractions.
• Try each of our four operations, then compare the answers.
• Try all four operations might not be necessary.

Find the missing operation that makes the following true:

$$5 \_\_\_ 2\frac{3}{4} = 7.75$$

This might be obvious, but let’s try all four operations to see what we get.

Addition: $5 + 2\dfrac{3}{4} = 7.75$

Subtraction: $5 - 2\dfrac{3}{4} =2.25$

Multiplication: $5 \times 2\dfrac{3}{4} = 13.75$

Division: $5 \div 2\dfrac{3}{4} =1.\overline{81}$

So, we can see that our missing operation here must have been a +.

We don’t always have to try every operation though, if we remember some key points:

Addition:        When adding a positive number to another number, it makes it bigger.

Subtraction:   When subtracting a positive number from another number, it makes it smaller.

Multiplication: When multiplying by a number bigger than 1, it makes it bigger.

When multiplying by a number smaller than 1, it makes it smaller.

Division: When dividing by a number bigger than 1, it makes it smaller.

When dividing by a number smaller than 1, it makes it bigger.

The numbers and fraction were quite nice, but how about if they are a bit tougher?

Find the missing operation that makes the following true:

$$\dfrac{13}{48} \_\_\_ 3 = \frac{13}{16}$$

Neither of these fractions are particularly nice, but we can estimate find estimate for them to make them a little nicer. We can estimate fractions by changing the numerator and denominator into numbers that let us simplify them down.

$$\dfrac{13}{48}\rightarrow\frac{12}{48}\rightarrow\frac{1}{4}$$

$$\dfrac{13}{16}\rightarrow\frac{12}{16}\rightarrow\frac{3}{4}$$

Now that we have our estimates, let’s put them in.

$$\dfrac{1}{4} \_\_\_ 3 = \frac{3}{4}$$

Our estimate is now asking “How do we use 3 to turn $\dfrac{1}{4}$ into $\dfrac{3}{4}$?” Well, we multiply of course!

$$\dfrac{1}{4} \times 3 = \frac{3}{4}$$

So, our missing operation was a multiplication!

$$\dfrac{13}{48} \times 3= \frac{13}{16}$$

## Example Questions

$$\dfrac{21}{100}\rightarrow\frac{20}{100}\rightarrow\dfrac{1}{5}$$

$$\dfrac{21}{50}\rightarrow\frac{20}{50}\rightarrow\dfrac{2}{5}$$

Our estimate is now asking “How do we use 2 to turn $\dfrac{1}{5}$ into $\dfrac{2}{5}$?” Well, we multiply, so this must be our missing operation!

$$\dfrac{21}{100} \times 2 = \dfrac{21}{50}$$

$$\dfrac{9}{40}\rightarrow\dfrac{10}{40}\rightarrow\dfrac{1}{4}$$

$$\dfrac{9}{10}\rightarrow\dfrac{10}{10}\rightarrow 1$$

Our estimate is now asking “How do we use 4 to turn $\dfrac{1}{4}$ into 1 whole?” Well, we multiply, so this must be our missing operation!

$$\dfrac{9}{40} \times 4 = \dfrac{9}{10}$$


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## LATEX-L@LISTSERV.UNI-HEIDELBERG.DE

 Options: Use Classic View Use Proportional Font Show Text Part by Default Show All Mail Headers Topic: [<< First] [< Prev] [Next >] [Last >>]

 Re: ideal future document processing "Michel Lavaud Orleans (France)" <[log in to unmask]> Wed, 25 Jun 1997 18:08:20 +0000 text/plain (41 lines) > > The structure of the document in its first stages is indicated > > with punctuation marks, and typesetting indications such as > > \smallskip, \hspace{2cm}, \bf or whatever extend the set of > > punctuation marks available, and provide a convenient visual way > > to refine the structure. This mimicks exactly what one does when > > writing an article by hand. > um, speak for yourself. I have tried to train myself *not* to write > articles simply as a mimic of how i used to write by hand. on the > contrary, when i write by hand these days i put in generic markup' You have cut this citation from its context. I did not say that I never use structured commands and use only \hspace{2cm}. I said that in the early stages of _creating_ a scientific article, it is totally unproductive to try to separate structure from formatting, because the structure is largely undefined: when one does research work, one does not know the final results _by definition_ so one knows still much less the final structure of the document. If the history of a scientific article would be done from conception to publishing, I would say it starts as a pure Plain TeX document, as there are mostly mathematical formulas plus some unstructured comments explaining what the formulas are about, separated by commas, points and blank lines. That's all. Then, during the refinement process, as mathematical / physical / whatever new results are found and accumulate, the document evolves toward a more and more structured document, either with LaTeX or with Plain TeX/AMSTeX and personal macros (for many mathematicians). Ultimately, when the document reaches _you_the_publisher_ in its definitive form, then it may be transformed into SGML or anything else you like (PostScript or PDF?). But trying to write a scientific article from the scratch directly into SGML would be absurd, in my opinion. A secretary, maybe, but not a researcher. I hope that only those who never used \noindent or \medskip in an article to tune the aspect of the output, instead of modifying the class of their document (or the DTD for SGML addicts) will throw me some electronic stones at the head :-) Michel Lavaud ([log in to unmask]) `


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# Set is Subset of Union

(Redirected from Subset of Union)

## Theorem

The union of two sets is a superset of each:

$S \subseteq S \cup T$
$T \subseteq S \cup T$

### General Result

Let $S$ be a set.

Let $\mathcal P \left({S}\right)$ be the power set of $S$.

Let $\mathbb S \subseteq \mathcal P \left({S}\right)$.

Then:

$\forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$

### Set of Sets

Let $\mathbb S$ be a set of sets.

Then:

$\ds \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$

### Indexed Family of Sets

In the context of a family of sets, the result can be presented as follows:

Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a family of sets indexed by $I$.

Then:

$\displaystyle \forall \beta \in I: S_\beta \subseteq \bigcup_{\alpha \mathop \in I} S_\alpha$

where $\displaystyle \bigcup_{\alpha \mathop \in I} S_\alpha$ is the union of $\family {S_\alpha}$.

## Proof

 $\ds x \in S$ $\leadsto$ $\ds x \in S \lor x \in T$ Rule of Addition $\ds$ $\leadsto$ $\ds x \in S \cup T$ Definition of Set Union $\ds$ $\leadsto$ $\ds S \subseteq S \cup T$ Definition of Subset

Similarly for $T$.

$\blacksquare$


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# Math Help - help with specific trinomials.

1. ## help with specific trinomials.

right, so i know how to do trinomials in a normal case with only one pronumeral.

a^2 + 2a + 1
(a+1)(a+1)

and i can do non-monic trinomials no problem too, problem for me is when i discover a trinomial which has 2 pronumerals.

a^2 + 2ab + b^2
now i have no idea how i'm meant to work it out.
i know that this simplifies to (a+b)(a+b) because i can do the expansion for it.
but i have no idea how to simplify it and if it were slightly more complex
eg. k^2 + 8kp + p^2 i would not know what to do in the exam.

help please! could someone please show me how to work out these expressions a^2 + 2ab + b^2

2. Originally Posted by Zellus
right, so i know how to do trinomials in a normal case with only one pronumeral.

a^2 + 2a + 1
(a+1)(a+1)

and i can do non-monic trinomials no problem too, problem for me is when i discover a trinomial which has 2 pronumerals.

a^2 + 2ab + b^2
now i have no idea how i'm meant to work it out.
i know that this simplifies to (a+b)(a+b) because i can do the expansion for it.
but i have no idea how to simplify it and if it were slightly more complex
eg. k^2 + 8kp + p^2 i would not know what to do in the exam.

help please! could someone please show me how to work out these expressions a^2 + 2ab + b^2

Complete the square:

$k^2 + 8p + p^2 = (k + 4p)^2 - 15p^2 = (k + 4p)^2 - (\sqrt{15} p)^2$.

Now use the difference of two squares formula to get $k^2 + 8p + p^2 = (k - [\sqrt{15} - 4] p) (k + [ \sqrt{15} + 4] p)$.

3. Originally Posted by mr fantastic
Complete the square:

$k^2 + 8p + p^2 = (k + 4p)^2 - 15p^2 = (k + 4p)^2 - (\sqrt{15} p)^2$.
how did you know that k^2 + 8p + p^2 when has it's square completed = (k + 4p)^2 - 15p^2 without having to go through a slow, laborious process of trial and error?

4. Originally Posted by Zellus
how did you know that k^2 + 8p + p^2 when has it's square completed = (k + 4p)^2 - 15p^2 without having to go through a slow, laborious process of trial and error?
The key is to know that ${\color{red}A^2 + 2AB} + B^2 = (A + B)^2$. So ${\color{red}k^2 + 8p} + 16p^2 = (k + 4p)^2$.

Hence $k^2 + 8p + p^2 = k^2 + 8p + 16p^2 - 15p^2 = (k + 4p)^2 - 15p^2$.


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# Failed attempts at fitting nonlinear Hill function (biochemistry) to data

I am trying to fit some data in Matlab to a Hill function of the form $y = \dfrac{1}{1+(K/r)^n}.$ I have data for $r,y$ and I need to find $K,n$.

I have tried following the approach shown here in Demo $3$ at this link: http://www.math.ubc.ca/~keshet/MCB2012/SlidesDodo/DataFitLect3.pdf

but I'm getting wrong results. Please help!

Here is my code:

Function file:

function response = func(x,dose)
EC50 = x(1);
n = x(2);
response = 1./(1+(EC50./dose).^n);
end

Script file:

xdata = (logspace(-2,2,101))';
ydata = [0.0981 0.1074 0.1177 0.1289 0.1411 0.1545 0.1692 0.1852 0.2027 ...
0.2219 0.2428 0.2656 0.2905 0.3176 0.3472 0.3795 0.4146 0.4528 ...
0.4944 0.5395 0.5886 0.6418 0.6994 0.7618 0.8293 0.9022 0.9808 ...
1.0655 1.1566 1.2544 1.3592 1.4713 1.5909 1.7183 1.8537 1.9972 ...
2.1490 2.3089 2.4770 2.6532 2.8371 3.0286 3.2272 3.4324 3.6437 ...
3.8603 4.0815 4.3065 4.5344 4.7642 4.9950 5.2258 5.4556 5.6833 ...
5.9082 6.1292 6.3457 6.5567 6.7616 6.9599 7.1511 7.3347 7.5105 ...
7.6783 7.8379 7.9893 8.1324 8.2675 8.3946 8.5139 8.6257 8.7301 ...
8.8276 8.9184 9.0029 9.0812 9.1539 9.2212 9.2834 9.3408 9.3939 ...
9.4427 9.4877 9.5291 9.5672 9.6022 9.6343 9.6638 9.6909 9.7157 ...
9.7384 9.7592 9.7783 9.7957 9.8117 9.8263 9.8397 9.8519 9.8630 ...
9.8732 9.8826]';
guess = [1 1];
betaHat = nlinfit(xdata,ydata,@func,guess);
semilogx(xdata,ydata,'ro','MarkerSize',8)
hold on
plot(xdata,func(betaHat,xdata));

I am getting the Warning: Imaginary parts of complex X and/or Y arguments ignored

and here is my abysmal graph: Red = data, blue = bad fit

I have used nlinfit from the Statistics toolbox.

• Can you figure out which line of your program triggers the warning (by just running each line one after the other interactively)? You are probably doing something simple wrong. – Ian Feb 3 '16 at 22:30
• Ok, I will try running it in debug mode. – user289771 Feb 3 '16 at 22:32
• Ah, I see the problem now: mathworks.com/matlabcentral/answers/… So instead of using the function $f(r;K,n)=\frac{1}{1+(K/r)^n}$, try the function $g(r;a,n)=\frac{1}{1+a/r^n}$. Then at the end you can recompute $K$ from $a$ and $n$. – Ian Feb 3 '16 at 22:42
• It's not exactly that your math was wrong, it's that you needed the assumption that $K$ remains nonnegative throughout the computation. But nlinfit does not guarantee this, which created the problems that you saw here. You can bypass this by switching to a parametrization of the problem in which no constraints are necessary, like the one I suggested. – Ian Feb 3 '16 at 22:49
• Actually, I think you have a problem in your setup: in the Hill equation, the $y$ values are between $0$ and $1$, but your $y$ data is not between $0$ and $1$. The only way for the $y$ data to not be between $0$ and $1$ is to have $K<0$. nlinfit figured that out, but also was trying to use floating point values of $n$, which is where the complex numbers came from. Is it possible you need another parameter, such as a factor in the numerator? Or is it possible that your data is wrong? – Ian Feb 3 '16 at 22:56

## 1 Answer

Try transforming your equation to a different form? Often fitting can go wrong because of errors piling up in working with very small/large numbers. Sometimes, it can be fixed by transforming the equation to a different form.

I would suggest going in for fitting the logarithmic form of this equation, i.e. getting rid of the exponent. This is a pretty common practice in biochem.

the new form would be -

$log(y/1-y)=nlog(r)-nlog(k)$


text:
# A neutron, a proton and an electron are placed in a uniform electric field

Q: A neutron, a proton and an electron are placed in a uniform electric field.
(A) The forces acting on them will be equal.
(B) Their accelerations may be equal.
(C) Magnitude of acceleration may be equal.
(D) Magnitude of acceleration will be different.

Solution :Force on neutron = 0 (as charge = 0)
Force on proton = qE
Force on electron = qE
Masses of proton and electron are different.
Hence their accelerations will be different.
 Hence (D) is correct.

## Author: Rajesh Jha

QuantumStudy.com for CBSE|IIT-JEE|NEET


text:
## Trigonometry (11th Edition) Clone

$-\sqrt3$
Convert the angle measure to degrees to obtain: $=\frac{5\pi}{3} \cdot \frac{180^o}{\pi} = 5(60^o)=300^0$ Thus, $\tan{(\frac{5\pi}{3})} \\= \tan{300^o}$ 300 degrees is in Quadrant IV so : (i) its reference angle is $360^o-300^o=60^o$ (ii) its tangent is negative From Section 2.1 (page 50) , we learned that: $\tan{60^o} = \sqrt3$ This means that: $\tan{300^o} = -\tan{60^o} = -\sqrt3$.


text:
# Spin in elementary particles

by TheAnalogKid2
Tags: elementary, particles, spin
 P: 3 Looking at the Standard model, I noticed that all the quarks and leptons have a spin of 1/2 and all of the gauge bosons have a spin of 1. Can someone give me a general definition of what "spin" is and the difference between a particle that has a spin of 1/2 and a particle that has a spin of 1?
 P: 976
 Sci Advisor HW Helper P: 11,863 Since the Standard Model comes from putting QED + EW together, the particles involved (rather their fields) must have spin 1 tops, as spins 3/2 and 2 are coming from SuperGravity theories. The elementary leptons are fermions and must have spin 1/2. The gauge particles are bosons with spin 1.
P: 11

## Spin in elementary particles

Wanted to ask the same thing. The question is what exactly IS this mysterious thing called spin? Answers above only state what are the spin values of different particles, not what this "spin" is, or WHY these particles have such spin values.
 P: 427 Spin is an intrinsic property of fundamental particles. It arises from relativistic QM but I am not an expert in that area, so I won't comment on how. Long before that was shown, spin was used in non-relativistic QM. It is called spin because it acts like spin angular momentum of a large body. If you don't think too hard about it, you could probably convince yourself that it is the particle spinning. In an atom, the spin and orbital angular momentum add to make a total angular momentum. For a while, it seemed reasonable that electrons were really rotating, but E&M would not allow this (you can probably find a nice explanation of why via Google). Spin is now considered a fundamental property of fundamental particles. I assume that you want more than that. Specific questions might help, but you should probably just get a decent QM book and spend a lot of time with it. I doubt you will be able to get much deeper from a layman's perspective.
 P: 123 "What is spin?" is a question of the species "What mass, charge, or color is?" and the answer is the same: "A fundamental property of elementary particles". I think that this property had been observed for the fist time in the famous "Stern–Gerlach experiment" or, according to wikipedia, "in the context of the emission spectrum of alkali metals". The first attempt to theorize spin was done by W. Pauli, although, except than a two-valued degree of freedom, he had no idea what was it exactly. Some physicists thought that the fundamental particles were actually spinning, so they had some internal angular momentum, but this idea was abandoned at once, because the velocity required to have the observed value of angular momentum exceeds the speed of light. The correct theoretical treatment of spin arises from the combination of QM and special relativity. Specifically, when you calculate the angular momentum of a particle in the rest frame of that particle, you find that it can have a non-zero value. That means that that the particle carries some angular momentum that is not due to its motion, i.e. spin is an internal amount of angular momentum. Spin is quantized, that is it can have only certain values, and these values can be integer or half-integer multiples of $\hbar$. According to QFT and specifically because of the “causality principle”, particles of half-integer spin have to be “fermions” and those of integer spin have to be “bosons”, that is, a large number of identical particles have to obey “Fermi-Dirac” or “Bose-Einstein” statistics, respectively. Those are some general information about spin and I hope will help you clear up the concept of spin.
P: 492
 Those are some general information about spin and I hope will help you clear up the concept of spin.
Nice summary Dust.

 Related Discussions High Energy, Nuclear, Particle Physics 12 Quantum Physics 1 Quantum Physics 5 High Energy, Nuclear, Particle Physics 20 High Energy, Nuclear, Particle Physics 6


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## During the fusion process, how is mass converted into energy?

Waleed A.
Featured 9 months ago

$E = m {c}^{2}$

#### Explanation:

This is calculated using the famous equation of Einstein,

$E = m {c}^{2}$

In Fusion reaction like the ones taking place in the core of a Star, there is enough pressure to fuse hydrogen nuclei to form one helium nucleus.

So, 4 hydrogen nuclei are fused together to form one Helium nucleus. But, where does the energy come from that keeps the Sun from collapsing?.

When 4 Hydrogen nuclei are merged together they show a certain discrepancy in the mass when a Helium atom is formed, i.e the mass of 4 Hydrogen atoms before Fusion is less than the mass of the Helium atom after the reaction this mass defect is converted into energy by $E = m {c}^{2}$.

$\text{mass of a hydrogen atom " = " 1.00794 u}$

$\text{mass of one helium atom " = " 4.002602 u}$

Where $u = 1.6605 \times {10}^{- 27} \text{kg}$.

The mass defect, $\Delta m$, will be

$\Delta m = 4 \times {m}_{H} - {m}_{H e}$

$= 4 \times \text{1.00794 u" - "4.002602 u}$

$= \text{0.029158 u }$ or $\text{ "4.8416859 * 10^(-29)"kg}$

In the equation $E = m {c}^{2}$, $c$ is the speed of light in a vacuum, approximately equal to $3 \cdot {10}^{8} {\text{m s}}^{- 1}$.

This means that you have

$E = 4.8416859 \cdot {10}^{- 29} {\text{kg" * (3 * 10^8)^2"m"^2"s}}^{- 2}$

$E = 4.35751731 \cdot {10}^{- 12} \text{ J per reaction}$

## What is the Aurora Borealis?

Hunaid L. Hanfee
Featured 9 months ago

Aurora Borealis are lights caused due to collision between charged particle & the air molecule in Earth's magnetic field.

#### Explanation:

Aurora Borealis are the dancing light which can be seen at sky in night. This light can be seen above the magnetic poles of the northern and southern hemispheres. They are known as 'Aurora Borealis' in north & 'Aurora Australis' in south.

For more picture please visit this site - Aurora near Akureyri

This light is actually the collision of charged particle from sun between air molecule in the Earth's magnetic field. When they strike the molecules to become excited. This molecules gives off the light as they calm down.

This lights are also know as Northern Lights.

Aurora displays appear in many colors although pale green and pink are the most common. But sometimes red, yellow, green, blue, are also visible. People call it dancing light because it appears in many form like arc, cloud, shooting rays etc.

Aurora shows different colors because of different gases present in the air.
Oxygen - Pale green or orange red.
Nitrogen - Blue or red.

## Does the Oort cloud exist? If so, why can't we see it?

Parzival S.
Featured 8 months ago

The Oort Cloud is theorized to exist but due to many factors we can't see it.

#### Explanation:

First let's talk about what the Oort Cloud is (and what it isn't).

The first thing to know is that it is theorized that the cloud exists. While definitive proof will be difficult to come by, the theory helps explain many questions in astronomy (where do long-period comets come from and how to describe their orbits, etc)

From http://space-facts.com/oort-cloud/, it's a region of space that surrounds the solar system where ice, rock, and the occasional larger body (sometimes called Dwarf Planets) exist. The region is a spherical shell and starts about 2,000 AU (Astronomical Units) from the Sun (and to put that into perspective, 1 AU is the average distance from Earth to the Sun. Pluto is 40 AU from the Sun.) and extends well out towards the closest star (perhaps as much as 100,000AU away from the Sun or roughly 1/4 of the distance between the Sun and the closest star).

What it isn't is a cloud in the sense we think about clouds on Earth - big puffy things made up of water vapour that are easy to spot in the sky. Clouds in the astronomical sense are far less dense but are much more vast so that we see them when the light of stars and galaxies is interfered with.

So what would make the cloud hard to see?

Things that we observe in space are observable only if we can see them.

So what does it take to be able to see something? Light.

And where does the light come from? It either is produced by the object (ex. the Sun, a lamp, etc) or the light is reflected off the object (ex. the Moon).

When light is reflected, all the light that strikes the object doesn't bounce back - it gets scattered, absorbed, and the like, so only a tiny fraction of what strikes the object is reflected back (which is why the Moon, even though it can be quite bright at night, isn't anywhere near as bright as the Sun. The Moon absorbs some light and what is reflected is scattered off in all sorts of directions - only a fraction reaches Earth).

Clouds in the Earth's sky are visible because there is a strong light source (the Sun) that allows the clouds to scatter the light and yet still have plenty left for us to look up and see them. There is no light source like that for the Oort Cloud, so what light it gets, when scattered, doesn't come back towards Earth but instead flies off in random directions.

How much light does the Oort Cloud receive? Light diminishes exponentially with distance. A strong light source at distance 1 will only be $\frac{1}{4}$ as strong at distance 2, $\frac{1}{9}$ at distance 3, etc. And this is for both the journey of the light to the object and the reflection back! What that means is that for an object at the closest edge of the Cloud, it receives ${\left(\frac{1}{2000}\right)}^{2} = \frac{1}{4 , 000 , 000}$ the amount of light we get (remember that Earth is at distance 1 AU and the edge of the Cloud is at distance 2000AU)! And then light that is reflected back towards Earth and so we see $\frac{1}{4 , 000 , 000}$ of that!

The Oort Cloud does not make its own light, so we need reflected light to see it. The Cloud is made up of small rocks and pieces of ice with huge distances between them, so most of the light that does manage to make it there passes right through and never reflects. Space rocks and ice are notoriously "dirty" - or covered in a dark coloured "space dust", so most of the light is absorbed by the rocks in the Cloud. And the distances involved are massive, so much so that to see Pluto and a few of the other large objects in the Cloud, it takes very powerful telescopes and a lot of luck in finding them.

The only other way of seeing the Cloud is with the use of a deep space probe. However, because of the vast distances, it's difficult to build a spacecraft that can make the journey. According to https://en.wikipedia.org/wiki/Oort_cloud, Voyager 1, the farthest and fastest moving probe we have and launched in 1977, will be another 300 years before it reaches the Cloud and by then will have no power to explore or send back data. Other probes also making their way out of the solar system will also run out of power long before making contact with the Oort Cloud.

## The Martian satellite Phobos travels in an approximately circular orbit of radius 9.4 * 10^6 m with a period of 7 h 39 min. What is the mass of Mars?

Parzival S.
Featured 8 months ago

${M}_{m a r s} = 6.479 \times {10}^{23}$ kg (I used Google calculator so you may get a slightly different number if you approximate $\pi$).

#### Explanation:

The equation that relates the mass of Mars with the orbital information provided about Phobos is the following:

${T}^{2} / {R}^{3} = \frac{4 {\pi}^{2}}{G \cdot {M}_{m a r s}}$

T is the period. Let's express that in seconds:

$7 \cdot 3600 + 39 \cdot 60 = 25200 + 2340 = 27540$ seconds

Let's plug in and solve for M:

$M = 4 {\pi}^{2} {R}^{3} / \left({T}^{2} G\right) = 4 {\pi}^{2} {\left(9.4 \times {10}^{6}\right)}^{3} / \left({\left(27540\right)}^{2} \cdot 6.673 \times {10}^{-} 11\right)$

$M = 6.479 \times {10}^{23}$ kg (I used Google calculator so you may get a slightly different number if you approximate $\pi$).

A quick check of "the internet" tells me that the actual number for the mass of Mars is $6.39 \times {10}^{23}$ kg, so we're pretty close.

Thanks to the following link for the equation: http://www.physicsclassroom.com/class/circles/Lesson-4/Mathematics-of-Satellite-Motion

## What is the string theory? How is it related to space time?

Azhar A.
Featured 7 months ago

String-Theory is the theory that replace the particles(atoms, electrons, photons) to vibrating strings.

#### Explanation:

In the early Twentieth century, two frameworks of physics were published:

$1.$ Theory of Relativity by Albert Einstein
$\textcolor{w h i t e}{0000000000000000000}$and
$2.$ Quantum Mechanics

Both these theories try to explain the formulated laws of Physics that we use now. But the problem arises when we try to reconcile Quantum Mechanics with Theory of Relativity because of Gravity.

String Theory try to solve this problem by replacing particles with strings. That's why it is a member for Theory of Everything.

On large scale distances, the string have normal properties like mass, charge, etc. It is a vibrating string that vibrates in all different ways.

When we get electromagnetic field in space-time, it is called as Quantum Field Theory. As we know that, quantum mechanics is based on probabilities, we can compute that with the help of Perturbation Theory.

But how string theory relates to space-time? If a closed string is traveling in a curved space-time, then the coordinates of the string in spacetime feel this curvature as the string propagates. The answer lies on the string worldsheet or stringy space-time. In order for their to be a consistent quantum theory in this case, the curved space in which the string travels must be a solution to the Einstein equations.

This was a very convincing result for string theorists. Not only does string theory predict the graviton (A hypothetical gravitatioal force carrier) from flat spacetime physics alone, but string theory also predicts the Einstein equation will be obeyed by a curved spacetime in which strings propagate.(Credit)

The problem with string theory is that it is not experimentally discovered yet. If we will get results, we can be able to reconcile gravity with quantum mechanics.

## Mars has an average surface temperature of about 200K. Pluto has an average surface temperature of about 40K. Which planet emits more energy per square meter of surface area per second? By a factor of how much?

JMicheli
Featured 6 months ago

Mars emits $625$ times more energy per unit of surface area than Pluto does.

#### Explanation:

It is obvious that a hotter object will emit more black body radiation. Thus, we already know that Mars will emit more energy than Pluto. The only question is by how much.

This problem requires evaluating the energy of the black body radiation emitted by both planets. This energy is described as a function of temperature and the frequency being emitted:

$E \left(\nu , T\right) = \frac{2 {\pi}^{2} \nu}{c} \frac{h \nu}{{e}^{\frac{h \nu}{k T}} - 1}$

Integrating over frequency gives the total power per unit area as a function of temperature:
${\int}_{0}^{\infty} E \left(\nu , T\right) = \frac{{\pi}^{2} c {\left(k T\right)}^{4}}{60 {\left(\overline{h} c\right)}^{3}}$

(note that the above equation uses $\overline{h}$, the reduced Planck's constant, rather than $h$. It is difficult to read in Socratic's notation)

Solving for the ratio between the two, then, the result is incredibly simple. If ${T}_{p}$ is Pluto's temperature and ${T}_{m}$ is Mars' temperature then the factor $a$ can be calculated with:

$\frac{{\pi}^{2} c {\left(k {T}_{m}\right)}^{4}}{60 {\left(\overline{h} c\right)}^{3}} = a \frac{{\pi}^{2} c {\left(k {T}_{p}\right)}^{4}}{60 {\left(\overline{h} c\right)}^{3}}$
$\frac{\cancel{{\pi}^{2} c {k}^{4}}}{\cancel{60 {\left(\overline{h} c\right)}^{3}}} {T}_{m}^{4} = a \frac{\cancel{{\pi}^{2} c {k}^{4}}}{\cancel{60 {\left(\overline{h} c\right)}^{3}}} {T}_{p}^{4}$
${\left({T}_{m} / {T}_{p}\right)}^{4} = a = {\left(\frac{200}{40}\right)}^{4} = {5}^{4} = 625$ times as much

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## G = S3×C2.C42order 192 = 26·3

### Direct product of S3 and C2.C42

Series: Derived Chief Lower central Upper central

 Derived series C1 — C6 — S3×C2.C42
 Chief series C1 — C3 — C6 — C2×C6 — C22×C6 — S3×C23 — S3×C22×C4 — S3×C2.C42
 Lower central C3 — C6 — S3×C2.C42
 Upper central C1 — C23 — C2.C42

Generators and relations for S3×C2.C42
G = < a,b,c,d,e | a3=b2=c2=d4=e4=1, bab=a-1, ac=ca, ad=da, ae=ea, bc=cb, bd=db, be=eb, ede-1=cd=dc, ce=ec >

Subgroups: 800 in 330 conjugacy classes, 131 normal (12 characteristic)
C1, C2, C2, C2, C3, C4, C22, C22, C22, S3, C6, C6, C2×C4, C2×C4, C23, C23, Dic3, C12, D6, C2×C6, C2×C6, C22×C4, C22×C4, C24, C4×S3, C2×Dic3, C2×Dic3, C2×C12, C2×C12, C22×S3, C22×C6, C2.C42, C2.C42, C23×C4, S3×C2×C4, S3×C2×C4, C22×Dic3, C22×C12, S3×C23, C2×C2.C42, C6.C42, C3×C2.C42, S3×C22×C4, S3×C2.C42
Quotients: C1, C2, C4, C22, S3, C2×C4, D4, Q8, C23, D6, C42, C22⋊C4, C4⋊C4, C22×C4, C2×D4, C2×Q8, C4×S3, C22×S3, C2.C42, C2×C42, C2×C22⋊C4, C2×C4⋊C4, S3×C2×C4, S3×D4, S3×Q8, C2×C2.C42, S3×C42, S3×C22⋊C4, S3×C4⋊C4, S3×C2.C42

Smallest permutation representation of S3×C2.C42
On 96 points
Generators in S96
(1 35 9)(2 36 10)(3 33 11)(4 34 12)(5 60 55)(6 57 56)(7 58 53)(8 59 54)(13 67 39)(14 68 40)(15 65 37)(16 66 38)(17 89 87)(18 90 88)(19 91 85)(20 92 86)(21 71 47)(22 72 48)(23 69 45)(24 70 46)(25 79 50)(26 80 51)(27 77 52)(28 78 49)(29 63 81)(30 64 82)(31 61 83)(32 62 84)(41 75 93)(42 76 94)(43 73 95)(44 74 96)
(1 29)(2 30)(3 31)(4 32)(5 80)(6 77)(7 78)(8 79)(9 63)(10 64)(11 61)(12 62)(13 17)(14 18)(15 19)(16 20)(21 75)(22 76)(23 73)(24 74)(25 59)(26 60)(27 57)(28 58)(33 83)(34 84)(35 81)(36 82)(37 91)(38 92)(39 89)(40 90)(41 71)(42 72)(43 69)(44 70)(45 95)(46 96)(47 93)(48 94)(49 53)(50 54)(51 55)(52 56)(65 85)(66 86)(67 87)(68 88)
(1 31)(2 32)(3 29)(4 30)(5 78)(6 79)(7 80)(8 77)(9 83)(10 84)(11 81)(12 82)(13 91)(14 92)(15 89)(16 90)(17 37)(18 38)(19 39)(20 40)(21 95)(22 96)(23 93)(24 94)(25 56)(26 53)(27 54)(28 55)(33 63)(34 64)(35 61)(36 62)(41 69)(42 70)(43 71)(44 72)(45 75)(46 76)(47 73)(48 74)(49 60)(50 57)(51 58)(52 59)(65 87)(66 88)(67 85)(68 86)
(1 2 3 4)(5 6 7 8)(9 10 11 12)(13 14 15 16)(17 18 19 20)(21 22 23 24)(25 26 27 28)(29 30 31 32)(33 34 35 36)(37 38 39 40)(41 42 43 44)(45 46 47 48)(49 50 51 52)(53 54 55 56)(57 58 59 60)(61 62 63 64)(65 66 67 68)(69 70 71 72)(73 74 75 76)(77 78 79 80)(81 82 83 84)(85 86 87 88)(89 90 91 92)(93 94 95 96)
(1 77 65 71)(2 5 66 44)(3 79 67 69)(4 7 68 42)(6 85 41 29)(8 87 43 31)(9 27 15 21)(10 55 16 96)(11 25 13 23)(12 53 14 94)(17 73 61 59)(18 48 62 49)(19 75 63 57)(20 46 64 51)(22 84 28 90)(24 82 26 92)(30 80 86 70)(32 78 88 72)(33 50 39 45)(34 58 40 76)(35 52 37 47)(36 60 38 74)(54 89 95 83)(56 91 93 81)

G:=sub<Sym(96)| (1,35,9)(2,36,10)(3,33,11)(4,34,12)(5,60,55)(6,57,56)(7,58,53)(8,59,54)(13,67,39)(14,68,40)(15,65,37)(16,66,38)(17,89,87)(18,90,88)(19,91,85)(20,92,86)(21,71,47)(22,72,48)(23,69,45)(24,70,46)(25,79,50)(26,80,51)(27,77,52)(28,78,49)(29,63,81)(30,64,82)(31,61,83)(32,62,84)(41,75,93)(42,76,94)(43,73,95)(44,74,96), (1,29)(2,30)(3,31)(4,32)(5,80)(6,77)(7,78)(8,79)(9,63)(10,64)(11,61)(12,62)(13,17)(14,18)(15,19)(16,20)(21,75)(22,76)(23,73)(24,74)(25,59)(26,60)(27,57)(28,58)(33,83)(34,84)(35,81)(36,82)(37,91)(38,92)(39,89)(40,90)(41,71)(42,72)(43,69)(44,70)(45,95)(46,96)(47,93)(48,94)(49,53)(50,54)(51,55)(52,56)(65,85)(66,86)(67,87)(68,88), (1,31)(2,32)(3,29)(4,30)(5,78)(6,79)(7,80)(8,77)(9,83)(10,84)(11,81)(12,82)(13,91)(14,92)(15,89)(16,90)(17,37)(18,38)(19,39)(20,40)(21,95)(22,96)(23,93)(24,94)(25,56)(26,53)(27,54)(28,55)(33,63)(34,64)(35,61)(36,62)(41,69)(42,70)(43,71)(44,72)(45,75)(46,76)(47,73)(48,74)(49,60)(50,57)(51,58)(52,59)(65,87)(66,88)(67,85)(68,86), (1,2,3,4)(5,6,7,8)(9,10,11,12)(13,14,15,16)(17,18,19,20)(21,22,23,24)(25,26,27,28)(29,30,31,32)(33,34,35,36)(37,38,39,40)(41,42,43,44)(45,46,47,48)(49,50,51,52)(53,54,55,56)(57,58,59,60)(61,62,63,64)(65,66,67,68)(69,70,71,72)(73,74,75,76)(77,78,79,80)(81,82,83,84)(85,86,87,88)(89,90,91,92)(93,94,95,96), (1,77,65,71)(2,5,66,44)(3,79,67,69)(4,7,68,42)(6,85,41,29)(8,87,43,31)(9,27,15,21)(10,55,16,96)(11,25,13,23)(12,53,14,94)(17,73,61,59)(18,48,62,49)(19,75,63,57)(20,46,64,51)(22,84,28,90)(24,82,26,92)(30,80,86,70)(32,78,88,72)(33,50,39,45)(34,58,40,76)(35,52,37,47)(36,60,38,74)(54,89,95,83)(56,91,93,81)>;

G:=Group( (1,35,9)(2,36,10)(3,33,11)(4,34,12)(5,60,55)(6,57,56)(7,58,53)(8,59,54)(13,67,39)(14,68,40)(15,65,37)(16,66,38)(17,89,87)(18,90,88)(19,91,85)(20,92,86)(21,71,47)(22,72,48)(23,69,45)(24,70,46)(25,79,50)(26,80,51)(27,77,52)(28,78,49)(29,63,81)(30,64,82)(31,61,83)(32,62,84)(41,75,93)(42,76,94)(43,73,95)(44,74,96), (1,29)(2,30)(3,31)(4,32)(5,80)(6,77)(7,78)(8,79)(9,63)(10,64)(11,61)(12,62)(13,17)(14,18)(15,19)(16,20)(21,75)(22,76)(23,73)(24,74)(25,59)(26,60)(27,57)(28,58)(33,83)(34,84)(35,81)(36,82)(37,91)(38,92)(39,89)(40,90)(41,71)(42,72)(43,69)(44,70)(45,95)(46,96)(47,93)(48,94)(49,53)(50,54)(51,55)(52,56)(65,85)(66,86)(67,87)(68,88), (1,31)(2,32)(3,29)(4,30)(5,78)(6,79)(7,80)(8,77)(9,83)(10,84)(11,81)(12,82)(13,91)(14,92)(15,89)(16,90)(17,37)(18,38)(19,39)(20,40)(21,95)(22,96)(23,93)(24,94)(25,56)(26,53)(27,54)(28,55)(33,63)(34,64)(35,61)(36,62)(41,69)(42,70)(43,71)(44,72)(45,75)(46,76)(47,73)(48,74)(49,60)(50,57)(51,58)(52,59)(65,87)(66,88)(67,85)(68,86), (1,2,3,4)(5,6,7,8)(9,10,11,12)(13,14,15,16)(17,18,19,20)(21,22,23,24)(25,26,27,28)(29,30,31,32)(33,34,35,36)(37,38,39,40)(41,42,43,44)(45,46,47,48)(49,50,51,52)(53,54,55,56)(57,58,59,60)(61,62,63,64)(65,66,67,68)(69,70,71,72)(73,74,75,76)(77,78,79,80)(81,82,83,84)(85,86,87,88)(89,90,91,92)(93,94,95,96), (1,77,65,71)(2,5,66,44)(3,79,67,69)(4,7,68,42)(6,85,41,29)(8,87,43,31)(9,27,15,21)(10,55,16,96)(11,25,13,23)(12,53,14,94)(17,73,61,59)(18,48,62,49)(19,75,63,57)(20,46,64,51)(22,84,28,90)(24,82,26,92)(30,80,86,70)(32,78,88,72)(33,50,39,45)(34,58,40,76)(35,52,37,47)(36,60,38,74)(54,89,95,83)(56,91,93,81) );

G=PermutationGroup([[(1,35,9),(2,36,10),(3,33,11),(4,34,12),(5,60,55),(6,57,56),(7,58,53),(8,59,54),(13,67,39),(14,68,40),(15,65,37),(16,66,38),(17,89,87),(18,90,88),(19,91,85),(20,92,86),(21,71,47),(22,72,48),(23,69,45),(24,70,46),(25,79,50),(26,80,51),(27,77,52),(28,78,49),(29,63,81),(30,64,82),(31,61,83),(32,62,84),(41,75,93),(42,76,94),(43,73,95),(44,74,96)], [(1,29),(2,30),(3,31),(4,32),(5,80),(6,77),(7,78),(8,79),(9,63),(10,64),(11,61),(12,62),(13,17),(14,18),(15,19),(16,20),(21,75),(22,76),(23,73),(24,74),(25,59),(26,60),(27,57),(28,58),(33,83),(34,84),(35,81),(36,82),(37,91),(38,92),(39,89),(40,90),(41,71),(42,72),(43,69),(44,70),(45,95),(46,96),(47,93),(48,94),(49,53),(50,54),(51,55),(52,56),(65,85),(66,86),(67,87),(68,88)], [(1,31),(2,32),(3,29),(4,30),(5,78),(6,79),(7,80),(8,77),(9,83),(10,84),(11,81),(12,82),(13,91),(14,92),(15,89),(16,90),(17,37),(18,38),(19,39),(20,40),(21,95),(22,96),(23,93),(24,94),(25,56),(26,53),(27,54),(28,55),(33,63),(34,64),(35,61),(36,62),(41,69),(42,70),(43,71),(44,72),(45,75),(46,76),(47,73),(48,74),(49,60),(50,57),(51,58),(52,59),(65,87),(66,88),(67,85),(68,86)], [(1,2,3,4),(5,6,7,8),(9,10,11,12),(13,14,15,16),(17,18,19,20),(21,22,23,24),(25,26,27,28),(29,30,31,32),(33,34,35,36),(37,38,39,40),(41,42,43,44),(45,46,47,48),(49,50,51,52),(53,54,55,56),(57,58,59,60),(61,62,63,64),(65,66,67,68),(69,70,71,72),(73,74,75,76),(77,78,79,80),(81,82,83,84),(85,86,87,88),(89,90,91,92),(93,94,95,96)], [(1,77,65,71),(2,5,66,44),(3,79,67,69),(4,7,68,42),(6,85,41,29),(8,87,43,31),(9,27,15,21),(10,55,16,96),(11,25,13,23),(12,53,14,94),(17,73,61,59),(18,48,62,49),(19,75,63,57),(20,46,64,51),(22,84,28,90),(24,82,26,92),(30,80,86,70),(32,78,88,72),(33,50,39,45),(34,58,40,76),(35,52,37,47),(36,60,38,74),(54,89,95,83),(56,91,93,81)]])

60 conjugacy classes

 class 1 2A ··· 2G 2H ··· 2O 3 4A ··· 4L 4M ··· 4X 6A ··· 6G 12A ··· 12L order 1 2 ··· 2 2 ··· 2 3 4 ··· 4 4 ··· 4 6 ··· 6 12 ··· 12 size 1 1 ··· 1 3 ··· 3 2 2 ··· 2 6 ··· 6 2 ··· 2 4 ··· 4

60 irreducible representations

 dim 1 1 1 1 1 2 2 2 2 2 4 4 type + + + + + + - + + - image C1 C2 C2 C2 C4 S3 D4 Q8 D6 C4×S3 S3×D4 S3×Q8 kernel S3×C2.C42 C6.C42 C3×C2.C42 S3×C22×C4 S3×C2×C4 C2.C42 C22×S3 C22×S3 C22×C4 C2×C4 C22 C22 # reps 1 3 1 3 24 1 6 2 3 12 3 1

Matrix representation of S3×C2.C42 in GL6(𝔽13)

 0 12 0 0 0 0 1 12 0 0 0 0 0 0 0 12 0 0 0 0 1 12 0 0 0 0 0 0 1 0 0 0 0 0 0 1
,
 0 12 0 0 0 0 12 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 12 0 0 0 0 0 0 12
,
 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 12 0 0 0 0 0 0 12
,
 8 0 0 0 0 0 0 8 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0
,
 5 0 0 0 0 0 0 5 0 0 0 0 0 0 8 0 0 0 0 0 0 8 0 0 0 0 0 0 12 0 0 0 0 0 0 1

G:=sub<GL(6,GF(13))| [0,1,0,0,0,0,12,12,0,0,0,0,0,0,0,1,0,0,0,0,12,12,0,0,0,0,0,0,1,0,0,0,0,0,0,1],[0,12,0,0,0,0,12,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,12,0,0,0,0,0,0,12],[1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,12,0,0,0,0,0,0,12],[8,0,0,0,0,0,0,8,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,0],[5,0,0,0,0,0,0,5,0,0,0,0,0,0,8,0,0,0,0,0,0,8,0,0,0,0,0,0,12,0,0,0,0,0,0,1] >;

S3×C2.C42 in GAP, Magma, Sage, TeX

S_3\times C_2.C_4^2
% in TeX

G:=Group("S3xC2.C4^2");
// GroupNames label

G:=SmallGroup(192,222);
// by ID

G=gap.SmallGroup(192,222);
# by ID

G:=PCGroup([7,-2,-2,-2,-2,-2,-2,-3,120,387,58,6278]);
// Polycyclic

G:=Group<a,b,c,d,e|a^3=b^2=c^2=d^4=e^4=1,b*a*b=a^-1,a*c=c*a,a*d=d*a,a*e=e*a,b*c=c*b,b*d=d*b,b*e=e*b,e*d*e^-1=c*d=d*c,c*e=e*c>;
// generators/relations

׿
×
𝔽


text:
# Canonical equal time commutation relations in QED

I understand that to quantize the classical electromagnetic field one needs to impose commutation relations and express the field in terms of creation and annihilation operators. I notice that the canonical equal-time commutation rule comes in to use here. How do you derive this? What is it meant to represent?

Also, how do you prove the following commutation relation?

In classical Hamiltonian mechanics one has the generalized coordinates $q^{i}(t)$ and momenta $p_{i}(t)$. The Poisson bracket is defined as, $$[F,G]_{PB}=\frac{\partial F}{\partial q^{k}}\frac{\partial G}{\partial p_{k}}-\frac{\partial F}{\partial p_{k}}\frac{\partial G}{\partial q^{k}}\ .$$ Using the q's and p's in place of $F$ and $G$ one has the fundamental PBs, $$[q^{i}(t),p_{j}(t)]_{PB}=\delta^{i}_{k}\delta^{k}_{j}=\delta^{i}_{j}$$ $$[q^{i}(t),q^{j}(t)]_{PB}=0$$ $$[p_{i}(t),p_{j}(t)]_{PB}=0 \ .$$ In order to set up a quantum theory, Dirac says that the phase space functions $F$ and $G$ are changed to operators $\hat{F}$ and $\hat{G}$ and the PB becomes the commutator, $$[\hat{F},\hat{G}]_{-}=\hat{F}\hat{G}-\hat{G}\hat{F}=i\widehat{[F,G]_{PB}} \ .$$ So, going over to the quantum theory in this way, the fundamental PBs become commutators, $$[\hat{q}^{i}(t),\hat{p}_{j}(t)]_{-}=i\delta^{i}_{j}$$ $$[\hat{q}^{i}(t),\hat{q}^{j}(t)]_{-}=0$$ $$[\hat{p}_{i}(t),\hat{p}_{j}(t)]_{-}=0 \ .$$ These three commutators are the ones in the question. This is because the generalized coordinates for the classical electromagnetic field are, $$q^{i}(t)\rightarrow q^{(\mu,x)}(t) \rightarrow A_{\mu}(t,x) \ .$$ The generalized momenta $p_{i}(t)$ will be the values of a vector field $\pi^{\mu}(t,x)$ and so the first fundamental commutator becomes, $$[\hat{A}_{\mu}(t,x),\hat{\pi}^{\nu}(t,y)]_{-}=i\delta^{\nu}_{\mu}\delta(x-y)$$ and this is the first commutator in the question. The other two commutators in the question follow from the remaining two commutators for the q's and p's with a bit of raising/lowering indices of four-vectors.


text:
calculus online - Free exercises and solutions to help you succeed!

# Gradient – Calculate scalar field gradient and direction – Exercise 4257

Exercise

Calculate the gradient of the scalar field

$$u=xyz$$

At point (2,1,1)

And calculate its direction at this point.

Final Answer

$$\nabla u(2,1,1)=\vec{i}+2\vec{j}+2\vec{k}$$

$$\hat{\nabla} u(2,1,1)=\frac{1}{3}\vec{i}+\frac{2}{3}\vec{j}+\frac{2}{3}\vec{k}$$

### Solution

Coming soon…

Share with Friends


text:
Tximport/DESeq2 Normalization Process
1
0
Entering edit mode
johnsony • 0
Last seen 13 months ago
United States

Hi all,

I've been working on a meta-transcriptomic pipeline for a set of incubation experiments (for phytoplankton) from a 2019 research cruise in the California coast, and am at a stage now where I am needing to convert my alignment output (Salmon) into a matrix file (containing raw counts) used for DESeq2. I already have my phylodb and keggannot BLAST outputs and am hoping to merge these with the counts, lengths, and abundances file that was the result of Tximport.

I've already calculated TPM from the raw counts using the count columns (from each sample) and divided them by the avg_gene_length (derived from the length columns). The result gave me a TPM matrix which I then merged with my phylodb and kegg outputs. I did all that on R on the cluster since my files are too big (25G) to use on R itself, but had realized that DESeq2 does not require me to normalize my counts and only requires raw counts. My questions are:

1. Would calculating the TPM be considered normalizing in this case? If so, would I have to go back and only merge my raw counts output (directly from tximport) with the phylodb/kegg outputs before feeding my counts into DESeq2?
2. Does DESeq2 perform the normalization steps for me? If so, why would I not need to incorporate the length columns or the avg_gene_length column and only the raw counts?

Here is my tximport R script that I submitted as an sbatch job on the cluster

BiocManager::install(‘tximport’)
library(DESeq2)
library(tidyverse)
library(RColorBrewer)
library(pheatmap)
library(tximport)
library(ggplot2)
library(ggrepel)

samples <- list.files(path = “/pine/scr/j/o/johnsony/PUP_quants”, full.names = T)
files <- file.path(samples, “quant.sf”)
names(files) <- str_replace(samples, “/pine/scr/j/o/johnsony/PUP_quants”, “”) %>%
str_replace(“.salmon”, “”)

txi.keggannot.transcriptlevel <- tximport(files, type = “salmon”, tx2gene = tx2gene[,c(“query”, “KO”)], txOut = TRUE)
txi.keggannot.transcriptlevel.csv <- write.csv(txi.keggannot.transcriptlevel, “txi.keggannot.transcriptlevel.csv”)

Here is my TPM calculation script (which I'm wondering if I need to perform before DESeq2):

rownames(kegg_tl) <- kegg_tl$X colnames(kegg_tl)[colnames(kegg_tl) == ‘X’] <- ‘query’ x <- kegg_tl %>% transmute(kegg_tl, avg_gene_length = rowMeans(select(kegg_tl, starts_with(“length”)), na.rm = TRUE)) rownames(x) <- rownames(kegg_tl) y <- x[36:69] / x$avg_gene_length

tpm.mat <- t( t(y) * 1e6 / colSums(y) )
tpm.mat <- data.frame(tpm.mat)
tpm.mat$TrinityID <- rownames(tpm.mat) tpm_mat.csv <- write.csv(tpm.mat, “tpm_mat.csv”) Sorry if these all seem like beginner's questions, but I do hope to have more clarification in regards to DESeq2 before I can properly perform statistics on my samples. Please let me know if you have any advice or suggestions. Thank you! tximport Normalization DESeq2 • 562 views ADD COMMENT 0 Entering edit mode @mikelove Last seen 8 hours ago United States Are all the samples quantified against the same reference (e.g. the same Salmon index)? Usually, you would just let tximport and DESeq2 take are of all these steps, simply with: txi <- tximport(files, ...) dds <- DESeqDataSetFromTximport(txi, coldata, ~condition) The point of tximport is so users don't have to do the guesswork on import and scaling factors. ADD COMMENT 0 Entering edit mode Hi Mike, Yes all of the samples were quantified against the same reference (Salmon index). And I ended up with a quant.sf file for each individual sample, which I then used tximport for with the first R script shown above and created the txi.keggannot.transcriptlevel.csv file that includes the abundance, counts, and length columns for all of the samples combined. Is it just the counts columns that I use to feed into DESeq2? I think another problem I have is regarding my file size. Since this is a meta-transcriptome, I'm not sure if it is capable of running using RStudios, or would I have to run DESeq2 through the cluster? The output I have is 25G and the only way I can manipulate it and create a file that's able to be fed into DESeq2 would be through the LongLeaf cluster. Thank you! ADD REPLY 0 Entering edit mode re: "Is it just the counts columns that I use to feed into DESeq2?" No the line of code I show above (dds <- ...) uses more than the counts column. It also uses the lengths. I'd recommend minimal filtering from the start because you probably have a lot of transcripts that are not expressed, and you can drop these even before importing into DESeq2. E.g.: # X or more samples with TPM > threshold: keep <- rowSums(txi$abundance >= tpm_threshold) >= X
filt <- lapply(txi[1:3], function(mat) mat[keep,])
filt$countsFromAbundance <- txi$countsFromAbundance # keep this metadata
0
Entering edit mode

Hi Mike,

I've tried using the script, and had made a sampleTable first from the colnames of my txi file, and it looks something like this:

sampleTypewideshelf
countsQ1          wideshelfT0
countsQ2          wideshelfT0
countsQ3          wideshelfT0
countsQ4      wideshelfT1Ctrl
countsQ7        wideshelfT1Fe
countsQ10      wideshelfT1DFB
...

But whenever I use this script:

ddswideshelf <- DESeqDataSetFromTximport(txi.mat.wideshelf, colData = sampleTypewideshelf, ~condition)

I get an error saying:

Error in round(txi\$counts) :
non-numeric argument to mathematical function

my txi.mat.wideshelf dataset looks like this, with each sample represented by a Q number, and the counts containing the raw counts from my quant.sf output:

[1] "Domain"              "Phylum"              "Class"
[4] "Order"               "Family"              "Genus"
[7] "Species"             "Organism"            "TrinityID"
[10] "KO"                  "countsQ1"            "countsQ2"
[13] "countsQ3"            "countsQ4"            "countsQ7"
[16] "countsQ10"           "countsQ13"           "countsQ14"
[19] "countsQ15"           "countsQ16"           "countsQ17"
[22] "countsQ18"           "countsQ19"           "lengthQ1"
[25] "lengthQ2"            "lengthQ3"            "lengthQ4"
[28] "lengthQ7"            "lengthQ10"           "lengthQ13"
[31] "lengthQ14"           "lengthQ15"           "lengthQ16"
[34] "lengthQ17"           "lengthQ18"           "lengthQ19"
[37] "countsFromAbundance"

I checked online to see what the error's about and it seems like I'm trying to pass these data frames into a function that only recognizes vectors? If so, how may I convert my raw counts into a vector? Thanks!

0
Entering edit mode

See here:

https://bioconductor.org/packages/release/bioc/vignettes/tximport/inst/doc/tximport.html#DESeq2

The input to DESeqDataSetFromTximport is the output from tximport. Above you are trying to provide something else to this function. The paradigm is:

txi <- tximport(...)
dds <- DESeqDataSetFromTximport(txi, ...)


text:
#### 6.NS.C.7.c. Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation. For example, for an account balance of $-30$ dollars, write $|-30| = 30$ to describe the size of the debt in dollars.

• No tasks yet illustrate this standard.

#### 6.NS.C.7.d. Distinguish comparisons of absolute value from statements about order. For example, recognize that an account balance less than $-30$ dollars represents a debt greater than 30 dollars.

• No tasks yet illustrate this standard.

#### 7.NS.A.1.a. Describe situations in which opposite quantities combine to make 0. For example, a hydrogen atom has 0 charge because its two constituents are oppositely charged.

• No tasks yet illustrate this standard.

#### 7.NS.A.1.d. Apply properties of operations as strategies to add and subtract rational numbers.

• No tasks yet illustrate this standard.

#### 7.NS.A.2.c. Apply properties of operations as strategies to multiply and divide rational numbers.

• No tasks yet illustrate this standard.


text:
# Lecture 1: Tic-Tac-Toe theory -- an escape from the combinatorial chaos

## Lecturer:

Jozsef Beck, Rutgers

I just finished a more-than-600-page long book (Tic-Tac-Toe theory) about combinatorial games, focusing on tic-tac-toe like games. In this talk I will discuss the main theorems and the main techniques in a rather informal way. I will attempt to answer basic questions like:

1. why are games so difficult?
2. what are the hopeless problems?
3. what can contemporary combinatorics do?
4. what are the surprising consequences of tic-tac-toe theory"?

## Date:

Thu, 16/03/2006 - 16:00

Lecture Hall 2


text:
Kattis

# Disgruntled Judge

Once upon a time, there was an nwerc judge with a tendency to create slightly too hard problems. As a result, his problems were never solved. As you can image, this made our judge somewhat frustrated. This year, this frustration has culminated, and he has decided that rather than spending a lot of time constructing a well-crafted problem, he will simply write some insanely hard problem statement and just generate some random input and output files. After all, why bother having proper test data if nobody is going to try the problem anyway?

Thus, the judge generates a testcase by simply letting the input be a random number, and letting the output be another random number. Formally, to generate the data set with $T$ test cases, the judge generates $2T$ random numbers $x_1, \ldots , x_{2T}$ between $0$ and $10\, 000$, and then writes $T$, followed by the sequence $x_1, x_3, x_5, \ldots , x_{2T-1}$ to the input file, and the sequence $x_2, x_4, x_6, \ldots , x_{2T}$ to the output file.

The random number generator the judge uses is quite simple. He picks three numbers $x_1$, $a$, and $b$ between $0$ and $10\, 000$ (inclusive), and then for $i$ from $2$ to $2T$ lets

$x_ i = (a \cdot x_{i-1} + b) \bmod {10\, 001}.$

You may have thought that such a poorly designed problem would not be used in a contest of such high standards as nwerc. Well, you were wrong.

## Input

On the first line one positive number: the number of testcases, at most 100. After that per testcase:

• One line containing an integer n ($0 \le n \le 10\, 000$): an input testcase.

The input file is guaranteed to be generated by the process described above.

## Output

Per testcase:

• One line with an integer giving the answer for the testcase.

If there is more than one output file consistent with the input file, any one of these is acceptable.

Sample Input 1 Sample Output 1
3
17
822
3014

9727
1918
4110


text:
List of unequal random numbers [duplicate]

How can I generate a list of 10 random integer numbers in the range between 0 and 100 so that all random numbers are unequal?

• Jul 12 '16 at 14:32

2 Answers

RandomSample[Range[0, 100], 10]

• I was to much focused on RandomInteger, thanks.
– mrz
Jul 12 '16 at 14:33
• @happyfish if you saw my comment ... I'm sorry, that was unprofessional Jul 12 '16 at 16:05
• @Young I didn't, but I am curious now :P Jul 12 '16 at 16:09
• @Young I am also surprised by the votes and I agree with you. I feel guilty since I have almost never posted an answer that requires real effort in this community. Perhaps because as a beginner, my knowledge is a subset of many active users. It will still take me sometime to learn before I can make real contributions. Jul 12 '16 at 16:21
• @Young I'm of the opinion that for a lot of duplicate questions, especially from first time posters (not that that applies here), it's often better to answer and let others worry about marking duplicates. I feel like it may encourage them to come back more, and our community here is not so large as to discourage newbies Jul 13 '16 at 1:14

happy fish gave the right answer, but sometimes it's fun to go oldschool. You can use rejection sampling and a While loop

list = {};
While[Length@list <= 10,
list = DeleteDuplicates@Append[list, RandomInteger[100]]]

• I like this approach as you don't need to generate the full list. A small modification might save a lot of iterations: While[(l = Length[ids]) < n, ids = DeleteDuplicates[Join[ids, RandomInteger[max, n - l]]]] Jan 27 '21 at 9:11


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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

Hamiltonian S^1 actions with isolated fixed points

I have in mind the following question for some time. Is there an example of a compact symplectic manifold with a Hamiltonian S^1 action with isolated fixed points, that does not admit a compatible S^1 invariant Kahler strucutre? One would say, of course there should be such an example. But I have not seen any...

Added. Apperently this is an open problem.

-
 wiki'd, per policy for known open problems. – Scott Morrison♦ Nov 16 2009 at 21:56

Lerman has constructed such an example see http://arxiv.org/abs/dg-ga/9601012v1 , well it has an isolated fixed point but also some fixed connected submanifolds, so maybe not exactly what you want.

The construction has been later revisited by Kim http://www.mathjournals.org/mrl/2002-009-004/2002-009-004-002.pdf

(note to editors: I guess the creation of 'torus-action' tag migh be useful.)

-
 Isn't it toric-action? – Ilya Nikokoshev Oct 24 2009 at 18:17 Actually, may be circle-action or localization? – Ilya Nikokoshev Oct 24 2009 at 18:18 It's definitely 'torus-action', see e.g. this famous book by Audin books.google.fr/books?id=uKXO9BDsk_kC That said we might also need 'toric-variety' or 'toric-geometry' at some point, but that's something else. – Thomas Sauvaget Oct 24 2009 at 18:22 Thanks a lot for both refferences! But still I want an action such that all fixed points are isolated. Though indeed it is nice to know that there are non-Kahler examples with at least one isolated point. – Dmitri Oct 24 2009 at 18:28

Sue Tolman has shown that there are non-Kahler 6-dimensional manifolds admitting Hamiltonian 2-torus actions all of whose fixed points are isolated ( http://arxiv.org/abs/dg-ga/9511007 ) . This might be a good first place to look to see if one of the components of the action satisfy the criteria you want.

Interestingly, Karshon has shown ( in dg-ga/9510004 [sorry, I can only post one link as a new user]) that if a 4-manifold admits a Hamiltonian circle action, it must be Kahler.

-
 I now this article. But surprisingly, as far as I understand, you can not just take a subgroup S^1 in this example to get something that does not admit a compatible Kahler metric. At least this is what I understood from Dusa McDuff. Apperently you do need a T^2 in this example... – Dmitri Oct 24 2009 at 18:49 Surprising and interesting! Perhaps the next thing to do is do a literature search for papers that cite Tolman's paper. – C. Lee Oct 24 2009 at 18:56


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# A dishonest shopkeeper uses a weight of 800 gm for a kg and professes to sell his good at cost price. His profit is - Mathematics

MCQ

A dishonest shopkeeper uses a weight of 800 gm for a kg and professes to sell his good at cost price. His profit is

• 20%

• 21%

• 24%

• 25%

#### Solution

25%

Explanation:

Let CP of 1000 g item = ₹ 1000

CP of 800 g item = ₹ 800

SP of 800 g item = ₹ 1000

P%=(1000-800)/800xx100

200/800xx100=25%

Concept: Profit and Loss (Entrance Exam)
Is there an error in this question or solution?


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# 6.11. Glossary¶

Barabási-Albert Model: The Barabási–Albert (BA) model is an algorithm for generating random scale-free networks using a preferential attachment mechanism.

Complementary CDF: $$CCDF(x) ≡ 1 − CDF(x)$$

Cumulative Distribution Function (CDF) A function which maps from a value, $$x$$, to the fraction of values less than or equal to $$x$$.

Explanatory Model: Is a model that gives a useful description of why and how a phenomenon is the way it is.

Growth: Instead of starting with a fixed number of vertices, the BA model starts with a small graph and adds vertices one at a time.

Heavy-tailed Distributions: In probability theory, heavy-tailed distributions are probability distributions whose tails are not exponentially bounded.

Preferential Attachment: Is any of a class of processes in which some quantity is distributed among a number of individuals according to what they already have.

Probability Mass Function (PmF): A function that maps from each value to it’s probabilities.

Power Law: A distribution follows this law if $$PMF(k) ∼ k−α$$ where PMF(k) is the fraction of nodes with degree k, α is a parameter, and the symbol ∼ indicates that the PMF is asymptotic to k−α as k increases.

Scale-Free Network: A network whose degree distribution follows a power law, at least asymptotically.

Standard Deviation: A quantity calculated to indicate the extent of deviation for a group as a whole.

WS Model: A model that has characteristics of a small world network, like the data, but it has low variability in the number of neighbors from node to node, unlike the data.


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Probabilistic-Numerics.org

# Tübingen Manifesto: Probabilistic Numerics and Probabilistic Programming

### By Michael A Osborne, 2014-09-01 00:00:00 +0200

We in Probabilistic Numerics face many unanswered questions in growing the field. Our roundtable in Tübingen aimed to bring together our new community to begin to address some of these questions. This is another of a sequence of posts that attempt to collate some of what we spoke about, and to, hopefully, provoke further discussion.

We were very fortunate to benefit from the coincidental presence of Noah Goodman in Tübingen, who generously spent an afternoon at the roundtable talking with us. Noah, of course, is a founding and deeply committed member of the Probabilistic Programming community. Noah had many fascinating reflections on developments within Probabilistic Programming (ProbProg), and how they might connect with Probabilistic Numerics (ProbNum).

ProbProg seems to be largely about allowing the design of complex generative models, and then ensuring that uncertainty is properly propagated in producing posteriors using the model. This is certainly not the same as the propagation required to manage uncertainty introduced through the use of finite-precision or approximate numerical procedures, but there are some commonalities. Below are a number of items drawing out some of the links between the fields.

## Meta-Reasoning

ProbNum offers the attractive potential of performing decision-theoretic management of systems of probabilistic numerical algorithms. That is, ProbNum could be used to select which part of a numerical pipeline to refine, that is, to decide when to stop a numerical algorithm achieving accuracy you don’t need. Noah was interested in this process, which he likened to meta-reasoning, and recommended making the connection explicit.

Noah also posed the excellent question of how much computation it was worth spending to perform this meta-reasoning. Of course, this is a question we couldn’t readily answer. Would a greedy selection of the numerical algorithm to spend the next unit of computation on be sufficient, or would more sophisticated strategies be required? At this point, Noah quoted Stuart Russell in recommending that, as a rule, “you should only do as much meta-reasoning as regular reasoning”. This seems sensible enough to me!

## Lazy Evaluation

Noah also mentioned links between ProbNum’s approach of returning numerical results of flexible degrees of accuracy to the notion of lazy evaluation common in functional languages like Haskell. In either case, only as much computation is performed as is absolutely required. For lazy evaluation, what is required can be determined exactly, whereas the ProbNum approach would treat this as a question to be answered with decision theory.

Another fundamental concept in languages that we discussed was that of overloading. Treating, for example, a probability distribution over integers as a type that generalises the type int, ProbNum might well benefit from being implemented using overloading. That is, functions could be overloaded to permit the optional input and output of variances in addition to the usual input and output estimates.


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### 3.7.227. Schur number

Denotes that a constraint was used for solving Schur problems. Given a non-negative integer $k$, the Schur number $S\left(k\right)$ is the largest integer $n$ for which the set $\left\{1,2,\cdots ,n\right\}$ can be partitioned into $k$ sets ${S}_{1},{S}_{2},\cdots ,{S}_{k}$ such that $\forall i\in \left[1,k\right]:i\in {S}_{i}⇒i+i\notin {S}_{i}$.


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topos theory

# Contents

## Separated presheaf

### Idea

The condition that a presheaf be a sheaf may be seen as a condition of unique existence. A presheaf is separated if it satisfies the uniqueness part.

### Definition

Let $S$ be a site.

Recall that a sheaf on $S$ is a presheaf $A\in {\mathrm{PSh}}_{S}$ such that for all local isomorphisms $Y\to X$ the induced morphism ${\mathrm{PSh}}_{S}\left(X,A\right)\to {\mathrm{PSh}}_{S}\left(Y,A\right)$ (under the hom-functor ${\mathrm{PSh}}_{S}\left(-,A\right)$) is an isomorphism. (For an arbitrary class of morphisms $V$, the corresponding condition is called being a local object.)
It is sufficient to check this on the dense monomorphisms instead of all local isomorphisms. This is equivalent to checking covering sieves.

###### Definition

A presheaf $A\in \mathrm{PSh}\left(S\right)$ is called separated (or a monopresheaf) if for all local isomorphisms $Y\to X$ the induced morphism $\mathrm{Hom}\left(X,A\right)\to \mathrm{Hom}\left(Y,A\right)$ is a monomorphism.

More generally, for a class $V$ of arrows in a category $C$, an object $A\in C$ is $V$-separated if for all morphisms $Y\to X$ in $V$, the induced morphism $\mathrm{Hom}\left(X,A\right)\to \mathrm{Hom}\left(Y,A\right)$ is a monomorphism.

###### Remark

As for sheaves, it is sufficient to check the separation condition on the dense monomorphisms, hence on the sieves.

For $\left\{{p}_{i}:{U}_{i}\to U\right\}$ a covering family of an object $U\in S$, the condition is that if $a,b\in A\left(U\right)$ are such that for all $i$ we have $A\left({p}_{i}\right)\left(a\right)=A\left({p}_{i}\right)\left(b\right)$ then already $a=b$.

###### Remark

The definition generalizes to any system of local isomorphisms on any topos, such as that obtained from any Lawvere-Tierney topology, or equivalently any subtopos.

### Properties

###### Proposition

The full subcategory

$i:\mathrm{Sep}\left(S\right)↪\mathrm{PSh}\left(S\right)$i : Sep(S) \hookrightarrow PSh(S)

of separated presheaves in a presheaf category is

Being a reflective subcategory means that there is a left adjoint functor to the inclusion

$\left({L}_{\mathrm{sep}}⊣i\right):\mathrm{Sep}\left(S\right)\stackrel{\stackrel{{L}_{\mathrm{sep}}}{←}}{↪}{\mathrm{PSh}}_{S}\phantom{\rule{thinmathspace}{0ex}}.$(L_{sep} \dashv i) : Sep(S) \stackrel{\overset{L_{sep}}{\leftarrow}}{\hookrightarrow} PSh_S \,.
###### Definition

For $A\in {\mathrm{PSh}}_{S}$ the separafication ${L}_{\mathrm{sep}}A$ of $A$ is the presheaf that assigns equivalence classes

${L}_{\mathrm{sep}}A:U↦A\left(U\right)/{\sim }_{U}\phantom{\rule{thinmathspace}{0ex}},$L_{sep}A : U \mapsto A(U)/\sim_U \,,

where ${\sim }_{U}$ is the equivalence relation that relates two elements $a\sim b$ iff there exists a covering $\left\{{p}_{i}:{U}_{i}\to U\right\}$ such that $A\left({p}_{i}\right)\left(a\right)=A\left({p}_{i}\right)\left(b\right)$ for all $i$.

This construction extends in the evident way to a functor

${L}_{\mathrm{sep}}:\mathrm{PSh}\left(S\right)\to \mathrm{Sep}\left(S\right)\phantom{\rule{thinmathspace}{0ex}}.$L_{sep} : PSh(S) \to Sep(S) \,.
###### Proposition

This functor ${L}_{\mathrm{sep}}$ is indeed a left adjoint to the inclusion $i:\mathrm{Sep}\left(S\right)↪\mathrm{PSh}\left(S\right)$.

###### Proof

Let $A\in \mathrm{PSh}\left(S\right)$ and $B\in \mathrm{Sep}\left(S\right)↪\mathrm{PSh}\left(S\right)$. We need to show that morphisms $f:A\to B$ in ${\mathrm{PSh}}_{C}$ are in natural bijection with morphisms ${L}_{\mathrm{sep}}A\to B$ in $\mathrm{Sep}\left(S\right)$.

For $f$ such a morphism and ${f}_{U}:A\left(U\right)\to B\left(U\right)$ its component over any object $U\in S$, consider any covering $\left\{{U}_{i}\to U\right\}$, let $S\left({U}_{i}\right)\to U$ be the corresponding sieve and consider the commuting diagram

$\begin{array}{ccc}\left\{\left({a}_{i}\in A\left({U}_{i}\right)\right)\mid \cdots \right\}& \to & \left\{\left({b}_{i}\in F\left({U}_{i}\right)\right)\mid \cdots \right\}\\ ↑& & ↑\\ A\left(U\right)& \stackrel{{f}_{U}}{\to }& B\left(U\right)\end{array}$\array{ \{(a_i \in A(U_i)) | \cdots \} &\to& \{(b_i \in F(U_i)) | \cdots \} \\ \uparrow && \uparrow \\ A(U) &\stackrel{f_U}{\to}& B(U) }

obtained from the naturality of ${\mathrm{PSh}}_{S}\left(S\left({U}_{i}\right)\to U,A\stackrel{f}{\to }B\right)$.

If for $a,a\prime \in A\left(U\right)$ two elements that are not equal their restrictions to the cover become equal in that $\forall i:a{\mid }_{{U}_{i}}=a\prime {\mid }_{{U}_{i}}$, then also $f\left(a{\mid }_{{U}_{i}}\right)=f\left(a\prime {\mid }_{{U}_{i}}\right)$ and since the right vertical morphism is monic there is a unique $b\in B\left(U\right)$ mapping to the latter. The commutativity of the diagram then demands that $f\left(a\right)=f\left(a\prime \right)=b$.

Since this argument applies to all covers of $U$, we have that ${f}_{U}$ factors uniquely through the projection map $A\left(U\right)\to A\left(U\right)/{\sim }_{U}=:{L}_{\mathrm{sep}}\left(U\right)$ onto the quotient. Since this is true for every object $U$ we have that $f$ factors uniquely through $A\to {L}_{\mathrm{sep}}A$.

## Biseparated presheaf

### Idea

Often one is interested in separated presheaves with respect to one coverage that are sheaves with respect to another coverage. These are called biseparated presheaves .

This typically arises if a reflective subcategory

$C\stackrel{\stackrel{}{←}}{↪}\mathrm{Sh}\left(S\right)$C \stackrel{\stackrel{}{\leftarrow}}{\hookrightarrow} Sh(S)

of a sheaf topos is given. This is the localization at a set $W$ of morphisms in $\mathrm{Sh}\left(S\right)$, with $C$ the full subcategory of all local objects $c$: objects such that ${\mathrm{Sh}}_{\left(}S\right)\left(w,c\right)$ is an isomorphism for all $w\in W$. A $W$-separated object is then called a biseparated presheaf on $S$ and their collection $\mathrm{BiSep}\left(S\right)$ factors the reflective inclusion as

$C\stackrel{←}{↪}\mathrm{BiSep}\left(S\right)\stackrel{←}{↪}\mathrm{Sh}\left(S\right)\phantom{\rule{thinmathspace}{0ex}}.$C \stackrel{\leftarrow}{\hookrightarrow} BiSep(S) \stackrel{\leftarrow}{\hookrightarrow} Sh(S) \,.

### Definition

###### Definition

A bisite is a small category $S$ equipped with two coverages: $J$ and $K$ such that $J\subset K$.

A presheaf $A\in {\mathrm{PSh}}_{S}$ is called $\left(J,K\right)$-biseparated if it is

• a sheaf with respect to $J$;

• a separated presheaf with respect to $K$.

Write

${\mathrm{BiSep}}_{\left(J,K\right)}\left(S\right)↪{\mathrm{Sh}}_{J}\left(S\right)↪\mathrm{PSh}\left(S\right)$BiSep_{(J,K)}(S) \hookrightarrow Sh_J(S) \hookrightarrow PSh(S)

for the full subcategory on biseparated presheaves.

### Properties

###### Proposition

Biseparated presheaves form a reflective subcategory of all sheaves

${\mathrm{BiSep}}_{\left(J,K\right)}\left(S\right)\stackrel{\stackrel{{L}_{\mathrm{sep}}^{K}}{←}}{↪}{\mathrm{Sh}}_{J}\left(S\right)\phantom{\rule{thinmathspace}{0ex}}.$BiSep_{(J,K)}(S) \stackrel{\stackrel{L^K_{sep}}{\leftarrow}}{\hookrightarrow} Sh_J(S) \,.

See quasitopos for the proof.

## References

The general theory of biseparated presheaves and Grothendieck quasitoposes is in section C.2.2 of

A concrete description of separafication appears on page 43 of

• Angelo Vistoli, Notes on Grothendieck topologies, fibered categories and descent theory (pdf)

category: sheaf theory

Revised on March 6, 2013 19:42:33 by Zoran Škoda (161.53.130.104)


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## Tuesday, September 28, 2021

### Science/tech consulting in creative arts

I've watched the first two episodes of the new adaptation of Foundation.  It surely looks gorgeous, though there are some script challenges (even apart from the challenge of trying to adapt an enormous book series that was always long on ideas and short on character development).   The issues I've spotted seem mostly to be ones of poor script editing for consistency.  (The emperor of the Galactic Empire says in the first episode that the imperial population is 8 trillion, and then in the second episode a character says that the core worlds alone have a population of 40 trillion.  The latter number is more reasonable, given the size of Asimov's empire.)

Watching this, I again think it would be great fun to do scientific/technical consulting for TV, movies, and even books. I'm on the list for the Science and Entertainment Exchange, though all I've ever done is give a tiny bit of feedback to a would-be author.  (My expertise probably looks too narrow, and not living in southern California seems to be a major filter.)

It feels like there are some similarities to the role of science in public policy.  In the creative productions, science can contribute (and these media can be a great way of getting scientific ideas out into the public), but in the end plot and what can practically be implemented will always drive the final product.  In policy, science and technical knowledge should definitely factor in when relevant, but fundamentally there are social and political factors that can overwhelm those influences in decision-making.  Now back to our regularly scheduled psychohistorical crisis....

## Wednesday, September 22, 2021

### DOE Experimental Condensed Matter PI meeting, Day 3

Here are some tidbits from the last day of this year's meeting.  (I didn't really get to see the other posters in my own poster session, so apologies for missing those.  For the curious:  the meeting attendees alternate between posters and 15 minute talks from year to year.)

• It's been known for a while that combining magnetism with topological insulator materials can lead to a rich phase diagram.  Tuning composition is a powerful tool.  Likewise, the van der Waals nature of these systems mean that it's possible to look systematically through a family of related materials.
• Tuning composition in flat-band kagome metals is also of interest.
• I had not appreciated just how important specific crystal growth approaches (e.g., rapid quenching vs. slow annealed cooling) are to the properties of some magnetic/topological materials, such as Fe5GeTe2.
• Strain can be a powerful tool for tuning electronic topology in some materials such as ZrTe5, and driving certain phonon modes via laser offers the potential of controlled switching of topological properties.
• Quantum oscillations (e.g., magnetization as a function of 1/H) are a conventional way to learn about Fermi surfaces, and it is always bizarre when that kind of response shows up in a correlated material that is nominally an insulator, or in thermal transport but not electrical transport.
• Speaking of quantum oscillations in insulators, how about thermal transport in the spin liquid phase of $\alpha$-RuCl3?  Looks like some kind of bosonic edge mode is responsible.
• If transition metal dichalcogenides are starting to bore you, perhaps you'd be more interested in trichalcogenides, which can be grown as individual 1D chains within carbon and boron nitride nanotubes.
Thanks to everyone for making the meeting enjoyable and informative, even if we couldn't get together in a random Marriott in Maryland this time.

## Tuesday, September 21, 2021

### DOE Experimental Condensed Matter PI meeting, Day 2

More highlights from the meeting.  Office hours for my class conflicted with a couple of the talks, so these are briefer than I would've liked.

• It is possible to use coherent x-ray scattering to look at time variations in the domain structure of an antiferromagnet.  In the magnetic diffraction pattern there is speckle near the magnetic Bragg spots that bops around as the domain structure fluctuates.
• Amorphous magnetic alloys can show some really wild spin textures.
• By growing a few nanometers of a paramagnetic metal, Bi2Ir2O7, on top of an insulating spin ice, Dy2Ti2O7, it's possible to get enough coupling that field-driven spin ice transitions can generate magnetoresistance signatures in the metal layer.
• Square planar nickelates can look a lot like the copper oxide superconductors in terms of band dispersion and possible "strange" metallicity.
• Some rare-earth intermetallic compounds can have an impressively rich magnetic phase diagram.
• I learned that some pyrochlore iridates can exhibit a kind of topological metallic state with giant anomalous Hall response.
• I had not previously appreciated how wild it is that one can engineer ferroelectric response in stacks of 2D materials that are not intrinsically ferroelectric, as in hBN or even WSe2.
• Ultrasound attenuation can be a heck of a tool for looking at superconductivity and other electronic transport properties.
• Strontium titanate remains a really interesting test case for understanding exotic superconductivity, with its superconducting dome as a function of doping, very low carrier density, and incipient ferroelectricity.  Phonons + paraelectric fluctuations + spin-orbit coupling appear to be the big players
• A related system in the sense of near-ferroelectricity and low carrier density is at interfaces of KTaO3.
• This experiment in graphene/hBN/graphene stacks (with encapsulating hBN and graphite top and bottom gates) is an extremely pretty, tunable demonstration of superfluidity of bilayer excitons, an effect previously seen in one limit in GaAs systems.

## Monday, September 20, 2021

### DOE Experimental Condensed Matter PI meeting, Day 1

Somehow I found the first day of the virtual meeting more exhausting than when we do these things in person, probably because I had to go teach in the middle of the event.  A sampling of highlights:

• Rather than relying on relatively crude methods to create defect centers in diamond for quantum sensing (or qubit purposes), one can use chemistry to build transition metal complexes with designer ligand fields (and hence energy level structures), as demonstrated here.
• I know understand better why it has historically been so difficult to demonstrate, experimentally, that the quasiparticles in the fractional quantum Hall effect obey fractional (anyonic) statistics.  In an interferometer, it's critical to use screening (from top and bottom electron gases that act like capacitor plates) to reduce Coulomb interactions between edge states and the bulk. Once that's done, you can see clear evidence of fractional (anyonic) phase slips.
• Some truly exceptional people can still do research even while being a university president.  At very low energies in an Ising ferromagnet with an in-plane magnetic field, hyperfine interactions can lead to hybridization of magnetic levels and the formation of "electronuclear" spin excitations.
• Ultraclean ABC-stacked graphene trilayers can show remarkably rich response, dominated by strong electron-electron interaction effects.
• High quality crystal growth can drastically lower the defect densities in transition metal dichalcogenides.  That makes it possible to construct bilayers of WSe2, for example, that can host apparent excitonic condensates.  Similar physics can be seen in MoSe2/WSe2 bilayers, where it is clear that exciton-exciton interactions can be very strong.
• Pulling and pushing on a sample can lead to elastocaloric effects (like when a rubber band cools upon being stretched), and these can reveal otherwise hidden properties and phase transitions.
More tomorrow.  (One fun idea from a colleague:  Perhaps the program officers have hidden a secret easter egg token somewhere in the Gather virtual poster area, and whoever finds it gets a bonus award supplement to support a summer undergrad....)

### DOE Experimental Condensed Matter Physics PI meeting, 2021

Every two years, the US Department of Energy Experimental Condensed Matter Physics program has a principal investigator meeting, and I've written up highlights of these for a while (for 2019, see a, b, c;  2017, see abc; for 2015 see abc; for 2013 see ab).

The meetings have always been very educational for me.  They're a chance to get a sense of the breadth of the whole program and the research portfolio.  It is unfortunate that the covid pandemic has forced this year's meeting to be virtual.  I'll do my best to summarize some tidbits in posts over the next three days.

## Friday, September 17, 2021

### Moiré materials and the Mott transition

There are back-to-back papers in Nature this week, one out of Columbia and one out of Cornell, using bilayers of transition metal dichalcogenides to examine the Mott transition.  (Sorry for the brevity - I'm pressed for time right now, but I wanted to write something....)

As I described ages ago in here, imagine a lattice of sites, each containing one electron.  While quantum statistics would allow each site to be doubly occupied (thanks to spin), if the on-site electron-electron repulsion $U$ is sufficiently strong (large compared to the kinetic energy scale $t$ associated with hopping between neighboring sites), then the interacting system will be an insulator even though the non-interacting version would be a metal.  Moving away from this half-filling condition, you can get conduction, just as having an empty site allows those sliding tile puzzles to work.

As discussed here, in bilayers of 2D materials can lead to the formation of a moiré lattice, where the interlayer interactions result in an effective periodic array of potential wells.  The Columbia folks got a moiré pattern by using a 4-5 degree twisted bilayer of WSe2, while the Cornell folks instead used an aligned bilayer of MoTe2 and WSe2 (where the moiré comes from the differing lattice constants).  In both cases, you end up with a triangular moiré lattice (encapsulated in hBN to provide a clean charge environment and protection from the air).

The investigators are able to tune the systems in multiple ways.  With overall gate voltage, they can capacitively tune the "filling", the ratio of number of "free" charges to number of moiré lattice sites.  By adjusting top gate vs. back gate, they can tune the vertical electric field across the bilayer, and that is a way of tuning interactions by pushing around localized wavefunctions for the lattice sites.

Both groups find that they can tune in/out of a Mott insulating phase when they're at one carrier per moiré lattice site.  Interestingly, both groups see that the Mott transition is continuous (second-order) - there is no sudden onset of insulating response as a function of tuning either knob.  Instead, there appears to be quantum critical scaling, and regions of linear-in-$T$ temperature dependence of the resistivity (a possible indicator of a strange metal) on either side of the insulating region.  The Cornell folks are able to do magnetic circular dichroism measurements to confirm that the transition does not involve obvious magnetic ordering.

This is very pretty work, and it shows the promise of the moiré lattice approach for studying fundamental issues (like whether or not the Mott transition in a triangular lattice is continuous).  I'm sure that there will be much more to come in these and related systems.

## Monday, September 06, 2021

### What is the spin Hall effect?

The Hall Effect is an old (1879) story, told in first-year undergraduate physics classes for decades. Once students are told about the Lorentz force law, it's easy to make a handwave classical argument that something like the Hall Effect has to exist:  Drive a current in a conductor in the presence of a magnetic induction $\mathbf{B}$.  Charged particles undergo a $q \mathbf{v} \times \mathbf{B}$ force that pushes them transverse to their original $\mathbf{v}$ direction.  In a finite slab of material with current perpendicular to $\mathbf{B}$, the particles have to pile up at the transverse edge, leading to the development of a (Hall) voltage perpendicular to the direction of current flow and the magnetic induction.  You can measure the Hall voltage readily, and it's used for sensing magnetic fields, as well as figuring out charge carrier densities in materials.

The spin Hall effect, in contrast, is a much newer idea.  It was first proposed by Dyakonov and Perel in 1971 as an extrinsic effect (that is, induced by scattering from impurities in a material), and this was revisited in 1999 by Hirsch and others.  It's also possible to have an intrinsic spin Hall effect (proposed here and here) due just to the electronic structure of a material itself, not involving impurities.

So what is the SHE?  In some non-magnetic conductors, in the absence of any external magnetic field, a charge current (say in the $+x$ direction) results in a build-up of electrons with spin polarized up (down) along the $z$ direction along the positive (negative) $y$ edge of the material, as shown in the bottom left drawing of the figure.  Note that there is no net charge imbalance or transverse voltage - just a net spin imbalance.

The SHE is a result of spin-orbit coupling - it's fundamentally a relativistic effect (!).  While we static observers see only electric fields in the material, the moving charge carriers in their frame of reference see effective magnetic fields, and that affects carrier motion.  In the extrinsic SHE, scattering of carriers from impurities ends up having a systematic spin dependence, so that spin-up carriers are preferentially scattered one way and spin-down carriers are scattered the other.  In the intrinsic SHE, there ends up being a spin-dependent term in the semiclassical velocity that one would get from the band structure, because of spin-orbit effects.  (The anomalous Hall effect, when one observes a Hall voltage correlated with the magnetization of a magnetic conductor, is closely related.  The net charge imbalance shows up because the populations of different spins are not equal in a ferromagnet.)  The result is a spin current density $\mathbf{J}_{\mathrm{s}}$ that is perpendicular to the charge current density $\mathbf{J}_{\mathrm{c}}$, and is characterized by a (material-dependent) spin Hall angle, $\theta_{\mathrm{SH}}$, so that $J_{\mathrm{s}} = (\hbar/2e)\theta_{\mathrm{SH}}J_{\mathrm{c}}$.

There is also an inverse SHE:  if (appropriately oriented) spin polarized charge carriers are injected into a strong spin-orbit coupled non-magnetic metal (say along $+x$ as in the bottom right panel of the figure), the result is a transverse ($y$-directed) charge current and transverse voltage build-up.  (It's this inverse SHE that is used to detect spin currents in spin Seebeck effect experiments.)

The SHE and ISHE have attracted a lot of interest for technological applications.  Generating a spin current via the SHE and using that to push around the magnetization of some magnetic material is called spin orbit torque, and here is a recent review discussing device ideas.

## Wednesday, September 01, 2021

### Rice University physics faculty search in experimental quantum science and technology

The Department of Physics and Astronomy at Rice University invites applications for tenure-track faculty positions in the broad area of experimental quantum science and technology. This encompasses quantum information processing, quantum sensing, quantum communication, quantum opto-mechanics, and quantum simulation in photonic, atomic/ionic, quantum-material, and other solid-state platforms. We seek outstanding scientists whose research will complement and extend existing activities in these areas within the Department and across the University. In addition to developing an independent and vigorous research program, the successful applicants will be expected to teach, on average, one undergraduate or graduate course each semester, and contribute to the service missions of the Department and University. The Department anticipates making appointments at the assistant professor level. A Ph.D. in physics or related field is required.

Beginning September 1, 2021, applications for this position must be submitted electronically at apply.interfolio.com/92734 .

Applications for this position must be submitted electronically. Applicants will be required to submit the following: (1) cover letter; (2) curriculum vitae; (3) statement of research; (4) statement on teaching; (5) statement on diversity, mentoring, and outreach; (6) PDF copies of up to three publications; and (7) the names, affiliations, and email addresses of three professional references. Rice University, and the Department of Physics and Astronomy, are strongly committed to a culturally diverse intellectual community. In this spirit, we particularly welcome applications from all genders and members of historically underrepresented groups who exemplify diverse cultural experiences and who are especially qualified to mentor and advise all members of our diverse student population.We will begin reviewing applications November 15, 2021. To receive full consideration, all application materials must be received by January 1, 2022. The expected appointment date is July, 2022.

Rice University is an Equal Opportunity Employer with commitment to diversity at all levels, and considers for employment qualified applicants without regard to race, color, religion, age, sex, sexual orientation, gender identity, national or ethnic origin, genetic information, disability or protected veteran status.