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# IIT question help me

1. Apr 30, 2005

### nomorevishnu

IIT question......help me....

hi all.....

I have a block in my hand.....i am pressing it against my bedroom wall
i have applied enough force so that the block doesnt fall down......now...i have been asked a question based on this by the great IIT people....they asked me.....which force is capable of producing a torque if i draw a free body diagram.....i couldnt think about torque here at all....but the answer they gave is that the normal force is capable of producing torque here...how?
can anyone explain??????

vishnu here

2. Apr 30, 2005

### whozum

As far as I can see,theres no reason the normal force can produce a torque and not your hand.

3. Apr 30, 2005

### nishant

this question came in this year's screening,although it was not mentioned but the force was aplied symmytrically on the cube due to which the force could not produce a torque about the centre of mass,and out of all the options the answer which uwrote was the only one relevant.

4. Apr 30, 2005

### xanthym

$$\setlength{\unitlength}{0.002cm} \begin{picture}(6000,6000)(1500,-6000) \thinlines \linethickness{1pt} \color{red} \put(1801,-361){\line( 1, 0){2100}} \put(3901,-361){\line( 0,-1){5400}} \put(3901,-5761){\line(-1, 0){2100}} \put(3901,-1561){\line( 1, 0){1800}} \put(5701,-1561){\line( 0,-1){3000}} \put(5701,-4561){\line(-1, 0){1800}} \put(4801,-3061){\vector( 1, 1){900}} \put(7201,-2161){\vector(-1, 0){1500}} \put(6001,-2011){\line(-2,-1){300}} \put(5701,-2161){\line( 2,-1){300}} \put(5701,-2161){\line(-2,-1){300}} \put(5701,-2161){\line(-1,-2){150}} \put(4930,-2600){\color{blue}\vec{R}} \put(6451,-2050){\color{blue}\vec{F}} \put(4801,-3361){P} \put(5400,-2050){Q} \put(7351,-2250){HAND} \put(6800,-3100){\color{blue}\textsf{\LARGE Total NET Torque = 0}} \put(6800,-3450){\color{blue}\Large However, HAND force F applies torque \,  \vec{\tau}_{ \left ( F \ about \ P \right ) }  about point P.} \put(7351,-1600){\color{blue} \vec{\tau}_{ \left ( F \ about \ P \right ) } \ = \ \vec{R}\times\vec{F} \ \neq \ 0 } \put(4400,-1411){BRICK} \put(2401,-2161){WALL} \put(1500,-6000){.} \end{picture}$$

Of course, the Total NET Torque is ZERO (0). However, any individual force can apply a torque about a particular point. Shown above is Point "P" near the Brick's center about which HAND Force "F" applies torque "τF about P". No matter where Force "F" is applied, a point can generally be found about which the torque from that force is non-zero.

~~

Last edited: Apr 30, 2005
5. Apr 30, 2005

### nomorevishnu

well i m not able to picturise why the normal force produces torque....

vishnu....

6. Apr 30, 2005

### xanthym

Because torque "τ" is mathematically determined by TWO points: 1) Point "Q" where Force "F" is applied, and 2) Reference Point "P" about which "τ" is calculated. Force "F" might be normal to the Brick's surface, but it's not normal to the vector "R" pointing from Reference Point "P" to Application Point "Q".

~~

Last edited: Apr 30, 2005
7. Apr 30, 2005

### whozum

Isnt that an unclear question?

8. Apr 30, 2005

### nishant

well vishnu read my post ,you will understand?

9. May 1, 2005

### xanthym

Torque vector $$\vec{\tau}$$ is only defined when 3 items are specified: 1) Reference Point "P", 2) Application Point "Q", and 3) Force vector $\vec{F}$. Speaking generally about a "normal force" doesn't provide enough information to properly determine the torque vector. If you completely specify all 3 items above, you should understand the torque produced.

~~

Last edited: May 1, 2005
10. May 1, 2005

### siddharth

First of all, take the torqure about the center of the block. It was not specified in the IIT question paper about which point you should take the torque, but the answer assumes you take the torque about the center of the block.

Further the Force applied on the block by the hand is acting along the center of the block. Also there is friction between the wall and the block (Otherwise the block would have fallen).

Now the question is clear, draw the free-body diagram of the block and equate torques about the center of the block. With a little thought the answer should be obvious.

11. May 1, 2005

### Dr.Brain

nomorevishnu and siddharth did u clear IIT screening?

I couldnot...Now hopes on BITSAT DCE and AIEEE

12. May 1, 2005

### nomorevishnu

i just wanna know this....what all can the normal force do in the present situation....explain relating normal force and possibilty of toque....please

13. May 1, 2005

### siddharth

Take torque about the center of the mass. Let the side length of the block be 'a'.
Now the torque due to friction is (a/2)*f. The torque due to gravity and the applied Force is 0. The only other force is the Normal force. We also know the net torque is zero because the object will not rotate. Therefore there exists a torque due to the Normal force which opposes the torque due to friction. In fact the line of action of the Normal force will not be through the center.

14. May 3, 2005

### nishant

in the original question{in screening} it was not mentioned that you had to take torque about the centre of mass,and may be that was the only thing that was confusing

15. May 4, 2005

### nomorevishnu

well y is the normal force not acting through the centre of mass????

16. May 4, 2005

### nishant

it is absolutely not necessary for the normal force to always act about the centre of mass.
By the way r u in 11th

17. May 5, 2005

### nomorevishnu

then how is the normal force acting when the force is acting through the centre....???? how does it balance the other force if it doesnt act through the same line opposite to it????or is it acting in such a way that its component cancels "F"??????.....doubtful yaar..please help

18. May 5, 2005

### nishant

ncert khlolo,and see the newton's law

19. May 6, 2005

### nomorevishnu

well nishant

i joined this website because i wanted someone to help me....if u are gonna tell me to read the book...i would just ask my senior abt a good book and obtain it.......i thought for the question i ask i would get a 100%
answer in a way such that a teacher is helping me....and let that be anyone but convincing......so if i could get an explanation for my previous question about normal it would be useful yaar.......


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# Evaluate the equation: ${{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}-8$ .A.$(a-b-2)({{a}^{2}}+{{b}^{2}}-2ab-4a+2a-2b)$B.${{(a+b)}^{3}}$C.${{(a-b)}^{3}}$D.$(a+b-2)({{a}^{2}}+{{b}^{2}}+2ab+4+2a+2b))$

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Hint: Take ${{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}-8$ and use the formulae ${{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}$ and ${{x}^{3}}-{{y}^{3}}=(x-y)({{x}^{2}}+{{y}^{2}}+xy)$ . Try it, you will get the answer.

In Mathematics, an equation is a statement that asserts the equality of two expressions. The word equation and its cognates in other languages may have subtly different meanings; for example, in French an equation is defined as containing one or more variables, while in English any equality is an equation.

Solving an equation containing variables consists of determining which values of the variables make the equality true. Variables are also called unknowns and the values of the unknowns that satisfy the equality are called solutions of the equation. There are two kinds of equations: identities and conditional equations. An identity is true for all values of the variable. A conditional equation is only true for particular values of the variables.
Single variable equation can be $x+2=0$.

An equation is written as two expressions, connected by an equals sign ("$=$"). The expressions on the two sides of the equals sign are called the "left-hand side" and "right-hand side" of the equation.

The most common type of equation is an algebraic equation, in which the two sides are algebraic expressions. Each side of an algebraic equation will contain one or more terms. For example, the equation,
$A{{x}^{2}}+Bx+C=y$
has left-hand side $A{{x}^{2}}+Bx+C$ which has three terms, and right-hand side $y$, consisting of just one term. The unknowns are $x$ and $y$ and the parameters are $A,$ $B,$ and $C$.

An equation is analogous to a scale into which weights are placed. When equal weights of something (grain for example) are placed into the two pans, the two weights cause the scale to be in balance and are said to be equal. If a quantity of grain is removed from one pan of the balance, an equal amount of grain must be removed from the other pan to keep the scale in balance. Likewise, to keep an equation in balance, the same operations of addition, subtraction, multiplication and division must be performed on both sides of an equation for it to remain true.

We have been given the equation ${{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}-8$ .
Now let us solve the equation.
${{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}-8$
$={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}-{{(2)}^{3}}$
$={{(a+b)}^{3}}-{{(2)}^{3}}$ ……………………(${{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}$ )
$=(a+b-2)({{(a+b)}^{2}}+4+2(a+b))$ ………….( ${{x}^{3}}-{{y}^{3}}=(x-y)({{x}^{2}}+{{y}^{2}}+xy)$ )
\begin{align} & =(a+b-2)({{a}^{2}}+{{b}^{2}}+2ab+4+2a+2b)) \\ & =(a+b-2)({{a}^{2}}+{{b}^{2}}+2ab+4+2a+2b)) \\ \end{align}
So we get, ${{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}-8=(a+b-2)({{a}^{2}}+{{b}^{2}}+2ab+4+2a+2b))$ .
The correct answer is option (D).

Note: Read the question and see what is asked. Your concept regarding the equation should be clear. A proper assumption should be made. Do not make silly mistakes while substituting. Equate it in a proper manner and don't confuse yourself.

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# 12 A Adjustable Step Down DC-DC Module

12 A adjustable step-down voltage source which is useful to build your own power adjustable laboratory. It has high power which is suitable for your experiments.

SKU: OTH43758334684

Price: 4.06 $Old Price: 4.75$

Product in stock

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# Overview

This product is a step-down voltage source, high power (maximum 300W provided with additional dissipation). This module can be used to obtain voltages of 0.8 V to 28 V, 7 V from 32 V, with maximum current of 12 A, providing a suitable cooling.

This is small in size  with a very good efficiency, up to 95%, and very little heat dissipation, as compared with the power control. With this circuit you can build your own source laboratory with adjustable voltage and current or you can include it in your projects. Moreover, you can control LED power or recharge batteries safely.

Specifications

• Input Voltage: 7 V to 32 V
• The input voltage must be greater than the minimum 1 V output
• Output Current: 0.2 A to 12 A (for maximum current must improve heat dissipation)
• Maximum Output Power: 300 W
• Efficiency: up to 95%
• Operating Frequency: 300 kHz
• Ripple: 50 mV (measured at 24 V –>  12 V, 5 A)
• Leakage Current: 20 mA (at 24 V ->12 V)
• Adjustable output voltage and current
• Dimensions: 65 x 47 x 23.5 mm


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Synonymous series?

Algebra Level 3

In school, Timmy was told to calculate the value of the series

$a + a^2 + a^3 + a^4 + \cdots ,$

for some known $$a$$. But Timmy misheard, he instead calculated the series

$a^{-1} + a^{-2} + a^{-3} + a^{-4} + \cdots .$

Surprisingly, Timmy gets the correct answer!
Assuming that he did his work correctly, what was his answer?

Submit you answer as -1000 if you think that this is an impossible scenario.

×


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# Homework Help: Is the given set path connected? A Domain?

1. Nov 4, 2014

### Bashyboy

• Member warned about posting without the Homework template
Hello everyone,

Here is the definition of path connected and domain my textbook provides:

Definition 171. An open set S is path connected if each pair of points in S can be connected
by a polygonal line (e.g. a finite number of line segments connected end to end). A domain
is an open set that is path connected.

I am asked to determine whether "The set given by $r>0$, $- \pi < \Theta < \pi$" is path connected and is a domain.

I just want to make sure I understand exactly what the set is. Is the set they describe $S = \{(r,\Theta)~|~r >0 \wedge \Theta \in (- \pi, \pi) \}$; and does this set consist of all the points in the complex plane, other than those along the negative real-axis?

If that's the case, it would seem as though it would be path connected.

2. Nov 4, 2014

### Dick

Sounds correct so far.

3. Nov 4, 2014

### Bashyboy

I would argue that it is connected because any two points in the plane can be connected by a certain number of polygonal lines, just so long as these lines never pass through the negative real axis.

4. Nov 4, 2014

### haruspex

But you have not shown that can be done.

5. Nov 4, 2014

### Dick

Fine. That wouldn't be hard to show in detail. Is it a domain?

6. Nov 5, 2014

### Bashyboy

Yes, it is a domain, because the set is an open one, and we have just agreed that it is path connected.

What do you mean by showing in detail? How would you be more detailed in answering this question? I am curious to know.

7. Nov 5, 2014

### haruspex

Not sure, but I think Dick is making the same point I made: your approach is ok but what you have written so far does not prove there exists any such path. Try constructing one: "Given two points in the plane, not lying on the negative X axis or at the origin, with co-ordinates (etc.), the following path contained in the set connects them ..."

8. Nov 5, 2014

### Bashyboy

Oh, I don't believe the book is asking for a rigorous proof, but rather an intuitive consideration of the problem; additionally, I wouldn't really know how to construct a proof because our textbook has not covered it.

9. Nov 5, 2014

### haruspex

If you say so, but then I wonder what was meant by "determine". Anyway, the construction is pretty easy.

10. Nov 5, 2014

### Dick

A detailed proof wouldn't be very interesting. But you should still think about how you might make one.


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# Kernel of a homomorphism from a free group into $\mathbb{Z}$.

Let $F$ be a non-abelian free group, that is, a free group of rank at least $2$, and let $\phi: F \rightarrow \mathbb{Z}$ be a nontrivial group homomorphism. How to prove that the kernel of $\phi$ is not finitely generated?

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This follows since the rank of any infinite-index subgroup of a nonabelian free group is infinite. One way to prove this is by covering space theory. An infinite index subgroup corresponds to a cover of a wedge of circles where every point has infinitely many preimages. Then, if the cover is not a tree, there is at least one loop, $L$. Acting on $L$ by the infinite group of deck transformations gives infinitely many copies of $L$, but some of them may intersect or even coincide. However, since each loop has finitely many edges, only finitely many copies of $L$ can intersect $L$. So we can produce infinitely many disjoint loops. Thus the graph has infinite rank.

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I think there might be a simpler proof, using elementary concepts. Anyway, thanks Jim Conant for the interesting answer. I am thinking about it. – rla Apr 19 '12 at 13:05

There is a useful result you could use to prove this.

Let $F$ be a free group and $E \leq F$ with $|F:E| = \infty$. Suppose that $\exists \: \{1\} \neq N \unlhd G$ with $N \leq E$. Then the rank of $E$ is infinite.

The proof is as follows and assumes familiarity with the theory of Schreier generators of subgroups of free groups.

Let $F$ be free on $X$, let $U$ be a Schreier transversal of $E$ in $F$ and, for $g \in F$, denote the element in $U \cap Eg$ by $\overline{g}$.

Let $1 \neq w = a_1 \cdots a_l \in N \leq E$, with $a_i \in X^{\pm 1}$. For $u \in U$, $Euw = Euwu^{-1}u = Eu$, since $uwu^{-1} \in N \leq E$. So $\overline{uw} = u$, and $uw \not\in U$, so there is a least $k$ such that $ua_1 \cdots a_k \notin U$. Let $u_k := ua_1 \cdots a_{k-1}$. Then $u_k \in U$ and $u_ka_k \notin U$, so $u_ka_k\overline{u_ka_k}^{-1}$ is not trivial. Since $U$ is infinite and $l$ is fixed, there is an infinite subset $V$ of $U$ and a fixed $k$ with $1 \le k \le l$, such that $k$ is minimal with $u_ka_k \notin U$ for all $u \in V$. Then $\left\{ u_ka_k\overline{u_ka_k}^{-1} : u \in V \right\}$ is an infinite subset of the set of Schreier generators of $E$, and hence $E$ has infinite rank.

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For a completely different proof, that uses only the universal property of free groups, you could factor your map $\phi:F \to {\mathbb Z}$ as $\phi:F \to H \to {\mathbb Z}$, where $H$ is the wreath product ${\mathbb Z} \wr {\mathbb Z}$, and observe that the kernel of $H \to {\mathbb Z}$ is infinitely generated.

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# Evaluate $\int\frac{\cos x}{\sin x + \cos x}\,\text{d}x$. [duplicate]

I'm trying to figure out how to take this indefinite integral:

$$\int\frac{\cos x}{\sin x + \cos x}\,\text{d}x.$$

I tried simplifying and rearranging it, and this is the best I got: $$\int\frac{1}{\tan x + 1 }\,\text{d}x.$$

But I still can't figure out how to integrate from there. I know that it's integrable, as Wolfram Alpha indicates that the integral is $\frac{1}{2}\big(x+\ln{(\sin x + \cos x)}\big)+C$, but I can't figure out the steps to deriving it. Does anyone know how to evaluate this integral?

## marked as duplicate by Martin Sleziak, Amzoti, rschwieb, Asaf Karagila♦, Mark BennetMay 30 '13 at 17:21

• This is basically the same as this question. (To be more precise, if you know one of the two integrals, it is easy to calculate the other one and vice versa.) – Martin Sleziak May 30 '13 at 16:42

Let $I=\int\frac{\cos x}{\sin x+\cos x}\ dx$ and $J=\int \frac{\sin x}{\sin x+\cos x}\ dx$.

Then $I+J=\int\frac{\cos x+\sin x}{\sin x+\cos x}\ dx=x+C_1$ and $I-J=\int\frac{\cos x-\sin x}{\sin x+\cos x}\ dx=\ln |\sin x+\cos x|+C_2$.

Hence $I=\frac{1}{2}((I+J)+(I-J))=\frac{1}{2}(x+\ln |\sin x+\cos x|)+C$.

• This is very elegant in that it requires no substitutions. – user54147 Feb 9 '13 at 12:20
• @user54147 You are using a substitution, $u = \sin(x) + \cos(x)$ in the end – Amad27 Aug 11 '15 at 19:58

A general (but not necessarily efficient) tool is the Weierstrass substitution $t=\tan(x/2)$.

Then $\cos x=\frac{1-t^2}{1+t^2}$, $\sin x=\frac{2t}{1+t^2}$, and $dx=\frac{2}{1+t^2}$. Do the substitution. We end up needing the following ugly integral: $$\int \frac{2(1-t^2)}{(1+2t-t^2)(1+t^2)}\,dt.$$ We are integraing a rational function, and it can be done using partial fractions. But it takes some work.

The same idea works in principle for any rational function of $\sin x$ and $\cos x$.

Notice that $$\sin x+\cos x=\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x\right)=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$$ Therefore $$I=\frac{1}{\sqrt{2}}\int\frac{\cos x}{\sin\left(x+\frac{\pi}{4}\right)}dx$$ Now let $x+\frac{\pi}{4}=t$ $$I=\frac{1}{\sqrt{2}}\int\frac{\cos \left(t-\frac{\pi}{4}\right)}{\sin t}dt=\frac{1}{2}\int\frac{\sin t + \cos t}{\sin t}dt=\frac{1}{2}\left(t+\ln {\left|\sin t\right|}\right)+C_1\\=\frac{1}{2}\left(x+\frac{\pi}{4}+\ln {\left|\sin x+\cos x\right|}\right)+C_1=\frac{1}{2}\left(x+\ln {\left|\sin x+\cos x\right|}\right)+C$$

$\displaystyle \int \frac{1}{1+ \tan x} \ dx$

$\displaystyle = \int \frac{1}{1+u} \frac{1}{1+u^{2}} \ du$ (let $u = \tan x$)

$\displaystyle = \frac{1}{2} \int \left( \frac{1}{1+u} + \frac{1-u}{1+u^{2}} \right) \ du$

$\displaystyle = \frac{1}{2} \int \left( \frac{1}{1+u} - \frac{u}{1+u^{2}} + \frac{1}{1+u^{2}} \right) \ du$

$\displaystyle =\frac{1}{2} \left(\ln(1+u) - \frac{1}{2} \ln(1+u^{2}) + \arctan u \right) + C$

$\displaystyle =\frac{1}{2} \left(\ln(1+u) - \ln(\sqrt{1+u^{2}}) + \arctan u \right) + C$

$\displaystyle = \frac{1}{2} \left( \ln(1+\tan x) - \ln (\sqrt{1+\tan^{2}x}) + x \right) + C$

$\displaystyle = \frac{1}{2} \left( \ln(1+\tan x) - \ln (\sec x) + x \right) + C$

$\displaystyle = \frac{1}{2} \Big( \ln \big(\frac{1+ \tan x}{\sec x}\big) + x \Big) + C$

$\displaystyle = \frac{1}{2} \left( \ln (\cos x + \sin x ) + x \right) + C$


text:
## A local Riemann hypothesis. I.(English)Zbl 0991.11022

This is a joint review for Parts I and II (see below).
Consider the integral $$Z(f,\nu,s) = \int_{F^\times}f(x)\nu(x)|x|^{s} d^{\times}x$$, where $$F$$ is a local field, $$\nu$$ a character of $$F^\times$$ and $$f$$ a Schwartz function on $$F$$. It is a meromorphic function of the complex variable $$s$$ and for suitable $$f$$ it is a local $$L$$-function. The authors discovered a set of Schwartz functions for which all zeros of $$Z(f,\nu,s)$$ lie on the line $$\text{Re}(s) = {1\over 2}$$, if $$Z(f,\nu,s)$$ is not identically zero. These Schwartz functions are the eigenfunctions of the Weil representation of the unitary group $$\text{U}(1) = \{z\in E^{\times}:\text{N}_{E/F} (z) = 1\}$$, where $$E = F(\sqrt{-1})$$, assuming that $$-1$$ is not a square in $$F$$. The case $$F = {\mathbb R}$$ is treated in Part I, the case of a non-archimedean $$F$$ with residual characteristic $$\neq 2$$ in Part II. In the latter case there are more quadratic extensions of $$F$$, for the other three the assertion about the zeros of $$Z(f,\nu,s)$$ does not hold.
The proof in Part II relies on two facts: all eigencharacters of $$\text{U}(1)$$ have multiplicity 1 and Fourier transform is in the image of the representation. The latter is true for $$E = F(\sqrt{-1})$$ only. The corresponding split case is also treated.
In Part I an extension to the $$n$$-dimensional harmonic oscillator is also proved.

### MSC:

 11F70 Representation-theoretic methods; automorphic representations over local and global fields 11S40 Zeta functions and $$L$$-functions 11M38 Zeta and $$L$$-functions in characteristic $$p$$ 22E50 Representations of Lie and linear algebraic groups over local fields 11F27 Theta series; Weil representation; theta correspondences

Zbl 0991.11023
Full Text:


text:
# Easiest way to find the screen resolution of all machines on a windows network?

I'd rather not run round checking each machine individually, is there some server-side software I can use? It's a mixed linux/windows network (clients and servers) but I'm only interested in the windows clients.

-

For Windows clients you can use a WMI script to obtain this. The info here should help get you started.

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Looks good, but how do I get it to run on all client machines in the domain and report back to me? –  Draemon Feb 17 '10 at 11:29
Have a script run at logon that outputs the results to a text file (named with the client computer's name.txt) with the results to a everyone-write-access share? –  Bart Silverstrim Feb 17 '10 at 12:33
It should be easy enough to adapt once you have a list of computer names to start with. Loop through the list and call a sub containing most of the code in that link. Results could go to a file share, be sent via email or logged to a database. But we're getting more into the realm of "how to write VBScript" here, which is probably better covered in a separate question once you've made a decision on which way to go. –  Darth Melkor Feb 17 '10 at 13:22
@mh not really. I know how to write VBScript, but I don't know WMI. What I really meant was "how to run at logon", but I've figured that out. What I didn't realise is that you can do this from one machine given a list of the other machines. –  Draemon Feb 17 '10 at 13:30

Thanks to @mh for the link to WMI. Here's the script I ended up with.

On Error resume next

strDomain = "domain.local"

Set objFSO = CreateObject("Scripting.FileSystemObject")
if err.number <> 0 then Wscript.quit

set domObj = GetObject("WinNT://" & strDomain)
domObj.Filter = Array("computer")
For Each objComputer In domObj
err.clear
strComputer = objComputer.Name
computerName = "UNKNOWN"
loggedOnUser =""
sp = ""
ram = ""
cpu = ""
cpuMhz = ""
xres = ""
yres = ""

Set objWMIService = GetObject("winmgmts:\\" & strComputer & "\root\cimv2")

if err.number=0 then
' Query "BIOS" properties
' computer name
Set colItems = objWMIService.ExecQuery("Select * From Win32_BIOS")
For Each objItem in colItems
computerName = objItem.Path_.Server
Next

' Query Processor properties
' Family
' ...
Set colItems = objWMIService.ExecQuery("Select * From Win32_Processor")
For Each objItem in colItems
cpu = objItem.Manufacturer & " " & objItem.Name
cpuMhz = objItem.MaxClockSpeed
Next

' Query Operating System properties
' SP Level
' RAM
Set colItems = objWMIService.ExecQuery("Select * From Win32_OperatingSystem")
For Each objItem in colItems
sp = objItem.ServicePackMajorVersion
ram = objItem.TotalVisibleMemorySize
Next

' Query Computer System properties
' User
Set colItems = objWMIService.ExecQuery("Select * From Win32_ComputerSystem")
For Each objItem in colItems
Next

' Query Display properties
' Screen resolution
Set colItems = objWMIService.ExecQuery("Select * From Win32_DisplayConfiguration")
For Each objItem in colItems
xres = objItem.PelsWidth
yres = objItem.PelsHeight
Next

strFile = "\\server\data\inventory\" & computerName & ".txt"

Set objTextFile = objFSO.OpenTextFile(strFile, 2, True)
if err.number <> 0 then Wscript.quit

objTextFile.WriteLine("Computer Name: " & strComputer)
if err.number <> 0 then Wscript.quit

if loggedOnUser <> "" then objTextFile.WriteLine("User: " & loggedOnUser)
if cpu <> "" then objTextFile.writeLine("CPU: " & cpu)
if cpuMhz <> "" then objTextFile.writeLine("CPU MHZ: " & cpuMhz)
if ram <> "" then objTextFile.WriteLine("RAM: " & ram)
if sp <> "" then objTextFile.WriteLine("Service Pack: " & sp)
if xres <> "" then objTextFile.WriteLine("Horizontal resolution: " & xres)
if yres <> "" then objTextFile.WriteLine("Vertical resolution: " & yres)

objTextFile.Close

end if
next

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You can do this via SCOM but it's a big hammer to crack a small nut to be honest.

-


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# What is wrong with violation of locality (EPR paradox)?

When studying the EPR paradox, at some point we must resign ourselves that Reality and Locality can not be both true in the current theory of quantum mechanics.

A lot has been said in Physics.SE about the EPR paradox and about entanglement, but my question is quite different. To me, the violation of Locality doesn't look like a big deal: through entanglement you are not actually transmitting anything, neither physical quantities (energy, momentum,...) nor just information. And indeed that form of violation of Locality doesn't contradict Special Relativity. It really looks to me that nothing is being transmitted at faster than light speed. We are just gaining information about something which could even be at a space-like distance, but what is wrong with this?

Is there actually a theory that gets contradicted by this kind of violation of Locality? Any experiment? Or are scientists reluctant to give up on Locality just based on common sense and a classical view of the world?

• note that, Bell or not Bell, quantum mechanics is local, in the sense that superluminal communication is not possible (and the accuracy of quantum mechanics has been tested countless times). Philosophy aside, I'd argue that having to use a nonlocal theory/formalism to describe strictly local phenomena would be a bit weird.
– glS
Jun 21 at 9:02

Or are scientists reluctant to give up on Locality just based on common sense and a classical view of the world?

That's exactly it.

The EPR paradox is not really a paradox. It is explained by entanglement and by (the Copenhagen interpretation of) quantum mechanics. There is no transmission of anything, since there is only one wavefunction that spatially extends to encapsulate both bodies, and its collapse happens at the same time everywhere. So it's even a locality problem per se.

Also, in the usual EPR thought experiment, the probability of getting spin up or down is exactly 50-50, that is purely random. So you cannot even use this to transmit information. If the probability were not 50-50, then you must have had information about the particular decay process prior to the event...

The whole point of the "paradox" is that, classically, you expect things that are infinitely far apart from each other to be independent of each other. The whole question is "how far do the two bodies, that briefly interacted with one another at time 0, need to be in order to not require inclusion of the other one to describe the first?". Quantum mechanically, the answer is never.

• "Never" doesn't really answer "how far?". This is a bit confusing. I guess you meant "however far, the objective won't be fulfilled". Jun 19 at 22:39
• @Ruslan I don't really understand the criticism. Would you have wanted me to answer "at infinity"? But that assumes the degree of entanglement is exponentially decaying which, albeit maybe true, I didn't want to assume. My logic was like answering "I am not selling the car" to someone asking "How much for this car". Jun 20 at 5:49

There is a computer analogy to explain QM that I do not think anyone will disagree. It is the computational version of quantum mechanics, that says that an entangled state in QM cannot be simulated by two separate computers that model each particle separately. That is, if you simulate two entangled particles that move away from each other, you cannot dedicate a separate computer to each particle, even if separated. For more details see here. In this sense, QM is non local. Many physicists, for some reason, still remain unconvinced that this computer analogy adds anything to the discussion. To me, it is a proof that non-locality is unavoidable.

We are just gaining information about something which could even be at a space-like distance, but what is wrong with this?

Nothing. Simply there are to points of view how to deal with entanglement. The usual is that the superposition is a temporarily but real thing, even over distance. And the entanglement is finished after production and only our knowledge about the components is zero until measurement. The complication comes from three fact that destroying the states but gaining the right result for both particles at less than 50 percent.

• Even so, it is still possible to correlate (not entangle) two objects And send them to two separate testers with results that match quantum mechanics. The measurements will follow Malus Law cos2theta. Jun 22 at 23:18

We are just gaining information about something which could even be at a space-like distance, but what is wrong with this?

I think you're underselling the paradox a bit here. Of course you can imagine a classical situation wherein you gain information about an object space-like separated from you, in a way that would not surprise anybody. For instance, if you draw one ball from a bag with a red and blue ball, without looking at the color, and then travel to the Andromeda galaxy and look, instantly you also know the color of the other ball.

Bell showed, as a simple mathematical fact, that this is not sufficient to explain entanglement in quantum mechanics. One easy way to describe it is the CHSH game, where Alice and Bob (space-like separated throughout the game) are given independent random bits $$x, y$$ and they must output bits $$a, b$$ such that $$a + b = xy \mod 2$$. Classically it's not hard to convince yourself the best they can do is guess $$a = b = 0$$ and win $$75\%$$ of the time. But if they share an entangled pair of particles they can win $$\approx 85\%$$ of the time. This is much harder to come to terms with than the situation in the previous paragraph.

At the end of the day, sure, there's nothing wrong with this, and it has been conclusively demonstrated by experiments that this is really how the universe works. But it's definitely something surprising and counter to our classical intuition.


text:
PLANET Discussion: Daeridune

Discussion in 'The Manaverse Wiki Project' started by Eebit, Mar 27, 2014.

1. Discussion of and questions pertaining to the collaborative world, Daeridune, go in here.

#1
2. There has been some discussion going on in the IRC as to whether or not Daeridune warrants a page, as it is strictly created to house the chatplay Algorithm of Fate. Additionally, little has been done in the way of development of the world, and there has been hardly any contribution in the way of 'worldbuilding discussions.'

Would any of the members partaking in AoF as a chatplay like to champion this page? And otherwise, should we mark it as a candidate for deletion? Please weigh in below.

@Shadow @CerberusLycan @Starrie @Kudamon @"The Kakuzato"

#2
3. Once we actually develop it some, it would warrant its own page. As just a skeleton though, it shouldn't be there yet.

#3
4. Okay I'm going to toss out random notes and ideas for Daeridune because this needs to be a thing.

- Barren, wastelandish world
- Government placed a taboo on magic under the pretense that it was magic that destroyed much of the planet in the first place
- We have established there are two suns and one moon
- Major city, as far as AoF goes, is Antriquel
- There are small towns and such as well

- We have not established any sort of pantheon; we've been sort of assuming Daeridune doesn't have one
- Very little has been developed in the way of native species. Humans are the only one confirmed.
- An idea was tossed around that there might be a government-condoned slave trade

Let's figure out what to do with this information and try to toss more ideas out there.

#4
5. Continuing off my last post here because it doesn't account for newcanon;

- High tech level
- AoF and newcanon may or may not be in the same timeline- if they are, newcanon is before the apocalyptic event
- Individuals from other planets are present in both canons (Newcanon: a pair of Ereva, a Diclonius, a Vindrite, what I'm assuming is a Feronite, and who knows what else; AoF: a Kaichronan, a Vindrite, a Phantasian, and what I'm assuming is one of Starr's races)
- G, from AoF, can possibly be related to the high tech level before the planet got gg wp no re'd (my own personal connection, this hasn't been discussed)

#5
6. This is canon until someone definitively tells me otherwise. It's also WIP and I will link my dump forum post here: http://keileon.createmybb3.com/thread-83-post-479.html#pid479

Daeova
• Energy weapons (Dazenth's sword, Damien's railgun)
• Possible genetic experiments
• Religion (god of progress) + government = Exelism + radical fascism
• Treats dissenters and disruptive individuals as criminals
• Very order-oriented
• Possible interplanetary travel

Daeova can be considered Fascist by Terran standards, but deviates from this form of government by integrating religion more firmly into it. The country worships the God of Progress, Exelixus, and refer to their own system of government and religion as Exelism. This particular form of government and religion is Daeova-unique and unpopular in the eyes of less radical societies.

Daeova is ruled by one man, named Silas Jasper Payne. He keeps the company of what he refers to as Exelist Enforcers, which include Damien Thrage.

Daeova's currency is based around coins called Tael, which come in copper, silver, and gold variants. It is fifty Copper Tael to a Silver Tael and fifty Silver to a Gold. Daeovan economy is good to those able to work in military, construction, or scientific fields, and poor to the rest of society. This leads to anarchic crime in larger cities (which is quickly and methodically silenced by Payne's enforcers) and rampant poverty in the smaller, weaker cities (which Payne ignores and allows to be picked off for the sake of the rest of the country's progress).

Daeova's capital city is named Aeficeon and houses military advancements. Daeova also includes the city of Antriquel (the main setting of Algorithm of Fate), and Kevarus (a large city which houses the scientific and technological advancements). It is a part of the continent of Auvresh, which has the highest rate of technological and societal progress of the three main continents of Daeridune.

#6
7. This is what I developed a while ago:

Mendia
Theme: Monk-themed who live in the mountains. Their spelltypes tend to be focused around improving their bodies and physical ability.

Architecture: Their buildings are all completely flat. They only live in the peaks of mountains, but their buildings are made on boarding extending off the sides. Magic allows them to keep everything completely level and balanced. The look of their buildings is inspired by traditional Japanese buildings, and most rooms are connected by outdoor walkways.

Ideology: They believe in the purity of the human form. They do not accept tampering with or modification to the human form. They react violently to non-humans, but are calm and wish to exercise diplomacy when confronted by “equals.”

Pragara
Theme: Chemists and Pressure engineers. They have extremely good medicine and command complex and powerful water-pressure machines.

Architecture: They live on the coast and build cities of labyrinthine pipework. Buildings are all wrapped up in complex piping or cables like webbing around the cities. This makes their cities hard to navigate, which leads civilizations to get frustrated with visitors.

Ideology: They are generally communist. Government strongly insists everyone keep working, which often leads to rushed or forced projects, and is the general contributing factor to the mess that is the faction’s infrastructure. Aside from this, they strongly value life, and many funds are contributed to health and the environment.

#7
8. I led a discussion tonight in #Manaverse about Daeridune with the intention of rekindling the worldbuilding effort there. I wanted to make sure the topics discussed didn't get lost, so I am posting here for the sake of reference, discussion, and posterity.

First, here's a link to the current Daeridune map.

On discussing where the currently developed nations were situated, I commented that I had the headcanon that Daeova was above the Sandsea, and Pragara was to the south. However, a comment written on the wiki states "The southern hemisphere of Daeridune is the hub of the technological progress", which indicates to me that Daeova might have been intended to be on the part of the continent below the Sandsea. Cerby pointed out that "there was also a comment somewhere that said the south, I believe, builds upwards to escape the Sandsea due to its limited landmass, which we know Daeova does, while Pragara was stated to build labyrinthine structures, which would indicate more lateral building". Clarification from Kuda or Flem (or both) to this end would be appreciated.

For the Daeridune wiki page, we have some structural differences from the way the rest of the planet pages have been penned. On the majority of our pages, we have:

Code:
==Geography==
===Continents===
====[CONTINENT_NAME]====

'''CONTINENT_NAME''' is situated [blah blah blah] and comprises the nations of [foo], [bar], and [spam].

but on Daeridune's page, we have created fifth-level subheadings for each nation that exists within the respective continents. The fifth-level of subheading looks quite similar to the fourth-level, so without a glance at the Table of Contents, it is difficult to decipher whether Auvresh is a continent or a nation. I know that this structural difference came about as a result of simply trying to get the established information on Daeridune in one place. At some point in the future, we should create a continental subpage for Auvresh and move the information on the continent (and Daeova and Pragara) there. In its place, we summarize the information on Auvresh. I am particularly commenting on Auvresh for now since it is at least named, and is the most developed of our Daeridite continents.

NAMES! Speaking of names, we could use some for these continents. We have a lot of nations ending with 'a,' so I think future developments might want to avoid that if possible. One name that I had mentally mulled around was the name "Eslistan." Originally I thought it might suit the green (i.e. magic/idealism-focused) continent, but upon further consideration I thought it sounded a little too similar phonetically to Exelixus, the name of the Deity of Progress. Cerby mentioned that it sounded slightly fitting for the blue islands. Perhaps it could refer to either the islands as a collective, or just be the name of one of them.

Another thing that has come up is pinpointing the primary "themes" of Daeridune, as a planet. Cerby and I discussed this briefly, and he pinpointed them as follows (quoting him because I can't possibly summarize it as succinctly and elegantly as he already did):

[hr]

[GREY_CONTINENT]

I started a discussion about the grey continent in the north. Shadow once mentioned to me that he envisioned Daeridune as having an "arctic continent." Geographically, it would seem to be sensible to put that up in the north.

With the idea of the themes in mind -- namely, Idealism and Progress -- Cerby said that he figured the grey continent might be tangential to those two opposing forces. He likened them to the Yin and the Yang of Daeridune, and then went on to draw the parallel of the icy grey continent's "theme" to be similar to the Taoism and the concept of wuji, the "absence of a complementary force. In the frozen north, there is no societal progress to be found or ideals to foster, there is only the cycle of nature, repeating itself in its own form of balance."

I myself have had a thought about placing a floating, autonomous (and likely man-made) city-state in the north, beyond the grey continent. To clarify, I mean floating on the water rather than in the sky.

I had originally thought that it would be tied to one of the Auvresh nations, but I feel like neither of the presently developed ones would want to tie up their resources in keeping a sustainable project going up there. It doesn't seem to fit with Pragara, at least - I would expect that their communism would have them focused on their own people.

My thought that it would be floating on the water beyond, rather than being actually on the grey continent is because rule of cool and the founders knew of just how inhospitable the grey continent was, and thus deemed it unfit for developing a city.

In addition, my current thinking has led me to believe that the residents of this floating city are primarily scientists, and potentially expatriates/defectors of the other nations. This means that the city itself doesn't have any ties to any one ideology that dominates Daeridune.

So the main thing we've yet to work out would be why exactly the gray continent is inhospitable, and what exactly the researchers of the floating city are, well, researching.

#8
9. Daeova builds high and is intended to be in the southern part of the continent.

#9
10. So, as I mentioned briefly in the chat the other night, I've started to delve into some more ideas for Pragara. It all started because I wanted to bring my old roleplay, Liquid Salvation, to life. Pragara with its connection to water-pressure machines seemed like a good setting for a steampunk roleplay. And on top of that, having a depletion of water and a lasting drought wouldn't have repercussions on Daeridune in quite as severe a fashion as it might on Terra.

To begin, what do we know about Pragara?

We know that Pragara has a lot of sprawling architecture. To me (and to Cerby), this indicates that Pragara is a much more "sprawling" nation than its neighbour, Daeova, to the south. Presumably, Pragara doesn't cover the ENTIRE north Auvresh landmass, but I picture it as being in the northern hemisphere, but still close to / bordering the Sandsea.

Passage to Daeova from Pragara is probably fairly restricted by cost or willingness to cross the Sandsea. I wouldn't expect that crossing the Sandsea would be a particularly easy journey, and it would probably require some specialized navigation by some hardy veterans (or otherwise, fools). I imagine the passage of such vessels would be restricted to those who are willing to pay the price... or who are willing to take a chance with an untested crew. Not a journey one would tend to want to make alone or in small numbers.

Although it isn't explicitly mentioned anywhere, the absence of mention of Pragara having any sort of reverance for Exelixus makes me assume that they are not particularly inclined to worship him. My first thought is that Pragara might be much more inspired by Daeova's progress rather than the ideology of one of Daeridune's patron deities. Pragara's name, itself, has made me think of "pragmatism," also, so maybe that is the root of the nation getting its name.

I also feel like, politically, Pragara might have had its regime installed by Daeova in order for Daeova to have access to Pragara's superior medicine and medical capabilities. And maybe also its land. So Pragara previously / historically might not have been a particularly advanced nation, but contact with Daeova has led to a rapid attempt at modernization or industrialization. And this, to me, is the cause of Pragara's projects seeming so "rushed" -- they're attempting to play catch up with Daeova. However, they are doing it without the help / inspiration of a deity.

e: Might be worth noting that I internally view the relationship between Daeova and Pragara as being comparable to the real-world relationship between China and North Korea, respectively... except less volatile (maybe?). Particularly, that the latter nations are subservient to the former nations, and the former often provides them with aid and direction.

#10
11. For the sake of having it written down in a place that is easy to access:

#11
12. This is extremely important information.

#12


text:
# Properties

 Label 4600.2.e.c Level $4600$ Weight $2$ Character orbit 4600.e Analytic conductor $36.731$ Analytic rank $0$ Dimension $2$ CM no Inner twists $2$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$4600 = 2^{3} \cdot 5^{2} \cdot 23$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 4600.e (of order $$2$$, degree $$1$$, not minimal)

## Newform invariants

 Self dual: no Analytic conductor: $$36.7311849298$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-1})$$ Defining polynomial: $$x^{2} + 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{7}]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$

## $q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of $$i = \sqrt{-1}$$. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + 2 i q^{3} + i q^{7} - q^{9} +O(q^{10})$$ $$q + 2 i q^{3} + i q^{7} - q^{9} -5 q^{11} -i q^{13} -4 i q^{17} -7 q^{19} -2 q^{21} + i q^{23} + 4 i q^{27} -5 q^{29} + 2 q^{31} -10 i q^{33} -2 i q^{37} + 2 q^{39} + 11 q^{41} -i q^{43} -8 i q^{47} + 6 q^{49} + 8 q^{51} -14 i q^{57} + 14 q^{59} + 10 q^{61} -i q^{63} -8 i q^{67} -2 q^{69} -10 q^{71} -7 i q^{73} -5 i q^{77} -7 q^{79} -11 q^{81} + 15 i q^{83} -10 i q^{87} -10 q^{89} + q^{91} + 4 i q^{93} -4 i q^{97} + 5 q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q - 2q^{9} + O(q^{10})$$ $$2q - 2q^{9} - 10q^{11} - 14q^{19} - 4q^{21} - 10q^{29} + 4q^{31} + 4q^{39} + 22q^{41} + 12q^{49} + 16q^{51} + 28q^{59} + 20q^{61} - 4q^{69} - 20q^{71} - 14q^{79} - 22q^{81} - 20q^{89} + 2q^{91} + 10q^{99} + O(q^{100})$$

## Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/4600\mathbb{Z}\right)^\times$$.

 $$n$$ $$1151$$ $$1201$$ $$2301$$ $$2577$$ $$\chi(n)$$ $$1$$ $$1$$ $$1$$ $$-1$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
4049.1
 − 1.00000i 1.00000i
0 2.00000i 0 0 0 1.00000i 0 −1.00000 0
4049.2 0 2.00000i 0 0 0 1.00000i 0 −1.00000 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
5.b even 2 1 inner

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 4600.2.e.c 2
5.b even 2 1 inner 4600.2.e.c 2
5.c odd 4 1 4600.2.a.c 1
5.c odd 4 1 4600.2.a.n yes 1
20.e even 4 1 9200.2.a.i 1
20.e even 4 1 9200.2.a.bd 1

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
4600.2.a.c 1 5.c odd 4 1
4600.2.a.n yes 1 5.c odd 4 1
4600.2.e.c 2 1.a even 1 1 trivial
4600.2.e.c 2 5.b even 2 1 inner
9200.2.a.i 1 20.e even 4 1
9200.2.a.bd 1 20.e even 4 1

## Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(4600, [\chi])$$:

 $$T_{3}^{2} + 4$$ $$T_{7}^{2} + 1$$ $$T_{11} + 5$$ $$T_{13}^{2} + 1$$

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$T^{2}$$
$3$ $$4 + T^{2}$$
$5$ $$T^{2}$$
$7$ $$1 + T^{2}$$
$11$ $$( 5 + T )^{2}$$
$13$ $$1 + T^{2}$$
$17$ $$16 + T^{2}$$
$19$ $$( 7 + T )^{2}$$
$23$ $$1 + T^{2}$$
$29$ $$( 5 + T )^{2}$$
$31$ $$( -2 + T )^{2}$$
$37$ $$4 + T^{2}$$
$41$ $$( -11 + T )^{2}$$
$43$ $$1 + T^{2}$$
$47$ $$64 + T^{2}$$
$53$ $$T^{2}$$
$59$ $$( -14 + T )^{2}$$
$61$ $$( -10 + T )^{2}$$
$67$ $$64 + T^{2}$$
$71$ $$( 10 + T )^{2}$$
$73$ $$49 + T^{2}$$
$79$ $$( 7 + T )^{2}$$
$83$ $$225 + T^{2}$$
$89$ $$( 10 + T )^{2}$$
$97$ $$16 + T^{2}$$


text:
# Why does the normal force go down in an downward accelerating elevator?

When standing in a stationary elevator the force due to gravity and the normal force cancel each other out hence the net force is 0. When the elevator starts accelerating downwards there's a net force in the downward direction meaning that the body in the elevator now experiences three forces, the force due to gravity,the force that is causing the acceleration of the whole elevator and the normal force, accordingly the downward force is now of a greater magnitude than that of the normal force the body experiences. Then my question is, why does the normal force need to reduce if the downward force is still greater than the normal force even if it doesn't decrease. The body feels the force of gravity and the force which causes the elevator to accelerate these two are greater than the normal force and therefore the elevator accelerates in the downwards direction, why should the normal force get smaller?

• You're mentally putting the acceleration and force on the same side of the equal sign when they are on different sides in F=ma: See physics.stackexchange.com/questions/632735/… The other way to think is if the elevator is accelerating downards, there is a net force downward, meaning one of two things needs to have changed from the case where normal force and force of gravity cancel to zero: Either gravity has increased or the normal force is reduced. Presumably, gravity has not increased. Feb 27 at 19:17
• The "problem" would seem to be that "The elevator now experiences three forces…" Sorry; that's not a problem; it's simply a mistake. Mar 1 at 0:25

When the elevator starts accelerating downwards there's a net force in the downward direction meaning that the body in the elevator now experiences three forces, the force due to gravity,the force that is causing the acceleration of the whole elevator and the normal force

This is not correct. There are only two forces acting on the body in the elevator. One is the downward pointing force of gravity, and the other is the upward pointing normal force. The force of gravity is constant, but the normal force can vary. There is not a third force involved. The net force is just the sum of the two forces, not a separate force on its own.

So if the body is not accelerating then the net force is 0 so the two forces are equal. If the body is accelerating downward then the net force is downward so the upward pointing normal force must be reduced so that the magnitude of the downward pointing gravitational force can exceed it.

• I think I understand it. If the net force was zero the body would slam on the incoming accelerating elevator's ceiling. If the body is accelerating downwards it means that the body has a net force downwards, hence the normal force mist be less, so the person is not pushing on the floor as hard because it's constantly acceleraring in falling motion and therefore not pushing as hard, if It was it wouldn't accelerate right Feb 27 at 19:20

The body feels the force of gravity and the force which causes the elevator to accelerate

The body does not feel the net force, it only feels the normal force. This is sometimes called apparent weight. In a free fall, the body does not feel any weight.

Common misconception is that normal force equals weight. This is not true, and the normal force is usually determined from some assumptions. For example, for a body sliding on a ramp, the assumption is that the body does not move in the direction perpendicular to the ramp surface. This is one simple example in which normal force does not equal weight, even when the body is at rest.

To understand the accelerating elevator, you should start from the net force on the person. If an elevator accelerates downwards at $$a_\star$$, an observer on the ground (inertial reference frame) would see the person standing on the elevator floor accelerate downwards at the same acceleration. The net force on the person is then

$$\vec{F}_\text{net} = m \vec{a}_\star = m \vec{g} + \vec{n}$$

from which it follows

$$\vec{n} = m (\vec{a}_\star - \vec{g})$$

If downward acceleration is $$|\vec{a}_\star| < g$$ then the normal force is negative, i.e. it points upwards. But if downward acceleration is $$|\vec{a}_\star| > g$$, then the person in the elevator ends up on the ceiling and the normal force is positive, i.e. it points downwards.

The force that the elevator acceleration excerts on you is in opposite direction of the direction of acceleration. If the lift accelerates upwards you feel a bigger force downwards and if it accelerates down the force on you will diminish. If it accelerates down fast hard enough you can land up on the ceiling. The normal force reacts accordingly.

It looks like you're imagining some kind of mechanism that actively accelerates the elevator downwards. But that's not what happens.

An elevator essentially1 hangs from a (counterbalanced) cable that's on a pulley. When the elevator needs to start moving down, the pulley is allowed to start rolling in a controlled way, which briefly reduces the tension (upward force) in the cable, letting gravity win. So it's gravity (diminished by the cable tension) that makes the elevator (and the passenger) accelerate downwards.

1 In reality, elevator designs may be quite a bit more complicated. There are also other kinds of elevators (e.g. hydraulic elevators).

When the elevator is standing still, you experience the normal force because you are being accelerated by gravity (meaning, and I'll come back to this later, that gravity is trying to increase your current instantaneous speed), but your surroundings (the elevator car) are fixed, so you sort of "run into" the floor. The normal force is essentially just the molecules of the floor not letting you pass through them, counteracting gravity.

When the elevator is accelerating, you still "run into" the floor, but since the floor itself is accelerating (at a smaller rate than 1g), you don't push into it as much, so the reaction force from the floor is lessened. So in your local (non-inertial) frame of reference, it feels like the strength of gravity has dropped (remember, if you were free falling, you'd feel weightless).

Now, this lasts only a short while, because the elevator soon starts moving at uniform speed, and the tension in the cable exactly counteracts gravity once again. So, now you're moving down, but there's no net force on the elevator. Remember, in absence of a net force, things want to continue moving uniformly in a straight line. Both you and the floor are moving down, but neither experiences a net acceleration, meaning that the floor fully counteracts gravity's pull on you once again - so you feel your usual weight. Since the elevator is now moving uniformly, it's an inertial frame of reference, so the force balance situation is the same as if the elevator was standing still! Gravity is trying to increase your current instantaneous speed - it doesn't matter if that speed is 0 m/s or 20 m/s or 1000 km/h; in fact, the exact number you put on your speed depends on your choice of reference frame. Gravity makes you "run into the floor" just the same, and the floor then reacts in the same way.

When you reach the bottom, the cable tension has to be greater in strength than gravity, because a net upward acceleration is required just to stop you and the elevator car. The floor is now trying to stop, and you aren't (not on your own), so you push into it harder, making the reaction force on you greater. In your local reference frame, it briefly feels as if the gravity has increased in strength.

In the other direction, the sequence of events is reversed; for the elevator to start moving up, the cable tension has to briefly overcome gravity, then it has to counteract it (net force of zero for the elevator car + passenger system) while the elevator is traveling up, and finally, it has to briefly let gravity win in order for the elevator to stop.

The body feels the force of gravity and the force which causes the elevator to accelerate

There is a false assumption here that the person experiences the downward force on the elevator. This would happen if the elevator accelerates faster than gravity, which might cause the person to slam into the ceiling and then be pushed down by the ceiling, hence experiencing that force.

But that would still just be the normal force, which would have flipped directions, hence reducing below 0 and becoming negative. And somewhere before that, the lift would accelerate at the rate of gravity, the floor would fall away from the person, and they would experience free fall with a normal force of 0.


text:
# compute hma command

## Syntax

compute ID group-ID hma temp-ID keyword ...

• ID, group-ID are documented in compute command

• hma = style name of this compute command

• temp-ID = ID of fix that specifies the set temperature during canonical simulation

• keyword = anharmonic u p Pharm cv

anharmonic = compute will return anharmonic property values
u = compute will return potential energy
p = compute will return pressure.  the following keyword must be the difference between the harmonic pressure and lattice pressure as described below
cv = compute will return the heat capacity

## Examples

compute 2 all hma 1 u
compute 2 all hma 1 anharmonic u p 0.9
compute 2 all hma 1 u cv

## Description

Define a computation that calculates the properties of a solid (potential energy, pressure or heat capacity), using the harmonically-mapped averaging (HMA) method. This command yields much higher precision than the equivalent compute commands (compute pe, compute pressure, etc.) commands during a canonical simulation of an atomic crystal. Specifically, near melting HMA can yield averages of a given precision an order of magnitude faster than conventional methods, and this only improves as the temperatures is lowered. This is particularly important for evaluating the free energy by thermodynamic integration, where the low-temperature contributions are the greatest source of statistical uncertainty. Moreover, HMA has other advantages, including smaller potential-truncation effects, finite-size effects, smaller timestep inaccuracy, faster equilibration and shorter decorrelation time.

HMA should not be used if atoms are expected to diffuse. It is also restricted to simulations in the NVT ensemble. While this compute may be used with any potential in LAMMPS, it will provide inaccurate results for potentials that do not go to 0 at the truncation distance; pair_style lj/smooth/linear and Ewald summation should work fine, while pair_style lj/cut will perform poorly unless the potential is shifted (via pair_modify shift) or the cutoff is large. Furthermore, computation of the heat capacity with this compute is restricted to those that implement the single_hessian method in Pair. Implementing single_hessian in additional pair styles is simple. Please contact Andrew Schultz (ajs42 at buffalo.edu) and David Kofke (kofke at buffalo.edu) if your desired pair style does not have this method. This is the list of pair styles that currently implement single_hessian:

In this method, the analytically known harmonic behavior of a crystal is removed from the traditional ensemble averages, which leads to an accurate and precise measurement of the anharmonic contributions without contamination by noise produced by the already-known harmonic behavior. A detailed description of this method can be found in (Moustafa). The potential energy is computed by the formula:

$\left< U\right>_{HMA} = \frac{d}{2} (N-1) k_B T + \left< U + \frac{1}{2} F\bullet\Delta r \right>$

where $$N$$ is the number of atoms in the system, $$k_B$$ is Boltzmann’s constant, $$T$$ is the temperature, $$d$$ is the dimensionality of the system (2 or 3 for 2d/3d), $$F\bullet\Delta r$$ is the sum of dot products of the atomic force vectors and displacement (from lattice sites) vectors, and $$U$$ is the sum of pair, bond, angle, dihedral, improper, kspace (long-range), and fix energies.

The pressure is computed by the formula:

$\left< P\right>_{HMA} = \Delta \hat P + \left< P_{vir} + \frac{\beta \Delta \hat P - \rho}{d(N-1)} F\bullet\Delta r \right>$

where $$\rho$$ is the number density of the system, $$\Delta \hat P$$ is the difference between the harmonic and lattice pressure, $$P_{vir}$$ is the virial pressure computed as the sum of pair, bond, angle, dihedral, improper, kspace (long-range), and fix contributions to the force on each atom, and $$k_B=1/k_B T$$. Although the method will work for any value of $$\Delta \hat P$$ specified (use pressure units), the precision of the resultant pressure is sensitive to $$\Delta \hat P$$; the precision tends to be best when $$\Delta \hat P$$ is the actual the difference between the lattice pressure and harmonic pressure.

$\left<C_V \right>_{HMA} = \frac{d}{2} (N-1) k_B + \frac{1}{k_B T^2} \left( \left< U_{HMA}^2 \right> - \left<U_{HMA}\right>^2 \right) + \frac{1}{4 T} \left< F\bullet\Delta r + \Delta r \bullet \Phi \bullet \Delta r \right>$

where $$\Phi$$ is the Hessian matrix. The compute hma command computes the full expression for $$C_V$$ except for the $$\left<U_{HMA}^2\right>^2$$ in the variance term, which can be obtained by passing the u keyword; you must add this extra contribution to the $$C_V$$ value reported by this compute. The variance term can cause significant round-off error when computing $$C_V$$. To address this, the anharmonic keyword can be passed and/or the output format can be specified with more digits.

thermo_modify format float '%22.15e'

The anharmonic keyword will instruct the compute to return anharmonic properties rather than the full properties, which include lattice, harmonic and anharmonic contributions. When using this keyword, the compute must be first active (it must be included via a thermo_style custom command) while the atoms are still at their lattice sites (before equilibration).

The temp-ID specified with compute hma command should be same as the fix-ID of Nose-Hoover (fix nvt) or Berendsen (fix temp/berendsen) thermostat used for the simulation. While using this command, Langevin thermostat (fix langevin) should be avoided as its extra forces interfere with the HMA implementation.

Note

Compute hma command should be used right after the energy minimization, when the atoms are at their lattice sites. The simulation should not be started before this command has been used in the input script.

The following example illustrates the placement of this command in the input script:

min_style cg
minimize 1e-35 1e-15 50000 500000
compute 1 all hma thermostatid u
fix thermostatid all nvt temp 600.0 600.0 100.0

Note

Compute hma should be used when the atoms of the solid do not diffuse. Diffusion will reduce the precision in the potential energy computation.

Note

The fix_modify energy yes command must also be specified if a fix is to contribute potential energy to this command.

An example input script that uses this compute is included in examples/USER/hma/ along with corresponding LAMMPS output showing that the HMA properties fluctuate less than the corresponding conventional properties.

## Output info

This compute calculates a global vector that includes the n properties requested as arguments to the command (the potential energy, pressure and/or heat capacity). The elements of the vector can be accessed by indices 1-n by any command that uses global vector values as input. See the Howto output doc page for an overview of LAMMPS output options.

The vector values calculated by this compute are “extensive”. The scalar value will be in energy units.

## Restrictions

This compute is part of the USER-MISC package. It is enabled only if LAMMPS was built with that package. See the Build package doc page for more info.

Usage restricted to canonical (NVT) ensemble simulation only.

## Default

none

(Moustafa) Sabry G. Moustafa, Andrew J. Schultz, and David A. Kofke, Very fast averaging of thermal properties of crystals by molecular simulation, Phys. Rev. E [92], 043303 (2015)


text:
## Radiation vector for a short dipole near a perfect magnetic conductor

Hi there.

If I wanted to calculate the radiation vector (in z > 0) produced by a short dipole with uniform current Io (+z direction) on a infinite perfect electric conductor (plane z=0), I'd have to apply the images method. So I'd have to calculate the radiation vector produced by two short dipoles, one in z > 0 and one in z < 0, with currents Io.

However, if we change the perfect electric conductor by a perfect magnetic conductor, how can I apply the images method?

Thank you.
 PhysOrg.com engineering news on PhysOrg.com >> Sensitive bomb detector to rove in search of danger>> PNNL-developed injection molding process recognized with emerging technologies award>> How soon could car seats enter the 3-D comfort zone?

Mentor
 Quote by Bromio Hi there. If I wanted to calculate the radiation vector (in z > 0) produced by a short dipole with uniform current Io (+z direction) on a infinite perfect electric conductor (plane z=0), I'd have to apply the images method. So I'd have to calculate the radiation vector produced by two short dipoles, one in z > 0 and one in z < 0, with currents Io. However, if we change the perfect electric conductor by a perfect magnetic conductor, how can I apply the images method? Thank you.
What's a perfect magnetic conductor?
 It's an idealization. PEC (Perfect Electric Conductor): $\hat{n}\times\vec{E}=0$ and $\hat{n}\times\vec{H}=\vec{J}_s$. PMC (Perfect Magnetic Conductor): $\hat{n}\times\vec{E}=-\vec{M}_s$ and $\hat{n}\times\vec{H}=0$. Thank you.

## Radiation vector for a short dipole near a perfect magnetic conductor

At very low frequency, mumetal of permalloy are good magnetic conductors.

It would change the sign of the current in the image. Opposite current achieve zero magnetic field at the magnetic conductor, while same currents achieve zero electric field.
 Hi. I understand what you say, but, in that case, radiation vector is 0 (because currents are opposite), isn't it? Thank you.


text:
## AbstractA failed attempt at explaining Entropy,and one zombie, coming right up...

And so it was, on a nice evening, much like this one, that we had all sat around the table, and a question popped up...

The question was along the lines of "what is the Entropic principle?", and it was asked by my brother, a brilliant man, and a science-fiction aficionado, who unfortunately for the physicist community, never had the chance to dabble with physics, and so they have to find a poor substitute in the image of your humble servant here...

At first I asked if he meant the Anthropic principle, but he just wanted to understand Entropy.

Thus I found myself trying to explain Entropy, and the 2nd law of thermodynamics to the uninitiated, in layman terms, and, after a fashion, follow the Einstein grandmother rule - "You do not really understand something unless you can explain it to your grandmother." (A.Einstein).

Incidentally , you could probably calculate the period time $$T$$ of Einstein's turning over in his grave, by someone misquoting him, or otherwise deifying him, and justifying a falsehood or plain ol' stupidity by attributing something to old Einei that he never would have meant in a million light years.

We'll start with some observational data - I have around 400 people in my human network, and on average I get an Einstein quote which falls under the aforementioned category, maybe once every two weeks.

Now, suppose only a third of the world's population leads a somewhat western lifestyle (either connected to facebook, google+, twitter etc. or alternatively reads the paper and or listens to the radio at least once a day), we have about 2.3 Billion people.

let's be harsh and assume each of the 400 people sub-networks are non-connected between them and so we neglect back-propagation we can put a lower limit of
$\frac{2.3\cdot 10^{9}}{400}=5.75\cdot 10^{6}\,\text{instances in 2 weeks}$
Divided by the number of seconds in a two-week period we get:
$\frac{5.75\cdot 10^{6}}{14\cdot 24\cdot 60\cdot 60}\approx 4.75\,\text{times per second}$

And so the period time of Einstein's turning in his grave would be $$T\approx 0.21 \,sec$$.

So basically even the lower limit states that Einstein, by now, is a zombified Olympic athlete, even considering the initial rigor mortis....

 By now, he would have a solid six-pack.

Anyway, I digress, I was going to explain Entropy and then a random rant stole my attention... sorry for that.

In a nutshell, Entropy is a measure of disorder, and I will explain.

<Failed attempt at an explanation : but is still worth a read>

Imagine a group of four coins, each with two sides - heads, and tails - right? (we'll have non of that Two-Face shenanigans here!)
Now suppose every coin is perfectly balanced so there's a fifty-fifty chance of getting heads or tails for each coin flip.

So, now, what are the chances of getting all 4 heads, when you flip 4 coins?
if you do the experiment enough times, you get an average of 1/16 chance.
the state of all heads, or equivocally all tails is the most "ordered" result, why?
because it is the most homogenous result (and we humans like homogeneity, symmetry and by the same token order).
Now, what's the most plausible result?
that's easy - the result where two coins are tails up, and two coins are heads up (regardless of their locations), happens ideally $$\frac{3}{8}$$ of the times you flip (almost half of the times you flip the coins, you'll get this result).

That is the least "ordered" result, since we don't care about locations, and the coins show the most diversity in results.

Now, I won't go through the whole derivation, but if you'll try the same logic with 6 coins and then 8 coins you'll get a breakdown of $\frac{1}{64},\frac{6}{64},\frac{15}{64},\frac{20}{64},\frac{15}{64},\frac{6}{64},\frac{1}{64}\,\text{for six coins}$
and a breakdown of $\frac{1}{256},\frac{8}{256},\frac{28}{256},\frac{56}{256},\frac{70}{256},\frac{56}{256},\frac{28}{256},\frac{8}{256},\frac{1}{256}\,\text{for eight coins}$
And so on and so forth, the reason I'm sticking to even numbers is because it LOOKS more clear that way, but really, it makes no difference. You could go on and on until kingdom come, and you'll find the middle, most unorganized result will be the most common.

It turns out that this kind of dynamic is best approximated by a gaussian function called the $$g$$ function (or the multiplicity function) and I'll spare you the details in favor of a graph:
 Probable results graph
So, what you see here, is basically an overlay of 4 graphs that show the relative probability of results as they stray from the middle "disorganized" and probable result.
What is interesting, is the bigger the experiment is (i.e. instead of 8 coins, let's say a 100 or 1000 coins) the sharper the peak is, meaning it's narrower, and higher in respect to other possible results. that means by the way that the most probable result is highly probable, and the others highly improbable.  Now imagine an experiment of $$10^{23}$$ coins, every result other then the most probable and it's immediate neighbors is SO improbable, it virtually is IMPOSSIBLE (in the sense that it would take a ludicrously impossible amount of experiments to perform to actually get a significant chance to get such a result).

A word of caution though - this is probability we're talking about, so in theory a highly organized result MIGHT happen, in actuality - yeah, not so much...

By the way, there are roughly no more than $$6\cdot 10^{14}$$ coins in circulation today in THE WORLD, meaning even if you took all the coins in the world today you couldn't perform such an experiment, even once!

Incidentally the ridiculously high number of participants in a single experiment, makes all the difference between "hard sciences" even if they are statistically oriented, and "soft sciences".

Even if we take all the people in the world, and get them to participate in one of our experiments, the result will produce some correlation that may, or may not apply to a single participant.

In physics, while the same is true, you could say a statistic result applies and be absolutely correct on a macro level (with deviations so small as to be insignificant for most purposes), and be correct almost every time on a micro level as well!.

So anyway, Entropy is defined as the logarithm of the multiplicity function.

The reason for taking the logarithm is for the sake of defining a cumulative quantity, as opposed to multiplicative.

< /Failed attempt at an explanation : but is still worth a read>

So anyway, obviously I failed at this attempt but let's try it in a simpler manner:

Entropy is a quantity that signifies how probable a result is.
by a fluke of chance, which isn't a fluke at all, more of a deep connection really, the most probable result is also the most diverse one, or differently put, the most disorganized.

Thus, Entropy becomes a measure of disorder of a system.

Entropy is a cumulative property in the sense, that when two non-interacting experiments are done the combined entropy is the sum.
However, when systems are allowed to interact, the combined entropy is typically larger than the sum of individual entropy.

it is by that sense, that entropy tends to increase over time (and interactions).

<example of entropy increase>

Suppose, we have two systems, each of 4 coins.

The most probable state is given by 2 heads, and 2 tails for a single experiment right?
as was explained in the above failed attempt, the chance for that happening is $$\frac{3}{8}$$.

Now, what is the chance of each of the experiment to get the most probable state independently? you guessed it - the product of the two independent probabilities i.e. $$\frac{9}{64}$$ right?

OK, but now, let's put all the coins in a single experiment, an flip all of them, the chance of hitting 4/4 division of heads/tails is given by $$\frac{70}{256}=\frac{35}{128}$$ which is almost double the size of the product of individual probabilities.

So what happened here, really?
in essence, the combined system has more places to choose from, meaning more diversity of scenarios that lead to the same end result, thus the combined system is more "disordered" thus bigger probability that leads to bigger Entropy.

</example of entropy increase>

OK, that wraps it up for this time.
Obviously I don't understand Entropy enough, since I feel I have failed at explaining it,
but I will try again ("if at first you don't succeed" etc...).

Oh by the way, for all us $$\LaTeX$$ geeks out there, isn't it cool that every time we use a fractional, we immediately get a reference to Battlestar Galactica?

Anyway, till next time...

P.S. some quick ideas to utilize Einei's incredible turnover time of 0.21 seconds :
1. Attach him to a turbine and generate electricity.
2. Display him as the 8th world wonder - the quickest man on the planet (faster than Usain Bolt).
3. Use him as the engine for a horse-ride carousel for my kid.
4. Use him to refute the 2nd law of Thermodynamics as he is both dead (Entropy supposed to get bigger), but in the greatest shape of his life (which requires work and decrease in local Entropy)...

Just sayin'...


text:
Hurewicz theorem related to Galois group (or Tannakian categories)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T15:12:52Z http://mathoverflow.net/feeds/question/56431 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56431/hurewicz-theorem-related-to-galois-group-or-tannakian-categories Hurewicz theorem related to Galois group (or Tannakian categories)? Jakob 2011-02-23T19:06:45Z 2011-03-05T02:35:43Z
Is there a proof of the Hurewicz theorem $\pi_1(X)^{ab} = H_1(X, \mathbf Z)$ ($X$ a connected topological space) expressing $\pi_1(X)$ as the "Galois" group of $X$, i.e., group of deck transformations of the universal cover?

(as opposed to a proof using the construction of $\pi_1$ as the group of paths up to homotopy).

Thank you.
 http://mathoverflow.net/questions/56431/hurewicz-theorem-related-to-galois-group-or-tannakian-categories/57428#57428 Answer by Denis-Charles Cisinski for Hurewicz theorem related to Galois group (or Tannakian categories)? Denis-Charles Cisinski 2011-03-05T02:35:43Z 2011-03-05T02:35:43Z
For $X$ a $0$-connected nice space (say, a CW-complex), and for any group $G$, there is a natural bijection of the following shape

$$[X,BG]\simeq Hom(\pi_1(X),G)$$

which can be proved roughly as follows (if you like tannakian-like arguments): maps from $X$ to $BG$ correspond to $G$-torsor over $X$, which correspond to maps of topoi from the topos of sheaves over $X$ to the topos of $G$-sets; but, as any $G$-torsor is locally constant, this also corresponds to the maps of topoi from the topos of locally constant sheaves over $X$ to the topos of $G$-sets. As, by Galois theory, the topos of locally constant sheaves over $X$ is canonically equivalent to the topos of $\pi_1(X)$-sets, we conclude from the fact that, given two groups $A$ and $B$, exact and colimit preserving functors from $B$-sets to $A$-sets correspond to homomorphisms of groups from $A$ to $B$.

To be precise, $[X,BG]$ means the set of homotopy classes of maps from $X$ to $BG$, while for $G$ an abelian group, $Hom(\pi_1(X),G)$ means the set of group homomorphisms (for a non abelian $G$'s, we have to quotient a little bit, but we won't care here).

For $A$ an abelian group, we thus get bijections $$H^1(X,A)\simeq [X,BA]\simeq Hom(\pi_1(X)^{ab},A) .$$ By the Yoneda lemma, to prove that the map $\pi_1(X)^{ab}\to H_1(X,\mathbf{Z})$ is bijective, it is sufficient to prove that, for any abelian group $A$, the map $$<\star> \quad Hom(H_1(X,\mathbf{Z}),A)\to Hom(\pi_1(X)^{ab},A)$$ is bijective. But, instead of checking this for all $A$'s, it is sufficient to prove this in the case where $A$ is an injective object in the category of abelian groups (because there are enough injectives). In this case, as $Hom(-,A)$ is an exact functor, we have a bijection $$Hom(H_1(X,\mathbf{Z}),A)\simeq H^1(X,A) .$$ Therefore, for any injective $A$, the map $<\star>$ is bijective.

If you like topoi and pro-groups, you may play the same game and prove this for locally $0$-connected topoi with essentially the same argument.


text:
# The meaning of gauge-fixing in covariant quantization of the electromagnetic field

I am having trouble wrapping my head around the idea behind the covariant quantization for the electromagnetic field that is usually done in textbooks (I'm currently following Mandl & Shaw and David Tong's notes in my QFT course). The starting point for quantizing the electromagnetic field is to talk about gauge freedom and how it can be useful for treating electromagnetism, and then you conveniently choose the Lorentz condition as an auxiliary constraint that you want the 4-potential $$A_\mu$$ to satisfy; after quantization, however, this condition becomes a constraint on the space of acceptable physical states, and not on the field operators themselves: "physical states" $$\left|\Psi\right\rangle$$ are those such that $$\partial_{\mu}A^{\mu+}\left|\Psi\right\rangle = 0$$ (Gupta-Bleuler condition). Now, my question is twofold:

1. Is there any quantization procedure that does not rely on gauge fixing? I am still not very comfortable with the idea of you having to fix a gauge in order to consistently quantize a theory that you formulated explicitly as gauge invariant from the get-go, and it seems to me that this is done at a high cost -- gauge fixing manifests itself as a constraint on physical states!
2. (Somewhat related to the 1st question) How does the Gupta-Bleuler condition exactly fixes the gauge, since it is not a condition on the operators, but on the physical states? Am I supposed to think of these two things as being totally equivalent?

Firstly, you have to fix the gauge since the 4-potential $$A_\mu$$ would have 4 degrees of freedom instead of the 2 polarizaton of the photon.

In general there are 3 different ways to quantize a theory which presents a first class constraints (which is the case for the Electromagnetic field). To whom is not used to first class constraints, they are those constraints $$C_\alpha$$ which generates gauge transformations as $$\delta z^A=\{z^A,\epsilon^\alpha(z)C_\alpha(z)\}$$ where $$z^A$$ are the coordinates in the phase space and $$\epsilon^\alpha$$ are the gauge parameters.

Now the methods are:

1. Reduced Phase-Space method;
2. Dirac-Gupta-Bleuer method;
3. BRST method.

## Reduced Phase Space:

The first method fix a single representative in the gauge orbit through a gauge fixing condition $$F^\alpha(z)=0$$ with $$det\{C^\alpha,F^\beta\}|_{C=F=0}\neq0$$ (this leads to have a new set of constraints which are now of the II class, and there can be defined Dirac parenthesis as new poisson parenthesis in order to quantize in the canonical way).

## Dirac-Gupta-Bleuer

The second method consist in quantize all the phase-space, leaving in all the non-physical variables. Since we have quantize everything, now we must require that our constraints, when acting on physical states, give zero: $$\hat{C}_\alpha(\hat{z})|\psi_{ph}\rangle=0$$ This the prefered method to quantize the electromagnetic field, in this way there can be removed all the non-physical degrees of freedom after the quantization.

## BRST Lagrangian

This method is the more complicated one, but in my opinion is the most fascinating one. The idea, here, is to add new degrees of freedom, called ghosts (with grassman parity opposite to the constraints' one) and their momenta (with same parity as the ghosts) (In order to understand this procedure, you will have to be familiar with some basic notions of supersymmetry).

Now, it can be proved that there exists a BRST charge $$Q$$ with some properties, in particular it is nilpotent ($$Q^2=0$$), which generates the BRST transformation.

In this contex, a physical state is defined if it is BRST-invariant, i.e. $$\hat{Q}|\psi_{ph}\rangle=0$$ But thanks to the nilpotency of $$Q$$, one can consider the Cohomology class of the physical state, which means that two sates are equivalent if they differ by an exact term, i.e. $$(|\phi_{ph}\rangle\sim|\psi_{ph}\rangle)\iff(|\phi_{ph}\rangle=|\psi_{ph}\rangle+\hat{Q}|\chi\rangle)$$ for an arbitrary $$|\chi\rangle$$.

This seem just an overcomplicated procedure, but when one consider the functional integral with all these new terms, one has the freedom to add a new term to the action which is an exact term in order to make a lot of simplifications and to arrive at the right result in a more sofisticated way.

I leave here a reference (textbook) which cover the quantization of gauge theories: Marc Henneaux and Claudio Teitelboim "Quantization of gauge systems".

1. Why do we gauge-fix the path integral in the first place? If we were doing lattice gauge theory, we didn't need to gauge-fix. But in the continuum case, (the Hessian of) the action for a gauge theory has zero-directions that lead to infinite factors when performing the path integral over gauge orbits. To avoid this, we gauge-fix.

2. The Gupta–Bleuler condition is not a gauge-fixing condition per se, but rather a condition to determine physical states. In general there are also conditions on an observable $$\hat{\cal O}$$. This is perhaps easiest to see in the BRST formulation. A physical state $$|\Psi \rangle$$ and an observable $$\hat{\cal O}$$ should both be BRST-invariant $$\hat{Q}|\Psi \rangle~=~0 \qquad\text{and}\qquad [\hat{Q}, \hat{\cal O}]~=~\hat{0}.$$

• I'll try to check BRST formulation, I had never heard of it before. So regarding your first point, are you saying one motivation to gauge-fix is to not "overcount" physical states more than once when doing the path-integral? And so there is no formal reason for you to do that in canonical quantization except if you reach out to the path integral formulation? – Bruno De Souza Leão Oct 9 '18 at 12:36


text:
I'm writing a document which includes a list of URLs that I want to be clickable, so I'm using hyperref to achieve this. This has the awkward side effect that my equation numbers, some of which I don't want to be clickable links, to also be clickable. I've hidden this by giving hypersetup the options colorlinks, linkcolor={black}

But this is suboptimal, in that the link is still clickable. I tried the following MWE, but it didn't work (the * just displays above the environment.)

\documentclass{article}
\usepackage{xcolor}
\usepackage{hyperref}
\title{This is the title}
\begin{document}
\let\oldref\ref
\let\ref\ref*
$$\label{eq:one} e^{i\pi} + 1 = 0$$
This is a link to the equation \ref{eq:one}, when it shouldn't be...
\let\ref\oldref %% Changing it back to normal behaviour
\end{document}

I also tried using renewcommand, but that failed too.

It's got something to do with the *, since \let\ref\autoref works fine.

I can't just find and replace the ones I want unlinked, because they are inside LTXexample environments, and I want the code to display as \ref not as \ref*

So I have two questions:

1. How do I get this temporary redefinition of \ref to work?
2. Or, how do I turn off equation links globally, but still have URL links work? (And if possible, is there a way to then have some equations linky with a different command)
-
Just to explain why \let\ref\ref* can't work: * is not letter, so the command is not \ref* but \ref, and this takes an optional argument *. So what you did was \let\ref\ref (not helpful :-)) and then typset *. –  Hendrik Vogt Nov 29 '10 at 14:27
Would it be that bad to have clickable links all over if they didn't have any special colors or boxes around them in the PDF? –  MSC Apr 4 '14 at 17:49

Instead of \ref* use \@refstar:

\makeatletter
\let\ref\@refstar
\makeatother

Why? hyperref redefines \ref this way.

\DeclareRobustCommand\ref{\@ifstar\@refstar\T@ref}%

\@ifstar is a LaTeX command which tests if the next character is a *. In this case the * would be gobbled and the first of the following commands will be executed (\@refstar), otherwise the second (\T@ref).

Thus, \ref* is essentially the same as \@refstar und you could use that instead.

If one would like to go deeper: the source code documentation of hyperref (or hyperref.dtx on CTAN or on your local TeX system) tells that \HyRef@StarSetRef belongs to \ref* and would not be linked, and \@refstar is defined this way:

\def\@refstar#1{%
\HyRef@StarSetRef{#1}\@firstoffive
}

-
thanks for the explanation! –  Seamus Nov 29 '10 at 14:45


text:
MathSciNet bibliographic data MR733302 (85d:22015) 22D25 (43A20 46K99 46M99) Leptin, H. The structure of \$L\sp{1}(G)\$$L\sp{1}(G)$ for locally compact groups. Operator algebras and group representations, Vol. II (Neptun, 1980), 48–61, Monogr. Stud. Math., 18, Pitman, Boston, MA, 1984. Links to the journal or article are not yet available

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