Two distinguishable particles in a box

Question: I would like to determine the number of energy states two free, distinguishable particles in a box of length $L$ have. I would then like to determine the number of states two free, indistinguishable particles, with spin $3/2$ each, have in that box at the elementary level. Finally, determine the number of states in case these two particles with spin $3/2$ each are distinguishable.
I am familiar with the following formula for the energy states
$$E_n=\frac{(\hbar)^2 \pi^2n^2}{(2mL^2)}$$
but am not sure how to proceed. If the particles are distinguishable, does that entail that one is a fermion whereas the other is a boson? I am not sure.
Furthermore, if the two particles have spin $3/2$ each, that means they are both fermions, right? If that is correct, then, due to Pauli's exclusion principle, the two could not be at the same state. I also know that for spin $3/2$ there could be $4$ particles per energy level. But again I am not sure how to coherently process the given data and would appreciate some guidance.
Comment: As of right now, your question is not very well-defined. Can you show what you have tried and what exactly you are trying to do/got stuck on?
Comment: @Danu Is the question itself not well defined or the attempt at solution? I am very new to this topic so I am not sure what else I could possibly add. I have written everything which I deemed relevant to solving this problem, to the best of my understanding.
Comment: In case the spin of each particle is $3/2$, would the number of energy states be $\binom{4}{2}$ for indistinguishable particles and $\frac{4!}{2!}$ for distinguishable particles?
Comment: Distinguishable:  all permutations of available states.  Indistinguishable: all combinations of available states.
Comment: @CarlWitthoft Is what I wrote incorrect, namely $\binom{4}{2}$ and $\frac{4!}{2!}$?
Answer: "If the particles are distinguishable, does that entail that one is a fermion whereas the other is a boson?"
No, the particles being indistinguishable from each other does not imply that one is a fermion and the other a boson, they can be any types of particles as long as they aren't the same type.
Eg: One could be an electron and the other a muon (both spin-$\frac{1}{2}$ fermions). One could be a hydrogen atom and the other an alpha particle (both bosons). Or one could be a boson and the other a fermion.
From this question I believe you have a misconception about indistinguishable particles---you seem to imply you think that all fermions are indistinguishable from one another and that all bosons are indistinguishable from one another; this is not the case. The Pauli Exclusion Principle applies to fermions of the same type, not all fermions (and similarly with the exchange symmetry for indistinguishable bosons).
"Furthermore, if the two particles have spin 3/2 each, that means they are both fermions, right?"
Yes, it follows from the spin statistics theorem in quantum field theory that half-integer spin particles are fermions and integer spin particles are bosons. Without going into quantum field theory, however, you just have to take this as given.
http://en.wikipedia.org/wiki/Spin%E2%80%93statistics_theorem
"If that is correct, then, due to Pauli's exclusion principle, the two could not be at the same state."
No, as stated above the particles can both be fermions and yet be distinguishable and, hence, not obey the Pauli Exclusion Principle.
So to answer what I think your main question is:
For a box in which the spectrum of a single particle is $E_n$ where $n$ is a positive, non-zero integer, the energy of two non-interacting, distinguishable particles is $E^{(1)}_n + E^{(2)}_m$ where $n$ and $m$ are both positive, non-zero integers, the superscripts refer to the first and second particle, their spectra may be different because, for example, they may have different masses.
As you've given the spectrum for an infinitely deep well, there are infinitely many single particle bound states and, hence infinitely many bound two particle states. Were you talking about, for instance, a finite potential well there would be only be finitely many single particle bound states. If the number of bound states for the first/second particle is $n^{(1/2)}_{tot}$ the total number of bound states for the two (distinguishable) particle system would be $n^{(1)}_{tot}\times n^{(2)}_{tot}$; these numbers could be different, again, because of the possibility of different masses.
Comment: First of all, thank you very much! Second, did you read my attempt of the second and third parts of this question, referring to two free, indistinguishable (then distinguishable) particles with spin $3/2$ each (at the elementary level)?
Comment: Is that attempt correct?
Comment: Are you talking about your comments?
If so I'm not sure what (4,2) and 4!/2! are meant to be the answer to.
Is the question: "for two spin 3/2 free particles, how many different spin configurations are there for each value of energy in both the distinguishable and indistinguishable case?"?
Comment: I am not sure whether the formulation actually implies spin configurations. The original question is: how may I determine the number of states two free, indistinguishable particles, with spin 3/2 each, have in that box at the elementary level?; also, how may I determine the number of states in case these two particles, with spin 3/2 each, are distinguishable? Are the answers I wrote in my comments correct?
Comment: Are you quoting the question exactly?

If the particles are in the box they are not free; free means no potential term in the Hamiltonian, but the presence of the box is a potential term.

Further, if you count all of the possible energy states there are infinitely many in either case.

Just talking about a single particle: for a free particle there is one energy state for each value of the momentum (a real number) for particles in the infinite square well, there is one energy state for each value of the principle quantum number, n, a positive, non-zero integer.
Comment: Regardless of whether the pair of particles are distinguishable or indistinguishable there will still be infinitely many energy states for the two particle system, if there are infinitely many energy states for the single particle system. I don't know what is meant by 'the elementary level'.
Comment: The elementary level is n=1. I believe the answers should respectively be 4!/(2!2!) and 4!/2!. Are you certain these aren't correct?

Two distinguishable particles in a box


How do I change the weapons of the other characters I hang out with?

Question: In GTA V's story mode, you are able to call certain other characters and ask them to "Hang Out."
How do I change the weapons of these other characters I hang out with? (Because when I take out a weapon, they only take out a Thompson.)
Answer: Sadly, there is no way to make them change the weapon that they are using.
Internally, the game decides what is the best weapon for the other characters (being considered pedestrians when not controlled), so for some reason the game decided that the Gusenberg Sweeper was the best weapon for them.
If you still want to change the weapon, your only option would be to get rid of all of the weapons that you don't plan to use.


How to setup Appium in Ubuntu for android

Question: I am new in appium(automation testing technology).
I'm using a PC running Ubuntu Linux.
I have searched about this topic but I have not got any useful tutorial. Can anyone point me to the right documentation?
Comment: Appium is a Unit Testing tech. The way you worded it, it sounds like it's a technology designed at testing automation. You don't seem to be trying to add Arduinos to your house and test them, so you might want to rephrase a bit. Also, by looking up Google I found a few tutorials. You might want to search Google too?
Answer: Do not install nodejs through apt-get, which will need sudo rights and appium will not work if node is installed as sudo user. If you have already installed remove it using
sudo apt-get remove nodejs
sudo apt-get remove npm

Download latest nodejs linux binaries form http://nodejs.org/download/
Extract into a folder that doesn't need sudo rights to access, for example your home folder.
tar -xvf <downloaded_binary_tar.gz>

Add the following line to your ~/.bashrc file.
export PATH=$PATH:<full_path_of_the_extracted_node_folder>/bin

Open a now terminal and do
npm install -g appium
appium

Answer: I'm sure you will find plenty of tutorials on this (and this will only work for android since you would need an OSX box to do iOS) but here is what we did:
Install nodejs/npm:
sudo apt-get update
sudo apt-get install -y python-software-properties python g++ make
sudo add-apt-repository ppa:chris-lea/node.js
sudo apt-get update
sudo apt-get install nodejs
UPDATE: use homebrew to install node
Install grunt-cli:
npm install -g grunt-cli

If you run into an issue about it not being able to install in a directory, do this.
Install Appium:
npm install -g appium

Set up a symlink in your .bashrc file for Appium:
ln -s /path/to/appium.js /usr/bin/appium

Test to make sure it can run by running appium in your terminal. The output should be something like:
info: Welcome to Appium v0.16.0 (REV 292d265edd9c7aaf96f165009285c814b218363d)
info: Appium REST http interface listener started on 0.0.0.0:4723
   info  - socket.io started

Install Java JRE 6
sudo apt-get install openjdk-6-jre

Install Android SDK:
Download the SDK and extract it to your home folder.
Launch the Android SDK Manager:
~/path/to/android-sdk/tools/android

Install the packages that you'll need in the new window:

Android 4.X
Android Support Library
Android SUpport Repository
Google Play services
Everything under Tools
Everything under Extras

You can also create a symlink for the Android SDK Manager by doing:
ln -s /path/to/android-sdk/tools/android /usr/bin/android

Comment: This approach won't work. Appium won't run if node/npm are install as a privileged (sudo) user. Use Homebrew for Linux to install it instead as mentioned in one of the answers.
Comment: @plosco How to configure appium UI as it able to do on windows? How does  set  android `apk` at this case? Because appium is running at terminal.
Answer: Steps to get appium working on ubuntu pc :
Install ruby: Paste the below command at terminal and hit enter
sudo apt-get install build-essential curl git m4 ruby texinfo libbz2-dev libcurl4-openssl-dev libexpat-dev libncurses-dev zlib1g-dev

Install linux brew:Paste the below command at terminal and hit enter
ruby -e "$(curl -fsSL https://raw.githubusercontent.com/Homebrew/linuxbrew/go/install)"

set path for brew
Type: gedit .bashrc at terminal and copy paste following into the .bashrc file
export PATH="$HOME/.linuxbrew/bin:$PATH"
export MANPATH="$HOME/.linuxbrew/share/man:$MANPATH"
export INFOPATH="$HOME/.linuxbrew/share/info:$INFOPATH"

Install node:Paste the below commands one by one  at terminal and hit enter
brew update
brew install node
brew link node

Install appium
npm install -g appium

npm install wd

To start appium: Paste the below command at terminal and hit enter
appium

Comment: I'm using Appium version 1.5.1 in Mac OS. I want to change appium configuration when it is installed by npm command (npm install -g appium)   Is there a way to change it like changing the device name, command timeout, platform version ... Thanks.
Comment: you can specify device name , platform etc in your script itself  desired_caps['platformName'] = 'Android' desired_caps['platformVersion'] = '4.4.4' desired_caps['deviceName'] = '9e3484f6'
Comment: This is really best solution for Ubuntu.
Comment: @krishnachetan, I am trying to install appium using above instructions, it went well till "brew install node", only error I got is "Error: Failed to download resource "python"", Now I am trying to run "brew link node" on terminal but I am getting error "Error: No such keg: /home/sonal/.linuxbrew/Cellar/node". Can you please help?
Answer: Steps to install Appium on Linux.
Basics setup from the link.
Now install linuxbrew taking help from the link.
Now use the following commands from the terminal
Make sure to start a new terminal:
brew update
brew install node
brew link node
npm install -g appium@version

Hope this helps those who face the issue installing with "sudo".
Answer: Just to save somebodies time: on my Ubuntu 14.04 I was trying to install npm both via sudo apt-get and using binaries. Both approaches were wrong but everything is Ok after I istalled npm via HomeBrew
Answer: Many people face issue while installing appium on ubuntu because we install using sudo we should install appium using npm
Detailed step by step explanation to install appium on Ubuntu using npm
CLICK HERE

To install node.js and appium without sudo we use linuxbrew.
To install linuxbrew these are the dependencies
Ruby 1.8.6 or newer

GCC 4.2 or newer
Git 184.108.40.206 or newer
Linux 2.6.16 or newer
64-bit x86 or 32-bit ARM platform

Install Ruby using below command
sudo apt-get install build-essential curl git m4 python-setuptools ruby texinfo libbz2-dev libcurl4-openssl-dev libexpat-dev libncurses-dev zlib1g-dev

Install Linux Brew using below command
ruby -e "$(curl -fsSL
https://raw.githubusercontent.com/Homebrew/linuxbrew/go/install)"

Set Path for brew

first type sudo gedit.bashrc in terminal  and copy the below in the .bashrc file.export
PATH="$HOME/.linuxbrew/bin:$PATH"exportMANPATH="$HOME/.linuxbrew/share/man:$MANPATH"export INFOPATH="$HOME/.linuxbrew/share/info:$INFOPATH"
7 .Install Node:Open new terminal and copy the command and press enter
brew update
brew install node
brew link node

8.Finally Install Appium using below command
npm install -g appium

9.Let's Check if the installation is successfull. Open the terminal and type "appium" and hit enter.You should see something like this in the terminal
 info: Welcome to Appium v1.4.12(REV 8db2d00b9afcf2c50a09a80a2e8d56b05a902caf)
 info: Appium REST http interface listener started on 0.0.0.0:4723
 info: Console LogLevel: debug

Answer: It is very simple and easy, Please check the below steps!.
1) Install latest Node.js
Terminal commands:
curl -sL https://deb.nodesource.com/setup_8.x | sudo -E bash -
sudo apt-get install -y nodejs

More information please check official website.
2) Install latest Appium
Terminal commands :
npm install -g appium

or
sudo npm install -g appium

More information please check official documentation.
Answer: I've installed the NVM on CentOS 7 and it works perfectly.
curl https://raw.githubusercontent.com/creationix/nvm/v0.13.1/install.sh | bash

I will strongly recommend going for that with which you will have control of NodeJS versions.
Btw, Linuxbrew doesn't work for me.
Answer: I can tell about the 14.04 version

Download the file from https://nodejs.org/en/

2.Now for extracting the file path press
Ctrl+l , it will give the exact path , copy it
3.open bash.src file and it has to be open with gedit, this file is hidden
Press Ctrl+h to see the hidden files
4.now in the bash.src file,paste the below commands
   export PATH=$PATH:"somepath u copied upto bin"

   export NODE_PATH=$PATH:"somepath u copied upto bin node_modules'"

It will surely work


Why they use "would" in this case?

Question: If it's to practice a specific writing style--business-related letters in this case--the easiest way would be to read various examples. Afterward, or while reading, compose new business letters about any topic.
Comment: It's the "would" that occurs in the apodosis of remote conditionals where it indicates modal remoteness. Contrasts with the "will" that occurs in open conditionals.
Comment: ie it's merely a more 'hedged' version of '– the easiest way will be to read various examples.' Here, 'hedged' covers distancing, as BillJ says, which in turn involves less abruptness, and something of a disclaimer if the suggested course of action doesn't seem to work. So subsets 'oiling discourse ... politeness' and 'back-covering-.
Comment: Is the whole body text a quote from somewhere?
Comment: Using ***would*** instead of ***is*** or ***will be*** can reflect "deference" ***and / or*** "uncertainty". But sometimes it doesn't really carry either of those implications - it's just an "established" format that people are used to using.
Comment: "If you were to do it" is implied.
Answer: Either 'is' or 'would be' are appropriate in this example. However, 'would be' will often be preferred in this type of sentence because it implies a possible course of action in the future, while using 'is' is more of a judgement that that way is in fact the easiest possible way.
However, especially in spoken conversation someone might choose 'is' without actually meaning the extra implication of it being the easiest possible way.
The difference could be used by a writer as a subtle way to contrast personalities, but it would be very subtle unless it was part of a larger pattern.

Why they use "would" in this case?


Is this character too cliche?

Question: So I'm writing a story and my main antagonist is Death (At least from my MCs point of view). Is he as a antagonist too cliche? He owns a casino, his headquarters, and the setting is in 1700, London during a plague.
Comment: I'm imagining my Death as literal death personified and I am planning to later reveal that he was never the villain and in fact is more important than Life (opposite of Death)
Comment: Not Hades or the Devil, for Hades is portrayed as the Lord of The Underworld and the Devil is the King of Hell.
Comment: If anything he is more like Letum, Roman form of Thantos
Comment: Hmm... Interesting. But I don't think it is cliche at all. You should be fine! :)
Comment: This question is being discussed on [Meta](https://writing.meta.stackexchange.com/questions/2401/specific-cliche-requests-and-similar-questions-off-or-on-topic).
Comment: Thank You For Answering!
Answer: Your character sounds a little like Randall Flagg from The Stand: Flagg makes Las Vegas his headquarters, and a plague is present. From the information given, I would imagine you have read the Stand.
Now, that doesn't make it cliché, as one can still only claim a similarity between the two characters (not a cliché). However, seeing how people are still reading the book, going on half a century after it's publication (it's still in almost every bookstore I've been to), how a TV-series was made of it 30 years ago, and how they made another TV-series out of it last year or so, you might risk people seeing as a cliché, even if it isn't one.
Comment: First off, I haven't read the Stand, sounds like an amazing book though, second off, I'm really invested into this story- how do I make it so my character doesn't seem too much like Randall Flagg?
Comment: That's for you to figure out; Where Randall's nuance lie, what you want to take from him, what not, etc. Either way, if the story you have in mind is freshly imagined (originality is extremely rare, to non-existent in narratives these days), then I see no reason why you shouldn't write it, regardless of what people might think of it in the end.

Is this character too cliche?


tax_input in wp_insert_post partly not working

Question: I'm inserting a post via wp_insert_post() and in the $args the tax_input looks like this:
<code>'tax_input' => array(
    "programmas" => 'Modern Family',
    "zenders" => 'ABC',
),
</code>
But what is happening: the "programmas" is correctly inserted but the "zenders" is not. I'm 100% sure that I've spelled these custom taxonomies correctly. When running the function, a post is correctly inserted except for the "zenders" part.
I have all the capabilities, because I am the administrator and I've inserted my user id in the "post_author" part.
Answer: Try to append it in array, like this:
see this: Insert post with custom taxonomy and post meta data
<code>'tax_input' => array(
    "programmas" => array('Modern Family'),
    "zenders" => array('ABC'),
),
</code>


Produce decision version of the problem

Question: An optimisation problem requires minimising some function $f(x)$, where $x$ is a
vector of integers. What is the corresponding decision version of the problem?
Comment: what are the steps of creating decision versions?
Answer: "Given $f(x)$, what is $\min(f(x))$?" becomes "Given $f(x)$, and $k$, decide if there is an $x$ such that $f(x)< k$?"
Comment: Indeed. Your question is very basic. Please consult some introductory books on computability or complexity theory.
Comment: I see from the edit @blk that a decision must be a yes or no answer question?

Produce decision version of the problem


Does scaling apply if I don't level up?

Question: I am currently saving my perk points until I meet the skill requirements for the perk I want.  Let's say I am at level 16 and have enough XP to get to level 20.  Are the enemies scaled as though I am at level 20?
Comment: When leveling up, you can go to the perks screen, choose between magicka/health/stamina and stop there, keeping your points until you reach the level you want in a given skill to spend your points.
Answer: No.
Once you reach the next level, you do not gain a level for scaling purposes until you have taken that extra boost to Health/Magicka/Stamina. However, since you said "saving my perk points," keep in mind that you are not required to pick a new perk every time you level up; you can save them until you meet a particular requirement. In fact, it's actually quite common for perk points to exceed desirable perks if you have a particular build in mind.
In my opinion, it's generally not worth saving a level up for a few levels just so you can, say, take four perks you want all at once. The exception I can think of is if the intermediate perks are really awful and you're struggling with difficulty. Barring that, I prefer either taking the level up at once or using it opportunistically (you get a complete refill of health, magicka and stamina), especially early game.
Comment: +1, my immediate reaction when reading this question was "Saving level ups *for what?*" This isn't Oblivion, and there should be no reason to bank them.
Answer: I do not think so.
Although they continue to accrue through more skill use, the level ups are only applied when you open your char menu and assign the compulsory stat point. Before that point you have not levelled. Assigning the perk point is not necessary for a level-up.
All level related things are on hold until this point including your level, training points and level-dependent quests.

Does scaling apply if I don't level up?


GFCI Protection for Garage Lights?

Question: I know all receptacles in a garage require GFCI protection. But what about garage lights? I plan to hard-wire some new wiring to new garage lights and I'm wondering if this circuit needs to be on a GFCI breaker.
Answer: NEC 210.8.2 requires all outlets in a garage to be GFCI protected so if your lights are connected by an outlet they need GFCI protection if they are hard wired they do not require GFCI protection. Exhibit 210.10 provides a illustration of this.
Comment: Yes receptacle , not hard wired.
Comment: Better check current code @phillippnagel, the code now states all outlets and the exhibit I referenced shows a gfci for a corded connected door opener. There used to be an exception over 5-1/2' but that was removed several code cycles back.
Comment: sorry, you are right, I just checked, I misread, the exception is for some very specific devices, not as general as I remembered. I'll delete my comment.
Comment: when you say outlet do you mean receptacle? Technically, a hard-wired light fixture is an outlet
Answer: Find out if your jurisdiction has adopted NEC 2014 yet.  Many haven't, because the requirements for AFCI and GFCI are really, really over the top, especially since the requirements for $40 AFCI basically to cover lazy builders who insist on using backstabs.   Pre-2014, garage lights whose receptacles are on the ceiling, and therefore are not readily accessible,  do not require GFCI protection.
Most especially, lights should not share GFCI protection with common receptacles, because a tool tripping the GFCI will plunge you into the dark.  Now you have a dreadful situation: you are blind, the saw blade is still turning with a lot of energy, and the SawStop's contact detection has lost power, so it will not stop the blade if you get your hand in it.
Comment: According to the wording and exhibit I cited they do require GFCI protection got hit on plugged in lights on the house my company built last year. In the handbook the commentary specifically states there are no exceptions all receptacles are required no matter there location in both the 14 & 17 code.
Comment: @EdBeal cite added.  Either he considered your ceiling outlets accessible, or he has a local variance, or he's wrong.
Comment: @EdBeal hold on, I'm not sure.  I just realized my cite is rather dated.
Comment: I agree with your comments both on the reason they have been required and that the lights should not be on the same circuit for safety reasons.
Answer: Here in Italy is mandatory for all (home and home-related) circuits to be RCD-protected.
That's because here we use TT distribution and short-circuit current usually isn't enough to trip the magnetic breaker (short circuit protection). Also because RCD let trough little lower current than MCB so there's big chance of not being killed in case of touch of an active conductor compared with MCB-only protection.
So my suggestion is to install it anyway.

GFCI Protection for Garage Lights?


Finding B(t) in the Vasicek model relating to the bond equation, more specifcally from the initial condition

Question: In the Vasicek model for derving bond prices, we have the ODE $$\frac{dB}{dt}=\gamma B-1$$ which gives rise to the general solution $$B(t)=C_1 e^{\gamma t}+C_2$$My problem is that we have the "initial" condition $B(T)=0$, but apparently this one initial condition is sufficient for us to arrive at $$B(t)=\frac1 \gamma(1-e^{-\gamma (T-t)}) $$and I cannot see how this one condition allows us to realise both unknown constants.
Comment: Please check the equations. How the 1st equation follow from Vasicek model? Are you sure the 2nd equation satisfies the 1st? The bond price at maturity is face value, thus check the initial condition.
Answer: B(T,T)=0 implies the constant is zero. Here are the steps. Notice I have suppressed dependence of B on t and T in the beginning for clarity, and also ignore the dt in the 4th step:


Resource Scheduling in Arena Simulation Software

Question: I have a big problem in a simulation model. Entities arrive before and wait until 6.00 AM. And entities arrive after 6:00 AM, don't get any problems and walk inside. How does one model this in Rockwell Arena?
Answer: This course's site contains some examples and lecture notes that might be helpful for Arena. Specifically, this lecture covers scheduling resources in Arena which start on slide 30.
Your best bet is to do some Google searching. A lot of class notes/lectures and possibly some examples show up:

https://www.google.com/search?q=arena+resource+schedule
https://www.google.com/search?q=arena+resource+schedule+filetype:ppt
https://www.google.com/search?q=arena+resource+schedule+filetype:pdf

Finally, here is the full blown reference guide for Arena (pdf). I think more official documentation can be found here:
http://literature.rockwellautomation.com/idc/groups/public/documents/webassets/browse_category.hcst
Answer: I can't speak specifically to Arena, but in most discrete simulation engines tasks have a field where you can define criteria for a task to start.  In your case, you wouldn't allow a task to start until 6.
Answer: You can use the "based on schedule" element in the create module. Then at the schedule data module you can enter the arrival rates that is zero until 6 AM. However, if you are implying that you must also take into consideration the entities that arrived but waited until 6 AM then taken into process; you can possibly use a dummy create module that handles those arrivals that arrive before 6AM and terminate it at 6AM and then take that number as a extra arrival in your original system.


Is "Legs would carry them no longer" a metaphor?

Question: Is it? And if it is or is not, can you include some explanation why? And in which case it would/wouldn't be?
Answer: Definition: Metaphor
This phrase has been used in the "The Complete Grimm's Fairy Tales"

"And as they were so weary that their legs would carry them no longer, they lay down..."

You can literally read the phrase and it makes perfect sense, so it isn't a metaphor. The answer may vary depending on the source you are referring the phrase from. You'll have to add more detail to your question.
Answer: It isn't a metaphor. Nothing is being compared to their legs. It is arguably anthropomorphisation, since legs have no volition ("would"), only ability, so "could" is a more accurate verb.
Answer: Yes and No
The word "metaphor" has a strict meaning and a broad meaning. Strictly, it's a figure of speech that equates one thing to another, which is not literally true, such as, "I am the sun."
More broadly, "metaphor" can mean any figurative language. (Note, this broad usage is not just a layman's definition; it gets used this way in literary scholarship, too.)
So, strictly speaking, your example is not a metaphor. More broadly speaking, the expression is a personification: his legs were not literally "carrying" him, nor did they "refuse" to carry him at the end.
The expression evokes a feeling of the walker being "abandoned" by his own legs, which is, of course, not literally true.


How to recognize the 2nd edition of Railways Of The World?

Question: The 2nd edition includes cards for the Mexico map, which as far as I understand are rather indispensable for the 2 and 3 player variant of the game. My question is: is it possible to differ between the first and second edition on the box's outsite?
Answer: Besides for the Mexico variation, the reprint have slightly grayer art and a photo of the game in use on its backside..
Here is the two games for comparison :
Answer: Aparently, someone has already asked on BoardGameGeek. As I expected, it is very hard to see.


Ubuntu 16.04 does not start after installation next to Windows 7

Question: Installation ends succesfully, computer restarts and... Windows boots up.
   My computer: old laptop HP Compaq 6710b (no UEFI, no secure boot).
Two hard drives:
 - 1st is 240 GB SSD and 2nd is some old 200 GB HDD.
 - 2nd drive is installed in special adapter which fits in optical drive bay.
SSD is partitioned like this:

HDD has one partition:
Comment: Does the GRUB menu (OS selection) show or the Windows just boots immediately?
Comment: No menu pops up.
Answer: Your problem is probably that, during installation, it / you picked the wrong option in "device for bootloader installation". (See here for a screenshot)
Your laptop presents the hard drive as /dev/sda, and the internal SSD as /dev/sdb (From the screenshot). Thus, the installer installed the bootloader onto /dev/sda by default (Assuming you didn't tell it to do otherwise).
The problem is that the laptop isn't looking on the HDD for the bootloader - it's looking on the SSD, and thus only sees the Windows 7 bootloader (Which is on the SSD, and not the Ubuntu one (grub), which is on the HDD.

See How can I repair grub? (How to get Ubuntu back after installing Windows?) for details on how to fix this, making sure you intall to /dev/sdb, or you can use Boot-Repair for this.
Another, possibly easier, option is to set the boot order in the BIOS set-up, so that it is set to boot from the hard drive.
Let me know if it helps.
Comment: Also he could set the HDD before the SSD in the boot order.
Comment: Indeed, that's also an option, if the laptop supports booting from the HDD in the optical drive bay.
Comment: Boot-Repair did the trick. I quess it places grub on SDD drive.
Comment: Boot sequence setting - on the other hand - was usueles for me, cause there was no option to choose between HDD and SDD. When I invoke boot menu (using F9 on my computer), a see "USB Hard Drive" (when pendrive is connected), "laptop hard drive" (something like this) and "Ethernet" (or LAN something). It doesn't seem to notice that there are two drives installed.
Answer: I finally managed to regain access to my Ubuntu with following set of commands:
sudo mount /dev/sdb8 /mnt
sudo mount --bind /dev /mnt/dev/ &&
sudo mount --bind /dev/pts /mnt/dev/pts &&
sudo mount --bind /proc /mnt/proc &&
sudo mount --bind /sys /mnt/sys
sudo chroot /mnt
grub-install /dev/sdb
grub-install --recheck /dev/sdb
update-grub
exit &&
sudo umount /mnt/sys &&
sudo umount /mnt/proc &&
sudo umount /mnt/dev/pts &&
(All commands executed using Ubuntu 16.04 launched from USB drive.)


Desktop Windows Composition and Windows 8.1 x64

Question: I stumbled upon a problem when switching from Windows 7 x64 to Windows 8.1 Pro x64.
I have the EXACT same setup, same laptop, same hardware, same programs, same configuration, everything.
Windows 7 never did anything like the problem I now have in Win 8.1. Here's the problem;
Whenever I run more than 1 3D application, after 1-5 minutes the screen blinks, and hangs the graphics of one, some or all the programs (randomly). It doesn't hang the actual program (or game), it just stops rendering the graphics -- I can still hear the software's sound and if I click around, I can hear myself interacting with it.
I have ALL the latest drivers for the computer, motherboard, ALL devices including graphics cards etc. Also ALL Windows Updates installed, nothing more to install.
All of the programs I use is specifically set to use only the nVidia adapter, forcibly through the nVidia Control Panel, so it's not a switching graphics problem because it's a laptop. Also, never was a problem in Windows 7 either. I checked all the other threads about this, but it wasn't relevant to my specific problem.
Here's what I gathered from event viewer right when it happened;
A logon was attempted using explicit credentials.
Subject:
    Security ID:        SYSTEM
    Account Name:       PIXIELIGHT$
    Account Domain:     HYPERUNKNOWN
    Logon ID:       0x3E7
    Logon GUID:     {00000000-0000-0000-0000-000000000000}
Account Whose Credentials Were Used:
    Account Name:       DWM-1
    Account Domain:     Window Manager
    Logon GUID:     {00000000-0000-0000-0000-000000000000}

.
An account was successfully logged on.
Subject:
    Security ID:        SYSTEM
    Account Name:       PIXIELIGHT$
    Account Domain:     HYPERUNKNOWN
    Logon ID:       0x3E7
Logon Type:         2
Impersonation Level:        Impersonation
New Logon:
    Security ID:        Window Manager\DWM-1
    Account Name:       DWM-1
    Account Domain:     Window Manager
    Logon ID:       0x412659B
    Logon GUID:     {00000000-0000-0000-0000-000000000000}
Process Information:
    Process ID:     0x304
    Process Name:       C:\Windows\System32\winlogon.exe

.
Special privileges assigned to new logon.
Subject:
    Security ID:        Window Manager\DWM-1
    Account Name:       DWM-1
    Account Domain:     Window Manager
    Logon ID:       0x412659B
Privileges:     SeAssignPrimaryTokenPrivilege
            SeAuditPrivilege
            SeImpersonatePrivilege

.
An account was logged off.
Subject:
    Security ID:        Window Manager\DWM-1
    Account Name:       DWM-1
    Account Domain:     Window Manager
    Logon ID:       0x3E3290A
Logon Type:         2

.
So as you can see, DWM is logging in and out some account that can be all the way up to DWM-7 from what I've seen so far. It just logs in, sets some parameters and the directly logs out.
This is a ONE-user system, Only my user account is on it, and remote login has been disabled.
In Windows 7, at least DWM could be disabled - not that I ever had to, because it worked with all my stuff, but seeing as DWM is screwing up totally in Win8, I thought I'd just disable the server, but it's nowhere to be found.
Also, it seems Microsoft changed how it works, the system has been redesigned to generate an OK even if it's still on, like this example below:
DwmEnableComposition( DWM_EC_DISABLECOMPOSITION ) does not work on Windows 8, even though it returns S_OK.  Confirmation is by calling DwmIsCompositionEnabled(), which shows the former function had no effect.
There's only 1 dwm.exe running.
There's only 1 user logged in (me).
There's only 1 user in the userlist (administrative tools -> computer management) except for the Administrator account.
Running only 1 of the softwares at a time is not an option for me, either for work or for home use, so how can I fix this?
Comment: DWM can't be disabled without hacks: http://msdn.microsoft.com/en-us/library/hh848042%28v=vs.85%29.aspx
Answer: In the interest of information sharing, I just thought I'd post the resolution.
I didn't have time to experiment with this at all for another 2 weeks but then I eventually solved it.
Solution:
Set all the games to use the nvidia card.
Set all processes that has to do with obs/open broadcaster to use the integrated intel card.
Problem solved, with full FPS everywhere.


Is the print on eggs safe for consumption?

Question: I notice that many eggs nowadays have print on them that states the expiration date of the egg.  I have only see the print in pink.  Are these print safe for consumption, for example, if I boil an egg with such printing on it along with other food in the same pot, is it safe to eat the food boiled along with it?  I am wondering what's the content of the print, is it just food coloring, which should be safe to eat?
Comment: It'd be ironic if the eggs were marked with a date to enhance food safety, but the vendor used an ink that's not food-safe.
Answer: This is an excerpt:

Regulatory compliance. Inks used for egg coding must comply with
  government regulations governing food marking inks. These broadly state
  that inks printed on food items must be safe for human consumption,
  both when the food item is raw and when it is cooked. Egg coding must
  not weaken the egg shell or penetrate through the shell. Once dry, the ink
  should remain on the egg shell and not contaminate the egg during the
  cooking process.

Now, that is taken from this link which is actually an advertising pamphlet by a company selling their printing service, but it should be true for commercial products. If a small farm is stamping their own, then it might be worth questioning them.
I would expect to see more and more eggs getting the inkjet treatment as more countries are making it a requirement, not just acceptable.  One would assume this is to prevent practices like carton swapping and relabeling of old product.  Myself, the only eggs I have purchased with it on also happened to be in red and it was only logos, not date and tracking info.  They did not do a good job with the fast drying aspect as the logos were smeared on some eggs, but when the eggs were boiled it did not fade or further smear.  There are multiple food safe inks, often meat and meat products are inked thought typically on fat, skins or casings, but candies, baked goods and such are expanding that practice steadily.  I am not sure what type of inks they are using as soy based food grades any I have seen tend to smear and wash off.  I can only hope since these are food agency approved and even required in some places they were reasonably tested.
Comment: Videojet is a reputable authority for print technologies in the industrial food production sectors, so they are a good source to reference from.  However, this answer could be improved by the following: 1) reference to code for the reg cited, 2) archival or cached copy of the hyperlink included (which is now dead), 3) specifications on food-safe / GRAS inks in common use, which I suspect could be obtained from VideoJet if you have a rep who's willing to dig it up for you.


Prove that for each Fibonacci number $f_{4n}$ is a multiple of $3$.

Question: The Fibonacci numbers are defined as follows: $f_0 = 0$, $f_1 = 1$, and $f_n =
f_{n-1} + f_{n-2}$ for $n \ge 2$. Prove that for each $n \ge 0$, $f_{4n}$ is a multiple of $3$.
I've tried to prove to by induction. So, my basis is $f(0)$, which is true. Then, i assume that it's true for some integer $k$, and I am stuck at this point, can't really get what to do
Comment: A generic hint for proving things about Fibonacci numbers by induction: formulate the conjecture in such a way that it involves all of them. As induction hypothesis assume that it holds for two previos values (the recurrence relation refers to both of them so this is natural). Observe that the base case needs to check it for two first values of the parameter.
Answer: The question is indeed covered by the material at the link in the comments, but that's overkill and perhaps not most conducive to learning how to attack such a problem at an elementary level. Start by examining the first few Fibonacci numbers modulo $3$;
$$\begin{array}{rccc}
n:&0&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16\\
F_n:&0&1&1&2&3&5&8&13&21&34&55&89&144&233&377&610&987\\
F_n\bmod 3:&0&1&1&2&0&2&2&1&0&1&1&2&0&2&2&1&0
\end{array}$$
These numbers suggest the possibility that the sequence $\langle F_n\bmod 3:n\in\Bbb N\rangle$ is periodic with period $8$, repeating the block $\langle 0,1,1,2,0,2,2,1\rangle$ indefinitely. This would mean that
$$F_n\bmod 3\equiv\begin{cases}
0,&\text{if }n\equiv 0\pmod 8\\
1,&\text{if }n\equiv 1\pmod 8\\
1,&\text{if }n\equiv 2\pmod 8\\
2,&\text{if }n\equiv 3\pmod 8\\
0,&\text{if }n\equiv 4\pmod 8\\
2,&\text{if }n\equiv 5\pmod 8\\
2,&\text{if }n\equiv 6\pmod 8\\
1,&\text{if }n\equiv 7\pmod 8\;.
\end{cases}\tag{1}$$
This is quite straightforward to prove by induction. Let $P(n)$ be the following statement:
$$\begin{align*}
&F_{8n}\equiv0\pmod3,F_{8n+1}\equiv1\pmod3,\\
&F_{8n+2}\equiv1\pmod3,F_{8n+3}\equiv2\pmod3,\\
&F_{8n+4}\equiv0\pmod3,F_{8n+5}\equiv2\pmod3,\text{ and}\\
&F_{8n+6}\equiv2\pmod3,F_{8n+7}\equiv1\pmod3\;.
\end{align*}$$
We saw in the table above that $P(0)$ is true. The induction step is to assume $P(n)$ and prove $P(n+1)$.
Note that this is an example of a phenomenon that is actually quite common: it's actually easier (or at least more straightforward) to prove the stronger statement that $(1)$ is true for all $n\in\Bbb N$ than to prove directly that each $F_{4n}$ is divisible by $3$.
Comment: well, that seems to make sense. But then, how do I prove P(n+1)? like F8(n+1) = 0?
Comment: @user2847624: In eight pieces. The first thing that you have to prove is that $F_{8(n+1)}\equiv0\pmod3$. Your induction hyp. tells you that $F_{8n+6}\equiv2\pmod3$ and $F_{8n+7}\equiv1\pmod3$, so $$F_{8(n+1)}=F_{8n+8}=F_{8n+6}+F_{8n+7}\equiv2+1\pmod3\;,$$ and hence $F_{8(n+1)}\equiv0\pmod3$, as desired. Now keep going in this fashion for $7$ more steps.


What does running on your last click mean?

Question: In this GDC talk the speaker says

If you run on your last click you're doing it wrong.

What does running on your last click mean? What are some example use cases?
Answer: It seems to be slang relatedto the game Netrunner; all Google searches seem to lead there. It's impressive how obscure the language is, I thought I'd understand things easily by context but I can't even figure out how the game works. The best clue I found to its meaning is here:
https://boardgames.stackexchange.com/questions/15589/what-are-some-common-tactics-to-make-runs-against-servers-with-unrezzed-ice-in-n
The most relevant section:

Run with a click (and two credits) to spare, especially against NBN (and some Weyland). Tags are a genuine threat, and Corps won't ever try to tag you without having something nasty round the corner. Scorched Earth is a game-defining card. Your last click should be reserved for clearing a tag until you know you won't need to.

Soo, it looks like "running" is a game action, and "clicks" are resources required to do actions, and "running on your last click" appears to be ill-advised because you might need that last click for some other purpose after you run ?
The video you link to doesn't say much (otherwise you wouldn't be posting here presumably), other than paraphrasing this principle as "Instead of spending energy trying to be smart, try not to be stupid".
Putting all this together I might suppose the speaker is saying, don't make super-elaborate, utterly optimized plans (I figure people are tempted to "run on their last click" because they use all of the clicks before on very smart strategic preparation ?) that risk falling apart once something unexpected happens (like getting "tagged" by a Corp ?).
That's all I've got, you can probably get better info on a Netrunner forum, on Stack Exchange or Reddit.


How do Ethernet cables/USB cables send binary 0?

Question: I was wondering how your computer sends binary zero as an electrical signal. Is there a certain delay or does it do something unique.
Answer: The most basic CMOS/TTL logic uses a voltage within a specified min/max range to represent either a "logic low" 0 or a "logic high" 1. These include the discrete logic gates like 7400 / 7402 / 7432 that were used in the 1970's (and are still used sometimes on solderless breadboards), as well as more modern higher-integration chips. The exact voltage range is listed in the device's datasheet Electrical Characteristics table:
- VOH = voltage output high; specified as a minimum limit
- VOL = voltage output low; specified as a maximum limit

The device that is driving the output, is guaranteed to drive a logic low 0 as some voltage between GND and VOLmax; and also to drive a logic high 1 as some voltage between VOHmin and VCC (power supply rail). The gap between VOLmax and VOHmin is a dead band where the output is undefined -- this is what provides the noise immunity of digital signalling as compared to analog signals.
- VIH = voltage input high; specified as a minimum limit
- VIL = voltage input low; specified as a maximum limit

The device that receives the input, will interpret a voltage between GND and VILmax as meaning a logic low 0, and interpret a voltage between VIHmin and VCC as meaning a logic high 1. Any input between VILmax and VIHmin is not valid. And any input below GND or above VCC may violate the Absolute Maximum Ratings (i.e. permanently damage or degrade the device).
For CMOS, the VOH/VOL and VIH/VIL thresholds are usually a percentage of the power supply, like 30%VCC / 70%VCC. For TTL, the thresholds are absolute with 2.4V the usual VOLmin voltage.
Timing is a separate concern. If the logic is just implementing some Boolean equation ("glue logic"), the output signal simply follows the input after some specified propagation delay time. If the logic implements a state machine or a CPU, there will be a clock signal that determines the system's timing.

You also mentioned the Ethernet and USB communications protocols; these are a lot more complicated. It's a lot harder to even frame the question in terms of sending a single binary bit, since there is a lot more information that is required (such as host IP address, frame number, etc.) These build on the basic idea I described above, but add a lot more layers that are specific to each standard.
Ethernet has several layers of communications protocols; the datalink layer is different even for different types of Ethernet -- 10Mbit and 100Mbit are not just different speeds but different signalling protocols. This is described in IEEE standard 802.3
The USB protocols are described in the USB Standard, as well as on Jan Axelson's USB Complete website.
If you've never read a standards specification document before, I'd recommend starting with USB -- it's comparatively a bit simpler than Ethernet.
Answer: The simple answer is it depends...
Generally computers transmit information as bits, either a 1 or a 0.
To indicate a 1 most transmit devices/controllers raise the voltage on a line/output pin to a set value. Eg 3.3v and sink the voltage to grnd. eg 0v when they wish to indicate a 0. The receiver device then monitors the input line/pin voltage and when it crosses a threshold voltage (eg from low to high) it registers the change as a 1. The threshold will be a volt or so below the max (in this case say 2.5v) and the same happens when the voltage drops towards 0. When the receiver detects it below say 1v it registers a 0.
This covers a simple on/off yes/no signal.
To get more information, timing is used to let both ends know how long a "bit" is. Think of morse code and dot vs dash.
The circuit designer(chip dev) decides on the thresholds and protocol (timing)
When you start looking at USB or ethernet it works on the same principle but follows a recognized protocol specification. This governs the exact voltages etc to be used and allows interoperability between devices from different manufacturers.
They also include info on error checking and packet sizes etcetc but thats a whole other can of worms :)
Comment: I will add that the thresholds at the receiver are there to give a bit of leeway and attempt to filter out any stray voltages/resistance on the line.


There is an alternative to launchpad for a deb repository?

Question: I'm curious to know if there is a real alternative to Launchpad for a developer who wants to build his own repository.
Comment: Are you looking for a piece of software to create and publish your own repository yourself? Or are you looking for an online service providing this for you?
Comment: @gertvdijk a service, I would also like to set the steps used to build the software.
Answer: A repository is just series of directories. It's commonly on a web server (over http) but it doesn't have to be. The full process of doing this is long:

Create and upload a GPG signing key
Build and sign your packages
Set up a web server to host these things
Use something like Dak or Reprepro to set up the directory structure and export that to the webserver.

The result is a real repository that you control. The difference from a LP repo is there's no shortcut for adding your signing key to the client computers. You'll have to get them to use the old-fashioned method:
wget -q http://path/to/key.asc -O- | sudo apt-key add -

And then add the repo (add-apt-repository can add http://... addresses).
And even after that, you need to guarantee uptime. If you can't keep the repo up you're going to cause 404 warnings on clients' machines. Nobody likes that.
Debian has the best documentation on how to do this:

https://wiki.debian.org/HowToSetupADebianRepository
https://wiki.debian.org/SettingUpSignedAptRepositoryWithReprepro
Comment: thanks but I'm looking for a service, not a guide.


Stopping people from ssh(ing) into my Ubuntu machine

Question: I want to be able to stop people from been able to SSH onto my machine but still need to SSH out, is this possible?
Comment: Welcome to SuperUser!  I have suggested an edit to your question to remove the string "Is there a way of encrypting the whole drive instead of reinstalling the whole OS?".  If you want to know the answer to that, please ask it as a separate question.  And please elaborate:  make your question at least 50 words.  Include an explanation of why you want it.  :)  Enjoy the site, and again, welcome!
Answer:
i want to be able to stop people from been able to ssh onto my machine but still need to ssh out. is this possible

Yes, just disable the SSH daemon sshd.
Comment: Or just uninstall the server: `sudo apt-get remove openssh-server`
Answer: You have many options:

disable SSH daemon on boot with sudo update-rc.d ssh disable and reboot the machine (or stop the SSH daemon with sudo service ssh stop)
disable SSH daemon until machine is rebooted: sudo service ssh stop
select the users (or groups) authorized to ssh to your machine:
sudo nano /etc/ssh/sshd_config
add lines for each user:
AllowUsers user
or
AllowGroups group
then restart ssh: sudo service ssh restart to activate the filter
Answer:
I want to be able to stop people from been able to ssh onto my machine but still need to ssh out

If you do not want anyone (yourself included) to be able to ssh to your host, simply do not run sshd.  That has no influence in your ability to ssh from your computer to other computers.
E.g. via sudo update-rc.d -f ssh remove (This will ntt remove the sshd software package, but it won't automatically start again.).
I am curious though: Why not simply do not give them an account?
Answer: "I want to be able to stop people from been able to SSH onto my machine" - even with SSHd (the SSH daemon) turned on, this should in general not be possible unless:

You have a guest user, with no password, that has SSH access turned on
You have a regular user account that has a simple password, a password that others know or something that can be easily guessed.

Yes, you should definitely turn off sshd as others have recommended, but even with the default setup, it shouldn't be possible for the average Joe User to SSH into your box.
Answer: anybody who wants to ssh "in" to your system would need to have appropriate authentication to come in, which would be in your control. So, I believe your question is pretty much self answered, dont give anybody such access.
further, like many have suggested here, you can stop running the ssh daemon itself, which would mean that you yourself wont be able to ssh-in to your system, if you need to.
so really, you need to figure out, what exactly do you want to achieve, block others but not yourself, or block everybody (including yourself), and there goes your response.
another trick could be to change the port of your ssh daemon, which will basically make it harder for your system to be addressed by someone unknown. (for example, read here - http://techie-buzz.com/foss/change-default-ssh-port-in-linux.html)
Comment: thanks for all the negative votes.. would anyone care to comment why did you downvote this ?


Kähler potentials that depend only on geodesic distance

Question: Hermitian symmetric spaces of constant curvature have the property that the potential for their Kähler metric can be expresed as some function of the geodesic distance. Does anyone know if there are any general results concerning Kähler manifolds with this property?
Comment: Could you elaborate on this question? How can a metric fail to be a function of geodesic distance?
Comment: @Igor: He means that a potential for the metric is of the form $f(x)=h(d(p,x))$ where $p$ is a fixed point and $h$ is some function of one variable.
Comment: For a hermitian symmetric space of constant curvature, this holds for any point $p$ (as defined in Misha's comment). It should not be difficult to give an example where it holds for only one particular choice of $p$ but not for others. So presumably the question is about manifolds where the potential can be written as a function of distance from a point $p$, for any choice of $p$?
Comment: Also, I am under the impression that the statement is not literally true for a compact manifold, since the Hessian of the distance function has to degenerate at the cut locus. It seems to me that the question needs to be stated more carefully.
Comment: @Misha: Correct Misha. The Fubini-Study, Begman, and the Euclidean metrics all have potentials of this form.
@Yang: I'm speaking locally here; obviously the cut locus will need to be avoided. However, the base point $p$ is arbitrary; you simply avoid the cut locus of $p$.

I've thought about this some more and I realized that the potential is also a function of $|z|^2=\sum_i|z_i|^2$. Here $(z_1,\cdots, z_n)$ are local coordinates. Rotationally symmetric metrics have this property.
Answer: Update:  I have had a little time to think about this further and have been able to show that if a smooth Kähler metric $g$ on a complex $n$-manifold $M$ has a potential $f$ that is a function of the distance from a point $p\in M$, then the metric must be locally rotationally symmetric about $p$, i.e., the group of local $g$-isometries that fix $p$ is isomorphic to $\mathrm{U}(n)$.  In particular, this shows that, if it has this property with respect to every point $p\in M$, then it must have constant holomorphic sectional curvature.  I'll describe the steps in my argument, but I won't put in the details (which are somewhat long) unless there is interest.  I'm leaving my original answer below, for notational purposes, even though that preliminary analysis was inconclusive.
Fix notation as in the original answer: There is an open $p$-neighborhood $U\subset M$ on which there exists a Kähler potential $f$ for $g$ that is a function of the $g$-distance $r:U\to\mathbb{R}$ from $p$, i.e., $f = h\bigl(r^2\bigr)$ for some function $h$ and $\frac{i}{2}\partial\bar\partial f = \Omega$, where $\Omega$ is the Kähler form of $g$.  (Clearly, one can assume that $h(0)=0$, and it is not hard to show that $h$ must be smooth on $[0,\epsilon)$ for some $\epsilon>0$ and must satisfy $h'(0) = 1$.)  Because $r$ satisfies $|\nabla r|^2 = 1$, it follows that $|\nabla f|^2 = 4 r^2 h'(r^2)^2 = \phi(f)$
for some function $\phi$ that satisfies $\phi(0)=0$ and $\phi'(0)=4$.  Elementary identities then show that $f$ must satisfy the equation
$$
\partial f \wedge \bar\partial f \wedge \bigl(\partial\bar\partial f\bigr)^{n-1}
= \frac{1}{4n}\phi(f)\ \bigl(\partial\bar\partial f\bigr)^{n}.
$$
Conversely, if $f$ is any solution of this equation on a neighborhood of $p\in M$ that satisfies $f(p) =0$, $df_p=0$, and $\frac{i}2\partial\bar\partial f >0$ at $p$ (in the sense of $(1,1)$-forms, then the Kähler form $\frac{i}2\partial\bar\partial f$ deinfes a Kähler metric $g$ such that $|\nabla f|^2 = \phi(f)$ and, from this it easily follows that $f$ is a function of the $g$-distance from $p$.  It remains now to show that such an $f$ must be rotationally symmetric in the above sense with respect to some $\mathrm{U}(n)$-action on a neighborhood of $p$.
This problem can be simplified in the sense that, if such an $f$ exists, it can be shown that there is a function $\psi$ defined on a neighborhood of $0$ such that $\psi(0)=0$ and $\psi'(0)=1$ and such that $s = \psi(f)$ will satisfy
$$
\partial s \wedge \bar\partial s \wedge \bigl(\partial\bar\partial s\bigr)^{n-1}
= \frac{1}{n}s\ \bigl(\partial\bar\partial s\bigr)^{n}.
\tag1
$$
In other words, one can reduce to the case $\phi(t) = 4t$ (which is the same as saying that one reduces to the case in which $f = r^2$).  Thus, we assume that $s$ satisfies (1) from now on.
The next step is to show, using a straightforward Taylor series argument, that, for any $k\ge2$, there exists a local, $p$-centered holomorphic coordinate chart $z_k:U_k\to\mathbb{C}^n$ in which $s - |z_k|^2$ vanishes to order $k$.  This shows that the Kähler metric associated to $s$ is flat to infinite order at $p$, but, since we do not (yet) know that $g$ is real analytic, this does not show that $s$ is flat everywhere.  (Also, one does not (yet) know that the sequence of coordinate charts $z_k$ has a convergent subsequence.)
Finally, one uses the fact that each geodesic through $p$ lies in a totally $g$-geodesic complex curve passing through $p$ to show that the $(2,0)$ part of the Hessian of $s$ restricts to each such curve to be generated by pullback of a holomorphic map from the complex curve to the Hermitian symmetric space $\mathrm{Sp}(n,\mathbb{R})/\mathrm{U}(n)$.  Since this map must have its differential vanish to infinite order at $p$ (by the argument in the previous paragraph), it follows that it must be a constant map, which implies that the $(2,0)$ part of the Hessian of $s$ must vanish identically, i.e., that $\mathrm{Hess}(s) = g$.  From this, it is elementary to conclude that $g$ must be flat and hence $s$ must be invariant under the $\mathrm{U}(n)$-rotations.  Since the original $f$ is a function $s$, it, too, must be invariant under this $\mathrm{U}(n)$-action, and, in particular, must be of the form $f = h\bigl(|z|^2\bigr)$ for some function $h$, as desired.
Original Answer: This is only a preliminary answer, but it's too long to go into a comment, so I'm putting it here. I'll add to it when I have the time to figure out more about it.
It seems that there are two questions here, a pointwise one and a local one:  First, let's say that a (smooth) Kähler metric $g$ on a complex manifold $M$ is polar at $p\in M$ if there is an open $p$-neighborhood $U\subset M$ on which there exists a Kähler potential $f$ for $g$ that is a function of the $g$-distance $r_p:U\to\mathbb{R}$ from $p$, i.e., $f = h\bigl({r_p}^2\bigr)$ for some function $h$ and $\frac{i}{2}\partial\bar\partial f = \Omega$, where $\Omega$ is the Kähler form of $g$.  (Clearly, one can assume that $h(0)=0$, and it is not hard to show that $h$ must be smooth on $[0,\epsilon)$ for some $\epsilon>0$ and must satisfy $h'(0) = 1$.)  Let's say that such an $f$ is a polar potential for $(g,\Omega)$ at $p$.  If a polar potential at $p$ exists on some $\epsilon$-ball about $p$, it is unique on that ball.
Now, a polar potential satisfies a natural differential equation, namely, the differential equation that states that the $g$-gradient lines of $f$ are $g$-geodesics.  This is, typically, an overdetermined equation for a pseudoconvex potential $f$, so one can hope to get some information by using this.  In fact, calculation shows that this implies that each of the geodesics emanating from $p$ lies in a unique, totally geodesic complex curve passing through $p$.  Moreover, the induced metrics on all of these complex curves (one tangent to each complex line in $T_pM$) are isometric and, moreover, they are each possess an isometric rotation about $p$ within the complex curve.
In complex dimension $1$, this implies that the metric is rotationally symmetric about $p$, so the polar potentials are the rotationally symmetric ones.  In particular, in dimension $1$, a Kähler metric that is polar at every point has constant curvature and hence is locally symmetric.
In higher dimension, since the curves are totally geodesic and isometric, they all have the same curvature at $p$ and hence it follows that the ambient metric has constant holomorphic sectional curvature at $p$.
In higher dimensions, it appears that, while rotationally symmetric (i.e., $\mathrm{U}(n)$-invariant) pseudoconvex potentials are polar with respect to the center of rotation, these are not the only solutions of this equation.  Already in dimension $2$, it appears, from the structure equations, that there are (local) pseudoconvex polar potentials that are not rotationally symmetric.  I haven't had time to integrate the structure equations, though, so I can't yet say what these solutions look like.
However, by the above argument, a Kähler metric that is polar at every point must have constant holomorphic sectional curvature, and hence the only metrics that have this property at every point are the complex space forms.
Comment: Robert, is the term \textit{polar} your terminology? If so, what motivated it?

I found a paper by Gromov entitled "K\"{a}hler Hyperbolicity and $L_2$-Hodge Theory" in which he states that for Hermitian symmetric spaces the K\"{a}hler potential is expressible as a function of geodesic distance. I didn't see any restrictions on curvature. Have I missed something?
Comment: @Oliver:  Yes, I chose *polar* arbitrarily so that I'd have something to call the property; I was thinking of 'pole of rotation', nothing deeper than that.  I'm surprised that Gromov would claim this; it's not true for the product of the Poincaré disk and the complex plane, which is an Hermitian symmetric space, albeit a reducible one.  Perhaps Gromov meant to write 'rank 1 Hermitian symmetric space' and inadvertently omitted the *rank 1* part?  I can put in a sketch of the argument if you are interested.
Comment: @Oliver: Here's a way to see that the irreducible higher rank Hermitian symmetric spaces *can't* be polar (at any point): The polar property is clearly inherited by any totally geodesic complex submanifold (with the induced Kähler structure).  So, for example, if the real Grassmannians $Gr(2,n)=SO(n)/(SO(2)SO(n))$ were polar for $n>3$, then $Gr(2,4)=S^2xS^2$ would be polar, but it's not.  The Kähler potential for $S^2$ is $f(r) = 8\log(\sec(r/2))$ and the distance function for $S^2xS^2$ is $r = \sqrt(r_1^2+r_2^2)$, but, clearly, no function of this $r$ can be a Kähler potential for $S^2xS^2$.
Comment: @Robert: Gromov had the proviso that the Hermitian symmetric space has no Euclidean factor. That takes care of your product example. Sorry for the omission. His paper is available online or I can send it to you if you're interested.
Comment: @Oliver: I looked at Gromov's paper; the example you quote is on the second page.  He clearly intended the polar claim for all Hermitian symmetric spaces, just not the claim of boundedness for $df$ (which is what he really cared about) if it has an Euclidean factor.  In spite of his claim, Hermitian symmetric spaces of rank bigger than one do *not* have the polar property (whether it has an Euclidean factor or not).  The argument that I gave above for $S^2xS^2$ works for the product of two Poincaré disks as well, which shows that this product, which has no Euclidean factor, isn't polar either.
Comment: @Robert: OK, so you're saying Gromov is wrong. I'll need time to study your comments.
Comment: @Oliver:  Yes, he slipped on this.  Consider the Poincaré disk metric in polar coordinates: $g=dr^2+\sinh^2rd\theta^2$ and the area form is $\Omega=\sinh rdr\wedge d\theta$.  The Kähler potential is $f=8\log(\cosh(\frac12r))$.  On the product of two copies, $g=dr_1^2+\sinh^2r_1\ d\theta_1^2+dr_2^2+\sinh^2r_2\ d\theta_2^2$ and $\Omega=\sinh r_1dr_1\wedge d\theta_1+\sinh r_2dr_2\wedge d\theta_2$ while $r = \sqrt(r_1^2+r_2^2)$ is the geodesic distance from the point $(r_1,r_2)=(0,0)$.  However, there is no function $h$ of 1 variable so that $f=h(r)$ is a Kähler potential for this metric.
Answer: @Robert: This was too long for the comment field and so I'm putting it here.
I have another example which backs up what you're saying. Hermitian symmetric spaces of compact type admit a K\"{a}hler embedding into complex projective space. If we consider two points $x$ and $y$ on the embedded submanifold $M$ then there are two distances; the geodesic distance $d_M(x,y)$ in the submanifold and the Fubini-Study distance $d_{FS}(x,y)$ in the projective space. An implication of your result is that it is not possible to express $d_{FS}(x,y)$ as a function of $d_M(x,y)$ (except for complex space forms). I looked at the case of the complex Grassmannian for which there are explicit formulas for geodesic distance in terms of stationary angles. It's clear that you are correct; a simple calculation shows that it's not possible to express the Fubini-Study distance solely as a function of the geodesic distance. Thanks very much for your answer.
Comment: @Oliver:  Well, since the Kähler isometric embedding of a compact type Hermitian symmetric space into a complex projective space that you speak of is not totally geodesic in the higher rank case, the argument I was mentioning above doesn't really apply.  I don't see how you are drawing your nonexistence conclusion.  I can see that this says that there is a Kähler potential that is expressed in terms of $d_{FS}$ and it's clear that $d_{FS}$ is not a function of $d_{M}$, but how does that prove that there's not a *different* Kähler potential that *is* expressible in terms of $d_{M}$?
Comment: @Robert: yes, good point. When formulating the example above I had in mind Calabi's diastasis potential. This has a nice property with respect to Kahler submanifolds. However, the diastasis is constructed from an arbitrary potential and I don't think it would be difficult to show that if there existed a polar potential then the diastasis must also be polar. I'll check.


Why Ctrl+Tab stops working when opening a flash embedded page?

Question: This a huge pet peeve I have. Everytime I watch a video in youtube.com or megavideo.com, I can't just Ctrl+Tab to change tabs. I'll have to clic in some portion of the page without video to get it to work, but even that, isn't always the case.
Answer: The flash player takes over the Tab keybinding when it has focus. If the flash player inside Firefox has focus, when you press Ctrl+Tab, Firefox just sees the Ctrl keypress because flash ignores the fact that Ctrl is being held down and intercepts the Tab keypress.
Unfortunately, I haven't found a solution to this problem yet.
Answer: The trick is to bring the focus of the keyboard away from the Flash and back to the browser:

Click on the page background (outside the Flash area)
Now Ctrl+Tab works!

Dan's reasoning is correct:
when the Flash app has the focus (for instance because you clicked its Play button)
then your key presses go to the Flash app (useful if it's a game) and not to the browser.
By clicking outside the Flash app, you put the focus back to the browser, so the browser will interpret your key presses again. This is not limited to Ctrl+Tab but to any normal keypresses (meaning that the Winkey or Ctrl+Alt+Delare still handled by Windows as you'd expect, but that's because these keys aren't intended at the browser in the first place).
Answer: If you hit alt-tab then alt-tab again to get back to the same window the player loses focus. I use the keyboard more than the mouse though.


How to get started with image processing and photogrammetry?

Question: I am sorry about the Title for this question, but really I don't know how to name it.
There is something that I have seen sometimes, and I ask myself how to do that kind of software. It's about the PitchTrax that MLB uses in its TV-games. I think that is did it using cameras or something like that... I would like to read (or to learn) how to do little applications using this kind of technology, but I don't know anything about that, really I don't know where "to start" studying for this... Do you know something about this??
I am sorry about my English.
Comment: I think what the OP is trying to ask when he refers to PitchTrax is when the pitcher throws the ball, the TV displays the ball's path of travel from the pitchers hand to the catchers mitt
Answer: Looks like it uses cameras to track the ball as it is pitched. From here:

The UIS uses QuesTec's proprietary
  measurement technology. Quite
  different than "video insertion"
  technology that simply adds graphics
  to the broadcast video, QuesTec
  technology actually measures
  information about interesting events
  during the game that would not be
  available any other way. This
  technology is so innovative it
  appeared in a Scientific American
  article in September of 2000. The ball
  tracking component uses cameras
  mounted in the stands off the first
  and third base lines to follow the
  ball as it leaves the pitcher's hand
  until it crosses the plate. Along the
  way, multiple track points are
  measured to precisely locate the ball
  in space and time. This information is
  then used to measure the speed,
  placement, and curvature of the pitch
  along its entire path. The entire
  process is fully automatic including
  detection of the start of the pitch,
  tracking of the ball, location
  computations, and identification of
  non-baseball objects such as birds or
  wind swept debris moving through the
  field of view. No changes are made to
  the ball, the field of play, or any
  other aspect of the game, to work with
  QuesTec technology. The tracking
  technology was originally developed
  for the US military and the company
  has adapted it to sports applications.

So if you are looking to learn something about this, image processing would be a good place to start.
Answer: There's a bit of detail about how the technology works in this article in the NY Times.
It's not a lot, but it may help you get started with where to look. It mentions the company that wrote the PitchTrax software, and a bit about how they use stereoscopic camera technology to figure things out. There also seems to be some related info in this blog.
You may also want to buy (rather than develop) stereoscopic imaging - it's not a simple field.
Answer: I don't know pitch trax specifically (try google or their site)
But the general field of object tracking and image processing then openCV is probabyl a good place to start (as a programmer)
There is a good book at http://www.amazon.com/Learning-OpenCV-Computer-Vision-Library/dp/0596516134/ref=sr_1_1?ie=UTF8&s=books&qid=1262807256&sr=8-1
Answer: I have no idea.
But I can make stuff up that sounds good.
I'm guessing they're using several cameras (at least two, perhaps three) and image processing to "follow the ball".
Once they can identify the ball (not super difficult if you think about it), then each camera can know "it's space" in a 3D realm. Multiple cameras will fix the ball in space, and from that they can calculate the track.
In 3D Graphics, there's a concept called "Match Move" that is used to place 3D models in the space observed by cameras. You need to know the camera angles, etc. plus the focal lengths of the lenses to figure that out.
I imagine the underlying tech is related. One extracting objects from space, the other putting them in.
Answer: In general, you're looking for information about Augmented Reality- the wikipedia article isn't a bad place to start. Processing has some third-party OpenCV bindings, which could make it easy to play around with the technique.


Is there a way to force Thunderbird to start with the default colums for the Inbox every time?

Question: I have several elderly (75 and 80+) family members that are reasonably able to use a computer.
(I'm actually quite proud of them that they got so far. They all never touched a computer until well in  in their 70's.)
But they all struggle with the same problem (and as the family Helpdesk I get to deal with it).
Using Thunderbird for email goes reasonably well, but there is one aspect they somehow keep managing to mess up.
The Inbox view: They manage to hide columns/change the sort-order and they don't know how to set it back to normal.
From observing them using the computer it is clear they do it by accident. None of them has very good eye-hand coordination with the mouse. They move the mouse over the top bar of the Inbox view while accidentally keeping a mouse-button pressed. This changes sort-order when they hit any of the column-headers (left mouse button), hides a column (right mouse button) and sometimes they manage to hit the button that activates the column-selection menu and they accidentally toggle entries in that menu.
Thunderbird retains the last used configuration over restarts, so the only way to get it back to "normal" is to select "Reset to default" from the column-selection menu. Even though they sometimes manage to find this menu "by accident", I have found it next to impossible to have them be able to reset it by themselves.
Any action that is not in plain sight and requires another action to bring it into view (like clicking a button to call up the column-selection menu) is just to complicated. I can explain and demonstrate it to them again and again and they keep forgetting how to do it.
So that leads me to my question: Is there a way to make Thunderbird "forget" any user-changes to the columns of the Inbox view on startup so it always starts with the default columns and sort-order?
If there is such a possibility I can setup their Thunderbirds to always startup with the desired layout.
I have been looking through all the advanced config settings and there doesn't seem to be any way, but I might be missing something.
Please note: Use another mail-client isn't an answer. Getting them to Thunderbird (from Outlook Express and Windows Live Mail) was painful enough a couple of years ago. The Windows 7 to Windows 10 transition didn't help either. I'm not going to force another transition on them.
Comment: A non-technical solution, but you could always just write them a note that explains how to reset the view. That way they have something to refer to if they forget.
Comment: @MattiVirkkunen Done that. Full instruction with pictures. That works a couple of times. Then they think they know it by heart and throw it away. 3 days later my phone rings again...
Comment: At that point I would probably tell them to go get the instructions back from the trash...
Comment: Workaround - compacting all folders it seems to reset them to the default columns, maybe they could learn to click file, compact when starting up?
Answer: Unfortunately, no.  Not today, and nothing coming soon that I know of.


How to prevent a IIS7 Rewrite rule from loading ASP.Net Ajax Client Framework?

Question: I'm developing a webapplication where users will have a custom url just like in Twitter (twitter.com/holiveira). I've created a redirection rule that points to a page where I use the string after the domain name to search in the database.
The problem is that this rule is preventing the Scriptresourse.axd files used by Asp.Net Ajax Client Framework from loading properly.
Any idea how to solve this issue?
This is the rule I'm using:
<code>            <rule name="RewriteDropbox" stopProcessing="true">
                <match url="^([^/]+)/?$"/>
                <conditions logicalGrouping="MatchAll">
                    <add input="{REQUEST_FILENAME}" matchType="IsFile" negate="true"/>
                    <add input="{REQUEST_FILENAME}" matchType="IsDirectory" negate="true"/>
                </conditions>
                <action type="Rewrite" url="dropbox.aspx?clienturl={R:1}"/>
            </rule>
</code>
Answer: I figured out the problem. The rewrite rule was to loose, since I was not adding the domain to it, and this caused the requests fo the ScriptResource.axd file to get lost, because it was redirected to my dropbox page.
When I added the domain to the rule it worked perfectly.


How to keyframe all the objects viewport/render visibility inside the selected collection via python code?

Question: How to keyframe all the objects viewport/render visibility inside the selected collection via python code? And there's a boolean there that you can just type "False" or "True" to decide whether unhide or not. I only have a little bit of knowledge of scripting, especially in blender python library. I was thinking of this after knowing collections viewport/render visibility can't be keyframed, because I want to keyframe the object collections that aren't needed in certain camera shots. But trying to make it myself takes long time now, that I have to meet a certain deadline to pass my project.
Answer: After several moments I just got it, but I know there maybe a cleaner approach
<code>import bpy

for ob in bpy.context.collection.all_objects:

    ob.keyframe_insert(data_path="hide_render")
    ob.keyframe_insert(data_path="hide_viewport")
    if ob.hide_render == True:
        ob.hide_render = False
    else:
        ob.hide_render = True
    if ob.hide_viewport == True:
        ob.hide_viewport = False
    else:
        ob.hide_viewport = True
</code>


Raspberry Pi file sync

Question: Okay so I have a cluster of pi's version 3b running up to date raspbian.  I need a quick way to make changes (upload/sync) files to 20 pi's preferably on boot as I have a startup script that will launch the app but would like the newest version of the app to be run.  However the boot script is run at rc.local as it need root permissions.  So how can I sync files + run as root on startup?  Any method will work github/dropbox/nfs/etc. (nfs will be tricky as the network is huge with many subnet so I prefer to avoid it as I am not the network manager and its a pain to get him to set things up for me) so for that reason I prefer some kind of remote file sync.
Comment: sync application updates.  All the data gets synced to a db on a server.  Yes its a cluster as all the pi's are working on the same thing but with a different section of data.
Comment: Do you want to synchronise application or data? And do you really have **a cluster**? What clustering solution do you use that leaves you wondering how to sync files?
Answer: Try GlusterFS: https://blog.godatadriven.com/rpi-gluster.html
I have never tried it myself on Pi, but we are using it to connect CentOS workstations for the same purpose.

Raspberry Pi file sync


Can webcams be turned on without the indicator light?

Question: I've made a series of penetration tests in my network and one of the things I've tried was to record webcam and microphone.
Recording an end-user's microphone seems to be a stealth thing, but what about the webcam?
In my tests, the indicator is turned on and I can't figure out a way to do this without turning on the light.
So far, I'm assuming that if someone broke into my computer and turned on the webcam, I'll know that.
But, if that's possible, which of the available hardwares on the market are vulnerable to that kind of attack?
Comment: What kind of access? And what about an employees computer controlled by their company?
Answer: Most definitely,  but depending on the manufacture you may have to patch the camera's firmware and then flash it.   Similar attacks have been used to disable the "shutter sound" on cameras.
Some camera manufactures have a hidden option to disable the camera light, which maybe desirable by some (perhaps unusual) customers.
Comment: The answer is absolutely wrong! You're forgetting about the fact that it doesn't take firmware, but only a simple change in system configuration (as malware would be able to change). Check http://forums.logitech.com/t5/Webcams/Can-I-turn-off-red-LED/m-p/277305#M52816 for one of the many examples. LOGITECH WEBCAMS as an example DO NOT need firmware changes to have their light disabled! And note that I'm not yet talking about former Adobe Flash security issues that enabled camera use without indicating activation (read: while leaving the activity-indicating light off). Firmware only? You wish!
Comment: I'm not trying to troll this answer, but I don't really see why this isn't just hyperbole.  If the camera's power circuit includes an LED then how could one power the camera without powering the LED?  I know there is a link in the above comment, but that is for a peripheral device, not an integral laptop camera.  Has someone actually looked to see if they are on separate circuits controlled independently by firmware? Lights were added by manufacturers to solve the RAT problem, so I'd be surprised if they didn't solve this simply by pulling the LED in parallel from the power circuit.
Comment: @user6373 that is one way of accomplishing the same task,  it depends on the device. But, at the end of the day, the lights on the device is controlled by the firmware.
Comment: @Keyne doubtful,  its probably firmware. in that the device has a set reactions to an incoming command like "start recording".
Comment: @rook most cameras (including many if not all Logitech cameras) have the light controlled by firmware, yes. However, this firmware has a SOFTWARE INTERFACE that offers application-based control over the camera's indicator light. We actually use this at work to disable the LEDs on our C920 cameras, just takes a command from `uvcdynctl` to set the camera to "LED Off" before activating the camera, and the LED never comes on. I suspect many other cameras have similar interfaces with this potential for abuse; however, there are also legitimate uses for it.
Comment: Does drivers have something related?
Comment: @AN6U5 Some developers find real value in connecting the dots.
Answer: Im not sure about built-in webcams, but I think it is most likely possible.
I've found info on Logitech Webcams, where you can turn off the LED in the registry keys..

For QuickCam versions 18.104.22.1689 thru 18.104.22.1683, LVUVC_LEDControl is located in the following registry key:
HKEY_LOCAL_MACHINE\SYSTEM\CurrentControlSet\Control\Class\{6BDD1FC6-810F-11D0-BEC7-08002BE2092F}\0000\Settings

Note - If more than one camera is installed, you will have a "folder" for each device (i.e., 0000, 0001, 0002, etc...).
It has a default Data Value of REG_DWORD = 0x00000005 (5).
Based on your comments, I will assume that the following information is true:
0x00000000 (0) = LED Off
0x00000008 (8) = LED On
Please note that only certain cameras support this feature, so the mere addition of this key will not cause the LED Control buttons to appear in the QuickCam® Advanced Settings.

It is taken from: http://forums.logitech.com/t5/Webcams/Can-I-turn-off-red-LED/m-p/277305#M52816
Hence, I do not see why it shouldnt be possible for built-in webcams, since they also require drivers (mine is for example up side down after a fresh install)
Comment: @Keyne Computerworld just brought an article about a flaw in flash that could allow webcam spying. http://www.computerworld.com/s/article/9221052/Adobe_to_fix_Flash_flaw_that_allows_webcam_spying?source=CTWNLE_nlt_security_2011-10-21&utm_source=feedburner&utm_medium=feed&utm_campaign=Feed%3A+computerworld%2Fs%2Ffeed%2Ftopic%2F17+%28Computerworld+Security+News%29
Answer: It really depends on the type of camera, and how it is built. On Apple MacBook Pros (at least recent ones), the camera light is directly connected via the power supply to the camera module, so the light can't be circumvented even via a firmware hack. Older (really old) Logitech cameras had a similar design.
But just to be safe, I'm one of those guys with a dark tape over the camera.
Comment: Any teardown evidence to support this claim?
Comment: A few years ago, some researchers found out a way to turn on the webcam on a Mac without the light being turned on: https://jscholarship.library.jhu.edu/handle/1774.2/36569
Comment: I'm not familiar with 2011 Macs, but I'm pretty sure the camera is just connected via USB like in any other laptop and the LED is controlled by the firmware. According to your description the LED would be on all the time because the camera would be powered on constantly.
Comment: @user2428118 This was on pre-2008 Mac models only.
Answer: It seems as if the design of my ASUS notebook is good in this respect.
It uses a hardware video shutter, that cannot be turned on electronically.
That is a great solution to the privacy issue (excepting sound recording as you said).
Comment: This is what I have on all my web cams - it is a prereq in my buying decision:-)
Comment: A piece of electrical tape also works and looks stylish.
Answer: One other thing that I haven't seen mentioned was popularized during the Lower Merion District School spying scandal last year is that the software that was used (LANRev TheftTrack) got around the Apple MacBook's camera light protections (camera light is hardwired to camera power) by simply turning it on briefly for a snapshot. Thus the camera isn't on all the time, but only at random intervals for a split second.
Comment: This is a common feature on metasploit framework and it's used to take pictures.
Answer: Yes it can be done. Many web-camera control programs give you the ability to turn off the light. So it's definitely possible.
So as to know if the camera is active at a given time, I guess the best thing you have to do to ensure it stay disable, is to turn it off in the device manager. Off-course, you have to trust the Operating System to obey your commands. But that is a different question.
Answer: Under Linux it's possible to control the LED on a number of USB cameras using the uvcdynctrl libwebcam command line tool (available via most package managers) e.g. to turn off the LED on device video1:
uvcdynctrl -d video1 -s 'LED1 Mode' 0

On Windows a number of cameras come with control software that allows for disabling of the LED in a number of cases (e.g. some Logitech models).
Comment: I use `uvcdynctl` at work for our fleet of systems, it's quite handy for us -- but yes, this demonstrates that many cameras allow the LED to be controlled by software, thus rendering the status light an inadequate indicator of compromise as OP was asking.
Comment: For reference, our users are aware the camera may be recording (it's stated in our signed agreements with them, and also in a short-form FAQ we have too, just to ensure they're aware), but the light would be incredibly annoying and distracting to them so we shut it off.
Answer: At least on older MacBooks / MacBook Pros from 2007 and 2008, there is a way to disable the LED as demonstrated by researchers from Johns Hopkins University and published in a paper.
They have also created a kernel extension for Mac OS X to prevent such an attack, called iSightDefender, available as source code on GitHub.
So, looking at those Logitech and iSight webcams' security, you can safely say that your privacy is not safe with a webcam aiming at you. You might also want to look at those laptop webcam cover stickers from EFF.
Answer: Aside from uvcdynctrl mentioned in other answers, it's also possible to use v4l2-ctl in Linux:
v4l2-ctl --device=/dev/video0 --set-ctrl led1_mode=0

Answer: I used a neat little trick that will go unnoticed by the unsuspecting. I use a very small round rubber that often comes on the bottom of small devices like Vonage VoIP router. It provides a little lift off the surface to allow better ventilation and to reduce sliding of the device. I just use it to cover the LED and persons will think it came with the laptop.
Alternatively you can use the very small kitchen cabinet door cushion. They are normally used to reduce the banging when the cabinet door is close. Some are plastic or synthetic cloth with adhesive on one side. Pick up a pack of 100 for a a dollar or two at the hardware. Whatever colour they are just use a black permanent marker and that did the trick. My laptop is black.


Colour coding of wires on USB connector

Question: I am new to circuits and electronics.
I was building a portable mobile charger(Battery Bank) an had to connect a USB port.
The tutorial I was watching had 2 wires , Red for +ve and White for -ve.
The USB connector I have has 4 wires , Green , Yellow , Blue and Black.
Can you help tell me which one +ve and which one is -ve.
It would be of great help !
Thanks !
Comment: Use a multimeter or continuity tester to identify to which pin of the USB connector ("plug") each wire belongs. Then you can identify what each pin is using this: http://en.wikipedia.org/wiki/USB#Pinouts
Answer: Beyond red and black (and even then not all the time - sometimes black is hot and not ground) there are no standards. If it's cheap USB cable you are not even guaranteed the same coding between batches - the machine that makes the cable doesn't care what colour is in spool #2.
If you are buying a USB cable and cutting the B end off, your options are limited to getting a multimeter and checking the continuity. Every time.
Electronics supply shops just might have breakout cables available - A-connector on one end, red and black tinned leads on the other, ready for assembly.
Comment: +1 for cheap cables not bothering with colours - I took one apart for a project way back and it used Red for GND, Green for +5V and Black/White for the D+/D- pair!

Colour coding of wires on USB connector


Nexus card interview questions, not living in Canada yet?

Question: I got my Canadian PR 6 months ago but haven't moved there yet. I am living in US since last 8 years and this makes me eligible for Nexus card. I will be giving my interview this week. I would like to know people's experiences with these interviews and the kind of questions they ask. My main objective in getting the Nexus card is the possibility of using fast lanes during land/air travel. I am already conditionally approved but worried about CBSA asking residency questions (I still have 2.5 years technically to live outside Canada and still satisfy RO).
Answer: Are you a permanent resident of the US?  If so, you also qualify for NEXUS, and should have no problem:  https://www.cbp.gov/travel/trusted-traveler-programs/nexus/nexus-eligibility ... just be sure to notify the NEXUS enrolment office of your change of circumstances once you relocate to Canada.
It sounds to me like you qualify now, and you will qualify once you move, so enrol now based on your current qualifications.  Answer any questions they ask honestly.  Your eligibility will, for now, be determined outside of your intent to immigrate to Canada.
Comment: @CJH the worst case scenario is that your Canadian permanent resident status will expire and you'd have to reapply.  That wouldn't invalidate your eligibility for NEXUS.
Comment: No, I am in the US on a visa that expires next year so I will be moving to Canada latest by next year (so I will be satisfying my RO anyway, I just hope the CBSA officer doesn't reject my NEXUS application or says "oh you should've moved there already". To be fair there's no law that states when I should move, as long as I satisfy the RO).
Comment: Thanks for your reply. I will be making sure it doesn't expire :). I will be moving there next year at the latest and I still have 2.5 years to go (out of the 5 years) and still satisfy RO.
Comment: how did your interview go ? I am in a similar situation and have an interview tomorrow .


Should company protect DB schema at the same level of the data itself?

Question: If a company holds DBs that has access to different users. If the company has to decide how much to invest in protecting the data itself and how much to protect the DB schema (the structure and inner relations between tables and fields). Should the company invest only on protecting the data itself? Is there any threat for DB schema penetration. Is the schema has valuable meaning at all or can it increase the DB vulnerability for future attacks?
Comment: I'd treat the database structure similar to the source code. As an IP issue, not as a security issue.
Comment: Having a schema makes it a lot easier to pull out useful data from a database if you can gain access. It can also be useful when identifying what data might be available, in the case of table level restrictions (e.g. can see there is a relationship to x table, but not the contents). However, if someone can access either unintentionally, it's not good!
Comment: Ha, **Information Security in its original, information theoretic sense**. I like it! See my answer below.
Comment: If you can automate obfuscating the production schema and it doesn't kill your productivity, then might as well
Answer: No
Protection of the database schema is a type of security by obscurity.
Your software should be designed and configured in such a manner that none of its security is compromised when a hacker learns its design.  In fact, in the best case your design is made public so that security professionals can critique it and you can continuously improve it.
Answer: @John-Wu's is the right answer, investment should not be made to prioritize protection of schema above protection of the data.
Data is most important and is a compromise target for attackers. Schema is at best an information leak.
That said, databases themselves typically treat schema metadata at a different and higher privilege level than the data, not without reason. As @marcus-muller says, schema provides attackers with useful hints. And a user with privileges to see schema can almost always see data, while users with privileges to see data may not be able to see schema.
So schema should not be considered unnecessary to protect. But data is most important.
Comment: wouldn't agree with you on this: *while users with privileges to see data may not be able to see schema.* Users with access to data can very often infer schema.
Comment: Right- I was, perhaps poorly, making a more abstract point, but concretely- certainly it is difficult to prevent a user with a database client from being able to see or infer schema. However, a user who simply has access to a webapp that renders the data can not necessarily infer the schema. That is, I think the circumstances in which data protection has to be considered vastly outnumber those in which schema exposure may be an issue, even though schema can still be considered privileged information.
Answer: Although I agree with @JohnWu's answer that Security By Obscurity is a bad concept, there's one thing:
Yes. In a way.
Complex database schemes typically have complex relations. But let me come up with a simply example:
You've got a table
+----+------------+-------+
| ID | CustomerNo | Order |
+----+------------+-------+
|  0 |          1 |  Nuts |
|  1 |          1 | Bolts |
|  2 |          2 |  Nuts |
|  3 |          1 |  Wire |
+----+------------+-------+

Where ID is an auto-increasing primary key, CustomerNo refers to the primary key of your customer table, and Order refers to the primary key of your order table. This way, you map your orders to customers.
You, of course, only let each customer see the rows that "belong" to themselves.
Still, do you see the problem?
Exactly, through the linearly increasing ID field, customer 1 has an easy way of figuring out when the competition does their orders, simply by not acting like a "normal" customer, placing orders in bulks, but by automatedly placing orders every minute.
That way, they figure out that every Tuesday afternoon at 4:30pm, the competition orders. Now, knowing the market for hardware very well, they put a big sell option on the market every Tuesday 4:20pm, thus increasing the market price of nuts, harming their competitor, and also, financially profiting from knowledge they should never have gotten in the first place.
This is a question of data cleanliness, in principle. A monotoneous primary key gives you (partial) information you shouldn't have. Hence, your customer API would need to make sure the information leakage is zero. And I mean that mathematically: Information has a pseudo-unit, bit, and there's a whole mathematical discipline called information theory, that describes how much information is in the observation of a certain event based on randomness (for example, if there's more than two customers in that table, you can't know which of your competitors did the ordering at 4:30pm, but you still get probabilities out of this, and hence, non-zero information).
It's very unlikely you can hide such information from your customers if they have access to the database scheme. In fact, that scheme doesn't help them without you giving them info they probably shouldn't be getting. In that sense, well, your scheme doesn't need to be secret, but some of the data that you probably still only consider "relational" within a customer's accessible records should, in fact, be secret from the customer.
Comment: The values of a primary key is part of the data, not the schema. So the need to keep the primary key secret (or random) to stop enumeration attacks is unrelated to whether or not to keep the schema secret.