Proof of the superposition theorem in linear circuit analysis

Question: I couldn't find a proof of the superposition theorem from circuit analysis anywhere online. I thought it might be helpful to ask and provide my proposed proof as an answer to gather feedback and improvements.
The superposition theorem I am referring to is the one that says if you only have "linear circuit elements", then the solution to the overall circuit, in the case of DC circuits at least, is the sum of solutions you obtain to variant circuits where only one source is left on (voltage sources become short circuits while current sources become open circuits).
Comment: It is a fundamental linear time invariant property. A +B = C= B+A. with scalar properties.  https://dsp.stackexchange.com/questions/19765/prove-using-superposition-that-an-lti-system-is-linear
Comment: I believe it is different from the superposition theorem in circuit analysis where the theorem states that someone can solve a whole bunch of variant circuits by leaving on only one source and add all the solutions to get solution to the overall system.
Comment: No. It is the same proof
Answer: A linear system in electronics means a system that can be expressed as a linear differential equation. Hence
$$\dot{x} = \mathbf{A} x + \mathbf{B}u,$$
where \$x\$ is the state vector, \$u\$ is the input vector, and \$\mathbf{A}\$ and \$\mathbf{B}\$ are matrices. Let \$x_1\$ be the state from \$u_1\$ and let \$x_2\$ be the state from \$u_2\$. Thus
$$\frac{\mathrm{d}}{\mathrm{d}t} ( x_1 + x_2) = \dot{x}_1 + \dot{x}_2 = \mathbf{A} x_1 + \mathbf{B}u_1 + \mathbf{A} x_2 + \mathbf{B}u_2.$$
Therefore
$$\frac{\mathrm{d}}{\mathrm{d}t} ( x_1 + x_2) = \mathbf{A} (x_1 + x_2) + \mathbf{B}(u_1 + u_2).$$
This means that applying both inputs together is equivalent to adding the states for each input applied individually. For superposition of more than two inputs apply induction.
Comment: Would the input vector contain the current and voltage sources in my above example?
Comment: @ShuhengZheng Yes, the input vector holds the current and voltage sources, i.e. the inputs to the system.
Comment: In general, could x the "state" represent either voltage or current?  I can see how Kirchhoff's two laws be casted into the above form although there would be no derivative term.
Comment: @ShuhengZheng There will be derivative terms, if you have capacitors / inductors. Have a look at state-space representations for more information.
Comment: @ShuhengZheng the state holds enough information to uniquely determine the state of the system, e.g. the state vector for resistors in series is the current, since if you know the current, you can determine the voltages across each resistor.
Comment: Are there any real world elements that have higher derivatives?
Comment: @ShuhengZheng in a way, there are. Capacitors and inductors are first order, but a system of capacitors and inductors can be of higher order. An example for that would be a filter, e.g. low-pass Butterworth. A simple LC network is second order.
Answer: It is generally said that the superposition theorem is a result that the circuit is linear.  I think that is not correct and actually a bit circular.  Because one describes the linearity of the input/output relationship of the circuit while the other talks about how to solve the circuit by turning off sources.
Here I prove that the superposition theorem in circuit analysis holds if you restrict to certain elements.
Motivating Example

simulate this circuit – Schematic created using CircuitLab
Let's motivate the general proof with a specific example first.
Take a look at the attached circuit.
I'm going to use \$I_C\$ to denote the current of the current source and \$V_V\$ to denote the voltage of voltage source.
Small letter \$v_1, v_2, i_1, i_2\$ denote voltage and current across the resistor elements.
Systems of eqns are:
$$
V_V = v_1 + v_2 \\
I_C+i_2=i_1
$$
We switch the current source with open circuit and get the equations
$$
V_V = v_1'+v_2' \\
i_2'=i_1'
$$
Switch the voltage source with closed circuit and get equations
$$
0=v_1''+v_2'' \\
I_S+i_2''=i_1''
$$
Implicitly, there are also the Ohm's law equations such as
$$
v_i=R_ii_i
$$
The key thing to notice is that solutions to the "turned-off" circuits, I will now call them the basis circuits, the solutions are all homogenous except for the equations that involve the one element that remains.
That means when you add one solution from each basis circuit, most solutions do not affect the "in-homogenousness" of other solutions.
For example, if you add the primed variables to the double primed variables, you will get solution to the original circuit.  The primed solutions are homogenous  in the second equation, i.e. it won't "touch" \$i_i\$  while the double primed equations won't alter the voltage variables.
General Proof
In general, we can say translate all equations that govern our circuit into several forms, and see that this line of thinking works.
Kirchhoff Current Law
For each node in the circuit not directly attached to a voltage source, The Kirchhoff current law gives us equations of the form
$$
i_1+i_2+...i_k=I_{S1}+I_{S2}+...+I_{Sn}
$$
where the lowercase i on left hand side denotes unknown currents while right hand side denotes any current source connected to the node.  If there are no current source, it would just be zero.
The basis circuits obtained by turning off all sources except for one would either leave the right hand side with 0 or with one constant left.  The basis circuits obtained by keeping S1, S2, .. etc. would each contribute to each of the term on the right hand side.  When you add the solutions of all of them, you can see that the final solution satisfies the above equation.
This holds for all nodes.
Kirchhoff Voltage Law
For closed loops that do not have current source, Kirchhoff Voltage law has form
$$
v_1+v_2+...+v_k=V_{S1}+...+V_{Sn}
$$
The same argument as above holds.
Ohm's law
Ohm's law is homogenous and relates v and i.  So adding solutions to the basis circuits would also solve the Ohm's law equations for the full circuit.
This argument also holds for the other linear circuit elements such as inductors, capacitors, and could be extended if newer elements are discovered/created.  As long as the equation governing the element is a linear function in voltage, current.
Comment: Kirchhoff. One k, one c, two h, two f. There is a built-in schematic editor you can use to add schematics to questions and answers...that is one of the worst schematics I have ever seen on this site.


Who is the biological mother of Gon?

Question: It is not Mito obviously.
When Gon is listening to the tape that Ging leave him, Gon stop the taped when Ging ask if Gon want's to know about his mother. I've waited so far but until the last episode of the anime it is not yet revealed.
If Gon just listen to the tape, then the mistery should already solved. Is there any answer that can be found in the Manga? I'm prepared for spoilers.
If Leorio was mad about how Ging acted when he knows the condition of Gon, I'm sure Leorio will be more mad if he knows that there's a mother to Gon.
If she is ever alive, then who is she.
Comment: i bet his mother is still alive and  a high ranking hunter like his father as well.
Answer: Unfortunately, this question cannot be answered right now.
When Gon was younger, Mito told him that both of his parents were dead. However, it is later found out that it was a lie, because his father Ging was obviously alive and well.
Before the Greed Island Arc, Gon got a tape from Ging. At the end of the tape, when Ging was going to tell him about his mother, instead of listening to it to the end, Gon just stopped the tape and stated that Mito was his mother.
There is a lot of speculation as to whom his mother could be, but until Togashi supplies the answer, we will never know.
Comment: Even in the manga, but i think i read somewhere that even ging stated that her mother was dead and in the tape he only wanted to describe her, but i can be completely wrong the series is advancing at snail speed
Comment: Even in the manga? I'm not updated in the manga series.

Who is the biological mother of Gon?


she has a pretty face vs she has the pretty face

Question: My student asked me why we say "she has a pretty face" when describing someone. Is it possible to say "she has the pretty face" (we obviously have only one face)? And if not, how can we explain it? Can we apply the rule "The indefinite article is used with singular countable nouns
to refer to a person or a thing that you are mentioning for the first time"?
Comment: She has the pretty face would imply that there is only one pretty face in the universe and she has it.
Answer: To say

she has the pretty face

would suggest that there is only one face, or only one pretty face. In fact there are many possible faces. So one says:

she has a pretty face

which indicates that there are many possible pretty faces, and she has one of them.
In general when describing aspects of a person, one says in standard English

S/he has a/an {adjective} {noun}

and not a sentence using a definite article.


Is nonce of HMAC secured message considered secret?

Question: Consider app using say, HMAC-SHA2, with securely pre-shared symmetric key. Are the nonces of exchanged messages considered secret as well?
 Or in other words, does its easy predictability or even outright knowledge(plaintext) of nonce to attacker compromise/reduce HMAC security?
Some sources suggest using cryptographically secure PRNG for nonce generation as if the PRNG algorithm/seed was part of preshared secret which confuses me.
Comment: nonces are not secret, but hmac doesn't use nonces, so it's a trick question.
Comment: @dandavis hmm, true. Wondering how should I rephrase the question - when using nonce for messages it is in fact more of a protocol(?) secured by HMAC rather than HMAC itself, isnt it?
Answer: NIST (National Institute of Standards in Technology) defines an nonce as a time varying, non-repeating value to help prevent 'replay attacks.
The nonce doesn't necessarily have to be secret but should be a random sourced non-repeating value used in block ciphers for symmetric and asymmetric encryption.
Please reference NIST SP 800 series of documentation for additions details of best practices, implementation & reasons for use.
Comment: There seem to be conflicting definitions, e.g. NIST SP 800-63-2: ...`a nonce is not necessarily unpredictable.` while  NIST SP 800-44 Version 2: ...`Because the sender randomly generated the nonce, this defeats playback attacks because the replayer cannot know in advance the nonce the sender will generate.` Which would imply that predictable nonce, like timestamp, does not prevent replay attacks(why use such?). Could you clarify this a bit?
Comment: While not necessarily unpredictable an nonce should never be used more than once. The nonce can be best described as the equivalent of a salt when hashing passwords. While not always a value from a PRNG, it should never be used more than once for each key. Some uses of an nonce don't necessarily require the security a truly random sequence of bytes offers for keys


Why use 'step down' instead of 'resign'? Is there any difference?

Question:
Ozzie announced his plans to step down from his role at Microsoft on October 18, 2010
Comment: If it's not clear, note that "step down" can only be used for people in very senior positions (board members, royalty, etc). If you're an ordinary salaried worker, blue collar worker, or junior worker, you'd never use "step down" - you'd just use "resign."  Really, it is a somewhat **pretentious** term.
Answer: Compared to "resign", "step down" has a connotation of an amicable parting with the possibility of a continuing relationship. The word "resign" is often used when employee parts employer over some disagreement or other negative cause.
If I recall correctly, Bill Gates' active role at Microsoft changed several times without him severing all activity at the company. Saying he resigned might not give an accurate impression unless carefully qualified by the particular role he was relinquishing.

Bill Gates stepped down as chief executive officer of Microsoft in January 2000. He remained as chairman and created the position of chief software architect. In June 2006, Gates announced that he would be transitioning from full-time work at Microsoft to part-time work.

Wikipedia
Answer: Step down is just a euphemism for resign.  It could indicate, as Tragicomic said, that the position being left was one of importance, but that need not be the case to make use of the euphemism.
Comment: One could argue that since *resigned* is a euphemism for *was fired*, it was necessary to create another euphemism to cover the original meaning which was thereby left without a valid term to describe it!
Comment: "Resigned" is not the same as "was fired".  It is quitting.
Comment: @theidiotbox: The nature of euphemisms is that they are often technically inaccurate. People often say things like, "Sally and I decided we needed some time apart" when what he really means is "Sally dumped me". There was no "we decided" involved. That's why it's a "euphemism" and not a "literal statement of fact".
Comment: Reminds me of a press conference I saw once where company A bought out company B, and the president of company A was asked if they would be laying off people from company B. He replied, "That's hard to say. Staffing may increase, or it might fluctuate." So the opposite of "increase" is "fluctuate"? Oh, okay.
Answer: While both step down and resign mean the same thing, step down is used as a phrasal verb to speak of someone resigning from an important position. Ray Ozzie held the position of chief software architect at Microsoft.
Answer: To resign is to "quit." To step down is to "climb down" from a high position.
It is possible to "step down all the way," as Ray Ozzie did at Microsoft. In this regard, the two are synonymous.
On the other hand, Bill Gates initially "stepped down" from CEO to Chief Software Officer, thereby taking a lesser post, without actually quitting or "resigning." (He finally fully "resigned" several years later.)
It's also possible to resign as janitor or low level person without "stepping down," because there is no lower level to step down to.
Answer: 'Step down' is to 'resign' as 'bow out' is to 'lose'. Using 'emeritus' for 'retired' is another fancy term. My wife forgot to renew her driving licence yet was still telling me how to drive. She's a 'driver emeritus'.
Comment: Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer).


What is the device taken from Alex Trusks at the end of Hypercube?

Question: The Cube series is a really complicated movie series.
At the end of the Hypercube, Kate gives a device to some serious guys (military guys maybe). This device was taken from Alex Trusks' neck.
What is the purpose of this device?
Answer: The only information in the original version of the film is the last line spoken by an unknown African American in a suit, who presumably works for Izon or the military:

We'll take that to Darcy and see if anything recorded on it

The fact he said "recorded on it" implies it's some kind of sensor device, such as a camera, microphone or other more high-tech information gathering system rather than a storage device, as usually you'd say "saved on it" or "stored on it" in that case.
There's more information given in the alternate ending which I can't seem to find, though Wikipedia says:

The longer alternate ending included in the special features on the DVD reveals the "owners" to be the government; in the shorter version it is unclear who they are, but it is assumed they are Izon. Kate is executed in both versions, but she is praised for being the first operative to make it out alive. In the alternate ending it is revealed to Kate that she was in the Hypercube for just six minutes and fifty-nine seconds. It was an experiment used for quantum teleportation.

The fact that this was an experiment would seem to back up the idea that it was a sensor as the scientists working on the project would need a way to get data back from the experiment and there was no other machinery shown in the "real world" which could be monitoring what happened inside:
Answer: It was some sort of storage device with confidential information on it.  I don't remember many other details though as to why she was looking for it or who had it initially.
Comment: Hi DForck42, do you have any evidence to support your theory?
Answer: It may be much more than a simple recording device. As shown, time moves differently inside, yet all of Jerry's watches are perfectly synchronized. We know they were taken from different versions of Jerry at different times, so if they were functioning as normal watches, you'd have wildly different times on each watch, as well as much more time having passed.
The fact that they all line up to the "6:06:59" reading with Kate suggests that Alex Trusk's necklace was some kind of stabilizing device. The watches (and time) stabilized and ran normally when in proximity to the device. This is also supported by the fact that at the end, when the room is disintegrating, the only one that remains is Kate, who took the device from Alex's neck after her death.
This would explain why Simon ran into multiple versions of Jerry and the others and experienced more time, but Kate did not.
Of course, that doesn't explain how what seems like hours passed for Kate after she met up with Alex (and was in proximity to the device), yet the watches still only show 6 minutes and 59 seconds have actually passed.
I suppose it's possible that the device had SOME stabilizing effect, but not completely. And as the cube imploded, normal time from the outside universe began leaking in.
Or, it could just be that Jerry's watch only counts time for the outside universe. Modern watches count time using the oscillation frequency of quartz when exposed to an electric charge, so if somehow that were unaffected by the weird time frame of the cube, it would explain it. In effect, the watches would only tick when exposed to time from the outside universe. If normal time seeped through to the cube at predictable intervals (and they were increasing), you could probably use that to calculate when the cube would implode. Hence, the scientist making that "6:06:59" calculation.
In fact, it could be both of those explanations together. If so, and Alex Trusk were hiding in the tesseract, it would make sense that they would send in an operative to retrieve it. She developed a way to stabilize (and ultimately control) the tesseract, and based on her hiding, I doubt she sent the government instructions. They'd need to retrieve it to figure out how it worked.
Or it could be just weird, timey-wimey stuff. In the end, WHO KNOWS. :-P


How to push through transaction that has been pending for 2 weeks using Blockchain.info?

Question: I'm using blockchain.info and was trying to send some bitcoin to Binance. It's been 2 weeks and it is still pending because I put too low of a fee. I'm wondering what method I can use to push this through and how to do it using blockchain.info Thanks!
Answer: Normally, I would refer you to:
Q: Why is my transaction not getting confirmed and what can I do about it?
But I'm going to make it simple for you.
You basically got two options:
1. Wait it out.
Easiest solution. Wait until the transaction gets picked up or dropped.
2. Ask miners for help
Some mining pools and miners offer services to allow you to prioritize your transaction in their mempool so that it is chosen sooner for inclusion in a block. One such service is https://pool.viabtc.com/tools/txaccelerator/. Also you can find others here.
Some members on Bitcointalk.org offer free transaction accelerator services, some of them have access to several pools. If urgent, you may consider the paid services(Beware of the scammy ones though).
3. Rebroadcast
You can try to manually broadcast the transaction, but this probably won't help you at all because your fee is low.
4. Replace by Fee (RBF)
Won't work with blockchain.info.
5. Child Pays for Parent (CPFP)
Won't work with Binance.


Speed of the butterfly effect

Question: Consider the simplest abstract model of a double pendulum. What is the maximum possible amount of change in the initial conditions so that the trajectory doesn't visibly change in the first minute, i.e. the maximum change in the angle is less than 1 degrees? I'm not looking for an exact figure, I just want to know the order. Is it in order of $10^{-3}$ degrees, or $10^{-15}$ or even less?
Comment: We still need some unit of measurement to set the timescale - your pendulum could be a kilometre long or a nanometre long. Looking at papers online also give a range of claimed Lyapunov exponents at least between 0.1-10 depending on experimental setup. That gives a rough estimate between $10^{-261}$ and 0.00247 for the required initial angle deviation - a *lot* hinges on assumptions and timescales.
Answer: The hallmark of chaos is exponential divergence of nearby trajectories. This means that if you start the system in state $\pmb{x}+\pmb{\epsilon}$ the divergence between the unperturbed trajectory $\pmb{f}_0(t)$ and the perturbed one $\pmb{f}_\epsilon(t)$ will grow exponentially as $\epsilon\rightarrow 0$: $||\pmb{f}_0(t)-\pmb{f}_\epsilon(t)||\propto e^{\lambda t}$ where $\lambda$ is the Lyapunov exponent. In fact, one can speak of several Lyapunov exponents describing the divergence of trajectories in different directions, but it is the largest exponent that describes how mixing the system is.
So we can restate the question like: given a chaotic system with largest exponent $\lambda$, after time $T$, what is the initial deviation $\epsilon$ that produces a given deviation $\Delta$? The above gives $\Delta \approx \epsilon e^{\lambda T}$, or $$\epsilon \approx \Delta e^{-\lambda T}.$$
The problem is finding the Lyapunov exponent of the double pendulum. The exponent is energy dependent: if you start with a nearly vertical pendulum it will just be a perturbed non-chaotic pendulum, but with enough energy it will be chaotic. And of course, it depends on the length and mass of the pendulum and how you measure time. Plus, the exponents are averages over long time which means that for a particular trajectory the deviations behave differently. In short, published values rarely replicate. Anyway, this paper gives 7.5 or 7.9/s depending on whether it is simulation or measurement. This paper gives 0.179/s. This one 0.02-0.04. This one 0.41/s. This question has about 0.27. So I think any answer between 0.02 and 10 is possible for some configuration.
Anyway, if we try with $\lambda=0.1, 1,$ and $10$, $T=60$ and $\Delta=1^\circ$
 I get $\epsilon=0.0024, 8.756\times10^{-27}$ and $2.65\times 10^{-261}$ degrees, respectively. The really small exponent 0.02 gives a modest 0.30 degree perturbation (for modest values of $\lambda T$ one can Taylor-expand the exponential and make a linear estimate).
In short, this is somewhat of a "how long is a piece of string?" question. The key thing to note is that it really matters how wildly the pendulum is swinging: if $\lambda T>1$ nearly any deviation becomes macroscopic.
Comment: You can get positive Lyapunov exponents if the dynamics is just unbounded, but for systems that stay inside some bounded volume, then having at least on positive Lyapunov exponent is a sign of chaos. The size of it tells you how strongly trajectories diverge. You can also have several positive exponents and get what is sometimes called "hyperchaos" where there is mixing in more dimensions. But usually bigger exponent means more chaotic.
Comment: Is every system either chaotic or not, or instead, there is a conventional threshold that the more system is beyond that, it's more chaotic?
Answer: You would consider that the error would accumulate linearly here.
If you have 60 oscillations in the minute, and the divergence wanted as 1 degree, then it would travel through 60*360 degrees, making a divergence of 21600 degrees.  This would mean that an angular velocity of the order of 10 seconds, would not show for the first minute, and one of 2 seconds in the first 5 minutes.
Thus it's of the order of 10^-4 degree.
Comment: The error accumulates exponentially, not linearly, right? That's why it's chaotic.
Comment: Yes, exponential divergence of nearby trajectories is the classic hallmark of chaos. This answer is wrong.

Speed of the butterfly effect


Own android app can't access internet on emulator

Question: I'm modifying an android app and I want to test it with an emulator on my machine so I can examine the apps data.
I'm experiencing some internet connectivity issue with my own app but the browser on the emulator works without any problems.
Now I already read a dozen of articles regarding this subject but so far there has not been a solution that works for me.
I tried:
- adding dns-server to startup params.
- Disable LAN, only enable wifi of my laptop
- Toggle 3G connectivity in the emulator
- Add GSM Modem Support to emulator
- Restart Eclipse
- Restart Laptop
- Double check the internet permissions in the manifest
I'm running out of options here and hope some people could provide me with some wisdom.
I'm running all of this on
- Mac OS X - 10.6.8
- Eclipse Indigo Service Release 1 - Build id: 20110916-0149
- Android MinSdk 7
- Emulator is using SDK level 16
A small footnote, I don't have any issues regarding internet problems on a real device. I'm able to run the application on a Google Nexus with Android 4.1.1
Any information would be much appreciated!
Thanks in advance.
Comment: Do you have the INTERNET permission in your manifest? Without code all we can do is guess
Comment: Yes like I mentioned in my post above I double checked the permissions, they are in the manifest. Otherwise I wouldn't be able to get the app running on a real device.
Comment: can You suggest me the solution for this? I am facing the same problem..
Answer: Have you tried to set Command Line Artument in Run Configuration of Application.


Multiplayer Xbox One - Xbox 360

Question: My son just got an Xbox One, and I want to play multiplayer games with him using the Xbox 360 in the same house on the same WiFi network. Are separate Xbox Live accounts needed to play multiplayer games on the same WiFi system? We are running the 360 version of COD.
Comment: which COD? [Only COD 2, 3, Black Ops 1, and World At War are backwards compatible](https://majornelson.com/blog/xbox-one-backward-compatibility/).
Comment: We are running the 360 version of COD.
Answer: You can't play cross platform with the exception of the Xbox One running select backwards compatible games that were downloaded.  To that extent, it won't matter if you have separate accounts or not if the game isn't backwards compatible for the Xbox One.
But yes, you can share gold across multiple users as this states:

Multiple members of a household will be able to share a single account

The article also states:

"We want to make sure that you and your son both have your own account," Kilgore said, in describing the case of a father and son sharing an Xbox Live Gold membership. "We want you log in to get your stuff, and when he logs in to get his stuff. On that console, if you have Gold, he can use Gold as well."

That sounds just like your situation.  The only thing is, certain games are only backwards compatible. See this list for all the current Xbox 360 games that Xbox One can also run and play with other Xbox 360s.
Comment: I'm not sure about having multiple accounts share one Gold subscription. I don't know how that would work. I noticed that that article you cite is from 2013. It sounds like he's describing the system they would like to design, not the system that is actually in place.

Multiplayer Xbox One - Xbox 360


Would Gandalf know Sauron personally?

Question: The thought just occurred to me that the two had co-existed for eons. Even not taking the music of the Ainur into account, wouldn't they have had ample opportunity to interact with each other on Arda in some way? Prior to Sauron's fall, wouldn't it have been productive for two of the most powerful Maiar to corporate?
Comment: That big roar from the movie actually translates to "Mithrandir, you shifty bastard! Where's that £20 you owe me?"
Comment: Maybe, but Ainur & Co. has an indeterminate amount of employees, and they had different supervisors.   And they worked in different sales regions for _ages_.
Comment: Nowhere is either Gandalf nor (to my knowledge) any Balrog stated to be one of the most powerful Maiar. Olórin may have considered one of the *wisest*, and a Balrog is simply more powerful than any Man and or Elf like *any* Maia would be.
Comment: The wording is a bit confusing here - in the title, you have "Would Gandalf know Sauron/Durin's Bane?", and in the body "Would Durin's Bane know Gandalf/Sauron?" Are you actually saying "Did any of these three characters know each other?" Would it be easier to answer if you focussed on a single pair of characters, rather than a triangle of three possible relationships?
Comment: Very little is written pre-music.  I would be very surprised if there was an answer that was not just guessing.  After the music there was little reason to think there was any interaction between Ainur who were and were not aligned with Morgoth, outside of the early battles between them.
Comment: @IMSoP Sorry for the confusing question. What I meant to ask was if Gandalf knew either personally. However, I now realize that any potential question that applies to either would most likely apply to both. Therefore, I have edited my question to be more clear.
Comment: @chepner I was referring to the power of Sauron and Gandalf, not the Balrog in regards to being unusually powerful for Maiar.
Answer: On first glance, it seems surprising that they don't know each other, but there are three good reasons why they might not.
First, as was noted in the comments, we know nothing about the number of Ainur. In the first Music, they do sometimes sing together in small groups, but the Music is mostly a succession of solos. The Second and Third Musics are more choral.  That's all we learn about their domestic arrangements during the era before the creation of Arda, and there's nothing in it that suggests that their existence involved much socializing.  Had Olorin (the earliest name we have for Gandalf) and Sauron sung together, they might have known each other, but this apparently didn't happen.
After Arda was created

some abode still with Iluvatar beyond the confines of the World; but others, and among them many of the greatest and most fair, took the leave of Iluvatar and descended into it.

The most powerful of the Ainur were few, but nothing, really, is said about the number of lesser Ainur.
It's also worth remembering that Tolkien says that the Ainur were literally angels. Since he was well versed in Medieval studies, he knew that they believed that there were a very large number of angels, and most probably he did also. (The old bit about "How many angels can dance on the head of a pin" was not foolish at all, but actually was asking if the number of angels was finite or infinite.) There's every reason to believe that Tolkien may have imagined a lot of lesser Ainur in Arda.
In the Days of Bliss, Gandalf lived in Lorien

Wisest of the Maiar was Olorin. He too dwelt in Lorien, but his ways took him often to the house of Nienna, and of her he learned pity and patience.

while Sauron -- later Gorathaur the Cruel --

was of the Maiar of Aule, and he remained mighty in the lore of that people.

There is no reason at all to think that they encountered each other in their daily life.
So based on what we know of their early histories -- not much -- all we can say is that there is nothing that demands that they knew each other.
Answer #1: They simply may not have interacted much.
On to things more speculative!
First, while Tolkien is clear that the Ainur in Arda were powerful beings, he never shows them as being mentally of a different order than Elves or Men -- spiritually, yes, and of personal power, yes but, for instance, he never hints that they can contemplate the Cosmic All like Mentor of Arisia.
Yet it would take a very different mind than ours to remember with any clarity the entirety of a life outside of time followed by a timeless interval before the Years of the Trees followed by many, many thousands of years afterwards. (I can barely remember what I had for lunch yesterday!)
Even if Olorin met Sauron many times in gatherings, would he still remember him? And if his memory was clear enough that he did could he possibly be the sort of person that he was? Someone who remembers a hundred thousand years (some of them timeless if that makes sense) of experiences perfectly would be very, very alien to us.  Gandalf was not, and so he probably didn't.
Answer #2: Even if they met many times, they almost certainly didn't remember it.
There's a third point, though, which applies to Olorin, though not necessarily to Sauron: The process of becoming one of the Istari almost certainly affected his memories of life before then.
We know that when the embodied Maiar Melian loved and married Thingol, she was bound to her embodied flesh and could not leave it and return to Valinor until Thingol's death.  Something similar seems to have been true of the Istari, since they were clearly bound to their human bodies and could not leave them at will. And this doubtless affected their memories of the time before.
At one point he told people in Gondor that

Olórin I was in my youth in the West that is forgotten

While this could be taken to say that it is the West that is forgotten, we know that the Men of Gondor rise before meals to remember Numenor and the lands beyond, so it must be Olorin's youth that is forgotten.
Gandalf is saying that his memories of his life in Valinor are mostly or entirely forgotten.
Embodiment as an istari gave Gandalf a human-like memory if he did not already posses one.  As the Fellowship stands before the Doors of Durin, he says of the markings on the stone:

'They are wrought of ithildin ... and sleeps until it is touched by one who speaks words now long forgotten in Middle-earth. It is long since I heard them, and I thought deeply before I could recall them to my mind.'
...if I am allowed a little peace from foolish questions, I will seek for the opening words.
'I once knew every spell in all the tongues of Elves or Men or Orcs that was ever used for such a purpose. I can still remember ten score of them without searching in my mind.

That could have been any of us.  Gandalf's memory was human, not divine.
Later, after he dies in the fight with the Balrog and is resurrected and meets Aragorn, Gimli and Legolas, he says

'Gandalf,' the old man repeated, as if recalling from old memory a long disused word. 'Yes, that was the name. I was Gandalf.'

Answer #3: Even if Olorin the Maiar remembered meeting Sauron, Gandalf the Istari did not.
Comment: Excellent! Given the paucity of information, and extrapolating from what is known, I think this is as good answer as we're going to get.
Comment: This brings up an interesting follow-up question: did Sauron and Saruman know each other from before? I know there's probably not any canonical sources mentioning it. Both were maiar of Aule and it would make perfect sense for Valar to assign the leadership of Istari to someone who knew Sauron well.
Comment: I think your example very much suggests that Gandalf's mind is mentally quite far from human. It suggests that he is familiar enough with *every* language spoken by three different species (thousands, if the real world is anything to go by) over some tens of thousands of years that, for any given purpose, he can remember about two hundred obscure spells in those languages, spells that he probably has not thought about for hundreds or thousands of years, without even having to think about them at all, and that if pressed, he could probably remember all of them.
Comment: @Adamant Well, that is at least what Gandalf boasts, but then he cannot recall the word "mellon" that is even carved into the very door of Moria, a place he must surely have visited before in the thousand years since he became an Istári. So maybe we cannot take hom at face value in that instance, but clearly he is more than human in most regards -- he is immortal, wields magic and could defeat a balrog -- so it stands to reason that his memory would be better too.
Comment: @Mark Olson   Plus one for mentioning Mentor of Arisia.

Would Gandalf know Sauron personally?


Can I create a link that when clicked creates a new list item.

Question: I dont have access to SharePoint designer :(. My goal is to make a button that says something like "Create new item" and when clicked pulls up the form to create a new item in a list. Is this posible?
Answer: By default your "new list" form will open as what is called a modal window (this is how it overlays your list) but is actually a separate page.  To make it easy to get the URL of the form, you can:

Open your list, and in the ribbon select List Settings
Selected Advanced Settings Scroll to the bottom of the page and find
the "dialog" heading and set to no.
go back to the list and select "new item" and it will open a full page view
Copy the URL and use that for your link.

Be sure to turn the dialog option back on if you want that when working in the list directly.
Comment: Thank you this worked! One more question. Is it posible to copy the adress or code for creating a new word document from the ribbon toolbar? For example when i click create new document instead of upload in a document library it goes straight to opening up the template.
Comment: I'm not sure how you would accomplish this off the top of my head, There is a SharePoint.Opendocument javascript method (from CSOM) that you can probably leverage, but I have not done anything in that scenario.  I would go ahead and ask this as a new question and see if anyone has other inputs.
Answer: Sounds like you need a custom button that opens a new form dialog.  Pop this into a Content Editor.  Make sure you click the 'Edit HTML source' in the Content Editor settings:
<code>    <script type='text/javascript'>
function OpenNewForm() {
   var listName = 'MyListUrl';
    var options =
            {
                url: L_Menu_BaseUrl + "/Lists/" + listName + "/EditForm.aspx?ID=1&IsDlg=1",
                dialogReturnValueCallback: Function.createDelegate(null, newFormClosedCallback)

            };
    SP.UI.ModalDialog.showModalDialog(options);

}

function newFormClosedCallback(result, target) {
    if (result === SP.UI.DialogResult.OK) {
        SP.UI.Notify.addNotification("List item was added.", false);
    }
    if (result === SP.UI.DialogResult.cancel) {
        SP.UI.Notify.addNotification("Add new item was cancelled.", false);
    }

}
</script>
<input value='add a new item' type='button' onclick='OpenNewForm();' id='btnNewForm' />
</code>
I added a callback that adds a notification when the form is closed. You can remove that if you want.


Where to download and install Oracle Directory Manager?

Question: Would someone please tell me specifically where can I download Oracle Directory Manager component? I know it's in one of the Oracle 10g discs but which one? Database, Client or Companion CD.
Answer: If you search for odm or oidadmin you get a few pointers, such as this forum post.  So it seems it's part of the client suite, which can be downloaded here


Tester reads "HOT / GRD. REVERSE"

Question: Everything was working as usual then NOT.  I used a 3 prong circuit tester in several outlets and got the same reading on all of them. HOT / GRD. REVERSE. how or why would this suddenly happen and how do I fix it?
Comment: You really need to provide more information on what you mean as NOT. Describe the scenario of working up to and through to the NOT situation. Something has changed for you to reach that determination. And that series of events may have next to nothing to do with what you are seeing on this three prong circuit tester that you pulled out of your pocket.
Comment: Please define what kind of indicators are on your tester and what indicators are NOT working as expected. What readings do you get on "good' outlets.
Comment: did you put a voltage tester and check what you have at your receptacle ... check your breaker if you're not getting 120
Answer: For some reason those plug-in receptacle testers show this result when there is an open neutral.
This means the neutral conductor for the circuit is open or compromised somewhere. Most often it is a poor connection at a receptacle where the wires are back-stabbed, or stuck into the "quickwire" holes in the back.
A quick test to do is to plug the tester into each receptacle on the circuit, working or not, and wiggle it. Many times the faulty connection will show as the circuit flickers or comes back on.
The best course is to at the very least move the wires from the quickwire holes to the side screw terminals. The best bet is to replace the receptacle entirely using the side screws.

Tester reads "HOT / GRD. REVERSE"


Curvature in space time during Big Bang and present scenario

Question: Space time in the presence of masses is curved. But during the time of Big Bang it's presumed that all the matter in this universe was at a single point, so it must have been super dense and had very high mass. So space time at that point would have curved very deep down but today it's almost flat. How can it be so flat today? Or is it due to our observation limit (like from surface our Earth seems flat, but in reality when observations are taken from far away we can see the curvature)?
Comment: *But during the time of Big Bang it's presumed that all the matter in this universe was at a single point* This is not true. In a cosmology where the universe is spatially infinite, it's spatially infinite at all times.
Comment: But if we look out into the universe, back into time, towards the big bang; wouldn't everything in every direction just curve back into the origin point of the big bang? Wouldn't there basically be an event horizon where it's impossible for us to look to far out into space, because it would look the same in every direction as it terminates at the same point at the same time in the past?
Answer: We don't known if it was ever a point, General relativity tells you that it must have been very very small, approaching a point, but if you take the theory too far in this direction it breaks down, and we say we have a singularity, one of the goals of physics is to produce a theory to cure this singularity.
Then you ask why spacetime is flat now, it isn't. Spacetime isn't flat. Space alone is, at cosmological scales, flat. That depends on the density of matter and energy present in the universe. It appears that our universe has the exact density to be flat.


How to derive Ehrenfest's theorem?

Question: $$\frac{d\langle p\rangle}{dt}=-i\hbar \int_{-\infty}^{\infty}\frac{d\psi^*}{dt}  \frac{d\psi}{dx}+\psi^*\frac{d}{dt}\Bigr(\frac{d\psi}{dx}\Bigr)$$
I didn't know the coding of partial derivative which are inside the integral
There is a question about this on site, but I don't have much experience with bra -ket notation.
Ehrenfest's theorem derivation. The first integrand makes sense but I'm completely at odds with the second one
Comment: That integral needs parentheses and a differential.
Comment: Are you asking where the second term on the right comes from?
Comment: \partial which goes to $\partial$
Answer: $$\langle p \rangle = -i\hbar\int\psi^*\frac{d \psi}{dx}dx$$
$$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\frac{d}{dt}\int\psi^*\frac{d\psi}{dx}dx$$
$$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\int\frac{d}{dt}\left(\psi^*\frac{d\psi}{dx}\right)dx  \ \ \ \ \text{By Leibniz's Rule}$$
$$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\int \left(\frac{d\psi^*}{dt}\frac{d\psi}{dx} + \psi^*\frac{d}{dt}\frac{d\psi}{dx}\right)dx  \ \ \ \ \text{By the Product Rule}$$

How to derive Ehrenfest's theorem?


Raspberry pi3 and odoo image

Question: Every time I use odoo posbox image on the raspberry, I can't use the sudo command, did I miss something?
Answer: I am using posbox image in pi in Linux. You are using wrong command. Don't need to use command : sudo apt-get update Here "update" command won't work. That's why you are getting base error. Try with some else command. Because "sudo" is working in pi.
Answer: In your picture it is the Pi part of the command that bash is complaining about.
The sudo command gives the current user root privileges and does not need to be told what user you are.  You can state the user with sudo -u user If you do not want to be root (for example www-data).
See here for a quick overview of sudo and here for an overview of users and groups.
One minor point - if you are using the command line to update, it's now better to use apt than apt-get as it gives more feedback.  apt-get is now being used for automated updates.
A simple way to bring the system up to date is:
sudo apt update && sudo apt -y upgrade

This updates the repository list and if (and ONLY if) that has run fine, downloads and automatically installs the latest versions from the repository.

Raspberry pi3 and odoo image


Does torbrowser leave details of bookmarks after deletion of bundle from windows 10

Question: I have read a lot of conflicting info regarding this.  If i delete torbrowser from my pc using windows 10, does tor leave behind traces of bookmarks?
Answer: Tor browser does not encrypt the bookmarks, so this means that if the tor browser folder just got deleted - and not overwritten by other data - then it can potentially be recovered with recovery tools.
This is why its very important to always use a tool that deletes and overwrites the bytes on the disk where the data was located.
Bleachbit is one such application, it also works on multiple platforms.
You can then use this to scrub empty space on the disk if you already deleted something and want to make sure its gone for good, this will then make sure any deleted files cannot be recovered.


Change Add to cart button text based on shopping cart price rule

Question: I am looking for a solution to Change Add to cart button text based on shopping cart price rule in product list page and product view page.
Answer: I don't think this is possible, unless you have other products in your cart before.

Shopping cart price rules create discounts for orders at the checkout
  level, based on a set of conditions. The discount can be applied
  automatically when the conditions are met, or be applied when the
  customer enters a valid coupon code. When applied, the discount
  appears on the shopping cart page under the subtotal. Once
  established, the price rule can be used as needed for a season or
  promotion by changing its status and date range.

Link: http://merch.docs.magento.com/ce/user_guide-Jan-29/content/marketing/price-rules-shopping-cart.html


Quantum Fourier Transform and Entropy

Question: QFT is a nonlocal unitary transformation and so can generate entanglement in a system. It means a separable pure state can be converted into an entangled pure state. Now since the presence of entanglement can be witnessed via an increase in the entropy of the subsystems. Since all the subsystems witness a positive entropy change ,does the entropy of the complete system also increase (it seems to increase since entropy is additive) ? Now if it does increase , It seems to violate reversible nature of Quantum algorithms. I am very confused.
Answer: Just because the entropy of the subsystems increases, that doesn't mean that the entropy of the whole system increases.  This is possible here because of entanglement: an entangled pure state has zero overall entropy, but the subsystems have non zero entropy.  A simple example is the state
\begin{equation}
\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle).
\end{equation}
Both qubits have subsystem states
\begin{equation}
\frac{1}{2}(|0\rangle\langle 0|+|1\rangle\langle 1|),
\end{equation}
and, if you compute the entropy of these states, via
\begin{equation}
S=-\textrm{tr}[\rho\log(\rho)],
\end{equation}
you get $\log(2)$ as the entropy for both qubits.  But, if you compute the entropy of the full system, you get zero.
Comment: what about additive nature of Entropy ?
Comment: @KishorBharti that's only true if the systems are independent.  In this case, if the two qubits have the overall state $\rho_1\otimes\rho_2$.  In the usual statistical mechanics derivations of the additivity of entropy, people are probably looking at that situation.

Quantum Fourier Transform and Entropy


Strawberry sauce fermented in fridge?

Question: I left a tiny plastic take-out container of strawberry or raspberry sauce for cheesecake from a restaurant in the fridge a little too long and it was bulging and leaking. I am not eating it but wondering about clean up. Since it should be acidic/sweet is this likely a fermentation thing, not a botulism thing? Thanks!
Answer: C. botulinum germinates when there is no oxygen, generates toxin when dying and doesn't like low pH (below 4.6), so you should be safe.
If the container is airtight, there is still a possibility of some spores (if they're present due to improper cooking and handling) germinating and then dying, so your sauce could still contain trace amounts of toxin, but it looks like a very remote possibility.
In case you want to be absolutely sure, toss the entire thing out and cover the spill with a 1:4 v/v bleach solution, paper towels and let it sit for 15 minutes. Then remove the towels and clean with water and soap to remove the bleach. Wash your hands thoroughly. More details here

Strawberry sauce fermented in fridge?


How should I break-up with a second author?

Question: My advisors told me he wants to help me publish the theoretical section of my dissertation. He offered it! I have written my part, but he hasn't done his section. It's been a year. I have reminded him a few times, but he doesn't keep his promise. I know I have to move on and publish this piece on my own. I just don't know how to tell him! Is this common? I know he helps others, so I feel ignored.
Answer: Has he offered a timeline when he might be able to do his part? Have you had any communication regarding this? If not, you should reach out in advance and offer your timeline, giving him a reasonable timeframe to complete his work.
If you've had such communication before, and he hasn't been able to stick to agreed upon dates, I think you should talk to your advisor and explain the situation. Then, you could write a very polite email to him, including your advisor and any other stakeholders, explaining that you need to submit this work soon. At this point you could ask him if he still has time and would be able to complete his section, say within one month. Mention that you understand he is busy and has multiple priorities to juggle, and that if there isn't time, you'd just like to concentrate on your part and submit it. If, at a later stage, you both have time, you'll be willing to work together if he'd like to submit the theoretical section separately, with him as the first author.
I think this will sound pretty reasonable and you'll be able to publish without antagonizing or destroying working relationships with anyone.
Comment: @Solar Mike a year probably would have been sufficient a year ago, but if the co-author hasn't done anything until today, inisiting now that the work should be ready "yesterday" doesn't help, as annoying as this may be.
Comment: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/86267/discussion-between-henning-and-solar-mike).
Comment: offer him more time? a year already... and give more... It'll be out of date before he gets to do his section...
Comment: If you have given him time earlier, and you communicated very clearly upto what time you could wait, then there isn't a decision to be made the and the question is isn't needed. If you haven't done so, then it makes sense that you do so now.
Comment: So you say the year is not sufficient...
Comment: @henning so what you mean is it is probably easier to drop that second author and publish or keep the second author and never publish...
Comment: @SolarMike Either that or do what OP so far seems not have done: Communicate clear expectations for the timeline *from now on* (and if these are not met, publish without co-author).
Comment: @henning so what was the "I've reminded him a few times"... stated by the OP?
Comment: @SolarMike true, but I didn't get from the question whether or not the reminders also explained that OP will submit alone if the coauthor doesn't collaborate.
Comment: @henning what it does show is a lack of responsibility on the part of the supervisor...
Answer: There's surely a lot of politics involved. You have to take these into account in deciding the best way out of this. Also weigh in what are your short-term plans. First of all, make sure your slow co-author isn't actually a good one but just lazy or with poor communication skills. Towards this goal you'll rely a lot on diplomacy and intuition for probing the person. When doing that, find out politically where he stands.
I will discuss a few potential scenarios below.

Your 2nd author is a younger student of a friend of your advisors'. In such case this person has booby-trapped you into providing him a gift authorship. Problem is, if you don't he'll start moaning to his advisor who will complain to yours, and then you get scolded and/or sabotaged in the backlash. In such a case, I suggest you diplomatically push and probe your co-author in the presence of advisors, directly asking this person whether he's no longer interested in the project so that you can move on.
Second author is a senior academician who's actively seeking to intrude in as many papers as possible for a glossy record at career start. This is a common situation. These guys go around fishing for "collaborations" when all they want is their name on another piece of paper. In this case you should evaluate how close & dear is this person to your advisor, and to you. If you're likely to rely on his signature in the near future, perhaps you should start digesting the fact that you'll write everything on your own and just add his name. If this person is still fresh in the department and not holding any influence towards you, just politely tell him you'll be moving on. Expect some veiled threat and/or yelling. If you find yourself forced to swallow him, at least try to get him to pay for publication fees and get the paper into a better journal somehow.
Your lazy coauthor works in the same lab. Regardless of position, in such case you should consider either taking this person in for free or stalling until the moment you leave the group. If you leave, just brush him aside. In either case, start writing his section immediately.

Whatever you decide, do it as diplomatically and clean as possible. You're likely impulsive and inexperienced at this stage, and any escalation of this situation may easily become a career burden or even a permanent stain.
Judge wisely, and good luck!
Comment: @KarMasia From what you describe then it's more like situation 2. If you'd like to get criticism on the technical quality of your work, ask input from 1-3 colleagues you trust. Don't rely on seniors taking part in it to judge because they've private interests in your actions, so they're prone to manipulate you. Once you find out how bad is the ms and improve it, you can test his reaction. If you think you're being used, find a diplomatic way around the situation. Unfortunately abuse of foreign scholars is fairly common in academia—read other threads in SE Academia. Good luck!
Comment: The second author is a senior academic. He told me he wants to be the second author to help me with publishing and that his name would put my name on the map as well! I start thinking that my work is SO BAD that he doesn't know how to get rid of me, so he is stalling...that's what I think.
Comment: Honestly, working as an international student in the US academia is by far the most difficult thing I've done in my life. The work doesn't bother me. The mental health aspect of it is the most difficult part. I never know if I am as qualified and valuable as the local students. It's exhausting!


Name of the Rope used to Tie Climbers Together

Question: Does anyone know if there's a technical term for the rope/method which climbers / mountaineers use to tie themselves together (i.e. climbers tie in, then if one falls the others self-arrest to prevent being pulled off / stop the faller from falling further than the rope allows).
Comment: Everything I've seen in the last five minutes seems to suggest it's called rope. Curious to see if someone comes up with something here.
Answer: Wikipedia has an article about rock-climbing gear that answers this (a search for mountaineering gear also leads to the link). The important part is here:

Ropes used for climbing can be divided into two classes: dynamic ropes and low elongation ropes (sometimes called "static" ropes). Dynamic ropes are designed to absorb the energy of a falling climber, and are usually used as Belaying ropes. When a climber falls, the rope stretches, reducing the maximum force experienced by the climber, their belayer, and equipment. Low elongation ropes stretch much less, and are usually used in anchoring systems. They are also used for abseiling (rappeling) and as fixed ropes climbed with ascenders.

The article also provides a link to the Wikipedia entry on rope, with a specific section on rock-climbing ropes, which basically says the same thing.
In all of this, along with a plethora of other sites, I cannot find a technical name for rope used in mountaineering or rock-climbing, although it appears sometimes "belaying rope" is used. The closest I have found are dynamic and static ropes. To reiterate, dynamic ropes are more elastic and exert less force on a falling climber; static ropes are used in anchoring systems because they have much less stretch.
The method for anchoring a climber is called belaying, which is a technique to limit the distance a climber may fall by having a partner anchor them with a rope. Wikipedia says:

Belaying refers to a variety of techniques used in climbing to exert friction on a climbing rope so that a falling climber does not fall very far. A climbing partner typically applies the friction at the other end of the rope whenever the climber is not moving, removing the friction from the rope whenever the climber needs more rope in order to be able to continue climbing.

In a technique where multiple climbers are tethered together in less steep climbs, the method is a combination of self-arrest and crevasse rescue:

Teams of two to five climbers tie into a rope equally spaced. If a climber begins to fall the other members of the team perform a self-arrest to stop the fall. The other members of the team enact a crevasse rescue to pull the fallen climber from the crevasse.
Answer: In Italian there is the term "Catena di assicurazione" that I can literally translate in "Belaying chain" where "chain" does not mean "a series of linked metal rings"1 but it means "a sequence of items", the items are:

Climber
Climbing harness
Rope (and knots)
Anchor(s)
Belay device
Quickdraw(s)
Carabiners

One of the the key of this model is to pay attention to the mixing of "rigid" item (like for example a metal made one) with "elastic" item like the so called "dynamic ropes".
1A chain can be indeed used as a part of an anchor.


empathy "Facebook account requires authorization" problem

Question: I am trying to connect to my Facebook account from Empathy on Ubuntu 13.04.
When setting up the account in Empathy everything looks OK, but then Empathy just shows the message:

Facebook account requires authorization

According to bug #1170832, this is due to a problem in the telepathy-gabble package which has been fixed in version 0.18.1.
However, I manually installed telepathy-gabble_0.18.1-1_amd64.deb from Launchpad and the problem persists. Has anyone else had any luck with this?
Answer: The telepathy-gabble 0.18.1 package in Ubuntu actually contained a work-around for the problem, not a proper solution. Quoting from the ChangeLog:
telepathy-gabble (0.18.1-1) unstable; urgency=low

  * New upstream stable release
    - work around Facebook server behaviour change so we don't consider its
      service discovery response to be spoofed (Closes: #721883)

 -- Simon McVittie  Mon, 09 Sep 2013 13:05:13 +0100
Apparently, this work-around does not work, and in fact the status of the bug has been reverted from Fix Released to Confirmed.
For the proper fix I'm sorry you will have to wait for the upstream bug to be fixed (unless someone proposes an another work-around in the meantime).
Answer: Had the same problem and been looking at the suggestions.
On Ubuntu 13.10
What worked for me:

Removed fb account from online account in settings.
Went to fb on firefox and the little gear>account settings>apps>removed access of ubuntu.
Re-entered fb account on online accounts on desktop this time using the fb "username" only NOT the email address.

That worked for me.
Hope this helps.
-Owie
Answer: Just a couple suggestions you might try...
No-1. Log into facebook and turn off notifications for log-ins...
Facebook > Account Settings > Account Security (change) >
Change "Would you like to receive notifications for logins from new devices?" to NO.
No-2. Try logging out of facebook and logging back in (with the "facebook username", not email)
More suggestion and info here → http://ubuntuforums.org/showthread.php?t=1487433
Hope this helps... Good Luck!
Comment: Yeah I tried that already, did it work for you ?
Comment: Didn't work for me


Whiteboard marker stain cleanup from textbook

Question: I accidently dropped white board marker on my textbook and unfortunately it landed on its tip. so now there is a big stain on the page. any ideas that how can i remove it. i tried eraser. actually textbook is pretty important.
Comment: Try a solvent, such as acetone or ethyl-methyl ketone (nail-polish remover), with a pad of paper towel underneath to absorb the ink. It might work, make the stain a bit lighter, or do nothing. *Caution!* Flammable.
Answer: If it is your own textbook, and you are in a study class where others have the same book:

photocopy the stained page front and back from a borrowed copy
if you can, copy to both sides of a single sheet of paper
cut off the damaged page leaving enough to tape to
use clear tape to overlay the stained page
cut off excess paper to be smaller than other pages, so it doesn't protrude
Comment: Ohh so no way to remove the stain? lol thanks btw


How do I bower install Raphael.js?

Question: I'm trying to install Raphael.js in my project in the most modular way possible. Raphael has a registered bower component, so that seemed like my best best.
With some guidance from the Snap.svg readme (which did eventually allow me to install Snap), I've gotten the farthest with these steps (starting in my project root directory):

bower install raphael --save
cd dev/components/raphael/
npm install -g grunt-cli
cd dev/
npm install
grunt

The last step appears to begin to build the component, but fails immediately:
Running "build:dist" (build) task
Warning: Unable to read "../eve/eve.js" file (Error code: ENOENT). Use --force to continue.

Aborted due to warnings.

The error makes sense, since there is no eve.js file at that path. I know that it's a dependency of Raphael, but where and how would I have installed that?
FWIW, I'm a bit green on bower & grunt, so I could just be missing a simple step. Unfortunately, an hour or two of experimenting and my best Google skills couldn't sort this out.
Answer: I think, bower install raphael is enough. You don't need to build raphael, because bower already delivers the compiled raphael.js and raphael-min.js.

How do I bower install Raphael.js?


Microsoft SQL Server 2008 - Catching user role changes with SQL Server Profiler?

Question: I'm trying to figure out why one of my SQL server users keeps losing sysadmin role when I uninstall a particular program. Can I use the SQL Server Profiler to track down the action that causes this to happen? If so which events do I need to check off on the Events Selection screen?
Comment: What program are you uninstalling? Does that user have anything relation to the program you are uninstalling?
Answer: The "Security Audit>Audit Add Login to Server Role Event" will capture role drops as well, not just role adds.  However, that one alone might not give you the information you're looking for, depending on what you need (such as the specific statement executed if necessary).  So you could also add the Stored Procedure>RPC: Completed event, and if you want to get really granular add the SP: StmtCompleted as well.
An alternative could also be to use a server-level DDL trigger.  I use a trigger on the ADD_SERVER_ROLE_MEMBER event so I can catch if one of my admins decides to throw someone into the sysadmin role without my knowledge.  In my case I have an admin database with a table designed to store DDL event info and I insert a record for this type of event, among others.  The full list of events you can create a trigger on is here: http://msdn.microsoft.com/en-us/library/bb522542.aspx and there are good basic examples of how to use these events in T-SQL in the trigger body here: http://msdn.microsoft.com/en-us/library/ms175941.aspx
Comment: I'm assuming that the "vmuser" account is registered with this software application in some way, such as a service account, or DB access account?  If so, it would appear that one function of the uninstall process for this software is to remove sysadmin permissions from the service account(s) used.  Yes, that is the SQL statement that would be executing, which is a call to a system stored procedure "sp_dropsrvrolemember" passing the parameters of [NPTAD\vmuser] (the user to execute against) and [sysadmin] (the role the user is to be removed from)
Comment: Thank you! I think I found where it happens, it's a SP: StmtCompleted event that reads `sp_dropsrvrolemember [NPTAD\vmuser], [sysadmin]`. Does this mean that somewhere this exact line of code is being executed or is this an output of the results of executing code?


Phone not booting after rooting attempt

Question: I tried to root my phone Samsung Galaxy J1 (SM-J100H) with Odin3 software following some steps, but after that now it's not working, I turned on the phone and it vibrates one time and I see only the Samsung logo and nothing else. I tried too wipe all data and factory reset but it doesn't do anything. What can I do?
Answer: A factory reset won't undo rooting since it only clears the /data partition and not /system.
What you should do to get your device working is flash a stock ROM with Odin.  You can get one from sammobile.com.  As usual, you would put the .tar file from there into the AP/PDA field of Odin and flash it while the phone is connected in Download mode.
Comment: I get this error: Complete(Write) operation failed.  http://i.imgur.com/6QBQg04.jpg Please so much help me to solve this.  can you suggest me a version please? 1.85 is good?
Comment: @Dimas The versions are in the PDA column, things like `J100HXXU0APB3` not `1.85`.  Just filter by your country/carrier and then try the second-newest, or maybe the oldest, etc.  I don't have this phone so I don't know what versions are the best. Maybe try Romania?
Comment: @Dimas No problem, glad it's working!

Phone not booting after rooting attempt


Discovery of Suction pump

Question: As I know that before Galileo,all the scientists believe that vacuum cannot exists. But how suction pump(which itself remove air)was discovered before Galileo. What was the steps involved in it to make suction pump?
Comment: "All scientists" did not believe that vacuum cannot exist, the common view was that nature "resists" its creation, see [Torricelli and the ocean of air](https://www.ncbi.nlm.nih.gov/pubmed/23455767). Siphons, syringes, and suction pumps were known long before Galileo or modern explanations, already in antiquity, see [Wikipedia](https://en.wikipedia.org/wiki/Siphon#History).
Answer: Aristotle believed that no void(vacuum) could occur naturally, because the denser surrounding material continuum would immediately fill any incipient rarity that might give rise to a void. Even in his Physics, book IV, Aristotle offered numerous arguments against the void: for example, that motion through a medium which offered no impediment could continue ad infinitum, there being no reason that something would come to rest anywhere in particular.
But much before Galileo, in the 13th and 14th century, considerable attention was given on these issues by European scholars such as Roger Bacon, Blasius of Parma and Walter Burley . Eventually following Stoic physics in this instance, scholars from the 14th century onward increasingly departed from the Aristotelian perspective in favor of a supernatural void beyond the confines of the cosmos itself, a conclusion widely acknowledged by the 17th century, which helped to segregate natural and theological concerns.
Using geometry, Ibn al-Haytham(965–1039) mathematically demonstrated that place is the imagined three-dimensional void between the inner surfaces of a containing body. According to Ahmad Dallal, Abū Rayhān al-Bīrūnī also states that "there is no observable evidence that rules out the possibility of vacuum". These steps led to the making of suction pump which appeared in Europe from the 15th century.
Thus, it is not true that scientists before Galileo did not believe about the existence of vacuum or void but were aware of it.
Source: Wikipedia

Discovery of Suction pump


Compact subset of locally compact $\sigma$-compact

Question: Let $X$ be a non-compact, locally compact space. Also suppose that there is a sequence of compact non-empty sets $\{K_n\}_{n\in N}$ such that $$X=\bigcup_{n\in N} K_n,\quad K_n\subset K_{n+1}.$$
Now could we say that for any compact subset $K\subset X$ there is a number  $n\in N$ such that $K\subset K_n$? If not, which conditions can help? or, is it possible if we choose sequence of compact sets? How about topological groups?
Answer: No, not as stated.
Let $X = [0,1)$ be the half-open unit interval, which is locally compact but not compact.  Let $K_n = \{0\} \cup [\frac{1}{n}, 1-\frac{1}{n}]$ which is compact. Then we have $K_n \subset K_{n+1}$ and $X = \bigcup_n K_n$.  But the compact set $K = [0, \frac{1}{2}]$ is not contained in any of the $K_n$.
However, under these assumptions here is something we can prove.

There exists a sequence of compact sets $K_n'$ such that $\bigcup_n K_n' = X$ and $K_{n-1}' \subset (K_{n}')^\circ$ for each $n$.  In particular, for any compact set $K$, there is some $n$ with $K \subset K_n'$.

Proof.  By local compactness, for each $x \in X$ there is an open set $U_x$ such that $x \in U_x$ and $\overline{U_x}$ is compact.  We construct $K_n'$ recursively.  To get started, let $K_0'=\emptyset$.  Now to construct $K_n'$, suppose $K_{n-1}'$ is already constructed.  Since $K_n \cup K_{n-1}'$ is compact, there exist $x_1, \dots, x_r$ such that $K_n \cup K_{n-1}' \subset U_{x_1} \cup \dots \cup U_{x_r}$.  Set $K_n' = \overline{U_{x_1}} \cup \dots \cup \overline{U_{x_r}}$ which is compact.  Then $K_{n-1}' \cup K_n \subset (K_n')^\circ$.
In particular, $\bigcup_n K_n' \supset \bigcup (K_n')^\circ \supset \bigcup_n K_n = X$.  So the $K_n'$ cover $X$.
Moreover, the sets $(K_n')^\circ$ are an increasing open covering of $X$.  So if $K$ is any compact set, we must have $K \subset (K_n')^\circ \subset K_n'$ for some $n$.
Comment: Yeah! You are right.
Comment: Thanks! Now I can make a bounded approximate identity for Banach algebra $C_0(X)$ of bound $1$ when $X$ is $\sigma$-compact.
Answer: I can't comment due to lack of reputation.
In the argument of @Nate Eldredge, do we need the space $X$ to be a Haussdorf space, in order to have closed compact neighbourhoods of each point?
Comment: Yes! It is Haussdorf space!


The usage of 'to'

Question: My problem is that I don't know whether I should use 'to' or not when I say the sentences I wrote below. Can anyone tell me which one is right and why, 'have you gotten New York safely?' or 'have you gotten to New York safely?'? I don't know what to say when it comes to those. I get confused. I often hear people say 'did you get home?' and 'did you go home?' and they don't use 'to'. But they also say '...go(get) to your office...' or '...go to school'. I want to know the reason.
Comment: You are on the wrong site. You need English Language Learners.
Answer: I think the sound of got-to run together may have left you thinking of gotten.
Only the animal catching its prey has gotten its meal.
Using simpler words the right meaning would come through more easily.
'Did or will you get to New York safely?'
'Have you got to New York safely?'
'Had I got to New York safely I would have called'
These are all about a destination of a city or place where to would be correct.
From to the state of New York to the city, to my street or to my house these all use to. Going to my home would be correct as well but it would be fine to say you went home. Or did you go home or get home. The nature of this exception baffles me.
I hope this helps.
Comment: Then, would you please tell me if it is 'I'm on my way back 'to' where I belong.' or 'I'm on my way back where I belong?
Comment: Either of those would work and not make a bad sound as English but I'd pick the second. Leaving 'to' in or not would then be important for keeping the number of syllables right for the song lyrics.
Comment: "Going to my home" is very commonly heard from non-native speakers (speakers of Romance languages especially), but I've never heard a native speaker say it. _Home_ is a sort of weird pseudo-preposition on its own and does not normally take any preposition to indicate movement-towards. Same with words like _(t)here_ and _back_. "*I went to home/there/back" is quite ungrammatical. (Note that _house_ is different from _home_: "Went to my house" is fine, but "*went to (my) home" is not.)

The usage of 'to'


Ubuntu Remove/Uninstall from single boot system

Question: My problem is I have ubuntu 11.04 installed on my system. I have no other os installed. I have tried to install windows xp using bootable cd of XP. But it doesn't work. Blank screen appears. How can I remove ubuntu then? I have tried many ways using gparted and all but no result.
Comment: "install windows xp using bootable cd of XP. But it doesn't work. Blank screen appears.". This is a windows problem and not Ubuntu; removing Ubuntu from your system does --not-- guarantee the install from the Windows CD works. It is more likely the copy you burned failed to do it correctly or that your forgot to set the option to boot from CD from within your BIOS.
Answer: 3 possible methods for removing an operating system from your hard disc (so it does not even have to be Ubuntu):

You can remove the current OS from your hard disc by using the Live CD you used to create the installation. It will have gParted installed and loaded into memory and all you need to do it delete the partitions gparted shows you. Mind you: you can only delete unmounted partitions. And not all operating systems will have a live CD.

You can also use the gParted Live cd that you can find here. You need to burn this though. So if you can not access your current OS you might need a 2nd computer to burn the disc.

You can also use another OS installation media to format your disc. All operating systems installation disc will have some sort of partition manager. Windows starts with it when you boot from the installation CD. Linux flavors tend to ask for some information 1st (like language settings etc) before you see the partition manager.

If you decide to stick to gParted then this link explains how to do that. Copy/paste from that link:

Deleting a Partition
To delete a partition:

Select an unmounted partition. See the section called "Selecting a Partition".

Choose: Partition → Delete. The application displays the delete partition operation in the Pending Operations pane.

Do not forget to read the "caution" part underneath these pointers.


Gravitational potential energy of mass between two planets

Question: Suppose I want to launch a rocket from earth to some point $O$ between the center of earth and the center of moon (on a straight line connecting their centers), where the gravitational force of the moon 'cancels out' the gravitational force of the earth (this point is located at $\approx 54 R_E$ from the center of earth where $R_E$ is the radius of earth). I want to know how much energy I should spend in order for the rocket to get there (neglecting the atmosphere and the rotation of the earth around its axis). So, I know that this is basically the difference between the potential energy at the start point and at the end point of the destination. However, $O$ is located not only in the gravitational field of the earth, but also in the gravitational field of the moon. And it seems that I cannot neglect the gravitational potential energy of the body at the moon's gravitational field. So my question is - how can I combine these two? How can I calculate the total GPE of the body in two (or even more) intersecting gravitational fields?
Comment: You are looking for terms "Delta-V required to get to the Earth-Moon Lagrange point no.1".
Comment: @DeerHunter - sorry, but I'm not familiar with Lagrange mechanics. I'm just starting with physics and Newtonian mechanics.
Comment: @grjj3 He's not talking about Lagrangian mechanics but about the Lagrange points. Same guy, different ideas.
Comment: @dmckee - Thanks for the remark. But for me - still no idea of what it is. Hopefully, one day I'll learn that too.
Answer: Gravitational Potential is a scalar quantity so can be added algebraically directly for both(or more) bodies.
Also GPE is just Gravitational potential times mass. $$E=\underbrace{\big(\sum P\big)}_{\text{due to all bodies in vicinity}}\times m$$
Now , rest of your aproach is allright ! Continue using this.
Comment: young man then :) I myself student too.
Comment: So, the gravitational potential energy of a **single** body is just the mass of body multiplied by the sum of the quantity $- G \frac{M}{R}$ for each planet that its gravitational field influences the body?
Comment: @grjj3 Potential due to a single body is $-GM/R$ at a distance $R$ from it's center if it's a sphere. Mass times this quantity is GPE of that body and M(planet) and not just GP.I don't understand what u mean by " that its gravitational field influences the body?"
Comment: I know. I just ask about the **sum**. I know that $-GMm/R$ is GPE. I'm asking how GPE will look for a **single** body in **multiple** gravitational fields. Is it $(-GM_E/R_{E,O} - GM_{m}/R_{m, O}) m$ where $M_E$, $R_{E,O}$ are mass of earth and distance from center of earth to $O$ and $M_m$, $R_{m,O}$ are mass of moon and distance from center of moon to $O$?
Comment: Yes , you are correct . They are scalars and can be added algebraically .
Comment: @grjj3 Please don't say sir........ `:D` . I'm a high school student (just cleared)


If a basis set is complete, are the elements in it mutually orthonormal?

Question: If a basis set is complete, are the elements in it mutually orthonormal? For example, we can express the field operator in the basis of the creation and annihilation operators.This basis is complete, are the elements in it mutually orthonormal?
Answer: The answer is  NO
Given any complete basis, one can construct another by taking independent combinations.
Assuming the original basis is complete and orthogonal, and contains $V_1$ and $V_2$, (which are thus orthogonal).
Replacing $V_2$ by $V_2'=V_1+V_2$ does not change the completude of the basis. But now two vectors of the new basis, namelt $V_1$ and $V'_2$ are not orthogonal.
If the questio is, whether it is always possible to find an orthogonal basis is one thing. But there is no need for a basis to be orthogonal to be complete.
Comment: Moreover, one can have overcomplete sets (v.g. cohérent states) which are not orthogonal in general.


Backup TO URL S3

Question: Can I do a SQL backup to AWS S3 URL from an EC2 machine
I know that Azure side is possible but I do not know if it is possible AWS side
thank you for your advice
Comment: I said an EC2 machine
Answer: it is possible with SQL Server 2022. here is the Amazon article on the subject:
https://aws.amazon.com/blogs/modernizing-with-aws/backup-sql-server-to-amazon-s3/
Comment: Perhaps you could add the salient points from the article to the actual answer, lest it becomes useless once the link goes stale.
Comment: Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer).
Answer: As Alexei mentioned, this is possible from SQL 2022.  If you are on an older version of SQL Server I would recommend using File Gateway. Using a file gateway will simplify your backup and restore operations vs writing to local storage and using scripts to push the files to S3.  Using command line tools to push large files to S3 can also be quite CPU intensive and will increase your overall backup time.
If you do need backups from S3 it can also be painful to have to pull them down to local storage to restore them.
File Gateway works as a file share that is backed by S3 storage. So you just write your backups to a UNC path and works with Ola's scripts.  Restores also work directly from the File Gateway.
It does have a cost associated with it though.  In our case, this was completely offset by reducing the amount of EBS storage we needed.  We didn't need to duplicate our backup storage on the passive node.  Also, you might not need quite as big a volume for the cache on the File Gateway.
Answer: No, at the moment there is no way to transfer backups directly to S3.
It seems to me that that option will not be added, since Microsoft and Amazon are competing in the cloud computing market. They are not interested in the mutual integration of products and services.
But in general, this is not a problem. Both SQL Server and AWS S3 support command-line interfaces, so you can write a simple backup script, or use a third-party utility.
To write a script, we need two utilities.
sqlcmd is a utility for interacting with SQL Server from the command line. It is installed with SQL Server.
aws is a command-line utility for interacting with Amazon Web Services. After downloading it, you need to configure it by running AWS configure. You will need to enter the access keys that you need to create in advance, the keys must have write access to the S3 bucket.
Here is a very simple script for backing up a SQL Server database to S3.
set day=%date:~7,2%
set month=%date:~4,2%
set year=%date:~-4%
set backup_name=%day%_%month%_%year%.bak

sqlcmd -Q "BACKUP DATABASE [AdventureWorks] TO  DISK = N'W:\temp\%backup_name%.bak'"

aws s3 cp W:\temp\%backup_name%.bak s3://my-backet-name/%backup_name%.bak

Note that the first three lines that receive parts of today's date to form the file name are guided by the position of the numbers in the %date% system variable, which depends on the regional settings.
To prevent your S3 storage from overflowing, configure in the backup bucket to automatically delete old files (or move them to AWS S3 Glacier)

Backup TO URL S3


Wave nature of an electron

Question: Quantum mechanics tells us that any quantum particle behaves as particle or waves in different situations.
Assume that the electron is behaving like wave in some situation, how can I visualise this situation.
Whether the electron is continuously vibrating about its position or not?
In internet I read that a single electron in hydrogen atom mostly behaves as wave .
In that it is compared to the stranding waves on the guitar string
I just want to know what it is doing when it is behaving as wave and as particle
Please clarify this,otherwise I simply waste my time by keep on searching in Google .....
Please...
Comment: "What it is doing" is evolving according to the Schrodinger equation.
Comment: Sorry sir, I'm just a college student. I know nothing about this Schrödinger equation. But wave means what? Is it lyk the standing wave of that string I mentioned?
Comment: Read the double-slit experiment for the electron in Feynman's Lectures in Physics at http://www.feynmanlectures.caltech.edu/I_37.html#Ch37-S1..... (experiment with bullets and waves) if you have not already.
Answer: It is a bit like the standing wave, but the "particle" (electron, in your example) would correspond to an excitation of the string, not the string itself. The string itself would represent the "wave function" of the electron.
But it's a string with a very strange property:  it is either vibrating with a particular amplitude, or it is not vibrating at all.  Vibrating means that the electron exists, not vibrating means it does not exist.  This strange property is how the Pauli exclusion principle would work in your model:  only one excitation allowed <=> only one electron per quantum state.
Believe me, it takes a while to get used to thinking about things this way.  There really is no adequate analog or metaphor in our everyday experience that can be used to develop the visualization that you seek.  In particular: don't think about a vibrating particle.  That picture is far enough removed from real electron behavior that it is fruitless.
Comment: Today I watched the video about Schrödinger cat. that means we can't really tell whether there is an electron or not in a particular place in an atom until we observe it at that place. From this what I understood is that 1).an electron can be anywhere in that atom. 2).it's not at all revolving 3).wave function is just a mathematical concept from which we can predict the position of electron 4). So finally wave nature means it can be in place around the atom. Are all these conclusions correct?
Comment: Yes, that's pretty much it.

Wave nature of an electron


The workings of a stateful firewall

Question: Say I have network A and network B.
I wish to SSH from network A to network B. Hence, I would set a route from network A to B, and in the stateful firewall, I will allow a rule for Network A to B at port 22.
In this case, do I still need to create a rule to allow Network B to Network A on all ports, since there is TCP handshake? I believe we do not have to, since it's "stateful"? But how does it work?
Comment: I think this is more of a basic networking question than an InfoSec question. It requires that you understand the "statefulness" of protocols like TCP.
Answer: Connections are 2-way. Opening a port in a firewall allows a connection to be opened.  Once opened it is 2-way.
Answer:
In this case, do I still need to create a rule to allow Network B to
  Network A on all ports, since there is TCP handshake?  But how does it work?

No, you do not need to do so because the traffic already took place in both directions. It is based on the stateful inspection principle (called also  dynamic packet filtering) in which the stateful firewall not only relies on the rules set for it to function but is able to memorize information within dynamic state tables about the incoming and outgoing of the network packets.
Such information is the one hold by the header of the packet such as  the destination, the source the protocol used and the port number in use.  These information is used by the firewall to decide whether to allow or not pass a packet. It monitors the exchange of packets in both direction (A and B),  opening holes in the firewall for each communications session on an as-needed basis , and then close the holes as soon as they're no longer needed for authorized traffic.
Note that this philosophy used by such firewall to function is not totally safe. Some attacks  are done by triggering an internal host to generate a TCP packet that (thanks to the dynamic inspection principle used by the firewall) will change the firewall's internal state and enable an attacker to establish a TCP connection to a filtered port through the firewall.

The workings of a stateful firewall


Why does the shadow of the hairs on my leg seemingly only appear when the leg is moving slowly?

Question: I happened to notice the shadow of my legs in the sunlight on the floor and I noticed that I could see the shadow of the hairs on my legs only when I moved my leg slowly (slowly moving onto my toes etc.) As soon as the movement stopped the shadow of the hair seemed to diffuse. Why? Many Thanks, Terry
Comment: As with most of this "illusion" questions this is probably not a physics question but a human-visual-apparatus question. I'm not closing, however, as I don't have a good explanation.
Comment: It could be a diffraction question, perhaps there is some kind of aliasing effect with the effective "frame rate" of the eye and the movement of the hairs, if it is dependent on the physics of the eye it is a good question, I never noticed this before myself...
Comment: Well, in direct sunlight, at about 25cms away from an A4page there is no difference in seeing the shadows of the fuzz on my arm. Individual hairs show up whether moving or not. At a distance of +50cm I can see nothing, whether moving or not, except a difractive pattern on the shape of the arm. Glasses?
Answer: I think that its related to how quickly your eyes can perceive small and fast moving things. Take an annoying mosquito for instance. While a mosquito is slightly larger than your leg hairs, its awfully difficult to keep track of them flying around you with your eyes. I'm willing to bet its related to this concept of human visual processing speeds.
According to the Wikipedia article on Frame rate:
http://en.wikipedia.org/wiki/Frame_rate
It states that at 48 frames per second is what's being used in the film industry to make things look more real. Taking this information and applying it here, it would mean that information that changes faster than 1/48 of a second, would probably be unnoticed by the human eye. The information would be the change of image for film, and for our case, the information would be what you see, which is the change of position of your leg.
So if you move your leg slowly, it makes sense that you'll be able to keep track of your leg hair shadows, where as moving it fast, you wouldn't because they are very thin/small and moves really fast like a mosquito.
If you have lots of leg hair like a forest, than I think the concept of looking through a fence while driving by quickly makes sense. Have you ever drove by a fenced off playing ground and noticed that you can kinda "filter out" the fence because you are seeing through the fence at different positions really fast? When your going slow, you're almost always obstructed, until a certain crucial speed.
Moreover, when you focus on the background, the image of the playground, you can see it, but if you focus on the foreground, the fence, then you'll see the fence and not the background moving really fast. It's difficult to see the foreground and the background at the same time.
I think this would be the same with your leg hair shadows too if there are a lot. It would make a cris-cross pattern like a fence and you can "see through" it if you move your leg fast, but you can't see through it if you move your leg slow.
Comment: Hi, Many thanks for the answer. It was interesting that today I was trying to take a photo of this effect - but my camera is not good enough to be able to see the effect. I did notice that while I was standing there looking at the shadow (of my leg) there is a tree outside the window and the wind was blowing and as the shadow moved up and down the shadow of my leg the shadow of the hairs appeared and my leg was not moving at all. It seems counter-intuitive - you would think that the shadow of the hairs would show when not moving instead of when moving. Again, Many thanks, Terry.
Comment: What your referring to probably has to do with reference frames. Although you are not moving, the tree is moving due to the wind being blown. Therefore, there are moments when sunlight directly hits you, and moments when it's indirect, because the tree is blocking the sunlight. When you don't have a bright enough light, you won't be able to make out small thin shadows.

Kinds like when your sitting inside a train adjacent to another train. When your train moves vs. the other train makes it look as though the scenario is the same.  Something is moving relative to another, in your case, the tree
Answer: The retina in the eyes detect changed impressions of moving objects. It cannot detect small variation in illumination compared to position change due to speed. If you are near a tiger in forest an advice to  escape danger is to be steady, and not move at all. (It is risky,but if you move, the tiger is faster :) )


How we can detect a magnetic monopole?

Question: How we can detect magnet monopole? How we can distinguish it from electric monopole? If we change E to B and B to -E in Maxwell's equations we get proper equations for charge and current density which we can call them magnetic charge and magnetic current. Even if magnetic monopole exits there is no distinction with electric monopole!
Comment: Possible duplicate of [How would I go about detecting monopoles?](http://physics.stackexchange.com/q/7540/)
Comment: I'm inclined to not close as a duplicate, since this gives a very specific line of reasoning that would seem to preclude measuring them in *principle*, while this other question is about *practicality*.
Answer: The symmetry between electricity and magnetism, though present in Maxwell's equations, is completely absent in the real world. Electrons have electric charge and disk drives have magnetic domains ... not the other way around. If you tell someone "actually electrons have a magnetic charge and disk drives have electric domains", that person will tell you that you're crazy.
Since things like electrons and bar magnets are readily obtained for our testing purposes, the upshot is that we can easily tell a magnetic monopole from an electric monopole.
More specifically: An electric charge interacting with a magnetic charge looks quite different than two electric charges interacting with each other, or two magnetic charges interacting with each other. So we can set up interactions between the alleged magnetic monopole and some electrons, and that way figure out whether it is truly a magnetic monopole.

How we can detect a magnetic monopole?


Vista: change clock tooltip to short date?

Question: Is it possible to change the tooltip that shows above the clock from the "Long date" format to show the "Short date" format?
Answer: The only solution I can think of is tho change the Long Date Format system-wide (!) in the Control Panel:

Open the Control Panel (Classic View)
Click on the Regional and Language Options icon.
Click on the Customize this format button.
Click on the Date tab.
Click on the drop down arrow to the right of the Long date: and click on the date format you want.
Click on the Apply button.

Here are some other options.
Comment: That doesn't work. All of the "long date" options say something like "19 July 2009" I want one that says "19/7/09" and there's no such options in the long dates combo.
Comment: Yes, that's true, but I think there is no other option. At least the two other formats are shorter...
Answer: Unless you particularly want to use the Vista built-in tray clock, you could try some of the free tray clock replacements that are around and see if any of them suit you better.


How can i find out which network interface managing solution is used on a given linux distribution in an easy and robust way?

Question: I frequently find myself sitting in front of various different Linux installations and need to check or change network settings.
They are different versions from different distributions (mostly debian and opensuse, but also ubuntu and derived variants) in varying states of customization, which means i can't make reliable assumptions about their configurations.
So how can i find out in a easy and robust (that is distro and version independent) way, which service is used to manage network interfaces and related network settings. (like NetworkManager, Systemd-networkd or ifup/down)
-> Ideally with basic system tools
The System can find out which service is to be used/started at system startup, so there has to be either a command that tells me or a config file where i can look it up.
I am aware, that similar questions have been asked numerous times, but all answers i could find boil down to one of two not very helpful gists:

either the author mentioned their distro/version and someone told them, which program is probably used
or they were told to basically try each possibility until they found the running manager

both of which i consider neither very reliable nor safe. (not to mention that trial and error is very time consuming)
Comment: /etc/*-release could be a starting point. But content is also distribution specific. IP setup is something where distributions differ a lot
Answer: Shows the currently enabled network service (Wicked or Network Manager).
<code>systemctl show -p Id network.service
</code>
source: SUSE LE Guide


Where does music downloaded from Google Play go on my phone?

Question: New Nexus 6 with Lollipop. I downloaded some music from Google Play, but cannot for the life of me find where it is on the phone. I've searched through the entire thing using Astro, and can't find my music. (The Play app sees it, but I want to make one of my songs a ringtone, and apparently with Lollipop, you need to copy your file into a certain location for it to be available.)
Answer: I'm afraid that's not possible. Like with iTunes, Google uses DRM (Digital Rights Management) technology in order to prevent you from illegally copying music. So, the music is stored on your phone in an encoded way and thus you can only access the music through the Google Play app.