Will the Graviton discovery mean the end of the General Relativity

Question: I know that this question has been addressed in one form or the other however I would appreciate some other thoughts - other then "GR works so it doesn't matter what's behind it". In fact it does matter because everyone agrees to few conclusions such as the theory that in early universe all FOUR fundamental forces were one and the same. GR says gravity is not a force; it's a consequence. So if it's not a force then why is String Theory hypothesizing closed curve strings propagating the FORCE of gravity? In other words, if gravity is just a result from the spacetime curvature, then this means there is really no point in looking for GUT because gravity will never be combined with the other three forces. I find it hard to digest an idea where gravity is left to just "be there" because one man stated it and experiments proved the effects.
Comment: Did general relativity mean the end of Newtonian mechanics? Of course not. The least useful of all theories will be the theory of everything. It will explain everything, but calculating even the most trivial problem will be such a hard thing to do that nobody even among those who will understand it will want to do it more than once in their life. Gravity, by the way, was never a force, not even in Newtonian mechanics.
Comment: If you can muster at least 20 reputation you can discuss this in the chat. I'm afraid your question is too vague to be usefully discussed here.
Comment: Hi Tom, cheers for your question. I understand this is your first question, but I think this might be a bit off topic. There is no clear answer to your question, except "GR will probably still be there". Maybe you could edit the question to be something along the lines of: "Does graviton contradict GR" and you might get a fairly clear answer with references.
Comment: @CuriousOne "Gravity, by the way, was never a force, not even in Newtonian mechanics" I'm sorry, but ultimately I came into physics through mathematics and, although it seems a really ignorant question, what do you mean by that? I'm sure you're going to tell me something that I've misunderstood now since the age of 15 (when I first learnt to put $G\,M\,m/r^2$ on the left hand side of Newton's second law and thus "explain" Kepler's laws)
Comment: @WetSavannaAnimalakaRodVance: $g=9.81m/s^2$. That's an acceleration. It's independent of velocity and material composition, i.e. it fulfills the conditions of the equivalence principle. Did they lie to you about Newtonian gravity? I don't know. My physics teacher didn't, he pointed this out quite clearly. I have not seen a single classical mechanics text that treats $m_{gravitational}\neq m_{inertial}$ and discusses all the consequences properly. If it did, it would certainly lie to students because there is not a single experiment that indicates that the two quantities are not the same.
Comment: @CuriousOne Okay, I've got it. I've always been gobsmacked by the equivalence $m_g = m_i$ and wondered at it, so I'm relieved that I'm not missing anything. I guess I've always thought of the need to explain $m_g = m_i$ as something after Newton, although I'm the first to admit I'm not a historian.
Comment: @WetSavannaAnimalakaRodVance: Having said this, if gravity wasn't an acceleration but a real force, then Einstein would have been totally wrong and quantizing gravity should be near trivial at this point. I think Einstein's theory is quite alive and we still nowhere close to quantum gravity.
Comment: @WetSavannaAnimalakaRodVance: I do not believe that Newton was thinking about the gravity of Mars matter differently than about the gravity of Earth matter or the gravity of the sun's matter. Maybe I am wrong, but the equivalence principle was, more or less, already there with Galileo's (probably imaginary) fall experiments. The first real measurements had to wait until Cavendish, of course and precision confirmations are definitely 20th century experimental physics. All I am saying, though, is that ordinary Newtonian mechanics teaching is fully based on the equivalence principle.
Comment: @WetSavannaAnimalakaRodVance: OTOH, I would agree that discussing the difference between the two models of Newtonian gravity, i.e. the force vs. the acceleration model might be a wonderful teaching tool to explain hypothesis building and modeling in modern physics to gifted high school students. Average students would, as you say, be gobsmacked and have beaten their interest for science out of them for a lifetime if we actually tried to explain these finer points of physics to them... it would be an absolute educational disaster, which might explain the conventional high school approach.
Comment: Comment to the question (v1): It seems OP assumes gravity is part of [GUT](http://en.wikipedia.org/wiki/Grand_Unified_Theory). It is by definition not.
Answer: I'd like to add to Anna V's complete answer, and CuriousOne's gem of a comment:

Did general relativity mean the end of Newtonian mechanics? Of course not. The least useful of all theories will be the theory of everything. It will explain everything, but calculating even the most trivial problem will be such a hard thing to do that nobody even among those who will understand it will want to do it more than once in their life. Gravity, by the way, was never a force, not even in Newtonian mechanics.

These can be summarized by saying that General Relativity must always emerge as the continuum limit of any realistic theory of gravity. Its predictions have been now confirmed in full, so any deviation in future theories from GTR will be smaller than today's instruments and experiments can detect.
General Relativity is a very simple[1] symmetry and geomety motivated model that doesn't really delve into the underlying physics of spacetime much at all. That is, it describes what behaviors of spacetime must have to be like given very broad constraints without giving any description of the "machinery" of spacetime that gives rise to these behaviors. So there's a great deal more to the story: we want to find out the mechanics of a putatively quantized spacetime that gives rise, in the continuum limit, to general relativity. The substance of General Relativity can be summarized:

Spacetime is a twice differentiable manifold, a statement that encodes the principle of equivalence by postulating that spacetime is locally flat and therefore that an observer always has a momentarily comoving inertial observer whose motion you will undergo in the absence of force;
The gauge group of the manifold's frame bundle is the Lorentz group: in less technical language this means that locally the co-ordinates of different inertial observers at any spacetime point must be linked by proper Lorentz transformations. Thus, locally the speed of light is always $c$ and this situation allows us to impose constraints on solutions that will guarantee causality (by postulating that supraluminal signalling is forbidden as I discuss in my answer here) just as we can do in special relativity;
The geometry of spacetime is influenced by the distribution of stress energy within it, so we seek a simple generally covariant relationship $G = f(T)$ where $G$ is some tensor that describes the geometry of spacetime and $T$ is a tensor describing stress-energy distribution. As a first approximation, we postulate $G\propto T$;
We postulate energy-momentum are locally conserved, so that the divergence $T^\mu_{{},\,\mu}=0$ of the stress energy tensor must vanish;

So then we ask, what is the most general rank two tensor describing geometry alone whose divergence automatically vanishes? The most general such tensor involving second derivatives alone can be shown to be the Einstein tensor, whose divergence vanishes by dent of the general geometric principle that the boundary of a boundary of spacetime volume is always the empty set. The Einstein tensor defines the part of the Riemann Curvature Tensor that defines how the volume of a test region in spacetime changes (as I describe here) and well-defining boundary conditions of a given problem fix the Weyl tensor, which is the part of the Riemann curvature tensor that describes how the shape of the test volume evolves. Boundary conditions and field equations both fully define the Riemann tensor, which in turn fully defines the metric properties of spacetime.
So we are, through very broad considerations, forced to the Queen equation $\mathbf{R} - \frac{1}{2}\,R\,\mathbf{g} = \kappa\,\mathbf{T}$ which describes geometry by telling us how holonomy properties of spacetime are linked to the stress energy within it. Einstein showed that this reduces to Newtonian gravity in the nonrelativistic limit as long as we choose the value of $\kappa$ right, so we use Poisson's equation and Newtonian gravity to "calibrate" General Relativity to force it to reduce to Newtonian gravity in the appropriate limit.
In exactly the same way a future theory of quantum gravity will reduce to GTR in the appropriate, continuum limit (although that reduction will likely be a great deal more complicated than simply adjusting one constant!).
I hope I have shown that, even though it involves subtle concepts, General Relativity is indeed a simple and elegant "sketch" of how gravity must be and the whole picture behind that sketch is yet to be found. I hope I have also shown that the notions that would need to be falsified to falsify GTR are very fundamental, well tested ones, so that you can glean something of the motivation for believing that GTR must always be the limit of any sensible theory.
Why Can String Theory Talk In Terms of Forces
You also ask about the absence of "force" in GTR, aside from that we think of GTR as defining the geodesics of spacetime and therefore inertial motions, i.e. that it is a kind of "application note" for when to apply Newton's first law and what "force" one needs to deviate from geodesic motion. I know nothing of physics beyond GTR, but it is worth saying that many physical theories can be formulated in both "force mediators on empty flat background" terms as well as in geometric terms. The four-potential $A$ in electromagnetism defines the gauge covariant derivative and the notion of parallel transport through electromagnetic configuration space. So "force" and "geometry" can be complementary and not contradictory. Having said this, I have never really understood the need to make gravity of exactly the same character as the other forces simply because it's not "choosy" - it affects things equally independent of composition, or at least seems to in the continuum limit. It's quite possible that, at the Planck scale, this is no longer true.
[1] When I say very simple, I mean conceptually, not in the practicalities of solving the equations or applying it to systems we have little everyday intuition for. One needs to understand that simple is not always the same as easy.
Answer: The history of physics shows that there is not really an  end of physical theories, but it is a matter of regions of validity of the models in the space and time and energy momentum phase space. This is because physics theories are not just mathematical theories, but have extra postulates/laws which connect physical observables to the mathematical functions. A physics theory which is invalidated by new data, will just have a reduced region of validity in the phase space variables.
Of course physics theories that are mathematically well grounded may fall out of fashion, because one of the aesthetic requirements for a good physics theory is "simplicity" .Even the epicycle model expressing the geocentric system is still valid, as can be seen in any online planetarium . It has as many parameters as the heliocentric but the heliocentric is much simpler and allows for an extended region of validity with Newtonian gravity.
Newtonian theory is valid in planetary dimensions and can be shown to emerge from the General Relativity  theory. In the same way, quantization of gravity (and hence the existence of gravitons) will be successful when the proposed theory shall demonstrates that General Relativity emerges macroscopically , out of string theory, for example, or whatever the final quantization of gravity theory will be.
If gravitons are observed , it will be a great discovery pointing to the validity of the hypothesis that the underlying framework of all nature  is quantum mechanical. It will not affect the framework of validity of GR,
Answer: I am not sure if GR explicitly says that gravity is not a force. It explains it in terms of curving of space and I assume, the curving of space provides the necessary force.
But if GR explicitly says that "gravity is not a force", then, obviously forces can not be unified with something that is not a force itself. Therefore in order to have a unified theory, some correction will be necessary. And that correction can be either in GR (that gravity is a force), or in theory of other forces (that they are also curving of space).
To unify the four, they have to be expressed as same phenomena, a force, or a curving, or something totally different, as long as all four are expressed as same type of entity.
If GR explicitly said that "gravity is not a force", then in this sense, GR may have acted as barrier to unification even though it is the most accurately tested quantification of gravitation till date.
Comment: Gravity is not even a force in Newtonian mechanics. The equivalence principle makes it an acceleration and that's how we treat it. Naive unification of gravity with the standard model forces is not possible and no such attempt has been even borderline successful. The unification of forces is, by the way, a very naive idea and it doesn't work well, either. The standard model is just a fraction of physics and it can be used with some success above the nuclear energy range, but it mostly neglects nuclear physics, chemical bonds, ionic forces etc. with little hope for further "unification".


Is it kosher to splice wiring in the "can" portion of recessed LED lighting?

Question: I've recently installed some 4" recessed fixtures (IC rated if it makes any difference) that had an incompatible plug for the lights I bought.  The cans came with a two-pin orange colored plastic connector/harness, and the lights (Cree branded) came with an Edison style screw in connector.  In order to make this work, I bought a female edison-style screw-in adapter (with a rubber shroud), cut off the orange connector on the cans, and used wire nuts to splice the female screw-in adapter into the can's wiring  (see photo).  The photo makes this apparatus look huge but it fits very easily into the cavity above the light.  Can I get thoughts on whether this passes muster with the electrical code?  Or is it a bad idea for any other reason?
Comment: Chris, that's my understanding -  Cree put all of their circuitry in this model in that male Edison adapter (perhaps to limit the height of the actual light portion).  This is an admittedly stupid, roundabout way of fixing my mistake of buying a light with an Edison bulb connector. Hopefully it's code-compliant stupid, if there is such a thing.  Thanks for your time my friend.
Comment: What voltage???
Comment: So the transformer is in that male edison connector?
Comment: Well, it's not Kosher unless it's blessed by a Rabbi.  But I think it's safe.  Just make sure those connections are good and solid.  I find splicing wires of different gauges to be difficult at times.   Were your connections outside the can, then we'd have a clear code violation.
Comment: @ChrisCudmore 120VAC
Answer: You are in code as long as there is enough room for every thing and you do not have to modify the can. The can would be considered the junction box. Also you need to make sure you have installed the correct cans for the type of insulation that is around the can, you should be okay. ie If the can is installed in an area that is surrounded by insulation the can must be rated for it, or a shroud must be built as an air gap for those that are not.

Without the correct can, if the can has a thermal overload, the thermal overload will over heat and shut down power to the light and become a nuisance.
Answer: I see no reason that you would not be able to disconnect the Cree device (white connector) from the black socket.  Then cut and strip the red and black wires about a 1/2 inch each.
Make sure that power is turned off, remove the wire nuts which will free the black socket.
Twist each wire from the Cree fixture a wire that had a wire nut on it. That should free up space in the can. You perhaps will need someone to hold the fixture while you do the work doas to prevent dropping it.


What does x (execute) permission do on unix sockets?

Question: By my experience unprivileged user can not access unix socket he/she does not own without x bit set. What does this bit do exactly?
Comment: @ВладимирТюхтин, `0644` means just read permission for the group members and "others", so only the owner would be able to access the socket (since the write permission controls it). `0666`, or just `0222` as read permission is not needed, would be enough to give access to everyone. But, rather than give permissions to everyone, I'd look into using groups to allow access.
Comment: Is it the permission on the sockets or on the directory?
Comment: On a socket. Directory has mode 0755, so user can observe content. Socket has mode 0644, so logically user (which belongs to "others" because he/she does not own socket a belongs to group owning socket) should be able to write to the socket. But user gets permission deny message unless mode 0777 is applied to the socket.
Comment: I can confirm you do not need execute in order to connect to a socket. You do need write if you want to write to it.
Answer: Nothing, as I can see.
The Linux man page unix(7) says that the permissions of the directory containing a socket apply normally (i.e. you need +x on /foo to connect to /foo/sock, and +w on /foo to create /foo/sock) and that write permission controls connecting to the socket itself:

On Linux, connecting to a stream socket object requires write permission on that socket; sending a datagram to a datagram socket likewise
         requires write permission on that socket.

Apparently some other systems behave differently:

POSIX does not make any
         statement about the effect of the permissions on a socket file, and
         on some systems (e.g., older BSDs), the socket permissions are
         ignored.  Portable programs should not rely on this feature for security.

unix(4) on FreeBSD describes similar requirements. The Linux man page didn't say if socket access on some systems ignores the directory permissions too.
Removing the x bit from the socket seems to have the effect of giving a different error for trying execute the socket, but that's not much of a practical difference:
$ ls -l test.sock
srwxr-xr-x 1 user user 0 Jun 28 16:24 test.sock=
$ nc -U ./test.sock
Hello
$ ./test.sock
bash: ./test.sock: No such device or address
$ chmod a-x test.sock
$ nc -U ./test.sock
Hello
$ ./test.sock
bash: ./test.sock: Permission denied

(I did also test that indeed only the w bit seems to matter for accessing the socket on Debian's Linux 4.9.0.)
Perhaps the sockets you meant had all permission bits removed from the user, or you meant the x bit on the directory?
Comment: @AlexisWilke, because someone asked what the corresponding execute permission bit does on one.
Comment: Why on Earth would you try to execute a socket?!


How do you make the salutation on Person Accounts mandatory?

Question: I tried this using the validation rule:
ISPICKVAL(Salutation, "")
but this then also applies to the standard Account object. I just need this for Person Accounts so can this be restricted even more?
Answer: There should be a boolean field IsPersonAccount which can be used to check if the current record is a person account or not.
Comment: i.e. `IsPersonAccount && ISPICKVAL(Salutation, '')`.


Does elemental damage cause any extra effects?

Question: Is there any extra benefit to added elemental damage on weapon attacks? Are certain creatures vulnerable to certain types or do certain types have certain effects (such as shock damage stunning)?
Answer: Yes, here are the damage types in the game and their effects. From IGN's game guide:
(The grey parts are from IGN)

Arcane
This is the type of damage which is inflicted by Death in his Reaper form. There are also attack combos that will allow Death to inflict extra Arcane damage by briefly unleashing his Reaper Form.

Any Wrath abilities used from the Necromancer Tree are more powerful the more Arcane stat you have as well. Strength affects Harbinger Tree attacks.

Fire
This damage type will set enemies ablaze, dealing extra damage and steady, continuous damage over time.
Frost [my edit, originally they had "Ice" here]
This damage type will inflict freezing damage and will slow them down, making it easier to avoid their attacks and land counterattacks.
Lightning
This damage type will inflict extra damage and may stun enemies, leaving them very vulnerable for a few seconds.
Piercing
This damage type can go through enemy defenses and deal significant damage. This is particularly effective against heavily-armored targets.

I have done testing in game and have seen these effects myself.
Then there is Inferno, Ice, and Shock damage. These are upgraded versions of their fire/frost/lightning counterparts.
Inferno does the same thing as Fire except the DoT can spread to other enemies in close proximity to the original burning target.
Ice doesn't just slow enemies down like Frost, it has a chance to freeze them in Ice for a short time as well.
Shock has a greater chance to stun than Lightning.
There is also another stat that deals damage called Thorns, which many of us know as gamers, reflects damage back to enemies that attack you.
Comment: Do you know what Inferno, Frost and Shock does or are they just 'other' names for Fire/Ice/Lightning ? And what about Thorns ? Edit: i *think* i tested ice vs. frost damage once and ice would put them in an icecube and frost would slow them but not the other way around..
Comment: @Aerus - I noticed the same with ice vs frost. I'm curious about fire vs inferno, though. And is there an inferno/ice equivalent for lightning?
Comment: @SaintWacko I think that's Shock, but too be honest, i always get Ice because it's damn OP so i haven't really tested the others :)
Comment: Some testing against the wooden training dummies at Tri-Stone shows that both Strength *and* Arcane affect Reaper Form damage.
Comment: @bwDraco Yes that is correct.
Answer: Ooooook. The list is like this:

Frost -> Ice
Fire -> Inferno
Shock -> Lightning

The first ones are the first to encounter and weaker versions, having small effects:

Frost slows enemies down
Fire lights them on fire, adding a DoT
Shock leaves a condition effect that has a chance to ministun enemies affected by it

Now as for the upgraded versions, they all carry the same effects of the juvenile versions, but add a few extra effects and goes as it follows:

Ice adds the chance to encase enemies in an ice cube [ like those skeletons at the start and in Deposed King's lair ] and works on any type of enemy, except bosses.
Inferno, ontop of having the burn effect on the target you struck, spreads to adjacent enemies and also has a small splash damage effect on hit, like Teleport Slash`s mini-AoE.
Lightning adds to the 1 second "twich" effect of Shock, a paralyze effect that lasts 2 or 3 seconds [ I can't remember exactly the exact duration ], kind of like the Ice element's effect, but you'll still be able to target it if you`re in focus mode [ a lot of people complained about that downside of Ice, so Lightning is an alternative to it, but with its own drawbacks -- no slow ]

Also, since someone mentioned Arcane and .... got it wrong ... even if it doesn't directly answers your question, I`ll cover that aswell here: It only affects the damage of the skills in the Necromancer tree and Strength is what affects the Harbinger skills -- you can test that easily by equipping and unequipping a str or arcane item and hovering the skills in the menu to check what changed and by how much.
Comment: Thanks. I corrected my answer based on yours. I'm not sure why I wrote it, I must have read it from a reputable source since I am simply relaying information in my answer. Thanks again.


Are there penalties for dying in FFXIV?

Question: I played FFXI for a long time, and whenever you died, you lost experience points and could potentially level down. It looks like that may have changed in FFXIV because I don't think I lost anything after recently dying. Is there some penalty that I am not seeing?
Answer: By default, getting KO'd in FFXIV:ARR means that you get a small durability hit against your gear first.  Any spells/abilities/status changes are also removed (So Protect or class abilities will need to be recasted again [NOTE 1]), but any Food Effects on you will stay active.  There's no other permanent damage done against your character much like EXP loss in FFXI.
While KO'd, you can choose to return to your home point for free, or wait for a revive. (there's no "time till HP" like in FFXI). If you're in a Duty (Instanced Dungeon or Leve), you can chose to return to the start of the map. (In an instanced dungeon, a shortcut is created after every large boss).  This doesn't cost too much beside running time, but if you "return to start" in the middle of event battle (One where walls are created to block people from running away) then you cannot return to that battlefield until the battle is concluded. (Either the rest of the party falls or the boss is taken down.
If you accept a revive while KO'd, you'll return with a 1 minute "Weakness" debuff that reduces all of your stats by 15%. (Max HP/Max MP/and physical stats like STR and MND)
If you happen to get KO'd while under the weakness Debuff, then "Brink of Death" will be applied when you're revived again. (This replaces the "Weakness" debuff from here on out.)  Brink of Death reduces all of your stats by 30% for 1 minute, and getting KO'd while under Brink of Death will just reapply the same debuff over again. (So if you have 30 seconds left on BoD and get KO'd again, the BoD debuff goes back to 60 seconds on reviving)
Once either debuff expires, your stats return back to normal.  Likewise, getting KO'd again will only afflict weakness again as that debuff chain has been broken.

Edit: [Note 1] In a recent update (patch 2.3?) the abilities Shield Oath and Defiance will not be lost on being KO'd.  These constant abilities are used by the Paladin and Warrior Jobs in order to enhance their stats for taking on the Tanking role.  As you can guess, whenever a tank gets KO'd and raided without their tanking buff, they're stuck in a very underpowered situation to get the control of the fight again.
Comment: Update: The Sword/Shield Oath class ability explicitly *does* persist through death nowadays.
Comment: Does death impact equipment durability?  Like is there durability damage whenever you die?
Comment: By "Hit to durability" I mean that you'll lose durability points from your equipped gear.  I do not know how much is lost per KO (especially since the game reads the information out in percentages) but it is a steep enough to hurt after a enough falls. (my experience is around 10 KO's with heavy fighting before gear gets under 25% durability.)
Answer: Depends on where you die. If you die in the overworld, you either wait for someone to resurrect you, which inflict a 3 minute stat penalty debuff, or return your aethryte point, which can be really far away. If you die in a dungeon, you have the same option of being resurrected, or you can return the dungeon's starting point, necessitating a long walk back.
Other than that, they are no penalties for dying.

Are there penalties for dying in FFXIV?


Do different parts of body conduct current in different ways?

Question: When we touch anything(with small amount of current), like the end of our CPU cabinet, with our hands, it does feel like a small current is flowing through.But the same doesn't occur when touch the cabinet with our foot.Why does this happen?
Answer: I assume two things here:

You are mainly asking about why we feel the zap sometimes and don't at other times.
You are talking about a metal computer case and not a computer cabinet, which is usually wooden.

As to why you sometimes feel or don't feel the zap, this is primarily because of how densely packed some of your receptors are in your skin and how sensitive they are at that area. For example, your back is significantly less sensitive than your hands (palm, fingertips, back of hand, etc.) and your heel is even more insensitive than your back. You have many biological reasons why this is so but I'm not going into them here.
From a purely physical point of view, the body really doesn't vary much (nothing more than an order of magnitude) in conductivity or resistivity regardless of area chosen. Your body's composition is nowhere significantly different from the rest of it. Bone and fat have the greatest resistivity, muscle and blood the least. Surprisingly, skin isn't a great conductor because of its nature of having multiple layers of dry epidermis, dermis and fatty hypodermis. Things just do not flow and exchange as much when it comes to skin. Note that the thickness of the dead-cell layer and the wetness of skin are the greatest factors here (which is why capital punishment via electrocution needs a sponge wet with salt solution and why doctors use conducting gels for electrodes).
Note the difference between resistivity and resistance here; resistivity is an intrinsic property of the material (or composite, in this case) while resistance is an extrinsic property depending on various changeable factors like length, area, temperature, etc. Someone's resistance can be significantly changed based on where the current flows in and out. For example, someone grabbing a wire and standing on the opposite foot on the ground forces the current to travel a much longer distance through much more stuff than if the same person was insulated from the ground and touched on his skin with two wires of a circuit close together.
Comment: Yes, sweat does change resistivity significantly but I assumed dry skin throughout here, with you sitting indoors working on your computer and all.
Comment: I thought it's a kind of sweat effect and tongue being wetter, is more conductive than palms and palms being wetter than feet(due to more sweat) are more conductive than feet.However, thanks for assistance.Helped me alot
Answer: Yes, different tissues, and therefore body parts and organs generally have varying impedances.
Also, if it's really a current that you're sensing, then the grounding of your PC is compromised, and could pose a fire hazard.
In any case, the situation you're describing is, I believe, solved primarily by accounting for the thickness of the skin at the foot versus hand.  Perhaps you should give a little more details (where exactly are you touching the PC, a flat surface or a corner, and which part of the hand/foot, ...), and try touching with different parts of the foot to see if it makes a difference.
PCs emit vibrations and varying electromagnetic fields, either or both of which could be the physics underlying your sensation.  While in the latter case (EM fields) this could actually mean that your nerve cells are triggering as a response to the field, and therefore a current is involved, it doesn't imply that you're actually connected to any of the electric potentials within the cabinet.
Comment: Thanks.It helped me, really.BTW, I am touching it at the corners.
Comment: OK, corners give a clue that it could really be a current leak.  Whiskeyjack gives a good description at http://electronics.stackexchange.com/questions/205239/why-does-electrical-charge-concentrate-at-sharp-points

You might want to check the wiring of your PC.  There could be something like a PATA or SATA cable that's connected to the motherboard, but not the peripheral on the other end.  That could lead to charge buildup within the cabinet (via induction), and be the ultimate reason for your sensing a current.


Constructing image under homography from known information

Question: If $f$ is a homography of the real projective line, $f^2=id$ (is an involution), and $f$ has exactly two fixed points, how can I construct (geometrically) the image of an arbitrary point?
Answer: I assume you mean knowing where the fixed points are.
Assume for simplicity that the fixed points of $f$ are $0$ and $\infty$. Then it's not too hard to see (for example by the classification of Möbius transformations) that $f$ has to be $x \mapsto -x$. You know how to geometrically construct the image of an arbitrary point by that transformation.
In the general case, you can show that $f$ is the inversion with respect to the circle that contains the two points and whose center is on the real projective line.
Edit: I'm adding some details following OP's questions in the comments below.
Here I'm using the usual coordinate $x$ on the real projective line (defined by $[x_1, x_2] \mapsto x = \frac{x_1}{x_2}$) so that $\mathbb{R}\mathbf{P}^1 \approx \mathbb{R} \cup \infty$ and a homography acts like $x \mapsto \frac{ax+b}{cx+d}$.
In particular we can see homographies of the real projective line as special examples of homographies of the complex projective line $\mathbb{C}\mathbf{P}^1 \approx \mathbb{C}  \cup \infty$ (aka Möbius transformations), they are those with real coefficients, or equivalently those which preserve $\mathbb{R}\mathbf{P}^1 \subset \mathbb{C}\mathbf{P}^1$.
Circles (or should I say "circle-or-line"s) are well defined in $\mathbb{C}\mathbf{P}^1$ and there is a geometric transformation called inversion with respect to a circle (see link above). I am saying that in restriction to the real projective line, a "real" homography that fixes two points on the real projective line acts like the inversion with respect to the circle (in $\mathbb{C}\mathbf{P}^1$) whose center is on the real projective line and contains the two points.
Comment: Okay I understand some of what you're saying but not completely. First what do you mean by "know where they are"? What I meant is something like: let $P,Q\in\mathbb P^1$ (the projective line) be the two fixed points. Next, is $x$ a coordinate in a coordinate system? If it is then which? If not then what is $x$? A point in the line? In that case $-x$ doesn't make sense seeing as there's no addition of points, or at least none which can be considered an operation. Lastly how is a circle defined in a 1-dim. real projective space? All honest questions, no offence meant (in case it was taken).
Comment: No offense taken of course. I've tried to clear things up for you. Is it better now?
Comment: Hmm yes now I see completely what you meant, although I was hoping for an expression of $f$ as a composition of perspectivities (by including the line inside the projective plane). It's ok though, I should have been more specific


Bound on expectation of inner product of a Gaussian rv divided by product of norms

Question: Let $Y$ be a $n$-dimensional Gaussian random vector with expectation $0$ and covariance matrix $\Sigma$ (which explicitely is not diagonal). Let $\gamma$ be a vector in $\mathbb{R}^n$, let $\| \cdot \|$ denote the Euclidean norm in $\mathbb{R}^n$ and let $\langle \cdot, \cdot \rangle$ denote the Euclidean scalar product. My question is if it is possible to find a vector $\gamma$ such that it holds that
$$ \mathbb{E} \left[ \frac{| \langle \gamma, Y \rangle |}{\| \gamma\| \| Y \| } \right] \leq \frac{1}{n}$$
or if this inequality is false for every $\gamma$. Obviously, Cauchy-Schwarz gives the upper bound $1$ for any $\gamma$, but I would be content with finding just a single $\gamma$ that fulfills the stricter bound.
Answer: What you want to show is not true. Either the integrand has to be squared or the right-hand side needs to be $\frac 1 {\sqrt n}$.
To see this, choose a basis in which $\Sigma$ is diagonal and order its eigenvalues: $0\le\lambda_1\le\dots\le \lambda_n$. Let $\gamma=(1,0,\dots,0)$ (in this basis). Then for standard independent normals $N_i$ we have:
$$E_n^2=E\left(\frac{| \langle \gamma, Y \rangle |}{\| \gamma\| \| Y \| }\right)^2\le E\left(\frac{| \langle \gamma, Y \rangle |^2}{\| \gamma\|^2 \| Y \|^2 }\right) =E\left(\frac {{\lambda_1}N_1^2} {\sum_{i=1}^{n}\lambda_iN_i^2}\right)\le E\left(\frac {N_1^2}{\sum_{i=1}^n N_i^2}\right)=\frac 1 n$$
The last equality is achieved iff all eigenvalues are equal. However the first inequality is not sharp if $n>1$. Clearly it's the worst if all eigenvalues are equal so assume that. When $n=2$ we get:
$$\sqrt 2E_2=\sqrt 2 E\left(\frac {|N_1|}{\sqrt {N_1^2+N_2^2}}\right)=\sqrt 2E\left(\frac 1 {\sqrt {1+C^2}}\right)=\frac {2\sqrt 2} \pi\approx {0.90032}$$
where $C=\frac {N_2}{N_1}$ is Cauchy - already better than $1$. Asymptotically by SLLN we get:
$$\sqrt n E_n=E\left(\frac {\frac{\sum_1^n |N_i|}n}{\sqrt{\frac{\sum_1^n N_i^2}n}}\right)\to \frac {E(|N_1|)}{\sqrt{E(N_1^2)}}=\sqrt{\frac 2 \pi}\approx 0.79788$$
which is better than the original simple bound of $1$.
Comment: Thanks for your answer! Even if my original conjecture is wrong, this version is also very useful. Could you elaborate on the final step, however, i. e. how to obtain the bound $\frac{1}{n}$? If I use $\lambda_1 \leq \lambda_i$ for $i \neq 1$ in the denominator, I'm left with a sum of standard normals, which is a chi squared distribution - same in the numerator, but the two are dependent - is this even the correct way to go and what am I missing to obtain the bound?
Comment: @bhz $E(\frac {N_1^2}{\sum_1^n N_i^2})=\frac 1 n$ by symmetry.
Comment: Talk about thinking too complicated - I got it now. Thanks again.
Comment: @bhz Good. I also update the answer with some improved asymptotic estimate for the quantity you actually asked about. There might be some analytical way to track $E_n$ - either through chi-square distribution or transforming into Cauchy (I think they'll be independent), but I don't immediately see simplifications.
Comment: Those Cauchys certainly wouldn't be independent.


What is this bike model and year? Cannondale CAAD?

Question: I have an opportunity to buy this bike for $200 and was wondering what model and year the bike is? From one of the pictures I see CAAD and on another, I see seven, but I'm coming up empty handed when googling. The owner says he bought it last year.
Thanks!
Mike
Answer: Cannondale Quick 7: https://www.cannondale.com/en/USA/Bike/ProductDetail?Id=7adbe241-cc33-4ea5-b62f-9f32c8476f31
Not sure of the year; you can look up previous years and see if the color scheme is different, or if the components match what was spec'd on that year (derailleur, shifters, number of gears, etc.).
It's a fairly entry level hybrid from a major company, so bikes like these don't go huge overhauls very often -- it doesn't look much different than the 2018 visually aside from the black crankset in the figure below.
Image from Cannondale.com
Comment: It could also just be black in your market, or cannondale didn't update the photo from last year; a lot of ones that come up as 2018's on google image search have black cranksets.
Comment: Thanks, that's what I thought too but the crank on this is silver whereas on the photo is black. Maybe it was swapped out


Should I leave the U.K. the day before my work visa expires?

Question: I've a U.K. work visa that expires in several months.  Can I leave the country the exact day the visa is marked as "valid until"?  Or should I depart a day earlier?
Comment: I don't know the exact rules in the UK, I would assume leaving on the day itself would be OK but you should also consider the risk of delay, bad weather, etc. Planning to leave on the last day does not mean being able to do so.
Comment: Good point.  Ain't clear if they're forgiving about weather, missing your flight, etc.  I'm keep an eye on travel times to the nearest ferry too, thanks.  I'm asking about ideal scenario however.
Comment: @JeffBurdges On top of this check the immigration rules related to your visa.  There may be grace period allowed for you to move out of the country.
Comment: A question this serious should probably be addressed with immigration authorities rather than an arbitrary opinions on this site :)
Comment: I most often leave on the very last day of my visa validity and it's never a problem. I haven't done so in the UK though.
Answer: You are legally in the UK up to midnight of the expiry date of your visa.  If the date is 15 Feb 2014, for example, you have until midnight local time to leave.
Just adding...  If something happens and you can't depart, you can soften the effect by reporting to the Border Force (or in their absence, the police).  An Immigration Officer is empowered under Paragraph 31 and 31/A to extend, but he doesn't have to if he doesn't want to.  The police will slag you off, but at least you tried.
So it's midnight of the expiry date.
Adding:  For those interested in the 'grace period', the formal penalties start to kick in at 28 days.  Those are absolute and enforceable bans against entry.  But even for a time shorter than that, a subsequent application to renew or get a new work permit will be invariably refused on credibility grounds.


Is the baryon number consistent when it comes to annihilation?

Question: I've previously asked a question on here about if it was possible to change the barion number by radioactive decay, for example positron emission, and the answer was of course no, as the baryon number rmains consistent. I've also talked about this problem with some friends, and one of them came up with a process to turn baryonic matter into purely energy or leptons. Basically, their idea was, that, at first, a proton and an antiproton should annihilate each other, which should only leave energy, and the resulting photons should then, by the mechanism of pair production create an electron and a positron. Of course, to an amateur like me, this makes sense. But, as Baryon number is supposed to be consistent, this shouldn't work? Can someone explain why or why not this works? It would be greatly apreciated.
Answer: If you start with a proton and an anti-proton, the initial baryon number is zero, so there is no problem with ending up in a state with zero baryon number such as an electron and positron. According to wikipedia, typically mesons are an intermediate product in such a reaction, and you may also produce photons and neutrinos.
Comment: Interesting. In my original question I've talked about completely converting protons to electrons, so this means, this would only be possible, if, in a closed system, you'd have an equal number of particles and antiparticles?
Comment: The general rule of thumb is that "anything not forbidden is mandatory." In other words, a reaction is usually allowed to happen unless there's a symmetry (or equivalently a conservation rule) that forbids it. Focusing just on baryon number, a reaction $p^+ p^- \rightarrow e^+ e^-$ is allowed since the initial and final baryon number are zero. On the other hand, a reaction $p^+ \rightarrow e^+$ is not allowed, since the baryon number goes from 1 to 0. (There are also other things wrong with this reaction but we are just focused on baryon number).
Comment: In general you have to look up (or know) the baryon number for all the particles in the initial and final state, sum them up, and compare.


Does this setup give mechanical advantage?

Question: I've just started learning about pulleys in physics and wondering if below setup offers mechanical advantage. The forces balance nicely but it looks too simple to work. If I pull the rope with a force of $\dfrac{Mg}{4}$, would the masses move up?
(Assume the rope and the two horizontal bars are massless and frictionless)
Answer: What you have described is a double tackle assembly, which gives a W/4 (4 to 1) lift advantage. Frictionless, ideal rope and bars would work the same as ideal pulleys, assuming the ropes do not slip off of the bars. see https://en.wikipedia.org/wiki/Block_and_tackle
Comment: For example: In your ideal system pulling 4 feet of rope with 1 newton force would raise a 4 newton weight 1 foot. (discounting initial acceleration)
Comment: Wow so it works xD Thank you so much :) Looks magic how the input $\dfrac{Mg}{4}$ force copies itself $4$ times at the load!
Comment: It works, but it is NOT a question of pulleys, here. As Adrain Howard says, it has to do with tackles, NOT pulleys
Answer: My friend,
It's not about pulleys, it's about the force needed,
And the torque is in function of the diameter of the pulley, so you meed to have different pulleys, with different diameters in order to decrease the force needed,
Comment: For an ideal pulley system (frictionless pulleys and ropes) pulley diameter Is not part of the mechanical advantage calculation
Comment: Ohk thank you:) Then if the rope slides with out rotating the pulley, will my setup above work?
Comment: I'm not sure why you posted two answers saying something so similar, especially since torque had nothing to do with the question.
Answer: No, because in your setup,
The diameter is fix, it's not increasing,

Torque = radius$×$force,
  The force is always the same, since the mass is fix $F=mg$

But the torque decreases when $r$ increases,
That's why we use pulleys.
Comment: The lifting force of rope pulleys does not depend on the pully diameter.
Comment: Pulleys are objects that turn. Nothing turns here, the rope slips on the bars. If one neglects friction the rope will indeed pull the weights with a ratio 1/4 of forces. Don't think in terms of torque, since nothing turns !


Trigger that saves all the activities and events of a contact before a contact is deleted

Question: Trigger that saves all the activities and events of a contact before a contact is deleted. When a contact is deleted all the history related to that contact is deleted.
Comment: Welcome to SFSE. What did you try so far? Are you stuck with something?
Comment: I have created a trigger that is invoked before deletion. I am retrieving all the events data from the contact. Now I'm stuck at creating a dummy contact and linking this event data to that contact.
Comment: You should [edit] your question to include your trigger code.
Comment: i have included my code.
Comment: What you're doing doesn't make much sense looking at that code snippet, you should draw this out
Answer: Your trigger has a number of logical errors.

You are querying against the wrong Contact. You query against the WhoId of your dummy contact, while what you probably want to be doing is querying against a collection of deleted Contact Ids like oldMap.keySet() outside your for loop. These are the records whose Events or Tasks you wish to preserve.
You are not making any changes to the events, like reassigning their WhoId. (If that is in fact what you plan to do).
You are attempting to insert existing Event records, with a populated Id field, using Database.insert(). This will fail; you cannot insert records with an Id, and it looks like you probably meant to perform an update anyway.
You've named a variable event. Since Apex is a case-insensitive language, this looks the same as the class name Event to the compiler, which can cause very confusing errors. You do the same thing with contact.
You initialize events with an empty list, which you don't need to do since you populate it from a query.
You do not need to check Trigger.isDelete && Trigger.isBefore, since your trigger runs only on the before delete event.

I would suggest that you take some time to write out, in pseudo-code, what the logical flow of your trigger should look like, before you begin translating it into Apex. If you haven't already worked through Trailhead's trigger modules, especially Bulk Triggers, they might be a great help to you.


Why can't we downvote comments

Question: The title says it all, sometimes a person posts a really lame comment or even one that is just plain wrong and misleading. Why can't we downvote those just like we can download answers?
Comment: I'm sure somebody will pull some Meta.SE post out of thin air that gives some arbitrary reason.
Comment: Surely I've seen this question before somewhere on the stack.
Comment: @jippe I searched Meta before posting but couldn't find one -- Maybe I missed one though.
Answer: Because they're just comments.
Comments aren't supposed to contain meaningful information.  Their purpose is to ask for clarification, point out something the author should maybe add, remove, or edit, telling the OP he's a moron and why, explaining that we can't guess what you're babbling about without a schematic, etc.
Of course comments get abused, which is why you want to be able to vote on them.  However, the real problem here is that someone is trying to use comments to supply information, not that you can't downvote them.
Since comments can't be downvoted and are rarely policed, they are useful for, but not intended for:
Saying something unpopular you know would get downvoted in a answer.
Providing a quick answer that would otherwise require too much work to write properly, and that therefore would likely get downvoted.
Providing a answer that you're not sure is right.
Saying something off topic, possibly humerous.

Put another way, if you can't beat them, join them.
Comments do sometimes get cleared out by mods, so don't spend a lot of time on any comment.  Of course the character limit pretty much enforces that anyway.
Comment: Then why have up votes? Seems one-sided.
Comment: It keeps the dweebs busy, and also indicates which comments should be shown by default when there are too many to show all.
Answer: Google fu is weak
google magic: comment downvoting site:meta.stackexchange.com
Allow downvoting comments
Comment: Twenty-six answers!

Why can't we downvote comments


Is it OK to create a user account on website for user while he presses the 'Login with Facebook' button to view a website?

Question: After user has granted permissions to an app to fetch certain data from Facebook, while he does a 'Login with Facebook', is it OK for me to create an account for him on my app using his data fetched from facebook ?
Comment: isn't that the purpose of "Login with Facebook" ?
Comment: This could very well be a legal question in disguise. Certainly in Europe you can't store personal information without **explicit** permission. Facebook has that permission (else no Facebook account would exist), but that permission isn't transferable.
Comment: @MSalters: Is that still applicable to me if I dont host in Europe or I'm not a european company, but still engage european audience ?
Comment: Yes it is for users of your site that come from Europe.
Comment: I guess you can create your account but if you do that that account must be Facebook login only. If you want to have your account to be split from facebook it should be a user demands. Some site which want to keep users their own propose the fill the user form with facebook/g+/linkedin and then ask for login with third party as optional or the more classic user/pass
Answer: To be on the up and up, you should notify the user of your intention and explain the benefit of auto creating an account for them.
One benefit would be that if the user decides to delete their Facebook account, they should still be able to login with the account that you create for them, as long as the user is provided with login details.
Comment: The point of logging in with Facebook (or any other OAuth provider) is specifically not to have to have another set of login details, and not divulging any of the Facebook login details to the site allowing logins through Facebook. So how am I gonna provide them with login details when they delete their Facebook account? At that point I can no longer verify that they own that account, so I should not give them any site specific login details for that account. If someone wants to delete their Facebook account, they will have to decouple it...
Comment: ... from all sites where they used it before doing so. Decoupling meaning that on each site they login with their Facebook account and then set up a different means of authentication. Either by using a different OAuth provider, or site specific login details.
Answer: If you are going to store any state for the user (I can't imagine a useful service that doesn't do this), then you have to create an account in your own database that is linked to the Facebook account. Look at how Stack Exchange works when you log in with Facebook; it stores information about all your activity and lets you create your own Stack Exchange profile.
Comment: No, this is not a concern. Every website that uses Login with Facebook works this way.
Comment: But the thing is, I say "Login with facebook" (not "Sign up using Facebook"), & still create an account.. isn't this a concern. Do some others also do it this way ?
Answer: The users intent when they 'Log in with Facebook' is to create a relationship with your site, associated with their Facebook account. That is the entire function of the action, and so I do not believe that you are breaking the trust of the user by storing persistent data related to their choices and preferences on your site. The relationship is meant to be persistent, otherwise it is not much of a relationship.
However, you should provide users with a 'Log Out' control, that clears their session cookie and allows them to return to anonymity should they choose. But if they log in with Facebook again, the user will expect you to remember the preferences they chose previously; this is impossible without storing persistent user data in your application (and that's really the definition of 'account'... a persistent user token tied to their preferences).
Answer: I hate it when I choose "log in with XXX" and then the site makes me pick yet another password.
Often I give up using the site at that point.
I expect a site to remember my email address and Facebook ID and maybe my name but nothing else.   If the above is stored in an account it is not a problem for me, provided I never have to do anything to manage that account.
Comment: @ian that is bizarre as it's defeating the entire point of using a openID login. Are you sure they weren't phishing sites?
Comment: Where did the OP say anything about making you choose another password? Creating an account based on a Facebook login does not need another password as account access/login is always deferred to Facebook.
Comment: @MarjanVenema, I have had a few websites in the last month, do this to me with uisng my Google acount, so I expect they do the same with Facebook


Primary admin actions block on view dissapears if change path

Question: There is a "primary admin actions" block in the seven theme (provided by the Drupal 8 standard install profile) which only shows up if my view if the path is "admin/content"  but if I change the path to anything else, the block is gone.  where is this configured ?
Answer: The LocalActionsBlock is a block plugin that displays menu action links for a given route based on the LocalActionManager::getActionsFromRoute method.
The route that corresponds to the path /admin/content has action links, but whichever route that corresponds to the path you are using does not have any action links defined.
Action links are defined in MODULENAME.links.action.yml file.
If the desired outcome is to have the local actions that correspond to /admin/content on a different path, then a module should be created that defines additional action links for those routes.
Comment: Hi Can you give me a more detailed examplel of how to  add the node.add_page to a new url?


What hits Joker's thug in the bus scene?

Question: It seems that one of Joker's thugs at the beginning of The Dark Knight lost his consciousness because he was hit by something. However, there is nothing heavy enough to have caused such an effect in that scene (he got hit by a wood panel which weighed 5 kilograms tops). Is this a mistake in the film?
Comment: The wooden panel you're referring to seems to be part of the wall, likely weighs more, and is smashed against him by the bus. It seems quite reasonable the man is knocked unconscious.
Comment: I imagine it was meant to look like he got his by the bus, but IRL that would have killed the stunt man so, ya know, movie magic.
Answer: He got hit by the panel as you observed
BUT, it's not the weight of it that is important, it's the amount of kinetic energy that it carries.
A baseball doesn't weigh very much but at 90mph it will take you out if it hits you in the head.
A more extreme example is a bullet, weighs a couple of ounces but when fired from a gun it's lethal.
In this case, the panel (and I suspect it weighed considerably more than 5 kilos) has the force of a bus behind it and accelerates from zero to quite a speed before it hits the thug.
Comment: Don't forget the orangebox effect/trope!
Comment: In your example: a baseball bat with a bus attached to it!


non-compete clause in Canada

Question: I have a following clause in my employment contract that I signed when the company hired me: "For a period of ONE (1) year after the end of employment, the Employee shall not control, consult to, or be employed by any business similar to that conducted by the Company, either by soliciting any of its accounts or by operating within the Company's general trading area." I was on paternity leave during the last year. What are the probable consequences if I start working for another company that conducts the similar business? I live in Ontario Canada. Thank you.
Answer: You will have breached your contract. Common remedies that your ex-employer could seek are monetary damages or an injunction on you or your new employer preventing you from working.
Answer: You have a right to earn a living using your specialized knowledge or skills (and many people stay within the same industry when changing jobs).
From the link below:
Courts do not like to interfere with an individual's ability to earn a living any more than is reasonably necessary and will refuse to enforce covenants that are not. A non-competition clause, which prevents an individual from working in the same industry, is usually seen as an unreasonable restriction. A non-solicitation clause allows the individual to stay in the industry where they have their experience, contacts, and knowledge but prevents them from actively contacting customers for business opportunities that compete with their former employer. Courts are more likely to enforce such clauses so long as the terms are reasonable.
The top 10 employment law decisions of 2017

non-compete clause in Canada


Too silly to consider VS too silly to be considered

Question:
The strikers' demands are too silly to consider.
The strikers' demands are too silly to be considered.

What is the difference between these two sentences?
Answer: Even though one uses the active voice and one the passive voice, there is little difference in meaning because both use the infinitive form ("to consider" and "to be considered"), which does not define the subject. As a related example:

The book was too difficult to understand.

For who to understand?  The sentence doesn't say.  It's just a global statement about the overall difficulty of the book in the opinion of the writer -- or, to put it another way, it's a statement with which the writer thinks any reasonable person would agree.
Semantically there's little difference if written in the passive voice:

The book was too difficult to be understood.

Since we aren't told any details about the subject, the passive voice states the same thing in a different way.
It's much the same with your example.  The writer doesn't specify who thinks the demands are "too silly to consider".  That information might be obvious from context, or we might assume the writer feels this way, or it might just be a general statement which the writer thinks any reasonable person would agree with.
Comment: In **The book was too difficult to understand.**, we think there was a people who felt difficult to understand the book. But in **The book was too difficult to be understood.**, we think there is the book which was felt difficult by a people. So both are right. Am I right?


Resizing image from one field and save result in another field

Question: As mentioned on the comments on this question, on my node-editing-form there are fields for WIDTH, HEIGHT, and an IMAGE. I would like to resize the given image by the dimesions WIDTH and HEIGHT and to save the result in another image field. I hope someone could outline (or even concretise) how to build that module.
Answer: have you seen Imagefield Focus?
This would give your users a much nicer UI to edit and crop images, and more importantly choose the correct regions to crop.
Comment: Thank you, but that's not what i am looking for. Toi better understand the context I edited my question because the link to another, related question was missing. What I am looking for is indeed how to save the results from one field to another.


What do you call someone who studies Russia?

Question: This is my first post and I was wondering what do you call someone who studies Russia For a living like has an academic researcher or Scholar. A sentence would be I work has a -word-. Thanks in advance for any and all help.
Comment: Please see  [this guidance from Stack Exchange Management](https://english.meta.stackexchange.com/questions/2160/single-word-requests-crosswords-and-the-fight-against-mediocrity) about such questions.
Answer: The word Russologist is used to refer to persons who are experts on Russia.
Examples:

Stanford's most accomplished Russologist returns home, to cheers (...)
I took McFaul's Russian Politics class back in 1996/1997, when he was still a rising Russologist.
(...)Rice = Sovietologist (her PhD was actually on Czechoslovakia, but her subsequent specialization came studying the Soviet Union.) Mike McFaul = Russologist (his research has concentrated on the post-Soviet Union collapse, and specifically the transition from Communism to what he hoped would be a liberal democracy (...)

Source: www.thecardboard.org

Poet, translator and russologist Dr. Sajal Dey hails from Kolkata, West Bengal.

Source: https://milaplitfest.com/sajal-dey/

A distinguished Russologist, Jiri Valenta is a former consultant to the Reagan administration and among the few members (...)

Source: http://www.israelnationalnews.com/Articles/Article.aspx/20646

Most foreign observers who knew Russia well (...) shared the belief that Russia was not doomed to perdition. They included Otto Hoetzsch, the leading German Russologist of his generation, (...)

Source: google books
More examples of use online: https://www.google.co.in/search?q=russologist
Comment: Ah, for the good old days. Sovietologist and Kremlinologist are slight more euphonious to my ears than Russologist.
Comment: I suppose the word 'Russologist' gained (or regained) prominence after the collapse of the Soviet Union, @James McLeod.
Comment: Yeah but, does anybody *say* that? It also seems to be slanted towards politics? Why not just historian or sociologist? They're experts on *all* things Russian? There's only one man who's *that* good, but he doesn't *always* drink Dos Equis... a made up word from 1997? Is there Americanoligy? Americanolgists? ...
Comment: Very true @Mazura: it does seem slanted towards political studies. Maybe specialisation on one country in one field of knowledge is not even ideal? However I was mainly establishing that the word "Russologist" is actually being used to describe the 'Russia expert' and learned a new word in the process of answering OP's question! According to the answer by 1006a, the word 'Americanist' is used to describe a scholar (often political scientist) whose speciality is the USA.
Answer: The term I have heard in academia in the US is Russianist, which Wiktionary defines as

A person who studies Russian language, culture or history.

Oxford Dictionaries offers the example sentence

'The book is nothing less than "a celebration of personality," to cite the apt formulation of the distinguished Russianist John B. Dunlop.'

Presumably this was chosen to parallel the somewhat more common term Americanist, used for scholars (often political scientists) who study the US specifically.


Why most of the web applications use CRSF token as a prevention method instead validating through origin header?

Question: Why was crsf token discovered at the first place? One can always check from the the request generated by origin header
Answer: "the user agent MAY use the Origin header field to inform the server of the security context in which the script was executing when it caused the user agent to issue the request"
It is not obligatory for a browser to supply the header. In the absence of a valid Origin header, what assumptions should the application make?
"The Origin request header indicates where a fetch originates from. It doesn't include any path information, but only the server name"
Hence:

Your suggestion would only provide protection against attacks initiated from other vhosts, not from different applications on the same vhost

will cause legitimate requests to fail if the application implements domain sharding

requires the application to know the domain it is being served up by

Further, origin support is patchy.
Given that it provides a mechanism for user tracking/de-anonymization without relying on client side persistence, it is unlikely to be viable in the long term as a means of ensuring session integrity.
Comment: I think the "patchy" status of origin support would be the most persuasive reason. :)


How do I change my personal Mii on the 3DS?

Question: I imported my old Mii into the 3DS from my Wii. Now, I'd like to replace the "personal Mii" that I set up on the 3DS with the one from the Wii. Is there any way to do it, without editing the existing Mii to look like my old one?
A valid workaround would be to let me erase all my Mii data (but not the purchased games on the system), and start over with an import.
Comment: Can't you just import it from the Wii again?
Comment: @Ullallulloo I edited my answer and added more information. I hope it is better suited for what you wanted. :)
Answer: Hopefully, this describes what you want to do. From Nintendo:

How to Change the Mii Character Used in StreetPass Mii Plaza
Important Note! Only a Mii character that was created by the user can
  be used as the display Mii.
What to do:
Select StreetPass Mii Plaza on the HOME Menu, and tap "Open."
Highlight your Mii Character and press the A Button.
Select "Mii Settings."
Select "StreetPass Setting."
Select "OK."
Choose Mii from the list of Mii Characters. Select "Yes" to confirm.

Note: This is the only Mii you can change - you can edit your personal Mii, but you cannot delete it or switch it for a different Mii you have created. (It actually warns you about this when you create them, and also in the manual.) Also, if you attempt to delete them, you get a message that says "Your Personal Mii cannot be erased." You can re-format the 3DS, however, in order to create, effectively, a blank slate for yourself, but editing the Mii is really your only option if you do not wish to take that route.
If you want to reformat the 3DS, go to System Settings -> Other Settings -> Format System Memory (it is the last option on the 4th page.) The following information will be lost after formatting the system memory:

All data in the System Memory (including Friend List, Notifications,
and step count)
System Settings data
Built-in software save data
All photos in the System Memory
All other extra data (created by some games and applications)

It will not, however, delete your e-shop account.
Comment: Can you import from the Wii when you first make a Mii?
Comment: Unfortunately, that doesn't look like it is possible.
Comment: you could actually. It gives you an option to recieve a Mii on the Mii app on the Nintendo 3DS. You then go on your Wii. It'll tell you to press a combination of buttons.
Comment: @JerryRox yes you can recieve Miis, but you cant replace the original Mii you created.
Answer: Just edit your main Mii - that's what I did :p
Answer: You can change the creator of the Mii, and then erase it - that's what I did and it worked fine.
If not you can just edit your Mii!
Answer: You can reformat your 3DS, and it will let you make a new personal Mii. I did it, and it worked fine.


Justifications for import duty

Question: What is the political/economical justification for the existence for import duty? Is this purely a form of protectionism? Or are there other reasons as well? (Been thinking about something along the lines of the loss of income taxes from the workers or something like that)
Comment: What makes you think an [answer wouldn't](http://en.wikipedia.org/wiki/Tariffs_in_United_States_history) take a book? Perhaps you should narrow the focus to a period of time, a country, and perhaps a political philosophy.
Comment: My friend, without knowing you have answered my question :D . I am not a native English speaker, so I was searching the web using the term 'import duty' rather than the vaguer historically incorrect term 'tariff', but what I was looking for was exactly this: http://en.wikipedia.org/wiki/Tariff#Economic_analysis Not specific taxes, but rather the justification for import duty on a generic level.
Comment: @user1873 I'm sure that you can enumerate, and even briefly summarize the main justifications in a one page answer.  Sure, you can go deeper in detail and write as much as a book, but you can do that for virtually all answers here.
Comment: The "justification" might not matter much.Tariffs are much older than most of the concepts of modern economics. Historically, (import) duty was a major source of income for the state. Similar duties would also be imposed *inside* a country, when "importing" goods inside a province or city (cf. [Octroi](https://en.wikipedia.org/wiki/Octroi)). One reason for that is that it is easier to enforce (technically and politically) than, say, an income tax (which requires sophisticated accounting). It also feels like taxing foreigners, no matter what economists might have to say about efficiency.
Comment: Certainly, when Alexander Hamilton compiled the first lists of US import tariffs, there was a decided intent and targeting to provide protection to categories that they felt were important be robust, domestically.
Answer:
Is this purely a form of protectionism?

There are three reasons for trade tariffs:

Protecting and helping out your domestic producers. This can be especially crucial in the case of agriculture as food security is the number one task of any government.
An expectation of quid pro quo from other countries — as long as you tax our exports we are going to mirror the tariffs and tax your imports. This is why organizations such as the WTO exist — they help countries to come to a mutual agreement on how low their import tariffs are going to be.
Efficient revenue can be an important consideration for small countries or where tax collection is inefficient. In many island countries, for example, in the Caribbean, almost all goods are imported, so an import duty is functionally equivalent to a general sales tax with an exception for the tiny amount of locally produced goods, but is more efficient to collect at the dock than in the storefront. This source of revenue for government operations also permits these countries to be tax havens with respect to foreign investment by not having an income tax. (courtesy of @ohwilleke)
Comment: Efficient revenue is also an important consideration in some cases. In many island countries, for example, in the Caribbean, almost all goods are imported, so an import duty is functionally equivalent to a general sales tax with an exception for the tiny amount of locally produced goods, but is more efficient to collect at the dock than in the storefront. This source of revenue for government operations also permits these countries to be tax havens with respect to foreign investment by not having an income tax.
Answer: As SE is a Q&A site, let me answer my own question. In classical economics import duty is indeed seen as a form of protecting the internal market and even as form of independence declaration politically, which by neoclassical economists is seen as a distortion of the free market. Below I will quote from wikipedia the justification for the following, but in the end it can be economically understood that there is a gain called the terms of trade gain which is the economic gain from the internalization of some goods in the internal market which has to be bigger than the efficiency loss induced by the import duties (basically the protectionism). This gain, if I understand it correctly, depends on every specific market and there is no universal gain which is true for every single market.
In order to prevent link rot:

The diagram to below shows the costs and benefits of imposing a tariff
  on a good in the domestic economy, Home.

When incorporating free international trade into the model we use a
  supply curve denoted as Pw. This curve makes the assumption that the
  international supply of the good or service is perfectly elastic and
  that the world can produce at a near infinite q S amount of the good,
  but had a demand of D. The difference between S and D, SD was filled
  by importing from abroad. After the imposition of tariff, domestic
  price rises from Pw to Pt but foreign export prices fall from Pw to
  Pt* due to the difference in tax incidence on the consumers (at home)
  and producers (abroad).
At the new price level at Home, Pt, which is higher than the previous
  Pw, more of the good is produced at Home – it now makes S* of the
  good. Due to the higher price, only D* of the good is demanded by
  Home. The difference between S* and D*, StD* is filled by importing
  from abroad. Thus, imposition of tariffs reduce the quantity of
  imports from SD to SD.
Domestic producers enjoy a gain in their surplus. Producer surplus,
  defined as the difference between what the producers were willing to
  receive by selling a good and the actual price of the good, expands
  from the region below Pw to the region below Pt. Therefore, the
  domestic producers gain an amount shown by the area A.
Domestic consumers face a higher price, reducing their welfare.
  Consumer surplus is the area between the price line and the demand
  curve. Therefore, the consumer surplus shrinks from the area above Pw
  to the area above Pt, i.e. it shrinks by the areas A, B, C and D. The
  government gains from the taxes. It charges an amount PtPt* of tariff
  for every good imported. Since SD goods are imported, the government
  gains an area of C and E.
The net loss to the society due to the tariff would be given by the
  total costs of the tariff minus its benefits to the society.
  Therefore, the net welfare loss due to the tariff is equal to:

Consumer Loss – Government revenue – Producer gain

or graphically, this gain is given by the areas shown by:

(A + B + C + D) – (C + E) – A = B + D – E

that is, tariffs are beneficial to the society if the area given by
  the rectangle E more than offsets the losses shown by triangles B and
  D. Rectangle E is called the terms of trade gain whereas the two
  triangles B and D are also called efficiency loss, as this cost is
  incurred because tariffs reduce the incentives for the society to
  consume and produce.
The model above is completely accurate only in the extreme case where
  no consumer belongs to the producers group and the cost of the product
  is a fraction of their wages. If instead, the opposite extreme is
  taken by assuming that all consumers come from the producers' group
  and that their only purchasing power comes from the wages earned in
  production and the product costs their whole wage, the graph looks
  radically different. Without tariffs, only those producers/consumers
  able to produce the product at the world price will have the money to
  purchase it at that price.
Comment: So... care to expand what all the downvotes were for?

Justifications for import duty


Adding an old SDK to Xcode 4

Question: I have read that you can have multiple SDKs in Xcode 4 installed in the Developer/Platforms/iPhoneSimulator.platform/Developer/SDKs directory. What was not made clear is how you can get these different SDKs. I'd at least like to test for iOS 4.0 instead of the latest iOS 4.3. Though it would be cool to test for 3.2 also. My Xcode 4 installed only with the latest 4.3 SDK in this directory. Is there a convenient way to add them?
Answer: When you install Xcode 4.0, your previous version get's renamed to something like "Developer-3.2.5". If you still have a previous version of Xcode installed you can simply do this:
cd /Developer/Platforms/iPhoneSimulator.platform/Developer/SDKs
sudo ln -s /Developer-3.2.5/Platforms/iPhoneSimulator.platform/Developer/SDKs/iPhoneSimulator4.0.sdk .
sudo ln -s /Developer-3.2.5/Platforms/iPhoneSimulator.platform/Developer/SDKs/iPhoneSimulator4.1.sdk .

If you don't have a previous version installed, uninstall Xcode 4.x and then install Xcode 3.2 and start progressively from there. After you install Xcode 4.x your previous version of Xcode 3.x will get renamed on disk after which you can symlink the 4.0/4.1 SDKs under Xcode 4.0. Relaunching Xcode 4.0 will give you 4.0 and 4.1 in the drop down option list.
Comment: You might also want to install the corresponding simulator, though it's not strictly necessary...
Answer: No there is not a convenient way, unfortunately.  The easiest way is to install an earlier version and then upgrade it, progressively if necessarily, to the latest version.
Answer: The XCode downloads available from http://developer.apple.com/downloads appear to contain packages going back to before 4.0. I recently hit this same issue and was able to determine that:

From 4.1 onward (Maybe earlier - still waiting for the download for XCode 3.2.3 and SDK 4.0.1), the packages contain sub-packages named quite similarly
Despite the inner package name, it appears that the actual iOS SDK will be included for the version mentioned on the primary DMG file. For instance, xcode_3.2.5_and_ios_sdk_4.2_final.dmg may contain what appears to be SDK packages for a bunch of versions, but the iOS SDK is only fully included for 4.2.
I was able to get my hands on 4.1, 4.2, and 4.3 by downloading the DMG XCode installers, viewing package contents, and running the corresponding iPhoneSDK4_X.pkg installer.
After installing, I had to copy the stuff it dropped from /Platforms/iPhoneOS.platform/ into the new location at /Developer/Platforms/iPhoneOS.platform.
In addition to the SDKs for iPhoneOS, I also made sure the iPhoneSimulator.platform stuff was there. For me, it seems that XCode handles installing most device support for debugging..

I hope this helps! The tricky part for me was realizing that the package names may be the same, yet the contents may be different. Each download only seems to have included a single full SDK.
Once I did this, i was able to change my base SDK in XCode and build against the target SDK.
Cheers!
PS: Once you add additional SDK's, you will likely see additional destinations appearing in your schemes list with identical device names for the physical devices. It drove me nuts, as each one is associated with one of the SDK's, yet the SDK version isn't listed in the list. Not bad once you figure out which is which, usually the symptom being a failed build, since it targets another SDK based on which physical device is selected. More info on this post: duplicate device scheme in xcode 4

Adding an old SDK to Xcode 4


Installation of MSP430's IDE Energia in Ubuntu

Question: I am trying to install Energia for Ubuntu but until now the only thing that I have successfully managed to do is download the binary release version for Ubuntu! I have searched everywhere to find instructions and there some forums that try to provide them but they don't! I have started to lose my faith. Does any of you working with MSP430 microcontroller can explain simply how to install Energia in Ubuntu 14.04? Please don't provide links to other sites/forums because I am 100% sure that I have checked them all. Literally I have searched up to the 7th page in Google and every single page is purple as a result of visiting them all...
Answer: I don't believe it but I actually found a solution. Because, as I said in the question, I have searched extensively and most of the solutions were very messy and unclear I answer my question in the hope that it will help others and save them the hours I spent on this problem myself. This solution was hidden in a thread of a forum:
So in order to install Energia:

Download the binary release version from the energia site
Open the Ubuntu Software Center
Install the "Main Menu" package
From the first column select Accessories
Click New Item, name it Energia and select the file energia from the unzipped downloaded Energia binary release
Select Ok
Write Energia in Launcher and click on the Energia entry.

After that one must set up the serial communication with the Launchpad board. From another source I manage to find a solution to solve that problem doing the following:

sudo lsusb
sudo modinfo cdc-acm
Then create a file called 71-persistent-msp430.rules in /etc/udev/rules.d/ directory
Add this line in it:

ATTRS{idVendor}=="0451", ATTRS{idProduct}=="f432", MODE="0660", GROUP="plugdev"

sudo adduser user plugdev

Write your own username instead of user.

sudo service udev restart

Also it may need to change the Board from Energia's toolbar (Tools and after Board).
Comment: Looks like steps 2-7 of your first instruction is simply to let the launcher find the unpacked file. You could probably skip all of that and just open a terminal and run the program from there; or open the directory it contains and double click it; or make a shortcut on desktop.
Comment: @Shahbaz I am new to Ubuntu and this solution is a complete one composed out of others scattered in the internet.


Stability of web development packages?

Question: My experiences with web-frameworks was that they are relative "unstable". Not that they crash but that there are quite a few changes which then force one to reprogram ones code. I wonder what web development packages you've used and how much work it was/is to maintain that code?
Answer: "Changes are a part of the requirement."
I dont think that the web packages that we design changes a lot. If it changes, then it is a bad design. If we use external API's, a very few get deprecated, otherwise most of them are the same.
Some packages used as a JAVA/J2EE programmer:
 - MVC
 - Struts
 - few AJAX frameworks
These are very basic one used. Most the other ones are self developed and once the design of the web package is done, we don't change the design.
Answer: Any library which is under active development would be unstable. Look at .NET for example, every month there's a new better way to do old stuff. On the other hand open source libraries tend more to throw old deprecated methods away because it makes code better and that's what makes them happy.
But I wouldn't recommend to use anything old and unsupported anyway, you'll be on your own although the environment would be completely stable.
The best way possible seems to be just freeze the version of library you start using and switch to new one only there's huge benefit in doing so. At least that's how everyone is doing that.
Answer: Frameworks like .NET and jQuery have been largely backward compatible and allows you to use new features slowly.
Mootools however... the API broke so many things from 1.11 to 1.2 to 1.3. Upgrading was not straightforward in that case.
As a rule, I try to wait for something to be out of beta before embracing it in production code. An endorsement can go a long way too - since Microsoft has called jQuery the winner of client-side frameworks its been easy to encourage others to pick it up too.
Answer: Most of my experience is with Ruby on Rails, so I'll share what I've seen with it over the last few years.
Rails updates at a pretty good clip, but you don't really need to update unless you need features or the rare security patch. As an example, I have a rails app running in our company right now that was coded about 2.5 years ago that only needed to have some work done to it once this year to upgrade it to a new version to be compatible with apache mod_rails, it was originally written against Rails 1.2 I believe. Of course, that was an intranet app which didn't have any security requirements. All in all, it's been pretty pain free. If I had kept using mongrel + mod_proxy it wouldn't have only needed to be updated once for a security patch.
Rails is pretty secure, vulnerabilities are fairly far between. There have been a few more Ruby vulnerabilities than Rails vulnerabilities if memory serves me right, but all in all it's pretty solid, and upgrading your ruby shouldn't break rails, especially if you use a distro that backports security fixes.
Comment: Oh I've don my round of Rails I implemented somethign 2 years ago and it needed some extra week this  year to get from 1.1.6  to 2.2... So Rails is exactly what I meant with not that stable

Stability of web development packages?


DVD Player error 47

Question: I am unable to eject a disk from Apple DVD player. It shows "Error ~47" when I click on Eject button. Changed some DVD Player settings, such as cleared "Start playing disk" so that no files would be open when I connect the DVD player, but that did't help.
Searched Apple website; it has posts by other users about "Error ~47" in various settings, and comments about that being a file error, but no indication on how to fix that.
My system: MacBook Pro, Yosemite 10.10.3.
Please advise how to remove the DVD from the DVD player.
Answer: Got it: another user account from the same Mac had a file open on that DVD.

DVD Player error 47


How to submit documentation with plugin

Question: When I submit a plugin to the wordpress public repo, how am I supposed to submit documentation? What is considered standard or good practice? Should I just have all documentation on the be page that the plugin is mirrored on, or should I also include a readme of sorts? I know when submitting to CodeCanyon, it's standard to package documentation with the plugin, but it's not super important what file format you use as long as it can be viewed without paid software. I don't, however, know what the standard for submission to the public repo is. Insight anybody?
Comment: You can add it on plugin page. Or if it too large host it on your site.
Comment: Is there any sort of good resource on how to structure your documentation? I just want to ensure there isn't a set of rules I should be following.
Comment: I would wait for other comment to come in but Github Wiki is another good place for placing your documentation.
Comment: Utilize the plugin Description, FAQ, 'Other Notes' sections to its full extent. else as already suggested by Sisir, its best to host the documentation on your site and provide link in readme.txt. If you want to improve documentation by community/users, then Github Wiki is best.
Answer: This is my most popular plugin. Flowplayer HTML5 for WordPress
I did try to think through my documentation etc. All of the documentation should be in the readme.txt. It is the best place as that added to WordPress.org for easy access. You can use this link to validate your readme.
Not all plugins do need any documentation so it is not a must but can make it easier for users.
Another way of doing to it adding a link to your site in the readme where you have the documentation. Polylang does it this way.
You can update the readme.txt without needing to change the plugin version so it is no more difficult to update the documentation in the readme then it is on your site.
Here is also the the WordPress.org guidelines.

How to submit documentation with plugin


Endomorphism of Chow group induced by a birational map

Question: Let $\phi:X\dashrightarrow Y$ be a birational map between smooth projective $k$-varieties ($k=\bar k$) and $\Gamma$ be the closure of the graph of $\phi$. In Fulton's intersection theory example 16.1.11, it is said that $^t\Gamma\circ\Gamma$ is the sum of the identity correspondence and correspondences whose projections are contained in proper subvarieties of $Y$ but I cannot see why it is (formally) true.
Comment: We see that the restriction of $^t\Gamma\circ \Gamma$ to $U\times U$ (where $U$ is an open subset on which $\phi$ is a morphism) at like $\phi_*\phi^*$ on the Chow ring of $X$ but a priori the homomorphism from correspondence to endomorphism of Chow ring is not injective (isn't it?)
Comment: Related: https://mathoverflow.net/a/241907/82179
Answer: Let $U $ be the largest open subset of $X$ where $\phi$ is defined. Let $d=dim X$. Then the sequence $$CH_d(X\times X-U\times U)\to CH_d(X\times X)\to  CH_d(U\times U)\to 0$$ is exact. Now consider the cycle $\alpha:=\Gamma^t\circ\Gamma-\Delta$ on $X\times X$. Here $\Delta$ is the diagonal. Its restriction on $U\times U$ is 0. Hence $\alpha$ is supported on the complement $X\times X-U\times U$. Thus we get the required assertion.


Cheap guitar - steel strings

Question: I recently pulled out this really cheap guitar and tuned it up to start playing. My friend who is a guitarist recommended that I replace the cheap plastic strings with steel strings for a better sound. I bought .11 to .52 steel strings but I am struggling to insert them into the guitar. I have also removed the plastic strings and have no idea if and how to temporarily put them back so I can continue playing. Help! How do I re string this guitar! I haven't found any help at all on the internet, hope someone can answer my question.
Comment: Sorry to say, your friend gave you some bad advice. Probably your best bet at this point is to read Tetsujin's answer and then take your guitar to be strung correctly by a professional with **new** nylon strings - most stores that sell guitars will string a guitar for you, for a fee.
Comment: Thank you so much! I've literally been going crazy not being able to play :(
Comment: Just out of interest, how could you **NOT** find instructions on the internet?  google "how to string acoustic guitar with plastic strings" gives hundreds of hits, including very nice YouTube tutorials
Comment: Your friend's advice is a perfect example of how not enough information can cause a lot of damage. The guitar isn't designed for the additional stress of steel strings if it originally was designed for nylon strings.
Answer: General rule is - don't put steel strings on a classical guitar.

You'll ruin the neck, unless it has a truss-rod you can adjust.
If the bridge is meant to have the strings tied on rather than ball-end, you'll struggle.

Broadly speaking 'classical' & 'folk' guitars are different beasts & the strings cannot be interchanged.
BTW, the distinction is not "cheap plastic" vs "expensive steel" the two are made for entirely different purposes.

Cheap guitar - steel strings