How to interpret use of "status-declined" for a discussion question

Question: In reviewing Should we have a policy about "too much downvoting"?, I noticed that it was of type discussion but also had status declined. The documentation on How does Meta Stack Exchange work? says parenthetically that this status is usually used for feature-requests. I can understand how this status might also be used for bug or even support, but I'm not sure what to make of it for discussion. Can you help me understand the conditions under which moderators would "decline" a discussion question?
Comment: That discussion carried an implicit proposal for a new feature (an upper limit on the downvote to upvote ratio of a user), which will not be implemented.
Comment: Note that it was Jeff -- founder/CEO -- who both authored the question _and_ added that tag. This isn't a case of "moderators declining discussion" from a lay member.
Comment: I'm soooo tempted to downvote *just* for the abuse of tags in this question...
Comment: Also, it was a discussion started by the team, asking for input. (Rather than a feature request from a user, asking the team to implement it.)
Comment: @Arjan I've had numerous MSO questions edited by others to replace my plain text references to tags with actual tags. Is there some documentation you can point me to as what is appropriate? Should they never be used?  Only be used the first time in a question? They're a pain to type, relatively speaking, and I think they interfere with the flow, so I'd prefer not to use them given a choice.
Comment: @Asad You and a couple of others have indicated that any _reference_ to a potential change in the system within a question, even in the context of "what if" or "what are the pros and cons", should be treated as a "feature request" for all intents and purposes. If I understood you correctly, how would you go about asking for information about the impact of a hypothetical change without having it treated as an actual request to change the system? Would you have to do it in comments?
Comment: I'd only use them if you want to link somewhere. (Like when asking about a specific tag, which people might want to click to see all posts related to that tag.) Above, I don't think you want to link to anything? (Unrelated: for the very same reason [I dislike linking to Wikipedia for very common terminology](http://meta.stackexchange.com/questions/44405/for-well-known-terminology-how-useful-is-adding-links-to-wikipedia). Just makes folks wonder what's to see when following the link. Which is even worse on mobile, as many touch devices don't support hover.)
Comment: (And as an aside: I roll back edits made to *my* posts, that add `` or tags.)
Comment: @Arjan So this is just a matter of personal preference as far as you know? Do you do anything to denote the tags (quotation marks, italics, etc.) or just leave them plain text?
Comment: @PeterAlfvin *'You and a couple of others have indicated ... for all intents and purposes.'* That isn't what I've said at all. To quote the question we're talking about: *'Based on this data I am leaning towards enforcing a "no more downvotes may be cast" if...'* I'd hardly say that is an airy fairy hypothetical scenario, and is actually a proposal that might get implemented. It just happens to be mis-tagged discussion instead of feature request, and you seem to think that magically exempts it from any of the behaviors associated with a proposed change to features (eg. status declined).
Comment: Hmm. I guess we just look at this very differently. I'm not suggesting that discussion items should be "airy fairy". Indeed I would think that most discussion questions would be about things that someone has seriously considered.  I just think there's a difference between trying to have a dialog about a possible change (my view of "discussion"), even if you express your opinion, and actually making a formal request for that change (my view of "feature request").
Comment: @PeterAlfvin A "formal request" for change (aka a feature request) *is* an invitation for discussion and dialog on the merits and consequences of the change. Most answers to feature requests are tweaks, criticisms, insights etc. [discussion] is for discussions that don't fall under one of the narrower tag categories: it even says so in the description. I really hope the team implements the proposed change of making feature request questions immune to the effect of downvotes so that people finally stop mistagging feature proposals as bug/support/discussion out of fear of downvotes.
Comment: @Asad Thanks for clarifying your view. Ironically, it seems like we're looking for a similar end result, namely the free flow of ideas regarding improvements. I certainly think what you're proposing would be better than what we're experiencing today, but I think there's value in sometimes "preprocessing" feature request ideas as discussion items before submitting them, in order to separate the wheat from the chaff. In part, that's because I think responding to feature request, bug and support question is somewhat "mandatory" while responding to discussion questions is entirely optional.
Comment: @PeterAlfvin You can certainly submit feature proposals as discussion when you want to emphasize that the idea is half baked (possibly upgrading to feature request following input), but please don't try to demand that people don't vote on the merits/flaws of the idea, simply because you've disclaimed that you actually want it implemented. For example, [here's](http://meta.stackexchange.com/q/172783/200500) a feature that I think needs to be changed, although I'm not sure on the best course of action. I've tagged it both FR and discussion.
Comment: Again, I respect your view, although personally I like the policy as revised this month, namely that votes of agree/disagree should only be applied to feature requests. Whether those votes count against rep is a separate matter.
Answer: The discussion tag itself is often used as a "disguise" of sorts for a feature-request when the author isn't quite sure if the feature should be implemented and wants discussion from the community on whether it is a good idea or nailing down the specifics of the feature.
Sometimes the tag is omitted simply on the basis that it takes up another tag, and there are other, more prevalent tags that the author would rather use. Other times it may just be that the author didn't want to attach two required tags to their question. Generally speaking, the tag refers to the body of the question. If the question itself is requesting a feature in some way (even implicitly), then the question is subject to a status-declined tag like any other feature request.
Comment: I don't know if it was your intent, but the term "disguise" makes it sounds as there is subterfuge or misrepresentation involved when inviting discussion on a hypothetical change to the system. Other terms I've seen are "masquerading". When someone is genuinely unsure about the impact of some potential change, is there no way to neutrally collect feedback about it without it being treated as an actual request for that change?
Comment: @PeterAlfvin *"If I've got that right ... effectively from the start."* No, you can totally have discussions that aren't about feature changes without "agree/disagree" votes (mostly). The Meta community has become accustomed to using votes to express a trivial "yes please", "no thanks" response to discussions involving feature changes, regardless of what you tag them, especially when discussion of those features has already been played out extensively and/or there are existing answers that express why they feel the changes would be beneficial/harmful.
Comment: @PeterAlfvin Yes, there is. Present the proposed change neutrally, and consider both the pros and cons of the change in your question.
Comment: So, for example, in http://meta.stackexchange.com/questions/193762/should-downvote-without-comment-rate-remain-unpublished, would you say that my explicit request for "pros and cons" was simply inadequate and that I should have had to come up with some "cons" to balance my one sentence of "pros" or removed the "pros" entirely? What if I'd given no "pros" and just left the request for "pros and cons"? I frankly think I would have gotten the same reaction. Have you seen examples where a change was "presented", pros and cons offered, and people didn't vote based on agreement/disagreement?
Comment: @PeterAlfvin If you describe a change to the site, then expect the community to critique it: that's what is supposed to happen in a feature-request anyway. There is literally no difference besides the tag you've used and the fact that you've highlighted the change is hypothetical (as all proposed changes are). What I was suggesting is that you examine both sides of the picture yourself.
Comment: @Asad So to simplify this, if I just say "What are the pros and cons of 'x'?", I'm making the equivalent of a feature request and I should expect to get agree/disagree votes. If I say "'x' seems to have <...> pros and <...> cons, what do you think?" then I'm not making a feature request and I should be free of agree/disagree votes relative to 'x'. If I've got that right, then a corollary would be that you can't have "just" a discussion about something unless you can argue both sides effectively from the start.
Comment: @Asad Sorry, in all of this I've been referring solely to discussions about potential changes in features. I understand there is generally no issue with other types of discussions.
Comment: @Asad See my answer below for my attempt to explain some of our disconnect.
Answer: The mis-tagging here is that the question is marked as "discussion" despite the fact that it's a "feature-request".
Having a "status-*" tag is appropriate, because it is indeed proposing a new feature.
Comment: Actually, Jeff explicitly said he was _considering_ a change, which sure doesn't sound like a request. In any event, I'd be interested in your thoughts on my comment on the other question.
Comment: @PeterAlfvin Just because the poster isn't 100% positive that the feature should be implemented doesn't mean it's not a feature requests.  *All* feature requests should be made by someone willing to consider alternate solutions to the problem, modification of the suggestion, etc.  In short, all feature requests should result in a discussion of that feature request.  Feature request is, by definition, a subset of discussion.  Tagging it discussion alone is an insufficiently specific, using both is just redundant.
Comment: @PeterAlfvin Note that such "discussion" questions are treated as feature request questions anyway, so they may as well be tagged appropriately.  It's pretty clear to the readers that there is as feature being requested, so removing the FR tag clearly isn't creating the behavior you're describing, it's just causing confusion.
Comment: In many cases what's "clear to the reader" isn't in fact reality. I contend most _discussion_ questions are just that, with the questioner just looking for information. While I agree that all feature requests ought to be subject to review and feedback, that doesn't mean that all discussions ought to be treated as requests.
Comment: @PeterAlfvin I'm not saying all discussions are feature requests.  I'm saying that a noticeable number of people choose to tag their feature requests as discussions, even though they really are feature requests.  This is one example of it.  That doesn't mean that there aren't actual discussions out there.
Comment: I think I've got a handle on the disconnect here and I'm going to try to make a diagram ...
Comment: See my answer below. No diagram, but a table.
Comment: @PeterAlfvin What about it.  It wasn't really unclear to begin with.  I see no distinction between your options.  Every feature request brought up is intended to be discussed.  I see nothing to gain from tagging a feature request as "discussion", even if you aren't 100% sure that it should be implemented.  There is nothing to gain, and a fair bit of confusion created.  The post should just be tagged "feature request" and all of a sudden everything's better.
Comment: So you're saying that I'm alone in thinking the in *The intent takes precedence* interpretation is what we should be operating under, or rather, is how the current system was designed?
Comment: @PeterAlfvin I'm saying that feature requests should be tagged as feature requests, and that any existing feature request posts that aren't actually tagged as feature requests are mis-tagged, and should be fixed.  Trying to figure out why some people choose to use the improper tag doesn't seem like a productive discussion, just fix it when you see it.
Comment: Only the person asking the question knows their intent and I believe their intent is what should determine whether it's tagged/treated as a discussion or an implementation request. This will in turn determine how it's treated by tag-specific policies. I understand you view it differently.
Comment: @PeterAlfvin All feature requests are, by definition, discussing a feature request.  It's not that there is nothing being discussed, it's that it's redundant to add it.  To omit a key tag, namely "feature request" when requesting a feature is something that makes it harder to find the proper content for those who are looking for it.  Also note that tags are *very* frequently edited by users other than the author, mostly because most people *don't* do a good job of tagging their questions.
Comment: let us [continue this discussion in chat](http://chat.meta.stackoverflow.com/rooms/628/discussion-between-peter-alfvin-and-servy)
Answer: Summary: If this discussion item really morphed into a feature request at some point, then it should have been retagged as such and the status declined tag was appropriate. If the intent of seeking feedback did not change, then no matter what Jeff's role or "leaning" was, it should not have been given the status-declined tag.
Background: I think a lot of this and related consternation is traceable to different views of the discussion tag as it relates to questions about feature changes and, to help with the illustration, bug fixes.
Consider the following matrix with "topics" as rows and "intent" as columns:
<code>                 | Just want dialog  | Requesting implementation |
__________________________________________________________________
| Bug fix        |                   |                           |
__________________________________________________________________
| Feature change |                   |                           |
__________________________________________________________________
</code>
The question is, what tags do you associate with questions that fall into each of the cells above? I contend there is no controversy regarding the "Requesting implementation" column. Everyone seems to agree that the bug fix cell should get the bug tag and the feature change cell should get the feature-request tag. I think also that there is no controversy around the fact that although users should be prepared for some "discussion" when submitting either of those, these should not also be tagged with discussion.
The disagreement seems to rest with the "Just want dialog" column. I'm aware of two different positions on this:

The topic takes precedence - That is, if you just want to talk about a bug fix, tag it with bug and if you just want to talk about a feature change, tag it with feature-request.  The discussion tag should be reserved for discussions of other things.
The intent takes precedence - That is, if you just want dialog and you're not actually requesting implementation at this point, tag it with discussion.

There's a third possibility, namely that both the topic and intent tags should be included, but I haven't found that view promoted yet.
Perhaps I'm an outlier here, but until this is clarified, I think it's going to be difficult to have meaningful discussion about policies that are specific to these tag types.
As a final aside, I'm not aware of any controversy regarding the tagging of support questions or the tagging of discussion questions not related to bug fixes or feature changes.
Update: Just had a long chat with another user who is squarely on the "Topic takes precedence" side. For the record, I'm on the "Intent takes precedence" side.  Their primary argument was ease of finding questions related to features and was based on the assumption that people would agree/disagree vote feature-related discussion items anyway.  My primary argument concerned promoting free exchange of ideas and was based on the assumption that we'd figure out a way to greatly reduce agree/disagree voting on discussion questions.
Answer: As far as I'm concerned, the feature-request tag should be reserved for actual feature requests. You know, ones that you've put a lot of thought into, have a proposed implementation, and really think are in a finalized ready-to-be-implemented state. You're not really angling for community consensus, you're essentially demanding that the feature be implemented. (Although "demanding" sounds like a bad word, I don't mean it that way or to imply anything bad.)
On the other hand, a discussion question is one where you are looking for community input and trying to solicit a discussion. That discussion can be on anything related to the Stack Exchange network, including ideas for possible new features that have not yet been fully fleshed out. These are basically feature requests that are still in the incubator—you have a glimmer of an idea, something that you think should be done, but you're not entirely sure if you're right, and you have no ideas on how it should be done, but you want to open up a discussion about it anyway. Answers to this type of question would disagree that anything needs to be done, disagree that anything needs to be done now, propose a solution to the problem, provide additional data in support of the problem, etc. etc.
For example (and I pick on my own questions because they're handy and I know what I was thinking when I wrote them), this question is a lot like a feature request—I want something on the site to change. But I'm not proposing a concrete implementation, I'm just throwing an idea out there for discussion about something I'd like to see changed. I'm open for input from the community on my ideas, and hoping to get some good ideas on how to implement it. As such, I've tagged it simply discussion.
This one, I was pretty certain that I was right. I had a concrete feature I wanted to see implemented, and I knew exactly how I wanted it to work. It was simply a matter of writing the code and unit tests. In my mind, there wasn't any need for discussion about how it should work, just whether it will happen. Consequently, I tagged it feature-request. (And then that Shog9 guy came along and shattered my dreams.)
Of course, even questions tagged feature-request still often elicit discussion. It is perfectly reasonable for people to disagree with the inherent validity or proposed implementation of your feature request. They can post answers that explain the rationale behind their disagreement or propose alternative implementations, and they can downvote the question to signify their disagreement. This is a common and (in my mind) perfectly acceptable use of downvotes.
Both of these are perfectly valid uses of the tags. You could tag a question with both tags if you like, but you are not required to do so.
As far as the particular question you linked, it appears to fall into the second category: when originally posted, it was the beginning of a feature request, but not yet a full-fledged request. After the discussion took place, an actual feature proposal crystallized, which was subsequently declined. It would have probably been appropriate to substitute the discussion tag with a feature-request tag when it was marked status-declined, but that didn't happen—not a big deal in my mind.
The tagging system is not absolute. Just about every question here could be tagged discussion because this is a discussion site. Lots of discussion questions contain feature-request-like elements. Others are actually support requests in disguise (because someone doesn't know or assumes that a feature does not exist). Some even turn out to be bug reports. Categories are fluid, and that's why we allow questions to be retagged. Don't read too much into them. The content of the question itself—and the answers—is where the real magic happens.
Comment: What is the advantage of tagging some feature requests as "discussion" instead of feature requests?  How will it [positively] affect how other people find, view, or respond to the question, instead of tagging it "feature request"?  I, as someone responding to the proposal, don't really care if you think it's ready to be implemented or not; if I see problems with the feature I'll point it out, if I see a way it can be modified or improved I'll point it out, and I'll point out what I expect the consequences of it being implemented might be.  I would do exactly the same thing for either tag.
Comment: +1 Per my answer below, I think the way to look at _discussion_ tags and the rationalization for not adding them everywhere is that they mean _just_ discussion.


Is formatting recovery partition to update OS recovery safe

Question: So, I bought my HP Notebook 15 2 years back with Windows 8.1 installed. It came with a factory reset recovery partition. Last year, in the July of 2015, I updated to Windows 10, free of charge, as my software and hardware were compatible. But, long story short, I had to use this recovery partition, and it, as I expected, restored back to Windows 8.1. So I had to go through the upgrade process again. So, more to the point, I would like to know if it would be possible to format my current recover partition, and use it as a USB flash drive, and create a recover drive on that, and how.
Comment: Sorry, but it is hard to make sense of your question. *Format partition and use it as a USB flash drive*?
Comment: I apologize for the lack of clarity. I say this because, on the "Create a Recovery Drive" GUI, it displays "insert a USB flash drive." I'm wondering if this partition will be compatible.
Comment: Ok, so you want to replace your current Windows 8.1 partition with Windows 10 recovery partition. However when you try to do it, system does not allow you to select the existing partition and instead asks to insert a USB flash drive. Right?
Comment: Yes. But note I have not formatted it yet. But I know it's capable of formatting drives, as it also says in the GUI "The drive must hold at least 25 GB and *everything on it will be deleted*."
Answer: It is safe to format/delete an OEM recovery partition (which holds recovery tools from, say Lenovo or Dell) as long as you don't want any of it's offering.
But it is not recommended to delete/modify Windows recovery partition (and other reserved partitions), which are created during a fresh install of Windows onto a blank drive.
Also, whatever you do it's really unlikely you can have a HDD partition show up as removable USB drive. Even if you achieve so with sophisticated methods it's unlikely you can boot from it and install Windows without having drawbacks.
I would suggest creating a bootable USB with tools like Rufus or simply burn the DVD for beginners.
Answer: You can safely format the recovery partition once upgraded to Windows 10. You can do this from

Control Panel > Administrative Tools > Computer Management > Disk Manager

Then you can use it as a separate disk for storing files and not be troubled with going back to 8.1.
But if C: drive and that partition are actually on one single drive then  Windows won't allow you to create recovery partition.
Consider if your disk crashes. You would lose both recovery and the main C: drive. What's the point in creating recovery? It's like making backups of a project in the same file, so if the file gets destroyed, so are the backups destroyed as well.
If you have blank rewritable CDs or DVDs then you can create recovery boot disk there if you have no empty pendrives to spare.
Answer: Please do not take offense, I do not mean it. The question was "is it safe" and safety is relative.
Given your level of experience and confidence a safe way would be to leave partition as is, buy a USB drive, "play" with it (create recovery disk, restore from it). When you have tested the restore process, only then you can delete current partition and try to replace it with Windows 10.
If you started with deleting current recovery partition and something goes wrong, then you are left with no other option but getting physical installation media. Hence it is not safe in your particular case.
Comment: Are you implying that I am inexperienced and not confident?
Comment: How many times did you perform Windows 10 recovery from a system recovery partition?


Join meshes after rigifying one of them

Question: I had a model with eyes, gum and teeth and the basic human rigify was giving me errors when automatic weighting. So I separated those parts into another mesh. Now that I rigged one of the meshes, how can I join the two meshes so the entire head moves correctly?
Answer: You don't need anything bound to the armature but the body parts which need bone deformations. So eyes and teeth could most likely be parented (or Child Of constrained) to bones, rather than being skinned with automatic weights. In fact, you likely wouldn't want the eyes skinned in any case. Depending on how your rig is built, you will likely want eyes parented to eye control bones so that their orientation can be changed easily.
Note regarding parenting vs. skinning of the eyes: If you want the eyes controlled by bones that exist within your main armature, and don't want to use parenting or constraints to do it, then technically, you CAN bind them to the armature. But if you do this, the eye objects will need vertex groups. The name of the vertex groups must correspond to the name of the bones which will control them. The eyes technically don't even have to be separate objects from one another. They can be two joined spheres, provided they have unique vertex groups, with only the vertices for each eye assigned independently to each of those vertex groups. The eye bones will find the spheres if their vertex group names are the same as those bones, and control them. The important thing is to have only one vertex group assigned to ALL of the vertices for each eye (meaning no other vertex groups exist for the eyes, and so no other bones will ever influence the eye geometry). Then your eyes will move as if they are directly parented to the eye bones, even though they are technically bound to the armature in the same exact way that any other "skinned" mesh is bound to the armature. But the bind behaves as if it is a direct parenting relationship. Meaning rigid. This behavior results from only having one vertex group involved. If only one vertex group exists on an object, then only one bone can ever influence the vertices for that object.
Note on unjoining objects with modifiers: Whenever you unjoin objects, they keep all prior existing modifiers. These modifiers will now exist for ALL newly separated objects. This includes armature modifiers. Simply make sure that you only have the modifiers you need for all your objects as they now exist. Strange things can happen if you have the wrong modifiers in place, and the resulting issues can be hard to track down, since you might not be thinking of this separation step as a possible cause.

Join meshes after rigifying one of them


Tried referencing a Custom setting Text area(255) field inside a Formula in salesforce flow, but somehow it is erroring out saying field not exists

Question: Referencing Text area field belonging to custom Settings heirarchy object inside a condition formula of a Record Triggered Flow is giving following Error: Field $Setup.CustomSettingName__c.test__c does not exist. Check spelling.
Answer: Text Area fields are not supported in Formulas. You can vote for the idea for now.


In Hebrews 7:11 what does it mean that the Torah was "based on the priesthood of the Levites"?

Question:
[Heb 7:11 NLT] (11) So if the priesthood of Levi, on which the law was based, could have achieved the perfection God intended, why did God need to establish a different priesthood, with a priest in the order of Melchizedek instead of the order of Levi and Aaron?
Answer: The NLT is very paraphrastic. Here is my offering.

Hebrews 7:11-12 If indeed perfection was to be had in the Levitical priesthood (inasmuch as under the same the people were placed under the law), what need then was there that a priest arise after a different order, that of Melchizedek, and not rather one said to be after the order of Aaron? For with a change in priesthood, there is of necessity a change in law also.

The thrust of this argument is that in the Messianic psalm, there was to arise a priest who is said to be after the order of Melchizedek (Psalm 110:4), that is, not an Aaronic priest—his priesthood predated and was superior to the Aaronic priesthood by virtue of the Aaronic recieving blessing, via Abraham the progenitor of Aaron, from his (Genesis 14:18-20). The fact that God swore to make His Messiah a priest after the order of Melchizedek (argues the author of Hebrews), is sufficient cause for one to conclude that the Melchizedekian priesthood is superior in some sense to the Aaronic, by virtue of its being chosen over the latter for the Messiah to assume. Indeed, the Aaronic, being limited to a tribe, was a punishment for the disobedience of Israel when they prostituted themselves to the golden calf, and was not God's original intent; whereas Melchizedek, significantly, was a priest without respect to tribe or progenitors, inasmuch as in the text he appears apparently out of nowhere (something the author of Hebrews takes to typologically foreshadow Christ's perennial priesthood—Hebrews 7:3).
As far as the point about law as relates to the priesthood is concerned, the author draws attention to the fact that where there is no longer an Aaronic priest, but a priest after a different order, "there is of necessity a change in the law also," because the law was built around the Aaronic priesthood, not the "bread and wine" offering of Melchizedek, a clear pointing forward to the Eucharistic offering—one that will be made by Gentiles, called to partake in the priesthood of Christ, who is thus the highpriest (Luke 22:19; 1 Corinthians 10:16), and thus not by priests according to the order of Aaron (1 Corinthians 10:16; 11:29; Malachi 1:11-12)—meaning that the new Law necessarily follows from the different priesthood which is not Aaronic; and that it is preferred over the Aaronic in view of the Messiah, proves that the new Law is superior, and thus by extension that the former could not offer perfection (inasmuch as you cannot exceed perfection).
Comment: So while Jesus was still alive (as in Luke 22:19) and he "showed he was chief priest" by handing out a sacrificed piece of his flesh, did he feel any pain? Was the pita bread the flesh he was composed of or just pita made into the identical substance? If it was sacrificed in Luke 22:19 as you suggest I would think that if it were a piece of his leg he should have lost the ability to use that leg. And the same for a piece of arm. No? But most of the post was very excellent. +1
Comment: Your objection is the same as the Jews' when they heard Christ teach in John 6—mistaken for the same reason. The bread He gives for the life of the world is His flesh, but for spiritual nourishment, since to feed the flesh is pointless because "the flesh profits nothing"—not, notice, "my flesh," which He said, quite oppositely, is the life of the world. The Church used the term "change of substance," that is, while the physical properties, appearance, etc. remain, what it is to God and to the faithful is now the body and blood of the Saviour, mystically present, life-giving, and sacrificial.
Comment: No one believes that parts of Christ are removed from His body or destroyed in the reception of His flesh and blood, since it is a purely spiritually reception, not a physical one. Else what does He mean by, "this is my body," handing them bread? If it's bread it isn't His body. Unless the bread has taken on a new nature according to the miraculous words of God, "the words set down by Him."
Comment: Well that doesn't sound audacious to me.


What does the Basic Highlights adjustment do, and what if I want more than "100"?

Question: I have some excessively exposed images that show tremendous improvement in Lightroom Develop when I drag the Basic "Highlights" slider all the way to the left (to -100).  What I don't understand is why it stops there.  How can I get even more of the effect it is producing?
Answer: The sliders only work within a specific range where they can presume usable information within the image. Beyond that, there's a good chance of damaging the image by pushing the values too far.
That said, with images like yours, sometimes you need extreme measures. There are a couple of ways you can push the limits here. The easiest is to switch to the Tone Curve panel, where you can also adjust the whites and highlights, and that effectively allows you to double the change that you can make in the basic panel.
The other way to do this is to use either the graduated filter, the radial filter or the adjustment brush, cover the area (or the entire photo) with a layer, and then make adjustments there as well.
Note that doing this you're likely going to find it hard to get a GOOD photo out of it, but you may well get a usable one.
Answer: Lightroom simulates a photography darkroom processing, so the values of highlights, shadows measure in %, so you are applying a value up or down on the the highlight pixels in the image. If you want to add an extra effect you can combine with the exposure or the curves. Or for particular areas you can use the brush tool and apply more.
A great way to get over that limitation is at the CC Photoshop to apply that values and duplicate the layer and go to basic highlights and adjust again...in Lightroom you also can do it, replicate the image and process it again. But for extreme processing i recommend CC Photoshop or an equivalent.


Displaying PDF files white on black

Question: I couldn't find the option to enable it on Adobe Reader or Foxit Reader. I don't think Sumatra has it neither. Can someone help me? White pages are starting to hurt my eyes.
Answer: In Foxit
Tools -> Preferences -> Documents -> Document Colors Options

Answer: If someone's still looking for the option in sumatra, go to Settings -> Advanced options and change the following lines
FixedPageUI [
TextColor = #000000(make this #ffffff)
BackgroundColor = #ffffff(make this #000000)
SelectionColor = #f5fc0c
WindowMargin = 2 4 2 4
PageSpacing = 4 4
]

Displaying PDF files white on black


What would cause my sink to back up and my dishwasher pipe to overflow?

Question: So I installed a new disposal and connected the dishwasher directly to the disposal. I put a stopper in the second drain pipe the dishwasher was draining to and all has been ok. Recently, the sink isnt draining, I tried snaking, plunging, and removed and reinstalled the pipes. I removed the plug on the second drain pipe under the sink, but now everything is overflowing and spilling up under my sink. Any suggestions on what else I could do here? This is my last step before biting the bullet and calling a plumber.
Comment: do you have a street clean-out before the drain exits the building?
Comment: shop-vac it?  How long of a snake did you stick down it?  That's typically enough...does anything else not drain?
Answer: Call the plumber.  That kind of backing up indicates a problem down the line that is not easily addressed.


Program does not contain a static 'Main' method suitable for an entry point

Question: What do I do if I just want to create a project which contains a bunch of library functions?  In other words no Main method is required.  It seemed to be compiling a minute ago and then I added another .cs file and now I am confronted with this error message.
Answer: Create a .NET Class Library project if you only want a library project.  If this is a project that already exists, you can set the Project Output type to a DLL ("Class Library") instead of an Executable ("Windows Application"/"Console Application") in the project properties.
Answer: What type of project did you create?  It sounds like you meant to create a class library but accidentally created an executable assembly.  Ensure that you are in fact creating a class library assembly (i.e. ".dll" not ".exe").
If you aren't using Visual Studio and are compiling your code with csc.exe then make sure that you are specifying /target:library to compile your code into a library.
Comment: Note: for MSBuild, set the TargetType attribute on the Csc task.
Answer: You want to make the project a Class Library type. I believe you can change the type of project in the project properties settings.
Answer: or you could use the tried-and-true empty main method
Answer: I have the solution. Really simple. You wrote the static void main with lower case. You should write it like this: static void Main()
Answer: This problem is occured when we deleted App.xaml file from our project after the required method written due to that please ensure that your App.xaml file is in correct format with respective namespace and references, if it is not, create it and add it in your project.

Program does not contain a static ‘Main’ method suitable for an entry point


Is a magnet and bouncy balls a case of parity violation?

Question: This may well be a very odd question; however, I'm currently studying parity violation and it came to mind that, if a Cobalt-60 atom decaying by the weak force and emitting more electrons opposite the magnetic field direction is a parity violation (since in its mirror image, more electrons are emitted in the direction of the magnetic field), why could I not simply take a macroscopic magnet which I have designed to emit particles (e.g. bouncy balls) in one direction and not the other. Then, in a mirror, the magnetic field direction, since it is an axial vector, would not change, but the direction of particle emission would and so the mirror image would not be symmetrical and there would be a parity violation. Could someone tell me if and where I'm wrong?
Answer: Yes: you can do this, an no, it will not violate parity. Parity violation in Wu's famous experiment is not really about the magnetic field, it's about the nuclear spin, ${\vec J}$. The magnetic field ($\vec B$) is there so one can align a macroscopic number of  60-Co spins.
With that, the probability, $p(\theta)$, of decaying into an direction $\theta$ (as defined by the momentum direction $\hat k$) can be written in terms of the available physical quantities:
$$ \frac{dp}{d\theta} = a + b(\hat k \cdot \hat J)+ c||\hat k \times \hat J||^2 $$
Here $a,b,c$ are unknown numbers to be measured by experiment.
Under a parity inversion, $\hat k \cdot \hat J$ changes sign, while the $\hat k \times \hat J$ does not, so:
$$ \hat P\{\frac{dp}{d\theta}\} = a - b(\hat k \cdot \hat J)+ c||\hat k \times \hat J||^2 \ne \frac{dp}{d\theta}$$
Where the inequality holds of $b\ne 0$. In other words, any alignment of the decay electron with nuclear spin violates parity.
Note that, in terms of $\theta$:
$$ \frac{dp}{d\theta} = a + b\cos{\theta}+ c\sin^2{\theta} $$
so $a, b, c$ are monopole, dipole, and quadrupole terms, respectively, so even (odd) order terms conserve (violate) parity.
For your contraption, aligning $\vec p_{\rm ball}$ with $\vec B$, you are going to have to introduce other axial vectors into the problem (say, something that measures $\vec B$), and given Maxwell's equations and Newtons's Laws: it will not violate parity, whatever it is.
Answer: Here is another take. It does not matter if you can do it macroscopically. Sure you can build a bar magnet that only ejects particles from one side and not the other. But under a mirror that would not look suspicious as it is a macroscopical object, you could easily create another macroscopical magnet that ejects particles from the opposite end.
But when it comes down to elementary particles and your particle decays in such a way that it emits electrons in the direction of spin, then you should be able to find particles that emit electrons in the opposite direction. The mirror image of the phenomena should be equally valid. Yet this is not what happens.
Comment: The mirror image does not need to be symmetrical, think of a car rolling next to a mirror, the angular momentum of the wheels is not inverted in the mirror image but nobody is claiming that it implies some parity violation. If we replace the mirror with another car it is very similar to the previous mirror image and physics remains valid. But with the Co60, you cannot replace the mirror with another nucleus that reproduces the previous mirror image.
Comment: Thanks Mauricio, I understand the gist of your answer but I'm still not fully persuaded- surely, even if you can construct the mirror image of a microscopic bar magnet emitting particles, this mirror image is still not symmetrical since the direction of particle emission has changed but the magnetic field direction, being an axial vector, has not. Thus, given that the mirror image is not symmetrical, surely by the definition of parity, it is a violation regardless of whether or not this mirror image could be constructed in the non-mirrored world?
Comment: This macroscopic object has an electron ejector at $\vec r$, so the electron momentum is proportional to it: $\vec p =\alpha \vec r$, under parity inversion, that reads $-\vec p =-\alpha \vec r$, which is the same equation, aka parity symmetry. If you can build something such that $\vec p = \alpha \vec B$, that device violates parity.
Comment: @JEB sure but if you try that macroscopically $\alpha$ would most probably be a quantity that flips signs under parity.


How does a ceiling fan blades redirect air throughout a room to cool you off?

Question: I already know how a table fan works because of how the blades are angled. For a ceiling fan I am confused about how the blades cut through air to cool off things in a room?
Comment: I am mostly confused about the blades cutting through and blowing air part of the question.
Comment: Ceiling fans don't cool rooms. The evaporation of sweat on your skin cools you.
Answer:
For a ceiling fan I am confused about how the blades cut through air to cool off things in a room?

See, basically it is an application of Bernoulli's Theorem if you already didn't know it. Bernoulli's Theorem states that $P + \rho gh + \frac{1}{2} \rho v^2$ = constant where the symbols have the usual meanings.You can check it here if you want. For constant $h$, we can say that $P + \frac{1}{2} \rho v^2$ = constant i.e. $P$ varies as $\frac{1}{v^2}$.
Now consider the blades of the fan to be rotating. The air column above the fan can be considered to have streamline flow and same elevation. So as the fan blades rotate, they are specially angled to make the air column move and this air column attains a high velocity. So from Bernoulli's Theorem, we can say the pressure of the air column decreases i.e. it has low air pressure.
However the air column below the ceiling fan has high pressure since it is not agitated by the fan blades and has low or no velocity.And we know air moves from high to low pressure. So air from above the fan rushes down and creates winds which increase the rate of evaporation of any liquid on any surface, creating a cooling effect.
Comment: Kind of generates lift like an airplane.
Comment: @Ethan Yes you are right. It kind of acts like that


How is y=mx+c derived?

Question: I understand what the coefficients represent. I just can't figure out how Rene Descartes would have derived this straight line equation.
Comment: What do the coefficients represent? For surely a constant gradient can only describe a line. Its intercept rules out parallel lines. I suspect there is a gap in your knowledge about what gradient is.
Comment: You're correct. I'm having a hard time grasping the concept of straight line graphs. I'm trying to familiarise myself with them by trying to understand various aspects of the line itself. I confess, my understanding of gradients and x,y intercepts is still very low.
Comment: The gradient is the change in $y$ per unit change in $x $ whilst the $y $ intercept is the value of $y $ when $x=0$ the form you have displays these values in a convenient way. Note though that $x+y-3=0$ is a perfectly good line but the gradient and intercept are not so obvious.
Comment: I think I'm beginning to understand this. Thanks a lot Karl! :)
Answer: Think in terms of motion. You are at point A and you want to go in a straight line towards point B. Set up your rectangular coordinate system as convenient. You have to move a certain amount East (the "run", call it a) and a certain amount North (the "rise", call it b). To go the shortest distance and keep in a straight line, you want to move a  proportional amount of each direction with each step you take. So for example you travel half the run, you should travel half the rise at the same time; a tenth of the run with a tenth of the rise;  and so on.
If you list the coordinates of A as $(x_1, y_1)$ then starting at time $t = 0$ with t in fractions of a time unit (say parts of an hour), your position at any time $t$ should be $(x_1 + ta, y_1 + tb)$. Assuming you get to point B in exactly one hour, you get to B = $(x_1 + a, y_1 + b) = (x_2, y_2)$ which is correct, following the original assumption that run = $x_2 - x_1 = a$ and rise = $y_2 - y_1 = b$
(Note: when you learn about vectors and lines in 3D, this is exactly how they are calculated, with a point of origin A and a direction vector $(a,b)$ plus a parameter $t$ which it is often helpful to think of as time.)
If you draw a diagram of this and mark your position at a few different points between A and B, then draw a horizontal line along A and a vertical line through B and another vertical line through each point you noted, you see you have several similar triangles. For any point $(x,y)$ on the line, the "rise" $y-y_1$ over the "run" $x - x_1$ is equal to the total rise over total run (It has to be, since we said we are going to work proportionally) = $b/a$.
Then for convenience we name the fraction $b/a$ = slope. This is called $m$ in most US texts, $a$ in our local texts in Quebec, whatever letter your local books choose.
This gives us the formula $\frac{y - y_1}{x - x_1} = m$
Note -- very important -- that $(x,y)$ is ANY general point, moving up and down the line, a variable. The subscripts are quite different. $(x_1, y_1)$ is a fixed point, our given initial point A. A lot of beginners get muddled on this so take some time to get it clear.
We multiply the above equation by the denominator and get
$y - y_1 = m(x - x_1)$
There is one of the standard forms of the line, developed through a simple idea of proportionality and our knowledge of motion in space.
Comment: excellent answer.
Answer: If you're familiar with the other forms of straight line equations (two point and point-slope) you should realize that this form, slope-intercept, is just a case of point-slope, whose point is the y intercept. Converting from two point to point-slope is simple: the slope is the rise (difference in y values of the points) over the run (diff x values). The point for point slope is either of the 2 point points.

How is y=mx+c derived?


Are the Drinfeld compact quantum groups simply connected ?

Question: To fix notations : let G be simply connected simple compact group, and $U_q(\mathcal{G})$ the Drinfeld-Jimbo universal algebra quantization of its complexified algebra defined as usual, with q not root of unity. Recall that it's a *-algebra, and I denote $G_q$ the unital $C^*$-algebra dual to it as usual (thus completion of matrix coefficient of representations...).
My question is very naive and loose, so all suggestions are welcome. Is there a general notion of fundamental group (e.g. using like étale), such that $G_q$ has trivial fundamental group?
Comment: Presumably you mean that the "noncommutative space" associated to the algebra $G_q$ should have trivial fundamental group? (Algebras are not spaces, to repeat one of my usual bits of nitpicking.)
Comment: Yes either the $C*-$algebra or the dense algebra of matrix elements, your choice ;)
Answer: This is a negative answer to your question, or at least a partially negative one.  I don't know if you've thought about things in this way, but there is a naive way to formulate the idea of the fundamental group of a space in terms of its $C^*$-algebra of continuous functions.  The problem is that, at least according to my quick calculations, the noncommutative space associated to the quantized function algebra $\mathcal{O}_q(SU_2)$ does not have trivial fundamental group.  This may mean that this notion of fundamental group doesn't make much sense for noncommutative spaces, or that quantum phenomena arise which just make these quantum groups not simply connected, or something else.  My suspicion is that the first is true, my reasoning being that if there was a good notion of fundamental group for noncommutative spaces, we would have heard of it already.  Perhaps somebody can chime in on that?
Anyway, let me tell you what I've been thinking about.  The fundamental group of a space $X$ is defined as homotopy classes of based continuous maps $S^1 \to X$.  Continuous maps $S^1 \to X$ are in bijection with unital $*$-homomorphisms $C(X) \to C(S^1)$, at least for $X$ compact.  So for a compact noncommutative space with corresponding unital $C^*$-algebra $A$, loops should be defined as morphisms (of unital $C^*$-algebras) $A \to C(S^1)$.
I believe that the correct notion of homotopy of morphisms is the following: two morphisms $\Phi_0, \Phi_1 : A \to C(S^1)$ are homotopic if there is a morphism $\Psi : A \to C([0,1]) \otimes C(S^1)$ such that $\mathrm{ev}_0 \circ \Psi = \Phi_0$ and $\mathrm{ev}_1 \circ \Psi = \Phi_1$.  This is spelled out in Chapter 1 of Elements of Noncommutative Geometry, by Varilly, Gracia-Bondia, and Figuera.
To say that a loop takes basepoint to basepoint is a little trickier for noncommutative spaces.  However, for the quantum groups this is ok because there is still one classical point, namely the identity element, which corresponds to the counit $\varepsilon : \mathcal{O}_q(G) \to \mathbb{C}$.  For a loop/morphism $\Phi : \mathcal{O}_q(G) \to C(S^1)$, then, we can say that $\Phi$ preserves basepoints if $\Phi(x)(1) = \varepsilon(x)$ for all $x \in \mathcal{O}_q(G)$.  Or in more fancy terms we could say that $\mathrm{ev}_1 \circ \Phi = \varepsilon$.
Finally, we have to define the correct analogue of the constant loop.  Again this may be difficult in general, but for the quantum groups we have the counit, so the constant loop is the morphism $x \mapsto \varepsilon(x)1$ of $\mathcal{O}_q(G)$ to $C(S^1)$.
So this gives us most of the elements to define the fundamental group of a noncommutative space with associated $C^*$-algebra $A$: we take based morphisms $A \to C(S^1)$ modulo homotopy, and say that a loop is trivial if it is homotopic (through based morphisms) to the constant loop.  I haven't said how you're supposed to compose loops, but this at least defines the fundamental group as a set.

Now the bad news: for $\mathcal{O}_q(SU_2)$, this doesn't give the right answer.  For the sake of completeness,  $\mathcal{O}_q(SU_2)$ (for $0 < q < 1$, say) is the unital $C^*$-algebra generated by elements $a,b$ with relations
$$ a^*a + b^*b =1, \quad aa^* + q^2 bb^* = 1,   $$
$$ bb^* = b^*b, \quad ab = qba, \quad ab^* = qb^*a.  $$
The first two relations show that if $\Phi : \mathcal{O}_q(SU_2) \to A$ is a morphism to a commutative $C^*$-algebra $A$, then $(1-q^2)\Phi(b^*b) = 0$, so $\Phi(b) = 0$.  (Thus we really need $q \neq \pm 1$ for this to work.)
Now $C(S^1)$ is the universal $C^*$-algebra generated by a single unitary element $u$.  This is a quantum subgroup of $\mathcal{O}_q(SU_2)$ via the map $\pi : \mathcal{O}_q(SU_2) \to C(S^1)$ defined by
$$ \pi(a) = u, \quad \pi(b) = 0.   $$
Combining this with the preceding statement, any morphism $\Phi : \mathcal{O}_q(SU_2) \to C(S^1)$ must factor through the quotient map $\pi$, so homotopy classes of morphisms $\mathcal{O}_q(SU_2) \to C(S^1)$ are in bijection with homotopy classes of morphisms $C(S^1) \to C(S^1)$.  These clases correspond to homotopy classes of maps $S^1 \to S^1$, so this means that the fundamental group is $\mathbb{Z}$, rather than the trivial group.
Again, my guess is that this notion of fundamental group is just too rigid to make sense for all noncommutative spaces.  One needs at least one classical point in order for this to make sense, which excludes any simple algebras, such as noncommutative tori.  The question is interesting, though, and I would be glad to know if anybody else has thought about other notions of a fundamental group for noncommutative spaces.
Comment: Yes that's exactly the kind of things I had in mind. For extension of abelian $C^*-$algebras, I only know of the paper of "On covering spaces of $C^*$-algebras and Galois extensions",Childs, but to quote it, most of the ideas were known to Grothendieck.
Comment: Hello,
thanks for your thoughts, but you should know that already in abelian algebraic geometry, using $S^1$ is proscribed, this is the origin of étale theory, so I wasn't even thinking of such down-to-earth approach, and you showed that it isn't a good idea. Also, you should know that for $C*-$algebras, people don't really use $S^1$ to study coverings, but rather use extension theory, and that's what seem to be missing for quantum groups, as far as I know :).
Comment: Yes, I thought that this approach was too simple-minded.  I guess you are saying that rather than define the fundamental group directly, one can take the point of view of coverings, and then the fundamental group is something like a limit of groups of transformations that preserve the fibers of coverings?

Do you know a good reference where extension theory for $C^*$-algebras is discussed from the point of view of coverings?


Why don't FCC metals have a brittle-to-ductile temperature transition?

Question: I initially thought that it had something to do with the number of slip systems in FCC vs. BCC, but they're both the same.
Answer: You are mistaken to imply that FCC metals do not undergo a transition from brittle to ductile behavior. FCC metals can fracture at sufficiently high stress - at some temperature.  Please provide more information in your question if you believe otherwise.
Answer: For what it's worth (http://materialiaindica.ning.com/forum/topics/ductile-brittle-transition ):
"Let us compare BCC and FCC here. At high temperatures, both of these have mobile dislocations, and thus they can sustain large plastic deformations without undergoing fracture.
At low temperatures however, while dislocations in BCC are no longer mobile, dislocations in FCC can still move very quickly. This lack of dislocation movement makes BCC brittle, while FCC stays ductile...
Now the key question is thus why do dislocations in FCC stay mobile at low temperatures while dislocations in BCC find it increasingly difficult to move as the temperature is lowered.
This is easy to understand: in FCC there are closest-packed planes belonging to each slip system and slip means a corner atom being moved to centre of the face.
In BCC though there are as many total number of slip systems as in FCC(12), the movement of dislocations happen only as a line of atom jumps from one potential energy valley to another (so-called Peierls valleys) - a process that could be enhanced by application of stress, or by thermal activation. The planes are not as close-packed and any slip means a corner atom being moved to centre of the cube. Thus one would see kink nucleation and propagation in BCC dislocation movement, since cost of moving an entire row of atoms from corner to centre at the same time is too high.
Thus to summarize, BCC dislocations' movement is thermally activated while (relatively) FCC dislocations' movement needs significantly smaller activation. This leads to BCC materials becoming brittle at low temperatures while FCC staying ductile irrespective of temperature - but probably not at 0 Kelvin since an atom still has to be moved. I think however that there is an activation energy for FCC slip, only it is much less than kbT."
Answer: The poster above is incorrect: FCC metals do not have a ductile to brittle transition temperature and instead remain ductile at low temperatures. This is because the stress required to move dislocations is not strongly temperature-dependent in FCC metals, and thus failure occurs by plastic flow instead of crack propagation.
In BCC metals, the stress required significantly increases at low temperatures while the cracks propagation stress is not strongly temperature-dependent. Thus BCC metals fail by crack propagation at low temperatures. More info here: http://www.doitpoms.ac.uk/tlplib/BD6/ductile-to-brittle.php
The only time that FCC metals experience brittle fracture is in the final stage of fatigue failure.
Comment: Which other answer? There are two other answers there.
Answer: The presence of a ductile to brittle transition temperature implies there are insufficient (ductile) deformation modes at low temperatures to support plastic deformation and therefore fracture occurs to release energy/load.
In FCC materials, dislocation slip of both edge and screw dislocations is relatively athermal and due to the number of active slip systems it is relatively homogeneous. Furthermore, there are more than 5 independent slip systems that can activate to accommodate arbitrary plastic strain (a requirement of von Mises' deformation criteria).
In BCC materials, screw dislocation motion is not athermal. In particular, the cores of the screw dislocations tend to 'unzip' into a sessile configuration when the material is not under load. Under load, the dislocations cores, with some thermal activation, may reconfigure into a glissile dislocation core. These dislocations can then move and enable plastic deformation. Deformation in BCC materials without screw dislocations is not sufficiently independent to accommodate arbitary shape changes and so the material often deforms through fracture. The critical temperature for screw dislocation mobility (the thermal contribution to the rearrangement of the BCC screw dislocation core structure) is the root cause of the DBTT in the BCC materials.


Chainring use of teeth

Question: At any instant how many teeth in a (not new) 52 chainring are actually doing the chain pulling when torque is being applied.
Comment: i think this question is a bit vauge
Comment: Depends on 1) how not new the chain ring is and 2) the wear (i.e., stretch) of the current chain.  Also it will not be a binary engaged/not engaged but the force experienced by each tooth will be slightly different so we would need criteria for what constitutes engaged/pulling.
Comment: Given that the calculations we've made depend on the rest of the drivetrain, you should update your question with more information. As it is, it's too broad for a definitive answer
Comment: I'm voting to close this question as off-topic because it's a question about physics, not cycling.
Answer: Depends how well the chain and chainring mesh.  Assuming perfect alignment, exactly half the teeth are engaged.
In the real world, the chain will be somewhat elongated.  So your worst case is one single tooth is taking all the strain.  At this point, wear is accelerated.
You should use "27 teeth" and "1 tooth" as inputs for your needs.

EDIT ChrisH calculates that assuming a 52 tooth chainring and an 11 tooth rear cog, there will be an 11 degree angle between the horizontal and the chain, on either  side.
So 22 degrees over 180 gives 202 degrees of the chainring are touching the chain.    202/360 shows 56.1%of the chainring, which is 29.18 teeth, so 29 teeth in contact with the chain, highest possible value assuming a 400 mm chainstay length.
Comment: Note that this assumes there's no derailleur: with one (especially a long-arm), the chain is likely to be in contact with less than half the ring.
Comment: Assuming the sprocket isn't the same size as the chainring it won't necessarily be 50% (even with a derailleur, which we don't know the OP has).  Imagine a single-speed with small wheels (short chainstays increase the angles) and 52/11.  You'll get well over 50% engagement
Comment: Following http://www.stem2.org/je/pulley.pdf I get just over 29 teeth engaged (11° where the chain arrives and the same again where it leaves).  So close to your guess
Comment: @ChrisH good point - Assuming a super short 400mm chainstay and an 11 tooth rear cog, what's the effective contact length on the 52 tooth chanstay?   My maths isn't up to it, but I suspect it wouldn't be more than 7 degrees which one tooth.

Chainring use of teeth


What are you doing in your free time? or What do you do in your free time?

Question: As I understand second variant should be correct, as question is not about current time. But when I tried to google the first phrase it appeared quite common as well. Why?
Comment: Why do you think it is not about current time?  It could easily mean "What are you doing in your free time *right now*?"
Answer: It is quite common, because the question can be about the current time.

What are you doing in your free time (during this winter holiday)

There is an understanding that the free time is limited and temporary, and the questioner is only interested in how the free time that is currently and temporarily available is being used.
However, although the second form is probably more common.  This exact phrase is the sort of phrase that is more common in textbooks than in real life.  When was the last time you asked someone about their free time, in your native language.
Comment: "It is quite common, because the question can be about the current time." Actually it's the opposite, as most of times second phrase is asked during an interview. Obviously it is asked not about current time.
Comment: Hello @MarkShor  Feel free to write your own answer to the question. What do you mean by "during an interview". That information doesn't seem to be in the question.
Comment: Hello @James, When I am in doubt I often try phrases in Google. "What do you do in your free time?" returned more than a million results. "What are you doing in your free time?" returned 500,000 results, most of them as question during an interview or similar. Some of them as part of the sentence, like: When you are not working, what are you doing in your free time? But some just as it is, which is a puzzle to me.
Comment: I think I have answered the question:  sometimes it *is* about the current time, as in "when you are not working, what are you doing..."
Comment: Thank you James.


How to define the order status after creating an shipping?

Question: I have searched a lot with google but somehow I'm using the wrong search terms or there is no answer to my specific problem yet.
In Magento I'd like to have the following procedure: Customer creates an order, I generate an invoice (Status switches to processing) which is fine. I create shipment against the order ( Status swiches to Complete) I don't want it to switch to complete, so I created a - New Status ( Cust_ Confirmation ) Now the order stays in Cust_Confirmation State. How do it mark it complete?
This is what i want: Pending-->Processing-->Cust_ Confirmation-->Complete
I am stuck now between Cust_Confirmation & Complete
Answer: I would suggest you to follow this link to make custom order status and use it without any problems:
http://www.dckap.com/blog/magento-order-state/


What happens to light clock at speed of light?

Question: Consider a spaceship traveling (hypothetically) at the speed of light with respect to an inertial observer. What happen to a light clock in the spaceship according to time dilation time? Does it stop in the spaceship with respect to the inertial observer, so that the same thing happens to a light clock, so the light ray in the light clock stops? How can I imagine this? I'm totally confused.
Comment: There are no clocks, light clocks or otherwise, that have speed *c* relative to any inertial reference frame (IRF) according to special relativity (SR).  As has been stated here *many, many* times, an entity with speed *c* in an IRF has speed *c* in *all* IRFs which is to say the entity is *not at rest* in any IRF.  Stating that the spaceship is (hypothetically) at the speed of light is to state that (hypothetically) SR is wrong.  What theory then shall we use to answer your question since you've discarded SR?
Comment: the spaceship traveling at the speed of light with respect to anybody means division by zero regarding the period of ticking of the light clock.
Comment: @Alfred Centauri However at no point in the question does he mention anything from the pov of some "light speed inertial frame". He says "does it stop _with respect to the inertial observer_?" where that "observer" is an _outside_ one looking at it.
Comment: @mike4ty4, what is *it*? A light clock with speed *c*?  It doesn't exist.  Does the question *what does something that doesn't exist look like to an inertial observer* make sense?
Answer: A "ship" that could go at the speed of light would have to be itself made of light, or something similar to it (e.g. some other massless particle). So a "light clock" at the speed of light would just be three photons in succession, essentially (the other two serving as useless "mirrors"). This simply travels as a unit, the photon in the middle does not bounce against the others but keeps a constant distance to them as seen to the outside observer. You could thus say it records no time.
It is right as the other comments have said, you cannot transform to an inertial frame moving along with the object, but you can talk about what happens from the pov of the outside, and the above is just that, it records no time. What this goes to also suggest is that nothing which travels at the speed of light is capable of undergoing any internal change or evolution. And in fact, this is observed with particle physics as well: all massless particles are completely stable and impervious to decay.
I'd think this is a legit to say "time is frozen at 'c'" -- even though there is no inertial frame possible to move at the speed of light. After all, time dilation is not measured in the object's own frame, but to an outside observer. Time essentially measures the evolution of something, that is, the progression of change, if it is not evolving at all, it is not experiencing time. There is no need to assign a rest frame to make sense of that.
Furthermore, because of this, you cannot also ask what a "being" would "see" that moves at the speed of light because nothing can evolve, meaning no consciousness and so no information processing is possible. Were your body's particles somehow to "magically" loose mass and move at the speed of light, yet maintaining coherence in their shape, you would be dead, transformed into the most rigid corpse imaginable.
The above also shows that motion at the speed of light is qualitatively different from motion at other speeds, in particular, there are things that can happen (internal evolution) for motion at slower speeds that cannot happen at this speed, and furthermore, there is no rest frame for this speed. Indeed, there are actually 3 qualitatively different domains of speed in relativity theory: $v < c$, $v = c$, and $v > c$. It is impossible to accelerate something between these three domains, and what domain a particle inhabits depends on what its mass is: respectively positive, zero, and imaginary. Only the first two appear to be realized in the physical universe: the third leads to difficulties with causality and the causality of our universe looks to be rigid on top of not seeing anything that moves in that domain, so such a domain is strongly disfavored empirically. The "principle of relativity" which says there is no universal rest or motion only applies within a domain, or perhaps even better, only with the domain $v < c$ since that is the only one in which we can clearly make sense of an inertial frame. Things moving in the other domains (in our universe, just the domain $v = c$, apparently) are indeed in an absolute state of motion, again illustrating this qualitative difference.
Answer: From the inertial observer's point of view, the light clock is stopped. The light ray moves parallel to the spaceship and never reaches either mirror. Because light always travels at the same speed, if it moves along with the spaceship (at the speed of light), then it can't have any velocity perpendicular to the spaceship, since that would result in a larger total velocity.
From the spaceship passenger's view, everything is normal, including the light clock. They will observe that time has stopped outside the spaceship.
The important thing to remember about relativity is that it describes how to take observations in one reference frame and predict what a second observer ina different reference will observe.
Comment: There is no spaceship passenger's view, i.e., no inertial reference frame, no rest frame for the passenger, the light clock, or the spaceship if, as the OP stipulates, the spacecraft has speed *c* in an IRF.
Comment: Sometimes sensible conclusions can be drawn but not in this case.  The second paragraph isn't sensible at all because it's just plain wrong according to SR.  If the spacecraft has speed $c$ in an IRF, it has speed $c$ in *all* IRFs (invariance of the speed $c$), i.e., *there is no spaceship passenger's view where everything is normal including the light clock* because the spaceship is not at rest in *any* IRF; to all inertial *observers*, the spaceship is moving with speed *c*.
Comment: @AlfredCentauri True, but sometimes sensible conclusions can be drawn from taking the limit as velocity goes to c.


How can I apply the same change to multiple channels in Graph Editor?

Question: I'm a beginner and learning with a video course in Blender Cloud.
I would like to apply the same modification to two channels such as the X Scale and Z Scale in one action in the Graph Editor.
In the video course, Animation Fundamentals, 101 Simple Bouncing Ball (Part02), around 22:55, the instructor edit both the X scale and Z scale at once.
Answer: Use the eye icon to show only the curves you want to edit, then select all keyframes you want to edit, then press G to move, G X to move along time, G Y to move along values, S X to scale along time, S Y to scale along values. Confirm with enter or Left click.
You can set the pivot point of scaling with the usual pivot icon (Bounding box, individual center or 2D cursor).
Comment: Thank you. It worked. I tried to solve it only with mouse operation. With GY shortcut keys with multiple channel selection with shift key solved the problem.
Comment: Perfect, if it has been useful please mark the answer as accepted.


Help identify an SMD (diode?) component

Question: I suspect this is a diode: the board has PD13 marking next to it. I found a schematic for a similar product that identifies the PD13 as TVS_SMAJ20A. Would that make it a TVS diode?
Comment: I'd say it was a TVS. It seems there was enough fault current to destroy it so there's other collateral damage methinks. You can replace the diode, but it is not needed for the circuit to function.
Comment: Looking at how large the copper pours at the anode and cathode are, I wouldn't be surprised if they have used the diode as a reverse polarity protection.
Comment: If it's a TVS which seems likely, the part was perhaps destroyed since it is mounted backwards. It would conduct all the time. TVS tend to short whenever they go poof.
Answer: PD13 is a designator it doesn't really tell you the part number or the part characteristics. It probably stands for "protection diode" or something like that, and it's the 13th in some kind of ordering.
You need the second character in the code to be sure. It's plausible that's what it is, because the SMAJ20A (which is a part number) has the code BV. Since it has the lines on one end it's unipolar type. From the datasheet

Caution though- there are maybe 10 part numbers that have a code starting with 'B' so the ID hinges on how 'similar' your similar product is. Since it has the same designator I'd guess it's very close...
Comment: I would say that TVS is the most likely too, although in that case it is mounted backwards. Alternatively it could be a series diode for polarity protection and then it can be pretty much any part, likely a Schottky.

Help identify an SMD (diode?) component


Can an Delta Dirac function (not response) ever be sampled?

Question: This is purely theoretical question but can an actual delta-dirac function ever be sampled? To make it clear again, I'm not talking about an impulse response from the impulse function. Lets say we have a truly instantaneous impulse as defined with delta-Dirac that the impulse with infinite(or finite) height and no width, an instantaneous pulse. This pulse is to be detected by sampling, but to find this instant signal, wouldn't this require an infinitely fast sampling or infinite sampling rate? Thinking this in terms of frequency domain, I believe delta function is 1 or has infinite spectrum. So would delta function ever be sampled unless we have infinite sampling rate?
Answer: A Dirac delta impulse cannot be sampled because it is a distribution and as such it does not make sense to talk about the value of $\delta(t)$ for e.g. $t=0$ because that value isn't defined. However, sampling extracts values of a function for certain values of $t$. The Dirac impulse is no function in the conventional sense. It's only under an integral that $\delta(t)$ makes sense.
Furthermore, the mathematical description of sampling is a multiplication with a Dirac comb. However, the expression $\delta^2(t)$ is undefined, so sampling a Dirac impulse can't be described mathematically in any meaningful way.
Note that the discrete-time unit impulse $\delta[n]$ can be sampled without any problem, because it has well-defined values for any $n$, unlike its continuous-time counterpart.
Comment: Ok, that was a great explanation, thanks!
Answer: Yeah, I think your conclusion is right. According to Nyquist' sampling theorem, the sampling rate should be at least twice as many as the maximum frequency of the target signal. And the maximum frequency of delta function is infinite. If sampled, aliasing must occur. In fact, the discrete delta function $\delta(n)$ is a sampled version of continuous delta function $\delta(t)$. The spectrum is overlapped while sampling, of course.
Comment: The reason why $\delta(t)$ can't be sampled has nothing to do with the sampling theorem. Neither is $\delta(n)$ a sampled version of $\delta(t)$.
Comment: Not sure what you mean by "cannot be operated", but I stick with my claim that $\delta^2(t)$ has no generally accepted and useful meaning, especially in the context of sampling the distribution $\delta(t)$. If you apply $\delta^2(t)$ to some test function which goes (at least quadratically) to zero for $t\rightarrow 0$, please use it if you can make sense out of it. But there's no way you can handle $\delta^2(t)$ in the context of the OP. Furthermore, $\delta[n]$ is definitely NOT a sampled version of $\delta(t)$. You might want to consider correcting or deleting your answer.
Comment: @MattL. delta function is not a coventional function does not mean it can not be operated. And $\delta^2(t)$ is meaningful in some way. See this http://mathoverflow.net/questions/48067/is-square-of-delta-function-defined-somewhere


BIOS says 16gb of ram, windows says 12gb

Question: bios and cpu-z both say 16gb of ram installed windows and task manager both say 12gb installed and control panel has only 60mb reserved and also ram slot 3 and 4 dont work i have to go to asus bios and disable DRAMM b slot 1 and put the stick to slot 1 for some reason, if i have all ram spots enabled then the pc wont boot with 3rd ram stick in 3rd of 4th spot. I have 4x2 4gb sticks and 1 8gb stick all hyperx fury sticks, my motherboard is asus pro b250
Comment: You have an integrated GPU in your CPU right? Some of the system ram gets set aside for that GPU to use as video ram. You should be able to reduce how much in the bios.
Comment: Do all your sticks share the exact same characteristics in terms of clock speed? Could be one is slower and is being ignored by your mobo because it can't underclock the whole bus. Can you set the RAM clock speed manually on this mobo?
Answer: Check your motherboard socket for bent pins.
For what you say SPD (the protocol used to tell which slot is populated, with the manufacturer and specs) is working fine, but the data lines are not. That is also why CPU-Z show all the capacity.
Most common cause for this are:

Bent/damaged pins in the socket
Bad memory controller on the CPU (sometimes only one of the 2 channels fails)
Broken or badly plugged memory stick (we can discard this in your case, since you swapped them)

First two possibilities would also explain the other problem you had. It is not likely to be a case of the integrated GPU using the memory as one comment stated, since it would show as "system reserved" in the control panel aswell, and 4GB of lost RAM is far greater than the values that are common in these cases.
Answer: Both BIOS and CPU-Z use SMBus to access the Presence Detect on the RAM modules, so they will "see" them almost no matter what.
This does not mean that the motherboard MMU can or wants to use them. I do not know what kind of channel organisation your motherboard has, but try placing the three sticks this way:
slot 1: 4GB
  slot 2: 8GB
  slot 3: 4GB
  slot 4: empty (and possibly disabled)
If you put 4 GB in slots 1 and 2, and 8GB in slot 3, it might happen that only those lines on the 8GB stick that also are available on the 4GB stick will be used by the MMU, in order to best exploit dual-channel access strategy. In other words, the motherboard will privilege speed over capacity, and give you 4+4+4 = 12 GB of RAM.
Using a 4-8-4-0 configuration, you should have 8GB of fast RAM and 8GB of slightly slower RAM. Or you might have 16GB of slightly slower RAM, all timings in accordance to the lowest common denominator access method available. I'm betting on the first, because this is what you already should be having - (4+(8/2))GB of fast RAM and one 4GB stick accessed more slowly.
(But, of course, it might also happen that no RAM is seen at all).
There might be some configuration option in the BIOS on how to optimize memory, but if you have one of those newfangled super-duper gamer mobos, then chances are that it has a mind of its own, and will only do autoconfigure after testing the slots.


Can hydrogen atom exchange induce attractional forces between $e^-e^-$?

Question: G. Feinberg and J. Sucher showed that neutrino-pair exchange can induce long-range repulsive forces between electrons that is proportional to $r^{-6}$. In the paper, they also stated that  changing a particle to its antiparticle changes the sign of the force. However, in case of neutrino-pair exchange, one cannot induce attractive forces. So i was thinking about exchange of hydrogen atom (as it can be considered as a boson, particularly "composed boson" ), could it be induce attractive forces between $e^-e^-$? And if it couldn't induce attractive forces, which bosons do?
Answer: First, a hydrogen atom is not an exchange particle. It is a real particle. An exchange particle is termed a virtual particle. A virtual particle mediates an interaction between two other particles. The reason "virtual" is used is  because it is usually "off mass shell" and not observable. The lifetime  for such particles is ridiculously small and this amount of time $t$ and their energy $E$ are determined by the Heisenberg Uncertainty relation, namely
$\Delta E \Delta t \geq \frac{\hbar}{2} $
But if we worked with it anyway, and we applied this to the hydrogen atom (using the mass/energy of a proton) this places its range at roughly $0.11$ fermi or about 1% of its own 'radius'.
Comment: Thank you Dr jh for you useful corrections. So which properties a boson must have to induce stronger-than-electromagnetic forces between $e^-e^-$?
Comment: Hi again Elsayed. The photon, a boson, is the exchange particle responsible for the electromagnetic interaction. This includes electron-electron scattering. I'm not sure I understand what you mean by stronger-than-electromagnetic. cheers.


Distortion Removal=Parallax Removal?

Question: So, as the title suggests: does distortion removal caused by lenses means that the parallax is also removed from the pictures?
Comment: What do you mean by "parallax"? Accordingly the commonly accepted definition, parallax is the difference in perspective between two viewpoints, for instance human eyes. In a (non-stereographic) photo you only have a single viewpoint so talking about parallax is not meaningful.
Comment: @JohannesD Yes that's correct. What I meant was if it is possible to view a number of scenes stereographically after the lens distortion is removed from them.
Answer: No, these are unrelated. Distortion removal is a 2D mapping which moves pixels to remove barrel and pincushion distortion. Parallax correction requires multiple images or depth information for each pixel and is performed by completely different software algorithms.
Comment: This depth information is exactly what I am curious about. So, you say that it is not affected at all? From many projects -that involve 3D representation from pictures- that I have done, I get that it is not affected. I am just looking for the confirmation from someone that has this knowledge.
Comment: Ah, I think you are asking *if we perform distortion correction, do we get new depth info*. No, not really because the correction is 2D, so it does not generate new depth-info. If you have a depth-map though, then you would have to apply the same geometric processing to it too, so that it corresponds to the undistorted image.
Answer:
if it is possible to view a number of scenes stereographically after the lens distortion is removed from them

Yes.
In theory all distortion correction does is undo some distortion a lens introduces because it is not an ideal ( theoretcally perfect ) lens.  You will note this distortion is radially symmetrical and the effect is most notable in the curving of what should be straight lines in the image ( for rectilinear lenses ).
This should, in principle, make viewing two images as stereographic pairs better as the distorted images would be more likely to cause (minor) discrepancies from what the human vision system expects than the undistorted ones.
The correction should, again in theory, make it more accurate for a tool that estimates distance from stereographic pairs.  It is the distorted image that is more likely to cause problems.

Distortion Removal=Parallax Removal?


五年ぶり、ぶり返す、。。。What does ぶり mean?

Question: Where does the ぶり in 五年ぶり (Five years ago?) and ぶり返す come from, does it have a kanji, and what does it mean?
Comment: My guess is you probably first encountered 〜ぶり in お久しぶりですね!
Comment: Haha. I guess I did!
Answer: Spelling
In reply to istrasci's final comment, the buri in burikaesu is the same 振る as furu, and can be spelled in kanji as 振り返す.  That said, the buri is more often spelled in kana to make the voiced reading explicit.
Phonology
Changing an h- or f- on the front to the voiced b- variant is common in the latter element of compounds (see https://en.wikipedia.org/wiki/Rendaku), explaining why -fur- changes to -bur- in compounds.
When the fur- is in word-initial position, the shift to voicing can imply some kind of pejorative meaning, as noted by a number of linguists.  From Laurence Labrune's The Phonology of Japanese (2012), p.102:

The voiced obstruents which appear word-initially in Yamato words come from the following sources.  First, one finds a number of cases which result from secondary voicing of an originally unvoiced obstruent, whose phono-pragmatic function is to introduce a pejorative or expressive connotation.

Semantics (Meaning)
振る furu is ultimately cognate with the furu in classical furuu, modern 震{ふる}える "tremble, shake, vibrate, quake".  The furu on its own can mean "wave, shake", but it also has a range of other meanings, a number of which imply "the movement of a thing as a whole".  That underlying sense might be where derived term 振り furi came to include meanings such as "likeness; kind; one's appearance, style, or behavior".
-Buri as the latter element in compounds derives from this furi, and can be found in words like 身振{みぶ}り "body language, how one moves, appearance", or 女振{おんなぶ}り "a woman's face or appearance".
When appended to a time-related noun or phrase, -buri again carries stronger connotations of "whole, entire", in terms of "the whole XX amount of time has passed".  三週間ぶり would mean "it's been three whole weeks [since ...]".  久しい literally means "eternal; a very long time."  久しぶり is thus often an exaggeration, similar to English "it's been forever!" even though the speaker and listener might have seen each other recently.
For the buri in burikaesu, the meanings given for this verb are generally negative, suggesting that the voicing is indeed pejorative.  Whereas furikaesu can mean simply "to turn something back", burikaesu means "to get worse again, as said of a previously improving situation."
Answer: The ぶり in 5年ぶり comes from the verb 振る which is usually read ふる, but can also be ぶる.  The kanji is not often used (in my experience) in this context.  It means "shake" or "wave", but I'm not sure how that really plays into these time expressions.
As a verb ending (〜ぶり), it can mean "way of doing"; similar to 〜[方]{かた}, but I believe more subjective.  Here, 振る takes the meaning of "assume", "pretend", "put on", or "pose as".  For example, 高ぶる means to be arrogant -- "assuming" a high position.  So you can see the verb ending comes from this.

話し振り → Way of speaking → One's (assumed) mannerisms of speech (subjective)
話し方 → Way of speaking → How to speak (methodically; objective)

As far as ぶり返す, I cannot tell from all the definitions in my dictionaries if this is the same ぶり or not, so I'll defer that question to someone else.

五年ぶり、ぶり返す、。。。What does ぶり mean?


Nemo Randomly Makes Itself Default File Manager and Creating Desktop Shortcuts

Question: I've installed Nemo but haven't made it default and I have nothing on my desktop. But sometimes when I boot up I see Computer, Home and Trash icons on my desktop and suddenly Nemo is the default file manager as when I click any of my shortcuts or the new icons on my desktop Nemo appears. When I restart it all returns to normal.
My question is...well... WTF?! I was testing out Nemo and intending to replace Nautilus with something else but this stuff is really weird, can someone tell me what's going on? Should I uninstall Nemo? I'm really new to Ubuntu or Linux in general but I didn't find where those icons would be disabled when I look at my Desktop on Nemo via opening by the shortcut on my Desktop, it shows nothing is there.
When I look for Desktop Icons in Unity Tweak Tool appereantly nothing is there and they don't go away when I tinker with the settings.
But after I restart 1 or 2 times the problem disappers and everything goes back to normal.
So please, somebody help me make some sense out of this. Also could someone tell me how to uninstall a software thoroughly in Ubuntu? It always leaves a lot of junk behind from my experience.
Screen Shot 1, The Problem:
Screen Shot 2, Two Restarts Later every thing is back to normal:
Answer: To completely remove nemo:
sudo apt-get remove --purge nemo

Comment: Thanks, do you think removing Nemo is necessary/useful regarding my condition?
Comment: Well as you still have nautilus, you'll be fine. It's not gonna hurt anything. Remove it when the icons have gone to be safe.
Answer: Copy and paste the below command in terminal exactly to not show shortcuts on Desktop.
gsettings set org.nemo.desktop home-icon-visible false;
gsettings set org.nemo.desktop trash-icon-visible false;
gsettings set org.nemo.desktop computer-icon-visible false;
gsettings set org.nemo.desktop volumes-visible false;


ITS4060S-SJ-N overvoltage protection

Question: I (think I) want to use an ITS4060S-SJ-N (+datasheet) in a circuit that can't handle over 40V (an LM2576T-5), but will normally receive 8-30V with a ≤12W load. Its purpose would be to provide overvoltage protection, short circuit protection, thermal shutdown, and reverse battery protection*. But I'm having trouble making sense of the datasheet to find what voltage it would clamp at. From the datasheet, I see that:

Overvoltage protection \$V_\text{SAZmin} = 41\text{V}\$ (Product Summary, p2)
Max supply voltage \$V_\text{S} = 40\text{V}\$ (Absolute Maximum Ratings, p5)
Overvoltage protection \$V_\text{OUT}=V_\text{S}-V_\text{ONCL}\$, \$V_\text{SAZmin}\$ = 41V (Electrical Characteristics, p8)
Output clamp at \$V_\text{OUT}=V_\text{S}-V_\text{DSCL}\$ (inductive load switch off), \$V_\text{DSCLmin}\$ = 41V (Electrical Characteristics, p8)

Does that mean it would clamp to 41V? \$V_\text{ONCL}\$ is only mentioned once, without further explanation. I've searched for \$V_\text{SAZmin}\$ online and couldn't come up with anything either. The \$V_\text{DSCL}\$ line seems to indicate that the output voltage is at least 41V lower than supply (\$V_\text{S}\$) under inductive load switch off, so perhaps that means that for say \$V_\text{S}<41\text{V}\$ then \$V_\text{OUT}=0\text{V}\$. But what constitutes the maximum output voltage? It could be anything up to 41V in that case. I think there would have to be something more in the logic block on p3...
* Will include a 150Ω ground resistor and a PMOSFET to limit load's reverse current, see note 4 on p8.
Answer: So we have some ITS4060S-SJ-N, so I decided - why not just test it? Well, it passed up to 40V straight through to the OUT pin when the load drew around 20mA. That indicates that it won't disconnect the load if the supply voltage is too high. I suppose I would need something like http://www.ti.com/product/TPS2663 and the increased cost if I wanted that functionality. Oh well!
Comment: It says overvoltage is 41V. So it would make sense it's still fine with 40V.

ITS4060S-SJ-N overvoltage protection


Can a company restrict an employee from looking for a job while they are on a PIP?

Question: Can a company or employer restrict a person from looking for a job while he/she is put on a PIP?
Comment: Do you mean looking for another job while you're on duty for the current job? Or looking after hours, on your days off, etc.?
Comment: You need to provide more information. What does your contract say, what is your location, do you look for jobs on company time or your own time? As the question stands, it is not clear what exactly you are looking for.
Comment: How does the disability tag tie into your question?
Comment: A PIP cannot restrict you from doing anything outside of the time you are being paid by the company you currently work for. So, if they try adding a line in the PIP restricting your non-work activities, they are stepping out of their "jurisdiction". If your currently employer states that you cannot use company resources (computer, iPad, phone, etc) or while during work hours, they are within their rights. Were you job searching while on the clock? If so, they probably noted it when reviewing your computer activity and decided it needed to be said.
Answer: A PIP is usually a preparation to fire you. So think about it: If you are on a PIP, and you look for and find another job, what's the worst that can happen? They fire you (however, firing you for looking for another job would be something that might be used against them), which they wanted anyway.
Most likely the company is only too happy if you find a job elsewhere and leave on your own.
Answer: Unless your contract says otherwise, you can always look for another job ,-- on your own time, of course, and subject to any  non-compete clauses in that contract
Comment: You can always look for another job on your own time, REGARDLESS of what your contract says. This is especially true for PIP's which are widely regarded as a code-word for "you're going to be fired, look for another job."


Uniqueness of Exponential Objects up to Isomorphism in any Category

Question: I want to prove that for any pair of objects $a,b$ in a category $\mathcal{C}$, the exponential object $a^b$ of $a$ and $b$, if it exists, is unique up to isomorphism. It looks to be really simple, but I can't figure out the proof myself. (I need a proof that doesn't mention adjoints or Yoneda's lemma, since I'm working from a category theory textbook that hasn't defined these concepts yet)
Comment: It's the same as the uniqueness proof for any universal object: define a category such that the object you want is an initial or terminal object in that category.
Comment: Please define what _you_ mean by exponential object, because the one that immediately comes to mind uses adjoints.
Comment: $b^a$ is an exponential object of two objects $a,b$ in a category $\mathcal{C}$ if there is an associated evaluation function $eval:b^a \times a \rightarrow b$ such that for any object $c$ in $\mathcal{C}$ there is a unique arrow $\lambda g: c \times a \rightarrow b$ such that $eval \circ (\lambda g \times 1_a) = g.$
Answer: Shortest answer. Right adjoints are unique up to unique isomorphism.
Short answer. The definition of the exponential object $b^a$ implies there is a natural bijection between arrows $c \to b^a$ and arrows $c \times a \to b$; the latter does not depend on $b^a$, if $d$ also has the universal property of $b^a$, then there is a natural bijection between arrows $c \to b^a$ and arrows $c \to d$, hence, $b^a$ must be isomorphic to $d$, by the Yoneda lemma.
Long answer. Let's spell out exactly where the isomorphism comes from. Suppose $d$ has the universal property of $b^a$: so there is a universal morphism $e : d \times a \to b$ such that for any $g : c \times a \to b$, there is a unique $g' : c \to d$ such that $e \circ (g' \times \textrm{id}_a) = g$. But in particular we can take $c = b^a$ and $g = \textrm{eval}$, and this gives us $g' : b^a \to d$ such that $e \circ (g' \times \textrm{id}_a) = \textrm{eval}$. But by the universal property of $b^a$ there is a unique $\lambda e : d \to b^a$ such that $\textrm{eval} \circ (\lambda e \times \textrm{id}_a) = e$, so we get
$$\textrm{eval} \circ (\lambda e \times \textrm{id}_a) \circ (g' \times \textrm{id}_a) = \textrm{eval}$$
but $(\lambda e \times \textrm{id}_a) \circ (g' \times \textrm{id}_a) = (\lambda e \circ g') \times \textrm{id}_a$, hence $\lambda e \circ g' = \textrm{id}_{b^a}$ by uniqueness. Conversely, we have
$$e \circ (g' \times \textrm{id}_a) \circ (\lambda e \times \textrm{id}_a) = e$$
so by uniqueness again we conclude $g' \circ \lambda e = \textrm{id}_d$. Thus $d \cong b^a$.
Remark. All of these answers are actually the same, just expressed in different ways. As others have said: go learn the Yoneda lemma and representable functors.
Comment: Thanks for the worked out solution. Yeah I think I need to work with another text alongside Goldblatt's Topoi, one that introduces functors and adjoints much earlier.
Answer: Awodey's great book (2nd ed), chapter 6, has a good introduction to exponentials, without Yoneda, ect.
The proof works out in the same way that is used for products (for example) or any other universal contruction.
Remember: in the definition of an exponential E1 there is a unique arrow, so if you suppose that there is another exponential E2......in the end there must be a unique arrow from E1 to E2 and a unique arrow from E2 to E1, so E1 and E2 must be isomorphic.
Study the technique used for products, equalizers or pullbacks for inspiration (always in Awodey's book)
Comment: Thanks for the recommendation and link.
Comment: I am glad it helped :-)
Comment: Why is Awodey's book "great"?
Comment: To distinguish it from Awodey's little book (first edition only had 256 pages)
Answer: $b^a$ is by definition a representation (or just the representing object) of the functor $c \mapsto \hom(c \times a , b)$, i.e. we have a natural bijection $\hom(c,b^a) \cong \hom(c \times a,b)$. The Yoneda Lemma tells us that representations are unique up unique isomorphism.
PS: First learn the Yoneda Lemma, then the rest ...
Comment: @downvoter: An explanation or hint of improvement would be nice.
Comment: I think the downvote was me by an entirely accidental finger-slip! Sorry -- the totally unintended vote the wrong way now seems to be "locked". Many apologies!


Question about Bilinear form and inner product space

Question: This is a question I have stumbled upon in a test I found on the web, and I don't even know how to approach it:
Say $V$ is a vector space with an inner product above $\mathbb{C}$ (We don't know what is the inner product). Prove that if a linear operator $f\colon V\to V$ is adjoint to itself, the bilinear form $B\colon V\times V\to \mathbb{C}$ defined by $B(v,w)=(f(v)\mid w)$ is a Hermitian form.
Now, I understood that an adjoint operator $F$ is an operator that fulfills: $$\langle F(v),u\rangle = \langle v,F(u)\rangle$$
Now, what is $(f(v)\mid w)$ and what is the definition of an adjoint bilinear form? I only found the definition of an adjoint operator so that's why i'm asking.
Help would be very appreciated.
Answer: The bilinear form should be defined as
$$
B(v,w) = \langle f(v), w \rangle, \  v,w\in V,
$$
note that it uses the inner product! Now use the definition of self-adjointness
$$
\langle f(v), w \rangle = \langle v, f(w) \rangle \quad\forall v,w\in V
$$
to conclude
$$
B(v,w) = \overline{ B(w,v) } \quad\forall v,w\in V.
$$
To me it looks like the confusion comes from different notations for inner product: $\langle \cdot,\cdot \rangle$ vs. $( \cdot | \cdot )$.