Why don't atomic nuclei emit EM radiation which destructively interfere with the radiation emitted by the electrons?

Question: When atoms get heated up, they start emitting EM radiation because of accelerating electrons. But why don't the protons emit EM waves which destructively interfere with other emitted EM waves, because they're also accelerating just as the electrons do, because they're both jiggling as one (atom)?
Answer: The classical idea that the electrons emit energy because they are accelerating does not work in the atomic realm. This was the precise reason that Rutherford's atomic model was rejected by the classical physicists in the early 20th century.
The atoms emit light when the electrons from one energy level make a transit to a lower energy level due to the instability of the atom. The same happens with the case of the proton too, but the energy gap between the consecutive energy levels of the proton is so large that the thermal energy cannot excite them to the higher energy levels. Thus, when an atom is heated, the electrons excite to the higher energy level and then fall off to the ground state giving light.
Comment: Thank you. It's annoying that I find what I wrote everywhere on the internet, even in textbooks. :/
Comment: But it is still correct that accelerating charges create electromagnetic waves? If so, if we're accelerating an atom, is it true that electrons are creating their EM wave and protons are creating their counter EM wave, so they together emmit almost nothing? But wouldn't that mean that ions emmit EM radiation when accelerated, while atoms don't?
Comment: It is not quite right that the electrons accelerating always emit radiation (If this is so, the atom would continuously emit radiations and at the end, the electrons would collapse into the nucleus). You can find more about it here: http://www.pbs.org/wgbh/aso/databank/entries/dp13at.html.
Comment: Not sure what then decides if electron is going to emmit light. I thought this is happening because of electric charge moving, so it creates something simmilar to Doppler effect. But I'll try reading up more about it. It's annoying that there is so much contradictory information.
Comment: @MaDrung Regarding that 'contradictory information', have you tried looking in an introductory quantum physics textbook? Something along the lines of Beiser's *Concepts of modern physics* should be clear enough.
Answer: The existing answer is correct, but it's in a sense 'orthogonal' to the question, and it doesn't really address the core concern.
Atoms are indeed quantum mechanical objects, and the classical understanding in terms of accelerating charges no longer works, and it needs to be replaced with transitions between different states of the atom, but this does not make the problem go away.
The reason for that is that the transition has an amplitude - a number which governs how likely the atom is to change states, known technically as an electric dope transition matrix element - which depends on the position of the electron with respect to the center of mass, but also in the position of the nucleus, which therefore also contributes to the radiation, exactly as in the classical case.
So, why doesn't the radiation cancel out? Well the word dipole should give it away: what really matters for the radiation isn't so much the movement of charge, but the dynamics of the electric dipole, and here the contribution of the nucleus adds constructively with that of the electron: it has the opposite charge, sure, but it also moves in the opposite direction, so the global sign is $+$, and the two contributions to the radiation / acceleration / transition dipole / whatever are in the same direction.
Moreover, this is also the case for the classical dynamics of the Rutherford model of the atom: the nuclear charge is opposite to that of the electrons, but it moves in the opposite direction, and the contributions add constructively. This is why I said the answer "because QM" is orthogonal to the question - the answer doesn't really care whether you're doing classical or quantum dynamics.
Now, why do we normally ignore the nuclear motion? Well, it is much heavier than the electrons, by a factor of ~1800 and up, so it doesn't move much, but it still has a contribution and its effects can be seen in the atomic radiation if you do a careful enough experiment. However, because its effect is constructive, the answer wouldn't change even if you had two opposite charges of the same mass orbiting each other (as in, say, positronium).
Comment: Wow. This answer feels so satisfying. So bassicly the atoms still jiggle inside a material, but the majority of the movemen't is done by the electrons bouncing around? Is that correct? But accelerating free charges still radiate EM waves right? And so do the ions, but not atoms. How does that extra electron in an ion know to radiate EM waves while ion is accelerating, but the rest of electrons don't?
Comment: "Jiggle" is a terrible term, and it doesn't describe what goes on particularly well, but I feel you're misinterpreting this answer. It's impossible to define "the majority of the movement", but as far as the electric dipole of an atom goes, the bulk of the contributions (by a factor of 1800 to 1) comes from the electrons. If by "but not atoms" you mean that an acceleration of the center of mass of a neutral atom does not emit radiation, then that is correct, but that doesn't mean that atoms don't emit radiation at all due to the internal 'motion'. And seriously, pick up an intro-QM book.
Comment: I have worded it badly. I think I understand what you meant. I just had a follow-up questions to it. Which is to make it clear why accelerating atoms (not bound by material) don't emit EM waves, while ions with 1 extra electrons emit pretty much the same amount as a free electron if their accelerations are exactly the same in one unchanging direction. How does that one extra electron in an ion know to start emiting light? And yes I will start reading a QM textbook as soon as my 3 weeks of leave come this year. :)
Comment: It's not that 'the extra electron knows' that it needs to radiate. If you accelerate the center of mass, *all* the electrons radiate, but here the nucleus does emit radiation that interferes destructively with that of the electrons; for the neutral they cancel, but for ions the cancellation is imperfect. This is different to your original question, though: thermal emission is not due to accelerations of the center of mass, but rather to reorganisations of the internal motion (and, for neutral atoms, any thermal COM acceleration does not radiate, because of the cancellations.
Comment: ... this is why "jiggle" is a terrible term, among other factors. Handwavy language can carry you to a point, but without a more solid foundation it is subject to these sorts of pitfalls.
Comment: Ok, so it is as I have imagined about accelerating charges. One last question. Doesn't an accelerating atom get slowed down if all of its electrons and protons are emiting light? Even if destructive interference is made, they still emited the energy, that got countered by the other half of emited energy.
Comment: @MaDrung See [this question](https://physics.stackexchange.com/q/23930/) regarding the interference, but in this particular case, neither is actually radiating meaningfully (and particularly no energy is radiated) so the atom does not get slowed down.


Why didn't Snape see Dumbledore's Army in Occlumency lessons with Harry?

Question: In Harry Potter and the Order of the Phoenix, why didn't Snape see Dumbledore's Army in Harry's mind during Occlumency lessons?
Comment: Is it possible that he had, but decided to ignore it because it was part of the plan?
Comment: Similarly, it's possible he deliberately didn't go poking for unrelated information so that, should the Dark Lord ask, he'd have plausible deniability.
Comment: In the movie he saw Harry kissing Cho, that was in the room of requirement where they were organizing as Dumbledore's Army. So, in some sense, he did see Dumbledore's Army (at least part of it). I don't remember if this is in the book, though (which is why it is only a comment for now).
Answer: Because the memories Snape saw were pretty miscellaneous.
The memories which emerge during Legilimency appear to be fairly random, unless the Legilimens is actively searching for a particular memory or thought within their target's mind. When Dumbledore was questioning Morfin and Kreacher he was looking for specific information. When Snape was teaching Harry Occlumency he wasn't searching for anything in particular. He was just testing Harry's defences. As such, the memories which Snape sees are pretty sporadic and random.

He was finding it increasingly difficult to disentangle separate memories from the rush of images and sound that Snape kept calling forth.
(Order of the Phoenix, Chapter 26, Seen and Unforseen).

It's worth pointing out that Snape did see a DA lesson, even if he only saw the make-out session at the end rather than the lesson itself.

A hundred Dementors were closing in on him beside the dark lake...Cho Chang was drawing nearer to him under the mistletoe...
No, said a voice in Harry's head, as the memory of Cho drew nearer, you're not watching that, you're not watching it, it's private -
(Order of the Phoenix, Chapter 24, Occlumency).

We don't have a complete record of all the things Snape saw in the Occlumency lessons. Some of the lessons aren't described at all in the books. So he may have seen a DA lesson in progress. If he had done it wouldn't have been a big deal. We know that Snape was loyal to Dumbledore, not Umbridge. Snape wouldn't have betrayed Harry's secret to her. He may have passed comment on the DA, asking what it was and whether Harry was getting up to anything he shouldn't. It would've been fairly easy for Harry to lie at this point and pretend that the meetings had happened in the pre-Umbridge era.
We don't know if Snape told Dumbledore about the DA, even supposing that he saw any of the memories. Dumbledore probably knew about the meetings anyway, however. Anything that Snape told Dumbledore would've been in addition to the information he'd received from Aberforth and Mundungus. It wouldn't have told him anything new. Nevertheless, the most logical explanation is that, amongst the random stream of memories that Snape saw, the DA meetings never came up.
However, there is another explanation for why Snape may not have seen the meetings. Although the Legimilency process is fairly random it isn't totally random. Snape was more likely to see the memories that Harry was most eager to hide.

"Get up!" said Snape sharply. "Get up! You are not trying, you are making no effort, you are allowing me access to memories you fear, handing me weapons!"
(Order of the Phoenix, Chapter 24, Occlumency).

Most of the memories Snape saw were from Harry's childhood. It was less likely that Snape would see the DA lessons when so many of Harry's memories were of childhood embarrassments.

He had just been forced, yet again, to relive a stream of very early memories he had not even realized he still had, most of them concerning humiliations Dudley and his gang had inflicted upon him in primary school.
(Order of the Phoenix, Chapter 26, Seen and Unforseen).

There's no particular reason why Harry would fear Snape seeing the DA meetings (no more so than any other teacher discovering them) and they certainly weren't embarrassing or humiliating. These were the sorts of memories that Snape ended up seeing so it's probably unlikely that the DA meetings would've surfaced during the Occlumency lessons.
Answer: It's also likely that he already knew about it as Sirius and Molly Weasley already knew about it so the Order would have been notified:

"She [Molly] says on no account whatsoever are you to take part in an illegal secret Defense Against the Dark Arts group. She says you'll be expelled for sure and your future will be ruined. She says there will be plenty of time to learn how to defend yourself later and that you are too young to be worrying about that right now. She also" (Sirius's eyes turned to the other two) "advises Harry and Hermione not to proceed with the group, though she accepts that she has no authority over either of them and simply begs them to remember that she has their best interests at heart. She would have written all this to you, but if the owl had been intercepted you'd all have been in real trouble, and she can't say it for herself because she's on duty tonight."
Harry Potter and the Order of the Phoenix

Therefore, his loyalty remained... true-ish and he didn't do anything about it...

Note that this is an addition to @TheDarkLord's answer, but was far too long for a comment...


Is there statistics on the frequency of individual pinyin syllables in real texts?

Question: I'm interested in the frequency of pinyin syllables (both when including tones and toneless). This implies the usage in real text, weighted by the occurrences in the corpus, not by parsing a dictionary which would overblow some really rare syllables that occur in many bookish words. I might confidently assume that "de" of Netral Tone is the most frequent; but which one is the rarest (in everyday usage)?
Comment: What is the purpose of this study?
Answer: Jun Da has a lot of frequency data on his website, including syllable frequencies both with and without considering tones, based on frequency data for the 3,500 most common characters. This is, in other words, indirectly based on corpus data, but does not include rare characters.

Syllable frequencies irrespective of tones 不记声调的音节频率
Syllable frequencies with tones 记调音节频率

I'm not sure exactly why you're asking this question, so this may or may not be enough for you. Asking what's the rarest of anything is kind of pointless, in my opinion, as there probably are edge cases which might have an odd occurrence here and there. Where do you draw the line?
Answer: Any of the less common pinyin sounds would be rarest, ang2, chong3, or pie4 are possible contenders.  looking at the pinyin without the tone is counterintuitive to me, since the two aspects are not separable in chinese.  Of course, you could do it for fun, but it might be hard to get anything that makes sense as a result.
I think it is impossible to officially calculate the true rarest pinyin sound, since written text does not notate the pronounciation used for the characters.  You would need someone to actually go through all chinese text and recordings to manually check what was used most, which is certainly not feasible.
However, if you look up rare pinyins that only belong to a few characters you can see the contenders, especially those that are literary chinese or otherwise uncommon. (for constrast ka is not that common of a pinyin, but the characters it is used for are very common-- things like cards and coffee).
的 is definitely the most common character written in modern chinese, and its de pronounciation the most common as well.
Answer: I didn't find a frequency table based on pinyin. However, I found one Chinese character frequency list 汉字字频表 based on Chinese characters. This table includes pinyin and frequency of each character listed in it. You may want to import the table to Excel and manipulate it to get what you want.


Is it possible to change Russian Rubles to EUROs in Irkutsk?

Question: Possibly people in Irkutsk do not need much EURO?
Comment: Irkutsk is a major trans-siberia hub and there would be services to convert to EUR.  The drawback is a controlled rate.
Comment: People sometimes buy it to fly in Europe or to create bank-saving account in Euro.
Answer: It's of course possible to exchange currency in Irkutsk. It's a big city with a population of some 600 thousand, roughly the size of Helsinki, Stuttgart or Lisbon, so it certainly has a lot of banks and currency exchange points, any of which can be expected to accept Euros. A brief search shows 105 exchange points in the city, you will surely be able to find one close to your location in the city, especially if you arrive by rail, which is a major transportation method there.
This is a page showing the current exchange rates for EUR in Irkutsk. The three major numbers, left-to-right, are: the central bank's exchange rate, best buy rate in the city, best sell rate in the city. As of this moment, the best rate for selling EUR is at Гринкомбанк (Grinkombank), at 74.1 rubles per Euro, with the official exchange rate of the Russian central bank being 73.63.
Comment: @Max, buying euros or dollars are usually not a big problem, as long as you do not need small bills (say 20 or below). 50's, 100's and 500's are usually ready available. Your experience is unusual, maybe it was due to the area you were in (maybe in airport they have shortage of currency as everybody buys it, but that'll be still strange), or maybe you tried to buy in a currency rush — with all these changes to ruble exchange rate in past 1-2 years, on some days a lot of people head for exchanges buying dollars and euros fearing that the exchange rate will soar even higher.
Comment: My experience in Moscow (which you would think would be even easier than Irkutsk) was that it was easy to exchange euros to roubles, but significantly harder to go the other way... many exchanges just refused to exchange; in the airport I found a place that was willing to sell me 500€, but no smaller amount. Did I encounter a currency shortage that day, or did people just not like my face (or the fact that I didn't speak Russian -- not a problem converting the other way)?
Answer: From my experience, Euro and USD are easily convertible in major Siberian cities, in both directions (and you can get a decent rate with some searching, although exchange rates and commissions do vary a lot). All other currencies are much more of a problem.
Answer: You will not trade with people, you'll deal with banks. Most of them, probably all, will provide currency exchange service.
However, in the recent years more and more shops, hotels and restaurants began to accept international bank cards (Master Card and Visa). I was surprised when even a small shop in a tiny village on Baikal coast allowed to pay by card. Taking this into account, you may not need to exchange a large sum of money: currency exchange will be performed transparently.


Why does God the Father need to anoint Jesus with His Spirit if Jesus is fully God?

Question: Isaiah 42:1-4 referenced again in Matthew 12:18. Similar to the descending of God's Spirit on Jesus at his baptism (Matthew 3). I don't understand how to explain this in alignment with the Trinity doctrine.
Comment: Christ means anoint. Messiah is a transliteration of the Hebrew Masheach which means anoint. If Jesus had not been anointed He would not be the Christ, the Messiah, the Masheach.
Comment: The anointing of Jesus is essential to the doctrine of the trinity but that is not the purpose of this site.  This question, in it present for would be better in Christianity SX.
Comment: Welcome to BHSX.  Thanks for your question and joining the group.  Please remember to take the tour (link bottom left) to better understand how this site works.
Comment: It is The Christ who is the anointed. And The Christ is manifested in humanity. It would be incorrect wording to say that the Son of God is 'anointed'. Anointing. from above, presupposes _that which is below_.
Answer: Jesus was fully human

Traditions flowing from before the Greek-Roman split about AD 1054 believe that Jesus's divinity and Jesus's humanity are not mutually exclusive. Usually this is in the form of Trinity or Oneness traditions. That is not in question here; it is either presumed or rejected. My answer presumes this.
This question is about why Jesus needs to do things humans do if Jesus is fully God.
My answer uses this as another reason why Jesus's divinity and humanity are not mutually exclusive.

Jesus's obedience
Matthew 3:14-15 (NASB, emphasis added)

14 But John tried to prevent Him, saying, "I have need to be baptized by You, and do You come to me?" 15 But Jesus answering said to him, "Permit it at this time; for in this way it is fitting for us to fulfill all righteousness." Then he permitted Him.

...or to fulfill all justice, which addresses the sacraficial law matters at the Cross, among others.
...Paul answers the same question as the OP because NT readers were also as curious as John was...
Philippians 2:8 (NASB, emphasis added)

Being found in appearance as a man, He humbled Himself by becoming obedient to the point of death, even death on a cross.

...appearance as a man from σχήματι εὑρεθεὶς ὡς ἄνθρωπος (SBLGNT) could mean being found as man...
This cuts to the core: Jesus, as a human, was obedient.

In ideas about the "hypostatic union" Jesus wasn't here to be a "superman". He truly was one of us in every way. Nothing excluded him from everything that any human would need to do.
Comment: #Jesse. You are imposing on Jesus what he refused. How is God remain fully God if he emptied himself. How can he be one of us in every way if he is fully God?
Comment: @AlexBalilo If you want to post your answer, you may. If you want to talk about the arguments for and against your statement, do a search on Christianity.SE.
Answer: Surely the clue is in Matthew ch12 v28; "If it is by the Spirit of God that I cast out demons..." Jesus was walking around Judah as a man, and therefore the Spirit presumably provided the more divine powers that he needed in his work.
One aspect of the doctrine of the Trinity is that Christ is fully God AND fully man, in conjunction, so the humanity of Jesus and the attendant limitations does not conflict with it in the slightest.
Comment: I agree with Alex;  it doesn't say that anywhere.  And as a general principle, if you need a long, involved discourse to answer a short, simple question, chances are you don't have an answer.  Long discourses mainly divert attention from the thrust of a short, simple question, the goal being to confuse the asker of the question.
Comment: Where in the bible does it say the Christ is fully God AND fully man?
Comment: Start a thread Alex and I will be happy to tell and show you!
Comment: @Mr.Bond. I am fine commenting. Are you happy to tell and show me in a comment?
Comment: @AlexBalilo No, the comment section is not enough for my explanation. I would be happy to take this conversation to chat.
Comment: @MrBond. I don't chat. But at least I know now that you need a long answer to answer my comment "Where in the bible does it say the Christ is fully God AND fully man? .
Answer: Simple answer: Because Jesus is NOT God!
Nowhere in the Bible does Jesus ever say that he is God, or that there is more than one God. Here are 10 reasons among many more that prove the Unitarian truth of Scripture:

Because Jesus Christ is represented by the sacred writers to be as distinct a being from God the Father as one man is distinct from another.  "It is written in your law, that the testimony of two men is true.  I am one who bear witness of myself, and the Father that sent me beareth witness of me," John 8:17, 18.

Because he not only never said that himself was God, but, on the contrary, spoke of the Father, who sent him, as God, and as the only God.  "This is life eternal, that they might know Thee, the only true God, and Jesus Christ whom thou hast sent," John 17:3.  This language our Saviour used in solemn prayer to "his Father and our Father."

Because he is declared, in unnumbered instances, to be the Son of God. "And lo, a voice from heaven, saying, this is my beloved Son, in whom I am well pleased," Matt. 3:17.  Can a son be coeval and the same with his father?

Because he is styled the Christ, or the anointed of God. "God anointed Jesus of Nazareth with the Holy Ghost and with power," Acts 10:38.  Is he who anoints the same with him who is anointed?

Because he is represented as a Priest. "Consider the ….High-Priest of our profession, Christ Jesus," Heb. 3:1.  The office of a priest is to minister to God.  Christ, then, as a priest, cannot be God.

Because Christ is Mediator between the "One God," and "men." "For there is one God, and one Mediator between God and men, the man Christ Jesus," 1 Tim. 2:5.

Because, as the Saviour of men, he was sent by the Father. "And we have seen and do testify that the Father sent the Son to be the Saviour of the world.  1 John 4:14.

Because he is an Apostle appointed by God. "Consider the Apostle,…Christ Jesus, who was faithful to him that appointed him," Heb. 3:1, 2.

Because Christ is represented as our intercessor with God. "It is Christ that died, yea, rather, that is risen again, who is even at the right hand of God, who also maketh intercession for us," Rom. 8:34.

Because the head of Christ is God. "I would have you know, that the head of every man is Christ; and the head of every woman is the man; and the head of Christ is God," 1 Cor. 11:3.

The entire Bible is unitarian, and wholly perfect.
Jesus is the Anointed One, the Son of the Living God. He is the viceroy of God and the mediator between God and man, who intercedes for us before the Father.
Comment: **Comments have been [moved to chat](https://chat.stackexchange.com/rooms/151592/discussion-on-answer-by-joshua-b-why-does-god-the-father-need-to-anoint-jesus-wi)**
Comment: But, he claimed to be God the Burning Bush of Moses in [John 8:48-59](https://www.biblegateway.com/passage/?search=John+8%3A48-59&version=NASB1995) *Jesus said to them, "Truly, truly, I say to you, before Abraham was born, I am." Therefore they picked up stones to throw at Him*. These verses can be interpreted other ways. But, there is some reasoning behind this answer to explain how some people view this, so it certainly can stay.
Comment: **Dear all:** This answers a hermeneutica question with a Bible-verse theology idea. It could draw some disagreement. Keep comments few and short so they don't need to be moved again.
Comment: @Jesse, the Angel of the LORD (Jesus) is the viceroy of God, which means that he personally relays messages from God to man and can speak to man as though he is God despite not actually being God, because God has given his Son the authority to do so as his only-begotten. This is why Jacob was puzzled that he "saw God and lived"; because he saw the face of God though the face of his Son. Thereby he was given the new name of Israel, for he had died to the old and became alive in the new working of God in his life.
Comment: You're talking about [Arianism](https://en.wikipedia.org/wiki/Arianism), which is theology, which we don't discuss on this site. You can, however, discuss that on [Christianity.SE: What are the biblical arguments against Arianism?](https://christianity.stackexchange.com/questions/36128/).
Comment: I am locking comments on this because we don't want the site too chatty. You could continue ideas in the [chat for this post](https://chat.stackexchange.com/rooms/151592/discussion-on-answer-by-joshua-b-why-does-god-the-father-need-to-anoint-jesus-wi).
Answer: He had to receive a bigger portion of Elijah's share of power, to be the next Elisha. Elisha described his 12 yoke of oxen for a new way of living, following God's plan closer. 12 tribes, 12 oxen holding the bronze sea at the temple, 12 stars in Revelation. It's about the same lineal legacy but through a remnant and new covenant (same paradigm with new particulars).
Also Christ was human, he carried God's essence but was ordained to be the perfect human and a vessel to allow God to suffer and/or empathize with us beyond outside observation (he could manifest however he wanted before but didn't have "skin in the game". I think Christ was a human body for the Angel of the Lord, didn't have an autonomous human soul. God can't die and isn't corporeal, but if he made a corporeal conventional human body (he did as Messiah) and martyred it, then he saw and felt everything he demanded of his people and specifically copied the patterns of his prophets and champions (many were nazirite miracle babies and proto christs, he built his own lineage through a nation and specific bloodlines).
Answer: Here is something interesting that might help give a different understanding as to why God is putting His Spirit on him.
I have seen this same scripture say,

"Lo"!my boy who I prefer! My beloved, and whom my soul delights!  I shall be placing My spirit on him, ... CLT

This is what that word often translated as servant means.

pais ►
Strong's Concordance
pais: a child, boy, youth
Original Word: παῖς, παιδός, ὁ, ἡ
Part of Speech: Noun, Feminine; Noun, Masculine
Transliteration: pais
Phonetic Spelling: (paheece)
Definition: a child, boy, youth
Usage: (a) a male child, boy, (b) a male slave, servant; thus: a servant of God, especially as a title of the Messiah, (c) a female child, girl.

Now, if you tie that in to a few verses later, it might open up even more understanding as to why the Spirit is upon him.

Then was brought to him a demoniac, blind and dumb, and he healed him, so that the blind and dumb both spake and saw. And all the multitudes were amazed, and said, '

"Is this the Son of David?'

But the Pharisees having heard, said, 'This one doth not cast out demons, except by Beelzeboul, ruler of the demons.'

The story goes on to show that if it was the by Spirit of God, then the kingdom of God has come upon you.  Mathew 12:22
So, with all that being said, a few things that are revealed through God's Spirit upon this son, is God's soul delighted in him , that this son in front of them was chosen by God.
It also made the crowd wonder if this was the son of David that was being revealed to them.
Through Jesus own words he had demonstrated that it was by the Spirit upon him by healing, a demon possessed man who was both blind and dumb. His healing was evident because he could now could now speak and see.
It was to reveal to Israel at that kingdom of God, was right there in front of them, because God's Spirit was upon this son which we know  is God's not only God's son but as David son also.
God Spirit is upon him, and by overcoming the head of demons, was proof of who this son of David was.  They were having a foretaste of the kingdom of God.

>And the multitudes that went before, and that followed, cried, saying, Hosanna to the Son of David: Blessed is he that cometh in the name of the Lord; Hosanna in the highest.
Matthew 21:9

"Blessed is the kingdom of our father David that is coming in the name of the Lord!" "Hosanna in the highest!"
Mark 11:10

So going back to half of the question why did God the Father need to anoint Jesus with his Spirit in the context of Mathew 12:18.
I would say it was to show Israel that this was the son of David, who was anointed by God, by his spirit.
Here is something from got questions which gives more understanding as to why that is so important.

The Davidic Covenant refers to God's promises to David through Nathan the prophet and is found in 2 Samuel 7 and later summarized in 1 Chronicles 17:11–14 and 2 Chronicles 6:16. This is an unconditional covenant made between God and David through which God promises David and Israel that the Messiah (Jesus Christ) would come from the lineage of David and the tribe of Judah and would establish a kingdom that would endure forever. The Davidic Covenant is unconditional because God does not place any conditions of obedience upon its fulfillment. The surety of the promises made rests solely on God's faithfulness and does not depend at all on David or Israel's obedience.

The Davidic Covenant centers on several key promises that are made to David. First, God reaffirms the promise of the land that He made in the first two covenants with Israel (the Abrahamic and Mosaic Covenants). This promise is seen in 2 Samuel 7:10, "I will provide a place for my people Israel and will plant them so that they can have a home of their own and no longer be disturbed. Wicked people will not oppress them anymore." God then promises that David's son will succeed him as king of Israel and that this son (Solomon) would build the temple. This promise is seen in 2 Samuel 7:12–13, " I will raise up your offspring to succeed you, your own flesh and blood, and I will establish his kingdom. He is the one who will build a house for my Name."

But then the promise continues and expands: "I will establish the throne of his kingdom forever" (verse 13), and "Your house and your kingdom will endure forever before me; your throne will be established forever" (verse 16). What began as a promise that David's son Solomon would be blessed and build the temple turns into something different—the promise of an everlasting kingdom. Another Son of David would rule forever and build a lasting House. This is a reference to the Messiah, Jesus Christ, called the Son of David in Matthew 21:9.

The promise that David's "house," "kingdom," and "throne" will be established forever is significant because it shows that the Messiah will come from the lineage of David and that He will establish a kingdom from which He will reign. The covenant is summarized by the words "house," promising a dynasty in the lineage of David; "kingdom," referring to a people who are governed by a king; "throne," emphasizing the authority of the king's rule; and "forever," emphasizing the eternal and unconditional nature of this promise to David and Israel.

Other references to the Davidic Covenant are found in Jeremiah 23:5; 30:9; Isaiah 9:7; 11:1; Luke 1:32, 69; Acts 13:34; and Rev 3:7
Answer: Answer
It is rather a simple answer if we closely follow the pure Scripture instead of the "sophisticated and cunning" (Mat 11:25) philosophers of the world.
God the Father needed to anoint Jesus with His Spirit "because Christ Jesus EMPTIED HIMSELF" of His infinite Godhead and became FULLY a human being (Philip 2:6-8).
Explanation
FIRST (Phil 2:6)
Jesus Christ was the powerful Yahweh of the Old Testament before His Incarnation. God the Father never interacted with the human kind directly (John 1:18; 6:46; 1 John 4:12) except through Jesus Christ (Gen 19:24; Hosea 1:7).
In Genesis 19:24, we see Yahweh on earth raining brimstone and fire from Yahweh out of the heavens.
Hosea 1:7 is also intriguing. Here Yahweh says that He will save the Jews by their Yahweh God!
What is significant is that the verse does not talk about the house of Israel but only about the house of Judah, that is, the Jews. And Jesus came to the Jews!
Hosea 1:7:

"And I (Jehovah) will have mercy on the house of Judah (only
Jews; not the house of Israel – verse 6) and will save them by
Jehovah THEIR God.

Yahweh says He will save the Jews through their God Yahweh!
And Jesus came to the Jews.
SECOND (Phil 2:7-8)
But, one time in history, Christ Jesus emptied His great infinite Godhead and fully became a human being.
Do not misunderstand me: Jesus, even when He was a man, was God Himself but without any divine attributes. He left His Godly glory with the Father (John 17:5).
"Emptied" is not an empty word. It has a meaning. It means He left His power! Jesus was not a "hybrid" Being.
Now Jesus is fully a mortal man. Anybody could kill Him with a knife and the Romans did kill Him. He hungered (Mat 4:2), felt thirsty (john 19:28), became tired and slept (Mat 8:24) etc.
He prayed to God His Father with great cries and tears (Heb 5:7). He didn't have His former infinite knowledge (Mark 13:32). He did miracles with the power from God the Father (John 5:30). He possessed great authority (Mat 7:29; 9:6) but He did everything through the Holy Spirit (Mat 12:28).
THIS IS WHY JESUS NEEDED AN ANOINTING OF THE SPIRIT!
THIRD (Phil 2:9-11)
After the Resurrection, Jesus received back the same glory He shared with God the Father before the foundation of the world, that is, His "emptied" Godhead was restored to Him by God the Father (John 17:5).
What is funny is that God the Almighty Father does not have any problem in calling Jesus as the Creator God who created everything (Heb 1:10-12). But some people do not like this idea and disagree with God the Father making Him a liar (1 John 5:10)!
Conclusion
Jesus is God in flesh (1 Tim 3:16) but without the Godly attributes. He fully depended on God the Father as a faithful Son.
Hence, He, a complete human being, required an anointing of the Holy Spirit.
[Some people, without understanding the inspired Word of God ask funny questions, like: how God can die, How God has blood, why God prays to God etc. Well, Philippians 2:6- 11 gives the plain answer]


Merge overlapping rasters with ignoring no data values QGIS3

Question: I thought this would be extremely basic task but obviously it's not. In QGIS3 GDAL Merge (GUI), I am trying to merge two rasters which overlap and each of them has a different distribution of nodata values. I want the rasters to be mosaiced together such that nodata values are ignored if one the rasters does not have nodata value on that spot. In ArcGIS it is that simple, why not QGIS?
Answer: Probably best performed using the command line. Yes there is the gdal_merge script, but an alternative approach is to use gdalbuildvrt with the component rasters and use gdaltranslate to create the final raster. See the answers to this question:
Merge thousands of rasters with gdal_merge.py
For my application, I had a lot of no-data values and did not have any problems. I had to use custom merge scripting though. So I created a python script which updated the VRT and applied the required Python functions.
Comment: Thanks a lot and yes, I saw this coming. Thought there would be a GUI solution since this belongs to a basic map algebra and needs just sort of an "override no data with valid data" option.
Answer: https://gis.stackexchange.com/a/306293/22646 suggests to go via a virtual raster (gdalbuildvrt or gdal_merge.py -of VRT).
If you would like to go down this path in QGIS, you could try the GDAL Build Virtual Raster algorithm (found in the Processing Toolbox).
You can choose as many Input (raster) layers as you like (the ... button). For Resolution (in overlapping areas) you can choose between Average, Highest and Lowest.
The resulting virtual raster can (for instance) be saved as GeoTiff (Layer-> Save As...).


How do they produce air on the ISS?

Question: How do they provide air on the ISS?  I know it is produced by electrolysis but is that all they breathe, hydrogen and oxygen?  The air here on Earth contains other things like nitrogen.  Do the people on the ISS not need nitrogen?  Or do they take what is needed on each launch?  If so how much water do they take and how long does a certain amount of water last when used for air?  Seems like a whole lot of weight.  Now they are planning a yearlong trip.  How much water will they have to launch with?
Answer: The ISS is using electrolysis, in the "Elektron" oxygen generator,

http://en.wikipedia.org/wiki/Elektron_(ISS)#Elektron

to separate water to hydrogen and oxygen. NASA claims that those devices don't work well and prefers bottled oxygen delivered from Earth which the Russians only consider a backup option. Another backup option are solid fuel oxygen generator canisters "Vika" producing oxygen from a liter of lithium perchlorate.

http://en.wikipedia.org/wiki/Vika_oxygen_generator

The atmosphere at the ISS has 101,300 Pa, just like on Earth, and contains all the right composition of gases we know from Earth. A typical astronauts' alternative is pure oxygen which is OK to breathe for humans but poses a higher fire risk.
The mass of the air the space station needs is rather modest. A typical person breathes 4-12 times a minute and one breath is 0.5-6.0 liters. Take 10 times 2 liters a minute, 20 liters a minute, 1,200 liters an hour, 30,000 liters a day. It may sound like a lot but 30,000 liters is just 30 kilograms or so and only 1/4 of it is "consumed" (oxygen), so what one needs is comparable to a few kilograms per day.
The Vika system above claims to give 1 person enough oxygen for 24 hours just from 1 liter of the lithium perchlorate (whose density is 2.4 kg/liter). To have enough oxygen for a 100-day mission, you have to deliver about the same mass of the lithium perchlorate for the person as his own weight. The counting is comparable with other sources, including the bottled oxygen (that occupies more room because the density is lower). The astronaut will need twice his mass of water to drink and one times his mass of food for those 100 days. Estimates. At any rate, what you need to deliver with the astronauts to allow them life for many months is themselves plus a small multiple of their mass in various food/water/air reserves.
See also a question here on cooling mechanisms and suits:

What kind of cooling mechanism could be used in outer space?
Comment: https://en.wikipedia.org/wiki/Breathing#Components claims that only about 1/20 of the 'breath' is consumed oxygen (21% while breathing in vs. 16% while breathing out), which gets you from 30/4 = 7.5 kg to 30/20 = 1.5 kg.
Comment: If I understand Wikipedia, even the electrolysis solution doesn't get O2 from CO2.  That implies that none of the options are a closed Oxygen system.  The same could be said for food, of course, but I just want to be clear on the method.
Comment: Of course, electrolysis always means it starts from water. They have to have enough water, which is a rather unproblematic compound to store. They actually need more water for drinking. And it's almost always new water. The algorithms recycling water... continue to be speculations so far. Pretty much nothing this important is being recycled. There are packages "in" with food, drink, packaged air in some form; and they create "garbage".
Comment: Well, there's a brand new machine making water out of trash CO2, see http://www.theregister.co.uk/2010/04/16/iss_sabatier_reactor/ - I believe those things remain fringe possibilities that are not really economically justified.
Comment: Thanks, Claudius, this correction makes all the numbers much more compatible.
Answer: In addition to Luboš's answer regarding the O2/CO2 requirement, the N2 aspect isn't as much of a problem as you don't 'use' it like you do O2.
Once you strip out the expired CO2 from the air and replace with O2 the N2 is still there as it was before, so you don't need to take up vast quantities of it from Earth. Which is rather lucky, as it obviously makes up the majority of the air we breathe.
Answer: Not sure I'm convinced by the decision to use N2/O2 to reduce fire risk.
Is fire much more of a problem in 0.2atm of pure O2 than in air?
The Apollo I fire was because some idiot put it under a 1.2Atm O2 partial pressure, even if you had added another 0.8atm of N2 it would still have been lethal.
If having a 1atm pressure is necessary for comfort (maintaining humidity, sinus issues etc) wouldn't you choose a different gas than N2? Perhaps He or Ar, it's not like cost is a big driver.

How do they produce air on the ISS?


Analyzing Likert scale data

Question: I am doing an university project, and a big part of my data is ordinal (Likert scale). In some of the hypothesis I have Likert scales both as DV and IV, so for example IV is I shop on a tight budget (1-7) and the DV is I tend to buy price reduced suboptimal food (again level of agreement from 1 to 7).
So under other circumstances I would use one way ANOVA, however as the data is ordinal, this does not seem a safe choice.
I have looked into running the Kruskal-Wallis test, however it is stated that it is not as punctual and people sometimes run ANOVA on Likert scales anyway.
So my question is, what do you think makes more sense to do? Or maybe there is another third option?
Thanks in advance
Comment: You don't say what null hypothesis you are testing? Are you comparing DV and IV groups to see if their Likert means or medians are signif. different? // ANOVA and K-W are for comparing more than 2 groups. Maybe your choice is between Welch t and 2-sample Wilcoxon. // Please say what you are trying to find out and what null and alternative hypotheses you have in mind.
Answer: Here are some simulated data for DV and IV for purposes of
illustration.
set.seed(2020)
dv = sample(1:7, 100, rep=T, p=c(1,2,2,3,3,4,4))
iv = sample(1:7, 150, rep=T, p=c(1,2,4,3,2,1,1))

tabulate(dv); tabulate(iv)
[1]  6  8 11 21 13 21 20  # DV
[1] 13 27 49 20 18 11 12  # IV

summary(dv); summary(iv)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.  # DV
1.00    3.75    5.00    4.70    6.00    7.00
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.  # IV
1.00    2.00    3.00    3.56    5.00    7.00

Roughly speaking, DV Likert scores seem to be generally higher than those for IV. A two-sample Wilcoxon test shows that DV is significantly (P-value near 0) 'shifted to the right'
compared with IV.
wilcox.test(dv, iv)

        Wilcoxon rank sum test with continuity correction

data:  dv and iv
W = 10236, p-value = 7.337e-07
alternative hypothesis: true location shift is not equal to 0

An 'empirical CDF' plot of data is a stairstep plot based on sorted observations. Suppose there are $n$ observations. Starting at height $0$ at left, if there are $k$ observations at a value, the plot jumps up by $k/n$ at that value. At the right the ECDF plot reaches height $1.$
If the ECDF of one sample lies mainly to the right
of the ECDF of another, then the first sample is said to be
located to the right of the second. Here DV is located to the
right of IV.
plot(ecdf(dv), col="brown", xlim=c(0,8), lwd=3,
     main="ECDFs for DV (orange) and IV Scores")
  lines(ecdf(iv), col="cyan3", lty="dotted")

Doing a two-sample t test makes the assumption that Likert scores
are numerical (not merely ordinal). Among other things that implies
it makes sense to sum Likert scores in order to compute sample means.
Some researchers make this assumption routinely, and others feel
uncomfortable making it.
So doing a t test is somewhat controversial.
However, if the assumption is warranted, then there is no question
that the t test detects a significant difference in means for my
simulated data.
t.test(dv,iv)

         Welch Two Sample t-test

data:  dv and iv
t = 5.0286, df = 200.12, p-value = 1.095e-06
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 0.6929647 1.5870353
sample estimates:
mean of x mean of y
     4.70      3.56

Analyzing Likert scale data


Illustrator: which settings to design a logo for print

Question: What settings should I use when creating a document for logo printing? Looking to achieve the best quality for my logo design.
Comment: Illustrator has a "Print Document" preset in most versions, from at least CS4 and later.  Choose that preset when you create a new document.
Answer: It does not really matter, the settings in the dialog have no effect on quality of your output if your logo is all vector as it should. The only thing that might adversely affect your quality is the raster effects setting, which you carank up later anyway.
There is just no quality metric in the dialog.
Comment: @creator well, different printers operate differently, and how you package a file depends on where its used. So a file suitable for embedding and printing in say Word or latex is different from a file you would send a offset print. You would store the original logo in atleast 2 separate ai files, one for cmyk and one for RGB use. For Word you would say 2 hallelujahs and pray for divine intervention, for latex you use eps and for the offset printer you would send a suitable version of PDF-X that they can handle.
Comment: and what file type should I use to get the best quality for printing sir?
Comment: @creator depends on what you print with.
Comment: what do you mean? sorry (newbie)
Answer: When creating a new document for print, make sure the 'Color Mode' is set to 'CMYK' and 'Raster Effects' to 'High (300ppi)'. Normally these should be the default settings in AI's 'New Document' panel.
Comment: and what file type should I use to get the best quality for printing sir?
Comment: Save your document in AI format.


Terminal App not found

Question: I am new to Lineage but not new to Linux. I am running on a raspberry pi. How do I find the AppDrawer? How do I find the terminal App. Many thanks
Comment: Based on my personal experience for LineageOS on Nexus 5, LineageOS has an in-built Terminal app, but it should be enabled first from Developer options under the name "Local terminal".
Comment: Please keep in mind that Android may be Linux based it is not a Linux system you have worked before. The whole user-system and also the UI-system is totally different from any Linux distribution I know. Therefore I recommend to you to not see Android as a Linux system but as a new system unknown to you that re-uses some basic parts of Linux.
Comment: https://android.stackexchange.com/a/16244 https://android.stackexchange.com/a/218507
Answer: Not sure how LineageOS works on raspberry pi.
But on android smartphones there is no real terminal. the closest to terminal is terminal emulator which can be installed as usual apk (example)
The AppDrawer is probably the Android Launcher which can be installed as apk too (example)
There is also a built-in app list in Settings - Apps from where you can uninstall or clear data (and even start apps on older android versions), which is therefore some kind of AppDrawer too.
On Smartphones another terminal access is the Android Debug Bridge (adb) which is even accessible when system_server is crashed or restarted (assuming the initial setup is done from developer options). Most likely on raspberry pi this is already enabled (and indeed without physical display/keyboard the only way to access your device)

Terminal App not found


Rails LoadError when trying to start server

Question: I'm following SaaS from edX and I'm trying to run my rails server that I created using
rails s

and it keeps throwing a load error for
gem/ruby-1.9.3-p448/gems/ruby-debug-base19-0.11.25/lib/ruby_debug.so:
undefined symbol: ruby_current_thread

I'm not quite sure what I can do to fix this, or what other info would be required to help me solve this. I'm sorry for such a vague post but ruby on rails is rather new to me and I'm not sure why its throwing this
ive tried
gem install rails

and i get a permissions error
*Edit 1
Gemfile The debug19 is what i was told to put in per my instructions from edX 169.1
source 'https://rubygems.org'

gem 'rails', '3.2.14'

# Bundle edge Rails instead:
# gem 'rails', :git => 'git://github.com/rails/rails.git'

gem 'sqlite3'

# Gems used only for assets and not required
# in production environments by default.
group :assets do
gem 'sass-rails',   '~> 3.2.3'
gem 'coffee-rails', '~> 3.2.1'

# See https://github.com/sstephenson/execjs#readme for more supported runtimes
# gem 'therubyracer', :platforms => :ruby

gem 'uglifier', '>= 1.0.3'
end

gem 'jquery-rails'

# To use ActiveModel has_secure_password
# gem 'bcrypt-ruby', '~> 3.0.0'

# To use Jbuilder templates for JSON
# gem 'jbuilder'

# Use unicorn as the app server
# gem 'unicorn'

# Deploy with Capistrano
# gem 'capistrano'
gem 'haml'
# To use debugger
# gem 'debugger'
group :development ,:test do
  gem 'ruby-debug19' , :require => 'ruby-debug'
end

Comment: can you link to the blog entry you say you're following?
Comment: also, can you post the contents of your Gemfile? I'm surprised you have ruby-debug-base19 in there at all.
Answer: I don't know enough about your environment to give an exact answer, but you can try removing ruby-debug-base19 and ruby-debug from your Gemfile. You can also see if it's being installed by another gem by looking for references to them in your Gemfile.lock. If they are dependencies, remove the gems they depend on as well.
Comment: They were added in by me, as to the course im following the enviroment is the one i was given by them.


Diagnoses Live OS

Question: I was wondering if there is a live ubuntu or linux in general OS built for running hardware and software diagnostics no matter what condition the hardware is in.
Answer: Yes, there are several Live Linux to do hardware diagnostics. Such as system rescue cd or grml. System rescue CD is based on Gentoo. Of course, You can use Ubuntu Live CD also for using partitioning and other stuff. It has Gparted software installed.

There is also Ultimate Boot CD which can be used as diagnosis tool
You can also try parted magic
For network diagnostics Backtrack is always proved great.

But note that, In a regular Ubuntu Live CD, you can install any software from repository and use this same Live CD to solve any of your problem which can also be solved by other CD's. It is an invaluable feature of Live CD.
This page from distrowatch lists some of the rescue CDs or distros. You can take a look at there.
Comment: Alright thank you, if you look at my previous questions such as the one on super user, my computer has all but been killed by a update to windows vista on one of its partitions, I've never seen a update screw up so many things before. Also if you could list a few others besides SR and grml it would be awesome, I find it's good to have variety.

Diagnoses Live OS


Stop Action on Navigating event of WebBrowser Control

Question: I have created a little Win Form App in C# and added the WebBrowser component to it. What i am trying to achieve is a little app that can load a local html page from a file which has "custom" protocols in it and can of course also navigate to a web address.
For example i would have entry as follows in my webpage
'Close Company
' which would open a task in a program.
The way i tried to achieve this was via the Navigating event as shown below
       private void webBrowser_Navigating(object sender, WebBrowserNavigatingEventArgs e)
       {
        if ((webBrowser.StatusText.Contains("Special")))
         {
            //For some reason the stop doesn't do much it still tries to proceed to special:123
            //diplaying can not load page..
            webBrowser.Stop();

            //Launch program here.
            MessageBox.Show("Special Command Found");
         }
       }

Problem is that it still navigates and says it can't find of course the page.
I swapped Stop with GoBack which for some reason has the same issue the first time i run it and when i then select backward in the browser it works from thereon.
I also tried navigated and use of GoBack, besides having a flashing in the app from going back the event does not fire again after the first time anymore.
Has anyone any ideas how to solve this or what i am doing wrong here ?
Answer: Instead of using WebBrowser.Stop();
just set e.cancel = true;


Compile SD file in ArcGIS Pro

Question: Is there a way to build Service Definition files in Pro? I assume not since they have everything on lockdown to Portal. Thought I would ask here though.
Answer: You can build .SD (service definition files) in ArcGIS Pro. Only using Python, no "GUI" way to do it.
See CreateWebLayerSDDraft command.
Following the building of the SDDraft and Staging into a SD, you'd use the Upload Tool using the proper key word (My Hosted Services) to share the .SD file to your portal that you've authenticated too.
Comment: Interesting! Can I assume it will still require Portal or Online to use? Still a no-go for Server?
Comment: @Midavalo I haven't tested every service type that has a Create_X_SDDraft command, but for this one in particular, CreateWebLayer, I'm almost positive it only works with MY_HOSTED_SERVICES as the server type, as such, when using the UploadTool, only to My Hosted Services (Portal) - this command only exists in Pro. The other commands which exist at 10.x and Pro may work directly to Server, you'd need to experiment to know. (Although I'd say that it wouldnt be officially supported)
Answer: No.

ArcMap publishes data to ArcGIS Server to create web services. ArcGIS Pro shares content to a portal (ArcGIS Online or Portal for ArcGIS) and can create a web map, web scene, web layers, and their associated web services.

Therefore you would need to use this functionality to publish from ArcGIS Pro.
See this reponse from Esri's Derek Law on the Geonet forums, and his linked UC Q&A (you may need to scroll down to see the Q&A answer, but it should be highlighted on that page).
Answer: Thought I should update with a new feature in Arcgis Pro. This can be done with arcpy, and there is a simple way in the GUI right click layer, save as a service definition file (portal or standalone server)

Compile SD file in ArcGIS Pro


Rosemary turning yellow

Question:
I've moved the rosemary to this pot a month ago, and where I live, the temperature is around 28-30 degrees celsius these days and usually i water them on a daily basis.
Everything looks just fine and even great, except for the color. Is it a soil problem? What nutrient is lacking?
Comment: Possibly just adaptation of new pot. Which dirt did you use? "One month ago" should be so short that extra nutrients should not be needed. Try to reduce water (e.g. 4 or 5 day a week).
Answer: Rosemary is drought resistant, which means that it likes a bit less water than some other plants. Try watering only after you check by digging your finger into the soil and checking if it feels wet. You do not have to wait until it is bone-dry, just not really wet.  Also, rosemary is a Mediterranean type of plant and it might be somewhat dormant yet in those low temperatures.
Comment: 28-30 *celsius* isn't a low temperature. It's 80 to 85 F. Of course if the temperature was 28-30 *F*, the plant wouldn't be doing anything at all.
Comment: Oh, how silly of me!
Comment: Yeah, the thing is, the soil gets dry everyday in this season and sun and wind are very good at my terrace. That's why I water them everyday. Anyway, thanks for your advise! Everyone can be confused with Fahrenheit and Celsius. :)
Answer: So long as it is old leaves going yellow and there is new growth at the tops of the shoots, there is nothing to worry about. Rosemary likes full sun in summer (and more sun generates more essential oils) but it usually ends up looking a bit "weather beaten" by the end of the summer.
Comment: Thanks. But I wasn't sure about that because not the oldest leaves turned yellow  but the leaves that are not too old and not very new. And the overall color got brightened which was very dark green before.. Hmm... What you mean by "weather beaten" is the yellowish color?

Rosemary turning yellow


What do these red and blue icons mean in Adobe Premiere?

Question: What do these red and blue icons mean in Adobe Premiere?
Answer: The blue icon on the clips means your clip has a proxy file attached to it, and the red icon means the proxy file is missing from that clip.
Proxies are lower quality versions of your video clips that you can generate using Premiere and Media Encoder.  The advantage of using them is that when you are using very large video files, Premiere can become slower and less responsive.  When you generate smaller proxy files, Premiere uses those instead if you click the "Use Proxies" button in the transport controls:

If you can't see the "Use Proxies" button, you might need to click on the plus icon in the transport window to edit which buttons are visible:

There's a good tutorial explaining how and when to use them here:

and Adobe's page is here:
https://helpx.adobe.com/uk/premiere-pro/using/ingest-proxy-workflow.html


Reaction Order Greater than 1

Question: What is the chemistry behind why reactions of order >1 exist? And what role does the number of activated complexes formed have in determining reaction rate?
Comment: I believe you should breakdown your question into different questions. All of them are good questions, but together, the compound question becomes too broad to answer.
Comment: I agree with @SatwikPasani, breaking these down will help the answers stay organized.
Answer: The following answers might be vague or overtly simplified or basic, but the questions by nature are fairly open-ended, aren't they? feel free to inquire further if something isn't clear enough.

Take $$\ce{2NO2 \to 2NO + O2}$$ You can (extremely basically) think of this reaction as two $\ce{NO2}$ particles colliding, (forming a complex), then coming out as $3$ molecules, $2$ being $\ce{NO}$ and $1$ being $\ce{O2}$. But you needed two $\ce{NO2}$ molecules to collide. Collisions are unlikely, first the two molecules have to find each other, then they have to have enough energy, and be in the right orientation.
I'm not sure about multiple activated complexes, but I would only imagine that the reaction becomes even less likely if more than one activated complex is necessary for it; that basically means you need a second reaction from the first activated complex, which means a second collision. (likely slower rates? Someone else feel free to correct if I'm wrong)
Three molecules colliding at once ($3$ rd order reaction): way less likely, and that's why it's not studied so much, it's much harder getting $3$ molecules to collide (especially in the gas phase without) in the right orientation with enough energy.
By general rate law do you mean finding the constant $k$? Or deriving its form entirely? Finding the orders for each reactant and the constant can only be done experimentally by using different concentrations and seeing what comes out.
I think there is more than one way to derive the general rate law's form. One way is to take the steady state approximation. It has a useful problem that might explain more clearly. The steady state approximation is a commonly made assumption in various situations. Essentially, it assumes that the intermediate is consumed as soon as it's produced (so there's no net accumulation of intermediate – which makes sense for many situations). Using the info gained from this approximation you can use the intermediates and the relations that result to form the general rate law. Of course, it turns out the $k$ is composed of multiple constants.
It's true because they told you it was. This gives you details that you wouldn't have necessarily been able to deduce just by having the reaction equation there. Importantly, it's there to tell you straight up that the equation you have can't be broken down to further reactions... Again you can think of it like this: $\ce{a A}$ molecules have to collide with $\ce{b B}$ molecules to produce a single activated complex. That's one step. If they didn't specify that, maybe there could have been intermediate steps that could change the rate law.
Maybe it'll be easier to see if you consider extremes. Say Reaction $R$ can be divided into $4$ reactions, $r1, r2, r3, r4$. Now, say $r1, r2$, and $r4$ happen in microseconds, while $r3$ takes $100$ seconds. Already you're at the point where you can basically ignore the times of $r1, r2$, and $r4,$ because they're so small compared to the time $r3$ took (comparing $1\mathrm E-6$ to $1\mathrm E+2$ here). So as you can see in this example, $r3$ was literally the rate determining step.

http://www.science.uwaterloo.ca/~cchieh/cact/c123/steadyst.html
Comment: I have closed this as too broad, but when the OP breaks down the questions, you can rearrange your responses for each one and paste them as answers to the new questions.
Comment: Yes, for question 2 I do mean entirely. Why is each component of the rate law correct and significant in determining rate? I get the idea that higher concentration means higher rate, but why do we even have a constant?
Comment: Also, with regard to question 3, the answer was not helpful because it doesn't address why mole ratios in a single-step reaction equate to corresponding reaction orders in A and B. Please elaborate.
Comment: Hopefully someone else can answer #2 in depth.

For 3: The problem is specifying the conditions for you. You need experimental data to determine rate law for a chemical reaction. Just looking at the mole ratios isn't a valid way of determining the rate law. Hope that clarifies both a bit of 2 and 3.

Reaction Order Greater than 1


Modified By column Manipulation

Question: Is there is anyway so that the Modified By column is not updated at first when we create it but be updated when modified?
Comment: Could you give some details about what you are trying to achieve?  Why do you wish to prevent the Modified By column being populated?
Comment: I am creating a list for nominating employees for awards.
I have used the 'Created By' column as 'Nominated By'. I need a column which updates the one who approves this nomination.
I was hoping to achieve this by using 'Modified By'. But this column is updated when created and shows the Nominated By and Approved By with same names.
Answer: An easy way to do this would be to not worry about the ModifiedBy, (which you should never really mess with default anything in SP) and create you a multiline text field and make it required.  If you append changes to it, SharePoint will put the user below the comment.  Just have them put approved or whatever in the multiline text field and there you go.
Answer: You need to create a new column called 'Approved By'.  Go in the List Settings (assuming you are using a List).  Scroll down the menu and find the option to Create column>name your column>The type of information in this column is: Person or Group.
Before you save your new column review the other options on the page, under the heading Additional Column Settings.  There is something called 'Choose From' All Users or Groups - you might want to consider making an Approvers group to which you can add the names of your approvers, this will limit the range of possible approvers to what you specify (perhaps managers / supervisors?).
Besides this you might want to make a simple approval Workflow which emails an Approver when something needs their approval.
Comment: With this, the person has to manually type in his/her name as approver in list/Form. right?
Comment: Once you start typing Sharepoint will start making suggestions of names.  The list of names available to my Sharepoint system are based on Active Directory, though it's possible yours is configured differently (I'm no expert on the Active Directory side of things).  Also the names available to you can be determined within the field settings (i.e. your new Approvers group - if you only include 3 names in your Approvers group, there will only be 3 available to choose from).

Modified By column Manipulation


Disable xDB in Multisite Website

Question: Can we enable xDB for a specific website and disable it for other? I know we have this option for xdb tracking but what about Xdb.Enabled? can this setting be true/ false for a specific site?
Comment: What is your purpose? Means what do you want to achieve?
Answer: Change the Xdb.Enabled and Xdb.Tracking.Enabled settings to enable or disable the tracker and the xDB. Depending on your requirements, you can choose to:

Disable the xDB and disable tracking.

Disable the xDB and enable tracking.

Enable the xDB and disable tracking.

Enable the xDB and enable tracking.

You can control tracking by site definition like the below -
<site name="customwebsite" enableTracking="false" ... />

Note - If the Xdb.Tracking.Enabled is set to false, tracking is disabled for all sites - even if enableTracking is set to true on an individual site node.
The following table describes the effect of enabling/disabling the xDB and tracking in different combinations.

Follow this article for details - https://doc.sitecore.com/xp/en/developers/91/platform-administration-and-architecture/enable-disable-the-xdb-and-the-tracker.html
Update - You can Disable Analytics on an item. Try this to apply on Home items and see if this is also disabled on child items because I never used this feature.
To disable or enable analytics on an item:

In the Content Editor, expand the content tree and navigate to the content item that you want to change the analytics setting for.

On the ribbon, on the Analyze tab, in the Attributes group, click Attributes.

In the Attributes dialog, click the Settings tab and select or clear the Disable Analytics for this Item check box to assign this setting to the content item.
When you select Disable Analytics for this Item, the item is no longer visible in the visitor session trail or any other reports in Sitecore.

Click OK to save your changes and close the Attributes dialog.

Follow this - https://doc.sitecore.com/xp/en/users/93/sitecore-experience-platform/disable-analytics-on-an-item.html
Comment: What is your exact requirement, there is no OOTB way to disable xdb for a particular website.
Comment: I want to disable XDB not tracking, can that be done specific to website ?
Answer: You can enable/disable tracking per site:
To disable tracking for a particular site:

Create a patch file for the
\path\to\web\root\App_Config\sitecore.config configuration file.
In the site node, set the enableTracking attribute to true or false

<site name="customwebsite" enableTracking="false" ... />

If the Xdb.Tracking.Enabled is set to false, tracking is disabled for all sites - even if enableTracking is set to true on an individual site node.
Answer: XDB can not be disabled for a specific site in a multisite setup either it is on for all or off for all.

Disable xDB in Multisite Website


Does a particle's rotational kinetic energy change when it enters a uniform magnetic field?

Question: Consider a charged particle entering a uniform magnetic field with some constant velocity perpendicular to the magnetic field. I know that the particle will follow a circular path within the magnetic field, and the magnitude of the velocity will stay constant. Hence, the particle will have a uniform circular motion. Since the particle starts with zero angular velocity and ends with some nonzero angular velocity, does the particle's rotational kinetic energy change when it enters the uniform magnetic field even though the work done on the particle by the external magnetic field is zero?
Comment: "I know that the particle will follow a circular path within the magnetic field and the magnitude of the velocity will stay constant." That is wrong. The electron emits photons, follows a spiral path, exhausted its kinetic energy and came to standstill.
Comment: @HolgerFiedler Thank you for that comment. Although, I am considering classical mechanics here so the charged particle will not emit photons.
Answer: Your mistake is in saying that the initial angular velocity is $0$. When the particle enters the magnetic field it already has a non-zero angular velocity about the center of the circular path it will start following. Therefore, it's rotational kinetic energy does not change. This makes sense, since there is no torque about the center of the circle acting on the particle.
Comment: Thanks Aaron, you've given me the insight I needed. I calculated the angular momentum before and after the charged particle entered the magnetic field and found that the change in angular momentum is zero. Therefore, the net external torque on the charged particle is zero. Hence, the rotational kinetic energy does not change.
Comment: @SchwarzKugelblitz It has angular velocity before entering the field. An object doesn't need to be moving in a circle to have angular velocity about some point.
Comment: How did it obtain this initial angular velocity even? This seems strange as torque acting on it even during the moment when it entered the magnetic field's region is 0
Comment: @SchwarzKugelblitz It got it from whatever gave it its initial velocity to begin with
Comment: let's say it was projected straight by giving it an initial impulse into the magnetic field and it travelled some distance in a straight line before entering the magnetic field. Now upon entering the magnetic field, it gains an angular velocity. Now what would be the cause of the gain in angular velocity?
Answer: I am not sure if I understand what you mean by particle's rotational kinetic energy. If by particle, you mean an electron, the rotational kinetic energy is meaningless. If by particle you mean a conducting or non conducting sphere (hallow or solid), then I can answer the question.
If the sphere rotating as if it is attached by a string (or spring!) and some one is rotating it, at each moment you get an additional magnetic force component which is added to the force of the string. This can increase or decrease the angular velocity and radius of rotation. But changing the radius also changes moment of inertial and thus the total change in the rotational kinetic energy remains zero.
If by rotating you mean spin the answer will be as follow:
Conducting sphere: before the sphere enters the field, the electrons are rotating. As the sphere enters the field the electrons enter the field at different directions of velocity. So they will move on the surface of the conductor based in the direction of their velocities. If the sphere is very heavy and the magnetic field is weak the sphere moves on a straight line with some surface current. if the sphere is light and the magnetic field is very strong the sphere will stray from the straight line.
Non  conducting sphere: since the electrons cannot move (at least in small magnetic field) nothing happens and the sphere moves on a straight line. What happens to the electrons and atoms inside the non conducting sphere depends on the material and the bound of the atoms. Again if the sphere is light and the magnetic field is strong enough the sphere will stray from the straight line.
Comment: The OP is referring to motion about the center of the circular path. Not the spinning of the particle itself.


What's my confusion with the chain rule? (Differentiating $x^x$)

Question: When deriving $x^x$, why can't you choose $u$ to be $x$, and find $\dfrac{d(x^u)}{du} \dfrac{du}{dx} = x^x$? Or you could go the other way and find $\dfrac{d(u^x)}{du}\dfrac{du}{dx}$, giving $\ln(x)\cdot{x^x}$? Both methods seem to be equally wrong.
Comment: Short answer:  both wrong.  If you have two instances of the same variable you can not substitute just one of the out and then the other.  You must substitute them both out *simultaneously*.
Comment: @fleablood And that's why we don't take math variables to court.
Comment: So suing x for u means you must derive $\dfrac {d u^u}{d u}\dfrac {du}{dx} $ which helps you not one #@%&ing  bit.
Comment: $\dfrac {x^u}{du} =\ln x x^u$ implies the base $x$ is a constant with respect to $u $.  As $x=u $ this is not the case.
Comment: Coz there's an interplay between the two "x", you can't simply separate them up.
Answer: Both methods are wrong, but the fix is easy: the solution is the sum of the two proposals, and this is not by coincidence !
Naturally, turning a single instance of $x$ to a constant cannot be the way as that is not symmetric. The correct way is by differentiating on every instance in turn, and is justified by the chain rule with partial derivatives:
$$\frac{df(u,v)}{dx}=\frac{\partial f(u,v)}{\partial u}\frac{du}{dx}+\frac{\partial f(u,v)}{\partial v}\frac{dv}{dx}.$$ In other words, you keep one instance variable while the other remains constant and sum the two cases.
Here, $f(u,v)=u^v$ with $u=v=x$, and
$$\frac{dx^x}{dx}=\frac{du^v}{dx}=vu^{v-1}\cdot1+\ln(u)u^v\cdot1=x^x+\ln(x)x^x,$$ or with a more intuitive notation$$\frac{dx^x}{dx}=\frac{dx^v}{dx}\cdot1+\frac{du^x}{dx}\cdot1=vx^{v-1}+\ln(u)u^x=x^x+\ln(x)x^x.$$

This works with as many instances of $x$ as you like. For instance $x^{x+x^2}$ seen as $u^{v+w^2}$ yields

varying the first instance, $(v+w^2)x^{v+w^2-1}$;
varying the second instance, $\ln(u)u^{x+w^2}$;
varying the third instance, $\ln(u)u^{v+x^2}2x$.

Then globally
$$(1+x+\ln(x)(1+2x))e^{x+x^2}.$$
Comment: While your answer gives a very good strategy, it does use a complex tool (theory of functions of multiple real variables) to solve a reasonably simpler problem and it does not really help OP in alleviating the confusion about chain rule.
Answer: If you work with the formal definition of the chain rule, you'll see how what you're trying to do makes no sense.
But if you want to stick with the abuse of notation $\frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}$, I'd say that the heart of the problem is in your claim that $\frac{d(x^u)}{du}=x^u\log x$. This is only valid if $x$ is constant, and doesn't apply if $x$ is a function of $u$ (in our case, $x=u$).
That's the difference between a total derivative $\frac{d}{dt}$ and a partial derivative $\frac{\partial}{\partial t}$. The latter, $\frac{\partial f(s,t)}{\partial s}$, means, "change in $f$ when $s$ changes and nothing else does". Whereas $\frac{df(s,t)}{ds}$ means "change in $f$ when $s$ changes, and everything else changes accordingly". So you can't have $u$ depend on $x$ and calculate a total derivative in a way that assumes $x$ is constant.
Comment: Out of interest, why do you consider the Leibnizian version of the chain rule to be an abuse of notation?
Comment: @Joe: Normally, when people write $dz/dx$ they don't literally mean dividing $dz$ by $dx$ (defining differentials as actual objects is niche at best). It's just a piece of notation that means "the derivative of $z$ as a function of $x$". Since it's not really a division, you can't just assume that the $dy$ "cancels out" as if it were actual division. That said - being an abuse of notation doesn't make it incorrect. It is correct, and can be proven by replacing each notation with a more rigorous one.
Comment: That's true, and one other issue with Leibnizian notation is that it conceals the fact that the fact that we are trying to differentiate a composite function. $dz/dy$ means 'the derivative of $z$ with respect to the inner function $y$', but this is not made explicit in the notation. The problem with this becomes more apparent when we consider $f(x)=x^x$, since neither the base nor the exponent is an inner function. This problem disappears when we write it as $f(x)=e^{x \log x}$, since then it becomes clear that the actual inner function is $x\log x$.
Comment: (Ironically, I'm abusing the word 'function' in my above comment, but you know what I mean...)
Answer: Both are wrong, since in spite of choosing $u=x$, you are replacing only one variable $x$ by $u$ while leaving the other $x$ intact. And then again, you decide to differentiate with respect to $u$ by chain rule, initially treating $x$ as a constant in $x^u$ and in $u^x$, which again is wrong.
What you should do is:

Write $x^x$ as $e^{\ln x^x}=e^{x\ln x}$ and then you should
  differentiate with respect to $x$ using chain rule.
Comment: With dependent variables, you don't have to substitute all occurrences of one variable with the other; in fact it is often convenient *not* to do so and write expressions that use both variables.
Answer: Other answers answer this question well and, but I guess another way to find the error might be helpful to some. When you know you've done something wrong but don't know what it is, it's often a good idea to try to use the same method with simpler, even ridiculously simple, examples, and see where it goes wrong. At least I've used this approach with success.
Let's try. What about differentiating $x$? That's probably not instructive because you don't have two separate $x$s there so you really can't apply your idea. But let's "cheat" a bit by defining $f$ to be a constant function, say $f(y)=1$ for all $y$, and differentiating $x f(x)$ (which equals $x$ of course).
$$
\frac{dx}{dx} = \frac{d(xf(x))}{dx} =  \frac{d(xf(u))}{du}\frac{du}{dx} = x \frac{df(u)}{du} \times 1 = 0.
$$
But because $f(u)$ is a constant $1$, this would be equivalent to writing
$$
\frac{dx}{dx} = \frac{d(x\times 1)}{dx} = \frac{d(x\times 1)}{du} \frac{du}{dx} = 0 \times 1,
$$
or to simplify even more,
$$
\frac{dx}{dx} = \frac{dx}{du} \frac{du}{dx} = 0 \times 1,
$$
So what went wrong here and why? Ah! $\frac{dx}{du}$ probably shouldn't equal $0$ when we have defined $u=x$. So (as others have pointed out) in the original problem the corresponding place is when we calculate $\frac{d(x^u)}{du}$ as if $x$ were a constant.
Comment: Downvoter: What would you improve in this answer?
Answer: Your approach is fine, but your execution fails — you didn't post the calculation so I'm guessing, but it looks like you did two things wrong:

You assumed $x$ is constant with respect to $u$ and vice versa, which is clearly false if $x=u$
You mixed up the rules for $z^{(\text{constant})}$ and $(\text{constant})^z$.

What we do have is that
$$ \mathrm{d}(x^u) = u x^{u-1} \mathrm{d}x + \ln(x) x^u \mathrm{d} u $$
or equivalently
$$ \mathrm{d}(u^x) = x u^{x-1} \mathrm{d}u + \ln(u) u^x \mathrm{d} x $$
The nice thing about differentials is that equations like this remain true no matter how $u$ and $x$ are related. One method to derive such equations is that the coefficients can be viewed as partial derivatives; basically the same as the method described at this other answer.
If $x$ and $u$ are independent, it doesn't make sense to ask for things like $\frac{\mathrm{d}(x^u)}{\mathrm{d}u}$, because $\mathrm{d}(x^u)$ simply isn't a multiple of $\mathrm{d}u$. But if they are (sufficiently smoothly) related, it does make sense (because $\mathrm{d}x$ will be a multiple of $\mathrm{d}u$).
The other nice thing about differentials is that if we do something like impose the relationship $x=u$, then the differential of this equation is also true: in this case $\mathrm{d}x = \mathrm{d}u$.
Applied to the first equation, we'd get
$$ \mathrm{d}(x^x) = x x^{x-1} \mathrm{d}x + \ln(x) x^x \mathrm{d} x $$
or equivalently,
$$ \mathrm{d}(x^x) = x^x (1 + \ln(x)) \mathrm{d}x $$
which indeed leads to the correct formula for the derivative.
Comment: Aside: it's interesting to think of this calculation as differentiating $x^x$ with respect to one $x$ and then with respect to the other $x$. I wonder if that is the reason for the name "partial differentiation"?
Comment: @WillNess: Thanks: fixed.
Answer: There is an easier method.I tried this out.
$u=x^x,$
(taking log on both sides)
$\log u=x\log(x)$,
$\frac{d(\log u)}{dx}=\frac{d(x\log(x))}{dx}$,
$\frac{1}{u}\frac{du}{dx}=\log(x)+x\frac{1}{x}$,
so, $\frac{du}{dx}=u(\log(x)+1)$,
i.e. $\frac{d(x^x)}{dx}=x^x(\log(x)+1)$
The entire reason for substitution is always to make the differentiable item to be less complex. As you asked if you take both ways putting u as x for the exponent or base, the substitution doesnt make the work easier for us. But taking a log and then differentiating makes things far more easier.
Comment: How does this answer the question?
Comment: @JiK : This answer is useful because it shows what to substitute as u. See, as far as I know we substitute complex variables only to make our job easier else these substitutions hold no purpose. So basically we need to know what to substitute and I think thats was misleading the guy. If he gets to know what to substitute and what not to and why we need to substitute things would be more easier I suppose.
Answer: I have mentioned in one of my answers about identifying the building blocks of a composite function in order to correctly apply chain rule. In the current problem where $f(x) = x^{x}$ you should ask the question: How do I get from $x$ to $f(x) = x^{x}$ in a step by step manner?
Your restriction is that each step must be a simple function which is not itself a composite function (i.e it can be a function like $x^{n}, \log x, e^{x}$ or direct or inverse trigonometric function of $x$ or a combination of such functions via arithmetic operators $+,-, \times, /$, idea is that each step must be a function whose derivative you know beforehand). The answer to the question in previous paragraph is not that I put $x$ as the base and also as exponent and get $x^{x}$. You need to apply the basic functions listed previously on $x$ in step by step manner. The only way out is to realize that the function $f(x) = x^{x}$ is defined as $f(x) = \exp(x\log x)$.
Then we see that we first apply $\log$ on $x$ and multiply it with $x$ to get $x\log x$ and then apply $\exp$ on this result. We thus have $f(x) = g(h(x))$ where $h(x) = x\log x$ and $g(x) = \exp(x)$. Applying chain rule is now easy and left to the reader.

It appears (from the downvote) that someone is not really satisfied with the answer. I add some details on exactly why replacing one of the occurrences of $x$ by $u$ in $f(x) = x^{x}$ is a wrong application of chain rule. Replacing one of $x$ by $u$ we either get $x^{u}$ or $u^{x}$. Unfortunately we don't know how to differentiate any of these functions unless we are given that $u$ is a constant (remember that we are given differentiation formulas only for functions $x^{n}$ or $a^{x}$ which assume that either base or exponent must be constant). Working in this manner does not really express $f(x)$ as a composite function i.e. it does not express $f(x)$ as some function of $u$ (whose derivative we already know) where $u$ is another function of $x$ (whose derivative we already know).


How to consider electric potential energy for two moving point charges?

Question: this is my first question here, I'm sorry if I make any mistake. From what I read from my physics book, the expression for electric potential energy considering two point charges is: U=$\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r}$. This, however, was proven when one charge was stationary, and the other one had free movement.
If both charges are moving, then how do I consider this expression?
U accounts for both charges? Meaning that in the expression $U_1+K_1=U_2+K_2$, in $K_1$ or $K_2$ I must take into account both kinetics energies? Or I just use the previous expression for each charge separately?
Comment: Potential energy as a concept relies on the field being static, ie overcoming the E field from a stationary charge. With that being said, assigning PE for this situation doesn't make sense as the field is non conservative. To find the work needed to move a charge from infinity to the location, you need to manually compute the line integral for the specific situation.
Comment: However I think you may be wanting, is the total field energy,$$ \iiint [\frac{1}{2}\epsilon_0 \vec{E}^2 + \frac{1}{2\mu_0}\vec{B}^2] dv$$ but this formula doesn't work for point charges, only distributions where $\rho$ is finite
Comment: Actually you can use the general formula of total field energy for point charges, if you regularize it at small distances (eg assimilate the point by a sphere of small radius). The difference of regularized energy coincides with the difference in the usual Coulomb energy as the distance cutoff goes to $0$. More specifically, if the charges generate respective fields $E_1,E_2$, the cutoff will handle the constant self energy $\int E_i^2$ which cancels out in a difference, and the Coulomb energy is exactly $\epsilon_0\int E_1\cdot E_2$.
Comment: @jensenpaull So since these point charges are moving, the E field varies with time and there's no guarantee that the force will be conservative?
Answer: Electric potential energy of two charges is always (by definition)
$$
U = \frac{1}{4\pi \epsilon_0} \frac{q_1q_2}{r}.
$$
In other words, potential energy does not depend on charges' velocities.
If you're asking whether the relation
$$
U + \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 = const.
$$
holds universally, then the answer is it doesn't. In EM theory there is no guarantee of conservation of kinetic plus potential energy, because part of the energy can move away in the form of radiation. There is still local energy conservation, but it cannot be expressed using charged particles' positions and velocities, one has to use EM energy density and EM energy flux density, depending on electric and magnetic fields.
Comment: The potential, $\phi$, is one component of a relativistic four-vector.  Therefor, "potential energy does not depend on charges' velocities." is not correct.
Comment: @JerroldFranklin Relativistic four-vector is irrelevant in this question. Potential energy, by definition, is function of positions and not of velocities. In case of two charged point particles, it is given by the above formula.
Comment: @JánLalinský So just to make sure I got it. The U expressions stands always, whether the charges are moving or not. However, K + U=const, not necessarily. If I still not take into account any energy loss due to radiation, might it be correct?
Comment: The expression for potential energy stands, but K+U=const. is not universally valid. It is valid when magnetic energy is negligible, which is only when the motions are slow enough.


delete folder from trash in Ubuntu

Question: I delete folder from trash folder by mistake, but I want to retrieve it?
 I am using Ubuntu 14.10
is there any way to retrieve it again
How can I do that?
Comment: Check out this feed http://ubuntuforums.org/showthread.php?t=898835. But honestly, if you don't bother too much about those data, just consider them lost.
Answer: First, you should stop using the drive or partition.
Next, login from another partition, or from a live disk and have an extra flash drive handy to transfer the lost data to.
Finally, you can use photorec to recover different types of files. To install,run the following commands:
sudo apt-get update
sudo apt-get install testdisk

Then, run the following command to start photorec:
photorec

For more info on how to use, run:
man photorec

Note: use the USB to store recovered files to. Do not use the drive you are recovering from to store recovered files.
Other tools to look into (that I have not used):
ext4magic - recover deleted files from ext3 or ext4 partitions
autopsy - graphical interface to SleuthKit
scalpel - A Frugal, High Performance File Carver
sleuthkit - tools for forensics analysis on volume and filesystem data
For a complete list, run the following command:
apt-cache search forensics

Comment: the problem with photorec is that the name of the file will be lost but the file will be recoverable. Recovered files have random names and you have to go through the files one by one to find the one you want.
Answer: You can use some applications able to scan file tracks from partitions like test disk.
Comment: Hi! Welcome to Askubuntu and thanks for your answer! However, posting a link as an answer is not accepted here. You should provide the most important parts of the info on that webpage in your answer. Include things in your answer like how to install it and how to use it to recover the trash. If you do these things, comment on your answer with @galgalesh, and I'll upvote your answer.

delete folder from trash in Ubuntu


Apache2, configuration for microsites

Question: Please, I ned help. I'm trying configure microsites with Apache.
I have a main site "mainsite" (www.mainsite.aaa) with document root /a/b/mainsite, in advance ./mainsite for abbreviate.
I want serve some content "minisiteA", "minisiteB" and so on... with they particular domain "www.minisiteA.aaa", "www.minisiteB.aaa"... (there are about 80 minisites) , which principal content but not all, are under ./mainsite/microsite/minisiteA, ./mainsite/microsite/minisiteB. There are a lot references to scrips css, images,etc under ./mainsite/ccs ./mainsite/imgages... It is: "out of the pseudo-documentroot for minisiteX" (That is my main problem at this moment)
I try to resolve this creating a virtualhost to manage all the microsites (ServerAlias) with the same document root as the mainsite (I don´t whant to modify de mainsite vhost). But I don´t locate a soltution for this. I guess that it's a very common configuartion, but I don´t find it (I feel I don't user correct keywords).
I try with some configuration, here two of them, first one the most easiest but it don´t work with files below ./microsite/minisiteX, and the second one has the same problem. It is possible than I am far away of correct solution, so, tips and advisors are welcome.
<code>First config.
    RewriteCond   %{HTTP_HOST}                 ^www\.minisiteA\.com:8036$
    RewriteRule   ^/(.+)  /microsite/minisiteA/$1    [L]

Second config
    RewriteCond %{ENV:REDIRECT_SUBDOMAIN} =""
    RewriteCond %{HTTP_HOST} ^([a-z0-9][-a-z0-9]+)\.mydomain\.org\.?(:80)?$ [NC]
    RewriteCond %{DOCUMENT_ROOT}/subdomains/%1 -d
    RewriteRule ^(.*) subdomains/%1/$1 [E=SUBDOMAIN:%1,L]
    RewriteRule ^ - [E=SUBDOMAIN:%{ENV:REDIRECT_SUBDOMAIN},L]
</code>
Answer: I think 'dynamically configured mass virtual hosting' is what you want here.
Given that the sites follow the same structure you can have Apache map your domain names to your local structure, i.e. by mapping the domain names to folder names using a variable - 'minisiteA' can be mapped to /local/path/minisiteA, 'minisiteB' can be mapped to /local/path/minisiteB ...
http://httpd.apache.org/docs/2.2/mod/mod_vhost_alias.html
This has saved me before, having to write 1 config rather than scripting generating hundreds and hundreds of small, almost identical configs...

Apache2, configuration for microsites


Playing/Recording acoustic signal in matlab to tell difference between 2 sounds

Question: I will be using one central node/latop to play a tone and will have a second node/laptop to record and read the sound and be able to say this node is on the left or right. So trying to figure out the best way to distinguish between the two tones. I want to have a tone set for left and one set for the right. I want the sound to have 400 Samples at 44100 for 10ms between 18-22freq just outside the human hearing range so that the sound is not annoying etc. I want the sound to be different slightly but be able to tell the diff between them for the system to say the node that recorded and processed the sound is on the left of right. How would I go about doing this in matlab because I don't dont really code in this language much and only time I did was for image processing just doing this so I can graduate.
Comment: jojek- I want have a sound set so say the node recording reads it and says yes this node is on the left and have a sound for the right doing the same but how would I go about making it so that it plays that right way from a 2d array.
Comment: From my understanding of the question, the left/right might not refer to the spatial position of sound but just to two symbols that have to be transmitted? What kind of speaker and recording equipment will you use for that? Low-quality audio equipment is just not optimized for applications outside of the human hearing range - that would be a waste of engineering resources!
Answer: I understand that you want to tell if your second laptop is placed on the left or right (or maybe on the axis) of main node?
I suggest you then to play the two-tone stereo signal, i.e.: $15\;kHz$ from the left, and $16\;kHz$ from the right speaker at the same time. Assuming that you are going to use sampling frequency of $f_s=41000 \;  \text{Hz}$ and taking the average distance between two speakers in a laptop to be $25\;cm$, you can calculate that the maximum delay between two signals, if the second node is on the right, will be approximately: $\dfrac{0.25}{343}=0.7289\;msec$. This corresponds to $32$ samples at a given sampling frequency. It is therefore a hard task, but still possible. Then after finding the delay between two signals, from its sign you are able to tell which speaker is further away, ergo you are on the opposite side.
In MATLAB you just need to generate proper signals, given sampling frequency and play them back/record with: MATLAB - Record and Play Audio.
Few problems can arise:

You want to use very high frequencies, and this can be problematic as laptop speakers are usually low-grade gear and they possibly might have some problems- same goes to microphone. Also remember to turn any hardware noise cancellation, spatial enhancements, etc.
Detection of you signal onset could be problematic as the resolution of DFT will be to low. You can also use some kind of correlation detection, but this is a problem for later on.
Comment: I have to find the left or right answer differently than that because the article my professor gave me he wants me to try it different. So is possible I can use the the fact that one is being played at 15hz and 16hz...is there anyway to have the signal to read and decide which is which saying 15 is the left and 16 is the right. If that makes any sense.
Comment: @Asau7610: Well - instead of explaining it in your own words, maybe link to the article, state the purpose, or even add the picture? Because it's like you are copying and pasting same text. Secondly, if you want to transmit 15/16 Hz (not kHz) with laptop speakers, then good luck. No way to play and record ultrasound with such gear. Also distinguishing between them will be very hard. Not mentioning the fact that you originally proposed: `18-22freq`(Hz?) at 10ms, which makes it impossible to emit at least one period of this infrasound. Regarding sounds, you can even use sweeps for that purpose.


How to reduce leaf size on Ash bonsai

Question: Five years ago I asked here about ways to train this ash (Fraxinus excelsior) seedling into a bonsai.
The little ash bonsai has now grown and it's in his 4th year. See the photo below. Any suggestion on how I could limit leaf size at this stage? Or should I leave it to grow to its final trunk/branch shape first?
Comment: A simple google search brought me too: https://www.evergreengardenworks.com/leafsize.htm but I am not experienced with such matters (learning bonsai myself) so perhaps you have already seen this page.
Comment: What also helps is to put it in a bonsai pot (scale), the plant will get used to the small size and will react by keeping above ground also small (you'll have to help by pruning). And have patience... 4 years is nothing in bonsai terms.
Answer: Basically, there are only 3 ways to make small leaves.

One is to keep the tree very small, that is as a mame/mini bonsai, which clearly isn't what you are wanting to do.
Second is to cut the compound leaves back to just 2 leaflets, 4 leaflets, etc., and by then removing one more leaflet, you can make odd numbers
Lastly, you can defoliate. The new leaves are small and over time may enlarge to full size. Were you do defoliate in August, say, they may not attain normal full size before leaf drop. This, however, seriously weakens the tree.

It is also possible that the leaf size will reduce somewhat in bonsai culture. Many species leaves progressively 'shrink' over the seasons with roots confined in a small pot (what gardeners call 'root bound').
In brief, do #2.
Ed. in response to questions about what #2 means:
This whole thing is one leaf, the compound leaf of ash. I've marked red lines where you may cut it to illustrate.

Since the normal leaf has one leaflet at the end, cut leaves may look 'unnatural', which you can correct to a certain extent by also removing one of the leaflets on either side of the cut.
Comment: Can you describe in more detail #2? I am not sure I understand what to do...

How to reduce leaf size on Ash bonsai