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Avantgarde
Avantgarde
1:11 AM
Anybody?
 
vzn
vzn
1:25 AM
↑ 2 major wobbles in a 1 day. amazing :o o_O
 
 
1 hour later…
Jalapeno Nachos
Jalapeno Nachos
2:30 AM
anybody here?
:)
 
Physics Meta
Physics Meta
0
Q: Confused and frustrated by the silence of people who voted to close my question

a_sidMy question has been closed again. DavidZ has pointed out that it is acceptable for people to not give reasons for closing questions. I understand that the site does not require them to do that. However, if you care about the integrity of this website and genuinely want to help out users, should...

discussion specific-question
 
 
2 hours later…
user7777777
user7777777
4:32 AM
@danielunderwood Blatant posting of homework questions without any effort to solve the question/identify the problematic concept, or even type it into a proper format, is not acceptable here. Don't answer.
 
 
3 hours later…
Nobody recognizeable
Nobody recognizeable
7:05 AM
Temperature is the average kinetic energy of molecules. Then if we remove or put more particles in a container or fixed volume then the kinetic energy should increase per unit volume. Then why should temperature be intensive property ?
 
user7777777
user7777777
The pressure increases but the temperature doesn't.
 
Nobody recognizeable
Nobody recognizeable
7:23 AM
@user7777777 is the temperature the average kinetic energy of a single atom?
 
user7777777
user7777777
It is the average kinetic energy of the atoms
 
Nobody recognizeable
Nobody recognizeable
@user7777777 are you interested about black holes ?
 
Anonymous
@Nobodyrecognizeable No
 
Nobody recognizeable
Nobody recognizeable
@Blue are you?
 
Anonymous
@Nobodyrecognizeable What?
 
Nobody recognizeable
Nobody recognizeable
7:28 AM
@Blue are you interested about black holes? I have a question then.
 
Anonymous
@Nobodyrecognizeable I'm interested in black holes, yes. But that doesn't imply I know anything about them beyond what I read in popular science textbooks aimed at the general public. :P
 
Anonymous
I'm not really the best person to ask.
 
Anonymous
Wait for JR
 
Nobody recognizeable
Nobody recognizeable
@Blue are you the same guy who was doing be in jadavpur university?
 
user7777777
user7777777
Maybe
 
Anonymous
7:32 AM
@Nobodyrecognizeable Yes
 
Nobody recognizeable
Nobody recognizeable
@Blue I've met you here before . You continue on to change profile pictures so it's bit hard to recognise you :p
@user7777777 is this about your interest in black holes?
 
Anonymous
@Nobodyrecognizeable Heh
 
Anonymous
Indeed, we have spoken before
 
Anonymous
How're you doing?
 
Sir Cumference
Sir Cumference
Howdy
 
Nobody recognizeable
Nobody recognizeable
7:37 AM
@Blue perhaps you are changing profile pictures I'm changing names. Im fine . How are you?
 
Anonymous
@SirCumference 3:37 AM. Good night!
 
Sir Cumference
Sir Cumference
@Blue lol
I gotta get this assignment done first :P
 
Nobody recognizeable
Nobody recognizeable
@Blue in America?
 
Sir Cumference
Sir Cumference
a very very tedious problem, it's been taking like 8 hours
welp, how's life i guess?
 
Anonymous
@SirCumference In the "procrastinating before exams" stage now
 
Sir Cumference
Sir Cumference
7:40 AM
oh wait, what year are you again?
 
Anonymous
Debugged a monster program last night tho. So there's a sense of bliss for the time being XD
 
Anonymous
@SirCumference 2nd year
 
Nobody recognizeable
Nobody recognizeable
@Blue I'll likely to ask you about how to avoid parallax error in vernier calipers in travelling microscope it's indeed longer than my mouth and has fifty divisions.
 
Anonymous
@Nobodyrecognizeable I didn't read anything written after vernier calipers in that sentence...not interested now :P
 
Nobody recognizeable
Nobody recognizeable
@Blue ok then I'll ask later . Have a nice day goodbye.
 
Anonymous
7:43 AM
Bye!
 
Sir Cumference
Sir Cumference
wait @blue how'd you know my state?
 
Anonymous
@SirCumference I know your uni's name...
 
Sir Cumference
Sir Cumference
oh right. you were there when 0celo mentioned it.
i think at this point i'm just gonna head to sleep. leave tomorrows problems for tonight
 
 
1 hour later…
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
8:55 AM
I think there's a wrong answer at physics.stackexchange.com/questions/431522/… by HiddenBabel
 
Anonymous
9:11 AM
@coniferous_smellerULPBG-W8ZgjR It looks fine to me.
 
Anonymous
There is no need to be unnecessarily pedantic
 
Anonymous
@coniferous_smellerULPBG-W8ZgjR For the note, in all practical situations we consider the "net" power radiated because the actual power emitted is a useless quantity. Even Wikipedia frames it this way.
 
Arjun
Arjun
9:48 AM
Hi all!
 
Anonymous
@Arjun Hi
 
Arjun
Arjun
There's a simple question that I have got, which I think is too naïve for the main site.
Is it fine if I ask it here?
 
Anonymous
Yes, but there is no guarantee that anybody in this chat can/will answer it
 
Anonymous
Go ahead anyway
 
Arjun
Arjun
If heat transfer is directly dependent on temperature difference, why does radiation (which is a mode of heat transfer) take place at all temperatures (except at absolute zero) i.e. why is it only dependent on absolute temperature? Should it not be dependent on the temperatures of the other bodies of the system?
 
ACuriousMind
ACuriousMind
10:02 AM
@Arjun Of what system?
All things radiate. But for heat transfer, you also have to consider that the radiating body is also hit by the radiation from other bodies. If the body has exactly the same temperature as its surroundings, then it will receive exactly as much energy as it radiates away, meaning no net heat transfer.
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
@Blue I do not care about the usefulness, which is a subjective judgement. The question has been left entirely unanswered. It is question about the power emitted of a body, and this power does not depend at all on its surrounding. It is a legitimate question, because 2 different ways to calculate this power yield 2 different answers. and this point is not addressed at all in the already massively upvoted useless answer that doesn't give any clue whatsoever
 
iBug
iBug
Hello, I want to know if anyone can help determine whether this answer is a potential trolling post.
 
Arjun
Arjun
@ACuriousMind I see. I understand now. Thank you for your time :)
 
iBug
iBug
It have some characteristics of a known troll, particularly quickly deleting their user profile after submitting the answer.
 
ACuriousMind
ACuriousMind
@iBug Well, it certainly isn't an answer to the question (I've now deleted it)
 
iBug
iBug
10:09 AM
@ACuriousMind Can you figure out the username?
 
ACuriousMind
ACuriousMind
@iBug Yes· but there was no username set.
 
Anonymous
@coniferous_smellerULPBG-W8ZgjR "The question has been left entirely unanswered. It is question about the power emitted of a body, and this power does not depend at all on its surrounding." - then that isn't clear from the question. You may suggest an edit to the question.
 
iBug
iBug
@ACuriousMind You mean, before the user was deleted, it had the name of userxxxxxx?
 
ACuriousMind
ACuriousMind
@iBug Yup
 
Anonymous
If the usefulness of the concept of the net heat radiated is subjective, then so is the clarity of the question, in its current form. In fact, more so. Fwiw, I very much doubt that the OP is concerned about the actual emitted heat.
 
iBug
iBug
10:11 AM
Thank you for the information!
Would you mind temporarily undeleting the answer so I can archive it for future reference?
 
ACuriousMind
ACuriousMind
@iBug Is this for Smokey or something like that?
 
iBug
iBug
Yes, Smokey (accidentally edited the previous message)
 
ACuriousMind
ACuriousMind
Okay, I've undeleted it
 
iBug
iBug
Thank you. I've done.
 
Anonymous
@iBug Hi, iBug. How much do you cost?
 
iBug
iBug
10:18 AM
@Blue Math.NaN
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
@Blue It looks crystal clear to me... I won't edit it :) But it's not the first time anyway a wrong and useless answer gets massively upvoted despite comments below it showing it's useless and doesn't address the question
there's nothing I can do about it, that's how it works. Popularity does not correlate highly with good answer, or simply, with answers.
 
Anonymous
@coniferous_smellerULPBG-W8ZgjR Except that your comment wasn't too convincing this time. ;) The average human does believe in the notion of usefulness.
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
There's an accepted answer with over 30 votes stating that Ohm's law cannot be derived, while it can with Drude's model or by using Boltzmann transport equation under certain assumptions
people pointed it out, but this never reversed the vote popularity
 
Anonymous
Just buy SE for ~$40 million and re-code the whole damn thing.
 
iBug
iBug
@ACuriousMind Excuse me, can you confirm that 1. it's the only post of the user and 2. the user deleted themself soon after posting the answer (usually within a minute or two)?
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
10:31 AM
@Blue I think my choices are either just quit or live with it.
The choice is hard to make
 
ACuriousMind
ACuriousMind
@iBug Yes, less than a minute after posting.
 
iBug
iBug
Thanks, very valuable information!
 
Anonymous
@iBug On a troll hunt?
 
iBug
iBug
@Blue Yep.
Some known characteristics: Using a few consistent usernames (Anonymous Teacher is one), answering bountied or hot questions (in a bad manner), quickly deleting the user profile after posting the first answer, etc.
 
Akash. B
Akash. B
11:00 AM
@ACuriousMind what was the reason for my suspension?
@rob
 
user1732
user1732
for how long? @Akash.B
 
David Z
David Z
@Akash.B That would have been explained to you in the message you received when the suspension began.
 
ACuriousMind
ACuriousMind
@Akash.B You were told the reason for your suspension in a mod message around Aug 31.
 
David Z
David Z
We're not going to discuss it here.
 
user1732
user1732
11:25 AM
iirc, 0celo went from a month to a year
 
Anonymous
@user1732 Ron went from 10 years to a lifetime, beat that ;)
 
user1732
user1732
yup, no way to beat that pal
\o @NovaliumCompany
 
Novalium Company
Novalium Company
Hi @user1732
Guys, a quick question. How does fire sustain itself? I know oxygen plays a big role in the sustaining of a fire but how? For example, when you apply enough heat to a fuel (e.g wood), if there is oxygen, the wood will keep on burning until it's all burned out, but how having oxygen sustains the fire for it to burn?
 
ACuriousMind
ACuriousMind
@NovaliumCompany Burning is a chemical reaction - mainly oxidation, i.e. the molecules of the fuel fusing with the oxygen. If there's no oxygen, there's no burning. The reaction needs a certain amount of energy to start, and after that produces enough heat to sustain itself.
 
user1732
user1732
Fire triangle
The Fire Triangle or Combustion Triangle or ″Fire Diamond ″ are simple models for understanding the necessary ingredients for most fires.The triangle illustrates the three elements a fire needs to ignite: heat, fuel, and an oxidizing agent (usually oxygen). A fire naturally occurs when the elements are present and combined in the right mixture, meaning that fire is actually an event rather than a thing. A fire can be prevented or extinguished by removing any one of the elements in the fire triangle. For example, covering a fire with a fire blanket removes the oxygen part of the triangle and can...
 
Novalium Company
Novalium Company
11:33 AM
Got it, thanks. I'm just watching the new Skyscraper movie and started wondering about fire.
 
Anonymous
11:47 AM
@NovaliumCompany How's it?
 
Akash. B
Akash. B
hi
 
Novalium Company
Novalium Company
12:18 PM
@Blue Just finished it, pretty good actually. It's not like the best movie ever, but if you are bored, watch it. :D
 
Anonymous
@NovaliumCompany I had watched the trailer several months ago. The last scene was DJ jumping onto a building but the gap was too big iirc. Did he actually manage to reach the building in the movie? :P
 
Novalium Company
Novalium Company
@Blue Yep, he reached it. I mean, it would be kinda inconvinient for the main character to fall and die at the middle of the movie :D
So about the fire question, we need heat in order to knock out the wood molecules into the air so they can react with the oxygen, and when the reaction occurs, the products have high kinetic energy which when knocks off the next wood molecules and so on... This is how the fire sustains itself?
 
ACuriousMind
ACuriousMind
12:38 PM
@NovaliumCompany No, it's not about "knocking" any molecules out of the wood. The oxidization reaction simply needs energy/heat to get started, but after that sustains itself (it is exothermic).
 
Novalium Company
Novalium Company
It sustains itself because of the kinetic energy that the reaction produces? And with that energy the next oxidation can proceed? Also, oxidation is just the process of something loosing electrons, that's why the oxidizer is the one receiving the electrons and the one making the other something give it's electrons, but what has oxidation have to do in the process? Sorry, I'm a bit unfamiliar with oxidizing and reduction processes, they are a bit confusing and I always mess up the names.
 
ACuriousMind
ACuriousMind
@NovaliumCompany Ah, sorry, I didn't mean the redox meaning of "oxidizing", but rather that the fuel reacts with the oxygen to form some sort of oxide (e.g. carbon dioxide). The reaction is exothermic because there's a lot of energy in the double bond of the oxygen molecute which gets broken up in the process.
Technically, both reduction and oxidation happen during this reaction, but I didn't want to get that technical/chemical
 
Novalium Company
Novalium Company
@ACuriousMind Ok, thanks. And the combustion reaction sustains itself because it's exothermic and the released energy can go on the trigger the next set of reaction. (e.g continue to burn wood, methane...)?
 
ACuriousMind
ACuriousMind
Yes
 
Novalium Company
Novalium Company
@ACuriousMind Thanks for the help!
 
ACuriousMind
ACuriousMind
12:51 PM
you're welcome
 
user1732
user1732
I would highly recommend this @NovaliumCompany
he's trying to talk in the ye old English of faraday
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
1:24 PM
@Blue I have decided to answer physics.stackexchange.com/questions/431522/… based on the comments. Seems like everybody believed in the wrong answer...
it's wrong because it focuses on radiative heat transfer between human skin and its surrounding, which is astonishingly small compared to convection and even conduction in many cases
anyway I detailed why it's wrong. My job is done
I gave the answer to the question too, which is what mattered to me
 
Akash. B
Akash. B
btw I have a doubt
anyone?
@ACuriousMind In rockets you know ,they uses liquid oxygen for ignition ,right? .My question is " isn't the temperature of liquid oxygen -400 degree c ? how is it possible for it burn at high temperatures above 5000 degree Celsius ?
 
ACuriousMind
ACuriousMind
@Akash.B -400°C would be below absolute zero. Liquid rocket fuel isn't stored at low temperatures, but at high pressures. And I'm not sure I understand your question - car fuel is also stored at essentially room temperature, but it's burnt in the engine at higher temperatures. Same thing happens in rockets.
 
Anonymous
@coniferous_smellerULPBG-W8ZgjR Didn't read it thoroughly but it looks good. Usually writing an alternative answer is the best way to go about these situations. Future visitors can collect whatever they think is relevant from both the answers.
 
Anonymous
I still think the question is too vague though. I wouldn't have bothered to answer it
 
Anonymous
By the way, you might want to cite sources for the claim that most of the body heat is lost by conduction and convection. I haven't read such a thing anywhere (it might be true; I don't have the data).
 
Anonymous
1:42 PM
> The total energy radiated in one day is about 8 MJ, or 2000 kcal (source).
 
Anonymous
How much is lost by convection/evaporation on the other hand?
 
Akash. B
Akash. B
@ACuriousMind oh
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
so @Blue it means most calory intake is spent as radiation energy... doesn't change my point though, these 100W are not the full radiated power. Any body near room temp with the same surface emits 10 times more power.
that's the point I tried to make
 
Anonymous
Hmm, I don't know what you mean by "any body near room temp with the same surface emits 10 times more power", but anyway, I don't find myself too interested in this right now.
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
1:59 PM
@Blue I mean, if you take a carpet, or even a dead body, you'll find it emits 10 times more than the 100W one obtains if one assumes all calories intake is spent on heat emission.
the point is that thinking that calory intake accounts for all heat radiation is wrong. because it ignores the temperature of the body, and as Stefan boltzmann law shows, the temperatuer of the body is what matters
 
Anonymous
@coniferous_smellerULPBG-W8ZgjR "you'll find it emits 10 times more than the 100W one obtains if one assumes all calories intake is spent on heat emission." what formula are you using to conclude that?
 
Anonymous
Also, how does a dead body or carpet intake calories? :P
 
Akash. B
Akash. B
@ACuriousMind how does the high temperature is attained in space like region?
 
ACuriousMind
ACuriousMind
@Blue Some carpets are predators.
 
Anonymous
@coniferous_smellerULPBG-W8ZgjR It doesn't ignore the temperature of the body. In the other answer, they're clearly using $\epsilon\sigma(T^4-T_s^4)$
 
Anonymous
2:03 PM
Where $T$ is the temperature of the body
 
ACuriousMind
ACuriousMind
@Akash.B I'm no rocket scientist, but I imagine they just use something like a spark plug. It doesn't matter to an engine or a spark plug whether it's in space or not.
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
right @Blue and this other answer claims no net radiation if surrounding equals body temp
 
Lozansky
Lozansky
If the field lines of some vector field can be assigned (respectively) a begin point and an end point, then the vector field has a nonzero divergence. How is this formulated mathematically?
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
it would seem to indicate 0 radiation. but this is far from truth
i've edited my answer, it should be clearer now, i've removed a chunk of things
 
Anonymous
@coniferous_smellerULPBG-W8ZgjR Proof?
 
ACuriousMind
ACuriousMind
2:05 PM
@coniferous_smellerULPBG-W8ZgjR There's a difference between "net radiated power is zero" and "no radiation".
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
if T_env = T_body then P_net is 0 , but the fact that 1kW is still emitted is not mentioned
I know
I know @ACuriousMind and tat's why the current top answer sucks
 
ACuriousMind
ACuriousMind
Why? It computes the net radiated power.
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
I mean just read it, read mine, and figure out which one better answer the question. or simply which one does answer it
 
Lozansky
Lozansky
There is no heat transfer in thermal equilibrium?
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
because it doesn't answer the question at all @ACuriousMind
 
ACuriousMind
ACuriousMind
2:06 PM
You seem to think that one should, for some reason, care about the radiation emitted by the body, but it's not clear to me why
 
Anonymous
I'm pretty sure the other answerer is well aware of the difference between $P_{net}$ and $P_{emitted}$.
 
Anonymous
@ACuriousMind That
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
the question is why does a way to calculate emitted power differs from another way to compute it
 
Anonymous
Neither do I understand why you should care about $P_{emitted}$ when $P_{net}$ is what matters.
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
focusing on P_net doesn't help an iota
it doesn't @Blue
the dude wants to know why P_emitted differs if we take a body at 37C or if we consider calory intakes
 
Anonymous
2:08 PM
The calorie intake is proportional to net radiated power. Not the emitted power. That follows from the fact that the calorie intake helps you to maintain your body temperature above room temperature.
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
why is there a difference? my answer explains it
nah... if surroudning temp is same as body, net radiation is 0 but you still have a lot of calory intake
 
ACuriousMind
ACuriousMind
@coniferous_smellerULPBG-W8ZgjR That's not clear to me at all. The question asks "What am I doing wrong?". The answer you claim to be wrong points out one thing they're doing wrong, you point out another.
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
to keep the body at 37C will still require a lot of calories, though surely less than 2000 kcal a day
i point out that assuming that all calory intake accounts for all power emitted, this is what is wrong
simple as that, that' s the answer
this is a wrong assumption, that's all
I give the example of non living object at room temp, which has 0 calory input, but still radiates at almost 1kW
 
ACuriousMind
ACuriousMind
Of course, the fundamental mistake in the question is that the assumption that human calorie intake is completely converted to radiation to begin with, making it really more of a biology question - what OP is doing fundamentally wrong is not understanding biological organisms do different things with their energy than just radiating
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
no that's not it @ACuriousMind
that's only part of it, but the biggest error is thinking that all the calory input accounts for all the radiation while in reality it's more like 10% of it
 
Anonymous
2:11 PM
@coniferous_smellerULPBG-W8ZgjR Sure. That excess calorie intake is not used to heat up the body in that case!
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
as the asker shows, it's 1000W vs 100W
 
Anonymous
That's exactly why you lose weight faster in the winter
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
yes @Blue or even cool it down
exactly @Blue
that's my point 2 in my answer
the biggest reason is point 1.
that's all there is to it, it's not really complicated. but focusing in P_net is misleading and does not show that thinking that all calory input accounts for all radiation is wrong
 
ACuriousMind
ACuriousMind
At this point I'm not sure what we're talking about. There's numerous things wrong with the assumptions in this question. It doesn't strike me as very useful to try and claim one of them as "the answer" and all others as misleading.
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
i don't really care at this point. The truth is written, I can die happily even if nobody recognizes it.
a robot would
enough for me
 
Anonymous
2:19 PM
From a biological point of view, heating your body to keep it at a particular temperature, is extremely essential, or else the vital chemical reactions would cease and you'd die. So it is a good approximation if you're living in a considerably cold region that a majority of your calorie intake would be spent on keeping you warm (assuming you don't eat in excess)
 
coniferous_smellerULPBG-W8ZgjR
coniferous_smellerULPBG-W8ZgjR
I agree @Blue
so... the answer is that radiation of human body is 90% due to its temperature near room temp, + 10% of extra radiation heat due to calory input. A dead body lacks only the latter
again, assuming that the extra 100W is ALL the radiated power is wrong because it ignores the absolute temp of the body
it's only roughly 10%
this answers the question
 
Anonymous
On another note: Sorta missing the days when the chat used to abound with the mathematicians :/
 
Anonymous
@ACuriousMind gee....i s t h a t a v i d e o g a m e
 
Anonymous
> Smother: Melee Weapon Attack: +5 to hit, reach 5 ft., Medium or smaller creature. Hit: 0 (1d0) bludgeoning damage, and the creature is Grappled escape DC 13. Until this grapple ends, the target is Restrained, Blinded, and at risk of Suffocating, and the rug can't smother another target. In addition, at the start of each of the target's turns, the target takes 10 or 2d6 + 3 bludgeoning dmg.
 
ACuriousMind
ACuriousMind
@Blue Not really, while there are video games based on it, D&D is first and foremost a pen and paper RPG, i.e. meant to be played with actual people at a table.
 
Anonymous
2:30 PM
Wow, I see. Complete RPG noob boi here
 
Anonymous
So if I play the role of the carpet, I can smother others. Sounds legit.
 
ACuriousMind
ACuriousMind
Yes - but usually, all enemies are played by the same person
That is, you have a bunch of players who each play one character, and a game master who plays the rest
 
vzn
vzn
2:55 PM
main uses of calories (energy) by biological organisms is (1) muscles and (2) brain and there is some (3) heating (temperature regulation). human brain takes about 20% of energy.
 
blue_eyed_...
blue_eyed_...
suppose that a semiconductor contains equal number of donor and acceptor impurities do they cancel each other in the electrical effect if so what is the mechanism if not why not
 
Akash. B
Akash. B
3:41 PM
Hi guys
So silent when I am here 🙄🙄🙄
 
Secret
Secret
4:27 PM
Govindappa Venkataswamy
Govindappa Venkataswamy (1 October 1918 – 7 July 2006) was a visionary Indian ophthalmologist who dedicated his life to eliminating needless blindness. He is the founder and former chairman of Aravind Eye Hospitals, the largest provider of eye care in the world. He is best known for developing an innovative high quality, high volume, low-cost service delivery model that has restored sight to millions of people. Since inception, Aravind Eye Care System (a registered non-profit organization) has seen over 55 million patients, and performed over 6.8 million surgeries.Over 50% of the organization...
Wtf is needleness blindness. Is it really what common sense said it is...?_?
 
ACuriousMind
ACuriousMind
@Secret Read again. It's needless blindness, not "needleness"
 
Lozansky
Lozansky
lol
 
Anonymous
> The World Health Organization says that world wide, 45 million people are blind. And of these cases, 75 percent could be prevented or cured. One doctor, Dr. V., called this "needless" blindness.
 
Lozansky
Lozansky
4:42 PM
"Dr V" sounds like a Bond villain
 
dmckee
dmckee
4:57 PM
^ "I will cure a million people of their blindness and you will give me ten million dollars to do it!"
 
Anonymous
@dmckee Sounds like a good bargain to me. Just 10$ per person! ;)
 
danielunderwood
danielunderwood
Is a Lagrangian unique? Say could you have distinct physical systems that have the same Lagrangian?
 
ACuriousMind
ACuriousMind
@danielunderwood In what sense would they be "distinct", given that their equations of motion are the same?
 
dmckee
dmckee
@Blue Dang skippy it is.
In the US (and I suspect other rich nations) we sometimes overlook the scale of human suffering alleviated by people who figure out how to apply science to the problems of poor people.
Like this guy:
Norman Borlaug
Norman Ernest Borlaug (; March 25, 1914 – September 12, 2009) was an American agronomist and humanitarian who led initiatives worldwide that contributed to the extensive increases in agricultural production termed the Green Revolution. Borlaug was awarded multiple honors for his work, including the Nobel Peace Prize, the Presidential Medal of Freedom and the Congressional Gold Medal. Borlaug received his B.S. in forestry in 1937 and Ph.D. in plant pathology and genetics from the University of Minnesota in 1942. He took up an agricultural research position in Mexico, where he developed semi-dwarf...
 
danielunderwood
danielunderwood
@ACuriousMind Good point. Maybe you could somehow manipulate the choice of generalized coordinates to make them look the same?
I have my doubts on that though
 
ACuriousMind
ACuriousMind
5:05 PM
@danielunderwood Well, I guess you really have to more properly define what you mean by the Lagrangians being "the same"
 
dmckee
dmckee
You can add any constant to a Lagrangian or multiply it by any non-zero constant and still get the same equations of motion.
 
Anonymous
Agreed. To be able to substantially cut down on the cost of healthcare for the lower middle class/lower class would really go a long way, in terms of overall progress of a nation. I was a slightly embarassed though, that I didn't have their names (Dr. V, Borlaug,..) on the tip of my tongue. They're way less celebrated than they should be.
 
danielunderwood
danielunderwood
Hmm yeah I guess that would result in the same EOM for different systems
@Blue yeah I've actually never heard of either. That's kind of sad
 
Anonymous
I do somewhat take pride in the fact that healthcare in my country is a lot cheaper compared to most of the developed countries though.
 
Anonymous
But then, a lot of the good doctors are now leaving for a better salary in the elite private hospitals
 
Sir Cumference
Sir Cumference
5:21 PM
Have to wonder why mods on most sites seem away on weekends. You'd think they'd be more available than on weekdays
 
Anonymous
@SirCumference Most of them are not in their 20s and have a family.
 
Sir Cumference
Sir Cumference
@Blue Huh, now I feel like I'm in my own world...
 
danielunderwood
danielunderwood
Aren't most people more away on weekends?
 
Anonymous
Not me
 
Sir Cumference
Sir Cumference
Nor me
 
Anonymous
5:30 PM
On weekdays I spend most of my daytime at uni
 
Sir Cumference
Sir Cumference
^
 
Novalium Company
Novalium Company
5:43 PM
Guys, I'm having a chemistry test tommorow and the teacher told us to print out a periodic table with the charges. I don't know what that means but I guess the teacher wants us to know the charge of something in a reaction, but I don't know how to do that. I mean, I know that the total charge in the reaction is conserved. For example the teacher told us that magnesium has a +2 charge, but why?
In this reaction $Mg + Cl_2 $-->$ MgCl_2$ magnesium has +2 charge.
 
danielunderwood
danielunderwood
A fully ionized magnesium atom has +2 charge because it loses 2 electrons
I suppose it may be important to note that just magnesium itself doesn't have +2 charge
 
ACuriousMind
ACuriousMind
@NovaliumCompany Magnesium doesn't "have a +2 charge". But it is in the second column, meaning it has two outer electrons. Since atoms "like" to have full outer shells, it prefers to give these two outer electrons away, and when it has lost them, it is an ion with +2 charge .
Likewise, chlorine is in the 7th main column, missing 1 electron to a full shell of 8. So its most common ion has -1 charge
 
Novalium Company
Novalium Company
How do you know it misses 1 electron? Is there a way I can figure out?
 
ACuriousMind
ACuriousMind
The stable products of reactions are neutral, so you need two chlorine ions to combine with one magnesium ion
@NovaliumCompany The general rule is that 8 is a "full" shell, except for the innermost shell that only has place for 2 electrons. This works for all elements in the first three periods, after that it gets complicated.
Didn't your teacher explain any of this?
 
Novalium Company
Novalium Company
Absolutley not, welcome to Bulgaria's education system.
The teacher allows us to write down the charges, where can I get them from? Also about the magnesium, does it have 2 more electrons than protons and that's why it wants to get rid of them?
I see that in the outer shell it has 2 electrons, but why would it want to get rid of them?
 
ACuriousMind
ACuriousMind
5:51 PM
@NovaliumCompany No, it is the 12th element and therefore has 12 electrons in its neutral state. 2 in the innermost shell, 8 in the second, and 2 in its outer shell
@NovaliumCompany Best to just accept "elements like to have full outer shells because it's energetically favourable" as an axiom at this level. The actual physical explanations of the behaviour of chemical elements are rather complicated.
 
Novalium Company
Novalium Company
Ok, so in the $MgCl_2$ reaction, is this how the charges go:
$Mg^{-2} + Cl_2^{0}$ -> $Mg^{0}Cl_2^{-2}$ Hmm, doesn't look right ;\
 
Lozansky
Lozansky
Mg donates two electrons so it's $+2$
 
ACuriousMind
ACuriousMind
Um, well. The reaction would just be $\text{Mg} + \text{Cl}_2 \to \text{MgCl}_2$, where everything is neutral. But you can insert a middle step where the initial stuff is ionized: $\text{Mg} + \text{Cl}_2 \to \text{Mg}^{+2} + 2\text{Cl}^{-1} \to \text{MgCl}_2$
 
Novalium Company
Novalium Company
The teacher gave us this as an example:
$Ba^{0} + Cu^{+2}SO_4^{-2}$ --> $Ba^{+2}SO_4^{-2} + Cu^{0}$
 
ACuriousMind
ACuriousMind
Ah, right. I guess you could write $\text{Mg}^{+2}\text{Cl}^{-2}_2$ for the r.h.s. in the other case
My chemistry is a little bit rusty :P
 
Novalium Company
Novalium Company
6:05 PM
Hmm, but the question is, how do I know that $Cu$ is +2 and $SO_4$ is -2...? The teachers told us we can write them down and use them during the test, but where can I get them from?
 
ACuriousMind
ACuriousMind
@NovaliumCompany Well, you have to know that the $\mathrm{SO}_4$ compound works by the S giving away its outer 6 electron. This balances with the 2x4=8 electrons the 4 O lack to something that still wants 2 electrons, hence the ion is -2.
It's not really fair to expect any beginner to figure this out
 
Novalium Company
Novalium Company
Yes, but all I need for now is to get them for the test. So Mg is +2 and SO4 is -2?
 
ACuriousMind
ACuriousMind
Yeah. That Cu likes to be +2 you just have to know - the metals in the middle of the periodic table behave a bit erratically.
 
Novalium Company
Novalium Company
So Mg +2, SO4 -2, Cu +2. They are like this in every reaction right?
 
ACuriousMind
ACuriousMind
The honest answer would be no, especially for Cu, but for your purposes: Yes :P
6
 
Novalium Company
Novalium Company
6:11 PM
Ok, thanks. Where can I get the rest of them. For example, Ba, Ca, S, Na...?
 
ACuriousMind
ACuriousMind
As I said, for the first three periods looking at the table and counting how many electrons they need to gain/lose to either have 8 electrons in their outer shell or none works pretty well
 
Novalium Company
Novalium Company
How do I know how many electrons they have in the outer shell?
 
Leaky Nun
Leaky Nun
count them one by one
 
ACuriousMind
ACuriousMind
@NovaliumCompany The first element in each period has 1 electron in its outer shell. They gain one electron for each step to the right.
 
Lozansky
Lozansky
@Semiclassical Semi!!
 
Novalium Company
Novalium Company
6:16 PM
@ACuriousMind I don't get it ;/
 
Lozansky
Lozansky
I have a small E&M question :>
 
ACuriousMind
ACuriousMind
@NovaliumCompany Okay, example: Na is the first element in its period, so it has 1 outer electron. Mg comes after it, and so has 2. Al is the next, and so has 3
 
Novalium Company
Novalium Company
@ACuriousMind Oh got it. I accidentally confused period with a group. So that works only for the first 3 periods. That means F will have 7 electrons in the outer shell? Does that mean that it just won't want to give them and we'll write it neutural?
 
ACuriousMind
ACuriousMind
@NovaliumCompany No, it means it wants one more to get 8
F behaves the same as Cl - it shows up as -1
 
Novalium Company
Novalium Company
@ACuriousMind I though as we get to right, the number of outer electrons increases?
 
ACuriousMind
ACuriousMind
6:23 PM
@NovaliumCompany It does
And you were correct that F has 7 outer electrons.
But it's greedy and wants 8!
 
Novalium Company
Novalium Company
@ACuriousMind So Mg is greedy too and since it has 2 outer electrons, will try to get to 8?
 
ACuriousMind
ACuriousMind
@NovaliumCompany No, Mg is a wimp and just gives away its 2 out electrons for free
Generally, up to (and including) 4 outer electrons they give them away, after that they try to get a full set of 8
 
Semiclassical
Semiclassical
@Lozansky ?
 
Novalium Company
Novalium Company
@ACuriousMind So B will give away it's 3 electrons? And what about carbon (which has 4 outer electron)?
 
ACuriousMind
ACuriousMind
@NovaliumCompany They are both givers
 
Lozansky
Lozansky
6:26 PM
@Semiclassical It's about exercise 4.25 from Griffith's
 
Novalium Company
Novalium Company
@ACuriousMind N and P are not givers, they want to take 4 to get to 8, so they'll have +4 charge?
 
Semiclassical
Semiclassical
@Lozansky pic?
 
ACuriousMind
ACuriousMind
@NovaliumCompany N and P have 5 outer electrons.
 
Lozansky
Lozansky
 
Novalium Company
Novalium Company
@ACuriousMind Oh yep, my mistake, so they'll have +3 charge?
 
ACuriousMind
ACuriousMind
6:28 PM
No, -3!
Electrons are negative - if they take three electrons, they're at -3
 
Lozansky
Lozansky
@Semiclassical So for $z<0$ you have $\chi_e$ and for $z>0$ you have $\chi_e'$
 
ACuriousMind
ACuriousMind
Btw, this "charge" is more properly called oxidation state
If you go to the Wiki page of an element, there's a list of oxidation numbers, with one printed in bold. That bold one is the most common one, i.e. the one that's relevant in most cases.
 
Semiclassical
Semiclassical
seems like the obvious approach would be to try method of images again
 
Lozansky
Lozansky
Yes
So you have the charge $q$ at $z=d$
 
Semiclassical
Semiclassical
But, in the original method of images problem, you've got V=0 at the boundary
I can't remember if that's what you'd have in this case as well
 
Novalium Company
Novalium Company
6:31 PM
@ACuriousMind Oh, got it. So for Aluminium, the oxidation state will be +3 because they give out their 3 electrons?
 
ACuriousMind
ACuriousMind
@NovaliumCompany Yes
 
Lozansky
Lozansky
And since the charge $q$ is now in a dielectric material, there will be a polarization charge around $q$
Then you have the surface charge $\sigma_b$ on the top side from the lower dielectric and $\sigma_b'$ on the lower side of the top dielectric
 
Novalium Company
Novalium Company
@ACuriousMind Just to check my understanding:
H, Li, Na will be +1, Be, Mg will be +2, B, Al will be +3... But since Mg for example has 2 electrons and wants to get rid of them, at the left side of the reaction, magnesium will be -2 and the right, +2?
 
ACuriousMind
ACuriousMind
@NovaliumCompany You lost me at "Magnesium will be -2"
 
Lozansky
Lozansky
@Semiclassical Are you asking if the potential in the $xy$-plane is grounded?
 
Semiclassical
Semiclassical
6:35 PM
right
 
Novalium Company
Novalium Company
@ACuriousMind Well, at the left side of a reaction, the magnesium sill has those 2 electrons, before giving them away, so we should write it like -2? Or maybe we don't? Maybe we write it neutural and and the right side we write it like +2 becuase it has lost them. I think the second is true?
 
ACuriousMind
ACuriousMind
@NovaliumCompany Before it gives them away it is neutral
 
Lozansky
Lozansky
@Semiclassical No, why would it be? These are dielectrics, not conductors
 
Novalium Company
Novalium Company
@ACuriousMind Cool got it. Last thing. About those C and Si, do they give or take 4 electrons?
 
Semiclassical
Semiclassical
i guess not. in which case the method of images seems less pertinent here
 
ACuriousMind
ACuriousMind
6:38 PM
@NovaliumCompany You'll mostly see them giving them away, like in $\text{CO}_2$.
 
Lozansky
Lozansky
@Semiclassical Nah, it works
 
Novalium Company
Novalium Company
@ACuriousMind So CO2 will be: $C^{+4}O_2^{-2}$?
 
Semiclassical
Semiclassical
hmm
 
ACuriousMind
ACuriousMind
yep
 
Lozansky
Lozansky
But your boundary conditions are: $V$ continuous at $z=0$, the normal derivative of $V$ undergoes a discontinuity $-\sigma_b/\epsilon$ and $V \to 0$ at $\infty$
 
Novalium Company
Novalium Company
6:41 PM
@ACuriousMind And lastly, N and P will have 5 electrons and will want to take 3 to get to 8 so their "charge" (oxidation state) will be -3?, O and S will be -2, F and Cl -1, He, Ne, Ar neutral?
 
ACuriousMind
ACuriousMind
yep
 
Novalium Company
Novalium Company
@ACuriousMind Cool, thank you so much for explaning the things that every chemistry teacher should explain. Really appreciate the help!
 
Lozansky
Lozansky
Eh, not quite
 
ACuriousMind
ACuriousMind
@NovaliumCompany you're welcome :)
 
Loong
Loong
@NovaliumCompany Mg doesn't want to get rid of two electrons.
The ionization energy of neutral atoms is always positive.
 
ACuriousMind
ACuriousMind
6:45 PM
Uh-oh, an actual chemist is here!
3
 
Loong
Loong
Thus, the process Mg -> Mg^2+ + 2e- is endothermic.
No element wants that.
@ACuriousMind Oops, sorry, wrong room ;-)
2
 
Novalium Company
Novalium Company
@Loong Thanks, but I think for my chemistry level, just "getting rid of" will do the job. :D
 
Loong
Loong
Yes, that's probably what the text books expect from you.
 
Novalium Company
Novalium Company
Ahh, I wish I was in a school where they explain the things the right way... They just tell you: "just memorize the 'charges'"
 
Lozansky
Lozansky
@Semiclassical The discontinuity is given by $\sigma_b = \hat{n} \cdot (\mathbf{P}_1 - \mathbf{P}_2 )= \hat{n} \epsilon_0 (\mathbf{E}_1 \chi_{e_1} - \mathbf{E}_2 \chi_{e_2}) = -\epsilon_0 (\chi_{e_1} V_{1,n} - \chi_{e_2} V_{2,n})$ where $V_{i,n}$ signifies normal derivative at the boundary
 
Novalium Company
Novalium Company
6:51 PM
So in the case of $SO_4$, it's a bit weird and those 'rules' don't hold true, becuase S will give away it's 6 outer electrons, not receive 2. Is there some magical force that oxigen applies on S? (just out of curiosity)
 
Loong
Loong
@NovaliumCompany That magical force is "electronegativity".
 
Anonymous
Oxidation state
The oxidation state, sometimes referred to as oxidation number, describes degree of oxidation (loss of electrons) of an atom in a chemical compound. Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component. This is never exactly true for real bonds. The term oxidation was first used by Antoine Lavoisier to signify reaction of a substance with oxygen. Much later, it was realized that the substance, upon being oxidized, loses electrons, and...
 
Anonymous
It doesn't actually give away 6 electrons
 
danielunderwood
danielunderwood
Now that I think of it, I don't think any of my chemistry classes went into much of anything with atoms that had d/f-orbitals
 
Semiclassical
Semiclassical
@Lozansky hmm
 
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