The h Bar

user228700

02:01

Ello everyone :-)

user228700

How many of u believe that there is a subtle difference b/w the terms electric field and electric field intensity?

user228700

Anyone? _ /\ _

ACuriousMind

@Kaumudi I think it depends on the context and I don't think the difference is subtle: Either both mean the same or intensity means power per unit area in the context of radiation

heather

how would you install the graphicx tex package on ubuntu?

DanielSank

Hi, everybody

@heather install texlive

user228700

02:08

@ACuriousMind Hmm. I dunno how to provide even more context tho. I'm learning electrostatics and a lot of websites say that there is no difference but some don't agree with this :-|

ACuriousMind

@heather Any standard TeX distribution should have that one already included, you shouldn't need to install it individually

heather

@DanielSank, I have texlive/texmaker installed, is that all that is necessary?

oh, okay

DanielSank

@heather yes

Also, there is a very good Tex stack exchange.

ACuriousMind

@Kaumudi Provide an example of one that doesn't agree.

heather

@ACuriousMind, @DanielSank, okay, thanks =)

DanielSank

02:10

Also, at some point (maybe not soon), you may benefit from reading this.

ACuriousMind

In general, just put the \usepackage in your preamble and see if it compiles. Only if it doesn't you'll need to do something.

DanielSank

I'd be surprised if you need something not already in texlive.

user228700

> "Both of these terms are often used in place of each other. We do it quite nonchalantly. However, there exists a minute difference between these words."

user228700

> "Electric field is used to describe a region of energy present around charges. A more thorough and quantitative definition is easily available on the net. There is no notation as such to represent electric field."

user228700

> "However, when we talk about electric field intensity, we are actually describing the vector aspect of an electric field. That is we are talking about the magnitude and the direction of it."

user228700

02:14

I don't even understand the difference that well ^

DanielSank

I think I see what the author means.

ACuriousMind

Sounds as if the author is pretty confused to me :P

DanielSank

Imagine the ocean, completely still with no waves.

user228700

Erhm, OK...

DanielSank

This ocean can have waves. It is a medium in which waves might be produced, but the ocean itself is not waves.

user228700

02:17

Right...

DanielSank

I think, to this author, "electric field" is the ocean and "electric field intensity" are waves.

The "electric field" (or better, the electromagnetic field) is a thing in the universe that can be excited. Those excitations are called "photons".

What do you think @ACuriousMind?

user228700

Hmm, that makes a little sense to me. I think I found a better explanation of the difference (its lack thereof, really) b/w the two terms. Another person had gone on about the difference but then, he corrected his answer thus:

user228700

> "Ok, I have taken it in wrong context, as if we consider it as oscillating electric field, in that context the intensity can be given by a square of amplitude but in the case of electrostatic charge is not associated with any wave nature and the scenario is completely different. So you are right. If it electrostatic and the there is no oscillating charge then yes the electric field intensity and the electric field both are same."

user228700

^ Well, that is a comment that the OP posted in response to somebody pointing out the mistake in his answer.

user228700

But I haven't understood what the difference is, in different contexts. Can anybody please explain that to me?

ACuriousMind

02:24

@DanielSank I think you're somewhat confusingly mixing the classical and the quantum description

user228700

The better question to ask would be if anybody can please explain this:

user228700

> "Intensity is always given as the square of an amplitude, where field is the amplitude."

ACuriousMind

@Kaumudi In the context of radiation, "intensity" means "power per unit area". So when you've got electromagnetic radiation, one would talk about the intensity of the wave in that sense, which would be proportional to the square of the electric field.

The electric field is just the value $E(t)$ of the electric field at each point, while the intensity is proportional to $\lvert E_0 \rvert^2$ where $E_0$ is the amplitude of the electric field of the wave, i.e. the maximal value it attains.

user228700

It would be proportional to the square of the electric field? Okay, I'll go read more about this. Thanks very much :-)

DanielSank

@ACuriousMind indeed I shouldn't have mentioned photons.

@ACuriousMind that is not what I read in the first passage she quoted.

ACuriousMind

02:28

@DanielSank True, but it's what is meant in the "correction" she posted thereafter

user228700

And the correction is what I was looking to understand so I'm good @DanielSank :-)

ACuriousMind

I think the first passage indeed tries to imbue "the electric field" with an ontology that's different from "the value of the electric field at each point", but I don't really see how that's a good stance to take

DanielSank

Yes

Agreed

user228700

Although, @ACuriousMind, this is not rendering for me:

user228700

$\lvert E_0 \rvert^2$

ACuriousMind

02:30

Renders perfectly fine on my end

user228700

ACuriousMind

Which of the options to display ChatJax are you using?

user228700

Can u please type that w/o Latex?

ACuriousMind

I just wrote $| E_0 |^2$ there.

(If that doesn't render for you I am seriously confused :P)

user228700

Ohh. Okay :-)

heather

02:33

I'm very annoyed with libreoffice right now =P

I honestly don't know what possessed me to try to transfer my graph/diagram filled notes into latex.

user228700

@ACuriousMind I'm using my phone. If I'm unable to read it directly, I copy-paste the message into the box to make a question on the SE app. If I hit preview, the Latex renders but that didn't.

ACuriousMind

Strange, \lvert shoudl definitely render on the site

user228700

Anyhoo, thank u :-)

user228700

Also, @heather: Hello :-)

heather

@Kaumudi, hello!

=)

user228700

02:42

BTW, does anybody here experience chronic GERD?

heather

GERD?

user228700

a.k.a Acid reflux :-P

user228700

> "Digestive disease in which stomach acid or bile irritates the food pipe lining."

user228700

(Very common: More than 10 million cases per year [India])

heather

oh, nope. do you?

DanielSank

02:46

@heather libre office works ok but not great

@heather probably the fact that latex is a good way to write technical stuff.

@Kaumudi nope. Do you?

user228700

@DanielSank Yeah. Been having weird pains recently, wanted to know if it was part of the package.

DanielSank

@ACuriousMind yes, that is what I meant with the whole ocean thing.

DanielSank

03:04

::crickets::

user228700

03:17

I've also found another way to distinguish b/w the two terms:

user228700

> "Electric field=Region around a charged particle/body in which, if another charge is placed, it experiences electrostatic force"

user228700

> "Electric field intensity=Electrostatic force experienced by a unit positive point charge "

user228700

Now what?

user228700

Is the star indicative of the fact that what I've written is correct..? :-|

koolman

Yes

My sir also told to me that

user228700

03:22

Hmm, well, I hope I can get this clarified with @JohnRennie (Halp!) because too many variations exist.

Sir Cumference

How do I wake up fast?

Turns out from the moment I got home I slept until 10:30pm

Now I have all this work

user116211

03:42

@SirCumference Drink gallons of water before sleeping.

Sir Cumference

@MAFIA36790 That's a bit excessive...

user116211

I'm serious @SirC; this works provided that you set an alarm too.

user246160

Does anyone here know why supercooled water turns into ice on slight shaking ? I could'nt understand the concept behind it.

user228700

@SirCumference Hey :-) Dyou mean get out of bed as soon as the alarm rings, or dyou mean wake up earlier?

Sir Cumference

@Kaumudi I mean I need to wake up right now

I'm groggy

user228700

03:52

@SirCumference Get some exercise. Jog some, or do pushups or whatever it is that counts as exercise for u.

Sir Cumference

@Kaumudi All right, thanks

user228700

Let me know if that works :-P Also, a cold shower after exercising might help.

user116211

Or play the SU anthem loudly.

Sir Cumference

@MAFIA36790 Oh god

no more of that

user116211

good.

Sir Cumference

03:57

@SirCumference I get that enough as it is

user228700

What's the SU anthem?

Sir Cumference

@Kaumudi Look up "Soviet Union national anthem"

Soviet Union Anthem (Rock version)

►

dmckee

@Kaumudi Well, the units of electric field are not force units in any system that I've noticed, so it's a proportionality rather than an equality, but yeah. That's the basic idea.

user228700

04:12

@SirCumference :-P I'll check it out later, thanks.

user228700

@dmckee It's the force per unit charge, so it makes sense that its units are N/C...

dmckee

Well, yeah. But that's not quite what you wrote. A very minor nit, but in trying to form such definition it pays to be very careful. (Also, in E&M I hesitate to assume SI, other systems remain popular for the subfield, and people can get quite militant about the matter.)

user246160

@dmckee Hello. I am new to the chat and I see you are the moderator. Is this the right place to ask short questions like the one I asked above ?

dmckee

@TheStackExchange Yeah. You can ask almost anything here, but the number of people present varies, so your odds of getting a good response can be unpredictable.

The answer is that all kinds of things can nucleate the change for super-[verb]ed materials, and once it starts it proceedes quickly because it is very energetically favorable.

And of course it is way cool to watch a bottle of water you left in the car overnight freeze in a few seconds.

Sir Cumference

@SirCumference Testing something

dmckee

04:22

Not so cool to see a cup of cream you accidentally super-heated flash-boil and spatter you and your spouse with scalding dropplets.

But no permanent hard done.

user246160

@dmckee Sorry, but i could not understand the statement "all kinds of things can nucleate the change for super-[verb]ed materials".

user246160

Could you explain that in any other way ?

dmckee

@TheStackExchange bulk materials at rest are very uniform. Without something to break up that uniformity the phase change may not start.

Dust and scratches on the container surface and the like form places where conditions are sufficiently anisotropic for the change to starts.

user116211

@dmckee The whole Purcell is in Gaussian and he even gave a brief talk on why he preferred Gaussian to other system.

dmckee

And so can density variation from vibrations, stirring utensils stuck into the cream, and sometimes even cosmic ray ionization trails.

(The last is, after all, how cloud chambers and bubble chambers work.)

@MAFIA36790 I've heard a couple such talks, though not from so eminent a personage.

user246160

04:27

@dmckee Oh, I see now. Thank you so much !

Sir Cumference

@SirCumference Testing again

user116211

What are you testing @SirC?

Sir Cumference

@MAFIA36790 Some userscript I wrote

user116211

ooh.

Secret

For those interested, the ongoing progress of my zero term algebra are all documented here. Later on I will add the proof of the nonexistence of multiplicative inverse of nilpotents in associative algebras

user116211

04:41

I'll read it in my off-time.

DanielSank

04:56

o/

user116211

\o

Ramanujan

Any one know what is it?

user246160

@Ramanujan Looks interesting. I do not know what it is. From where did you get it ?

Ramanujan

Emilio Pisanty, Berlin, Germany

44.1k 8 99 215

Secret

http://www.gridgit.com/post_oauth-2-0-flow-diagram_540998/
A larger version of this diagram can be found here. Seemed to be a complex plot of something

Ramanujan

05:08

@Secret thanks

Secret

But otherwise, only Emilo will know what it is. Considering the resolution, it must be something from his experimental data

koolman

05:23

Can anyone explain me what is difference between gauss law in magnets and charges

@S007 @JohnRennie

user246160

@koolman May I explain ?

user116211

@koolman, please don't ping users unnecessarily; anyone interested will respond to your query.

koolman

@TheStackExchange yes please

@MAFIA36790 sorry , will not do it again

user116211

good; quick learner.

user246160

For charges the Gauss Law states that closed integral of E.dA is equal to the electric flux emanating from within the closed surface but other the other hand Gauss Law for magnetism states that the closed integral of B.dA is equal to the 0.

user246160

05:30

Gauss's law

In physics, Gauss' law, also known as Gauss' flux theorem, is a law relating the distribution of electric charge to the resulting electric field. The law was first formulated by Joseph-Louis Lagrange in 1773, followed by Carl Friedrich Gauss in 1813, both in the context of the attraction of ellipsoids. It is one of Maxwell's four equations, which form the basis of classical electrodynamics. Gauss' law can be used to derive Coulomb's law, and vice versa. == Qualitative descriptionEdit == In words, Gauss' law states that: The net electric flux through any closed surface is equal to 1⁄ε times the...

user246160

Gauss's law for magnetism

In physics, Gauss's law for magnetism is one of the four Maxwell's equations that underlie classical electrodynamics. It states that the magnetic field B has divergence equal to zero, in other words, that it is a solenoidal vector field. It is equivalent to the statement that magnetic monopoles do not exist. Rather than "magnetic charges", the basic entity for magnetism is the magnetic dipole. (Of course, if monopoles were ever found, the law would have to be modified, as elaborated below.) Gauss's law for magnetism can be written in two forms, a differential form and an integral form. These forms...

koolman

Why for charges it is not zero

user246160

E.dA is actually the flux emanating from any surface that is why. Any charges within the closed surface emanates a field which in turn results in a flux

user246160

But that is not possible for magnetism as magnetic monopoles do not exist.

koolman

Ok i understand it

@TheStackExchange thanks

user228700

05:48

@dmckee Hmm, yes. Thanks :-)

user228700

@MAFIA36790: I don't think S007 or JR mind if pinged. I can't say the same for other users, of course.

user116211

It's not whether they mind or not; but this is not a good practice to ping someone arbitrarily in the chatroom.

user116211

If they are willing to respond, they will do by themselves.

user228700

06:06

@MAFIA36790 It's not okay to arbitrarily ping random users in the chatroom. However, if there is no way to call their attention to a question, then pinging is necessary! Why else is the option even available? Speaking for myself, I don't usually ping anybody other than JR or S007 and I do this only because Ik for a fact that S007 doesn't mind and JR, well, he hasn't minded till now and I don't ping him every 2 seconds either.

2

user228700

I'm unable to understand why the notion of "pinging people is bad" exists. It's not like I'm pinging everyone /people whom I dunno. I ping just one/two people at most, if my question is that urgent.

2

John Rennie

06:24

@Kaumudi re the electric field/intensity: you often find words are vaguely defined in physics because we tend to us equations. For an electric field I'd normally use the field strength in volts per metre, which is usually denoted $E$.

The force on a charge Q is $F=EQ$, which tallies with what you were saying about force being used to define the field.

i don't recall seeing the word intensity used in connection with an electric field, but as I say the terminology is somewhat loose.

$E$ is defined as $E = \nabla \Phi$, where $\Phi$ is the potential (in volts). That's why the units come out as volts per metre.

user228700

Alright, thank you :-) Also, morning!

John Rennie

06:40

There can be a temptation to spend time trying to work out exactly what a particular author means, but the trouble is that doing so isn't necessarily teaching you anything about physics - just about that author's terminology.

Anonymous

@koolman You can join this room so that you can ping me when needed chat.stackexchange.com/rooms/48921/… . Here, sometimes pinging can disturb other users.

koolman

Oh thank you

Anonymous

@Kaumudi @Ramanujan You may join this room chat.stackexchange.com/rooms/48921/… if you wish :)

koolman

07:20

@JohnRennie please join the room

user228700

@S007 Alright, thanks :-) I quite like this room tho, so I'll stick around. As long as you don't mind me pinging u, I don't think it'll be a problem.

user228700

@JohnRennie Indeed.

David Z

08:22

@S007 I've never known pinging to disturb anyone except the person being pinged. So there's no point in having a separate room for it.

@Kaumudi I think the point is that when you ask a question which can be answered by anybody here (with relevant expertise), there is no need to ping specific members. If they are watching the chat, they'll see your question and respond to it. If not, it's presumably because they are busy doing something else. Pinging them is considered rude because you're dismissing the chance that someone else who isn't busy may be able to help you right away.

Actually in light of what I just wrote, maybe I should take back my previous reply to S007 :-P Pinging specific people could disturb others who would be equally capable of helping. But that seems like an awfully minor thing to get offended about.

John Rennie

Maybe we need to revive this idea:

17

Q: Opt-in to be always pingable to answer questions

Original idea: List of users with specialities who are willing to be pinged--should we do this? This is a place where users can opt in to be pinged by other users regardless of their chat status. What sort of pings is this for? Try to keep the ping related to P.SE . Some example uses of...

discussion questions answers chat

David Z

That was solving a different problem, I think. Though to be honest, I'm not sure I ever figured out just what problem it was solving.

Physics Meta

0

Q: To find the speeds of two skiers

So I was able to find the speed of one of the skiers in the following question but couldn't find the 2nd one, as it involved conservation of momentum. So here is the question. In a movie stunt, a 65 kg skier starts from a rest position at the top of a hill 30m high. She slides down the hill to t...

support physics

user228700

@DavidZ That's...a very different perspective.

David Z

08:37

Take that as an explanation of the reason why a person might get offended by pinging specific users to answer a general question - not as an official statement about the expectations for this room

Of course, like I said, it's a pretty minor thing. And of course, if you do specifically need someone in particular to reply (e.g. because you're referencing an earlier conversation that only they were involved in, or you're asking something about a post they made, or so on), it's absolutely fine to ping them.

user228700

@DavidZ Hmm, I hadn't thought if it this way before, thanks :-) The other day, there was a huge row over at The Periodic Table when person x pinged everyone in the room to ask for help.

user228700

@JohnRennie: Just to be on the safe side, you don't mind it when I ping u, do you? :-P

Ramanujan

in Mathematics, Oct 20 at 13:02, by Balarka Sen

@Ramanujan Please don't ping random people to answer your questions.

in Mathematics, Nov 13 at 10:41, by Balarka Sen

@Ramanujan Since you are part of the community, you're entitled to listen to the people in here, especially if you're going to keeping pinging them demanding to answer your questions. And it's not just me, this message was starred by 13 other chat-users.

John Rennie

@Kaumudi no you're welcome to ping me. In general I don't mind being pinged by people I've had conversations with before. I don't see it as any different from friends texting me. I'm not so keen on being pinged by people I've never heard of, though that happens only rarely.

user228700

Alright then :-) U can keep trusting me not to ping u for no reason.

Anonymous

09:40

@DavidZ I completely agree with you. But I would like to interact with some particular users in another chat room even on topics other than physics. Also, some of the unwritten rules followed by some users here on hbar and the continuous monitoring of messages by moderators doesn't suit me well. Anyway thank you for giving your views :).

David Z

@Kaumudi oy :-/

@S007 The rules are really all just specific cases of "be nice". Also be mindful that messages on all of SE chat are "continuously monitored" - it's not just this room, and not even just the rooms associated with sites. If you want to have chats that can't be seen by certain people (like moderators), you'll need to find somewhere separate from SE to do that.

You are, of course, welcome to make separate chat rooms, I'm just saying that pinging isn't a reason to do so. And I wouldn't want anyone to be misled into thinking that, say, pinging is forbidden here but allowed in other rooms.

2

user228700

Yeah, this pinging issue gets out of hand sometimes. That day, about 30 messages were moved to the trash from The Periodic Table.

user228700

People were very edgy :-P

David Z

wow

Often anything that breaks the norm will put people on edge

user228700

10:01

True.

Bernard Meurer

11:37

@JohnRennie Ping

John Rennie

Morning

Bernard Meurer

What up chap?

John Rennie

Quiet this morning.

I'm just going through my SF books and doing a bit of reorganising.

Bernard Meurer

Shhh. Don't jinx it man

John Rennie

Dammit, I've just realised I forgot to do something at work. Oh well. It'll wait until tomorrow.

Kyle Kanos

11:41

sigh Paternity leave is ending soon

John Rennie

@KyleKanos Until your sixth child :-)

Kyle Kanos

Hmm, having a 6th just for a few weeks (paid) vacation.....might be worth looking into

When we first started dating, my wife told me she wanted 14 kids. I'm reasonably certain she was just saying that to see if I was okay with a big family

Bernard Meurer

@KyleKanos Saying you want 14 kids is a dangerous move in a date :P

If someone even start saying something that remotely resembles "kid" I might just run away

Kyle Kanos

It'd weed out the guys who don't want several kids though, which is what my wife wanted

@BernardMeurer At some point in your life, this probably will not be true

Bernard Meurer

@KyleKanos If that changes I'll pay you a beer

Kyle Kanos

11:51

Hmm. I think I should have activated the boomer dojo instead of the heliport for this BTD5 level; it should still work

John Rennie

@BernardMeurer Most of friends have children and they've found it a surprisingly rewarding experience - though not without its trials :-)

Kyle Kanos

I get a good laugh out of my kids each day

They say & do lots of funny things

Bernard Meurer

@JohnRennie I was a kid not so long ago, I have a very clear memory of me as a kid. If I ever love a woman I do not want her to have to go through mini-me :P

Also as a kid I broke all the things

Kyle Kanos

My kids break crap all the time

Toys, dishes/cups, household items

Just part of life

Bernard Meurer

Jesus, you have 5 of them

Kyle Kanos

11:55

And which is why they're not allowed to touch my tools or the pool equipment without my being right there

Bernard Meurer

How do you still have a house?

John Rennie

@BernardMeurer these things are rarely logical decisions. Your brain employs all sorts of sneaky psychotropic chemicals to persuade you to have children. Evolution can be a bitch sometimes :-)

Bernard Meurer

Why is the pool equipment dangerous?

Kyle Kanos

It's mostly the younger side that breaks things. Once they pass 4 or 5, the rate at which they break things plummets

Bernard Meurer

@JohnRennie That's the true fear :P

The rate at which I destroyed things grew exponentially from the point when someone taught me how to use a screwdriver

Kyle Kanos

11:57

@BernardMeurer Most of the dangerous stuff is protected from even idiots like me, but it's majorly expensive to repair

Bernard Meurer

Than it plummeted at around age 13

John Rennie

@KyleKanos are you a QFT guy or was it GR? A QFT question has suddenly occurred to me ...

Kyle Kanos

@JohnRennie Sadly neither. I was a astrophysical fluids & astroparticle physics (sorta) guy

Bernard Meurer

@JohnRennie He's the debit collector for CitiBank

Kyle Kanos

^ am not

John Rennie

11:59

Ah yes. Oh well, I have to wait for ACM to wake up I guess.

Kyle Kanos

Wait...ACM sleeps?

3

Bernard Meurer

@KyleKanos He has to reboot every day, and his hardware takes a long time to POST

Kyle Kanos

I always assumed he was a robot and didn't need sleep

user228700

@JohnRennie: Are u familiar w/ fluid mechanics? Also, are u busy? :-P

Bernard Meurer

It's not quite "sleep". I told him I'd help him out by installing SSDs, but he's attached to his 1980's drive

John Rennie

12:02

@Kaumudi I would guess that any fluid mechanics you encounter will be pretty low level and I'm familiar with it up to that point. Doing fluid mechanics properly quickly gets hideously complicated.

user228700

Alright! I've only just started so please expect dumb (af) questions. I have one right now. The following diagram's given in my textbook:

user228700

John Rennie

OK

user116211

@KyleKanos I doubt; even last day I talked he was awake all through dawn; he was so involved in modifying his answer that he forgot to sleep.

John Rennie

Presumably this is working up to saying $dP/dy = \rho g$ ...

user228700

12:05

@JohnRennie You are correct! (:-P)

John Rennie

And you want to know how that works?

user228700

Anyhoo, alright, assuming that $dy$ is infinitesimally small and all, it makes a little sense that the pressure on top is higher.

user228700

Hang on, I'm not even sure how that's true. It makes 100% intuitive sense but I'm unable to reason it out...

Anonymous

@Kaumudi Pressure increases with depth.

user116211

@JohnRennie We once got tutored by a retired applied physicist who never got married. I wonder how they live such a secluded life.

user116211

12:10

Hey hey @Qmech.

John Rennie

@Kaumudi In fact it doesn't matter whether $dy$ is infinitesimal or not. We just make it infinitesimal so we can use integration.

user228700

OK..?

John Rennie

Well, it doesn't matter as long as we take $g$ to be constant (which we do). It would matter if $g(y)$ was a function that depended on $y$.

user228700

Right...

heather

@Kaumudi, hello!

user228700

12:12

Hi Heather :-)

Sir Cumference

Morning

heather

good morning @SirCumference

John Rennie

@Kaumudi If you have some element with an area A the force on the top of the element is $F = PA$. The force on the bottom is then $F = PA + mg$ where $m$ is the mass of the element. Yes?

user228700

Yep.

user228700

@SirCumference U didn't sleep?

Sir Cumference

12:14

@Kaumudi Funny you should mention that, I managed to get 10 more minutes of sleep than usual.

Which totals to 10 minutes. :)

heather

geesh, I think my computer has split personality disorder.

John Rennie

And it therefore the difference in the force is $mg$, and the difference in the pressure is $mg/A$. Yes?

user116211

Hello @ortho.

Sir Cumference

@Kaumudi But seriously speaking, I was up all night

orthocresol

@Kaumudi I'd mention that the messages I migrated to the other room were just people testing out the chatroom bot, which leads to the generation of a lot of large images.

John Rennie

12:15

And we haven't had to use any approximations that depend on $dy$ being infinitesimally small.

Sir Cumference

Almost no sleep ;-;

user228700

@JohnRennie Yes.

user228700

@SirCumference :-(

user228700

@orthocresol Perhaps u don't remember the day M.A.R and person x fought it out ._.

orthocresol

hey @MAFIA36790

@Kaumudi I didn't remove anything that day, though.

user228700

12:16

@orthocresol U did! Anyhoo.

orthocresol

There were one or two messages that were deleted, because they contained profanities, or were outright rude. Not 30.

John Rennie

@Kaumudi the mass is $m = \rho V = \rho A \,dy$, so the $A$ cancels out and we get $\Delta P = \rho g \,dy$.

orthocresol

Just wanted to clear that up. We don't delete messages just because we don't like what somebody is doing. Have a good week everybody :)

user228700

@JohnRennie Riight. Okay, lemme think about this some more...

heather

seriously, I think if the linux side interacts with the chromium side any more than it has too, it decides to be grumpy =P

John Rennie

12:18

@Kaumudi And as before we haven't had to assume anything is infinitesimal.

Anonymous

@Kaumudi I don't understand how pressure at the top of the block is higher than at bottom. It should be the reverse...

heather

ex: can't log into gmail on the linux side without there being "suspicious activity" when I log back in on the chrome side.

user228700

@orthocresol I realize I exaggerated but it was definitely more than just one or two.

user228700

@S007 That was my question too!

John Rennie

But it's nice to have a differential equation for the pressure so we can apply in more complicated circumstances e.g. where $g$ depends on $y$. So we let $dP$ and $dy$ become infinitesimals and we get $dP/dy = \rho g$.

Anonymous

12:20

@Kaumudi Your book is wrong. P+dP should be at the bottom as pressure increases with depth.

orthocresol

@Kaumudi There were three messages moved to Trash; you can see, because migrating messages leaves a note in the chatroom. chat.stackexchange.com/transcript/message/33114445#33114445

John Rennie

@S007 that would make more sense, but you can put P+dP at th top in which case dP is negative.

It depends on whether you take $y$ to be height or depth.

Anonymous

@JohnRennie If dP is negative it would be a really confusing notation for beginners :-P

user228700

@orthocresol Oh God. My brain was terribly wrong then, sorry! It was all so dramatic that I subconsciously exaggerated :-/

user228700

@JohnRennie Yeah, that too...

orthocresol

12:23

@Kaumudi mouse hovers over suspend button

2

John Rennie

@S007 Yes, I agree, I would have put P+dP at the bottom.

user228700

@orthocresol Noooooo :-P

orthocresol

@Kaumudi ;D have a good day. I'm off for lunch.

user228700

@orthocresol You too :-)

user228700

Awesome. Just my book being a moron again. Thanks, guys!

heather

12:26

got my first gold badge (::fist-pump::)

user116211

@heather great. well done.

heather

(::looks over at John Rennie's 32...::)

=P

user228700

@heather Awesome! ::pumps fist:: :-)

Kyle Kanos

@heather Well that means you're awesome

What did you get it for?

heather

@KyleKanos, close vote reviewing

next step: get on the leaderboard =P

Kyle Kanos

12:30

Pfft...good luck on that

heather

I have no idea how you got 10000+ reviews

Kyle Kanos

Obviously one vote at a time

:D

user116211

@heather full time dedication.

heather

I think my brain would melt out my ears or my mom would yell at me for spending so much time on it =P

Kyle Kanos

And also what MAFIA said

heather

12:33

=D

Kyle Kanos

I did that mostly during my grad school days, I had plenty of time for reviewing then

Not so much now that I have ~12 hour work days

user116211

13:05

> It is almost impossible for me to read contemporary mathematicians who, instead of saying "Petya washed his hands," write simply: "There is a $t_1<0$ such that the image of $t_1$ under the natural mapping $t_1 \mapsto {\rm Petya}(t_1)$ belongs to the set of dirty hands, and a $t_2$, $t_1<t_2 \leq 0$, such that the image of $t_2$ under the above-mentioned mapping belongs to the complement of the set defined in the preceding sentence."

user116211

I think Arnold had Bourbaki in his mind while saying that.

user116211

source

Ramanujan

13:17

Anyone here has installed chatjax of robjohn on android?

koolman

What is this

Ramanujan

When some one types $\sqrt2$ we can see it as √2

koolman

Ok

How oscillating electric field produces oscillating magnetic field

ACuriousMind

@KyleKanos Well, me becoming a mod means that only JohnRennie will possibly overtake you now since I now skip most items when I do visit the queue.

user116211

@ACuriousMind If a mod wants, can he clear the review queue like a usual non-mod reviewer or a close vote from mod close the post singly by himself? I'm still not aware about how mods review.

ACuriousMind

13:38

@MAFIA36790 I could clear the review queue - and my vote would be binding in every single case (but we don't do it because that defeats the purpose of the review queue).

Physics Meta

13:57

1

Q: Spam - gibberish or just promotional stuff

Let's say I come across a post which solely says: akdhlkfahsdkhfalsdhf Obviously this person is doing a little home-row bashing, but how are we supposed to flag this? My first thought was spam, but I wasn't sure if that was purely for promotional type stuff. My second thought was very low q...

discussion flagging spam

Secret

14:29

priceonomics.com/…

2

You may also like: A cat walk on keyboard detector

ACuriousMind

Hmm...if I take this formula literally the number of domain walls in a certain theory is -2.

::sigh::

Sanya

@Secret that's cool

Kyle Kanos

@ACuriousMind Actually looking at the "leaderboard," it's kinda nice to see the Top 20 all have Steward badges. That wasn't true when I left around April

Slereah

15:07

Hey

@ACuriousMind a bit low

Apparently QFT in multiply connected manifolds isn't too hard, but the hard part is finding the covering space

ACuriousMind

15:24

@Slereah Yeah. I think they just forgot to remore a $(-1)^k$ in the formula which they needed in another place but still, the prediction of having less than zero walls is intriguing

Slereah

Domain roofs

Secret

15:51

what does a negative domain wall even mean...?

Slereah

Nothing.

ACuriousMind

Arrrrgh. "Consider the $N=1$ theory at large $N$"...

Slereah

For large values of 1

ACuriousMind

It took me too long to realize those two $N$ are not the same. Would it have killed you to write the first as $\mathcal{N}$ as usual, Vafa? ::shakes fist::

It's actually rather obvious, but still unnecessarily annoying.

Slereah

16:08

Same thing happened to me with $\sigma^2$

I thought it was a square

So it turns out it was the second Pauli matrix

ACuriousMind

Oh yeah, I can see that, especially because the square, like $L^2$, would be a perfectly reasonable object to consider.

Secret

[Digging my old folders] Somewhere I found this article and I already forgot why I have downloaded it in the first place

Slereah

Well it is a more natural thing to expect in formulas than $\sigma_y$

Kyle Kanos

16:25

sigh still don't understand why crap questions that make little sense get upvoted :/

FrancescoS

Hi guys

Kyle Kanos

Hi

FrancescoS

Can you suggest some reference about the naturalness principle and hierarchy problem? Without too much calculations. I don't want to focus on the calculations, but on the physics. Why we need to introduce such principle, where the hierarchy problem arises, the SM alone (as ren. theory without considering gravity) is affected by this problem? Or it arises when we want to find the most general theory?

Kyle Kanos

Never heard of either

FrancescoS

*some gooood reference

Kyle Kanos

16:30

Wait, no, I've heard of the hierarchy problem, just forgot what it was

(and had to Google it, which reminded me)

ACuriousMind

@Slereah Hm, what am I to make of a formula that tells me there's one domain wall separating a vacuum from itself? :P

These papers are all terribly written, or I'm just not made for reading them :O

Slereah

The same ground state?

ACuriousMind

@Slereah Hm, wait!

It might be that there are two vacua at the same level here.

Or that I'm not supposed to use the formula for $k=0$.

Slereah

well then that's pretty normal :p

ACuriousMind

I'd be considering a $\mathrm{U}(0)$ theory in an intermediate step...welp, I guess I just can't plug in $k=0$ here

@FrancescoS The "naturalness principle" is just that it feels "unnatural" for the parameters to a theory to have dimensionless ratios far from 1 at specific scales. You won't find any "good references" on it because it's not a physical principle, but purely aesthetical

I've never been too convinced the hierarchy problem is actually a problem.

FrancescoS

16:41

@ACuriousMind thanks. I would like to read some discussions about this topic… I know it's not a physical principle, but it is due to a common sense probably

If you have in mind any possible reading about it… tell me :)

ACuriousMind

I haven't read much about it specifically because it seemed more like aesthetic dissatisfaction with our models than anything productive :P

FrancescoS

oh… ok :)

Slereah

Is $U(0) = Z_2$

Secret

ABSOLUTE ZERO - PBS NOVA DOCUMENTARY (full length)

►

I have an obssession of cold technologies

ACuriousMind

16:57

@Slereah Probably not, it's just $\{0\}$, since $\mathrm{GL}(0)$ is already just $\{0\}$ - there is just one linear map from the zero vector space to the zero vector space.

Slereah

I s'ppose

ACuriousMind

However, the 0-sphere is indeed two points ;)

At least, that way one can inductively define $S^n$ as the suspension of $S^{n-1}$ starting from $S^0$.

Alex

18:23

What exactly does it mean for a state to be in z spin up state $\chi_+ = \left(
\begin{array}{c}
1\\
0\\
\end{array}
\right)$
where it has eigenvalue $\frac{\hbar}{2}$? Is the following the correct interpretation: Does it simply mean that the spin around the z axis can take only two values $\frac{\hbar}{2}$ and $-\frac{\hbar}{2}$ which corresponds to counter clockwise spin around the $\hat{z}$ axis (spin up) so that the vector points up the z axis and respectively counter clockwise spin around the $- \hat{z}$ axis (spin down) so the vector points down the z axis?

ACuriousMind

18:49

@Alex The spin around any axis for a spin-1/2 system can always only take the values 1/2 and -1/2. Since the components of the spin operator do not commute with each other, there is no definite quantum analogon to the classical angular momentum vector - at least two of the three components of the vector are always uncertain.

To be in the z-spin up state simple means that measuring the z-spin component of this state will always return +1/2

There is no "clockwise" or "counterclockwise" spin here at all - nothing is actually rotating, so talking about a "clockwise rotation" does not make sense. Classical angular momentum up along an axis would indeed denote counterclockwise (iright hand rule) rotation around that axis, though.

Alex

@ACuriousMind Okay thanks, I was going to ask how the eigenvectors of the z component of spin $\chi_+ = \left(
\begin{array}{c}
1\\
0\\
\end{array}
\right)$ correspond to a vector up the z axis of rotation but as you said it doesn't. So there is no definite physical interpretation of what spin is in quantum mechanics?

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