The h Bar

DanielSank

04:05

@MAFIA36790 ಠ_ಠ

Sir Cumference

Is 0celo back yet?

user116211

No; why would he?

Sir Cumference

Was hoping the time passed by now

user116211

@DanielSank I know it sounds sexist; but it's a mnemonic after-all.

Sir Cumference

Turns out 3 months is actually quite long

user116211

04:08

@sirC, obe has changed his username to delete and has apparently requested for deletion as his profile description contains the infamous ~~death~~ phrase delete me.

Sir Cumference

@MAFIA36790 Wait is he trying to delete all of his accounts?

user116211

@SirCumference all of his? I think because all are signed in under same email account and he wants to get out of SE.

Sir Cumference

@MAFIA36790 Goddammit, I gave him 2/3 my rep just so he could get the association bonus

Now he goes and wastes it literally days after

You can see why I might be a bit frustrated

...besides the fact we're losing half of the cool people in the chat

user116211

@SirCumference You showed your modesty to him; that's it; now it's his matter whether his account exists or not; but you awarded the bounty to a reasonably good answer.

user116211

@SirCumference I know; chat is not that like those golden days where Chris, 0celot, JD interacted here.

Sir Cumference

04:12

@MAFIA36790 JD isn't around anymore?

Huh, never noticed

user116211

Ohh!!! How could I forget?

user116211

@DanielSank o/

user116211

But I hope like after all the suspensions, this time also 0celo7 will return; so am waiting.

Sir Cumference

@MAFIA36790 ACM used to be more chill when he wasn't a mod :/

Now I feel like the chat's becoming a safe space

user116211

Hmm.

Sir Cumference

04:19

@MAFIA36790 Have any jokes to tell?

user116211

I fear, no.

Alfred Centauri

04:34

@SirCumference safe space for whom and from what?

user116211

@AlfredCentauri Those days followed lots of flagging, trashing comments; and now they have decreased a lot to none.

user116211

I think @sirC meant that.

Sir Cumference

04:59

@MAFIA36790 @AlfredCentauri I mean now the mods are willing to delete anything that can remotely seem offensive

Even if the tone is jocular

user116211

@SirCumference Hmm; yeh; some are beyond reason; but will not extrapolate anything without knowing any exact reason(s).

Sir Cumference

@MAFIA36790 What do you mean, "extrapolate"?

user116211

@SirCumference Because we will never know any exact reason, that would be futile to further delve into it.

Sir Cumference

@MAFIA36790 Any reason for what?

Alfred Centauri

05:18

@SirCumference I don't hang around here much (so what do I knows?) but I don't think that your assessment is fair. Frankly, the fact that I exist is offensive to some and the mods haven't deleted me (yet). :)

DanielSank

@MAFIA36790 \o

John Rennie

06:12

@Kaumudi pinging me at 1 a.m.? There's optimism for you :-)

user228700

@Mew: Remember when u tried to explain the derivation of mean life to me? I didn't exactly return to it until today and I've realized that I still don't quite understand it, in that I don't understand why the expected value gives time, as opposed to...uhh, I dunno what .__.

user228700

@JohnRennie :-) Morning Sir. I was hoping u'd see the notification whenever u do come online.

John Rennie

@Kaumudi Well it worked :-) Did you want to ask about the mean life?

rob

Greetings, all

John Rennie

Morning Rob

user228700

06:21

@JohnRennie No, I was hoping u'd be familiar w/ the Ostwald-Walker Experiment to determine the lowering of vapour pressure...are u? (fingers crossed)

user228700

@rob: Hi :-)

John Rennie

I haven't heard of the Ostwald-Walker Experiment, but that doesn't mean I can't help with it. I may know it by another name.

user228700

OK, one second, let me show u a picture:

user228700

in The Periodic Table, 5 hours ago, by Kaumudi

John Rennie

OK, that looks straightforward.

user228700

06:26

@JohnRennie: I must have lunch. I'll be back in about 30 minutes!

John Rennie

@Kaumudi That would be my priority too :-)

rob

bmj.com/content/327/7429/1459

Conclusion: No double-blind studies of parachute effectiveness have ever been performed; people who think double-blind studies are the only way to learn in medicine are encouraged to participate in one.

John Rennie

The BMJ has a long history of publishing joke articles at Christmas :-)

When I was a lad our family doctor was also a family friend and he used to bring round the Christmas issue of the BMJ with the jokes in. It's typical British humour, and I heartily approve :-)

A quick Google found a copy of the article here

user228700

07:05

I'm 9 (whoops!) minutes late, sorry!

user228700

@JohnRennie: Are u still around..?

John Rennie

Yes, though right now I'm in the middle of something. Back in 5 minutes.

user228700

OK...

John Rennie

@Kaumudi I'm back!

user228700

Okay!

user228700

07:16

I don't understand what is happening in that experiment...

John Rennie

It's a lot simpler than it seems.

Suppose I have a solvent with a vapour press $P$ and I bubble air through it slowly enough that the solvent equilibrates with the air.

Then the amount of solvent that evaporates per unit volume of the air bubbled through is proportional to $P$.

user228700

When u say "equilibrates", dyou mean that since the pressure in the jar increases, some of the vapour will condense back?

John Rennie

No.

I mean it takes time for solvent to evaporate - it doesn't happen instantly.

user228700

We begin to pass dry air through those jars once those systems have attained equilibrium, no?

John Rennie

I think you are approaching the expt the wrong way.

user228700

07:22

OK :-| What dyou mean by this:

user228700

> "I bubble air through it slowly enough that the solvent equilibrates with the air."

John Rennie

Suppose we start the dry air flowing at a constant flow rate, then go away and have lunch.

When we get back the system has settled down so that the rate of solvent evaporation is enough to keep the air saturated with solvent.

i.e. the partial pressure of the solvent in the air is equal to the vapour pressure of the solvent.

This is what is meant by equilibrium.

user228700

Hmm. Why is this wrong, then:

user228700

> "When u say "equilibrates", dyou mean that since the pressure in the jar increases, some of the vapour will condense back?"

John Rennie

The pressure in the vessels doesn't change. The pressure is everywhere 1 atmosphere.

It's a constant flow system, so for every cubic metre of air that flows in a cubic metre of air flows out.

user228700

07:28

Alright, what's with the equilibration, then?

user228700

(I'm very sorry, please bear with me...)

John Rennie

Because if you flow air very fast through the solvent the solvent can't evaporate fast enough to keep the air saturated with the solvent vapour.

If the air wasn't moving then eventually it would become saturated in the solvent i.e. the partial pressure of the solvent vapour in the air is equal to the vapour pressure of the solvent. Yes?

user228700

Yep, I think I understand what u mean now...

John Rennie

This is what I mean by equilibrium. I mean the evaporation has time to reach equilibrium so that the solvent vapour in the air is equal to the vapour pressure of the solvent.

And that means the air has to be flowing slowly compared to the rate of solvent evaporation.

user228700

I'm finding it a little difficult to imagine that with the air above the liquid constantly flowing, the liquid equilibrates at all! :-|

user228700

07:33

@JohnRennie Ohh. OK...

John Rennie

Well, you're correct that a flowing system can never reach complete equilibrium, but if the flow rate is slow enough it can be very close to equilibrium.

user228700

@JohnRennie Yes, I guess...

user228700

@JohnRennie OK...

John Rennie

Was that your main concern about the expt?

user228700

I had several main concerns regarding that expt. :-P

user228700

07:36

I wasn't fully able to understand:

user228700

17 mins ago, by John Rennie

Then the amount of solvent that evaporates per unit volume of the air bubbled through is proportional to $P$.

John Rennie

Suppose the air is still i.e. not flowing at all. In that case the total mass of vapour in the air is proportional to the vapour pressure. Yes?

user228700

But I think I understand it now. Gimme one moment to ascertain if I definitely need help, please...

user228700

No matter, it will be good to make sure that I understand it properly (and I'm not so sure of this now).

user228700

@JohnRennie Yep.

John Rennie

07:41

@Kaumudi So if we quickly pump out all the air with solvent in it, pump in dry air and wait, then the mass of solvent lost by the liquid due to evaporation into the dry air is proportional to the vapour pressure. Yes?

user228700

Yeah...

John Rennie

That's all the experiment does. It flows dry air through and measures the weight lost by the liquid.

For every cubic metre of air that flows through the mass of vapour in that cubic metre is proportional to the vapour pressure.

user228700

But when the dry air comes out, it doesn't contain all of the vapour, does it?

John Rennie

@Kaumudi why not? Where would the vapour in the air go?

user228700

But what about the fact that the vapour saturates the air quickly enough and all? The liquid has to start from scratch?

user228700

07:45

Okay, that didn't occur to me.

user228700

I think the expt. makes sense now, thank you, guru! :-)

John Rennie

We start with dry air flowing in, so you're quite correct that the vapour has to saturate the air during the time the air is passing through it. That's why it's imporant we keep the flow rate slow enough for the evaporation to keep up.

user228700

Hmm, OK, I understand...

user228700

Dyou have the time for one more (I promise!) question?

John Rennie

Yes, of course.

user228700

07:51

Alright, I was wondering exactly how the term boiling is really defined. Every time this question came to mind before, we were in the middle of discussing something else so I had to let it go...

John Rennie

The term boiling point is well defined. It is the highest temperature that the liquid can exist in equilibrium with vapour.

The term boiling really just means evaporation

user228700

:-/

John Rennie

There's nothing special about the process by which a liquid changes to vapour at the boiling point as opposed to a lower temperature. In this sense the process of boiling is just evaporation.

user228700

But:

John Rennie

The thing that's special is that at the boiling point we can't raise the temperature as long as liquid is present.

user228700

07:55

> "The boiling point of a substance is the temperature at which the vapor pressure of the liquid equals the pressure surrounding the liquid and the liquid changes into a vapor."

John Rennie

Well liquid changes to vapour at all temperatures ...

user228700

Aarfgh.

John Rennie

I wonder if you're asking why there is a boiling point i.e. why is it that there is some magic temperature at which the liquid disappears completely?

user228700

No, no, I just happen to be extremely confused about its definition .___.

John Rennie

Well the term boiling is somewhat vaguely defined. That's why I made the distinction with the term boiling point, which is precisely defined.

user228700

07:59

@JohnRennie To be clear tho, it's not like Ik why such a temp./pressure exists. I'd like to know, but I'm afraid that it wouldn't be very useful to me at the moment.

John Rennie

The spectre of Gibbs free energy floats into view ... :-)

user228700

Ah, darn it.

user228700

I'm only confused about the definition, sir!

John Rennie

@Kaumudi the definition of boiling or the definition of boiling point?

user228700

The latter.

user228700

08:08

@JohnRennie And it is at this temp. that the vapour pressure equals to the "outside pressure"? (Whatever that is)

John Rennie

Don't worry about the vapour pressure. That's related to why there is a boiling point (and it's to do with the free energy). The boiling point is defined as the division between the liquid and vapour ona phase diagram.

user228700

OK...

user228700

Alright, you know what? Maybe it's best if I didn't care so much about this right now.

user228700

Thanks very much _/\ _

John Rennie

If you draw a line of constant pressure, i.e. a horizontal line on that diagram, then where it meets the gas/liquid dividing line is the boiling point.

Understanding why the boiling point exists is an altogether more complicated business. Indeed a boiling point doesn't always exist. Supercritical fluids have no boiling point.

user116211

08:18

@JohnRennie, help please.

user228700

(Y)

user116211

Could you check this post @JohnRennie?

user116211

The MathJax is messed up. I don't know why.

John Rennie

The OP doesn't know how to do the large vertical bar on the right and he's tried a method that doesn't necessarily render perfectly.

user116211

I have edited; but they are not rendering properly; I'm not getting why.

John Rennie

08:22

I think you do \left. (stuff) \right|

user116211

checking.

user116211

Removed the bars; they are not required here.

user116211

Thanks @johnR.

John Rennie

The \left. is an invisible left delimiter so it matches with \right for any symbol.

user116211

ahh.

John Rennie

08:27

Or of course \right. though there isn't much need to use that.

user228700

Is anybody familiar w/ the relation b/w the number of elements in a given period and the principal quantum number?

John Rennie

@Kaumudi in principle yes, though I'm not sure exactly what you're asking ...

user228700

Well, this is what my textbook says:

user228700

And I'm not sure I understand that ^ very well...

John Rennie

08:31

What isn't clear about it? How we relate the period to the 1s, 2s, etc shells?

user228700

Yes, exactly that ^

John Rennie

In the first period there are only 1s electrons - no 2s, 3s, etc

In the second period there are only 1s and 2s electrons - no 3s, 4s, etc

In the third period there are only 1s, 2s and 3s electrons - no 4s, 5s, etc

and so on

user228700

Ah, so so the whole shell isn't filled, just one orbital at a time. That added to the confusion.

DanielSank

@Kaumudi that's a neat emoticon

user228700

Alright, I understand, thanks very much :-)

user228700

08:34

@DanielSank Yeah :-P Also, hi! :-) How goes it?

John Rennie

@Kaumudi the connection to the principal quantum number is complicated because in the fourth period the 3d orbital isn't full

That's because the 3d energies are higher than the 4s energies

user228700

Yes! ^

John Rennie

How we define the periods is more a matter of convenience than anything fundamental. It's done to make the periodic table look nice and simple.

user228700

Hmm, OK...

user116211

I still didn't get though why it didn't work:

\frac{\delta P}{\delta V} -\left(\frac{\partial S}{\partial V}\right)\bigg|_T \frac{\delta T}{\delta V} =\left(\frac{\partial P}{\partial V}\right)\bigg |_T.

user116211

08:40

It gives this mess:

user116211

$\frac{\delta P}{\delta V} -\left(\frac{\partial S}{\partial V}\right)\bigg|_T \frac{\delta T}{\delta V} =\left(\frac{\partial P}{\partial V}\right)\bigg |_T.$

user116211

O.o

user116211

It worked here.

user116211

WTF ;/

user116211

I'm at the sea now.

user116211

08:41

Why didn't it render there this way? Hmm.

DanielSank

@Kaumudi it goes.

2

Slereah

What is even quantum field theory

I don't even know anymore

Solenodon Paradoxus

What do you mean? :)

Slereah

I am lamentong the mess that is the definition of QFT

user116211

I wonder if Bourbaki ever wrote something on QFT, what it would look like.

Solenodon Paradoxus

08:55

Why do you think that physical models have to be rigorously defined to yield predictions

Ok, I can see why you would think that about fundamental models, i.e. quantum gravity and unification of everything

But why does QFT have to be rigorously defined?

Slereah

Why not

Solenodon Paradoxus

I don't see how this is a meaningful question

Slereah

An effective model can be rigorously defined

Solenodon Paradoxus

Sure, but they don't have to be

Slereah

Also the problem with quantum gravity is

String theory is also not defined rigorously

Mew

08:59

what does defined rigerously mean?

Slereah

And covariant gravity is just QFT again

Mew

@SolenodonParadoxus do you have an example of any effective models that aren't rigerously defined?

Slereah

Axiomatically

Solenodon Paradoxus

@Mew QED/Yang-Mills in 4d

Mew

ok

so there you go, it has now been proven that effective theories don't need to be rigerously defined

Solenodon Paradoxus

09:01

@Slereah well, we have two points
1) String theory doesn't work (probably)
2) Perturbative gravity doesn't work
But it doesn't mean that something else wouldn't :)

Slereah

Which one works these days

Solenodon Paradoxus

None of them

Slereah

IIRC LQG doesn't either

Solenodon Paradoxus

You're missing my point

- What we have today is crap

- All possible theories are crap

See the difference?

And I mean "crap" in the most respectful way possible, of course :)

John Rennie

@Slereah what caused this sudden crisis in confidence? It's a bit early for your mid-life crisis.

Solenodon Paradoxus

09:03

Btw about LQG - it obviously doesn't work, at least because it doesn't include anything except for gravity. But it is rigorously defined.

Which suggest that something else could work and be rigorously defined at the same time

Slereah

Well yes, that's why I didn't mention it in non rigorous

What if the universe is unknowable

Solenodon Paradoxus

No its not

Slereah

Let's go back to an agrarian society

Solenodon Paradoxus

How about this argument.
If the Universe is in fact unknowable - then it won't hurt if you spend some time trying to make sense of things, unsuccessfuly.
But if it isn't - then by giving up you well... give up :)

Slereah

Oh no

It's Pascal's wager

Solenodon Paradoxus

09:08

Sure, do you see a problem with it?

(I am half-joking of course)

Solenodon Paradoxus

09:21

@MAFIA36790 They'd probably spend a couple of centuries defining the terms in the nomenclature

Slereah

What if QFT was inside us all along

Solenodon Paradoxus

@Slereah quite possible; QFT works in mysterious ways

@MAFIA36790 now that I think about it, Bourbaki's QFT papers would probably look similar to everything else they wrote. You'd have to be Bourbaki to tell the difference.

Slereah

"This master thesis should be readable to anyone with a healthy knowledge of real and complex analyis, measure and integration theory, functional analysis, operator algebras, Lie groups and Lie algebras."

Simple enough

user116211

10:22

It is noteworthy @Solenodon that in the book of Bourbakis, they have said no mathematical preliminary knowledge is required (except, might be, the prior volumes of Bourbaki, as there are many references to previous works). It is explicitly written in the Set Theory book.

user228700

10:43

@JohnRennie: I'm sorry, I've probably used up my quota for today, but can u help me with s'thing?

John Rennie

Yes of course

user228700

Okay. It's about the mean life (eek!) of a radioactive sample again.

John Rennie

OK

user228700

@Mew said that the expected value=mean value can be obtained by integrating the function $f(t)=N_{\circ} e^{-\lambda t}$ from 0 to $\infty$.

John Rennie

Yes

user228700

10:46

I don't understand how that works. Why is this value in units of time?

John Rennie

If you have a population of $N$ people, then to get the mean life you add up the lifetimes of all the people and divide by $N$. Yes?

user228700

Yep yep.

John Rennie

OK with radioactive decay the equation: $$N = N_0 e^{-t/\tau}$$ gives the number of atoms that still exist at time $t$. Yes?

user228700

Uhh, yes!

user228700

However, in the time frame b/w $t$ and $t + dt$, not every single one of these nuclei would've decayed, no? So how can $N$ give the lifetime of these nuclei?

John Rennie

10:52

Erm, yes, I've paused because I'm trying to think about it. You want the number of atoms that decay in the range $t+dt$, then multiply this number by $t$ to get the total lifetimes of those atoms. Then integrate wrt $dt$ to get the total lifetime.

user228700

Hmm, yes...

John Rennie

I might have to get back to you on this one. I need to go out in 5 minutes and I have to think this through.

user228700

Oh, OK...

user228700

I found this tho:

user228700

25. Physics |Radioactivity | Calculation of Mean Life of a Radioactive Element | by Ashish Arora

►

user228700

10:56

^ He assumes $dN$ to be the number of nuclei decayed from time $t$ to $t + dt$.

user228700

I'm still a little bit confused as to why we must multiply this with $t$, tho.

user228700

Alright, @Mew seemed to understand this so I think I'll ask him if he comes.

user228700

Thank you :-)

John Rennie

@Kaumudi Well $t dN$ is the total lifetime of the atoms thatd decayed i.e. you have $dN$ atoms, and each lives $t$ seconds, to their total lifetime is $t dN$. Yes?

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