The h Bar

> "The size of an anion will be larger than that of its parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in $Z_{eff}$"

John Rennie

6:32 AM

OK, that makes sense.

user228700

How is it that the increased repulsion cause a decrease in $Z_{eff}$?

John Rennie

Remember that every electron exists as a diffuse cloud that has a non-zero probability at all distances from the nucleus (except at a few nodes). So every electron screens every other electron to some extent.

user228700

Yes...

John Rennie

So if you take a neutral atom and add one more electron then that extra electron is going to provide some additional screening.

In effect the other electrons see a bit extra negative charge in between them and the nucleus.

user228700

Hmm, this makes sense and all, but I don't understand why this is described using "repulsions" rather than "screening effect" by my book.

user228700

6:37 AM

Alright, thank you! :-)

John Rennie

It's a slightly artifical distinction. Electrons are repelled by other electrons and attracted by the nucleus. We tend to talk about screening when we mean a change in the attractive force and repulsion when we mean a change in the electron-electron interactions.

But both are just changes in the overall electrostatic forces experienced by the electrons.

user228700

Hmmm, okay. Thanks!

John Rennie

You probably shouldn't star everything I say (I'm assuming it was you) or the star board will get a bit dull :-)

user228700

Nope, wasn't me. I think it was koolman. From what I've observed, he has a tendency to star stuff that he agrees with, in general :-P

John Rennie

Ah, sorry, a false accusation!!

Danu

7:04 AM

@DanielSank Me. 4/10

John Rennie

7:22 AM

@Danu You're being generous, though I suspect the universe doesn't care what score we give it :-)

user228700

@JohnRennie Yes :-(

DanielSank

@Danu wat?

Ohhhhh

You rate the universe 4/10?

koolman

7:44 AM

@JohnRennie sorry will not do it again

John Rennie

@koolman it's no problem and I didn't mean to criticise

Danu

8:30 AM

@DanielSank Indeed

Would not create if I had the choice

John Rennie

@Danu Damn right. I mean Navier-Stokes? Whose bright idea was that?!!!

Danu

It Ain't Me Babe - Bob Dylan (5/4/65) Bootleg

►

Mew

8:45 AM

alo

user228700

9:08 AM

Hey :-)

Swapnil Das

11:06 AM

Hi.

heather

12:06 PM

hello, anyone around?

Sir Cumference

Howdy

heather

hello =)

Secret

12:34 PM

[Ramblings]
$$\frac{d}{dx}x^n=nx^{n-1}$$
$$\int x^n dx=\frac{1}{n+1}x^{n+1}$$
Now
$$\int x^{-1} dx=\frac{1}{-1+1}x^{-1+1}=\frac{1}{0}x^{0}=\color{red}
{\textrm{undefined}}$$
$$\therefore\textrm{Define: } \ln x=\int_a^x t^{-1}dt$$
Where is the derivative analogue of this $x\neq -1$ 'gap' in the integral power rule? Actually it is right here:
$$\frac{d}{dx}x^0=0x^{0-1}=0x^{-1}=0 \color{red}{\textrm{ (Often skipped as it is straightforward: $x\neq 0)$}}$$
NB $0^0=\textrm{undefined, but in most applications, it is natural to set as 1}$

Secret

1:18 PM

(The following is a nonsensical scribble based on the above observation. In practice it may not work cause the partial orderings imposed by zero terms is likely to mess up the + structure so badly that calculus may be impossible)

However, at least formally speaking (assume we have taken care of the collapse problem and the ordering imposing of the zero terms problem), one major thing that need to be changed is that 1 is no longer a two sided identity in order for the process to be inverted

Sir Cumference

1:43 PM

@Secret Well your zeroes look like potatoes and your integral signs are nearly straight lines...

2

Then again my handwriting sucks too

Secret

well they are scribbles, thus obviously they are going to be potatoes

user228700

Wow. Very quiet here, these past few days. Maybe I haven't been asking enough questions xD

Secret

Yeah you and Johnrennie used to be the main activity in the h bar

The lack of activity is one reason my wavefunction is mostly residing in the maths chat atm

Sir Cumference

@Secret 0celo and obe are gone too

Same with Chris

That cut half of the activity here

Secret

I am not gone, cause ACM and co. are still here

And actually my activity is still here, it just become a lot more surreal

user228700

1:49 PM

@Secret During the morning time, yes :-P

Secret

you will know when there is significant activity in this chat room when you realise the reduction of my zero term algebra screen dumps

e.g. if other users are discussing something, I won't land dump here

Jim

2:22 PM

@Kaumudi I'm sorry, I was out sick the last few days. I let my guard down and things got too quiet in my absence. I take full responsibility for this failure

I should appoint a Designated Jim, to fill the role if I'm absent. I just need to choose carefully, lest they become a Jimposter

Lucas Gautheron

2:46 PM

hi, anyone here knows when the LHC Run 2 results are to be presented this year ? im interested about pp collisions

ACuriousMind

@Jim Well, a Jim poster would just be someone who posts stuff Jim would say, no?

0

A: Redirecting light beams from beam splitters

My neighbors are attacking my grill with lasers and I want to find a way to prevent it keeps heating up my grill and burning my turkey

wat

Secret

https://www.newscientist.com/article/mg23230990-500-beyond-batteries-this-technology-could-revolutionise-energy/
(Slightly old news: When it comes to the evolution and improvement of some component technology that play huge role in communities, various researches done by private companies signify a gradual revolution hatching)

Jim

3:23 PM

@ACuriousMind After that, I'll have to consider you as a front runner for the position

@ACuriousMind they have massive CO2 lasers and are using them for practical jokes

ACuriousMind

@Jim I approve.

Jim

Has the guy never head of mirrors? Use them to retaliate and use his neighbour's weapon against him

John Rennie

> Well your zeroes look like potatoes

Have you been near any of Bernardo's computers? :-)

vzn

3:58 PM

Tests confirm that Germany's massive nuclear fusion machine really works / science alert

ACuriousMind

@vzn That's a link to this chatroom.

Secret

qz.com/679335/…

cannabis in food in a 5 star restaurant

John Rennie

@ACuriousMind I would guess he means the Wendelstein stellerator though the news isn't as dramatic as vzn's post suggests.

ACuriousMind

@JohnRennie Everything as usual, then :P

John Rennie

I would really like the stellerator design to prove a success as it seems a clever solution to a fundametal design problem with torus shaped reactors. But it's a long, long way from commercially viable fusion.

vzn

4:08 PM

@ACuriousMind oops rats sorry correction

Tests confirm that Germany's massive nuclear fusion machine really works / science daily

John Rennie

"rats" - I thought I was the only person who actually used "rats" as an expletive. I'm oddly reassured to know that I'm not :-)

vzn

@JohnRennie also like cartoon swearing @#%&... goes good with misbehaving computers too :P

Scientists are close to building a quantum computer that can beat a conventional one / science

Secret

4:24 PM

Nice, so things are making progress

John Rennie

4:56 PM

It seems a bit harsh to close this:

0

Q: What magnet configuration produces these magnetic fields?

Because the poles do not interact with each other my best guess is two small magnets (likely of the rod type), with either their respective south- (top right) or northpole (bottom left) visible. Is this correct?

koolman

Hii @JohnRennie

I have a doubt

John Rennie

What is it?

koolman

In the above question should the answer be D

John Rennie

OK, which of the options do you know to be true?

koolman

According to me a,b,c are true

David Z

5:09 PM

@JohnRennie Floris has a point, it does appear to be a check-my-work question... I think

John Rennie

@koolman Which make D the answer. So what are you unsure about?

koolman

But answer is given as C

Or the answer given is wrong

John Rennie

Ah, yes, sorry I misread the question. I assumed D was the answer like you did.

koolman

There are chances of the answer given to be wrong

Just I want to confirm if I am correct or not

John Rennie

Hmm, I can think of (different) reasons why both C and D should be true. So obviously one of my reasons is wrong but I'm not sure which.

koolman

5:17 PM

Why you think C is true

John Rennie

To be honest I'm not sure what the answer is. I'd have to sit down with a sheet of paper and work out exactly how much work was done in each step, which isn't an easy calculation.

I think C is true because if we go right round the loop A-B-C-D-A then we return to the same temperature, and we know that the heat gained is equal to the heat lost because the lengths of BC and AD are the same.

koolman

Great

John Rennie

If some internal energy had been lost as work then it would require more heat to be pumped into the gas in BC than was lost by it in DA. Because some of the heat would have been converted to work.

koolman

Now I am also confused between C and D

ACuriousMind

What kind of diagram is that? "heat" is not a property of a system, so points $(Q,T)$ do not represent states.

koolman

5:22 PM

@JohnRennie does in this process change in internal energy is not zero

John Rennie

@ACuriousMind It's a perfectly good diagram: AB is an adiabatic compression, BC is an isothemal expansion, CD is an adiabatic expansion and DA is an isothermal compression.

ACuriousMind

$Q$ is a function of a process, i.e. it's a function on paths. Assigning heat to a point does not make any sense.

@JohnRennie I'm willing to believe that that's what the diagram intends to represent, but I don't see how it actually does so.

John Rennie

The reason I think D is true is because if you draw the cycle on a PV diagram then area inside the loop is the work done. So the only way D could be false is if the loop had no area i.e. it was just a line not a loop. I find it hard to see how this could be so.

Actually I find this the more compelling argument since as ACM says I suspect the diagram given could easily mislead. I am not used to interpreting diagrams like that.

ACuriousMind

I suspect it's actually meant to be a temperature-entropy diagram but the concept of entropy hasn't been introduced

koolman

according to me , For option C change in heat is zero for whole process . But change in internal energy is not 0 . So work will be non zero .

Am I correct

John Rennie

5:30 PM

Aah, the diagram looks like a loop but it isn't really i.e. as you go round the cycle ABCDA you don't return to the same state. That's the flaw in my argument in support of option C. Yes, option C is the false one for the reasons I gave above.

ACuriousMind

@JohnRennie So, what do the points A,B,C,D in the diagram represent? They're clearly not states, but...what are they?

koolman

@JohnRennie but change in heat is zero

ACuriousMind

I'd just give up trying to divine the meaning the author intended and declare the question as written non-sensical.

John Rennie

@koolman All the diagram tells us is that the same heat was added to the system as was lost by it, and that the system ended up at the same temperature. But it doesn't tell us that the system ended up at the same volume and/or pressure.

@ACuriousMind I'm not sure if it's a meaningless question or a subtle one.

koolman

So finally we can say option d is true

John Rennie

5:35 PM

You could calculate the pressure and volume change for each step AB, BC, etc. The calculation is a bit tedious but not hard. Once you had the PV curve then the answer would leap out at you.

koolman

Ohk

Thanks

John Rennie

@koolman I'm not going to venture an opinion without doing the calculation I mentioned above. And I'm not going to do that calculation because, well, it's tedious and unrewarding.

ACuriousMind

@JohnRennie I think you can make a question out of it if you say "the system undergoes four processes with the following heat/temperature changes:", but the diagram is non-sense.

John Rennie

@ACuriousMind the diagram is a concise way of stating how the system changes between each set of points, but it isn't a state diagram. I think that's the point. We're all used to seeing state diagrams and I must admit that's how I approached this diagram.

ACuriousMind

Well, but what are the points?

John Rennie

5:39 PM

Well for example AB is telling us: the temperature increased but no heat was added to the system

BC tells us: the temperature remained constant while an amount of heat $Q_C-Q_B$ was added to the system

And so on. So it's really just four statements in diagrammatic form. The Q axis exists only to show how much heat was added or removed in a step.

ACuriousMind

@JohnRennie What are A,B,C,D? I think drawing a line "from A to B" when you can't tell me what A and B actually represent is really not merely "misleading", but plain wrong.

That is, they labeled one axis of the diagram with "Q". When you label an axis, this means that you claim to be able to assign a value of "Q" to every point in the diagram.

John Rennie

Which you can't. Yes I agree that's a silly diagram.

Nevertheless, starting at some $P_0$, $V_0$, $T_0$ the diagram does contain enough information for you to calculate $P$, $V$ and $T$ at each point.

@ACuriousMind is shouting grounds for deleting a comment?

@koolman what were you going to ask?

koolman

In this what does they mean by stroke

And how they have solved it

ACuriousMind

@JohnRennie uhhhhhh

John Rennie

5:58 PM

@koolman I think they mean that a stroke increases the volume by $0.2V_0$. Kind of like this:

koolman

Ohk

Then how they have solved it

I could not understand it

John Rennie

Well if $P_1$ is the pressure after the stroke then Boyle's law tells us $P_0 V_0 = P_1 V_1$. Yes?

koolman

Yes

John Rennie

And rearranging gives $$P_1 = \frac{V_0}{V_0 + 0.2V_0} P_0 = \frac{P_0}{1.2}$$ Yes?

ACuriousMind

Physics is terrible: "M theory compactified on a K3 manifold is widely believed to be equivalent to the heterotic string compactified on a 3-torus"..."widely believed"? Is it equivalent or not?!

koolman

6:05 PM

Yes @JohnRennie

ACuriousMind

I can't use a "widely held belief", especially when claimed without citation.

John Rennie

@ACuriousMind widely believed = believed by wide boys?

ACuriousMind

"Wide boy is a British term for a man who lives by his wits, wheeling and dealing. According to the Oxford English Dictionary it is synonymous with spiv." Thanks, Wikipedia, two sentences in and I'm none the wiser.

John Rennie

@ACuriousMind :-) Google spiv

ACuriousMind

I swear, if the first thing I read is "synonymous with wide boy"...

Danu

6:08 PM

o/

> M theory

what is it?! :P

John Rennie

@koolman OK, the next step is less obvious. The piston returns to it's original position without changing the pressure, presumably by venting some gas.

ACuriousMind

@Danu I'm beginning to strongly suspect these jokers just mean that using M2-branes to get physics is "M-theory" as opposed to "SUGRA".

koolman

@JohnRennie why without changing the pressure

John Rennie

So now we start at $P_1$ and the second stroke decreases the pressure to: $$P_2 = \frac{P_1}{1.2} = \frac{P_0}{(1.2)^2} $$

Danu

@ACuriousMind Branes ain't no joke :P

John Rennie

6:10 PM

Likewise the third stroke decreases the pressure to: $$ P_3 = \frac{P_2}{1.2} = \frac{P_1}{(1.2)^2} = \frac{P_0}{(1.2)^3} $$ and so on.

So the pressure after the nth stroke is: $$ P_n = \frac{P_0}{(1.2)^n} $$

koolman

Ohk I got it

John Rennie

Whatever book you're getting those questions from it is absolute rubbish. They are terribly worded questions.

koolman

Yep

ACuriousMind

@Danu ::laughs manically::

koolman

@JohnRennie I have one more last question :p

John Rennie

6:14 PM

Yes?

koolman

How can I get the pressure of liquid

John Rennie

@koolman Just use Raoult's law

koolman

It will tell the vapour pressure

How can we get liquid pressure

Danu

@ACuriousMind hehe

John Rennie

@koolman on closer reading, I'm not sure I understand what the question is asking ...

ACuriousMind

6:28 PM

Arrrghh "Most of the flat connections arise from Wilson loops in the maximal torus of the gauge group" "Most"?!

Danu

Let's hope it's in the sense of Sard :D

koolman

@JohnRennie why

ACuriousMind

@Danu With what measure on the space of connections? :P

koolman

We have to just find that point when both vapour and liquid pressure are same

ACuriousMind

I'm going to go mad trying to read all those physics papers casually sweeping all the interesting calculations under the rug

Danu

6:30 PM

@ACuriousMind A generic one (in the sense of Sard)

ACuriousMind

If there are any explicit proofs/computations in these things, they always do that for the boring statements!

Danu

And thus, one invents $\infty$-categories

ACuriousMind

@Danu Is this going to result in the original most being in the sense of $\text{Sard}^\infty$?

John Rennie

@koolman Well using Raoult's law gives the partial vapour pressures as $P_a = 300mm$ and $P_b = 200mm$. So in th vapour $x_a = 0.6$ and $x_b = 0.4$. But I don't understand what they mean by the mole fraction in the liquid at the dew point.

ACuriousMind

Ah, I see you had the same thought ;)

Danu

6:31 PM

Can't believe you came up with something so close to what I said :D

John Rennie

I have to go now I'm afraid.

ACuriousMind

Hm, should that be $\text{Sard}^\infty$ or $\infty-\text{Sard}$?

koolman

@JohnRennie sorry dew point is when composition is same

Azeotrope

Danu

@ACuriousMind $\infty-\text{Sard}^\infty$

John Rennie

@ACuriousMind can you see deleted comments?

ACuriousMind

6:32 PM

@JohnRennie Yes, moderators can see deleted comments and comment edit histories.

@Danu mind=blown

John Rennie

What got deleted here? Or aren't you allowed to say?

Danu

@ACuriousMind someone call Urs!

koolman

@JohnRennie

Zeotropic mixture

A zeotropic mixture is a chemical mixture that never has the same vapor phase and liquid phase composition at the vapor–liquid equilibrium state. The dew point and bubble point curves in the graph shown do not touch each other over the entire composition range, with the exception of the pure components (the curve end points). Azeotropic describes mixtures in which the dew point and bubble point curves are touching each other at at least one point – indicating the same composition in the vapour and liquid phase. Vapor pressures and boiling temperatures of a zeotropic mixture are always between the...

Danu

Oh, damn I almost forgot about JD. Let's see hwat he's been up to...

John Rennie

@Danu you don't want to know. Trust me on this.

ACuriousMind

6:34 PM

@JohnRennie I'm not sure whether I'm strictly not allowed to, but why should I unless you give me a very good reason why you need to know?

Danu

CAUSE I WANNAAA

COME ON DADDY PLEEEAAAAASE

John Rennie

Malicious curiousity! That's a good reason - isn't it? :-)

Danu

right?

ACuriousMind

@Danu ...please don't call me daddy ;)

Danu

Whatever you say, mom.

ACuriousMind

6:36 PM

@JohnRennie I'm afraid in that case I see no benefit to disclosing what was deleted

John Rennie

:: John pulls a sulky face::

ACuriousMind

now, now, if y'all behave, there will be cake.

(read that in the voice of GLADOS)

John Rennie

Gosh, all sorts of stuff is kicking off in the Portugese Stack Exchange chat room.

I keep having to resort to Google Translate to find out how to vote.

Anyway I really must go. See you all first thing tomorrow.

Danu

@ACuriousMind Obvious :P

@JohnRennie Just don't.

John Rennie

@Danu actually it's all looking quite innocent, unless Google Translate is badly misleading me ...

So far I have voted all the flags as invalid.

They need to come the our chat room to find out how to get themselves banned :-)

Danu

6:41 PM

I'd say it's more prudent not to vote on stuff you might not understand.

John Rennie

And with that John became coherent long enough to tunnel into his armchair next to a glass of beer.

Danu

Good job.

ACuriousMind

Careful, large quantities of beer tend to lead to...loss of coherence ;)

Danu

Decoherence?

ACuriousMind

Might even be collapse if you go really wild :P

David Z

7:02 PM

@JohnRennie garden-variety rudeness, nothing noteworthy

ACuriousMind

8:39 PM

@JohnRennie I eventually decided to delete the comment, not really because "shouting is forbidden" but more because it started to generate more comments than the post itself.

Bernard Meurer

Metroid II is hard as hell

Danu

hurr durr young people easy games somethingsomething lazy

Bernard Meurer

@Danu Hm? Metroid II is from 1991

It's for the Game Boy

Danu

You're clearly too young to get it :P

Bernard Meurer

9:01 PM

@Danu :(

Slereah

@ACuriousMind So from what I can see

A cyclic vector forms a basis after application of an operator an arbitrary number of times

What is that operator for say, a free scalar theory?

Bernard Meurer

To those who heard my dilemma between a jacket for the winter and a Game Boy, I'd like to say the Game Boy is super fun

3

ACuriousMind

@Slereah Almost correct (that's the case for an algebra generated only by one operator). More generally a vector $v$ is cyclic for an algebra $\mathcal{A}$ if $\mathcal{A} v := \{Av \mid A\in\mathcal{A}\}$ is dense.

Slereah

What is such an operator here, then

I can guess that indeed if the vacuum is cyclical that's a problem if $\phi(x) \Omega = c \Omega$

ACuriousMind

In physics, the algebra is always the algebra of observables and for a field theory the vacuum should be cyclic.

Slereah

9:09 PM

Oh, so not for a single operator

I'm guessing that it would be something like $\{ a^\dagger_k\}$ for instance

Wait, subalgebra, maybe more $\{ a_k, a^\dagger_k\}$

Although those aren't part of the algebra technically I guess

Rigged Hilbert space and all

Emilio Pisanty

Hey @dmckee, speaking of the Lego Grad Student

Sensing his hands get warm on the active copy machine, the grad student is unexpectedly reminded of how lonely he i… https://twitter.com/i/web/status/805847302963240960

Tweeted by legogradstudent on December 5, 2016 at 6:52 PM

2

> Sensing his hands get warm on the active copy machine, the grad student is unexpectedly reminded of how lonely he is.

Emilio Pisanty

oh

before anyone starts complaining

13

Q: All Stack Exchange site logos are showing up as the 3D Printing one REDUX

All Stack Exchange site logos are showing up as the 3D Printing one This bug just started showing up for me again; I noticed it on both computer and phone.