The h Bar

dmckee

00:06

Comment not posted:

> For FSM's sake. Go learn something about orbital mechanics so you can answer these questions for yourself. Seriously. Science isn't some random collections of facts, and you don't learn it by accumulating facts. And you certainly don't make a contribution by asking question without understanding the subject you are asking about.

::bangs head on desk::

ACuriousMind

I feel you

DanielSank

01:05

@dmckee the secret lives of Physics SE mods.

Wdoctor123

hello

DanielSank

@Wdoctor123 hello

@dmckee who's FSM?

Wdoctor123

Is this chat room for asking questions?

Because i was wondering if particular floors of a building more likely to collapse than others in an earthquake.

?

@DanielSank you there?

DanielSank

01:22

@Wdoctor123 you can ask questions here.

If it's a good question, why not ask on the main site?

Wdoctor123

It's kind of urgent question

DanielSank

Usually people ask well-posed questions on the main site and more vague ones here.

Wdoctor123

i just need slight tips :)

DanielSank

Anyway, I don't know anything about building structural integrity in earthquakes.

Wdoctor123

oh ok

Should I ask at the main physics site?

DanielSank

01:26

@Wdoctor123 go for it.

Wdoctor123

thank you the advice

DanielSank

@ACuriousMind tasted the mead.

It's very good.

dmckee

02:01

Flying Spaghetti Monster

The Flying Spaghetti Monster (FSM) is the deity of the Church of the Flying Spaghetti Monster or Pastafarianism (a portmanteau of pasta and Rastafarian), a social movement that promotes a light-hearted view of religion and opposes the teaching of intelligent design and creationism in public schools. According to adherents, Pastafarianism is a "real, legitimate religion, as much as any other". Pastafarianism is legally recognized as a religion in the Netherlands and New Zealand – where Pastafarian representatives have been authorized to celebrate weddings and where the first legally recogniz...

DanielSank

02:23

@dmckee ah, yes, our Lord and Savior.

koolman

02:49

@S007 where is your room

dmckee

03:23

I always tend to doubt that things really are getting worse with "kids these days", because such complaints can be found from teachers in classical times and they read very much like a rant posted to the internet in 2016. But sometimes I have to ask "How did I not notice this going on around me when I was that young?" as the only possible alternative.

Was I that bad at communication? Were my peers?

I suspect the answer is "Yes, y'all were, chump." but rose-tinted memory gives nary a hint.

DanielSank

@dmckee I really like you.

Anonymous

03:54

@koolman I pinged you. Did you get it ?

koolman

Yeah

user116211

04:33

@DanielSank o/

DanielSank

04:53

@MAFIA36790 \o

user228700

05:54

Does anybody here have any sort of experience interpreting phase diagrams for binary liquid systems?

John Rennie

06:14

@Kaumudi everyone studies these at university, and I might even be able to remember something about them.

user228700

Ah, I'm frustrated now. I'll get back to them in the evening :-/

Anonymous

Anonymous

@JohnRennie Any idea how to solve the problem in the picture? If I assume the velocity of ring towards left is v, then do I also assume the velocity of bead w.r.t ring towards the right as v? After that how should I proceed ?

John Rennie

@S007 that's a poorly phrased question since just moving with a constant velocity won't make the bead move. I assume it means that the ring (but not the bead) is instaneously accelerated to a speed $v$, in which case relative to the ring the bead is moving at a speed $v$.

In that case the question reduces to: at what speed $v$ will the bead reach the top of the ring?

Secret

S007: Is the answer $\sqrt{4gR}$?

Anonymous

06:28

@Secret Yes! How did you get it ?

John Rennie

And to solve that you equate the potential energy difference between the top and bottom of the ring to the kinetic energy.

Anonymous

@JohnRennie In that case v=sqrt(5gR) but answer is sqrt(4gR)

John Rennie

@S007 No

Anonymous

I applied conservation of energy

Anonymous

@JohnRennie Why ?

Secret

06:29

$$\frac{1}{2}mv^2=mg(2R)$$

The "height" the bead need to travel is a diameter

John Rennie

@Secret I don't think it helps to just give the answer. S007 could have worked it out with some guidance.

Now you've denied him the opportunity.

Secret

Well, I typed too slow and you already gave him all the explanations he needs

user228700

I bring you all a quick question about electron capture instead. This is what my book says (and ai'm unable to understand it completely)

Anonymous

Well, I have a doubt. @Secret Did you assume that at top velocity should be equal to zero ?

Anonymous

Why should we assume that velocity at the top is zero ?

Secret

06:31

They ask for the minimum velocity needed, in that case, circular motion problems told us that the minimum velocity to just surmount the ring is when it is at rest at the top of the loop

user228700

Anonymous

But for minimum velocity to complete vertical circle at top it must have some velocity at least, isn't it?

Anonymous

@Secret

Secret

Think of it this way, the ball have just enough kinetic energy to reach the top, after that it rolls down on the otherside.

If the ball have nonzero velocity at the top, it means it has more than the kinetic energy needed to go over the loop

user116211

@S007 it's wrong; you need some non-zero velocity, no matter how infinitesimal it is to complete the circular motion; so you are right.

user228700

06:35

It says that if $X$ and $Y$ are taken as atoms, an amount of energy $Q$ and the characteristic x-rays of Y are emitted but no energy is emitted in case $X$ and $Y$ represent nuclei, but how come?! The neutrino will definitely have some energy, right?

Anonymous

@MAFIA36790 Wait, wait...who is right ?

Anonymous

sqrt(5gR) ?

Anonymous

or sqrt(4gR)

Secret

@MAFIA36790 So you mean the v=0 at the top case is just the limiting case while in practice you need nozero infintesimal velocity?

user116211

@Secret yes, correct.

user116211

06:36

@S007 Sorry, didn't check the problem.

Secret

ok, I don't know how to factor infintesimals on problems of this nature, cause as far I know, we often just asked to calculate the limiting case

Anonymous

@Secret Have a look at this youtube.com/…

Anonymous

From 5:10

Anonymous

You will see why sqrt(4gR) implies that block will leave circular track before reaching the top

Anonymous

Circular Motion inside a Vertical Circular Track (GA_CIR17)

►

user228700

06:38

Aaanybody?

Anonymous

But since it is a bead it cannot leave

Anonymous

the track

Anonymous

that is also another point

John Rennie

@Kaumudi what is $Q$? It isn't defined anywhere in the text shown in your picture.

user228700

@JohnRennie Oh, crap, sorry, $Q$ is supposed to be the energy released...which then manifests as the kinetic energy of the particles emitted.

Anonymous

06:41

@Secret Okay...now it seems correct. The velocity should be slightly greater than sqrt(4gR) so that a very small velocity still exists at the top which helps to complete the circle

John Rennie

@Kaumudi in that case I would have included $Q$ in the nucleus only equation as well. Why your book chooses not to I don't know.

user228700

Right, because the neutrino will zoom away with some energy, yeah?

Anonymous

Now I get it why sqrt(4gR) is the correct answer, normal reaction becomes zero at some point in between, but after that normal reaction begins to acts in opposite direction. BTW thanks a lot for the help :) @Secret

John Rennie

@Kaumudi The neutrino will head off one way at high speed and the nucleus will recoil in the other direction at a much lower speed. Both will gain KE as a result of the reaction.

Anonymous

@JohnRennie Thanks a lot :) I understood it now

user228700

06:44

And also, the energy of the emitted characteristic X-ray doesn't affect the energy of the nucleus, so that's not included in the nucleus-only reaction, yeah?

user228700

@JohnRennie Yes, okay...

Secret

I find it interesting your linked example, despite both being circular tracks with no friction, the one in the link need an extra $gR$ in the squareroot to reach the top. I might need to redo this problem some time later, it sems my mechanics still sucks

John Rennie

@Kaumudi I suspect they mean the X-ray comes from the rearrangement of the electrons in the new atom after the reaction. That is the transition to the element Z-1 leaves the electrons in an excited state and the decay to the ground state produces the X-ray.

user228700

@JohnRennie Yeah, that's what I figured as well...

Anonymous

@Secret No no...you were correct only...in the link they are using a block...but here we are using a bead....a bead can never leave the circular track

John Rennie

06:46

Your book does specialise in doing things the hard way :-)

user116211

@Secret hmm, it is not that the mechanics sucks; it's just the lack of practice; even I would be hesitating to solve a problem without revising the things.

user228700

@JohnRennie Never was there a truer word.

user228700

It never explains any of the steps!

user228700

Anyhoo, thank you :-)

Anonymous

@Kaumudi Take a free advice: CHANGE YOUR BOOK ...yesterday seeing them taking dP negative made me realize.....:-P

Secret

06:49

@MAFIA36790 Speaking about practice, I do lack quite a bit of that compared to my physics peers. I pretty much use the chemistry and maths mindset in studying physics, trying to get my head over the concepts before doing practice problems

user228700

@S007 :-P It's not all that bad. I mean, I do take pleasure in ranting about it, but it has great problems and besides, all this step-skipping improves my ability to think for my own.

user116211

@S007 hmm, the differentials are not taken negative; they are generally considered infinitesimal changes; be it positive or negative unless there is additional constraint and info.

user116211

@Secret same here; it takes huge time for me to get the concept first.

Anonymous

@Kaumudi I see :-P :)

user116211

hey, hey, I was reading Born, Wolf; I've few queries.

user116211

06:52

When discussing about the laws of reflection and refraction, they said, the time variation in the boundary $z~=~ 0$ of the incident, reflected and transmitted wavefunctions are equal.

user116211

$$t-\frac{\mathbf r\cdot \mathbf s^{(i)}}{v_1}~=~t-\frac{\mathbf r\cdot \mathbf s^{(r)}}{v_1}~=~ t-\frac{\mathbf r\cdot \mathbf s^{(t)}}{v_2}$$

Secret

In fact, for any past exams I have been through, I often do the past papers last, because I knew if I get the concepts right (As loosely speaking $<\textrm{set of concepts}>=\textrm{past papers}$) I can do any conceivable exam paper that can be generated by the examiner.

Does not work well for physics, through...

Anonymous

@Secret At what level are you studying ? Graduate ?

Secret

Currently waiting for my PhD next year

Anonymous

Oh I see...phd on which topic ?

Anonymous

06:57

@Secret

user116211

How did they write the relation above?

user116211

Are the incident, reflected, transmitted wavefunctions same? The equality should be $$\Psi_\textrm{i}~=~\Psi_\textrm{r}~=~\Psi_\textrm{t}$$ in the boundary.

Secret

@S007 Chiral at metal indenyl complexes. Basically theoretical calculation of ground state energies as a function of ligand electronic properties, and then experimental synthesis of them for characterisation

the aim is to establish a correlation between electronics and how far the indenyl ring will slip away from 5 coordinate bonding

which is known to control the chriality of the metal centre

user116211

Also, when do we get a non-linear relation between $\mathbf D$ and $\mathbf E$? The authors touched the fact but didn't discuss any.

Anonymous

@Secret Oh wow...seems great :-D...all the best for it :)

Secret

07:05

Right now, during the waiting period. I am working on my driver license revision (which is SOOO BORING thus I am not often seen doing it) and also my various personal science* and scifi^ projects that is sparked due to my curiosity:
*1. Division by zero and zero term algebras
^2. Negative mass properties
^3. The mathematical formulation of a tensorial momentum
^4. Time travel modelling with malleable histories
*5. Quantum mechanics revision and preparation to QFT
*6. Topology
*7. Abstract algebra (currently at group theory)

I also have concentration probelms thus like most people, I often end up dwelling on facebook doing nonsense for most of the time

Anonymous

@Secret Even I was addicted to Facebook once upon a time. Now I am addicted to stack exchange :-P

Sir Cumference

07:58

Math puns are best puns

user116211

So, you are awake @sirC.

Sir Cumference

@MAFIA36790 I'm away for Thanksgiving

user116211

aha!

user116211

Turkey, eh?

Sir Cumference

So time differences

@MAFIA36790 Somewhere in Europe ;)

Wdoctor123

08:01

is there anyone who can help me with some earthquake and resonance in buildings?

user116211

@SirCumference Was talking about Thanksgiving Turkey ;P

Sir Cumference

@MAFIA36790 Oh XD

Wdoctor123

anyone?

Sir Cumference

@Wdoctor123 Just ask

Wdoctor123

How should I go about doing an experiment/practical on "are particular floors of the buildings likely to collapse during an earthquake"

I did some research and all I got was some vague answers. So I thought I might ask here.

John Rennie

08:09

@Wdoctor123 I suspect you only got vague answers because it isn't possible to be more precise.

If your building has a resonant frequency that matches the frequency of vibrations in an earthquake then the system acts as a driven oscillator.

In that case the amplitude of the oscillations in the building can grow much larger than the amount the ground is moving, and this increases the chance the building will collapse.

But identifying resonant modes ina building is a difficult task that you'd probably have to do using some form of finite element analysis.

Wdoctor123

researched question I was given was "Are particular floors of a building more likely to collapse than others in an earthquake?"

I assume if the building was in resonance

there should be a standing wave

there for there are some vibration nodes where building are likely to collapse

I just don't how I should go about proving this with experiment

@JohnRennie thank you for the reply

John Rennie

@Wdoctor123 a quick Google found this site that seems to have lots of info on this subject.

Secret

08:40

yesterday, by Kaumudi

@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.

The way I handle pings is I have a list of people who have the expertise to some questions I want to ask, but I tend to use my pings very carefully thus most of the time I am not pinging at all

for example, if I have a quantum questions, I will wait for the quantum guys and ping them

If I have black hole, GR questions, I ping the black hole and GR guys

and so on

But most of the time, I don't waste pings. The times when I ping frequently is a sign I am actually going through a book in self study

In short, I knew who to ping for what questions, at what times

reply pings don't count because the occassional way to avoid them is often dififcult to pull off. But soemtimes, who I am talking to is clear in context

user228700

@Secret This is basically my life.

Sanya

is there a ping-ban going on?

Secret

not really but it seems there's a unsung convention that pinging too much and too many users is disruptive

Sanya

eh ... I mean, no one likes being pinged for nothing, but ... well, ... :|

Mew

what's wrong with pinging?

@DavidZ sup dude

Secret

08:56

As far I have read, koolman have been causing a lot of trouble in the periodic table and sometimes here in h bar due to massive pings of many people

Mew

so what tho

he's not banned there's no big deal

Secret

Ramanujan and a couple of users also recently caused some disruptions due to frequent multi pinigng in the maths room

(details can be found in their chat logs)

(didn't say they were banned though, else by SE rules I am not allowed to discuss them)

Mew

if they didn't get banned then doesn't matter hey

user228700

@Mew Actually, he is. (At The Periodic Table)

Mew

oh really

@Secret you just broke SE rules by discussing a banned person

user228700

09:02

I don't want to get into trouble for speaking about the ban tho, so ::exiting this particular conversation::

Secret

great, I did not knew he is banned

Mew

@DavidZ can you see if Secret needs to be banned for discussing a ban?

user228700

@Mew: Dude, chill! .__.

Mew

I kid i kid

u guys are paranoid

user228700

U kid too much :-P I wonder if u were kidding about all the money and famousness as well @Mew.

Secret

09:03

Speaking about multi pinging, I remember in the first few months I joined h bar, I sometimes multi ping. And then (details forgot), after that I seldom multi ping. I might need to search the chat log to refresh my memory

David Z

It is not against the rules to discuss a suspension. It's bad etiquette, but (if you can do it without running afoul of other rules) it is allowed.

2

Mew

sif

oh really

good to know thanks David :)

have you seen the new physics problems site?

@Kaumudi why do you only like me because I'm rich and famous?

Secret

David: But according to Acuriousmind and a couple others:

Nov 8 at 0:06, by ACuriousMind

Whatever your personal opinion is, this chatroom is not the place to discuss the suspension. We do not discuss specific suspensions in public without explicit consent and participation of the user in question, so everyone please drop the topic.

David Z

"we" = moderators

user228700

@Mew: I like u because u're funny+very helpful. I'm not even sure if u're rich or anything xD

user228700

09:06

Anyhoo.

David Z

For everyone else, it's recommended to avoid discussing suspensions because it's easy to go from there into unfounded speculation, which is against the rules (barely).

Mew

@DavidZ I thought people were in trouble for discussing a dude called 0celo7

but I tguess they were just trying not to speculate

David Z

@Mew They weren't trying not to speculate, that's the problem. If they were doing a better job of not speculating, there wouldn't have been any trouble.

Mew

true I don't really remember what they said

just that ppl didn't want to talk about it

@Kaumudi ty for not liking me for my fortune and fame :)

David Z

It certainly can be hard to discuss a specific suspension in an appropriate way. This is why I recommend (but do not require) that people avoid doing so.

This is also part of why the mods have our blanket no-discussion policy. We have access to nearly all the relevant information about suspensions, which makes it even harder for us to discuss them in an appropriate way. Saying nothing is an easy rule to stick to. (There are other reasons too - we've learned from experience.)

John Rennie

09:16

Cambridge University produces the UK's 'most sought-after graduates', new study reveals

Though in my case the only people seeking me are the police :-)

Anonymous

@JohnRennie What ? :-P What did you do ?

Mew

That's hardly a suprise isn't it John?

John Rennie

@S007 Crimes against good taste

Wearing a loud shirt in a residential area after dark.

Being in possession of an offensive haircut.

I can't think of any more - all suggestions welcome :-)

Driving (a computer) without due care and attention.

Mew

ol

user228700

@JohnRennie Wonder what an "offensive haircut" looks like on u :-P

Anonymous

09:21

@JohnRennie Haw! For such petty things police go after you in the UK? XD LOL!

Anonymous

The police there must be having a lot of free time :-P

John Rennie

I don't think I've ever been in trouble with the police ..

Mew

@S007 he was joking .___.

John Rennie

Actually, now I think about it, as a student I was fined £25 for cycling after dark without a light on my bicycle.

Sanya

@JohnRennie poor Oxford

although, I quote one of my professors here "I'd take a local Master student as willingly as an English PhD because they'll be around the same level" :p

Mew

09:31

@Sa

@Sanya were r u frm?

Sanya

@Mew germany

Mew

dats ryt

Sanya

but I actually think that statement by my prof is a bit too extreme and staying in the UK, I actually had the feeling that just the idea of how physics should be done and taught differs quite a bit

meaning that the set of skills aquired is just very different

Mew

wat lvl r u at in ur stdys?

in aus most peeps don't even do masters maybe that's the same in the uk too

Sanya

it indeed is

in the UK I was in my 3rd years Bachelor and also did some Master's courses

that's a few years ago though :D

Anonymous

09:42

@Sanya Is academia more sought after in the UK or are corporate jobs more popular ?

Sanya

@S007 hard to tell actually; because there seemed to be a bit of opportunities in academia - but at the same time, most people seemed to graduate with the Bachelor's degree according to my impression

I wouldn't want to give a qualified judgement though - I wasn't there long enough to be sure :D

Secret

0

Q: What's the relationship between temperature and density matrix?

As we know ,the definition of temperature can be traced back into the First Law and the Second Law in thermodynamics,which also can be explained by the more powerful statistical mechanics.$\\[0.5cm]$ And in many calculations concerning quantum statistical mechanics,I will meet the following formu...

Temperature always confuses me in the quantum level

user228700

I've a quick question about BE/A graph again.

Mew

go ahead

user228700

My book says that the nuclei having Z b/w 1 and 83 are stable (with only a few exceptions). I'm looking at the graph and it looks like the BE/A value for nuclei w/ Z b/w 1 and 20 is awfully small.

user228700

09:56

Plus, the nuclei that are actually radioactive (i.e Z value b/w 84 and 112) have BE/A values that are far greater than those for some of the lightest nuclei.

John Rennie

What is BE/A?

Mew

Binding energy per atomic number

i mean per nucleon

I assume

what is is b/w?

user228700

Yes. Sorry .__.

user228700

@Mew Damn. b/w=between.

Mew

ooo i see

ty

John Rennie

09:58

@Kaumudi whether a nucleus is stable or not depends on the energy barrier(s) in the various pathways for it to decay. This is not necessarily related to the binding energy.

user228700

Uhm, can u please elaborate? What are these "energy barriers" u're talking about?

Mew

Kaumudi, imagine a hill

you start at the bottom

user228700

(Just a very brief "elaboration" will do, BTW)

Mew

you go up the hill

you reach the top then go down the hill

now once you roll down the hill, it is possible that you end up a t alower level then you were before you started

thus going up and then down the hill overall was energetically favourable

because you end up ultimately in a lower potential energy state

user228700

@Mew No, dude, Ik what energy barriers are. I dunno why I put quotes around this term originally :-/ I was asking about the specific energy barriers in this case.

Mew

10:01

oh

that is the energy required to bring the nucelons together

I mean to bring two protons together costs energy

because +ve charges repel

so you have to overcome this electric repulsion first, and when the protons finally meet they can react using the strong nuclear force

few scientists have reported being able to perform cold fusion

user228700

Aii, but aren't we talking about fission/fusion? Well, OK, this makes sense with regard to fusion, but, erm, huh...

Mew

@Kaumudi weren't you talking about particles with low Z?

they are only going to undergo fusion

because they want to become more like iron

with lower BE per nucleon

user228700

Oh, right, but...

Mew

so they will fuse

user228700

Gimme a minute to think this through, please...

Mew

10:05

no p/

user228700

Ya know what, never mind. This a bit too confusing for me (esp. with the lower vs. smaller values and all). I doubt if I need to go into such depth for my exam anyway.

user228700

Thanks @Mew :-)

Mew

np

btw

user228700

Yeah?

Mew

is that wat u meant by low BE for low Z

user228700

10:09

Uhh, yeah..?

Mew

ok

user228700

@Mew@JohnRennie: Unrelated, but dyou guys remember ur 18th birthday?

Mew

yea

why?

user228700

Oh, crap, hang on, I'll be back in 10 mins.

Mew

when is ur 18th birthday Kaumu?

user228700

10:11

I'll be back in like, 38 mins .__.

Mew

aw

y

tv show?

k .__.

laterz

user228700

@Mew No, studying :-/

Mew

y so specific

i thort u were gonna talk about bdays

user228700

Wait, u're leaving?

John Rennie

@Kaumudi that's very precise :-)

Mew

10:12

yeah i was going to be back in 28 min

38 min

user228700

:-P Alright. See ya.

Mew

laterz

John Rennie

@Kaumudi If you take 39 minutes or longer can we ping you to complain? :-)

user228700

(In 38 mins)

user228700

@JohnRennie Sure! :-D

John Rennie

10:13

:: John sets his alarm for 10:52 ::

2

Secret

Last night dream:
Pretty slice of life theme
An analytical chemistry NMR assignment and on the day of presentation. Later a print error was found in the assignment distributed to students thus all presentation were suspended
3 friends of mine later have a social evening with a uni friend of mine at a campus restaurant. We then chat about stuff in a sketch book
The whole thing ends with the 3 friends performing on a play on stage and then left earlier than the audience when the play is finished

Slereah

10:39

"Quantum field theory on Clifford-Klein space-times"

The fuck is a clifford klein spacetime

user228700

I'm back! (Yes, I saw some numbers wrongly-hence the 38 instead of 28 :-/)

user228700

So, @JohnRennie: Dyou remember it?

John Rennie

@Kaumudi My 18th birthday?

user228700

Yes.

John Rennie

No. I recall we went out for a meal and drinks but I don't remember any details at all.

Why? Are you planning something special for yours?

user228700

10:42

@JohnRennie Pomodoro technique :-) Except, my "pomodoros" are usually much longer than 25 mins.

user228700

@JohnRennie Ohh, okay.

user228700

@JohnRennie Dyou know that exam I'm preparing for?

John Rennie

@Kaumudi Yes?

user228700

It's on the day after my 18th birthday :-( I was just wondering if people tend to remember these things. I realize that 18 is just another number, but still...

John Rennie

Are any of your friends doing the exam? If so I'd be tempted to postpone the birthday celebrations and hold a combined birthday + end of exams party!

user228700

10:46

@JohnRennie Sadly, no, none of my friends are writing it again .__.

John Rennie

Oh well, hold a combined birthday + end of exams party anyway

I think my birthday was a weekday, so the celebrations were fairly muted anyway. Just a quiet meal. Then we went out at the weekend for a proper party.

user228700

@JohnRennie I see :-) When u say "party", when does that mean..?

John Rennie

In any case it's good to relax the night before the exam, and a meal with your family then an early night would be just the job.

@Kaumudi for adolescent males party = excessive drinking

user228700

@JohnRennie :-P Just me and the 4 people in my family will be there, then. All my friends are attending college is far-off places.

user228700

@JohnRennie OK :-P Was just wondering if this was the norm 37 years ago, in a different country :-)

John Rennie

10:50

Sounds ideal. Finish the cramming at lunchtime then relax. Have a nice quiet meal, then ace the exams, then do whatever 18 year old Kerelan girls do to celebrate at the weekend.

user228700

@JohnRennie :-) I have another exam too tho, after the first one, but that one's like a month after that so, yeah.

user228700

I had an amazing birthday this year but I'll deal with the fact that not every birthday will be great-maybe that'll be best, since I am turning 18 and all, and I hear there's a lot of this not-greatness in adulthood :-P

user228700

Seems like @Mew's vanished. HELLO, dyou remember ur 18th birthday?!

John Rennie

@Kaumudi I wouldn't know, I haven't grown up yet :-)

user228700

:-)

Secret

10:58

sciencedaily.com/releases/2016/11/161121174027.htm

How is this breaking inversion symmetry, unless you mean inversion symmetry at each octahedral centre...?

Transcript for