The h Bar

@heather Do you have an idea of a date?

12:15 AM

@DanielSank, not quite yet, but I'll let you know as soon as my parents figure it out. =)

@heather k

@DanielSank, again, thank you so, so, so much!

12:41 AM

@heather Don't thank me until it happens!

Anyway, you can find my email through my PSE profile page.

1:01 AM

Hi, why is no. 42 special?

@0celo7 ^^?

1:15 AM

@Ramanujan, because it is the answer to life, the universe, and everything (A Hitch hiker's guide to the galaxy). It also has several other significant attributes: it is the atomic number of molybdenum, there is a Messier object M42 that is also known as the Orion Nebula, there is a red dwarf named Kepler 42, related to the three smallest known exoplanets, the second sphenic number is 42, and I could go on.

In fact, if you want some more of this, look [here](https://en.wikipedia.org/wiki/42_(number))

3:01 AM

Is the entropy increasing when a mixture of water and oil separates themselves to water and oil?

3:42 AM

@DHMO If it happens spontaneously, then the entropy increases.

@rob but why?

Like, are they not more ordered after the separation?

@DHMO If entropy doesn't increase, it doesn't happen spontaneously.

@rob I thought spontaneity is determined by both enthalpy and entropy (Gibb's Free Energy)

Note that the separation is never complete: if you were to mix water and, say, kerosene, you wouldn't want to drink the water even after "all" the kerosene had floated to the top.

@rob but it is still more ordered than before?

3:45 AM

@DHMO Minimal difference between entropy and enthalpy in this case, since the volume change is small. Enthalpy matters more for phase transitions where the density of the two phases is different, like water and ice.

or water and steam.

I haven't thought about oil and water from an entropy standpoint before.

Here's a discussion including enthalpy of mixing which I don't immediately follow.

@rob well, to mix oil and water together, you would need to break the bonds already present between water and between oil

but, water and oil do not form stronger bonds

Seems like for water and oil, the interface between the fluids has less entropy per unit volume than the bulk of each fluid

perhaps because there are fewer ways to orient the water molecules or the lipid chains? not sure here.

So for an oil-water mixture, the low-entropy configuration is the one with more homogeneous bulk and less fluid-fluid surface.

That makes the fluids separate into globules; relative densities determine whether one species floats to the top.

Q: expectation of position in fock space

4:02 AM

@rob another (half-relevant) question: when the enthalpy outweighs the entropy in Gibb's Free energy, and the reaction occurs spontaneously, is the entropy decreasing?

Physics Meta

0

What will be the vacuum expectation value of $x^2$ in fock space? If I write $\hat{x}=\sqrt{\hbar/2 m \omega}(\hat{a}+\hat{a}^{\dagger})$ i.e. in terms of harmonic oscillator and then use the relation $\omega=\sqrt{k/m} then $\hat{x}^2= \frac{\hbar c}{2}$. Is the answer correct? this is mass less...

discussion

@DHMO I suppose that's the mechanism by which one creates low-entropy regions.

My thermodynamics is rustier than I'd like.

@rob so entropy just decreased?? I thought the second law of thermodynamics is a law

Global entropy always increases, but you can have local decreases.

For instance, I can take a deck of shuffled cards, collect the four suits, and order each suit ace, two, three, etc.

Eventually I end up with a low-entropy deck of cards

The second law says that I must have increased the entropy of the rest my environment, by respiring and so on, by more than the entropy I removed from the deck of cards.

But not that the cards can't be un-shuffled.

4:19 AM

@rob where is the global increase in this case?

What, when I unshuffle cards? It's in my respiration, converting long-chain hydrocarbons to short-chain hydrocarbons and those sugars to carbon dioxide and water.

@rob I mean, when the enthalpy outweighs entropy and the reaction is spontaneous

Change in enthalpy ΔH includes a pressure-volume work term on the surroundings ΔV; look in the environment.

@rob the reaction must be exothermic right?

I'm not too familiar with these

Not 100% sure. Your question is threading a narrow route through the thermodynamics parameter space.

4:29 AM

@rob ?

By which I mean, I'm confused.

Alright

user228700

4:46 AM

@JohnRennie: Morning sir :-)

John Rennie

Morning :-)

user228700

I'm in dire need of help! :-( :-O

@rob @JohnRennie Are you familiar with radioactivity?

@DHMO Quite

Morning @JohnRennie

(Evening here)

The counts per minute of a sample is as below:
time/min, counts per min
0, 405
10, 270
20, 202
30, 166
(background radiation is already accounted for)

the problem is, the rate is increasing:
270/405 = 0.667
202/270 = 0.748
166/202 = 0.822

why? @rob

John Rennie

4:50 AM

@Kaumudi that sounds ominous! How can I help?

user228700

@JohnRennie CSE chat..?

@DHMO So 405 counts in first minute, 270 counts in eleventh minute, 202 counts in 21st minute, etc?

@rob yes

Real data or invented for a problem?

no idea

4:51 AM

Your data, or from another source?

From another source

which is?

A textbook

Okay

Might it be because of chain decay?

4:53 AM

Might be statistical fluctuations.

I can't imagine that much fluctuation..

Counting uncertainty goes like sqrt(counts), from Poisson statistics

@rob could you apply that to see if this fits?

You should read 400 ± 20, 270 ± 16, 200 ± 14, 166 ± 13

@rob is there any chance of fitting the exponential decay model?

4:56 AM

So those are only 5% relative uncertainties, smaller than your change in ratios. Not that.

@rob chain decay? any chance?

A chain decay or a mixture of sources with two time constants is a possibility, but not one that you can identify robustly with four measurements.

@rob what do you mean by "two time constants"?

From a real source, I'd guess you have un-subtracted background, but you say your problem discounts that.

@rob if you subtract anything, the ratio discrepancies become larger

5:00 AM

I'm apparently not doing this correctly in my head. I have to use paper.

@DHMO No, I think you've got it backwards here.

@rob ?

If I subtract 150 counts from each reading (thus 255, 120, 52, 16) then ratios get smaller with time.

@rob :o

Do a bisection search ... with background 120-125 counts/min each ratio is about 0.53

What I meant by "two time constants": in real life sometimes you find mixed sources.

@rob thanks

@rob I see

5:10 AM

For instance, neutrons on copper produces both Cu-64 and Cu-66; one is five minutes, one is twelve hours

@DHMO Which raises the question: what did you textbook mean, about background subtraction? It sounds like it needed to be done.

@rob not clear

Q: Origin of Selection Rules for Rotational Raman spectra

5:33 AM

Well, sweet dreams, all

user116211

6:00 AM

Check this classic joke: quora.com/What-are-some-famous-English-jokes/answer/…

The Dark Side

6:43 AM

In case this helps in getting a swift response, advertising my question here in chat (classic cheap strategy)

0

I am aware of the fact that selection rules for transitions between two different quantum states, are obtained via an assessment of the corresponding transition matrix elements [1,2]. When we consider the spectrum of a pure rotational spectrum, with different states identified by the rotational...

quantum-mechanics resource-recommendations spectroscopy raman-spectroscopy selection-rules

user228700

10:29 AM

@Ramanujan: U there?

11:48 AM

So empty...

school seems to make me miss all the busy times.

12:05 PM

@Kaumudi

@heather school on a saturday? :O

@Sanya, no, thank goodness. I was talking about during the week =)

Guess that was a little unclear

well during the week ... depends a bit on whom you want to meet I'd guess; I'm not sure anymore in which direction the time difference goes ... usually, during my European evenings here it's most crowded I'd guess

user228700

12:22 PM

@Ramanujan U still here?

user228700

Damn.

user228700

12:48 PM

..?

@Kaumudi you were asking some thing?

user228700

@Ramanujan Yeah. Thought we could try a problem? I'm not getting the correct answer :/

I will try @Kaumudi

user228700

@Ramanujan Alright. MSE chat?

newtonian-mechanics centrifugal-force string

Ok

Qmechanic

1:56 PM

2

Q: Skipping-rope in space

On a space station a person wants to play skipping-rope. Although they cannot jump, they turn the rope at a constant angular velocity. What would the shape of the skipping rope be, and could we set up an equation by taking a generalised case?

Duplicate of physics.stackexchange.com/q/22268/2451 ?

Sir Cumference

Howdy

Oh wait, was @Qmechanic a mod before?

If not, then congrats

user116211

@SirCumference ._.

user116211

You were always late.

user116211

But this is too much :(

Sir Cumference

@MAFIA36790 Yeah, nvm, I'm losing my mind...

Somehow I never realized it...

Drop the @Bass

user116211

2:09 PM

@Qmechanic Sure it is.

@MAFIA36790 Isn't there gravity involved in the other question as opposed to this one?

anyway, I'm a bit annoyed at the question for being like "please solve my problem for me", but that's another issue

user116211

It is implied here, isn't it?

user116211

@Sanya The user has asked many questions of this species.

user116211

Ironically, he does claim he knows the homework policy.

user116211

@loong o/

2:15 PM

well, people answer those kind of questions [I'm not innocent of that either to be honest], so why should anyone stop asking them? Even if they're closed, the answer is usually already there ...

Loong

@MAFIA36790 hi

@MAFIA36790 is it? well, then I guess it's a duplicate

user116211

@Sanya I mean here he writes space station. But he can get the motivation from other one Qmech linked which is dealing with the same although not exactly the same problem of the rope shape.

user116211

I got the old Wolf Born Optics book!

Qmechanic

@SirCumference : Take a guess.

Sir Cumference

2:22 PM

@Qmechanic Uh...belated congratulations?

user116211

@SirC: You need to get access of Speed Force.

@Sanya Yep.

user116211

The whole book is in Gaussian.

user116211

I like that ;)