The h Bar

user228700

10:00

Huh :/ U are very aware about me not knowing much about free energy...

John Rennie

My point is that the free energy of the water vapour increases with $X$. For low $X$ the free energy of the vapour will be less than that of water, so water will spontaeouly change to vapour.

For high $X$ the free energy of the vapour will be higher than that of water, so vapour will spontaeouly condense to liquid water.

user228700

OK..?

John Rennie

And somewhere in between there will be a value of $X$ for which the free energeries of the water and vapour are the same.

And that will be the equilibrium mole fraction of water in the air in contact with liquid water.

user228700

Riight...

John Rennie

Now, if you increase the temperature of the liquid water the free energy of the liquid water goes up.

To understand exactly why it goes up requires you to understand free energy, which you've just said you don't.

However for liquid water the free energy is roughly the same as internal energy. And obviously as you heat something its internal energy goes up.

Is this OK so far?

user228700

10:05

Uhh, I think so...

user228700

(God, I tried so hard to understand free energy. Never got the hang of it >.<)

John Rennie

As you heat water its free energy increases. And we've already established that the free energy of the water vapour increases with $X$.

So as you heat the water the value of $X$ increases i.e. the mole fraction of water vapour in the air increases.

Still OK?

user228700

Yeah, I still think so...

John Rennie

And there is a temperature at which the value of $X$ rises to one.

But $X$ can't get any bigger than one. You can't have a percentage of water in the air greater than 100%.

user228700

Yeah...

John Rennie

10:09

So once you reach this point you can't heat the water to a higher temperature because it just changes to water vapour instead.

And that's the boiling point.

The headline is that there isn't a difference between vaporisation and boiling.

Boiling is just an extreme of the vaporisation process.

user228700

Riight, OK...

John Rennie

The point is that steam at atmospheric pressure and $T=373$K has a certain free energy, and this creates an upper limit for the free energy of the water.

And therefore an upper limit for the temperature of the water.

user228700

OK...

John Rennie

I sense that you are not entirely convinced ...

user228700

(Ah! There's something wrong with either my mobile/the site! Stupid thing not allowing me to type if I switch back from another room!)

user228700

10:18

@JohnRennie Yes, not entirely, because like I said, the arguments involving free energy doesn't much that much sense to me. But it's alright, I think I understood what u were talking about, for the most part...

user228700

Thanks sir :-)

user228700

Do u mind answering a real physics question now? :-D

John Rennie

No, go ahead.

user228700

I've asked this question multiple times before, but not to you...

user228700

It's about integrating to find the center of mass for uniform mass distributions.

user228700

10:20

Haven't gotten any answers yet so asking again.

John Rennie

OK

yuggib

arxiv.org/abs/1610.07637

this does not seem a completely unreasonable paper on the philosophy of (quantum) physics

user116211

what are non-self adjoint observables?

user116211

It's new to me. Let me check.

user228700

Ahh, sorry about that.

user116211

10:27

Could you tell me @yuggib? I'm not seeing any wiki article on it.

user116211

I'm seeing there is a paper on it:

user116211

sciencedirect.com/science/article/pii/0375960188905919

yuggib

@MAFIA36790 observables that are not self-adjoint operators. The common practice among physicists is to exclude them

as physically reasonable observables.

user116211

okay. Why do we need them?

user228700

OK, so I'm confused about the different methods of integrating to find the center of mass of uniform mass distributions.

user228700

10:29

Damn, I said this already.

user116211

i have always seen books working with self-adjoint operators.

DHMO

@Kaumudi greetings

user228700

@DHMO: Long time! :-P

yuggib

It depends, there may be some situations in which are important

user116211

ohh.

user228700

10:30

@JohnRennie: What are the basic guidelines to doing that?

user228700

It gets pretty complicated for systems like hemispheres(hollow and solid) and all.

user228700

We have to assume some small angle $d\theta$ and everything. It varies for different bodies, obviously...

user116211

Any literature(s) dealing with non-self-adjoint operators @yuggib?

user228700

And my textbook doesn't explain anything (who's surprised?)

Secret

@yuggib, what type of cophanhagan interpretation is this?
1) wavefunction evolves determinstically
2) Measurement is probabilistic and such probabilities are given by the wavefunction (probabilistic outcomes). Meanwhile the wavefunction will be projected because of the measurement.
3) Before measurement there are no predetermined values exists in the system.
4) Interactions that establish the entanglement correlates some parts of the wavefunction of the subsystems to form a wavefunction of a composite system that spans spacelike distance, with relativity protected by no communication theorem.

John Rennie

10:33

@Kaumudi The general principle is the same for all bodies. the trouble is that in many cases it involves integrals that are hard to do so we resort to cunning tricks.

user116211

For hemisphere, you need volume integral @Kaumudi.

yuggib

@MAFIA36790 every book on operator theory...like Kato's perturbation theory of linear operators

user116211

thanks.

user228700

@JohnRennie Yes, these "cunning tricks" are ruining my life (only 'cause Idk how they work 'cause like I just said, my textbook doesn't explain anything!!)

yuggib

@Secret it seems pretty adherent to the standard Copenhagen interpretation

John Rennie

10:35

To be honest I'm not sure there are any general rules. The tricks required differ on a case by case basis.

user228700

So basically, we try and find mass as a function of position, yeah?

Secret

@yuggib but in copenhagan, isn't the wavefunction not treated as a real physical entity?

John Rennie

@Kaumudi Suppose we're working in Cartesian coordinates $x,y,z$

user116211

There is a nice post on it regarding that in Math.SE regarding that:

user116211

1

A: Center of mass of semi-sphere

$a > 0$: radius. \begin{align} \vec{r}_{\rm cm} &\equiv {1 \over V}\int_{V}\vec{r}\,{\rm d}V = {1 \over \left(4\pi a^{3}/3\right)/2}\int_{V}{1 \over 2}\nabla r^{2}\,{\rm d}V = {3 \over 4\pi a^{3}}\int_{S}r^{2}\,\hat{r}\,{\rm d}S \\[3mm]&= {3 \over 4\pi a^{3}}\left[% \int_{\Huge\frown}a^{2}\,{z \...

John Rennie

10:37

Then we have a volume element $dV=dxdydz$

user228700

Yeah...

John Rennie

And if the object is uniform (uniform density) then the mass of this volume element is $dm = \rho dV = \rho dxdydz$

user228700

Wait, hang on! You're explaining for a hemisphere?

John Rennie

No, I'm talking generally

This could be any shape.

What I'm getting at is that in cartesian coordinates the process is essentially straightforward. We want to sum up $\mathbf r dm$.

user228700

Yeah...

John Rennie

10:40

And as I said above $dm = \rho dxdydz$ and $\mathbf r = (x,y,z)$

yuggib

@Secret it is not observable, still encodes the information on the state of the system, and in that sense can be interpreted as a physical entity (essentially within Copenhagen)

Secret

hmm ok

John Rennie

@Kaumudi: the problem is that for objects like spheres, half spheres or whatever, Cartesian coordinates are usually complicated so we switch to polar coordinates instead.

user228700

@JohnRennie Yeah, that's what my book has done...

John Rennie

And that's where the fun starts because polar coordinates are unintuitive for beginners

user228700

10:42

(I'm so sorry; I think there's a bug in my phone >.<)

user228700

@JohnRennie Awesome!

user228700

^Sarcasm.

John Rennie

So I suspect what is causing you grief is not the general idea of calculating the COM, but having to do it in polar coordinates.

user228700

Yes, exactly.

John Rennie

The only solution for this is to get used to polar coordinates so you're comfortable using them.

Secret

10:43

Btw, regardless of interpretation, when is the bell correlations in an entanglment established. Is it
a) at the step when the two particles brought to interact and thus their observables become correlated, or
b) At the step of the measurement, which give the outcome, and the two parties/detectors whatever then realise there is a correlation when results were then brought together for comparison?

user228700

The problem is, most of the objects in my textbook are circular, so they've used $d\theta$. That is what u mean by polar coordinates, right?

John Rennie

@Kaumudi: yes. If you have an example of a calculation that you aren't sure about we could have a look at that. It's hard to say anything useful without considering a specific example.

user228700

OK. I'll post pictures. Here:

rob

@MAFIA36790 That particular answer uses $\vec r = \frac12 \nabla r^2$ to convert from a volume to a surface integral. Which is clever, but sort of an advanced dirty trick.

user228700

(This is for a semicircular ring)

user116211

10:46

@rob I loved this trick and it got my vote; don't know why no one upvoted the post.

user228700

user228700

@MAFIA36790 It was too advanced for me too.

user116211

@Kaumudi there are other posts too. You can check those.

user228700

That's it.

user116211

10:48

It was bookmarked and I copied-pasted the link here. I should have linked the question ;/

user228700

(Excuse the lovely apples by the side :-P)

user228700

@JohnRennie: Readable, yes?

user116211

@ManishEarth o/

John Rennie

@Kaumudi: that's using a mixture of polar and Cartesian coordinates, which is the sort of thing physicists sometime resort to.

user228700

OK..?

John Rennie

10:50

Do you want me to explain how the calculation works?

user228700

Please do..?

user228700

Wait, one second.

user228700

I'll tell u exactly what I don't understand.

John Rennie

OK

rob

@MAFIA36790 I wouldn't have noticed it was clever if you hadn't pointed it out. My muscles for skimming English-language text are much, much stronger than my muscles for parsing a big block-o-math.

user228700

10:53

So we assume the y coordinate remains same throughout this arc, yeah? Remains $\theta$.

user116211

@rob okay ;)

user228700

Oh wait...

user116211

@rob Irony is that yesterday I got confused with a similar trick Born and Wolf used in their book.

John Rennie

@Kaumudi No. $y$ is a Cartesian coordinate while $\theta$ is a polar coordinate. They are related by $y = r\sin\theta$.

user228700

Sorry! That's what I meant. Wrote it all wrong :/ But we assume that $d\theta$ is small enough for the y-coordinate to remain $Rsin\theta$, yes?

John Rennie

10:56

@Kaumudi Yes

user228700

OK. Can u pls explain that multiplication with the $\theta$ and the $M/(2\pi)$?

user228700

So $M/(2\pi)$ is the mass density, yeah?

user228700

Linearly distributed over the arc.

user228700

What's up with multiplying it with $Rd\theta$?

John Rennie

The total length of the ring is $\pi r$ because it's half the circumference of a circle of radius $r$. Yes?

user228700

10:59

Yep.

user228700

Uuugh.

user228700

Wait!

user228700

I made stupid typos again!

user228700

I meant "isn't $M/(\piR)$ equal to linear mass density!!

user228700

Not $2\pi$! >.<

ManishEarth

11:00

@MAFIA36790 o/

John Rennie

@Kaumudi yes that's right

user228700

OK. What I don't understand is multiplying that with $Rd\theta$.

John Rennie

So if you consider some short length of the ring $d\ell$ then its mass is just $d\ell$ times the linear density. So: $$dm = \frac{M}{\pi r}d\ell$$ Yes?

user228700

Pasting that ^ to the question window for it to render...

user228700

Yes.

John Rennie

11:04

Good. The next question is how $d\ell$ is related to $d\theta$. Can you see that, or do I need to go through it?

user228700

See, the problem is with $d\theta$.

John Rennie

Yes?

user228700

I don't understand that part alone. (I already had a good understanding of everything else)

user228700

Why $d\thets$?

user228700

OK.

user228700

11:07

So...

John Rennie

If you look back at my post above I wrote $dm$ as a function of $d\ell$. So our integral wrt to $dm$ can be written as an integral wrt $d\ell$. Am I making sense so far?

user228700

$Rd\theta$ is clearly the length of that arc.

user228700

But I don't get how...

user228700

In a mathematical sort of sense...

John Rennie

@Kaumudi Aha! Correct. $$d\ell = r d\theta$

user228700

11:09

Yeah, but I'm failing to see how that is.

user228700

Ik it is...but how?

user228700

@JohnRennie ell=l? :-P

rob

@Kaumudi Are you familiar with the radian?

John Rennie

Suppose you take that expression and integrate it all round a circle i.e. $\theta$ goes from zero to $2\pi$. The integral is:

user228700

@rob Yeah, I am.

John Rennie

11:10

$$ \int d\ell = \int_0^{2\pi} r d\theta $$ Yes?

user228700

Radians!!

user228700

God, I am stupid.

user228700

@JohnRennie: I get it now!

John Rennie

Cool :-)

user228700

@JohnRennie@rob: Thank you! :-)

user228700

11:12

OK, I'm going to try to go do this same sort of thing for other objects. Wish me luck! Braces herself.

user228700

Hey @heather: Remember u were trying to help me by asking other people to help me 'cause u couldn't? Well, that question got answered now. How long has it been? :-P

heather

@Kaumudi, oh, geesh, um...probably 2 weeks? Can't remember. A while, certainly. =)

Glad you finally got it answered!

user228700

Yeah, two weeks :-P

user228700

Yeah, me too :-D

user228700

I'm gonna go do stuff now. @heather: Have a nice day at school :-)

heather

11:16

@Kaumudi, thanks! Have a nice day. =)

ffahim

Hey everyone

I asked a question to u guys ... maybe u have noticed.. worries... could you guys pls tell about the proof of S={(u+v)/2}t

rob

@ffahim Context?

ffahim

Sorry?

rob

@ffahim What are S, u, v, t?

ffahim

S stands for displacement ,u is initial and v is last velocity and t was the time taken

user228700

11:21

@JohnRennie: Trying to do this for a semicircular disc.

user228700

Aw man, u left?

rob

@ffahim So the definition of average velocity is $v_{avg} = s/t$

user228700

OK, so my idea, integrate area by selecting a small sector and then u can vary $\theta$ from 0 to \pi$

rob

@ffahim And if the acceleration is a constant, so that the change in the velocity is linear, then the average velocity is the mean of the initial and final over the time interval.

ffahim

Let me open my laptop... since math formulas r not seen properly

rob

11:25

@Kaumudi This method works. You can also use the semicircle you just solved, and say that a disk is a bunch of semicircles with radius r and thickness dr

user228700

@rob It isn't working :/

ffahim

tell me one thing.. @rob (u+v)/2 = V..avg does it make sense? its not statistics? as i am learnt that V..avg= delta d/delta t isn't it?

user228700

I have the area of my small sector as $1/2R^2d\theta$

user228700

Where $R$ is the radius of my semicircle.

DHMO

@Kaumudi $\displaystyle \int_0^{2\pi} \frac12R^2 \ \mathrm d\theta = \piR^2$ ?

user228700

11:34

Ik how to integrate it! I don't think the other stuff is working.

DHMO

what is not working?

user228700

The method to find COM. So, basically, if $\sigma$ is the aerial mass density of the semicircular disc I'm trying to find the COM of, then $\sigma=M/(\pi(R/2)^2)$

Secret

Have a quantum question prepared for Acuriousmind. Will ask him about that when he gets on

user228700

Which gives $\sigma=4M/(\piR^2)$

user228700

Where $M$ is the mass of the disc and R is its radius, of course.

user116211

11:39

Give a space after $R$ viz. \pi R and not \piR.

user228700

Then I have:

user228700

@MAFIA36790 (Y)

user228700

Bless me!

user228700

Have I made a stupid mistake again, perhaps?

user228700

11:47

@rob See, ^. It's not working :/

DHMO

11:57

@Kaumudi what is the formula you used?

user228700

For what?

DHMO

for calculating the y-coordinate of the centre of mass

user228700

Same thing, y-coordinate.

DHMO

then where does $\dfrac{4m}{\pi R^2}$ come from?

user228700

12:04

U express mass as a function of coordinates.

heather

@Kaumudi, wait, so $M$ is the mass of your semicircle and $x$ is the x-coordinate of the center of mass, right?

Is $d$ diameter? and what is $m$?

DHMO

@heather no, $\mathrm d\theta$ means a very small angle

heather

@DHMO, $d$ is an angle? The angle of what?

DHMO

no

d means very small

heather

the derivative? i always thought that was rate of change of whatever

DHMO

12:06

no...

user116211

@heather It's a differential.

heather

@MAFIA36790, oh...

and $m$?

Oh, never mind, I need to go to the bus stop. Not like I had a chance of helping anyway =P

have a good day everyone

user228700

12:21

@DHMO: I can't figure out what's actually wrong :/

0celo7

12:32

What the fuck is going on my boys?

user116211

I'm doing undergrad Group Theory.

0celo7

You only talk about grad level books so I doubt that.

user116211

No.

0celo7

What book?

user116211

Books.

0celo7

12:39

What books?

user116211

Herstein, Seth Werner and Dummit & Foote.

0celo7

Ok I know one of those, and it's undergrad.

user116211

Yes.

user116211

I will read Bourbaki at night.

user116211

It has some awesome theorems on homomorphisms of monoids.

user116211

12:42

At least that part is interesting.

user116211

So many corollaries of Lagrange's Theorem ;/

ACuriousMind

@Danu Why the ':P'? :P

Physics Guy

Yo

"Higher categories are sheaves on manifolds"

0celo7

@ACuriousMind Which message?

Also hi

Long time no see

user116211

6 hours ago, by Danu

Hey @ACuriousMind https://arxiv.org/abs/1610.07864 :P

ACuriousMind

12:57

@0celo7 What? Click on the arrow!

@0celo7 heyhey

user116211

@ACuriousMind He has a ghost-phone. His fingers are fatty.

ACuriousMind

He's not on the phone right now because he directly replied to my message

Physics Guy

Hey, that Arxiv paper looks interesting.

user116211

Ah! okay. stares at 0celo7...

0celo7

@ACuriousMind phone

Hey screw you I am on my phone

ACuriousMind

12:58

oO

0celo7

Physics Guy

Hehe, that's me

ACuriousMind

lol, I would've believed you without proof

0celo7

In sitting in the analysis room waiting for my death

Physics Guy

Then you're not a mathematician

0celo7

13:00

@ACuriousMind I never believe anything without proof.

@PhysicsGuy correct. I don't know why people say I am.

ACuriousMind

I know :P

Physics Guy

@Ocelo7 I talked to ACM

Secret

Spontaneous parametric down-conversion

Spontaneous parametric down-conversion (also known as SPDC, parametric fluorescence, or parametric scattering) is an important process in quantum optics, used especially as a source of entangled photon pairs, and of single photons. == Basic process == A nonlinear crystal is used to split photon beams into pairs of photons that, in accordance with the law of conservation of energy and law of conservation of momentum, have combined energies and momenta equal to the energy and momentum of the original photon and crystal lattice, are phase-matched in the frequency domain, and have correlated...

@acuriousmind Is conservation of momentum the reason why the polarisation are correlated in this process?

^To prevent the message being split up by other messages, I often have to set a placeholder right after the link or image and type as fast as I can on the question or description

0celo7

ACM is more of a anthem avian than me

ACuriousMind

@Secret Here's a tip: Instead of posting an unnecessary oneboxing link, just ask your question and link to the Wiki article in it, if you absolutely must.

@0celo7 I have no idea what that means.

Physics Guy

13:04

@Ocelo7 Why do you claim you're not a mathematician ?

Aren't you ?

Secret

ACM: Tip noted

0celo7

@ACuriousMind you're more of a mathematician than me. What's hard to understand?

ACuriousMind

@0celo7 "anthem avian" means 'mathematician'? oO

0celo7

Typo.

Physics Guy

lol

ACuriousMind

13:06

@Secret no, why would it?

Secret

Then I have no idea how SPDC can ensure the pairs are always end up entangled in their polarisation states if there is nothing to force that to happen

Many links talked about this process, but I have not found any nor PSE link the explain the mechanism that SPDC generates entanglement. So unless it is "just is", I have no idea if there is a theoretical explanation on how it occurs.

user116211

Hey @arctictern. Welcome to the h bar.

ACuriousMind

@Secret I think it depends on the exact process that happens in the birefringent crystal. Note that it can't have to do with conservation because both the same and opposite polariazation can come out, depending on the crystal used

Secret

I see

Physics Guy

Hilbert had a student who one day presented him with a paper purporting to prove the Riemann Hypothesis. Hilbert studied the paper carefully and was really impressed by depth of the argument; but unfortunately he found an error in it which even he could not eliminate.

The following year the student died. Hilbert asked the grieving parents if he might be permitted to make a funeral oration. While the student's relatives and friends were weeping beside the grave in the rain, Hilbert came forward. He began by saying what a tragedy it was that such a gifted young man had died before he had had an opportunity to show what he could accomplish

But, he continued, in spite of the fact that this young man's proof of the Riemann Hypothesis contained an error, it was still possible that some day a proof of the famous problem would be obtained along the lines which the deceased had indicated. "In fact," he continued with enthusiasm, standing there in the rain by the dead student's grave, "let us consider a function of a complex variable...."

Lozansky

13:28

I don't get it

Physics Meta

0

Q: Withdrawal from site due to plagarism accusations

First, and most importantly, a serious thank you to everybody who has taught me a hell of a lot about physics over the last year or so, especially the mods. I have decided to delete my account because I have been accused of plagiarism. In fact I have already been found guilty if you count hav...