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user116211

conjure

Verb: conjure ‎(third-person singular simple present conjures, present participle conjuring, simple past and past participle conjured)

(intransitive) To perform magic tricks.
(transitive) To summon up using supernatural power, as a devil
(intransitive) To practice black magic.
(transitive) To evoke.
(transitive) To imagine or picture in the mind.

(23 more not shown…)

Noun: conjure ‎(uncountable)

(African American Vernacular) A practice of magic; hoodoo; conjuration.

It would have been much clearer if you had written "Magic!" I pretty much understand "conjure" as equivalent to "summon".

user116211

Are you sure you never shared the link to anyone?

6:34 PM

I can not conceive of a situation where I would have sent that to anyone

user116211

o.O

There aren't that many people I share SE links with anyway, I'm surprised I got 25 clicks at all, regardless of the question

But for this one, I don't even know why I would share it

@user36790 Just conjecture but one reason someone might want to do that is because (s)he fears that if the text of the problem is present a web search will reveal that they went to the internet to get help with the problem. Images generally aren't searchable, so it's a kind of security measure.

But Physics has no reason to care about the posters preference that their post not be discovered. We're going to make the best choice for us.

I'm not, however, gong to undelete the post because it's a "do my owrk for me" kind of question anyway.

@dmckee Sounds reasonable, the text does sound like a problem.

Doesn't it just. So do his previous two posts.

user116211

6:41 PM

@dmckee pics?

I'm baffled why they would not just stay userXXXXX to prevent being found out, though

They are also at the advanced undergraduate or graduate level, suggesting that if this is the case this person has been skating for a long time.

user116211

@ACuriousMind ip address?

@user36790 One is not deleted and you can look for yourself: physics.stackexchange.com/questions/219497/….

The other is similar.

@dmckee Sadly, a significant fraction of rather advanced technical questions here come from people who are hopelessly in over their head

6:44 PM

@user36790 Maybe. But that's not public info on Stack Exchange sites though.

user116211

@dmckee There maybe some John Skeet sitting somewhere ;/

user54412

@ACuriousMind Would you agree with the following statement? A significant fraction of people studying advanced physics are hopelessly in over their heads.

That said, (s)he does sometimes come from a major US university.

@ChrisWhite Not to the degree the askers I have in mind display, but to some degree, yes

@ChrisWhite What do you mean

6:46 PM

Hey- in the Boltzmann entropy equation, how do we figure out that boltzmann's constant is the right constant?

I'm often i over my head when it comes to advanced theory. I just know how to cope.

@Anthony By experiment. Like real scientists.

I was thinking in particular of a certain "run before you can walk" attitude that leads to some really confused questions.

@dmckee What about the fact that it's the same in entropy and the ideal gas law? Is the fact that they share this constant derivable from thermodynamic relations?

@dmckee I was in a lab earlier where a wall was black from smoke and fire. You know some real science happened there.

Boltzmann's constant is not the "right" constant. Temperature should be measured in energy ;P

A: Is the Boltzmann constant really that important?

6:49 PM

@ACuriousMind Could you elaborate on this? I've heard about this, but I'm a little confused on what the change would entail.

user116211

@Anthony

user116211

26

We can understand all of this business if we visit the statistical mechanics notion of temperature, and then connect it to experimental realities. Temperature is a Lagrange multiplier (and should have dimensions of energy) First we consider the statistical mechanics way of defining temperature....

@Anthony You just don't introduce the constant, or, in the usual physicist's parlance, you set it to 1.

Entropy is dimensionless because it's just counting states and temperature is energy because it is just an energy scale.

You don't have to change anything except the labels on thermometers ;)

How would temperature have units of energy? Isn't the energy in something relative? Isn't temperature absolute?

@Anthony The answer from DanielSank linked by @user36790 is really good, in my opinion.

6:57 PM

@ACuriousMind I read through it, still confused, reading through it again.

Also, what do you mean by relative? The average kinetic energy (squared, in the rest frame) of a collection of particles is not relative.

Yeah, nevermind I'm wrong lol.

@dmckee Is nuclear.exp so rooted in Word/PP because of its long history with the government?

The experimental condensed matter people at ORNL send your data in a Word doc for crying out loud!

Maybe they just suck.

@0celo7 All the nuclear experimenter I knew where unix/LaTeX guys. But that is about place. Facilities have culture of their own, and the people I knew in the field worked at Los Alamos, Brrokhaven and Fermilab.

user116211

7:02 PM

@ACuriousMind: Can I ask question on gibbs energy?

Yes, but you may not ask any more whether you can ask questions. Just ask them. Asking to ask a question is superfluous.

@dmckee The paper I gave a talk on was a Word document. Written by a LANL group.

Well, shows you waht I know.

user116211

@ACuriousMind Yes, I know but still whether anyone is interested on this topic. previously, I asked Yuggib and he told he last did thermodynamics back in 2002.

Q: What would these formulae be used for?

The lead author works here now. Maybe I should ask him. I feel like if I ask Lang about it, he'll say "what's TeX"

user116211

7:06 PM

-2

I'm an A2 Physics student and I bought this case for my MacBook. Whilst there are several equations that I am familiar with (e.g. e=mc^2 and ⍵=2πf), I've never seen at least 95% of them (and that's being generous!) I feel rather fraudulent for having such a snazzy case as, if someone asked me w...

soft-question

user116211

^^^ WTF!

Well, I'm not particularly good with thermodynamics either, I find all the strange historical paths typical courses on the subject take annoying and confusing.

@ACuriousMind I don't think they suck.

user116211

@ACuriousMind Still, plz listen

At least, they're good at physics.

I don't think TeX is a component of physics.

user116211

7:09 PM

If $\Delta G= -T\Delta S_\textrm{rev, universe}= 0$, then why for reversible process, $dG= w'$ where w' is the max. non-expansion work?

@ACuriousMind Oh, forgot to let you know. I had salsa on my eggs this morning. Don't bother asking me why I'm telling you this.

@0celo7 Neither is Word. Nor even Root or CERNLIB. Those are tools. You need good tools, but they don't tell you anything about the person using them.

@0celo7 If you're going to go that route do it right and make huevos rancheros. Hmmm. One of the Breakfasts of the Gods when done well.

user116211

I mean, if $G$ is measuring the entropy of the universe and when the entropy change of the universe is zero which happens in reversible process, wouldn't also $dG= 0$?

@dmckee I can't get all of the ingredients at breakfast time in our dining hall.

@user36790 Uh... $G = U+pV-TS$. Just apply $\mathrm{d}$ to that. $\mathrm{d}S= 0$ does not imply $\mathrm{d}G=0$.

...we have a limit on how many of their own posts users can delete per day?

user116211

7:24 PM

@ACuriousMind at const. temp and pressure.

@user36790 That still leaves $\mathrm{d}U$ and $S\mathrm{d}T$ in $\mathrm{d}G$.

user116211

@ACuriousMind $\mathrm dG= Vd\mathrm P -S\mathrm dT$

@user36790 Uh, no. Use the product rule.

That $\mathrm{d}$ is not some magic thermodynamic expression, it's just a differential.

@ACuriousMind One of a number of subtle defenses against various things that rage quitters do.

user116211

@ACuriousMind dG= TdS-PdV+ PdV+ VdP-TdS -SdT

7:32 PM

@user36790 No, how did you get that? If $G= U+pV-TS$, then $\mathrm{d}G = \mathrm{d}U + \mathrm{d}(pV) - \mathrm{d}(TS) = \mathrm{d}U + V\mathrm{d}p + p\mathrm{d}V - S\mathrm{d}T - T\mathrm{d}S$. Reversible is $\mathrm{d}S =0$, constant pressure is $\mathrm{d}p=0$ and constant temperature is $\mathrm{d}T=0$. So $\mathrm{d}G = \mathrm{d}U + p\mathrm{d}V$.

user116211

@ACuriousMind use $dU= TdS- pdV$

@user36790 Okay, then $\mathrm{d}G=0$ indeed. So what's that $w'$ which is "maximum non-expansion work"?

user116211

one term is missing in the last equation $TdS$?

What kind of process is this supposed to be, actually?

What process is reversible, doesn't change temperature, and doesn't change pressure?

user116211

@ACuriousMind reversible; temperature and pressure doesn't change

7:38 PM

That's not an answer

I'm asking you what is physically happening to this gas.

user116211

@ACuriousMind I'm in need of answer ;P

user54412

reversible means entropy doesn't change

@0celo7 and at any time you have to do nothing

user116211

@ChrisWhite yes. And that would mean $dG= 0$

user54412

@user36790 why?

user54412

7:39 PM

$dS \neq dG$

user116211

@ChrisWhite $dG= -TdS_\textrm{universe}$ at const temp. and pressure

@user36790 Yeah, but you're not making sense. I cannot right now imagine a thermodynamic process that keeps pressure, temperature and entropy constant.

What are you going to do to a gas that is all that?

user116211

@ACuriousMind I also indeed want to say that but why does that w' pop out?

You're not compressing it, you're not expanding it, you're not heating it, you're not cooling it.

@user36790 What is $w'$? where does it "pop out"?

user116211

@ACuriousMind The maximum non-expansion work.

Q: What is non-expansion work?

7:41 PM

I have no idea what that means.

user116211

:27782815

user116211

6

I learnt that Gibbs free energy is the maximum amount of non-expansion work. But the phrase non expansion work confuses me. Work is defined as pressure times change in volume. If there is no expansion, how can work be done?

thermodynamics free-energy

user54412

Indeed, I have never seen "$dG= -TdS_\textrm{universe}$," "$w'$," or "maximum non-expansion work" in thermodynamics.

@user36790 Uh, so who's telling you that this strange $w'$ (note there's no definition at the link, it's just a list of some things that may be called that) is non-zero in your situation?

@user36790 Thermodynamics is one of those things that I will hopefully and happily never deal with again in my life now that I am a mathematician

user116211

7:45 PM

@ACuriousMind books.google.co.in/…

lol, that's one circular definition if I ever saw one

He literally just hides all dependence of $w'$ on the state variables and then only sets those to zero where he can see the dependence.

user116211

@ACuriousMind the author uses $dU = dq + dw +w'$

user54412

Also, if this is in the context of chemical reactions (rather than the physics of a single perfect gas), there are $\mu dN$ terms lurking everywhere.

^

user116211

I've never seen such 1st law

7:50 PM

@user36790 I literally don't see a $U$ anywhere there.

Also, what ChrisWhite said. Your Gibbs energy (and consequently your "non-expansion work") has all those chemical potential terms in it if we're talking chemistry

user116211

@ACuriousMind chemical potential is partial molar gibbs energy

If there are no such terms, then your non-expansion work is zero

@user36790 What does that mean, and how is that relevant

You keep using all those strange terms, but where's the physics?

user116211

@ACuriousMind it's not strange term;

user116211

heard of partial pressure?

user116211

Dalton's law?

7:54 PM

Yeah, all those named laws are another reason I hate thermo

But looking it up, I again don't see how that is relevant

@ACuriousMind Not like everything in string theory or gauge theory is nameless

Your issue was that $w'$ is not zero. Apparently, it contains something that doesn't vanish at fixed volume, temperature, pressure and entropy

user116211

@ACuriousMind you mean I've to use $dU= TdS-PdV+\mu\,dN\,?$

user116211

But N is constant; it never changed.

user54412

@user36790 What, pray tell, has changed in the system you picture so well in your mind?

7:58 PM

@user36790 Uhhhh...if you use $\mu$ and $N$, then you are in the grand canonical ensemble, where $N$ is not fixed, but a state variable.

the very idea of having a chemical potential is that the damn particles keep reacting and thus the number of particles of one kind is not constant.

user116211

@ACuriousMind I'm really hating this Thermo.

user116211

Really maddening.

@0celo7 It's not as ridiculous as having like half a dozen names for special cases of the ideal gas law

user116211

@ACuriousMind I'm dealing with the system and the surrounding. Neither of them have N changed.

@user36790 "The system"

What is "the system"?

user116211

8:03 PM

@ACuriousMind anything you like.

Aha!

So I like having two reactive gases in a chamber.

user54412

Well I want T and P and S to change!

Then there is $\mu_1\mathrm{d}N_1 + \mu_2\mathrm{d}N_2$ in $\mathrm{d}G$.

$N_1$ and $N_2$ are not constant because they keep reacting back and forth

user116211

@ACuriousMind As in example by the author in that book, formation of water from the constituent elements was taken.

So why do you believe $\mathrm{d}G$ should be zero if I am free to choose reactive systems with chemical potentials?

user54412

8:05 PM

@user36790 Well then the numbers of oxygen and hydrogen particles have decreased, and the numbers of water molecules have increased. There are 3 N's, all of them changing.

user116211

@ACuriousMind You believe this $$dG= -T dS_\rm{universe}$$?

No

user54412

No

user116211

Why?

Why should I believe it?

user116211

8:07 PM

at const. temp and pressure?

That's not a reason to believe it.

Why don't you believe $\mathrm{d}G = \sum_i \mu_i\mathrm{d}N_i$ at constant entropy, temperature and pressure?

That would at least be correct :P

user116211

$$dS_textrm{total}= dS_\textrm{sys}+ \frac{1}{T}dU -\frac{P}{T}dV= -frac{1}{T}dG$$

Just look at the infinitesimal chemical form of the first law and then the expression for $\mathrm{d}G$ derived from it.

user54412

@user36790 everything there assumes constant N or 0 $\mu$, and thus cannot be applied to chemical reactions

user116211

I need break; call it a day; I need sleep; bye.

8:25 PM

@yuggib Today's PDE lecture was interesting for once. We talked about estimating the ground state energy of a 1D quantum system

Slereah

8:39 PM

2

@BernardMeurer Came across the functional $$R[u]=\frac{\left.-puu'\right|_a^b+\int_a^b[p(u')^2-qu^2]\mathrm{d}x}{\int_a^b u^2\sigma\mathrm{d}x}$$ in PDE today

@Slereah lol

@yuggib Today in PDE we proved that a regular Sturm-Liouville problem has a minimal eigenvalue and that this eigenvalue is $$\lambda_1=\min_{u\in S} R[u]$$ where $R[.]$ is defined above and $S=\{\text{continuous functions that satisfy the regular SL boundary conditions}\}$

@yuggib Then the prof said we can estimate $\lambda_1$ by finding functions in $S$ and plugging them into $R$.

He said one should find some trial function $u$ that depends on some parameters $\alpha_i$.

Then one does a minimization problem.

Suppose one chooses $u$ to be a polynomial, and writes $u_n=\sum_{k=0}^na_{n,k}x^k$.

Do $a_{n,k}$ converge to the Taylor series coefficients of the $\lambda_1$ eigenfunction in the $n\to\infty$ "limit"?

@Slereah Gotta love Goodstein.

@yuggib This was inspired by a similar question someone asked in class, and it was clear the lecturer didn't know for sure.

8:52 PM

@0celo7 What you haven't said is that you want each of the $u_n$ to be the respective result of the minimization procedure for polynomials of order $n$, right?

@ACuriousMind Oh, right, of course.

Question: Is the eigenfunction guaranteed to be analytic?

Otherwise the "In general, no" just follows from the eigenfunction to $\lambda_1$ just not having a Taylor series in general

@ACuriousMind Hmm.

We never talked about that.

Analyticity is a quite strong requirement for real functions, I'd be surprised if it was.

@ACuriousMind He said that $u\to \phi_1$ at least in a "weak" sense.

I suspect he meant a distributional sense.

($\phi_1$ is the first eigenfunction)

@ACuriousMind How does one prove analyticity?

8:58 PM

Do you know anything about its regularity? Integrable, continuous, differentiable, smooth?

@0celo7 huh?

You compute the convergence radius of the Taylor series, of course

Analytic functions are by definition those where it is infinite.

@ACuriousMind Ugh

I know that the defintion is

But like if you asked me if $f(z)$ is complex analytic, I'd strive to prove $\bar\partial f=0$

Is there a similar condition for real analyticity?

The easiest way to show analyticity is to show it is the restriction of a holomorphic function, though

@ACuriousMind $\phi_n$ are real.

So?

That doesn't tell you anything

@ACuriousMind They have to be $C^2$ because they're the solution of a second order ODE, right?

9:02 PM

@0celo7 Ahahahahaha...no

Or, well, maybe in your context, they have to

Oh crap

It's not an initial value problem

But in general, one looks for weak solutions to differential equations.

No Picardy Lindelhofer

@ACuriousMind The $\phi_n$ can be used for a generalized Fourier series

So?

I'm just telling you what I know about them!

9:04 PM

Functions in $L^2$ have Fourier transforms/series, they're not even continuous

I'm pretty sure the solution is not in general analytic and the question thus ill-posed

@ACuriousMind The prof did not mention anything about the regularity.

@0celo7 why do you want to approximate the eigenfunction with a polynomial?

@yuggib To estimate the smallest eigenvalue.

why with polynomials?

Because one can easily compute their derivatives and integrals?

9:09 PM

@0celo7 Okay, let's try something else then: Is the Legendre equation a problem of the type you are considering?

@ACuriousMind Not yet, but it's on the horizon.

(We're doing Laplace in spherical on Thursday)

Bessel?

On the horizon

Have you got any solved problems where you could test whether your ground state solution is analytic?

@ACuriousMind All eigenfunctions have been sines/cosines thus far.

9:11 PM

Sigh

Yeah, that's obviously analytic, but also obviously unhelpful because they're the most special functions in existence :P

We've solved: heat equation in one spatial dim, wave equation (with and without damping) in one spatial dim, Laplace equation in the plane

@yuggib are you thinking, got no clue, ?

@0celo7 I don't see why you should use polynomials

@yuggib What would you use?

first of all, you should say which is the space in which the operator acts

@yuggib Oh for fuck's sake

9:19 PM

no, that's exactly the crucial part

@yuggib The fact that he doesn't know anything about the regularity of the solution made me not ask that question as the answer was clear :P

Continuous functions on $[a,b]$ that satisfy $\beta_1 u(a)+\beta_2u(b)=\beta_3u(b)+\beta_4 u(b)=0$.

@ACuriousMind yeah, but then it is useless to talk about minimizers and so on

They pulled a physicist and just solved the equation :D

@ACuriousMind My prof only teaches applied math/math phys classes

But not math phys like @yuggib 's narrow understanding

9:21 PM

@0celo7 with which topology?

@ACuriousMind Sigh...is he trolling me?

@0celo7 No.

@ACuriousMind He has to know I have no clue what he's talking about.

to do an approximation, you need to show convergence

@0celo7 To talk about convergence you need to talk about topology.

9:22 PM

of the approximated function to the true function

@ACuriousMind I have no clue what you're talking about.

(I know that one needs a topology for that.)

(But this course doesn't go anywhere near topology, and @yuggib knows that.)

Since you're asking about topologies and whatnot, you don't know the answer outright. Is that correct, @yuggib?

@0celo7 but your question is a question of which minimizing sequence to choose

@yuggib Dude, I can't tell you what topology because I don't know what that means.

@0celo7 without more information, in my opinion there is no answer

and that's probably why the prof did not answer clearly to the question

And by "I don't know what that means" I mean "I don't know what topology you want me to give you"

9:25 PM

@yuggib Nah, he fixes the sequence (polynomials of order $n$ that minimize the expression among all polynomials) and wants to know if the coefficients of those polynomials converge (in $\mathbb{R}$) to the coefficients of the Maclaurin series of the lowest-lying eigenfunction.

@ACuriousMind Exactly. I thought this was clear?

@ACuriousMind you have no idea that the eigenfunction is smooth

That's not a particularly natural question to ask from the functional analysis viewpoint, but it doesn't need topology to be well-defined - it just needs existence of the Taylor series of the eigenfunction

@yuggib That was what I was also explaining.

@ACuriousMind Ok, now that objection is something I understand.

but if the operator acts only on smooth functions, and you have proved there is an eigenfunction, it has to be smooth

9:27 PM

@yuggib It acts on continuous functions.

btw, by the formula you gave, the eigenfunction is not even necessarily continuous

@yuggib What

if the "space of continuous functions with boundary conditions" is not complete, you can't guarantee that the minimizer is still in the same space

that's why it is absolutely a matter of topology

@yuggib Well...if they claim it is a minimum and not an infimum, it has to, no?

What?

Any polynomial is in that space

And we fix the boundary conditions on the polynomial before the minimization

9:30 PM

@ACuriousMind ah yes, they claim it's a minimum

anyways, every polynomial is in the space, but not every function in the space can be written as a polynomial, do you agree?

Yes.

so why would you expect that the aforementioned minimum realizing function can be written as a polynomial?

Because fucking magic

Are you not listening to me?

I said I understand that objection

so?

@0celo7 It's less an objection and more asking you to supply some motivation to expect what you're asking about being true. Have you checked for the wave equation if those polynomials look like they converge to the Taylor series of the sine?

9:36 PM

@ACuriousMind No, but we have to estimate $\lambda_1$ (for the heat eq.) using a third order polynomial for homework

Those two statements seem unrelated :P

what

@yuggib If $\phi_1$ is $C^\omega$, does the claim hold?

Having one third order polynomial is not going to tell you anything

@ACuriousMind I can plot the damn thing

Maybe it looks like the Taylor series to third order

lol

9:38 PM

I can't mathematica well enough to do higher

That's a very optimistic idea of the convergence

I'm an optimistic guy

@0celo7 I don't think it's necessarily true

@yuggib Do you have a special reason in mind?

yes, essentially related to this

9:42 PM

@0celo7 That's not a "philosophy" (and I don't know to which extent you are joking)

@yuggib uh, what does that mean

that there are minimizing sequences, but that converge to the minimizer with respect to "some topology"

@ACuriousMind In all seriousness, there's something very dissatisfying about using a counterexample to disprove a statement.

Uh, what?

I like its brutality.

You are a savage Saxon.

9:45 PM

All the theorizing and wishful thinking in the world won't help you against the wrecking ball that is a counterexample

And mathematicians hone those counterexamples carefully, like weapons, and pass them down to the next generation with glee

@ACuriousMind But the counterexample does not convey any idea of why the claim is wrong.

I am not arguing with the logical implications.

I never understood the question why something is wrong. Why is the sky not yellow? Why is a continuous function not differentiable? Why is a triangle not a square?

Those all look essentially and equally meaningless to me.

Suppose I have a claim about some objects $x$.

But I find an object $y$ for which the claim is not true.

So the claim is not true for all $x$, ok.

But what about all $x$ except for $y$?

@0celo7 That's almost the correct question. The correct question is "Which $x$ fulfill the claim?"

Where "which" doesn't mean you should enumerate them

@ACuriousMind Same thing in my mind.

@ACuriousMind Of course.

@ACuriousMind What?

Those are nonsensical questions, that's why they're meaningless :P

9:50 PM

Anyways, to develop my doubts, a sequence of polynomials that converges uniformly converges to a polynomial.

But it may not be true for other convergences, like the pointwise convergence

@yuggib The prof said he did not know about pointwise convergence.

And I doubt he's interested enough to think about it more.

It's well outside of the scope of the course.

@0celo7 Why is "why is a continuous function not differentiable?" a non-sensical question, but you can say that "why is this claim wrong?" is a meaningful question for claims disproven by counterexample?

Because nothing in the definition of continuity guarantees the existence of the difference quotient.

@0celo7 anyways, you can make any bold or unreasonable claim; that does not make them worth a disproving

I think you're just too smart to understand people's issues.

9:52 PM

they may be worth a proof, but they are probably false

And I don't mean that in a bad way.

Most people I've ever known ask "why is this wrong"

@0celo7 !!! And if you find a counterexample to a claim $X$ about $Y$, then nothing in the definition of $Y$ guarantees $X$ is true

Replace $X$ by "is differentiable" and $Y$ by "continuous functions" and you've got the exact form of my nonsensical question

@0celo7 not in mathematics

@yuggib I don't know many mathematicians.

And the average mathematian is certainly smarter than the average person I know.

however, conjectures in math are very popular

Q: Understanding the Faraday tensor

9:55 PM

@ACuriousMind It's not the same

0

I'm trying to get my head ahead understanding the Faraday 2-tensor. I first started by thinking about how I've understood the electric and magnetic fields in electro/magnetostatics so far. The electric field is just the force that a unit test charge would feel from a purely electric effect assu...

electromagnetism

@ACuriousMind Please answer with the full force of $G$-bundles.

@0celo7 Then where is the difference? What makes some questions of the form "Why is $X$ about $Y$ not true?" meaningful and others not?