The h Bar - 2016-02-23 (page 2 of 4)

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user116211

2:00 PM

@JohnDuffield Really, do you think my assumption was right?

user116211

I was corrected by Timeaus.

user116211

You didn't notice that?

user116211

Read his answer.

user116211

You can't bore a misconception of mine then in your defence now.

@user36790 btw a charge always creates an electromagnetic field either if it is moving or not (and interacts with it)

I don't see what's the problem with that, the equations of motion of the coupled system EM field + particle are rather easy to write

and globally well-posed in many interesting cases (where the charge distribution is extended)

2:09 PM

@user36790 : Timaeus said the EMF around a stationary wire is 100% due to the electric field. That's wrong. See the Jefimenko quote: Therefore, we must conclude that an electromagnetic field is a dual entity . The electron doesn't have an electric field, it has an electromagnetic field.

user116211

@yuggib yes; I was having wrong thinking then.

@user36790 that's not something to be ashamed of...everyone has wrong thinking now and then ;-)

If anybody thinks they have the right thinking now, try depicting the electron's electromagnetic field. If you cannot, ask yourself why not. You can probably depict a gravitational field, and a gravitomagnetic field. If you can't depict an electromagnetic field, there's some kind of gap in your knowledge.

This is a bit like the blind spot test. You can't see what you can't see, but there is a way to see that you can't see it.

OK, work calls. Gotta go.

@yuggib There might be a course next semester focusing on ODEs/PDEs in class mech, stat phys and QM.

2:42 PM

There's a mathematician names Quadrat.

@BernardMeurer You ready to learn some calculus

@0celo7 I'm ready to try

@BernardMeurer Mean value theorem: let $f(x)$ be differentiable on $[a,b]$. Then there exists some $c\in[a,b]$ s.t. $$f'(c)=\frac{f(b)-f(a)}{b-a}$$

s.t?

such that?

$user image$

@BernardMeurer Yes.

Okay, that one wasn't too difficult

2:45 PM

The RHS is the slope of the secant through the endpoints.

RHS?

@BernardMeurer Ok, we can now prove the fundamental theorem of calculus.

@BernardMeurer Right hand side.

The Riemann integral is defined by $$\int_a^b f(x)\,\mathrm{d}x=\lim_{\Delta x\to 0}\sum_i f(x_i)\Delta x_i$$

Alright, steady now

So we take $[a,b]$ and partition it. Each partition has width $\Delta x_i$ for some $i$ labeling which partition is it.

The limit is that we take the width of the largest partition to be zero.

user116211

@BernardMeurer: You didn't know MVT?

user116211

2:48 PM

@0celo7 Definite integral?

@user36790 What?

If I integrate over that same $f(x)$, what do the deltas stand for?

@BernardMeurer What?

$\Delta x$

user116211

@0celo7 You are defining definite integral?

2:49 PM

wait I'm fucking up

@user36790 Yes

I'm going to prove the fundamental theorem of calculus.

user116211

@0celo7 1 or 2?

@user36790 I don't know which one is which.

Why is it $\lim_{\Delta x \rightarrow 0}$

I'm going to show that that sum there is $F(b)-F(a)$.

user116211

2:50 PM

@BernardMeurer because the partition is going to zero.

@BernardMeurer We take the largest partition to zero.

Think of this in terms of rectangles.

@0celo7 sounds nice

$f(x_i)\Delta x_i$ is the area of the $i$-th rectangle.

Oh like when you try to define an analog curve in PWM?

PWM?

Only I may use acronyms.

@yuggib Perhaps, it will be application-oriented.

2:52 PM

user image

Yes.

@0celo7 I hope not too much

But then you make the rectangles infinitely thin

:-þ

@yuggib The prof who will be teaching it is an applied mathematician.

user116211

2:53 PM

@BernardMeurer this is the main purpose of $dx to 0$

@0celo7 name?

@yuggib My PDE prof.

user116211

lol

the engineer?

yuggib are you in any of the places I'm applying to?

2:54 PM

@BernardMeurer I'm not in the US

Somehow every Phd or postdoc student in the h-bar is in someplace I applied to

@yuggib LOSER. Jk, neither am I

are you applying in Europe?

user116211

moretti is in Italy.

user116211

yuggib is linguist.

@yuggib Nope, just to the US

user116211

2:56 PM

manishearth is in Mumbai.

@user36790 the chat is quite international

with a slight US bias

user116211

@yuggib unity in diversity?

@yuggib What?

Mathematicians who do PDEs but don't worry about functional analytic aspects are engineers?

@BernardMeurer Yes.

One can show that the limit is well-defined.

@0celo7 your PDE prof seemed quite engineering oriented in his works

but that's just my opinion

I can very well be wrong

@yuggib Ok, I'm an engineer.

3:01 PM

we know that

user116211

Is $dG= 0$ for reversible process?

user116211

G means free energy.

@BernardMeurer Shall we continue?

@0celo7 So, continuing, we define the $\int_{a}^{b}$ of $f(x)$ as being the limit of a sum of ?

Times a "small" distance.

But that's exactly what the area is.

3:04 PM

I mean that I'm having a little trouble getting that sum

T__T Riemann integration is boring

Note that $\int_a^b f(x)\mathrm{d}x$ is just the area, it has nothing to do with antiderivatives

I see how a range of infinitely small (thin) sections of well defined height will yield an area (right?)

user116211

Suppose the system is in contact with the environment; the system is in constant temperature and pressure. If the work done on the system is quasi-static, would that mean $dG$ is zero for the universe?

@BernardMeurer Yeah

user image

You then take $\lim_{\Delta x\to 0}$ and this is how we define area.

3:06 PM

Yeah, just like an ADC (or a reverse DAC)

@user36790 I do not think about thermodynamics since my first year of college...and that was 2002

Note that number of partitions also increases.

user116211

o,O

The number of partitions will increase (inversally) proportionally to how thin they are?

@BernardMeurer Yes.

Ok, so we have $\sum_i f(c_i)\Delta x_i$

3:08 PM

Gotcha, $\int_a^b f(x)\,\mathrm{d}x=\lim_{\Delta x\to 0}\sum_i f(x_i)\Delta x_i$

By the mean value theorem, $f(c_i)=\frac{F(x_i)-F(x_{i-1})}{\Delta x_i}$

Yeah that sum got me a little confused

Here, $F(x)=\int f(x)\mathrm{d}x$ is the antiderivative.

correct, why is it $c_i$ now?

@BernardMeurer I'm fickle with notation

3:09 PM

Alrighty

So we have $\int_a^b f(x)\mathrm{d}x=\lim \sum_i F(x_i)-F(x_{i-1})$

So it turns out that the limit does nothing because the sum does not depend on $\Delta x_i$ any more.

Didn't the MVT include a range $a \rightarrow b$?

@BernardMeurer It works in the range $[x_{i-1},x_i]$

why did you use $x_i$ and $x_i - 1$

Oh, okay I see it now

Those are the partions

Look at the picture above

$\Delta x_i=x_i-x_{i-1}$

3:12 PM

That height delta

So we have $\int_a^b f(x)\mathrm{d}x=\sum_i F(x_i)-F(x_i)$.

as marked on that picture that'll be my derivative!

So the sum is just a finite sum now.

It turns out to be a telescoping sum.

Telescoping?

i.e. it's of the form $a_n-a_{n-1}+a_{n-1}-a_{n-2}+\cdots -a_1+a_1-a_0=a_n-a_0$.

All the middle guys cancel out

3:15 PM

I see! Cool!

So, since $x_0=a,x_n=b$, we get $$\int_a^bf(x)\mathrm{d}x=F(b)-F(a)$$

user116211

QED

How did we define $F(x)$ again?

Now there's another fundamental theorem

@BernardMeurer The antiderivative of $f(x)$

i.e. $F'(x)=f(x)$

And we used the mean value theorem to bring it in

But isn't the antiderivate the integral?

3:18 PM

The other fundamental theorem is $$\frac{\mathrm{d}}{\mathrm{d}x}\int_a^x f(t)\mathrm{d}t=f(x)$$

user116211

@BernardMeurer no,

@BernardMeurer No

The only reason people can say that is because of what we just proved.

If you want to write $F(x)=\int f(x)\mathrm{d}x$, ok

Okay, I see what you mean

But then $$\int_a^b f(x)\mathrm{d}x=\left.\int f(x)\mathrm{d}x\right|_{x=b}-\left.\int f(x)\mathrm{d}x\right|_{x=a}$$

This is a nontrivial theorem but looks trivial because of notation

What's that weird line?

3:21 PM

$\int_a^b f(x)\mathrm{d}x$ has nothing to do with $\int f(x)\mathrm{d}x$ until you prove the fundamental theorem

@BernardMeurer "evaluated at"

One is an indefinite integral and the other is a definite one?

Indefinite integral is a terrible word, imo

It should just be "antiderivative"

Alrighty, I'm following

5 mins ago, by 0celo7

The other fundamental theorem is $$\frac{\mathrm{d}}{\mathrm{d}x}\int_a^x f(t)\mathrm{d}t=f(x)$$

You can prove that as an exercise.

Given the one we proved, this is trivial.

Working on it

Check iMessage; How is $F(a)$ going away?

3:29 PM

@BernardMeurer It's a constant wrt. $x$.

I WAS ABOUT TO SAY THAT

god dammit

got it

Hmm, well, you didn't.

there, proved

@BernardMeurer Never end a proof with $f(x)=f(x)$

What do I do then?

3:33 PM

@BernardMeurer You can write everything in one line

$$\frac{\mathrm{d}}{\mathrm{d}x}\int_a^bf(x)\mathrm{d}x=\frac{\mathrm{d}}{ \mathrm{d} x}[F(x)-F(a)]=f(x)-0=f(x)$$

So is proving theorems like coding high performance stuff where you should always strive for the least number of commands?

Not really

It's just that there's no reason to do it like you did

And $f(x)=f(x)$ is always true

You're tying to show that $stuff =f(x)$

Why did you put brackets there?

@BernardMeurer So it's clear that the derivative is acting on everything to the right

I see what you mean yeah, it is silly to say $a =a$

@0celo7 Got it

3:36 PM

@Danu: Another wonderful example of "physicist's math": "We have determined the kernel of the operator, so we have determined the operator uniquely."

@ACuriousMind Huh?

Can one uniquely obtain an operator from its kernel?

After like ten minutes of heated discussion, it turned out that in this specific case, we had determined the operator up to multiplication by an arbitrary function of time

@0celo7 only if the kernel is the whole space

@yuggib Uh, then the operator is the zero map.

@yuggib Why are you so allergic to applied math, anyway?

It's not as bad as physics math

@0celo7 I am not allergic to it...simply one should see applied math after having seen pure math

or at least after a sufficient amount of pure math

if else there's nothing to apply

3:43 PM

They've corrupted the beautiful purity of the entire field

@ACuriousMind Predictable.

And once again, "gauge theory" is not applied math.

@0celo7 tell that to a pure mathematician :-þ

@ACuriousMind Put down the chocolate-covered banana and step away from the European currency systems.

@BernardMeurer Uh

@BernardMeurer ...how is that related? That's a very strange way of saying someone asked the wrong question, nothing to do with pure/applied

3:46 PM

I just like it

@ACuriousMind Ok, what the hell is that equation on the bottom

Killing?

I want to make a tattoo that says that

right on my forehead

@0celo7 btw, I am an applied mathematician

@yuggib No, you aren't.

@0celo7 Yes, that's the pun.

3:47 PM

@ACuriousMind Only like 5 people who read that will recognize it

Abstruse goose is not read by more than those five people, I think.

@yuggib Did you get that pun?

It's niche, and it's proud of it.

@ACuriousMind A significant portion of the math and physics community would not get that pun.

Your point being?

3:49 PM

@0celo7 the whole branch of analysis is essentially considered applied mathematics in France

Abstruse goose is only read by differential geometers?

@0celo7 not immediately

@0celo7 Stop being literalist :P

1 min ago, by ACuriousMind

Abstruse goose is not read by more than those five people, I think.

@yuggib So?

Applied math has to have real-world applications somehow

Now don't go telling me that your functional analysis has raised the GDP

@0celo7 the application can be other mathematics ;-)

3:51 PM

Bull

@0celo7 And...finding solutions to differential equations which can occur as actual equations of motion of actual systems is not a real-world application?

@ACuriousMind No

Not unless those solutions are actually used for something

The "real world" is larger than what is dreamt of in your GDP

@ACuriousMind Not really!

@ACuriousMind an analyst almost never find solutions...he only proves they exist and are unique

3:53 PM

Does neatness of handwriting correlate (weakly) with intelligence?

To speak of correlation, you need to first define an objective measure of "neatness".

@ACuriousMind My eyeballs.

@yuggib Well, that's a start ;)

@ACuriousMind or the end, depending on the point of view

@ACuriousMind Ok, what's some applied math that does not raise GDP

Or have the potential to raise GDP

Someone could solve fusion in a basement, it won't raise GDP, but it could

3:58 PM

Well, potential is ill-defined. You never know.

at this moment

Before cryptography, most people would have sneered at the idea that number theory could have actual applications, for instance.

I still sneer

Unsurprising, but also irrelevant.

Chat session time?

4:00 PM

Chat session time!

All right! So, what shall we talk about today?

4

Q: Is this question about the meaning of derivations on topic?

Are derivations of physical laws less important than the laws themselves? This question is attracting a lot of attention, but there is a significant disagreement about whether it's on topic. I want to bring it up here to give the issue more exposure. I don't think I can give any summary that wou...

discussion scope specific-question

Ah, yes, good idea

What to do with questions where there seems to be absolutely no consensus on their on-topicness?

3 hours ago, by yuggib

http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.115.160505

poll: good physics, or bad physics?

@ACuriousMind I would use the words of wisdom of mother mary in that famous song

4:04 PM

Whoa the chat looks different on mobile

@yuggib I don't immediately see the particular significance (yes, that "Hilbert hotel shift" seems possible, but what's the big deal?), but it doesn't look as absurd as the Wallis formula to me

when did this happen

@yuggib that's weird

@ACuriousMind I essentially agree

@ACuriousMind Unsurprising?

4:06 PM

@yuggib I'm afraid I don't know what you refer to

@ACuriousMind "Let It Be"

Ah!

My fingers are too fat to click the new arrows

Well, but then, the question is which way to let it be?

Does anyone else have problems with that?

4:07 PM

@DavidZ That would have been my follow-up

@0celo7 No. The solution is to don't be on here with mobiles.

@ACuriousMind They updated the interface.

But it doesn't work.

I'm on mobile right now

Can you use the new arrows?

@GPhys test

@DavidZ We could just let the close and reopen votes play out. Unless someone feels really strongly about it, the question will stabilize either closed or open.

Yes

4:10 PM

@ACuriousMind that seems reasonable

@ACuriousMind yeah, I suppose so. In the absence of a consensus from meta I would probably just do that.

Is the current discussion about the particular meta question, or a meta meta discussion about questions without topic consensus?

What I don't like is the HNQ aspect

It's not good to have no clear policy for questions which have a higher-than-average potential to go hot

And, let's face it, philosophical waffling is more likely to go hot than hard physics :P

@GPhys the meta post was about that specific question

Maybe being on mobile does have its downsides

4:13 PM

I think the discussion we are having now is about the more general issue

Yes. Coming to a consensus (or rather, to a decision) about the specific question would also be nice, but that wouldn't resolve the general issue

I note a lot of downvotes on all answers to the meta question, but no arguing in the comments except for tpg2114.

that's strange

Having people anonymously express their disagreement is not very conducive to determining what exactly they disagree with

That's true, but it's the nature of these debates

Oh, other meta threads are far more...discussion-y!

Well, some of these debates

oh, I see you are talking about the meta-questions.

4:19 PM

"I better comment on all the other answers to let them know the opinions outlined in my meta answer is the correct one." -some meta "debates"

0

Q: Expectation value

I would like you can help me with something that I am not able to prove rigorously. That is: If two operators A and B have the same expectation value for any vector of the hilbert space, then A=B. Thank you in advance.

quantum-mechanics

the answer is probably subtler than the OP expects

@GPhys pretty much

@GPhys: this reminds me of this scene in Feynman's "Surely you are joking" where he is at a discussion with some hotshots and they all argue their point exactly once and in the end, the correct answer gets chosen. Well, this is not like that...

@yuggib Is it? (Note that "any vector of the Hilbert space" implies no domain issues)

@ACuriousMind yes, in my opinion it is subtle even with bounded operators

not extremely subtle, but subtle nonetheless

4:23 PM

@ACuriousMind I wish I could hear from the people downvoting @Kyle's answer in my meta question from a bit over a week ago, actually ( meta.physics.stackexchange.com/questions/7516/… )

@ACuriousMind how would you prove it (for bounded operators)? ;-)

For what it's worth, I've always placed more weight on well-reasoned arguments than on sheer vote count when it comes to meta posts. So the fact that the votes are balanced isn't necessarily a problem.

I think there are definitely more reasoned arguments in favor of keeping the question closed, at least last I checked.

My opinion the last time I read the meta post and question was that particular question was better off closed, but I did not subscribe strongly to a specific school of thought on why.

I'm still internally divided on the matter. I do think these are questions where good answers can be posted that can help people in learning and understanding physics (much like book recommendations), but I'm also unsure whether we would actually get the best answers in this community.

Should we try to refer more of those questions to philosophy? Is there a way to do that instead of flagging for moderator attention like with mathematics?

I guess the thing I take away from this is that nobody has made a strong argument for keeping it open, but several people have argued for keeping it closed. So on reflection, that suggests keeping it closed.

@Martin The system does allow us to set up a migration path to another site if we have enough successful migrations there, but I don't think that's necessary for philosophy. Honestly, in most cases, I think the best solution is to just close it as off topic and ask the OP to post on the other site.

4:32 PM

@DavidZ: Thanks. That sounds probably reasonable given the amount of questions.

Yeah, automatic migrations only work when our community already has a good sense of what is on topic at the other site, and I don't think that's the case for philosophy

There are a lot of questions we'd probably consider philosophical which the real philosophers wouldn't

That's true and I thought this was the reason why such an automatic migration didn't exist - but it can't hurt to ask in chat.

And I do agree that the part of my argument in favour of "keep open" is by far not strong enough to override especially your own argumentation.

@DavidZ I agree that the OP showed no great effort in his question; but how would you prove that given two bounded operators $A,B$ on $H$, $(\forall\phi\in H)\langle\phi, A\phi\rangle=\langle\phi, B\phi\rangle\Rightarrow A=B$?

@Martin I did get that sense from your answer (but again, it's nice to have)

@yuggib I would ask a mathematician :-P

:-D

4:38 PM

@yuggib Okay, thinking about it, it's not as simple as I intuitively assumed. For self-adjoint bounded operators, you get that the "expectation value" is a norm. Since you can uniquely reconstruct the scalar product from a norm obeying the parallelgram identity (and that's not trivial!), self-adjoint operators who agree in their expectation value are already the same. For general operator, split into self-adjoint and anti-self-adjoint parts.

Seriously though, off the top of my head, probably choosing a suitable basis and calculating matrix elements or something. But if you want a mathematically rigorous answer, or anything approaching it, you wouldn't get it from me.

At least in person, it is sometimes difficult for me to convince somebody a question is philosophy and not physics

@GPhys That's also the case here sometimes. People come to the site thinking a question is physics when it really isn't.

Not to mention, even physicists disagree about what constitutes philosophy.

I usually avoid these discussions

@ACuriousMind that's essentially one way, yes

4:40 PM

Wait...it's not a norm, is it? It can be zero or negative, after all.

Hmmmmm

it's a norm for strictly positive operators

Ah, okay, I have to split it also into positive, negative and null parts first

anyways there is a slightly easier way

Then do my argument for the positive and minus the negative part, and observe that the null parts straightforwardly agree.

the bounded operators are the dual of the trace class operators

(are isomorphic to)

4:44 PM

Ah, and on the trace class operators, $\mathrm{tr}(AB)$ is a scalar product, right?

therefore if $(\forall \rho\in \mathfrak{S}^1) \mathrm{Tr}(A\rho)=\mathrm{Tr}(B\rho)$, then $A=B$

now every trace class operator can be written as the sum of four positive trace class operators

What is $\mathfrak{S}^1$? Your notation for trace class operators?

yes, the so-called Schatten class

and each positive trace class operators $\rho_+$ can be decomposed as $\rho_+=\sum_{i\in\mathbb{N}}\lambda_i \lvert \phi_i\rangle\langle\phi_i\rvert$

so using the hypothesis and the definition of trace you get the result

@ACuriousMind (another notation is $\mathcal{I}_1$, because it is an ideal in the algebra of bounded operators)

OK, so I'll take care of that meta post shortly

what is free energy?

4:55 PM

@sharafzaman as in the Gibbs free energy, or other types?

Gibbs free energy

@yuggib I see

@DavidZ

Well, I'm not the best one to tell you about that, honestly

oh!!

4:59 PM

But have you looked at e.g. the Wikipedia article?

Oh, by the way, chat session time is over for now. Same time again in two weeks!

why?

Why do I ask if you've looked at Wikipedia? Because it's a good reference, and you should, if you haven't

i checked but i am still unclear

Well, what are you unclear about? That will help focus the answers you can get

5:36 PM

@yuggib I took one lesson from analysis today "sometimes you might have to do something"

New proposal for a site on "Theory of Everything" area51.stackexchange.com/proposals/95915/…

5:51 PM

Man, if I got a penny for every time someone makes a stupid dig at "ACuriousMind" being "closed-minded" because I disagree with them.

@HariPrasad And...what kind of experts is that supposed to attract? What would be in its scope that's not already in ours?

user116211

@sharafzaman: $G$ is thermodynamic potential which measures the entropy of the universe at constant temperature and pressure.

user116211

$$\mathrm dG= -T\,\mathrm dS_\textrm{universe}$$

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