The h Bar

user116211

09:00

as the wikipedia writes.

Slereah

Hm

Did not know that

user116211

So, Lanczos was a GR guy.

Slereah

Well I think he did a bunch of things

I seem to recall EM stuff from him

First guy to think about CTCs seem to be Weyl, btw

user116211

ohh.

Slereah

In the 20's he talked briefly about the notion that it might be possible for the metric to get twisted enough to allow it

it was just an idea though

He did not have a specific metric in mind

I have to say

The Anderson Institute might be a crazy crank

But very good page setting

I bet he is a powerpoint master

Bass

09:41

Hi guys, Nakahara (GTP p. 32) uses the following identity when talking about the path integral solution of the harmonic oscillator:

\begin{align}
\frac\delta{\delta x(t_1)}S & = \frac\delta{\delta x(t_1)}\int_{t_i}^{t_f}dt\,L\\& =\frac\delta{\delta x(t_1)}\int_{t_i}^{t_f}dt\left[\frac 12m\dot x(t)^2-\frac 12m\omega^2x(t)^2\right]\\& =-m\frac{d^2}{dt_1^2}x(t_1)-m\omega^2x(t_1)
\end{align}

How does he derive this? Note that $t_1$ is an arbitrary point in time (presumably $t_1\in[t_i, t_f]$). I tried it using some kind of Euler-Lagrange derivation, but I don't arrive at his identity:

\begin{align}
\frac\delta{\delta x(t_1)}S & = \int_{t_i}^{t_f}dt\left[\frac{\partial L}{\partial x(t_1)}+\frac{\partial L}{\partial\dot x(t')}\frac{\partial\dot x(t')}{\partial x(t_1)}\right]\\
& = \int_{t_i}^{t_f}dt\left[\frac{\partial L}{\partial x(t_1)}+\frac{d}{dt'}\left(\frac{\partial L}{\partial\dot x(t')}\frac{\partial x(t')}{\partial x(t_1)}\right)-\left(\frac{d}{dt'}\frac{\partial L}{\pa…

Slereah

The variation of the action is 0 at the boundaries

Bass

@Slereah But are we at the boundaries when we do $\frac{\delta S}{\delta x(t_1)}$? $t_1$ is any point in time, not $t_i$ or $t_f$.

Emilio Pisanty

@Qmechanic What's the deal with making this one CW?

2

A: Recent missed opportunities à la Freeman Dyson

There was an FQXi essay competition on this subject in Spring 2015: "Trick or Truth: the Mysterious Connection Between Physics and Mathematics" Here is the home page with links to winners and other entries: http://fqxi.org/community/essay/winners/2015.1 The competition is meant to encourage an ...

Slereah

09:57

@Bass The middle integral is between those, though

Bass

@Slereah Yes, and I expect $t_1\in[t_i,t_f]$. But how does the middle integral vanish?

John Rennie

10:35

Homework or conceptual?

0

Q: Pendulum force components at maximum displacement

If $\theta$ is the maximum angle for the pendulum's swing, then should $Tcos\theta=mg$ or should $T=mgcos\theta$. I think that the components must be done in the perpendicular directions $Tcos\theta=mg$ as the bob can't go any further. Can anybody please help me by telling me what would be the...

forces

Qmechanic

@EmilioPisanty : This policy was discussed most recently in this meta post, but if you think it should be non-CW, I can revert it. Btw. the current question formulation is not particularly clear unless you are familiar with the links already.

user116211

@JohnRennie It seems to be HW ;\

user116211

But it seems to me quite unclear....

Emilio Pisanty

10:55

@Qmechanic My bad. I thought you'd only made the answer CW and not the question, which would be weird if it wasn't actually completely incorrect.

Qmechanic

@EmilioPisanty : Ok.

John Jack

12:26

I have a simple question regarding spherical coordinates, why is the infinitesimal displacement of the polar coordinate given by $$dl_{\theta} = rd \theta,$$ is $rd\theta$ not the arc length of the sector rather than the magnitude of the vector in the $\theta$ direction?

Jim

12:37

@JohnJack the length differential has to have units of length. $d\theta$ ensures it is the unit magnitude, but whatever the radius is, it has to be multiplied by $r$ to properly represent the length in the theta direction

so yes, it's the differential arc length element

John Jack

@Jim Okay thanks.

Jim

@JohnRennie Brilliant. I'd be seriously surprised if most physicists didn't share that same reason for why they went into physics in the first place

0celo7

@Jim How many high schoolers know what "unified field theory" means

Jim

@0celo7 lots of them. It's a term commonly used in pop-sci. Any of them who have looked into physics beyond what school requires would probably have seen the term

0celo7

@Jim Hmm, well my graduating class (~600) had 0 math or physics majors so I'm probably a bad judge.

(yes, literally zero)

Jim

12:48

what about you?

did you not know the term in high school?

0celo7

@Jim I did, but after reading a few QFT books I decided physics was not for me.

Jim

Nevertheless, if you found the term, it stands to reason that other high school students could as well

0celo7

@Jim I found the term in a GR textbook that I was reading.

I've been told that's abnormal.

FrancescoS

Hi guys! I am trying to calculate the cross section of the process p p > jet jet via MadGraph. The cross section in higher than the usual one (of a factor 100). I cut the soft and collinear b-jets.. but the cross section remains high. What are other background processes I should consider to simulate the right crosssection?

Jim

in high school, I found the term in pop-sci

anyone who watches Cosmos has probably heard it too

0celo7

12:56

I've never experienced pop-sci I think it's midlessly boring

Jim

@0celo7 not to the uninitiated

0celo7

Hmm?

My first science book was Zee's GR text.

Jim

@0celo7 sounds like you jumped the gun a bit

0celo7

Perhaps.

ACuriousMind

@Secret There are no details to iron out. What you wrote there is $5\mathbb{Z}$, not $\mathbb{Z}/5\mathbb{Z}$, and is not finite. The only finite subgroup of $\mathbb{Z}$ is the trivial one consisting of only the identity.

@Danu I might be occasionally bored but not that bored :P

Jim

13:02

@ACuriousMind hey! My identity is not trivial!

0celo7

@ACuriousMind Went to the German place again, got a Schnitzel

spoke in German to the owner, impressed Rebecca

But I think the jig is up

I got the Artikel wrong on Schnitzel

No real German would do that

Jim

you used "der"?

0celo7

As soon as I said it I knew I fucked up

@Jim Yah

ACuriousMind

@Jim The important thing is that you believe that.

Jim

amateur

0celo7

13:03

@Jim :( I lived there for 7 yrs

The owner of this restaurant thinks I'm German

Jim

@ACuriousMind That's true. As the One True Jim, what I believe must be fact and/or law

0celo7

@ACuriousMind But the Schnitzel was not good. Too bready and greasy

Not very authentic -- but I'll keep going back for the Döners

ACuriousMind

@0celo7 I actually don't like Schnitzel that much anyway

0celo7

@ACuriousMind Not veal or pork?

ACuriousMind

@0celo7 Hmm? I don't find the breading all that tasty, I prefer meat without it.

0celo7

13:07

@ACuriousMind So you like Rahmschnitzel?

Bass

@ACuriousMind do you have a hint on this?

0celo7

My favorite traditional German dish

ACuriousMind

@0celo7 Ah, yes. I thought by "Schnitzel" you referred to the breaded variants.

0celo7

@ACuriousMind Yes, I was referring to the plain yellow breaded variant, specifically.

So you don't like Jäger?

ACuriousMind

Nope. That's also because I'm not a fan of most mushrooms, either

0celo7

13:10

there's peanut butter on GP :oooooo

@ACuriousMind ...how do you even function as a human being

ACuriousMind

@Bass Don't use that weird "chain rule", I'm not even sure that makes sense. Just use that $\frac{\delta}{\delta x(t_1)}$ and $\frac{\mathrm{d}}{\mathrm{d}t}$ commute - the variation of the derivative is the derivative of the variation, $\mathrm{d}_t \delta q = \delta \mathrm{d}_t q$ in physicists' notation should look familiar to you from the derivation of the Euler-Lagrange equations.

@0celo7 Barely

Then, to get the squared time derivative, integrate by parts appropriately before actually carrying out the differentiation.

0celo7

@ACuriousMind What's something only a German would know?

i.e. you have to have grown up in Germany to have reasonably heard about it

ACuriousMind

@0celo7 I will not reveal the secrets of my fatherland to you!

0celo7

@ACuriousMind my fatherland too

I have the birth certificate and everything

And since when are you a nationalist

@ACuriousMind So, how do humans factor into the EFE? Is free will free or not?

ACuriousMind

@0celo7 Pffff, I don't care for worthless paper!

@0celo7 (I'm joking because I have no idea how to answer your question :P)

@0celo7 ...what?

Define "free will".

0celo7

13:24

It would seem a human could cause the development of the initial data to "change"

Thus it would be nonunique

ACuriousMind

@0celo7 Why would it seem so?

Bass

@ACuriousMind Gonna try it that way, thx.

Slereah

humans are themselves deterministic classically, tho

0celo7

@ACuriousMind because unless all human actions are determined by the initial data, they could alter things

Slereah

They are

Sorry Thomas of Aquinas

Bass

13:28

To derive the EL equations, you use the chain rule too, like $\delta S=\int dt\left(\frac{\partial L}{\partial x}\delta x+\frac{\partial L}{\partial\dot x}\delta\dot x\right)$, or is there another way?

0celo7

And also how do the EFE account for probabilistic processes like radiation

Slereah

You don't need the chain rule

Also radiation isn't probabilistic?

or do you mean quantum radiation

ACuriousMind

@0celo7 They don't because they are not quantum?

Slereah

Just use the definition of $\delta$

ACuriousMind

@0celo7 Humans just belong to the matter content of the universe. Are you saying that some of the matter in the universe doesn't obey the equations of motion because it is organized into conglomerates we call "human"?

0celo7

13:31

@ACuriousMind Exactly.

Bass

@Slereah You mean some variation $x\mapsto x+\epsilon y$? But then again, to get the variation of $S$ under that variation, you need the chain rule, because $L$ depends on both $x$ and $\dot x$, no?

Slereah

$$\frac{\delta S[x(t)]}{\delta x(t')} = \lim_{\varepsilon \to 0} \frac{1}{\varepsilon} (S[x(t) + \varepsilon \delta(t-t')] - S[x(t)])$$

All the proper rules should be obvious enough from this

Just do a first order expansion of the first term and integrate it

ACuriousMind

@0celo7 If you postulate that, then it is of course true. What is the question?

0celo7

@ACuriousMind Why are you being like this?

ACuriousMind

I was born this way?

I'm not sure what you mean

Bass

13:35

@Slereah Hmm, never saw it that way. Is that a Dirac delta function, and an "infinitesimal" $\epsilon$ there?

0celo7

@ACuriousMind YOURE LADY GAGA NOT ALICIA KEYS

ACuriousMind

lol

I am finally unmasked!

0celo7

Makes more sense with your Johnson and all

@ACuriousMind I mean why do you always try to make me sound silly when I have an epistemological crisis

ACuriousMind

@0celo7 Because most epistemological crises are caused by silly things

Jim

Today I was able to used the sentence "One cannot simply explode a singularity into Mordor" naturally in a conversation. Quantum physics was right. Anything not forbidden must occur

ACuriousMind

13:43

And most of these issues go away when you actually try to formulate them carefully and consistently.

I apologize if you have the impression I am mocking you - I am not, my "making you seem silly" is intended to get you to phrase more carefully.

Jim

@ACuriousMind you can make me seem silly any time you want. <3

ACuriousMind

@Jim Careful, I might take you up on that :)

Jim

@ACuriousMind Not if I take myself up on it first

0celo7

14:09

@Jim @ACuriousMind get a room.

Jim

@0celo7 we're already in a chat room

Balarka Sen

14:44

I don't understand the Schroedinger equation.

Well, the time-dependent one.

ACuriousMind

@BalarkaSen Can you be more precise?

Balarka Sen

Of course the time independent one comes out from the time dependent one by separation of variables, but I get $E = i\hbar 1/\phi \partial \phi/\partial t$ along the way where $\phi$ is the time-dependent part of the wavefunction. What does that, physically, mean?

On a related note, instead of going from the time dependent equation to the time independent equation, how can I go from the time independent (the one I understand better) equation to the time dependent equation?

I feel an answer to the first question will give an answer to the second.

ACuriousMind

The "time-independent Schrödinger equation" is just a fancy name for an eigenvalue equation.

Namely, $H\psi = E\psi$, where $H$ is the Hamiltonian operator and $E$ the energy eigenvalue.

Balarka Sen

Sure, it says eigenfunction of $\hat{H}$ is $\psi$ and the correspondng eigenvalue is the total energy.

0celo7

@ACuriousMind Does "smallest open set of an uncountable collection of open sets" make sense

Probably not...

ACuriousMind

14:50

The time-dependent Schrödinger equation is the statement "The Hamiltonian is the generator of time translations"

Balarka Sen

"Smallest" in what sense?

ACuriousMind

^

Balarka Sen

@ACuriousMind Uh.

Not sure I understand what you mean by that.

ACuriousMind

Well, it says $\mathrm{i}\partial_t \psi = H\psi$, right?

0celo7

@BalarkaSen I have a set of the form $K\times[0,\epsilon)$ where $K$ is compact and I know that my function $\sigma$ is positive on some open $U\supset K$

Balarka Sen

14:52

@ACuriousMind Right.

(by $\partial_t$ I suppose you mean $i\hbar \partial/\partial t$)

0celo7

I need to be able to find an open $O$ such that $\sigma$ is positive on $O\times\{t\}$ for any $t$

ACuriousMind

@BalarkaSen Sorry, forgot the $\mathrm{i}$, and I just ignore the $\hbar$

Balarka Sen

Gotcha.

0celo7

$\sigma$ is positive on $K\times[0,\epsilon)$

$\sigma$ is smooth

ACuriousMind

A formal solution of this equation for initial condition $\psi_0$ is given by $\psi(t) = \exp(-\mathrm{i}H t)\psi_0$

0celo7

14:55

Now I'm pretty sure I can find an open $V_t\supset K$ such that $\sigma(.,t)$ is positive on $V_t\times\{t\}$

ACuriousMind

Where now $\exp(-\mathrm{i}Ht)$ acts as the operator that shifts the initial condition $\psi_0$ by $t$ in time.

Sebgr

Hey guys

0celo7

So I was thinking I should take $O$ to be $\bigcap V_t$ but that's not necessarily open.

Balarka Sen

Yes, I am not sure why or how that comes up. That was my original question. You need to solve $E = i\hbar \cdot 1/\phi \cdot \partial \phi/\partial t$, the geometric/physical significance of which I am unclear on.

ACuriousMind

@BalarkaSen I'm not sure what you mean by significance. It's the differential equation you get when you make the separation of variables ansatz. You solve it by using $\frac{1}{\phi}\frac{\partial \phi}{\partial t} = \frac{\partial \ln(\phi)}{\partial t}$ and then just integrating to find $\ln(\phi) = -\mathrm{i}\hbar t$.

0celo7

14:58

What separation of variables are you talking about

ACuriousMind

@0celo7 Write $\psi(x,t) = \psi_0(x)\phi(t)$.

0celo7

@ACuriousMind Oh, of course.

Now can someone pls help with my topology problem :<

Balarka Sen

I think of the wavefunction (well, it's modulus squared) as representing probability density of my particle. Why should the derivative of it's time-dependent part $\phi(t)$ have anything to do with the total energy of the particle?

I know the math; I just have no intuition for it.

Sebgr

@BalarkaSen You might want to check Ch. 3 of Ballentine's book

ACuriousMind

@BalarkaSen Oh, to understand that, you must first understand how energy and time are connected in classical physics: In the Lagrangian formalism, energy is the conserved quantity associated to time translation symmetry. It is the Hamiltonian formalism which we quantize, and in the Hamiltonian formalism, the energy operator is called the Hamiltonian and generates time translations in the sense that $\partial_t f = \{H,f\}$ for any function $f$, where $\{-,-\}$ is the Poisson bracket.

Balarka Sen

15:05

@Sebgr I'll have a look, thanks.

ACuriousMind

"naive"/canonical quantum mechanics essentially just takes the Poisson bracket and makes it into a commutator of non-commuting operators.

Balarka Sen

@ACuriousMind Hmm, alright.

ACuriousMind

In essence, that energy is related to the change of an object in time is tautological because the only way to abstractly define the energy is by declaring it to be the conserved quantity associated to time translation by Noether's theorem

And the $\partial_t\psi = -\mathrm{i}H\psi \iff \psi(t) = \exp(-\mathrm{i}Ht)\psi_0$ thing wonderfully generalizes to all sorts of other transformations/operators.

0celo7

@ACuriousMind He might understand Poisson brackets better in terms of symplectic topology :P

Balarka Sen

@ACuriousMind That sounds technical :\ Thanks though.

0celo7

15:10

@ACuriousMind Uhh, not in GR.

Sebgr

@BalarkaSen I think the idea (you can check the details in that book) is that you can create a unitary operator for time displacements ($U(t)=\exp(iHt)$) that acts on a state vector: $U(t)|\psi(t_0)\rangle=|\psi(t_0+t\rangle$, so infinitesimally you get Schrödinguer's eq. State vectors "rotate" in various ways in a Hilbert space thanks to unitary transformations

ACuriousMind

@BalarkaSen I firmly believe that quantum mechanics can only be understood through understanding its technical aspects. All "intuition" sooner or later has to face that our intuition is trained on classical scales where quantum effects are largely absent, hence we don't understand it intuitively.

The best "intuition" for quantum mechanics, to me, is firmly understanding classical Hamiltonian mechanics first.

Balarka Sen

Fair enough.

Sebgr

I have a question about "study" advice. I want to know more about particle physics, but I don't really like all the taxonomy, so I was thinking of skipping the Griffiths+PDG booklet road and going for a QFT approach. More theoretical, yes, but does one really have to memorize quark compositions and esoteric rules before learning the theory behind?

ACuriousMind

@Sebgr Not at all. Go for the theory if you don't want to bother memorizing the particle zoo.

0celo7

15:16

@BalarkaSen link.springer.com/book/10.1007/978-1-4757-2063-1

Sebgr

@ACuriousMind Does the "zoo" get more familiar once you tackle it with group theory + qft?

Balarka Sen

Thanks, @0celo7. Arnold is a great guy, I'll have a look at that book.

ACuriousMind

@Sebgr Somewhat. Sadly, many of the rules are a bit tricky to extract from the theory.

Balarka Sen

However one of the reasons I am not looking at a mathematically flavored book in QM is because I thought I wouldn't find the physical intuition in it.

ACuriousMind

@BalarkaSen Arnold isn't about QM, 0celo7 linked that because it's about classical Hamiltonian mechanics, but I don't think this level is necessary.

Balarka Sen

15:19

Ah, I see.

Sebgr

@BalarkaSen I strongly recommend Ballentine. He also wrote a nice review in RMP and he is very "responsible" and a good start for a mathematically trained reader

David Z

@Sebgr in fact one can learn QFT and the standard model without knowing anything whatsoever about what the real particles are called or what their quark content is

0celo7

Why would a book titled "classical mechanics" be about quantum mechanics

Balarka Sen

Because I misread. :P

vzn

@BalarkaSen it would seem you are asking about "interpretations" which is nearly a 4 letter word around here and a many-decades long perplexity of QM that has haunted its top minds incl maybe most notably (co)founder einstein, but there are some new povs on QM, pursued by a dogged minority, that seem to be slowly bearing fruit. much more in my blog, but not for the faint of heart :|

Sebgr

15:21

@DavidZ @ACuriousMind I have the impression that the typical "Introduction to Particle Physics" course is only useful for a very... dedicated would-be experimentalist. So I was thinking of skipping that class and concentrating in learning more QFT before sinking in a sea of "rules"

David Z

@Sebgr that doesn't sound like any "introduction to particle physics" class I've ever heard of

vzn

@BalarkaSen "physical intution" ~= "interpretations". re feynmans famous quote about QM. "I think its safe to say nobody understands QM..." (other similar famous quotes on my quotes pg)

0celo7

Topology!

ACuriousMind

@BalarkaSen I've been doing quantum mechanics and QFT for quite a while now and I don't think "physical intuition" is all that useful for it. What's necessary is to firmly grasp its mathematical structure - states as rays in Hilbert spaces, observables as Hermitian operators, taking expectation values, and so on - and to be comfortable in manipulating these to get the physical predictions. "Intuition" has rarely helped me in that.

David Z

I might be late to the chat session. Someone else should start it if I'm not around at the time. We can have a little update on the replacement for the homework policy but other than that I don't know of anything on the agenda for today.

ACuriousMind

15:24

In fact, I consider the "intuitive" exposition of many introductory QM resources that focus on "waves" and "wavefunctions" instead of the a bit more abstract picture of Hilbert spaces to be detrimental to actually understanding the structure of QM

vzn

@ACuriousMind the copenhagen interpretation is nearly "anti-intuition"... :(

0celo7

@BalarkaSen I have a set $K\times[0,\epsilon)\subset \Bbb R^k\times[0,\infty)$ where $K$ is compact. I know that the function $\sigma:\Bbb R^k\times[0,\infty)\to \Bbb R$ is positive on $K\times[0,\epsilon)$.

I need to find an open set $O\subset\Bbb R^k$ with $K\subset O$ such that $\sigma(.,t)$ is positive on $O\times\{t\}$

How do I do this?

vzn

@ACuriousMind nevertheless that is something of how it was discovered by its founders wrt the math/ povs of bohr, schroedinger, debroglie. waves etc

Balarka Sen

@ACuriousMind Nod. I am not really into physics - I like doing mathematics, and only recently started to take occasional peeks at this stuff because it intrigues me. So am naturally interested in what the formulas actually mean, because that's how I think in mathematics.

ACuriousMind

@vzn So what? We don't teach mechanics the way Newton introduced it in his Principia either.

vzn

15:26

@BalarkaSen lol strange comment from a mathematician. maybe you should read some hilbert :P

0celo7

@ACuriousMind We don't???

Sebgr

@DavidZ I might be judging from ignorance. I just wanted to see if its viable to tackle the SM once you have a good QFT background. Does the average theorist knows a lot of phenomenology by "exposure" rather than theory?

ACuriousMind

@0celo7 Have you tried to read the Principia?

It's gibberish to modern minds :P

0celo7

@ACuriousMind I do not speak Latin.

vzn

@ACuriousMind (as kuhn liked to point out) sometimes textbooks are rewritten =)

ACuriousMind

15:27

@0celo7 I mean, grab a translation. The language is not so much the problem as the frigging lack of modern calculus.

0celo7

@ACuriousMind Your calculus is my advanced analysis

Calculus is $\sim \sum f\Delta x$

Balarka Sen

@vzn To me intuition is an essential part of doing mathematics, and I have met actual mathematicians: I find that most of them think so. So I would not say it's strange.

Sebgr

I think Feynman's phrase about "understanding" QM is used too easily. QM is remarkably well understood.

0celo7

Limit is defined as "getting close"

Sebgr

It can be misleading to a person who is starting to read QM. We're not waving arms at all

ACuriousMind

15:28

@0celo7 I'm not talking about rigor

Bass

@ACuriousMind Isn't the part "be comfortable in manipulating these to get the physical predictions" some sort of physical intuition? You could be the expert on Hilbert spaces and self-adjoint operators, but if you lack the physical intuition, it's just abstract mathematics.

0celo7

@ACuriousMind Can you please even tell me if such an open set exists or if I'm on a wild goose chase

Balarka Sen

@0celo7 You mean $\sigma(., t)$ is positive on $O \times t$ for all $t < \epsilon'$ for some $\epsilon'$?

0celo7

@BalarkaSen No, I mean for fixed $t$ it is positive on the "slice" $K\times\{t\}$

ACuriousMind

@Bass I think you could say so, yes :)

0celo7

15:29

So there should be a relatively open set containing $K$ such that it is positive on it.

wait

Balarka Sen

By continuity.

@0celo7 If $\sigma$ is positive on a closed set, it's positive on a nbhd near it, no?

0celo7

I know that it is positive on $K\times\{t\}$

Balarka Sen

Uh-huh.

vzn

@BalarkaSen fully sympathy (it is maybe near my own leanings, and just playing devils advocate right now), but "what the formulas actually mean" is not nec the same as intuition. there can be some conflict on that in math... which reminds me, have you seen the latest ramanujan movie?

0celo7

@BalarkaSen No shit.

I'm asking how to find $O$

@BalarkaSen Yes

vzn

15:30

@Sebgr maybe there is something as yet undreamed of in your philosphy horatio

Terry Bollinger

@ACuriousMind If classical is emergent from quantum, a sort of breaking of quantum symmetries, wouldn't firmly understanding the very classical Hamiltonian mechanics first actually create a risk of throwing all of your intuitions off? Just a thought.

ACuriousMind

@Sebgr Well, the "basic" phenomenology you just pick up along the way. And any specifics to just learn when you need them

Balarka Sen

@vzn That's fair, my intuition comes from geometry. There are some algebra people who can think formulas. I don't understand how people do that.

But they do, I admit.

vzn

@BalarkaSen maybe intuition is something very multifaceted like a diamond, and something that never stops, and has many povs. maybe it is like the blind men and the elephant.

Terry Bollinger

@BalarkaSen geometry provides important insights, but formulas derived from such insights provide additional insights not available through geometrical thinking alone.

ACuriousMind

15:33

@TerryBollinger I'm not talking about such "ontological" intuition. My point is that classical Hamiltonian mechanics and quantum mechanics share in their formalism many features that need only little modification to be transferred from one to the other - like generation of symmetries by charges, or the commutator/Poisson bracket of observables.

Balarka Sen

@0celo7 $\sigma$ is $> 0$ on $K \times [0, \delta]$ for some $\delta < \epsilon$. That's closed. So there is an nbhd $U$ around $K \times [0, \delta]$ on which $\sigma$ is positive. Can't I set $O = U$?

Terry Bollinger

@vzn our minds were designed for surviving in this very physical world. We didn't start rotating equations, we started figuring out where food and safety are, and rotations and such help. An interesting cross-borrowing of organic machinery, that.

Balarka Sen

Ah, $O$ is an nbhd of $K$ inside R^k. Sorry.

Hmm, so I need to modify that.

vzn

@TerryBollinger lol rotations help you find food and safety? :P

0celo7

@BalarkaSen No no, I need $O$ to be in $\Bbb R^k$!

ACuriousMind

15:35

@BalarkaSen I'm afraid most of basic QM is rather devoid of geometry. :/ All the geometry only enters later

Balarka Sen

Right, I noted. But I think I can modify that, give me a minute.

Sebgr

@vzn It should be obvious to both of us that "undreamed possibilities" is not the way QM as a theory works.

Terry Bollinger

@vzn sure! Many of our "simple" mental translations help us imagine what might be on the other side of a tree, a rock, a possible source of food. The ability to make such predictions in our heads, to model the external world, is unbelievably valuable for survival.

vzn

@Sebgr "obvious" to 1 of us maybe :P

Sebgr

@vzn Ha ha, I just mean to say that I've read that phrase in all kinds of ideological positions. It doesn't mean much by itself

vzn

15:37

@Sebgr agree QM as it has been formulated is apparently quite "airtight" as far as consistency. personally have long thought/ suspected along with many greats it is "incomplete"

Balarka Sen

So I got an nbhd $U$ of $K \times [0, \delta]$ in $\Bbb R^k \times \Bbb R^+$ on which $\sigma$ is positive. I need to see if there is some open subset of the form $O \times I$ sitting inside $U$.

vzn

@Sebgr hmmm, shakespeare seems quite meaningful to me

ACuriousMind

@vzn And yet you consistently fail to make in any way precise what you mean by that.

Balarka Sen

It seems compactness of $K$ should definitely allow me to prove that any nbhd of $K \times [0, \delta]$ contains something of the form $V \times I$.

vzn

@ACuriousMind "time will tell" anyway there are no real/ serious "takers" on my implicit offer so far :P

ACuriousMind

15:39

For such a long period of thought you sure have little to say. I am incredibly tired of your repeated baseless assertions.

Sebgr

@vzn Don't tell anyone, but I secretly wish there was a great discovery or change of paradigm in my lifetime. Let's keep studying meanwhile.

vzn

@ACuriousMind :(

Balarka Sen

I feel like I should try to get the tube lemma to work.

vzn

@Sebgr yep, and maybe glimpses/ flickers of it are already here, right under our noses =)

Terry Bollinger

@ACuriousMind no disagreements! But everything you just said assumes a nicely classical level of information-rich navigation of a very classical mathematical world. Sometimes the maths need to go deeper, to see what you can be created starting from even simpler parts. Information is very, very hard to discard as a "given", even though the quantum world clearly doesn't use it like we do. Information is arguably the main differentiator between quantum physics and classical, I'd say (strongly).

0celo7

15:40

@BalarkaSen I used the Tube Lemma to get $K\times[0,\epsilon)$.

Because what I originally had is that $\sigma$ is positive on $K\times\{0\}$

Sebgr

@vzn I read that to The X Files theme song

0celo7

So there's an open set $U$ of $\Bbb R^k\times\Bbb R^+$ that contains that

vzn

@Sebgr lol so maybe theres hope for you yet =)

0celo7

and by the Tube Lemma $U$ contains $K\times[0,\epsilon)$

Balarka Sen

Right.

0celo7

15:41

(well, really $K\times[0,\epsilon]$ but I want it open on that side for reasons

ACuriousMind

@TerryBollinger Information is indeed crucial: The classical world has hereditarily pure states (every subsystem of a system that is in a pure state is in a pure state) and the quantum world has not.

Terry Bollinger

Question: If QM is fully defined, how exactly does observation work? Is there a truly good math framework for that one yet?

Balarka Sen

@0celo7 Alright, the generalized tube lemma proves what you want.

ACuriousMind

@TerryBollinger Depends on what you want. There is a rather rigorous formulation of measurement along the lines of von Neumann by Ozawa et al.

Terry Bollinger

@ACuriousMind that to me is absolutely fascinating. I often informally describe quantum physics to non-physicists (heh including me!) as the physics in which the history of the history of the system has not yet been set. It's a suprisingly powerful, and more importantly accurate, way of getting at the critical difference.

@ACuriousMind thanks, the question was serious and I'll check it out. I'm having great fun with a particular definition of "observer" that even has some predictive power, and avoids invoking "consciousness" and all of that (to me at least) kind of silliness.

vzn

15:47

@TerryBollinger "consciousness"! (acc to wigner & others!) :P

David Z

Actually I may not be around for the chat session after all this time. I'll post when I'm able to be here again, but I think I will have to put my update in a meta post later.

Terry Bollinger

@vzn Feynman caught the idea best in the voice version of his tapes when he emphazed the critical thought that there is "no information anywhere in the universe" to describe whether a neutron is reflected or captured. Alas, the real emphasis got edited out of the print version. I think Feynman "got" the idea a bit more profoundly.

Must go, nice chatting!

vzn

@TerryBollinger feynman was something of an unintentional zen master of physics =) ... (as for "neutron" one could easily replace with "photon")

0celo7

@BalarkaSen Hmm?

Balarka Sen

So, we took some $\delta < \epsilon$, looked at $K \times [0, \delta]$ in $\Bbb R^k \times \Bbb R^+$, and found an nbhd $U$ around it on which $\sigma$ is positive, right?

The generalized tube lemma says, since $K$ and $[0, \delta]$ are both compact on $\Bbb R^k$ and $\Bbb R^+$ resp, that there are nbhds $O$ and $I$ around it such that $O \times I \subset U$.

ACuriousMind

15:51

@TerryBollinger See you!

Balarka Sen

So $\sigma$ is positive on $O \times I$, hence in particular $O \times \{t\}$ for any $t \in I$, in particular any $t \in [0, \delta]$.

ACuriousMind

@TerryBollinger One introduction is this paper. However, be aware that something like the Born rule is still invoked as an axiom, so this explains how measurement apparati work when they, too, are described quantumly. That may or may nor be what you're looking for.

0celo7

@BalarkaSen I've never heard of that

Hmm

Sir Cumference

Anyone know the answer to this?

1

Q: Are white dwarves supported by proton degeneracy as well?

In general, fermions form a degenerate gas under high density or extremely low temperature. It's clear that white dwarves are supported by electron degeneracy pressure. However, there are still a significant number of protons in a white dwarf. Under those high densities, do the protons form a deg...

gravity white-dwarf density degenerate-matter quantum-mechanics

I got one answer but I'm not sure if I trust the answerer

Transcript for