The h Bar

0celo7

1:00 PM

@BernardoMeurer Balarka is ironically edgy

Ironically?

Indeed.

hello

can I use superposition rules for linear circuits with AC sources of different frequencies?

0celo7

1:18 PM

@heather so do you understand convergencece of sequences?

heather

@0celo7, I think so, yes.

0celo7

@heather Ok, can you compute $$\lim_{n\to\infty }\frac{1}{n}$$ and give a reasonable proof?

heather

I believe that would approach 0

now, as for proving it, hmm.

Slereah

HINT : it is 0

though you might need

The archimedean property~

0celo7

@heather Hint: $N=\lceil\epsilon^{-1}\rceil+1$.

damn

how do you do the ceiling?

heather

1:27 PM

try detexify

skullpetrol

^

0celo7

i already got it people

heather

i'm reading about how to do these proofs, and there seem to be a lot of mentions of the $\epsilon...\delta$ style of proof

is that what I should be using?

0celo7

no delta here

heather

okay

we know that $\epsilon$ must be greater than 0

by definition

okay, I really don't know how to prove this

everything I'm reading is for epsilon-delta proofs.

0celo7

1:41 PM

Let $\epsilon>0$ and $N=\lceil \epsilon^{-1}\rceil+1$. Then for all $n\ge N$, $$\left|\frac{1}{n}-0\right|=\left|\frac{1}{n}\right|=\frac{1}{n}\le\frac{1}{N}=‌\frac{1}{\lceil \epsilon^{-1}\rceil +1}<\frac{1}{\epsilon^{-1}}=\epsilon.$$

ACuriousMind

@heather Don't approach proofs by asking what technique you should be using. There's usually more than one way to prove something, and it does you no good to get used to someone telling you what to use. Start by writing down the definition of what you want to prove (in this case, convergence of that sequence), and then look at that and try to see how you might arrive at it.

(I don't expect you to come up with it if you've never seen a proof of convergence before, but any book that introduces convergence should show some examples before asking you to do it yourself)

skullpetrol

@heather proof in algebra is different than proof in geometry

0celo7

@skullpetrol if you're calling this algebra you will get banned

skullpetrol

::self-ban::

heather

what do those "ceiling" brackets mean?

ACuriousMind

1:47 PM

@heather Ceiling function

"round up", in less pretentious terms ;)

heather

ah. thanks

skullpetrol

There's also the "floor."

For completeness. :-)

heather

rounding down?

skullpetrol

Bingo.

"In less pretentious terms"

But no "walls" :P

Alex

In Sakurai he defines the operation $J$ which is the infinitesimal translation by $d \vec{x}'$ as $$\hat{J}(d \vec{x})| \vec{x}' \rangle = | \vec{x}' + d \vec{x}' \rangle$$
On an arbitrary state ket $| \alpha \rangle$ he then writes that $$\hat{J}(d \vec{x}')| \alpha \rangle = \hat{J}(d \vec{x}') \int d^3 x' | \vec{x}' \rangle \langle \vec{x}' | \alpha \rangle = \int d^3 x' | \vec{x}+ d \vec{x}' \rangle \langle \vec{x}'| \alpha \rangle.$$
How do we know that we can take the function $\hat{J}(d \vec{x})$ into the integral in the second equation?

If it was a countable sum we could use the linearity of $\hat{J}$ but since not how do we know we can take it into the integral?

AccidentalFourierTransform

2:06 PM

integrals are also linear, right?

Alex

@AccidentalFourierTransform Yeah, but how does that clear it up?

AccidentalFourierTransform

an integral is just a special case of a sum

0celo7

@Alex Don't try to use any kind of rigor here

AccidentalFourierTransform

also that ^

0celo7

@Alex And even if it's a countable sum, you'd need to know that the resulting sum from moving it inside converges too, which it might not

Sakurai is mostly heuristic, nothing he does is rigorous

@AccidentalFourierTransform :P

AccidentalFourierTransform

2:13 PM

I dont like your new avatar

0celo7

I never liked yours.

Alex

@Ocelo7 Okay, tricky studying a book without knowing when to be rigorous. A lot of people recommended this book as well...

0celo7

@Alex What physics is rigorous?

ACuriousMind

@Alex You never "know" when to be rigorous in physics unless you read expressly mathematical literature. Physics might use mathematics but simply rarely has the same standards of rigor.

0celo7

I'm assuming you want to derive $J(\mathbf a)=\exp(-\frac{i}{\hbar} \mathbf a\cdot\mathbf p)$?

The actually rigorous proof of that requires Stone's theorem. You probably don't know it if you're reading QM for the first time.

@ACuriousMind We're getting 10 inches of snow tonight .-.

Alex

2:18 PM

@Ocelo7 No I am just working through chapter one of Sakurai at the moment, haven't reached either of what you mentioned. I previously studied maths but am now doing physics.

0celo7

@Alex Unless you know Stone's theorem already, you won't learn it from Sakurai.

AccidentalFourierTransform

Kellyanne Conway suggests Barack Obama could have spied on Donald Trump through a microwave

2

hmmm

0celo7

Or any "physics" book. Quantum mechanics is a mess if you learn it from physicists

skullpetrol

lol @AccidentalFourierTransform

Slereah

@0celo7 does your QM book ever actually do QM?

Alex

2:19 PM

@Ocelo7 That's nice, wish I was getting snow...Drought here. Yeah I will eventually try to get into the mathematical physics part of QM, but for now I just need to learn some QM.

Slereah

Like do you ever solve the Schrodinger equation for a system

skullpetrol

How's the tonsils?

Slereah

I'm mostly fine now

0celo7

@Slereah What QM book?

Slereah

Just two days of antibiotics left

0celo7

2:20 PM

I own Sakurai

Slereah

@0celo7 the math one

0celo7

But I think Sakurai is a nice anecdote, as with all of physics

I take it with a grain of salt

Slereah

on what page does it for the first time say "Here's a solution to the Schrodinger equation"

0celo7

@Alex That's a zero in my name, not an O.

Slereah

Is it like O'neill where the EFE is 350 pages in

AccidentalFourierTransform

2:21 PM

0'neill

0celo7

Like 80 pages in.

Slereah

not too bad I guess

0celo7

The first two chapters are classical mechanics and some history

Slereah

ugh

I wish people stopped doing that

the first chapter of every physics book is a little walk through history

ACuriousMind

I'll take a brief history overview over a book retracing the historical steps throughout the entire book every day :P

Slereah

2:23 PM

I'd rather just skip it

0celo7

It's only 16 pages, and I did skip it

@Alex Do you want me to give a fully rigorous proof on your question?

skullpetrol

Skip nothing :P

Slereah

I've seen enough introductions to special relativity/classical mechanics/basic quantum mechanics to last me a lifetime

ACuriousMind

@Slereah It was interesting the first time you read it, right? It's just that you've read it so often by now that you can't stand it anymore

Slereah

yeah

Weinberg's introduction is pretty nice, actually

ACuriousMind

2:24 PM

You have to consider that these books are generally not written on the assumption that you already have read a dozen similar ones

Slereah

It goes over historical things not often brought up

0celo7

I just don't believe that the first time someone would see QM is via some functional analysis GTM

the author is kidding himself

Slereah

You know the worst offender?

Rovelli.

The first 6 chapters are reminders of various theories

ACuriousMind

@0celo7 Some mathematicians would

Slereah

He powers through GR, QM, QFT, classical mechanics and ADM formalism before actually going over the real topic

skullpetrol

2:26 PM

Mathematicians are rigours.

Slereah

It's a pretty intense ride

skullpetrol

By definition.

0celo7

@ACuriousMind I just don't buy that. Sakurai is pretty common for all learned people.

Slereah

I've never read Sakurai

My thesis advisor was a big russian beard man so he made me get Landau instead

ACuriousMind

@0celo7 What? Most mathematicians I know don't have the first clue about QM

skullpetrol

2:27 PM

Has penrose arrived? @Slereah

Alex

@0celo7 Yes thank you.

Slereah

It's not even shipped yet!

skullpetrol

WTF

Slereah

yeah it's on backorder

skullpetrol

Huge demand for that guy.

0celo7

2:29 PM

@Alex Ok I have to go somewhere, so I'll do it later. Do you know what a strongly continuous group action is?

Slereah

We all love Penrose

0celo7

@ACuriousMind Uh, 100% of mathematicians I know own Sakurai lol

Slereah

Penrose is pretty poorly done

skullpetrol

and to think he failed grade 5

Slereah

It's really just a big list of theorems

0celo7

2:30 PM

By "know" I mean "have talked to more than once"

Slereah

but it does the job

0celo7

Then again, I only know analysts.

@skullpetrol what a loser

Alex

@0celo7 Surely don't. I posted it as a question before everyone told me to not be rigorous. You can answer question, I will look up anything I don't know...

skullpetrol

Stop bullying me @0celo7

Qmechanic

-1

Q: Is the consideration of spacetime as a smooth manifold only an assumption?

General relativity (and already the two postulates of special relativity) seems to refer to timelike and spacelike worldlines of particles and fields. It seems to be impossible to apply special and general relativity to vacuum points of spacetime. One reason is the fact that the velocity-dependan...

Duplicate?

skullpetrol

2:32 PM

Yes.

0celo7

2:51 PM

@ACuriousMind Do you know if the formula for the translation operator holds on the test functions or on all of the domain of P? I guess that exponential wouldn't be defined on that.

ACuriousMind

@0celo7 I think the exponentials that come from Stone's theorem are defined on the entire Hilbert space.

Slereah

arxiv.org/abs/gr-qc/0411143

Whole paper on time functions

oh boy

another spiel I'm sick of is defining a spacetime

Secret

@slereah If I recall correctly, it is well known that there are many possible solutions for the cauchy problem in a CTC as seen from outside due to the cauchy horizon preventing any outside observer from seeing into the CTC region. But what about a CTC viewed from within, how does spacetime look like to someone or something that is within it. Is there some sharp notion of boundaries such as time just terminates and then loops back. How does light propagate within a CTC as seen from within a CTC?

Slereah

Locally spacetime is always causal

being inside a CTC doesn't look much different

Bernardo Meurer

@BalarkaSen Dude, Bowie is a classic!

@0celo7 He just likes good music

Secret

3:05 PM

So it is impossible to determine whether one is within a CTC by using telescopes to look far enough in space hence back enough in time?

Slereah

well there are non-local experiments you can perform

0celo7

@ACuriousMind They are. They're bounded and densely defined, so the BLT theorem extends them uniquely.

But you can't "expand in a power series" like physicists do

I wonder if that's just a happy coincidence that it works

Slereah

Hm, BLT theorem

0celo7

Bounded linear transformation

Secret

What nonlocal experiments one can perform to test whether the whole universe is a CTC. For example, if the universe is one giant CTC, how will the CMB differs?

AccidentalFourierTransform

3:22 PM

you'd see Zarathustra's face in the fluctations of the CMB

ACuriousMind

0

Q: Prove 3 + 5 = 8 with using Quantum Fourier Transform

I am taking Quantum Computing lesson in this term in my School. Our teacher told about Quantum Fourier Transform and its implementation. Then, He did want us to show 3+5 equals to 5 with Quantum Fourier Transform. I thougnt maybe i can use binary notation for this problem : |1000> = |0101> + |...

fourier-transform quantum-computer

...what?

@AccidentalFourierTransform Related

AccidentalFourierTransform

@ACuriousMind yeah, I read that one yesterday, from Davids post x)

Alex

@AccidentalFourierTransform Just to confirm is $d \vec{x}$ taken as am infinitesimal translation in the direction of vector $\vec{x}'$ is or is it just any infinitesimal vector?

AccidentalFourierTransform

it is an arbitrary infinitesimal vector

one does not assume that $\mathrm d\vec x\propto \vec x$

Alex

@AccidentalFourierTransform Okay understood.

0celo7

3:31 PM

@ACuriousMind what's the domain of an operator obtained from the functional calculus?

Slereah

3:42 PM

How many solutions are there for sort of free QFTs, anyway

Like QFTs with a classical potential

I can't think of all that many

Particles in boxes and with potentials

Hydrogen atom

electron around a black hole

Particles in various spacetimes

are there any other interesting ones

2017

3:53 PM

Is there any PSE answer which explains why surface tension increases/decreases when soluble impurities are added while it surely decreases when insoluble impurities are added ? None of my books have a good explanation!

samjoe

4:38 PM

hello i have a question!

in normal adjustment of telescope, (astronomical telescope), image is said to form at infinity. How can we see such an image? it is at infinity right?

sorry my internet connection is very bad, it said timeout several times.

John Rennie

@samjoe Your eye acts as part of the telescope optics. The real image is formed on your retina.

samjoe

for image refer to this:

skullpetrol

It says "virtual image at infinity"

2017

5:14 PM

@JohnRennie Are you free for a few minutes ? I'm badly confused about surface tension

I drew a rough picture of the mercury surface

1) Is surface tension same as cohesive force or it is the resultant of cohesive and adhesive forces ?

2) When we calculate excess pressure the do we have to take component of surface tension T along the walls or do we take T as a whole ?

I assume that the radius of the test tube is $r$ while radius of the bubble (mercury bubble) is $R$. $R\cos(\theta)=r$

samjoe

@skullpetrol @JohnRennie sorry for late response. One of my friend argued that the image is at infinity hence it will be blurred, but from your response, it seems that our eve lens will be relaxed, nad we can easily focus on the object right? we can see the image even though it forms at infinity? because our eye lens converges the parallel beams am i right?

2017

@samjoe That's right...

Our eye converges the rays

samjoe

thank you @2017

2017

This is how our eyes converge parallel rays

samjoe

@2017 for your question, we take component of T along the wall, the horizontal components cancel.

John Rennie

5:23 PM

@2017 surface tension is the result of surface energy i.e. an energy per unit area associated with the surface.

Physics Guy

Hello there.

samjoe

hello here!

Physics Guy

Did you talk about biological physics?

That seems interesting.

samjoe

eyes.. and how we see image at infinity.

Physics Guy

@Slereah I think if you exclude QFT's in various spacetimes then you exclude pretty much all of the QFT's.

I mean, theoretically you could do QFT on (nearly) any manifold, couldn't you?

2017

5:33 PM

$P_{in}=P_{out}+\frac{2T}{r}$ or $P_{in}=P_{out}+\frac{2T\cos(\theta)}{r}$....which one will be correct for the mercury surface?

@JohnRennie

($T$ is the surface tension)

John Rennie

In general the meniscus will not be a part of a sphere so the simple expression for the pressure inside a sphere does not apply.

2017

@JohnRennie Suppose we assume the the meniscus is an arc of the sphere?

Then which one would be correct?

John Rennie

I think it's $2T\cos\theta/r$

Oh hang on ...

samjoe

F net = 2 pi r T cos (theta) right?

John Rennie

The vertical force required to lift the meniscus would be $2\pi r T \cos\theta$, but you'd need to divide that by the area of the meniscus to get the pressure.

samjoe

5:42 PM

where r = R cos(theta)

2017

@JohnRennie In which direction does surface tension act? Tangential to surface or along the walls?

You took components along the wall

John Rennie

tangential to the surface

2017

@JohnRennie So taking $Tcos(\theta)$ would be wrong...isn't it?

as that is along the walls

samjoe

see its downwards right? the downward component

2017

@samjoe Why are you taking only the downward component ?

Surface tension acts along the surface

samjoe

5:45 PM

the other component cancels,

2017

So we should take T as a whole

samjoe

it is cylindrically outward, the horizontal components cancel

John Rennie

@2017 suppose the contact angle was $\pi/2$ so the meniscus was a flat disk. What then would be the pressure difference?

2017

0 ?

John Rennie

Correct.

Because $T\cos\pi/2 = 0$

2017

5:47 PM

Okay, that does make sense

Aha, you are correct

John Rennie

In this case the surface tension just pulls horizontally on the (rigid) walls of the tube so it doesn't cause any pressure change in the liquid.

It is the vertical component of the surface tension at the walls of the tube that causes the pressure change.

2017

@JohnRennie Ah, that makes a LOT of sense :-) I got it I think!

Thanks a ton

I was lacking the intuition all this while

@samjoe Thank you too :)

Bernardo Meurer

@TerryBollinger Heya, long time no see

@JohnRennie I maxed out the amp and almost tore my head open

John Rennie

@BernardoMeurer things not to do again, part 1 :-)

Bernardo Meurer

Hehehe

John Rennie

5:55 PM

I have my amp hooked up to a pair of Heybrook HB2s and I don't think I've ever come close to maxing out the amp!

Bernardo Meurer

The speakers can take 140W RMS, the amp put 100W RMS per channel on Pink Floyd's "Money"

Totally doing that again

Hmm the HB 2 is 90W@6 Ohms

these ELACs are 140W @4 Ohms

John Rennie

It's particularly dramatic on Money because the coin clinking noise has a high dynamic range. Also try the clocks on Time :-)

Bernardo Meurer

I did The Great Gig in The Sky too

I think I'll hear that screaming forever

John Rennie

Have you got a copy of Crime of the Century. That also has some massive dynamics that can cause abdominal bleeding if you ramp up the volume too far.

Bernardo Meurer

Do not, let me check discogs

John Rennie

5:57 PM

@BernardoMeurer Clare Torry

Bernardo Meurer

discogs.com/Supertramp-Crime-Of-The-Century/master/25374

discogs.com/Supertramp-Crime-Of-The-Century/release/6398036

There's a nice remaster!

John Rennie

Listen to track 2 - Bloody Well Right!

I'm not a huge Supertramp fan - Crime of the Century - is the only album I really like. It is an excellent album though.

Bernardo Meurer

I just popped Black Sabbath's self titled album in

John Rennie

Now that's a really, really good album.

Bernardo Meurer

It's one of the best sounding LPs I have

John Rennie

6:01 PM

Recorded in one afternoon!

Bernardo Meurer

The pressing is great, the mastering is great, the music is great

Really? One afternoon?!

The first track must be one of my favourite of all time

So good

Black Sabbath

John Rennie

@BernardoMeurer OK it was actually one day not one afternoon

Bernardo Meurer

Even so

John Rennie

Still an amazing achievement though.

That album is just so good. The other album I really like is Volume 4.

Black Sabbath - Supernaut HD

►

Bernardo Meurer

I need to listen to that one more

I like Paranoid a lot, mostly because I find War Pigs an amazing track

John Rennie

6:05 PM

Paranoid is a great album, but the lyrics are just a bit schoolboy in places.

The first album is great because they weren't trying to be clever with the lyrics and Volume 4 is great because they'd got it together by then.

I think Paranoid sits somewhat uneasily between them.

Have you heard Master of Reality? That's an underrated album. Children of the Grave is a fantastic track.

Not everyone will agree, but I think that after Volume 4 the drugs and alcohol began to take their toll and the later albums suffered as a result.

I grew up listening to the early Black Sabbath albums and can rant on and on about them for days :-)

2017

@JohnRennie The music is nice although I didn't understand most of lyrics. It has a sort of spooky tone.

I suppose metal isn't for me :)

0celo7

6:26 PM

@ACuriousMind So it turns out that if A is unbounded and f is bounded, then f(A) is defined everywhere.

I can't find any result in the case when f is unbounded.

Terry Bollinger

Hi @BernardoMeurer, @JohnRennie! Hmm. I'm a Glitch Mob and Pink Floyd fan... are these remotely albums might try giving a listen to?

John Rennie

@TerryBollinger the Black Sabbath albums?

Terry Bollinger

Used to listen to them all the time in high school. That gives me some feel for it; I like diversity, I'll try a sample.

2017

@TerryBollinger Hi, I didn't see you on the hbar before. Where are you from ? :)

John Rennie

6:41 PM

Oh well, I'm off to attempt to reread American Gods for the book group. I wasn't that keen on the first reading so this may not go well.

Terry Bollinger

DC area, I'm a cognitive sciences guy with a lifelong interest in and deep respect for fundamental physics. I also once helped keep the US federal government from shooting itself and us in the foot by banning the use of open source software.

2

Must go also, cheers all...

0celo7

6:58 PM

@TerryBollinger north or south of D.C.?

Qmechanic

@JohnRennie : Volume 4 is probably my favorite album.

0celo7

I don't even know what Black Sabbath is

It sounds offensive

Jaime Gallego

@JohnRennie My father introduced me to Crime of the Century. I concur, it is a great LP.

But my favourite song keeps being Fool's Overture

I shouldn't be talking right now because exams & cramming

AccidentalFourierTransform

right, you're bloody well right

get out

go study

:-P

Teacher tells you stop your play and get on with your work

Alex

@AccidentalFourierTransform When Sakurai states that $J(d\vec{x}') = 1-i\vec{K} \cdot d \vec{x}'$, am I right in stating that the $d \vec{x}$ is some constant n tuple number?

AccidentalFourierTransform

7:12 PM

yep

Alex

@AccidentalFourierTransform I just want to confirm something else. Will type it now.

AccidentalFourierTransform

ok

The Dark Side

@DavidZ Wow:

74

A: Is it ok to upload joke papers to arXiv?

There is a long tradition of posting joke papers to arXiv on or around April Fool's Day, especially in astro-ph. Examples: Superiority of the Lunar and Planetary Laboratory (LPL) over Steward Observatory (SO) at the University of Arizona On the utter irrelevance of LPL graduate students: an unb...

Alex

@AccidentalFourierTransform The commutation relation given as $$[\hat{x},J] = d \vec{x}'$$ which when considering $J$ as I wrote previously then we get $$-i\hat{x}\hat{K} \cdot d \vec{x}' + i \vec{K} \cdot d \vec{x}' \hat{x} = d \vec{x}'$$ and he states something in a strange way, he says "By choosing $d \vec{x}'$ in the direction of $\hat{x}_j$ and forming the scalar product with $\hat{x}_i$, we obtain $$[x_i, K_j] = i \delta_{ij}"$$ Do you follow how the working gets you to that equation?

AccidentalFourierTransform

why $[\hat{x},J] = d \vec{x}'$?

Terry Bollinger

7:27 PM

@0celo7 west, actually, Northern Virginia.

Alex

@AccidentalFourierTransform It's $[\hat{x},J]| \vec{x}' \rangle = d \vec{x}'| \vec{x}' \rangle$.

So the equation holds on the subspace of $| \vec{x}' \rangle$

AccidentalFourierTransform

hmm off the top of my head, I dont know where that expression comes from

probably something about $\hat n\times \hat\alpha$

or something similar

2017

en.wikipedia.org/wiki/Terry_Bollinger Ah, so you have a Wikipedia page :-)

AccidentalFourierTransform

I dont know, rotations are boring and confusing

2017

Terry Bollinger

Terry Benton Bollinger (born February 6, 1955, Fredericktown, Missouri) is an American computer scientist who works at the MITRE Corporation. In 2003 he wrote an influential report for the U.S. Department of Defense (U.S. DoD) in which he showed that free and open source software (FOSS) had already become a vital part of the United States Department of Defense software infrastructure, and that banning or restricting its use would have had serious detrimental impacts on DoD security, research capabilities, operational capabilities, and long-term cost efficiency. His report ended a debate about whether...

Impressive!

Alex

7:32 PM

The commutation relation comes from this.

2017

@TerryBollinger Are you a self-taught physicist? You wiki page mentions that you have a degree in Computer Science!

Sir Cumference

@DavidZ Thank you. I needed to read this before I die.

AccidentalFourierTransform

@Alex I guess you have to take $\mathrm d\vec x=\epsilon \vec x$, and simplify

Alex

@AccidentalFourierTransform It's not working out bro

AccidentalFourierTransform

thats not possible

did you break the QMs?

that's illegal

Alex

7:38 PM

@AccidentalFourierTransform I think he wants to take $d\vec{x}$ as $d\vec{x}j$ where the rest of the components are zero and then the position operator as $x_{j}$? Some bullshit of that persuasion...

AccidentalFourierTransform

what I usually do in your situation is to trust that the result is right and move on

and come back to it later on

so just keep on reading the book

try to understand the global ideal S. is trying to transmit

and come back to the boring details when you are sure you understand the big picture

Alex

@AccidentalFourierTransform Yeah I think that's good advice. But it's always irritating moving on in that way :)

AccidentalFourierTransform

or you could try to come back here later on, and ask someone else

Im stoopid and dont understand the QMs

Alex

@AccidentalFourierTransform One last thing, do you agree that $$-i\hat{x}(\hat{K} \cdot d \vec{x}') + (i \vec{K} \cdot d \vec{x}') \hat{x} =-i\hat{x}\hat{K} \cdot d \vec{x}' + i \vec{K} \cdot d \vec{x}' \hat{x} = d \vec{x}'$$

AccidentalFourierTransform

the first equality, yes

the second one, I dont see why that should be true

maybe it is, I just dont know why

Alex

7:50 PM

@AccidentalFourierTransform It is because he proves it for $J$ and then assumes $J$ is something else...Apparently that is sufficient :/

I think he motivates more in the next section.

Anyway thanks for the help.

AccidentalFourierTransform

np

good luck

0celo7

@TerryBollinger I consider NOVA to be south

Terry Bollinger

@2017 yep! I'm just a poor bewildered information specialist who prefers to apply search heuristics and cognitive theory to the problem of how to navigate the nearly infinite-width combinatorial tree of physics-related mathematics created by decades of often ill-chosen branch weighting, raw product expansions, ...

... and inadvertant creation of overly derivative images and reflections of relatively simple concepts, at least in terms of their actual information "density" (inherent, irreducible complexity). Very messy, but kind of fun.

2017

So you're basically a self-taught polymath :D "poor bewildered information specialist" lol XD

Anyway, really nice to see you here :)

@TerryBollinger

0celo7

@Alex I'm back, what are you stuck on?

@TerryBollinger I'm in PWC btw

0celo7

8:28 PM

@ACuriousMind I know that rigorously, we cannot make sense of $e^{iap}$ using a power series. But what about on $C^\omega(\Bbb R)\cap D(p)$?

I'm assuming that's not empty

ACuriousMind

There you probably can

Or, should I say, hopefully :P

0celo7

@ACuriousMind I'm writing @Alex a proof of $J(a)=e^{ipa}$ from the mathematician's point of view. The first paragraph is me explaining that the power series doesn't make sense. But I feel like, morally, it should.

It's true that writing $e^{iap}=1+iap+\cdots$ doesn't make sense on its own

but if $\psi$ is analytic, then it is indeed true that $J(a)\psi=\psi+iap\psi+\cdots$

I guess that's what I should write?

Bernardo Meurer

@TerryBollinger You're a hero

0celo7

@BernardoMeurer what did he do?

Terry Bollinger

@2017, nice to meet you also! @0celo7, I had no idea you were relatively local, perhaps when my health is better (treatable rare autoimmune issue) we could even try lunch sometime. (I'm getting off fully for above reason; back before Christmas some noodle-head shopper "shared" a virus with me that started a down spiral. Ah well!)

Bernardo Meurer

8:39 PM

2 hours ago, by Terry Bollinger

DC area, I'm a cognitive sciences guy with a lifelong interest in and deep respect for fundamental physics. I also once helped keep the US federal government from shooting itself and us in the foot by banning the use of open source software.

0celo7

@TerryBollinger I'm home for spring break, I normally live in TN

but thanks for the offer, hope you get better

Bernardo Meurer

@TerryBollinger If I'm ever by DC I'll bring you some Elmer's glue for us to share

0celo7

...what?

AccidentalFourierTransform

Dec 27 '16 at 16:33, by Terry Bollinger

@BernardoMeurer dude, as long as it's Elmer's, it's got better nutrition than some cafeteria meals... :)