The h Bar - 2018-01-21 (page 2 of 2)

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Captain Bohemian

2:06 PM

my eyes get so tired for reading on the computer screen.

Do we know any example of a theory that actually has the Hilbert space $$\mathcal{H} = \mathbb C \oplus \mathcal H_{p^2 = m} \oplus \mathcal H_{p^2 \geq M}$$

Sine-Gordon sounds like it should

But I still don't know if its Hilbert space is known

It seems to always be semiclassical stuff

2:40 PM

@ACuriousMind I too had a confusion that how come instantaneous acceleration be perpendicular to the velocity vector, that's why I asked whether it happens in general. I think it's applicable for only uniform circular motion and projectile motion.

Anonymous

It's not true for projectile motion

Anonymous

And for uniform circular motion tangential acceleration is 0, no?

Anonymous

There is centripetal acceleration, yes

Anonymous

And that is perpendicular to velocity vector

Projectile motion has its acceleration perpendicular to the direction of the velocity, isn't that?

@Blue

Anonymous

2:45 PM

Isn't the acceleration always $g$ downwards in projectile motion?

Oh. I think I got u. At maximum height there only remains horizontal component of projectile motion and the vertical component is zero, only at that point gravitational acceleration is perpendicular to the velocity vector @Blue, am I right?

Anonymous

@ffahim Yea

But I wonder why it's said in our textbook that instantaneous acceleration is perpendicular to the velocity vector. Where most of the Google searches showed that's not true in general.

Anonymous

Can you quote your book exactly word by word?

Anonymous

Or upload a picture of it

2:51 PM

Actually it's written in our Bengali language. So u can't understand for this. But that really meant that instantaneous acceleration is perpendicular to the velocity vector.

Anonymous

I know Bengali, go on

Really? Great

Anonymous

Yep

if the velocity is normalized, this is trivial

if it's not, then it's not true

are you talking about fluid mechanics?

user image

2:54 PM

what on earth is $\mathrm{Lt}$?

I can guess but it's horrifying

@0celo7, no only for kinematic

@Blue u wanted to see the picture

@0celo7 limit in Indian

Anonymous

@ffahim Yeah, I'm reading it. Give me a minute

@BalarkaSen is there a special reason why they can't be like the rest of the world

What means if velocity is normalized

2:57 PM

@ffahim arc length parametrization

@0celo7 Old Russian books use $\text{Lt}$

@BalarkaSen we both know I don't consider old Russian math worth considering

Sorry I am a beginner for learning calculus. I don't have enough idea what you are referring to :) @0celo7

Anonymous

@ffahim The sentence you underlined just says that at some point of the trajectory magnitude of instantaneous acceleration is equal to $|\frac{\Delta v}{\Delta t}|$ as $\Delta t \to 0$.

3:00 PM

I can't understand the mathjatex

Anonymous

It doesn't say anything about perpendicularity. It just says "barabar", meaning "equal" or "equivalent".

Anonymous

@ffahim math.ucla.edu/~robjohn/math/mathjax.html

No, barabar in hindi means equal, in bengali it means towards that direction

@0celo7 Here is some good Russian math

You're a PDE theorist, you'll appreciate

Anonymous

@ffahim Even then it just says that $\vec{a}$ is in direction of $d\vec{v}/dt$. Where did they mention perpendicularity?

3:06 PM

@BalarkaSen pls, I'm a geometric analyst

I also do gmt

@BalarkaSen oh my god

Told you

It's golden

The line means that Instantaneous acceleration at a given point is perpendicular to the direction of velocity of that point @Blue

Anonymous

@ffahim Which word in that sentence stands for "perpendicular"?

@0celo7 I mean it's certainly better than this fucking shit: youtube.com/watch?v=NPN4m0d6PvI

লম্ব stands for perpendicular @Blue

3:11 PM

@BalarkaSen your countrymen

@BalarkaSen can't do analysis hw pls help

The white guy doing the flabby thing at the back cracks me up tho

No, Balarka is from India, I am from Bangladesh, we're neighbors

Honestly Oppa nabla style should just be the rightful winner of Dance your Phd

I haven't seen a single good thing from the top ones

Anonymous

@ffahim I think they're using it in some other sense. But as far as that is concerned, instantaneous acceleration is perpendicular to velocity iff the velocity vector has a constant magnitude at all times (which is true in case of uniform circular motion as you noticed).

he mentioned complex but didn't really do anything about it

should have put the residue theorem in there

3:14 PM

he mentioned Sobolev spaces too

as a passing line

yeah I noticed

I see. So that's no true for an objective moving at straight line in uniform acceleration right? @Blue

@0celo7 whats your analysis hw

Anonymous

@ffahim Right

Anonymous

"constant magnitude of velocity vector" is the required condition

3:17 PM

Ok. Thanks a lot :) @Blue

@BalarkaSen If a function $f\in W^{2,p}_\text{loc}(R)$ has a second weak derivative in $L^p_\text{loc}$, then it has a first weak derivative in $L^p_\text{loc}$

Problem is, I have a proof, but it has to be wrong

clearly one should define $f'=\int f''$, but there's a constant missing

my proof doesn't seem to care about that constant, which is strange

integral of f'' over [something, x] i presume?

(I do not know these spaces)

well, fix a finite interval $(a,b)$ for definiteness

then define $f'(x)=\int_a^x f''(t)dt$

the idea is then that $df/dx=f'$ weakly

gotcha

but clearly this is only true up to a constant, because I didn't write constant in that integration

but moving on, for any test function $\phi$ some real analysis shows $d(\phi f')/dx=\phi'f'+\phi f''$

3:21 PM

uh, constant of integration appears only when you take indefinite integral? i dont see why you get fucked by constants here

might be stupid qn

Anonymous

@BalarkaSen omfg...what's that

@Blue truly a masterpiece

@BalarkaSen Well the correct formula for C^1 functions is $$f(x)=\int_a^x f'(x)dx+f(a)$$

Anonymous

That surely deserves an oscar, yeah.

@0celo7 Oh i see ok

3:23 PM

so I have $d(\phi f')/dx=\phi'f'+\phi f''$

then integrate to get $$-\int f'\phi'=\int \phi f''$$

mmk

and then use the definition of $f''$ to write $$-\int\phi'f'=\int \phi''f$$

so this is almost done, but the problem is that not every test function $\eta$ can be written as $\phi'$

to remedy this, let $\psi:(a,b)\to R$ be smooth with $\psi=0$ near $a$ an $\psi=1$ near $b$, and define $$\overline \eta=\int_a^x\eta-(\int_a^b\eta)\psi$$

Cool trick

this has the advantage of being in $C_c^\infty$, and the derivative is $$\overline \eta'=\eta -(\int\eta)\psi',$$ with both terms being test functions

True

3:27 PM

so you can plug this into above formula, and then use it again with $\phi'=\psi'$

so you end up with $$-\int f'\eta=\int \eta'f$$

but this cannot be true because $f'$ might start at the wrong value

@0celo7 what do you integrate this over?

@BalarkaSen $(a,b)$

my first reaction would be some constant gets ignored in your manipulation during integration

(of course i could be wrong)

well what's interesting is that in $$-\int_a^b \phi'f'=\int_a^b\phi''f,$$ you can redefine $f'$ by a constant and not change anything

Trying to detect it with limited understanding of things

Ah...

3:34 PM

because $$\int_a^b\phi'=\phi(b)-\phi(a)=0$$

so you start with something agnostic of the constant, and end up with something that detects it

Weird... \

4:05 PM

@Slereah you truly have the best countrymen

god bless the french

that we do

I understand this stuff now

do you mean french patriotism

Can anyone tell me how to share an image here ?

@SkyWalker are you on a phone?

4:13 PM

Yes

But desktop mode is on.

use imgur or something equivalent

Ah. Just click the upload button.

I can't see it.

user image

Anonymous

You need 100 rep for that I think

4:15 PM

On my phone screen, it is showing only the send button.

Anonymous

52+23+6+1+1=83

Anonymous

So get another 17 rep points, and then you can upload pictures here

:D

Anonymous

The workaround is to upload it to some other site like imgur and share the link.

@SkyWalker try refreshing the web page now and see if the button appears

4:19 PM

Actually, I am not going to post in physics meta. You people always downvote my questions. I am on the verge of getting banned from asking questions.

perhaps stop asking poor questions

Go on and please look at my profile.

3 duplicates!

Anonymous

Stop taking downvotes personally, really.

@JohnRennie No.

Anonymous

4:22 PM

Most newcomers get downvotes. That's the only way they learn to follow rules and ask good questions.

Anonymous

Or else it would become something like Reddit or Yahoo

@Blue I don't take personally. But, I am going to get banned. How can I edit my answers ?

I mean, how to improve them ? Specially the duplicate ones ?

Anonymous

Add more details. Add your research in trying to find the answers. Ask specifically what you couldn't understand even after searching the net.

@Blue I am not a spammer or a trash writer. I am a student who is interested in physics.

Anonymous

To be blunt, your questions look pretty straightforward and lack research efforts.

4:25 PM

rekt

Anonymous

For example:

Anonymous

1

Q: Resonance patterns in the video

https://youtu.be/wvJAgrUBF4w In this video, the sand particles form beautiful patterns on the vibrating metal plate on different frequencies. Can anyone explain what is going on there mathematically ?

acoustics harmonic-oscillator resonance vibrations

Anonymous

You just add a link to a video!

Anonymous

No mention of how you tried to find the answer

Anonymous

And where exactly you got stuck

4:27 PM

@Blue Yes. I ask questions without much details. I want to get the view of the answerer. I don't want him to think like me.

Anonymous

"Can anyone explain what is going on there mathematically ?"

Anonymous

That's one of traditional examples of lazy questions

Anonymous

@SkyWalker I don't buy that reason, sorry.

this is a beat down

@Blue Yes. I wanted the mathematical analysis as I lack mathematical skills to analyse the patterns.

Anonymous

4:30 PM

@SkyWalker No one has the time or energy to explain all the mathematics behind formation of resonance patterns. You are expected to search the net and read relevant books and ask where exactly you're getting stuck. Ask more specific questions and if you don't know the math used in those books, learn then from other books first and ask if you get stuck there.

Presentation is absolutely crucial in SE. No matter how good of a question/answer it is, you have to present it the right way to not be seen as a bad question

Recently there was a good question which should have a deep answer that got closed down in math.SE... nothing to be done about that

(To illustrate that it's deep, I'm still studying stuff to find out the "right" answer)

Anonymous

One good thing about SE is that they expect you to be very professional on the main site, which imo is very good. Laziness has no place here.

Anonymous

However, on chat, such questions may be fine once in a while.

It's not always very good.

@BalarkaSen You are from MSE. What do you know about Cleo ?

Anonymous

4:33 PM

@BalarkaSen Well, depends. But most of the times

There's no way to get around it, but it's certainly not the best system

Anonymous

I've had my share of bad experiences too. But it's the best that exists on the internet

Also it doesn't stop the tons of homework questions that barrages MSE

Anonymous

Personally I do have some issues with the policies, but I can live with that I guess.

Anonymous

@BalarkaSen True

4:34 PM

The question I am speaking of should be MO-level when appropriately generalized

@SkyWalker Nothing, like everyone else

quora.com/Who-are-the-best-living-mathematicians/answer/…

Please read the later part of the answer.

Cleo is indeed a mysterious user

I have never been fascinated with solving complicated integrals, myself, however

I know there's a group of people who are

Anyways, good bye

1 hour later…

5:49 PM

How to prove that a function is right-continuous?

Anonymous

@Yashas Just check if right hand limit is equal to value at that point

I need to prove that it is right continuous everywhere.

Anonymous

The logic doesn't change

Hmm, nevermind. I was stupid.

1 hour later…

7:16 PM

are you ready for some football? @dmckee

Haag talks about something called a "Cuntz algebra"

Fairly rude

that's similar to the objections that were raised against the name "blackhole"

arxiv.org/abs/1607.03120

maximum principle arguments are always scary

I have a hard time believing that harmonic functions can be nonconstant

it's so strange

> Title changed in journal by editor (reason unknown)

7:32 PM

:-D

But the new one isn't really better.

what's the new one?

doi.org/10.1103/PhysRevLett.117.211301

hey @0celo7 is the DLMF gonna shut down tomorrow?

@EmilioPisanty who knows

7:34 PM

=/

download it just in case

don't think it's possible

I've got the book, but it's not the same

is there a way to know how "large" a website is?

DLMF won't be particularly large

it's just text

admittedly with some reasonably complex markup thrown in, but still

screw it, I can't analysis

and I have fucking complex analysis homework

;_;

Anonymous

7:39 PM

@EmilioPisanty Found this site mentioned on Server Fault SE: httrack.com. Maybe it works (Haven't tried it)

@Blue oooof

no, it's not that important

as I said, I've got the book

@EmilioPisanty are government websites supposed to shut down?

if anything, it's the ASD that's trickier

@0celo7 dlmf shut down last time, as I recall

nice

it's still running, though, so I guess that's a good sign

7:41 PM

@skullpatrol I haven't really been paying attention to football for a few years. Especially not the pro game.

if the shutdown takes longer than a week, the world will probably explode

can u survive longer than a week?

@0celo7 unlikely

@0celo7 without the US government?

yes

I take it's time for the Game-that-exists-to-support-a-halftime-show-and-commercials again?

3

@EmilioPisanty perfect time for ISIS to pull some shit while everyone is unemployed

Anonymous

@0celo7 Where's this conversation even headed?

Anonymous

7:42 PM

:P

@Blue the world exploding

yeah, let's not do ISIS

also I forgot to eat lunch

damn

Anonymous

@dmckee lolol

8:23 PM

@dmckee yeah...pretty much.

Conference finals.

for the game-that-exists-to-support-a-halftime-show-and-commercials

8:54 PM

Help

How do you get from equation 6 on pdf page 9 here physik.lmu.de/lehre/vorlesungen/wise_07_08/tb3/vorlesung/… to the next equation on the top of page 10, seems like you just multiply across by $C$ on the right, but it doesn't work :\

9:24 PM

@BalarkaSen how would one integrate $\oint dz/(z-a)$ without Cauchy's formula?

parameterize the contour

@0celo7 you can in principle crank it out explicitly

z = a + re^itheta

pull it back, integrate

@BalarkaSen contour is not around $a$

@BalarkaSen if the path is a circle around $a$

9:25 PM

it's a circle enclosing the origin

@0celo7 radius?

i.e. bigger than $|a|$?

@EmilioPisanty not $|a|$

@EmilioPisanty I automatically assumed it is

you assumed WRONG

because 0celo7 didnt say it

you DIDNT TELL ME

9:26 PM

@0celo7 are you comfortable with $\oint_{\partial D} f(z)\mathrm dz=0$ whenever $f$ is analytic on the interior of the $D$?

^

it's Cauchy's theorem basically

If I were allowed to use any of the complex analysis I know, I would not be asking.

if you're not comfortable with deforming paths, then all you've got left is explicit integration

It has to be done via explicit integration, but the antiderivative doesn't exist because of logs being retarded.

well, that's sort of the point, no?

9:28 PM

log is well-defined on a disk outside of 0 and a half-line out of 0

the antiderivative does exist on any singly-connected domain that doesn't include $a$

The integral is around zero

and you can calculate it explicitly

@0celo7 ??? you just said it's not around zero

@EmilioPisanty well that doesn't do anything because I might be integrating around $a$

9:29 PM

@0celo7 doesn't matter

4 mins ago, by 0celo7

it's a circle enclosing the origin

OK, let's start with a simpler question

oh ok well it's outside of z = a

that's the pole of 1/(z-a)

is there some direction out of $a$ in which you can draw a ray that will only intersect the path once?

@BalarkaSen the function has no primitive in any neighborhood of its pole, so how do you want me to write it down?

this is literally a Cauchy formula problem but in the wrong section

it's pretty stupid

9:31 PM

What are you allowed to use

nothing

definition of the integral

@0celo7 it sounds like a pretty reasonable exercise to me

Well get fucked then. Compute it by hand

@0celo7 have you defined the logarithm yet?

@BalarkaSen How?

@EmilioPisanty nope

9:32 PM

ah

parametrize the contour by z = re^itheta, pull it back, work by hand

then you're screwed, yes

I know what it is of course, but like I said I would just use Cauchy and get it over with.

same

@BalarkaSen define "pull it back"

9:32 PM

change of variables

i'll be back in a bit

yes, so $$ri\int_0^{2\pi}\frac{e^{it}}{re^{it}-a}dt$$

I can't do that integral

@0celo7 if you can't use the explicit logarithm, then that's the mark at which I'd go to the instructor and ask them to clarify what they want exactly

@EmilioPisanty that's the plan

... assuming this is formal homework that you have to do

if you don't have to do it

don't

"have" is a strong word

9:35 PM

Just reprove Cauchy's theorem really

It's less work than this

lmao

^

pretty much

the last homework had a problem like this too

solution is "open mapping theorem" if you know it

a page of calculations if you don't

Anonymous

What's the problem with parameterizing it like 0celo did and trying to solve that integral? Cauchy's formula just provides a shortcut for that, no?

Anonymous

4 mins ago, by 0celo7

yes, so $$ri\int_0^{2\pi}\frac{e^{it}}{re^{it}-a}dt$$

@Blue You have to compute the integral 0celo7 wrote down

9:37 PM

if you can do the integral...bravo?

for real variables I would guess some log formula but here that's perilous

I mean you can prolly do it. Expand $1/(e^{it} - a/r)$ in power series as a geometric series in $e^{it}$.

@0celo7 well, an antiderivative is $\log(re^{it}-a)$

Then you have to integrate $\int e^{kit} dt$ stuff

Anonymous

@BalarkaSen Yeah, I mean it might be possible to solve it directly. I need to try

at least on fractions of the contour

9:38 PM

I am not going to try

but that's if you have the log in your toolbox

@BalarkaSen that's a possible way, but then you have to justify the convergence and bleh

@EmilioPisanty mhm I bet that works if you break the integral up into two parts and put the branches in the appropriate places

if you don't, then you'd need to rebuild the logarithm from scratch

@0celo7 yes, it works

so that they miss the two pieces of the contour

but that's a lot of work

@0celo7 yeah

9:40 PM

@0celo7 I would put my money that that's what they're asking for

it's a lot of work but it's a homework-problem-sized chunk of work

Who here would dare contradict xkcd.com ? :-) — StephenG 2 days ago

↑ lolz

the next problem almost certainly needs Cauchy's theorem

which one

compute $\int_0^\infty\sin(x^2)$

you can do it by real methods but then it's not a complex analysis exercise

Use that $\delta$ method

9:52 PM

this book is horribly imprecise

what the fuck does "continuously complex differentiable" mean?

complex derivative is $C^0$

$f'(z)$

@BalarkaSen that's true for any holomorphic function tho

yeah it's redundant but a lot of books assume it off the bat

"this provides a proof of Goursat's theorem under the additional assumption that $f'$ is continuous"

wot

yeah

9:54 PM

that follows from being complex differentiable...

not easily

I guess the point is that one used Cauchy's formula to prove that holomorphic functions are analytic?

Mhm

See, if you assume $f'$ is continuous, you can do easier arguments, like use Green's theorem

I realize that

the optimal version of Green's only requires $f$ be Lipschitz tho

I guess you won't have that in general

well, you do, but hard to prove without other things

oh well

yeah

10:02 PM

Simpler question: if $Q_a = C_{ab} \overline{Q}_b = C_{ab} Q_b^{\dagger} \gamma^0$ where $C^{-1} \gamma^{\mu} C = - \gamma^{\mu T}$ then how does $\{ Q_a,Q_b \} = - 2 (\gamma^{\mu} C)_{ab} P_{\mu}$ become $\{ Q_a,\overline{Q}_b \} = 2 (\gamma^{\mu})_{ab} P_{\mu}$?

@0celo7 see that big banner on the top of pubmed central? ncbi.nlm.nih.gov/pmc

10:18 PM

@EmilioPisanty yeah, most of the government isn't showing up monday

like I said, great time for ISIS

10:49 PM

@BalarkaSen fookin ell it took me like an hour to get the right estimate on the arc integral

How does this make sense, if
$$Q_a = C_{ab} \overline{Q}_b$$
and
$$\{Q_a,Q_b \} = -2 (\gamma^{\mu} C)_{ab} P_{\mu}$$
then using
$$C^{-1} \gamma^{\mu} C = - \gamma^{\mu}$$
we have
\begin{align}
-2 (\gamma^{\mu} C)_{ab} P_{\mu} &= \{Q_a,Q_b \} \\
&= \{Q_a,C_{bc} \overline{Q}_c \} \\
&= C_{bc} \{Q_a, \overline{Q}_c \}
\end{align}
so that
\begin{align}
\{Q_a, \overline{Q}_b \} &= (-2) C^{-1} (\gamma^{\mu} C)_{ab} P_{\mu} \\
&= 2 (\gamma^{\mu})_{ab} P_{\mu}
\end{align}
what happened to the whole lack of commutativity thing

... and apparently this is right because
\begin{align}
2 (\gamma^{\mu})_{ab} P_{\mu} &= \{Q_a, \overline{Q}_b \} \\
&= \{C_{ac} \overline{Q}_c, \overline{Q}_b \} \to \\
\{\overline{Q}_a, \overline{Q}_b \} &= 2 (C^{-1}\gamma^{\mu})_{ab} P_{\mu}
\end{align}
is right, what is going on?

11:06 PM

journals.aps.org/prl/abstract/10.1103/PhysRevLett.119.223003

hmm... putting quantum monte carlo and couple cluster together

11:21 PM

@0celo7 Attempt to download DLMF gives me 25000 files and still counting, increasing by 5 files /s

lol:

22/1/18, 10:20:33 am - ERROR: Download of "http://dlmf.nist.gov/about/bio/ABOldeDaalhuis",
referenced from "http://dlmf.nist.gov/13",
failed with error code -1100 (File does not exist.)

one of the references don't exist

jeez

at near 50000 files now, sitesucker just downloaded all the numbered theorems

now it is going through the bibilography table

how big is it?

I have no idea, on average each files is roughly 2-10KB, but some files can go to 118 KB

so less than a gig altogether?

11:30 PM

File numbers now peaked at 473?? and is now decreasing, guess it has finished crawling

22/1/18, 10:31:29 am - ERROR: Parsing of document "dlmf.nist.gov/10.3.F12.webgl.html"
failed because the closing ">" for the "MetadataFloat" tag could not be found

lol html blues

The following files don't exist in DLMF, looks like we have some deadlinks here...

22/1/18, 10:38:14 am - ERROR: Parsing of document "dlmf.nist.gov/1.2.html"
failed with error code 261 (The file “1.2.html” couldn’t be opened using text encoding Unicode (UTF-8).)
22/1/18, 10:38:14 am - ERROR: Parsing of document "dlmf.nist.gov/1.2.html"
failed with error code 261 (The file “1.2.html” couldn’t be opened using text encoding Unicode (UTF-8).)
22/1/18, 10:38:15 am - ERROR: Parsing of document "dlmf.nist.gov/1.2.html"
failed with error code 261 (The file “1.2.html” couldn’t be opened using text encoding Unicode (UTF-8).)

and similar urls still counting

11:55 PM

sup

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