The h Bar - 2017-10-03 (page 1 of 4)

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12:07 AM

Any mods in here? This mess could probably use comment clean-up:

2

A: Photons are faster than the speed of light because they travel in transversal waves?

Light is made of photons traveling in waves, doesnt this mean that photons are moving in longer distances than light itself because photons dont move in a straight line The thing that "waves" in the electromagnetic wave is not the position of the photon, it's the strength of the electric fie...

@ThePhoton Done; please use a flag next time, the probability that a mod sees it is higher that way.

@ACuriousMind Thanks

12:49 AM

Honestly Im amazed one of my answers has a score of 5

Looking back on it I have no idea how accurate my summarisation of a black hole is, especially given my lack of knowledge about GR. Seemed right tho, lol

1:15 AM

nature.com/nphys/journal/vaop/ncurrent/full/…

So, molecules next :?

journals.aps.org/prl/abstract/10.1103/PhysRevLett.119.132003

75000 jets is quite a large number

But more importantly, open access works

2:08 AM

@0ßelö7 are you there?

2:35 AM

@Icemybread I am now

look at skype plz.

1 hour later…

4:04 AM

Hi, everybody.

Prathyush Poduval

Hallo

tfw you can tell student HW was copied from the solutions manual

4:23 AM

@Semiclassical what did they do?

well, do you have a copy of Griffith's QM with you? that'll simplify this conversation

Nope

kk

@Semiclassical On a scale from 1 to 10, how likely is it I have a PDF already?

i dunno

4:25 AM

10, apparently.

part a of problem 2.11 tells them to compute the expectation values of x, p, x^2, p^2 for the ground state and first excited state of the harmonic oscillator

What page?

50

in the second edition

I only have the most up to date pirated books

lol. a key phrase in part (a): they're told to compute the integrals 'by explicit integration'

now, for x^2 and p^2, that's pretty much a necessity

4:27 AM

ugh

for x, though, you can use the fact that $\langle x \rangle = \int \psi^*(x)x\psi(x)\,dx$ with $\psi(x)$ being either even or odd

so it vanishes by symmetry

What is the difference between grad and undergrad QM

this looks exactly like my grad course

we did all of Sakurai

whether you start by doing bra-ket notation or not

@Semiclassical Ok I feel like I say this a lot, but how can that possibly be hard?

Griffiths delays bra-ket stuff til the third chapter

i don't mean that it makes it hard. I just mean that that's usually a pretty good gauge of whether it's a grad course or an undergrad one

4:28 AM

A part of me feels like there are some things that are litmus tests for fields

If you can't grok bra-ket in a day, you shouldn't do physics

it's a litmus test, yeah

meh. i won't say one way or the other

If uniform convergence confuses you, stop doing math

@EmilioPisanty There's nothing special about positive $t$ in the FLRW metric. You could perfectly well have universe collapsing from $t=-\infty$ towards $t=0$. So naively we can have a universe that collapses towards $t=0$, bounces and expands after $t=0$.

But what you can't do is connect the two regions $t \lt 0$ and $t \gt 0$ because at $t=0$ the geometry is singular to you can't evolve your geometry through that point. In fact we normally excise the singularity from the manifold so it literally doesn't exist.

but an undergrad course typically will start in position space, whereas in a grad course they'll be more likely to start with bra-ket notation from the get-go

anyways

In that case the two regions are completely disconnected.

4:30 AM

back to what I was saying

@JohnRennie Isn't FLRW formally time-symmetric

for p, one similarly can observe (going the integration route) that $\langle p \rangle =(-i/\hbar)\int \psi^*(x)\psi'(x)\,dx$

@Semiclassical that's the definition, where are you integrating there?

and since $\psi(x)$ has definite parity, $\psi^*(x)$ and $\psi'(x)$ have opposite parity hence odd and integral is zero by symmetry

Well, you're integrating

nvm

@Semiclassical right

4:31 AM

i'm stretching the notion of 'explicit integration' here, so as to be charitable to the students

@0ßelö7 there's no reason why the density has to be symmetric around $t=0$ because you can't match up the two sides at $t=0$. The universe could behave differently on either side.

however, the solutions manual for Griffiths---which is easily found online---doesn't compute <p>=0 in this way

griffiths instead notes that, since <x>=0 by symmetry, one has $\langle p \rangle = m \dfrac{d\langle x\rangle}{dt}=0$

@JohnRennie Yeah I guess. A topologically disconnected universe is...not a universe :P

now, this is rather clever. however, it's not explicit integration, and it's not something you expect a student to come up with on their own

...guess what fraction of the student solutions use that trick?

@Semiclassical No student is clever enough to come up with that

4:34 AM

it's something like 40 %, if I'm estimating it right

ouch

yeeep

I'm pretty good at QM and my immediate thought was symmetry

right

anything else I have to question

4:35 AM

to be fair, they were asked to show the above relation in a chapter one problem

and it is a good tactic. buuuuut

i don't believe for a second that that's why so many people did it

yeah

i'm going to chat with the other TA tomorrow on it and decide how hard to come down on it. the grey area here is that they're not forbidden from looking at the solutions manual, but they aren't supposed to copy from it

and the fact that they deviate from the problem instructions in this way is a pretty clear sign that they just copied it

@Semiclassical people did the same in my grad class

yeah

The prof got really angry

4:38 AM

it doesn't prepare you for the test

this prof isn't going to get angry about it---he expects it, since griffiths is such a popular book---but he has given us license to punish it when we see it

this prof was angry that a bunch of grad students were so lazy they couldn't be bothered to do homework

yeah, it's pretty pathetic

undergrads, I can understand

hey!

in the sense of workload.

oh yeah

some math grad students were complaining about taking 4 classes

I'm taking 7

4:40 AM

undergrads are typically taking a lot of courses at once, so I can better understand the inclination to cut corners

in a grad course, though, you should be able to stand on your own two feet

work with other people, sure. dig around online for insight if you have to? okay

but if you're a grad student i think copying crosses a clear line

add to that the fact that grad are generally TAs, and you've got an element of hypocrisy thrown in there as well

yeah

the slightly annoying thing here is that there were a few people who just said $\langle x \rangle =0\implies \langle p \rangle =0$ without saying why

if one was only judging by argumentation, that'd be worse than doing d<x>/dt=0

buuuut if you don't say why, you're not outing yourself as having copied from the text

so it's a bit of the old adage: "better to remain silent and appear foolish than to open one's mouth and remove all doubt." saying less might harm you in other ways, but it doesn't reveal the copying.

@Semiclassical if $\langle x \rangle = 0$ surely that automatically means the average momentum must be zero?

Because on average the position is constant

sure, it certainly does so. if it's zero at all times, then by Ehrenfest's theorem you know that <p>=m*d<x>/dt=0. no objection there

but that's not the point.

the point is that 1) the problem told them to do it by explicit integration, 2) half the class came up with <p>=m d<x>/dt=0 as the reason, and 3) that's exactly how Griffiths justifies it in the solutions manual

so that's a pretty clear signature that people didn't do that part on their own, but rather just glanced at the solutions manual and said "oh, so that's how you do it"

ugh latex

4:49 AM

had one or two students done that, i might've been able to chalk it up to cleverness.

but that many people is a clear pattern of copying

at a logical level, it's fine. but in context it's a sign of copying.

how pretentious is "lemmata"

well, everything greek sounds a bit pretentious already

@Semiclassical Will people think I'm insane if I book a whiteboard room in the library and practice these talks?

Do people do that?

nah, that seems typical enough

besides, you're asking the wrong question. the right one is: "does it matter if people think i'm insane for practicing in a whiteboard room?" :P

@0ßelö7 in industry presentations are always rehearsed.

2

From an industrial perspective it seems crazy that academics appear willing to try and wing it.

The point about presentations in industry being that money is usually involved. That kind of focusses you :-)

4:58 AM

capitalist pigs

Oink oink

Even in academia one should rehearse talks, though

If only to get the timing worked out

Though I personally find I can’t rehearse in an empty room

$$\mu_q=J_q(\varphi_q)=J(\varphi_2)||\varphi_q||^2_{2^\star}||\varphi_q||_q^{-2}‌\ge \mu ||\varphi_q||^2_{2^\star}$$

@Semiclassical it's remarkably hard to do :-)

I need to practice that one

it has a typo!

5:00 AM

The feedback loop is too strong. It’s only when I’ve got an audience that I can push past that

@Semiclassical I need to get an idea of how long what I have planned will actually take

I can wing a math talk. I can't predict how long it will take me to do so, however.

sounds like rehearsal is in order, then :)

is this the kind of talk where people feel free to ask questions in the midst of it?

yes, very informal

mmkay

that always complicates the question of timing

there are a few spots where I am afraid people will ask too many questions :P

5:03 AM

lol

are they spots which are important to the story?

@Semiclassical Morally? Yes. Actually? No.

lol

You need to find the norm of a certain embedding. It's relatively easy to get an estimate for the norm. Turns out the estimate is sharp.

Showing the sharpness is hard, and I have no clue how to do it.

But in the end it doesn't matter, because the estimate is all you need.

hmm

I don't suppose there's an example where you can check it explicitly?

The functions that show the estimate is sharp are known.

Hmm

I must have that wrong

5:07 AM

yeah, that seems strange

Is there a name in the literature for this, or is it more esoteric than that?

Oh it's very well known

Oh, the really hard part is finding all functions that give equality.

huh

link.springer.com/content/pdf/10.1007%2FBF02418013.pdf

RIP Tom Petty. Another one bites the dust.

@Semiclassical In any case, the computation of $L^p$ norms would bore everyone to death.

5:11 AM

has he actually passed now? there was an initial announcement of him having died, but that turned out to be premature (though it sounded pretty inevitable)

It's been confirmed by his manager

yeah, it's definitive now

Too many people dying in 2017

He was only ten years older than me :-)

@JohnRennie am I in your will?

5:17 AM

'Fraid not.

My will leaves everything to the home for distressed laptops.

@BernardoMeurer ????

Bernardo Meurer

What?

you are JR's heir

Bernardo Meurer

Yes

That is accurate

@BernardoMeurer did you get your question about the Big Bang sorted out?

Bernardo Meurer

5:21 AM

@JohnRennie Sorta, but I want you to answer it too because I love you

Assuming the universe is described by the FLRW metric we can subdivide it into 3D sections with $t$ constant everywhere in the section.

Technically this is called foliation

pretty lame one

it's just a product

So we can divide up 4D spacetime into an infinite set of 3D bits, each labelled with a value of $t$

And conventionally we label the section that corresponds to the big bang with $t=0$

So the Big Bang happened at time zero

Technically what I've called sections are submanifolds

@BernardoMeurer please note that my answer was not overloaded with technical jargon

going to the statement earlier re: walking south at the south pole, one can partition the 2D sphere into circles of constant latitude.

5:27 AM

But, just to be awkward, we usually say that the submanifold with $t=0$ is removed from the manifold that makes up the universe so it doesn't actually exist.

and then the south pole is by definition at a latitude of 90 degrees S, and it's not possible to talk about latitudes beyond than that

@Semiclassical see my reply to Emilio. I'm not keen on the what's south of the South Pole metaphor.

hmm

well, that seems rather akin to saying that one could put another sphere below the south pole, and agree that going south from the south pole moves you onto the other one

From a physical point of view I have no clue what negative t is supposed to be. I think the south pole thing is correct, if you remove the south pole from the Earth.

@0ßelö7 if quantum gravity does remove the singularity it's entirely possible the two sides are smoothly connected. in which case negative time is just the past.

5:31 AM

"If quantum gravity"

Bernardo Meurer

@0ßelö7 Why is it not part of the universe?

to push the analogy to its limit: maybe there's a piece of gum between the two spheres :P

Smoothly connected?

@BernardoMeurer By convention

There are ways of putting it in

i am far too uneducated on GR to have any business talking about this though

There's no problem in extending the manifold, it's just that the determinant of the metric is zero there.

5:33 AM

but hey, what about imaginary time :P

Er, zero or blows up. I think it depends on the matter.

@BernardoMeurer consider the graph of $y=1/x$. It has two disconnected parts for $x>0$ and $x<0$.

But its value isn't defined at $x=0$

To make the value of $y$ well defined everywhere we could simply exclude $x=0$ from the graph.

And that's pretty much what we do when we exclude the Big Bang from the manifold.

@JohnRennie $e^{-1/x^2}$ is a better example

you can continuously extend it

@0ßelö7 because $y(x) = y(-x)$ ?

@JohnRennie plot it

5:36 AM

Or do you mean extend it into the complex plane

it's undefined at $0$ but the left and right limits are equal and finite

Bernardo Meurer

@0ßelö7 So it's existensible by continuation

on the other hand, the limits along the imaginary axis go kablooie

@BernardoMeurer It might be. That's what JR was saying quantum gravity might show

@0ßelö7 ah ok, yes, I see

But in the FLRW metric the curvature invariants do go to infinity at $t=0$

5:42 AM

@JohnRennie then you'd better hope quantum gravity smooths it

@0ßelö7 I suspect it's more likely that the metric ceases to be a useful concept near the singularity

i.e. the spacetime geometry is no longer well described as a differentiable manifold

@JohnRennie Let's hope not. If metrics stop being useful I'm out of work

Bernardo Meurer

@0ßelö7 "work"

@JohnRennie Oh god I hope spacetime is really a varifold and physicists have to learn measure theory

Bernardo Meurer

What on earth is a varifold

5:44 AM

@BernardoMeurer A Radon measure on $R^n\times G(k,n)$

Bernardo Meurer

@0ßelö7 That tells me nothing

In mathematics, a varifold is, loosely speaking, a measure-theoretic generalization of the concept of a differentiable manifold, by replacing differentiability requirements with those provided by rectifiable sets, while maintaining the general algebraic structure usually seen in differential geometry. More closely, the varifold generalize the idea of a rectifiable current. Varifolds are one of the topics of study in geometric measure theory. == Historical note == Varifolds were first introduced by L.C. Young in (Young 1951), under the name "generalized surfaces". Frederick Almgren slightly modified...

Honestly, you guys need to get out more often

Bernardo Meurer

I get out plenty

I saw the sun this year like twice

More than that is teasing with skin cancer

The Sun is a deadly lazer

@0ßelö7 hi I have a question

Bernardo Meurer

@GPhys Question hi have a I

5:48 AM

what is it

@0ßelö7 if I'm computing the christoffel symbols from the metric and my metric is e.g. $ds^2=d\theta^2+\sin^2\theta d\varphi$

are my indexes just running between theta and phi?

yes

and raising and lowering indexes does nothing in this case @0ßelö7?

metric and inverse are not diagonal 1, so it should do something

$$\begin{pmatrix}1 & 0 \\ 0 & \sin^2\theta\end{pmatrix}$$

is my metric $g_{\mu\nu}$

Anonymous

5:56 AM

Where can I find a proof for the Fourier Series formula? i.e. $f(x)=a_0/2+\sum_{n=1}^{n=\infty}[a_n\cos(\frac{2n\pi x}{b-a})+b_n\sin(\frac{2n\pi}{b-a})]$

Anonymous

It's stated without derivation in my book

@GPhys so raising and lowering will involve powers of sine

@Blue Grafakos, Classical Fourier Analysis.

Anonymous

@0ßelö7 Checking

@0ßelö7 $$g^{\mu\nu}=\begin{pmatrix}1 & 0 \\ 0 & \sin^{-2}\theta\end{pmatrix}$$

?

yes

6:05 AM

@JohnRennie Meh. Serious academic group insist of rehearsals, too, except for talks that have been given by this same presenter several times already.

@dmckee Glad to hear it :-)

I've always felt it is discourteous to your audience not to rehearse and polish your talk. It's like going to a concert and finding the performer gave a second rate performance because they couldn't be bothered to rehearse.

are there any mathematicians here

@0ßelö7 so for example $\frac{1}{2}g^{\theta\theta}(\partial_\theta g_{\theta\theta}+\partial_\theta g_{\theta\theta} - \partial_\theta g_{\theta\theta})$ is 0 because e.g. $\partial_\theta g_{\theta\theta}$ is zero (i.e. I'm computing that right?

O__O

I've got the Christoffel symbols for a sphere somewhere if you want me to dig them out

@GPhys how is it zero

you're differentiating sine

6:18 AM

@0ßelö7 $g_{\theta\theta}=1$

oh the phi component has a sine

ok sure, what's the issue?

apparently there is none :)

I don't want to check that. Do you get the wrong answer in the end?

I will compute

@JohnRennie Most of the groups I've been part of expect young students to give two rehearsals with time to improve in between. Once they done well at a couple of conferences the expectation drops to one unless you smell up the room with that first one.

Other group members critique and offer suggestions.

It doesn't always make for great talks but it always help shore up the biggest weakness and generally results in a marked improvement.

Bernardo Meurer

6:49 AM

15 Sorting Algorithms in 6 Minutes

@dmckee Look how Merge sort is beautiful

I'm pretty proud I did that in pure C on doubly linked lists :P

And that I managed to implement it from memory during the oral exam

I regret not having gone through the trouble or making it parallel

@BernardoMeurer acceptable footnote? i.gyazo.com/96c4eddc01dd4d91ce7f0ae8cb9e0a8a.png

Bernardo Meurer

@0ßelö7 Yeah, that's pretty acceptable

DeCSS was one of the first free computer programs capable of decrypting content on a commercially produced DVD video disc. Before the release of DeCSS, Linux-based computing systems could not play video DVDs. DeCSS was developed without a license from the DVD Copy Control Association (CCA), the organization responsible for DVD copy protection—namely, the Content Scramble System (CSS) used by commercial DVD publishers. The release of DeCSS resulted in a Norway criminal trial and subsequent acquittal of one of the authors of DeCSS. The DVD CCA launched numerous lawsuits in the United States in an...

The first paragraph of this shows why you should make sure your DRM scheme works on Linux

Linux is pretty communist

Bernardo Meurer

Live free of die :)

DeCSS haiku is a 465-stanza haiku poem written in 2001 by American hacker Seth Schoen as part of the protest action regarding the prosecution of Norwegian programmer Jon Lech Johansen for co-creating the DeCSS software. The poem, written in the spirit of civil disobedience against the DVD Copy Control Association, argues that "code is speech". == History and significance == DeCSS haiku was created in the context of a series of protests, coming from the international hacker community, against the arrest of Norwegian programmer Jon Lech Johansen, and a series of related lawsuits against him and other...

Motherflippers implemented CSS breaking in a haiku

Holy moly

An illegal prime is a prime number that represents information whose possession or distribution is forbidden in some legal jurisdiction. One of the first illegal primes was found in 2001. When interpreted in a particular way, it describes a computer program that bypasses the digital rights management scheme used on DVDs. Distribution of such a program in the United States is illegal under the Digital Millennium Copyright Act. An illegal prime is a kind of illegal number. == Background == One of the earliest illegal prime numbers was generated in March 2001 by Phil Carmody. Its binary rep...

And in a prime number

Goodness

7:04 AM

@BernardoMeurer what?

Bernardo Meurer

What what

@BernardoMeurer look at these margins arxiv.org/abs/1705.06981

Bernardo Meurer

@0ßelö7 Disgusting

Yuck

@BernardoMeurer look at the last page tho

Bernardo Meurer

What about it?

7:08 AM

@BernardoMeurer look where the first two authors are from

I can educate them

Bernardo Meurer

Ah, lol

Please do

on proper formatting

Bernardo Meurer

I like wide margins and 1.5 spacing on drafts

because it's easy to add notes on the side and to underline and circle

but it's pointless on final docs

it just impairs reading

@BernardoMeurer 10 point font looks best

cf. i.gyazo.com/e010b2a136f00094f4ef7ff10e4c04bd.png

Bernardo Meurer

It's part of why I hate MLA

It just looks disgusting

I mean

Times New Roman 12pt, seriously?

Are you guys TRIPPING

7:14 AM

@BernardoMeurer Sadly those are just seminar notes. That proof looks really nice

Good equation/word ratio

Bernardo Meurer

No wonder people think the humanities are a meme

@0ßelö7 Prove Heine-Cantor

I have no idea what that is

Bernardo Meurer

"Every continuous function has a primitive"

$\int_{x_0}^x f(y)\, dy$

there

Bernardo Meurer

My book takes like two pages

7:18 AM

@BernardoMeurer shrug

I gotta sleep

cheerio

Bernardo Meurer

Seeya!

1 hour later…

Bernardo Meurer

8:47 AM

@JohnRennie Please tell me you're here

@BernardoMeurer I'm here!

Bernardo Meurer

copy.law.cam.ac.uk/cam/tools/request/…

@JohnRennie See if that link works for you

user image

Bernardo Meurer

Bah

I need a piece of archaic UK legislation

I need a copy of the Copyright Act 1710

from, well, 1710

D:\temp>telnet copy.law.cam.ac.uk 80
Connecting To copy.law.cam.ac.uk...Could not open connection to the host, on port 80: Connect failed

That server's not listening ...

Bernardo Meurer

8:52 AM

CALL THE QUEEN

Try: copyrighthistory.org/cam/tools/request/…

Bernardo Meurer

That one has the text though, I need something with just the original manuscript. I am trying to get revenge on a professor

By only citing manuscripts that are hard to read

Much like when I delivered all of my homework in x86 Assembly

But for my composition and literature class

@JohnRennie Are you not at least midly entertained?

@BernardoMeurer Perhaps you have the only professor in higher education who has never had that trick played on them before. Or perhaps not :-)

Bernardo Meurer

Oh come on

You can't possibly think other people are willing to go to this extent just for a little revenge

It's taking me all day!

I've been reading manuscripts for like 19 hours

because my paper is really good (IMHO) it just only cites early manuscripts and things like that

I think it will be quite unique :P

:-)

Mithrandir24601

9:10 AM

@BernardoMeurer This is both fantastic and hilarious :D

Bernardo Meurer

@Mithrandir24601 I do not got light on the professors I don't like :P

I am willing to put in a lot of effort to produce great work that is also horrible

Like now, writing about DRM and citing 18th century british legislation in manuscript :P

Mithrandir24601

Print off the PDF file and give that to him as the reference?

copyrighthistory.org/cam/pdf/uk_1710_1.pdf

Bernardo Meurer

Lol

This is what you get for taking points from me for "lack of clarity" when I cited Hegel's Philosophy of Right

10:03 AM

"A catastrophic region is composed of catastrophic points"

I love GR terminology

10:48 AM

@JohnRennie Centre of Mass is defined as $$\int \frac{1}{M}r^2 dm$$ right?

There's no derivation of the formula right?

I know that formula. My main question is there any derivation of it too?

I don't think there's an $r^2$ in there.

Oh yes. You are right.

$r^2$ would give the moment of inertia ...

I mean I did many rotational inertia problems yesterday so got too used to writing r^2

@JohnRennie Lol, I was typing that

Can you answer the main question?

How to derive it?

10:55 AM

@JohnRennie Today my teacher derived it...I argued with him that it's defined that way "why are you deriving it". He snubbed me and told me that I am wrong. Nobody saw a dream of this formula, he said.

I suspect your teacher is correct. It can be derived.

How?

He derived it using mass moment..

Suppose you put your object in a gravitational field. The centre of mass is the point abut which the object won't rotate.

Then even $F= ma$ can be derived from $F = dp/dt$ but that day semi said one step doesn't make a derivation.

yes..

That is, if you pivot your object around the centre of mass then it wouldn't rotate in a gravitational field.

10:58 AM

yes

And the condition for the object not to rotate is that all the torques on it must sum to zero.

$\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} )=0.$

$\mathbf {R} ={\frac {1}{M}}\sum _{i=1}^{n}m_{i}\mathbf {r} _{i},$

Right?

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