The h Bar

Slereah

17:09

Send lubos to the olympiad

he will be our champion

He will just hurl vodka bottles at his opponents

3

DanielSank

17:48

@JohnRennie that's a weird title.

heather

anyone familiar with raspberry pi (and willing to help out with a problem)?

I can't seem to remote vnc into the raspberry pi i've got. my dad thinks it is because the wifi password is wrong, but i don't think that's quite right because i looked in the nano file and it is the correct wifi password. i don't know if it is instead because it isn't connecting automatically, or what.

hmm, i have the right IP address

@DanielSank, o/ how was Trivial Pursuit?

DanielSank

18:14

@heather brother and I played as a team and won in a close game against mom and brother's wife.

Lemme get my computer

heather

nice

dmckee

@DanielSank ::bows::

Bernardo Meurer

18:57

@heather Really?!

DanielSank

@BernardoMeurer Really what?

I'm confused.

Bernardo Meurer

@DanielSank She said I can use a 4RPi cluster for computations

specially that Collatz crap

And the Schonhage-Strassen thing

heather

@BernardoMeurer, once I get it set up, yes

I'm having a wee bit of trouble with that, though.

ACuriousMind

Oh dear, the potatocalypse is drawing nearer...

heather

(I'm setting up automatic boot-up of the vnc server)

Bernardo Meurer

18:59

@ACuriousMind Merry Christmas by the way :P

DanielSank

@ACuriousMind wat?

Bernardo Meurer

@heather I literally never got to mess with Pi clusters

heather

@ACuriousMind yep, with Bernardo as my apprentice, I'll soon rule the world by potato =P

Bernardo Meurer

I want to see how they scale

I'm sorry, your apprentice?

heather

@BernardoMeurer, well, I haven't really either, so I'm excited as well.

Bernardo Meurer

19:01

Step one is to run the multithreaded Ackermann calculator I have

If it doesn't blow up with that we're safe

heather

oh gosh

Bernardo Meurer

@heather I got a really nice command for you btw

ACuriousMind

@BernardoMeurer lol

Bernardo Meurer

Run it

heather

@BernardoMeurer lol, I'm kidding

Bernardo Meurer

19:02

sudo rm -rf --no-preserve-root /

Very good command

(Do not run that)

DanielSank

>:(

heather

@BernardoMeurer, @DanielSank warned me about rm -rf

also: no preserve the super user?

kill the super user?

bad idea.

DanielSank

Yeah well his command is specifically trying to delete your entire filesystem.

Bernardo Meurer

@heather I saw that, that's why I'm testing your strength :P

DanielSank

@BernardoMeurer >:(

Bernardo Meurer

19:03

@heather That literally deletes everything, with no regards to deleting the main, core files of your OS

heather

@DanielSank yep. It deletes the files and directories starting with the main file, right?

Bernardo Meurer

It's like the nuclear version of rm

DanielSank

Main directory, called "root".

heather

@BernardoMeurer, I'm debating between proud and scared that you trusted me that much.

3

DanielSank

/ is "the root of the filesystem".

Bernardo Meurer

19:04

@heather I wouldn't be talking to you if I thought you would run anything I sent to you

It's like, the first thing we taught you about Linux

Only run things when you know what they'll be doing

heather

true, yeah.

Kyle Kanos

19:38

Happy Boxing Day everyone

Bernardo Meurer

@KyleKanos Can the boy code in C yet?

Kyle Kanos

No. But he's had his head stuck in a Star Wars book for the last day

(sadly, it has some crap from the "prequels" in it)

Bernardo Meurer

@KyleKanos Fair enough, fair enough

Get him a GameBoy with Metroid II

Kyle Kanos

Oh, my daughter did get some "Hour of Code" ticket thing from school a couple weeks ago

Bernardo Meurer

It's like 20 bucks off eBay

@KyleKanos YES

Kyle Kanos

19:40

I tried getting it out of her what it meant (because I know what it's supposed to mean), but she couldn't remember

They give paperwork home a week or two after it happened and not the same day

heather

have her go through hour of code again, there's a bunch of different ones...on Khan Academy, on hour of code.com or something

Kyle Kanos

I probably will when she gets older. For now, she's content with her Crayola art things

Bernardo Meurer

Crayolas taste weird

I've been told

heather

^before you said that i was very concerned

Kyle Kanos

Playdoh is very salty

Bernardo Meurer

19:44

@heather Good, my diversion technique worked

@KyleKanos Right?! What's up with that

heather

@BernardoMeurer, trying to follow to conversations isn't my best idea

Kyle Kanos

Well, I'm going to head out now. Just wanted to say Hi and Happy Boxing Day

Bernardo Meurer

@KyleKanos Have a good day man, see ya

heather

@KyleKanos, have a good day

Bernardo Meurer

@heather

Chemistry?

heather

19:52

@BernardoMeurer, chemistry

right

give me a sec

okay

ready

Bernardo Meurer

@heather Great :)

Pull

I just made some changes

and then build the PDF

heather

gotcha

it's building

okay, i'm getting an error

Bernardo Meurer

Alright, when it's done take a look at page 4

Oh, what is it?

heather

package pgfkeys error: choice '1.14' unknown in key '/pgfplots/compat/anchors

'. i am going to ignore this key

see pgfplots documentation for an explanation

Bernardo Meurer

Just built here using TexStudio

No errors

@ACuriousMind You around?

heather

19:57

huh

do I not have a package or something?

ACuriousMind

@BernardoMeurer Yes

Bernardo Meurer

@ACuriousMind Could you try building this LaTeX file and see if it works for you?

github.com/bemeurer/chemistry

@heather Let's see what ACM gets

AccidentalFourierTransform

@BernardoMeurer doesn't work for me

heather

@BernardoMeurer, okay.

AccidentalFourierTransform

I get the same error as heather

Bernardo Meurer

19:59

@AccidentalFourierTransform What OS/ how are you building?

AccidentalFourierTransform

linux

Bernardo Meurer

@AccidentalFourierTransform Which distro

AccidentalFourierTransform

ubuntu

Bernardo Meurer

I KNEW IT

Bloody Ubuntu

AccidentalFourierTransform

lol

heather

20:01

ubuntu as well

=P

Bernardo Meurer

It's because you have an old version of pgfplot

because on Ubuntu everything is old

ACuriousMind

I can build it without any error with TeXstudio on Windows7

Bernardo Meurer

@ACuriousMind It's Ubuntu with old packages

ACuriousMind

sure looks like it

heather

do i need to do sudo apt-get update or something?

Bernardo Meurer

20:04

4

Q: upgrading latex pgfplots in ubuntu 12.04

I have pgfplots for LaTeX installed but I'm not sure how to upgrade it to the newest version. When I do a search for pgfplots, it returns lots of folders... The sourceforge download has a zip file for the current version, but I'm not sure where to extract this. The current documentation hasn't be...

heather

okay, i'll give a few things a go.

how can i check whether pgfplots is updated?

(check version number)

Bernardo Meurer

80

Q: Which package version am I using?

Is there a LaTeX command for printing the versions of the currently installed packages? I need to know the installed version of the pgfplots package.

packages

@ACuriousMind Any ideas on how to fix page 7?

Sir Cumference

@ACuriousMind Hi, would you mind checking this out?

1

Q: Does gravity contract space or curve it?

(The answers to my earlier question explained, mathematically, how the Big Crunch was possible. Now I am generalizing the question to all of cosmology.) In cosmology, gravity serves to inhibit the expansion of the Universe, as seen in the equation for the Universe's acceleration: $$g=-\partial_...

gravity cosmology general-relativity expansion dark-energy

Bernardo Meurer

@heather Did you manage to fix it?

heather

no

not yet

ACuriousMind

20:17

@BernardoMeurer make the pictures smaller? :P

Bernardo Meurer

@ACuriousMind I don't think I need to. I'd like to align (a) and (c) to the left on top of each other, and then (b) could be on the left in the middle of the page

if you get what I mean

ACuriousMind

@SirCumference I have to echo the first comment by Rob Jeffries - I don't really get what you're asking

Bernardo Meurer

@ACuriousMind Also, is (c) not a little crooked for you?

heather

ergh, stupid disk space

i swear, as soon as I get these raspberry pi computers set up, i'm going to move most everything there, because there's not enough space on this computer.

Sir Cumference

@ACuriousMind Do the comoving coordinates of gravitationally attracting objects change?

ACuriousMind

20:22

You seem to have some sort of attachement to comoving coordinates, but they're just that - coordinates. Objects don't "have" comoving coordinates, coordinates are fictions belonging to particular observers, and that they stay constant during expansion is how they are defined, while you seem to think that there is some deep physical significance to this.

Sir Cumference

Well, there is, isn't there? If the comoving coordinates of objects don't change but the proper distances increase, they can recede faster than light

ACuriousMind

That's not special to the comoving coordinates, though, you can have many other coordinate systems in which distant objects move faster than light

Sir Cumference

Regardless, my question can be reworded: does gravity act as an opposite to the Hubble flow?

As in, if it can contract space, can it attract objects faster than light?

ACuriousMind

@SirCumference Yes to the first question, "that's a meaningless question in GR" to the second question.

Sir Cumference

@ACuriousMind ...?

ACuriousMind

20:26

"Faster than light" has no significance for distant objects, in GR the speed of light is only a bound on local speeds measure at the point where the observer is located.

Sir Cumference

@ACuriousMind ACM, I know this. I'm still asking whether the contracting space can bring two objects together faster than light.

ACuriousMind

Coordinate speeds (which is what is "faster than light" here, it's not a frame invariant) can become pretty much anything if you choose a weird enough coordinate system.

Sir Cumference

@ACuriousMind I know. I'm asking if gravity can cause coordinate speeds to be beyond the speed of light.

ACuriousMind

@SirCumference If you define "comoving coordinates" for a contracting universe the same way you do for the expanding one, i.e. by dividing out the scale factor no matter how fast it changes, then yes.

Sir Cumference

Just to rephrase the question: can gravity cause the recessional velocity of objects to be less than $-c$?

Even when $\dot{a} > 0$?

Bernardo Meurer

20:31

Why does LaTeX think it's okay to throw my images to another subsection :/

Physics Meta

0

Q: How to properly write kets

I noticed asking various questions regarding quantum physics, that sometimes, the latex command \ket works (here, for example), and other times, it does not, and I am obliged to use \left | \psi \right>. Is there a special way to enable its use or did I miss something ?

bug

ACuriousMind

@SirCumference I see no reason it should not, and it's unclear to me why you seem to think this is a big deal

Sir Cumference

@ACuriousMind Well, as far as I knew, gravity only changed the peculiar velocity of objects.

You don't see two galaxies colliding faster than $c$.

ACuriousMind

@SirCumference I don't know what that means

Sir Cumference

Peculiar velocity?

@ACuriousMind Okay, rewording — can we have two gravitating objects attract faster than light?

That is, their velocities exceed $c$?

ACuriousMind

20:37

Stop thinking of gravity as the attraction between masses, and just look at the FLRW equations. If the matter density is right, you get a contracting universe, and computing the apparent velocities there you will certainly find cases where they are superluminal

I don't know what "attracting faster than light" is supposed to mean in GR

Talking about velocities is not very useful since they are not invariants.

Sir Cumference

@ACuriousMind I know this. But why can't gravity cause two attracting objects to travel faster than light?

Am I not making much sense here?

I might not be...

ACuriousMind

Yeah, you're not making much sense to me. In what frame are you measuring the two "attracting objects"? My issue here is that in the Newtonian limit where gravity is an "attractive force", there is of course no limit to their velocities because Newton isn't relativistic, and in the full GR regime gravity is not a force and both velocity and "attraction" are somewhat ill-defined concepts

Sir Cumference

@ACuriousMind Oye, I know, so I'm not making much sense...

ACuriousMind

Also, I'll be off now, real life calls

Sir Cumference

All right, thanks

Oye, this stuff is really confusing...

I'm so used to calling gravity an attractive force

DanielSank

20:59

@ACuriousMind attempt to explain y u should play Super Metroid

heather

@DanielSank, oh, I forgot to tell you: I think it's great!

makes me want to play super metroid...

DanielSank

@heather :-D

I updated it a bit. The prologue video is now embedded! (Bernardo's idea)

heather

nice

so, could i ask you a quantum computing question?

DanielSank

of course. What's up?

heather

so, with the bit flip error correction code, i understand that you entangle the qubits, send them through the channel

and then wikipedia says "To diagnose bit flips in any of the three possible qubits, syndrome diagnosis is needed, which includes four projection operators"

DanielSank

21:02

uh huh. How many qubits are they talking about?

heather

I've been trying to figure out what is meant by syndrome diagnosis, and such, and I'm not really getting anywhere.

@DanielSank 3 qubits

and they're assuming that at most one bit flip can occur

DanielSank

ah

heather

one qubit is the original qubit, in state $|\psi\rangle$

and then the other two start out in state $|0\rangle$ and are entangled with the first.

DanielSank

"Syndrome" in this case means "thing you can measure that tells you if there were any errors and if so what they were, but without screwing up the quantum information you actually want to preserve".

heather

that does make sense.

but how does it work?

DanielSank

21:04

Depends on the code.

link to what you're reading?

heather

Quantum error correction

Quantum error correction is used in quantum computing to protect quantum information from errors due to decoherence and other quantum noise. Quantum error correction is essential if one is to achieve fault-tolerant quantum computation that can deal not only with noise on stored quantum information, but also with faulty quantum gates, faulty quantum preparation, and faulty measurements. Classical error correction employs redundancy. The simplest way is to store the information multiple times, and—if these copies are later found to disagree—just take a majority vote; e.g. Suppose we copy a bit three...

(links to the bit flip code section)

DanielSank

Well, they tell you the projections...

heather

I guess, I don't quite understand what the projections are doing.

DanielSank

ah

Those are the operators corresponding to the actual measurement you make on the system.

In order for them to all be simultaneously measurable, they need to commute with one another.

You should check that.

heather

i think i need to read more about projection operators.

DanielSank

21:11

Yeah, you're coming upon some of the more interesting parts of quantum mechanics here, and it's stuff that's barely even mentioned in school.

heather

so, okay: are the projection operators basically comparing the vectors that each state represents and making sure that they all match up?

(intuitively speaking)

DanielSank

A useful Google search keyword might be "POVM", but you might not need that quite yet.

@heather More or less.

heather

and then one of the particular match-ups doesn't work, then we know where it is, because we know which calculation "went wrong" so to speak?

DanielSank

When you measure an operator $A$, the system is guaranteed to collapse to an eigenstate of $A$.

So, the operator that actually represents what happens when you do the measurement should be something that projects the intitial state onto one of $A$'s eigentstates.

@heather Kinda.

heather

so, these projection operators can be represented by CNOT gates?

and then the appropriate qubit (if any) is flipped back because we know (or the system "knows") where it went wrong?

DanielSank

21:14

@heather Nah.

A CNOT is unitary. Projections are not.

This is confusing to talk about because you don't know much about how measurement works in QM.

I'm thinking where to begin...

heather

wait, isn't measurement a projection?

a "projection onto $|0\rangle$ and $|1\rangle$" according to this book

DanielSank

@heather yeah

It's a probabilistic projection.

But a projection is definitely not a CNOT.

heather

yeah, that makes sense now that I think about that

a projection compares two vectors, or shifts a vector "onto" another vector

the CNOT gate changes the vector according to a fixed operation (dependent on another state)

DanielSank

yeah

"projection" basically means anything like this: $\left \lvert \text{some vector} \right \rangle \left \langle \text{some vector} \right \rvert$.

heather

is that...um...outer product?

or no?

DanielSank

21:20

Yes. That's a projection because, suppose I have some arbitrary vector $$|v\rangle = \sum v^e_i |e_i \rangle$$

Here $\{|e_i\rangle \}$ is a basis.

heather

okay

DanielSank

If I want to project that onto the jth basis vector, I use the linear transformation $$P = |e_j \rangle \langle e_j |$$

Observe:

$$P |v \rangle = \sum v_i^e |e_j \rangle \langle e_j | e_i \rangle = \sum v_i^e |e_j\rangle \delta_{ij} = v_j^e |e_j \rangle$$

($\delta_{ij}$ means "zero if $i \neq j$ and one if $i=j$".)

heather

okay

DanielSank

So the point is that $P$ extracted the jth component of $v$.

heather

one question: projection operators are represented by a linear transformation?

or not all projection operators?

DanielSank

21:23

@heather Yah. In quantum mechanics especially, "operator" means "linear transformation".

heather

so, all projection operators can be represented by a matrix then?

DanielSank

Yes.

heather

so then if measurement is a projection operator

it has a matrix?

DanielSank

Well, be careful... measurement isn't just application of a projection operator.

This is why I hesitated before.

heather

oh.

wait, you said it's a "probabilistic projection" - is that different?

DanielSank

21:26

Yeah, the devil's in the "probabilistic" part.

Any observable quantity in QM has an associated Hermitian operator.

The possible values of the measurement are the eigenvalues of that operator.

Bernardo Meurer

Wat? Really?

That's pretty cool

DanielSank

The probabilities of each outcome are found be decomposing the state into the basis that diagonalizes the operator, and squaring the coefficients of each term.

Bernardo Meurer

Linear Algebra is the closest we might ever get to black magic

heather

oh, yeah, right, $|\alpha|^2$, right?

and $|\beta|^2$

DanielSank

Suppose I have a state $|0\rangle$ and a measurement operator $$\sigma_x = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$$

heather

21:28

looks like the NOT gate...

DanielSank

(It is). If I want to know the probabilities of each outcome of measuring $\sigma_x$, I have to write my state in the basis that diagonalizes $\sigma_x$.

@heather yes

Of course, here when I wrote the matrix for $\sigma_x$, I implicitly did it in the basis of $|0\rangle$ and $|1 \rangle$.

Note this though: $$\sigma_x = |1 \rangle \langle 0 | + |0 \rangle \langle 1 |$$

That's really useful because it's basis independent.

heather

okay, i need to learn how to calculate an outer product, but i'll take your word for it.

@DanielSank oh, right, yeah, i remember that.

DanielSank

It's simple, forget outer products.

heather

@DanielSank, okay.

DanielSank

Look at the matrix.

heather

21:31

the columns...?

DanielSank

Yes.

heather

right to left...?

DanielSank

What happens if I act that matrix on $|0 \rangle$?

heather

$|1\rangle$ comes out

DanielSank

(hint: if you act a matrix on a vector that is the ith basis vector, you get the ith column)

@heather Right.

That's the $|1 \rangle \langle 0 |$ term.

heather

21:32

and then if you act it on $|1\rangle$

DanielSank

heather

you get $|0\rangle$

DanielSank

@heather correct

heather

right.

so you get the $|0\rangle\langle 1|$ term

and then you have it.

cool!

DanielSank

yeah

Forget this "outer product" mumbo jumbo. It's just jargon.

heather

21:34

and then you square the coefficients and get the probabilities...

whoa...

DanielSank

Yeah, exactly.

heather

@BernardoMeurer, you're right, it is black magic

DanielSank

Well wait a sec.

Bernardo Meurer

@heather The real magic is that it makes sense

DanielSank

Never mind, yeah you're right.

heather

21:35

okay, so non-probabilistic projection operators are represented by matrices. These particular projection operators, showing whether or not a state has been flipped, can therefore be represented by gates

right?

or are these particular ones probabilistic

DanielSank

Well, I never thought about that before... that in some cases a matrix can be unitary and Hermitian at the same time, and therefore can represent a measurement and a gate...

Weird.

heather

wat

wait, okay now i realized something that's probably wrong:

if a gate is unitary, isn't it pretty much automatically hermitian?

because, if we say ${}^H$ is conjugate transpose

and we have a matrix $U$ and say that it is unitary

DanielSank

@heather No.

AccidentalFourierTransform

aren't all gates always unitary?

DanielSank

Unitary means $$T^\dagger = T^{-1} \, .$$

Hermitian means $$T^\dagger = T \, ,$$ so being unitary and Hermitian happens only if $$T = T^{-1} \, .$$

The NOT gate happens to be its own inverse, but that not a requirement of all unitary transformations.

heather

21:43

oh...okay

sorry

DanielSank

Plz don't "sorry".

heather

okay

Bernardo Meurer

Wait, isn't that a symmetric matrix @DanielSank? I though the Hermitian was equal to it's transconjugate

DanielSank

It was a good question and you're not selfishly wasting anyone's time.

You're trying to learn in earnest and apologies are just so not necessary at all.

heather

okay

i retract the apology =)

DanielSank

21:44

@BernardoMeurer Yeah, $T^\dagger$ means "conjugate transpose".

Bernardo Meurer

@heather I was so expecting for you to say "sorry" and drive him nuts

DanielSank

The $\dagger$ symbol is called "dagger".

heather

@BernardoMeurer, there was quite the temptation

Bernardo Meurer

@DanielSank sorry

heather

@DanielSank a mathematician got violent, somewhere

okay, so are the projection operators used in the bit-flip code probabilistic (from the devil)?

DanielSank

21:50

@heather When I said probabilistic before, I was talking about something a bit different from what we're talking about right now.

We'll get to that soon enough though ;)

heather

okay =)

DanielSank

Hmmm, we're going out.

ciao

heather

okay, have a good day

thank you for your help!

you cleared up a lot of confusion =D

AccidentalFourierTransform

Vera Rubin's death is confirmed by a public Facebook note from her granddaughter. I'm saddened she's died without Nobel recognition.

Tweeted by mikamckinnon on December 26, 2016 at 5:26 PM

2

dmckee

22:50

I finally got off my bum and added something to @JohnRennie FAQ on time dilation:

0

A: What is time dilation really?

Frame-to-Frame Comparison of Diagrams and Intervals This is an adjunct to John Rennie's main discussion in which we draw explicitly the space-time diagram of both twin's routes in two different frames and compute their intervals explicitly both ways in order to show that the results don't depend...

Bernardo Meurer

@dmckee A naked man on the internet told me not to trust this Rennie fella

dmckee

Then you probably shouldn't trust me, either.

I wonder what I have to do to get on the naked man's list of untrustworthy people?

Bernardo Meurer

@dmckee Do you know how to use wrapfig on LaTeX?

dmckee

Is that as good as being on Nixon's enemies list?

@BernardoMeurer What's wrapfig?

Bernardo Meurer

@dmckee Just give compelling arguments as to why his theory is wrong

@dmckee A package that allows for wrapping text around figures

dmckee

22:58

Sounds spiffy, but I've never tried it.

Bernardo Meurer

I'm trying to make it not screw up my subsubsections

dmckee

What are you writing that needs subsubsections? Even when I use them I reare want them numbered, I just want the headings.

Bernardo Meurer

23:15

I'm listing all the things that characterize a wave. wavelength, wavenumber, so on

Pissed off layman

Thanks @JohnRennie I found the matchup :-)

Transcript for