@0celo7 did u finish the lab report

nope

due in 2 hrs

As time goes on, I hate science memes more and more

Neil Tyson must die

yes!

@ACuriousMind Does boolean algebra form a vector space?

@BernardMeurer I don't understand the question

@BernardMeurer wat

wtf is "boolean algebra"

(in this context)

An algebra of boolean values

well if it's an algebra, what's the issue?

I think it does

Boolean algebras do not form a vector space, since boolean values do not form a field

If something is an algebra then there is always an underlying vector space

That's one of the axioms---but if it is not an algebra then there is not reason to connect it to vector spaces.

Varying definitions of algebra, danu

@Slereah How come it doesn't form a field?

@Danu highly depends on your definition of "algebra". In ring theory, an algebra may just be a ring with a homomorphism from another ring.

Well it's just two values

@ACuriousMind nopenopenope

GEOMETRY ONLY

> $\Bbb C$ is the only ring

to be a field doesn't it have to be associative, have an identity, & be closed under +,x

& inverses

@Slereah Why are two values not a field?

@Danu $C^\infty(M)$ is crying

@0celo7 Okay, okay

All smooth things allowed

:D

Oh wait, fields can be finite?

I guess it can be a field then

I dunno

@Slereah of course

Let's see

$\Bbb F_p$

$\wedge$ and $\vee$ are associative and commutative

@Slereah Yeah, in my head that makes sense

@Jim : your local measurements are not always a reliable indicator of what's actually happening. You continue to see your optical clock ticks normally. But when that gravitational wave comes through, they're going slower. Don't think they aren't because you're going slower too.

Truth is an additive element

And a multiplicative identity element as well

$\neg$ is an inverse

And it is distributive from De Morgan's law...

I guess it forms a field?

any field is trivially a vector space over itself

Wait

How could true & false have an identity?

Can you really have additive and multiplicative inverses of booleans?

what are the operations

Yeah I think

@Obliv OR, AND, NOT

@Obliv : $a \wedge T = a$

$a \vee T = a$

Wait no

The identity for or is false

Hm, let's see for inverses

$x+0=x$

$a \vee \neg a = F$

@JohnDuffield So which privileged frame then would you suggest is the correct frame? I say taking local measurements makes sense because that tells you what it is in your frame and location. Taking measurements from another frame is dubious unless you have a good reason to choose that frame.

That's... not true?

$x\cdot 1=x$

Is there a close reason to be used when someone apparently inadvertently asks a question based on a false premise?

Wait, F is the multiplicative identity element

So it's T

It checks out

I've seen Unclear used, but I don't know if that's right.

And... $a \wedge \neg a = F$

Yeah I think booleans form a field

^ Vote up

More importantly

I am investigating brewing alcohol

@Slereah So if booleans form a field do they form a vector space?

Gonna try to make some cider if I can get some proper equipment

@Jim : there is no correct frame. A frame does not actually exist. But those pulsars do, and so do your optical clocks. These are remote clocks and local clocks. When their relative rates change you know something is up. And it doesn't take you long to appreciate that all those pulsars didn't change.

I guess

@BernardMeurer you should check that any field naturally becomes a vector space over itself

@0celo7 Define check

Interestingly by the way

PROOF

There is a boolean derivative

@0celo7 Ah, okay, will do

It has little to do with actual derivatives

@Slereah what does that even mean

It's not related to topology or whatever

@0celo7 Will you teach me how to write maths today?

@BernardMeurer If you meet me in real life

I need it for this

It's just a boolean operator that you can apply to another

I say "proof" with an indian accent now

@0celo7 :/ okay

for....reasons

@Slereah I have experience in that.

@BernardMeurer you misunderstood

And it is = 1 if it changes signs or 0 if not

that was not a response

It is mostly used in circuit design

@BernardMeurer Not now. I watched the stupid debate last night and didn't get my lab report done

@DanielSank But I found scirate rather useless

doing it now...

BYE

There is Some Law of circuit design that you can only build circuits from MOS gates if the derivative is something something

@Danu Ok.

@JohnDuffield sure, but you aren't measuring the pulsars (at least, not in the hypothetical situation I posed that seems to have led to this). You measure the gravitational wave and you measure it when it's local. If you had a way of measuring it indirectly when it was at a random point in space, it wouldn't make sense to do so without converting those measurements into a local frame.

I don't remember too well

@Danu why did you find it useless and how could it be improved in your opinion?

@DanielSank It seems like nobody in HEP is using it.

@Danu Ah, a popularity / userbase issue.

Those can only be solved by more people adopting.

::shrugs::

I say ban everyone and let's restart from scratch

Perhaps some way of exposing the site to a larger audience... some form of visual display... an "advertisement", if you will...

hahaha

We say "In the pulsar's frame it is.....", which means we are making a local measurement of something around the pulsar. We don't say "In our frame, gravity waves around the pulsar are moving faster than $c$". Sure, it could be true, but who cares? The local measurement around the pulsar is the important stuff

So a few weeks back

I tried to write a little GR raytracer

For fun

So it turns out it's pretty bloody hard

@Jim : measuring the pulsars is measuring the gravitational wave. See articles like this. You're using those pulsars to work out that you and your local optical clocks are running slower. This is what gravity is all about. If optical clocks didn't run slower when they're lower, your pencil wouldn't fall down. And you can't measure your local clocks running slower using local clocks.

Without straight line geodesics it's hard to find out what is actually illuminated by a light source

At least not without doing tons of raycasting

@Danu You laugh...

So much for my idea to do a game in curved space

@JohnDuffield I am aware of the physics. Stop changing the subject. This isn't about clocks running slower, this is about the utility of reporting measurements in an arbitrary non-local frame. That is, local to the measured phenomenon, not necessarily local to us

Might be doable for very specific spacetimes, I dunno

Maybe doable for hyperbolic space

@DanielSank Why do you care about pushing SciRate, anyways?

@Slereah How do I prove that something forms a field

@slereah what's a 'raytracer'

Prove that its operations obey the axioms of a field?

@BernardMeurer Check the axioms!

A raytracer is what is used for generating 3D graphics

@Jim I humbly submit my opinion: As soon as "stop changing the subject" happens in a conversation with @JohnDuffield, it is courteous to the rest of the chat to either desist or move the conversation out of hbar.

@Danu Ah, brainfart. Will read on it :)

@BernardMeurer Because it could be interesting to use is as a crowdsourced review platform.

Every pixel on the screen is made from a ray that bounces on the various 3D objects until it reaches a light source

@Danu people.reed.edu/~mayer/math112.html/html1/node16.html Proper reference?

@DanielSank Wat?

I'll be talking with the devs on Thursday about tweaking the UI to support "issues".

@DanielSank I'm willing to wait for his reply and judge based on that

@Jim Fair enough, but let's keep established patterns in mind.

@DanielSank devs? Review platform? What's going on

@slereah light source?

I'm not advocating censorship, by the way, just that we've seen these nontechnical back-and-forths enough to know that they're rarely helpful.

a "lamp"

if u will

ah, I see. no I don't. So what would you be programming

@BernardMeurer arxiv is the de-facto place for preprint articles.

@DanielSank Which article?

@BernardMeurer Here's an example, one of my own papers

You see the title, abstract, pdf download link, and version history.

Great.

@DanielSank you keep replying to Bernard instead of me :P

@DanielSank Did you mean to ping @Danu regarding this?

@DanielSank But why did you mention me there?

@Jim : I'm not changing the subject. Pulsars have been suggested as a way of measuring gravitational waves. Restricting your measurements to local measurements only does not always give you the correct picture. When you don't measure a difference, it could be that both the thing you're measuring and the thing you're using to make the measurement have both changed. When you do measure a difference, it could be that the thing you're using to make the measurement has changed.

@Danu So, was that reference correct?

Those pulsars provide the extra information you need.

Yes, I meant to reply to @Danu. Sorry.

@DanielSank Is drunk

@DanielSank lol, I was pretty lost

I've said many times that I suck at chat.

Now we believe you...

@BernardMeurer Sure

@DanielSank ( ͡° ͜ʖ ͡°)

I don't think DS should be a mod -- he might accidentally ban someone on chat.

Anyway, @Danu, I think if scirate had an issue tracker for each article it could be useful.

@0celo7 lolwut?

@DanielSank will accidentally delete PSE while trying to mention @Danu

Also, I have no intention of being a mod.

@DanielSank issue tracker?

@Danu Yes.

Wasda?

In software, an issue tracker is essentially a comment system with a bit of metadata.

Oh, issue as in issues with the paper, or what?

An "issue" is a comment explaining a specific, fixable problem with the subject matter, which in this case would be an article.

(Usually issues are code bugs, or desired features, etc)

@slereah is it something like this? youtube.com/watch?v=FIiwGa2l9X4

No

So, folks could file issues saying e.g. "The x-axis on Fig 2a is missing a label". Then the author goes and fixes it and marks the issue as resolved.

@Jim : let's leave it at that Jim.

This is how everyone develops software. It's a time tested and proven workflow that is efficient and makes everyone happy.

@DanielSank Yeah, in principle that could be nice. But it's very very far from being well-known enough to do anything like that

^ Well, the only way to change other people's habits is to lead by example.

In mathematics, there is a very high quality but behind-closed-doors alternative in MathSciNet.

@0celo7

Every paper on there is professionally reviewed---not exactly what you're talking about but in the general direction.

@Danu Sounds great.

Of course, one cannot hope for such high quality for free---but an issue tracker could be interesting.

@Danu Yes. Exactly.

The main issue I see is that authors will probably not be willing to actually work on fixing small things in their papers.

In the HEP community, at least, it seems to me that people generally couldn't care less :P

I used github for my most recent paper. Three authors could all work together in a sane way, and we got issue tracking. The issue tracker was an amazingly useful tool, even for scientific writing!

@Danu That's their decision.

@DanielSank Writing is a different matter, I think that using something like an issue tracker (but more clumsily implemented) is already widespread in the writing process.

However, if someone uses issue trackers, they'll find (as I did) that it makes your paper better. If that happens, then you've got natural selection pushing more people to use it.

I could be wrong, of course, and that's totally fine. Just want to try and find out if it works.

Yeah, fair enough. Good luck :)

@Danu Emphasis on "more clumsily implemented".

@Danu Thanks. I'm talking with one of the devs on Thursday. He likes the idea but we have to discuss further.

Elon Musk is going to announce the SpaceX Mars project

@DanielSank Haha, that'll be funny.

@vzn just got a new AMA guest: C. Jess Riedel.

his blog

and CV

Neat!

@BernardMeurer I don't even known who that is

But I will watch it later.

@0celo7 How do I prove that every field forms a vector space?

Do you have a link?

@BernardMeurer it's an axiom

@Danu No, you have to prove it

It's easy, but not an axiom

Just take the definition of a vector field, and pick the elements of the field as the vectors

@BernardMeurer Just check that a field satisfies the definition of a vector space over itself.

Like Sam said, take the vectors to be the elements of the field.

Then it's pretty trivial to show that a field is linear wrt itself etc etc

@0celo7 What's a vector space over itself?

And the scalars are still the field elements.

@BernardMeurer $k$ a field, then $k$ is a $k$ vector space.

R is a vector field with R as its field

@Slereah I get that definition, but not @0celo7's

wait

@BernardMeurer Do you know what it means for $V$ to be a $k$ vector space?

Vector Field vs Vector Space

What's the difference?

@0celo7 Nope

Vector field is analysis/geometry

Don't worry about it now

@0celo7 You may find this room interesting.

inb4 element of $\Gamma(TM)$ @Slereah

Emphasis on "may".

@0celo7 In my algebra book it was an axiom.

@Danu What was? That a field is a vector space over itself?

Oh, wait, you're right @0celo7

@Danu That happens a lot :)

He didn't take it as an axiom---I mixed some stuff up in my head

@BernardMeurer The field we're talking about has nothing to do with vector field.

A vector field involves some kind of manifold (implicitly or explicitly).

I will argue before the rise of 3D printers, nobody will thought about making the replicator

@DanielSank cant hit his site yet but thx so much for suggestion. can you get him to sign up to the site? have you talked to him about it? etc... DZ said once he can arrange 1-rep new users chat access.

@vzn He's already a site user.

3D printers are way old man

They just used to not be called 3D printers

It used to be called prototyping machines

It's like 30 years old

@DanielSank ok credentials look fantastic! fine w/ me! do you know him? talked to him about the sessions? plz hand him meta links re sessions at least... plz have him join room asap & write anything... (cant ping him until he does...) DZ is talking about next chat session next tues as mod focused but maybe 2wk after that? oct 18th?

@vzn Yes, I know him. Yes, I talked to him about the session, as I already said. No, I will not have him join the chat room; that's up to him.

@DanielSank we're scattering something off of a step potential $V_0\Theta(x)$ centered at the origin

and the energy of the wave packet is $E>V_0$

@0celo7 3D?

no

1D

Then what does "centered at the origin" mean?

@DanielSank the step is at $x=0$

$\Theta$ is a well defined function.

Yes, ok what you said is redundant.

Go on.

I know it is

we have region II, which is where $V(x)=V_0$

so that's $x\ge 0$

Yes, I can read $V_0 \Theta(x)$.

Go on.

there we write $k_2=\sqrt{2m(E-V_0)/\hbar^2}$ and we write the energy eigenstate as $$u(x)=A_2e^{ik_2x}+A_2'e^{-ik_2x}$$

the prof then said $A_2'=0$ without any real explanation

It's a boundary condition.

You have an implicit time dependence $\exp(-i \omega t)$, right?

yes, if you want to add time-dependence to a stationary state you multiply by that

So then the $A_2'$ term is leftward moving. That can't exist if your boundary condition is that you have a source on the left but not the right.

Why not?

why can't the transmitted wave packet periodically send stuff back to the left?

@0celo7 Please use sentences where the antecedents are easy to identify.

Why not what?

@0celo7 You want math or physics answer?

@DanielSank I won't be convinced by a physics answer.

@0celo7 Ok, then the math answer is that differential equations are meaningless without boundary conditions. One of the boundary conditions in this problem is that the amplitude of the leftward moving wave has to vanish at $x \rightarrow \infty$.

If you ask why that's the boundary condition, the answer is "Because that's the boundary condition on this differential equation".

If you want to justify that, I refer you to physics.

@DanielSank I don't understand how either one is left or right moving

there is no time dependence

@0celo7 It's implied.

There's a $\exp(-i E t / \hbar)$ hiding everywhere.

if both have the time dependence $e^{-i\omega t}$, then both are moving the same direction

False.

explain

$\exp(i (kx - \omega t))$ is moving to the right (if $k > 0$).

Ok, @0celo7 imagine you have a wave $\psi(x,t)$. Freeze that wave at a particular point in time.

Say it's time $t=0$. That frozen time, the wave has a spatial shape which we write $\psi_{t=0}(x)$.

Ok?

yes

Ok now suppose the wave is moving right.

...at some speed $v$.

Now we freeze things again at time $T$. We now have a new spatial curve, $\psi_{t=T}(x)$.

Ok?

I don't have much time @0celo7. Are you still here?

@DanielSank yes

Ok do you follow?

@DanielSank what does that mean?

@0celo7 It means that $\psi_{t=T}(x) = \psi_{t=0}(x-vT)$ :)

yikes

What?

It means that the wave shape moves to the right.

yeah maybe but this is confusing

@0celo7 Why?

It just is!

You know that $f(x-x_0)$ is a rightward shifted version of $f(x)$, right?

So now the shift is proportional to time! That's what "moving to the right" means.

@DanielSank Yes, but that's very counterintuitive.

@DanielSank I'm not doubting what you're saying.

I just dislike it.

Why do you not like it?

I just don't like that $f(x-y)$ is a rightward shift.

The one thing I find confusing is that a wave is a function of two variables, i.e. $\psi(x,t)$.

@0celo7 Ok, why?

Is it because this is yet another example of physicists confusing variables and functions?

$f_\text{shifted by y}(x) = f(x - y)$

No, although I'm sure there's an active/passive transformation thing going on here.

^ That's much better notation.

@0celo7 Yes, that's the same thing as confusing functions and variables.

I just don't like it. No reason.

So anyway, waves are functions of two variables $\psi(x,t)$.

If the wave has no dispersion, i.e. its shape is preserved as it moves, then the arguments must appear as a group $x \pm v t$ where $v$ is the velocity.

So physicists get lazy and instead of defining a new function, they write $\psi(x \pm vt)$.

Ok, @0celo7?

I learned all of this in PDE

Then why are you asking me ;)

ok I gotta go

ciao

@IceLord he was probably at his computer

ok, one more analysis problem

@IceLord I've gotten instantaneous email from my advisor

I don't know why we just don't text

He texts with other students :(

So did 4 out of 8 candidates for mod just retract their nominations?

@dmckee @ACuriousMind remember the paper I submitted that solved a major outstanding problem in superconducting qubit state measurement but was deemed "too obvious" and "not of broad interest"?

Accepted.

Took you long enough.

@IceLord when we were writing a poster 15 mins before the deadline our emails were pretty quick.

@Danu Not...exactly retract. Look in the election room, afaik, only knzhou retracted voluntarily.

@DanielSank Nice, congratulations!

@ACuriousMind Oh, god.

Can somebody please tell me how to solve this problem? "You walk $62.4$ m at a speed of $1.3$ m/s and then run $80.2$ m at a speed of 3.26 m/s along a straight track. Find average velocity." I found total time is $72.6$s and distance is $142.6$m, but apparently the average velocity isn't $142.6\over{72.6}$ $= 1.96$

@IceLord Yes. Look in the election room and this room for details. The reasons for removal given are "not a serious nomination" and "was suspended in the last year".

@JoshuaLamusga Well, "average velocity" is a bit ambiguous. Is it supposed to be the velocity with which you could have run the entire distance in the same time at constant speed, or is it supposed to be the time-weighted average of the actual velocities run?