The h Bar

Anonymous

9:00 PM

@DanielSank That's an approximation....

Anonymous

Okay

Anonymous

If the density function has jumps I doubt that would be much useful

DanielSank

What?

I'm defining the process by saying that there's a probability per time of an event.

The whole point of the post was to show that a process defined that way has Poisson statistics.

Semiclassical

hmm. I'm pretty sure I'm making some obvious mistake, but these C-G coeffs are throwing me for a loop

I"m trying to do an example I set for myself as review so I could help my students

Emilio Pisanty

@Semiclassical is that the point at which you realize you're in for a rough ride?

or does that come later?

Semiclassical

9:05 PM

lol

DanielSank

C-G are confusing if you let them take over.

Gotta keep them in line.

Semiclassical

If I take a pair of spin-1 particles with $m_1=m_2=0$, then the C-G table gives me $|1,0\rangle_1 |1,0\rangle_2 = \sqrt{\frac23}|2,0\rangle_{12} -\sqrt{\frac13}|0,0\rangle_{12}$

which seems sensible enough.

Emilio Pisanty

btw here is an example of SE being awesome

note the authors of the last paper my answer cites

CooperCape

Oh wow that's mad

DanielSank

@EmilioPisanty I don't get it.

Emilio Pisanty

9:07 PM

@DanielSank then compare to the author of the accepted answer

Semiclassical

the comments are also worth noting...

CooperCape

Author of marked as answer question is... (dun dun dun...)

DanielSank

Ah.

Semiclassical

(continuing) the two particle case I just gave seems simple enough. But now I wanted to take a third spin-1 particle, with $m_3=0$ as well

Anonymous

@DanielSank My point is that, in a continuous probability distribution if the probability of an event happening in unit time (say $1s$) is $\lambda(1)$, then the probability of the event happening in any arbitrary $0.1s$ within that $1s$ may not be $\lambda(0.1)$. So if you are defining $\lambda$ for unit time, it's not totally correct to conclude that the probability of that event happening in $\Delta t$ time is $\lambda \Delta t$.

DanielSank

9:10 PM

@Semiclassical I'm reasonably sure that C-G coefficients do not directly answer that question.

Semiclassical

so I figured I could just use the table again to get $|2,0\rangle_{12}|1,0\rangle_3$ and $|0,0\rangle_{12}|1,0\rangle_3$ and in terms of $|s,0\rangle_{123}$

DanielSank

@Semiclassical I am pretty sure that doesn't work.

Semiclassical

You may well be right, because if I crunch that from the C-G table I seemingly get

$$|1,0\rangle_1|1,0\rangle_2|1,0\rangle_3=\sqrt{\frac25}|2,0\rangle_{123}-\left(‌\sqrt{\frac{4}{15}}+\sqrt{\frac13 }\right)|0,0\rangle_{123}$$

Anonymous

So, I'd rather talk in terms of probability density function rather than probability per unit time, which is quite an ill-defined term.

Semiclassical

which isn't normalized to one.

however, 2/5+4/15+1/3 = 1.

DanielSank

9:14 PM

@Blue IMHO, that is a willful misunderstanding of what $\lambda$ means.

Semiclassical

is the answer just "that's not what C-G is for"?

DanielSank

I already said that $\lambda$ is the probability per time of an event, as long as $\lambda \Delta t \ll 1$.

@Semiclassical I'm not sure. There's probably a way to do the 3-body case with CG coefficients, but I think it's not 100% trivial. Search the main site. There are posts on this.

Semiclassical

kk

ew, Racah W-coefficient: en.wikipedia.org/wiki/Racah_W-coefficient

though there is says you can j1,j2,j3 by first coupling j1,j2 to J12 and then coupling J12,j3 to the total J

which sounds a lot like what I was trying to do :/

CooperCape

Okay so if I have a wavefunction $\Psi(x,0)=Ae^{-\lambda x}$ and I want to normalize it I think I should do $|A|^2\int_0^{\infty} e^{-2\lambda x} dx$ but the answers says that it's $2|A|^2$ before the integral sign. Why is this? (A is a positive real constant...)

Semiclassical

probably they mean $\Psi(x,0)=Ae^{-\lambda |x|}$ for all real $x$.

bolbteppa

9:19 PM

14

Q: Why is Russell so critical of Aristotle?

In A History of Western Philosophy, Russell argues: I conclude that the Aristotelian doctrines with which we have been concerned in this chapter are wholly false, with the exception of the formal theory of the syllogism, which is unimportant. Any person in the present day who wishes to ...

'Where Aristotle says "Some B1 are B2" this is clearly to be read as an intersection of types, what in modern categorical logic is called a fiber product. '

Of course Aristotle was all about fiber products

Used to get the urge to read his physics until I realized it's written in language like this

thomism.wordpress.com/2014/08/22/…

CooperCape

@Semiclassical That... that would make sense.

DanielSank

@Blue I think you're confused about the meaning of "per unit time". This does not mean anything about time=1.

bolbteppa

'One way to express the critique of Aristotle’s physics in his own terms is to say that, while he was right that every action or change involves act and potency, he thought this meant there were agent-patient relations in the inanimate when in fact there are only interactions. '

DanielSank

The phrase "per unit x" is just a formal phrase that means the same thing as "per time".

It's a bad phrase. I shouldn't have used it.

Anonymous

It's way better to just say $\lambda$ is the average of the probability density function. Avoids a lot of confusion

Anonymous

9:23 PM

Also you don't have to hand-wave by saying things like $\lambda \Delta t \ll 1$

DanielSank

@Blue That's not hand-waving.

@Blue Ok fine, but how do you know the form of the distribution?

Balarka Sen

The usual way to obtain the distribution is to take a limit of the binomial distributions (which are more physically meaningful)

DanielSank

The whole point of the post is to show how you go from a physical principle of something having a probability per time of ocurring to the Poisson distribution.

Semiclassical

@DanielSank think i found the error upon looking at this note: web.pa.msu.edu/people/pratts/phy851/lectures/lecture26/…

Balarka Sen

I must say it's not entirely clear to me what $\lambda$ being probability per unit time means. It's the expectation of the probability density function to me (but I am open to enlightenment if someone wants to tell me what that means)

Anonymous

9:28 PM

@BalarkaSen Exactly. I've been trying to say that one should define $\lambda$ as the "expectation of the probability density function"

DanielSank

@Blue WHAT density function?

Balarka Sen

No I mean if he's saying it surely it means something I don't understand

Semiclassical

key point: "In order to couple the three angular momenta $j_1$ and $j_2$ were coupled to $J_{12}$ before $j_3$ and $J_{12}$ were coupled to $J$. Thus $J_{12}$ survives as a quantum number, which is necessary as the original state had six labels which requires six labels for the final state." (emphasis added)

DanielSank

@BalarkaSen Consider a process where some event may happen. In a small amount of time $\Delta t$, the probability that the thing happens is $\Delta t \lambda$, as long as $\Delta t$ is small enough such that $\Delta t \lambda \ll 1$.

Anonymous

@DanielSank We don't need to know the exact form of the density function.

DanielSank

9:29 PM

That's what I mean by "probability per time".

Anonymous

@BalarkaSen We're talking w.r.t Poisson distribution

DanielSank

@Blue Look, you can do what you're saying, but it answers a slightly different question than the one I asked.

My post explicitly asks "If I have a process where in each little bit of time there's a probability $\Delta t \lambda$ that a thing happens, please show that the probability of $n$ things happening in time $T$ is a Poisson distribution with mean $\lambda T$."

Balarka Sen

@Blue I am well aware

DanielSank

That is the question.

I am not asking a different question.

Balarka Sen

@DanielSank Hm.

DanielSank

9:30 PM

If you think the question is not well posed, then we can discuss that.

Anonymous

@DanielSank If you want to do it that way, then follow the Wiki definition: " The average number of events in an interval"

Anonymous

Poisson distribution

In probability theory and statistics, the Poisson distribution (French pronunciation [pwaˈsɔ̃]; in English often rendered ), named after French mathematician Siméon Denis Poisson, is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant rate and independently of the time since the last event. The Poisson distribution can also be used for the number of events in other specified intervals such as distance, area or volume. For instance, an individual keeping track of...

Anonymous

The word "average" is very important here

Semiclassical

those aren't the same thing.

Balarka Sen

@Blue You're not understanding Daniel Sank's premise at all

Semiclassical

9:32 PM

probability distribution of getting n events in small amount of time

Anonymous

@BalarkaSen Okay?

Balarka Sen

He's giving you a physical situation, giving you a $\lambda$ that comes from the situation, and telling you: it's the same $\lambda$ as in the Poisson dist.

DanielSank

@Blue you do realize that a real physical process always has a lower time limit below which the statistics aren't Poissonian, right?

Balarka Sen

You're begging the question by repeatedly telling him that he should define his $\lambda$ as the expectation of the Poisson dist.

DanielSank

@BalarkaSen Actually, be careful. Usually $\lambda$ is the mean of the Poisson distribution. That is not how I used it in my question.

@BalarkaSen Correct.

Balarka Sen

9:34 PM

@DanielSank That's literally what I said, I think

DanielSank

@BalarkaSen I think not :-\

> ...and telling you: it's the same $\lambda$ as in the Poisson dist.

Balarka Sen

You're obtaining that your distibution is a Poisson distrubution with mean $\lambda$ you're defining otherwise in the question, right?

DanielSank

@BalarkaSen No.

10

Q: Derive Poisson distribution from probability per time of event

Suppose we have a probability per unit time $\lambda$ that something (e.g. nuclear decay, random walk takes a step, etc.) happens. It is known that given a time interval of length $T$, the probability that $n$ events happen is given by the Poisson distribution $$P(n) = \frac{e^{-\lambda T} (\lamb...

probability stochastic-processes

Semiclassical

for one, they'd have different units. $\lambda$ as an expectation value would be a pure number. but then $\lambda\Delta t$ can't be a probability

DanielSank

Please read the post.

Balarka Sen

9:36 PM

Oh, mean $\lambda \Delta t$.

DanielSank

Yes

Balarka Sen

As Lang used to say, "Your notation sucks" ;)

But I understand now

Semiclassical

lol

Balarka Sen

Thanks

DanielSank

@BalarkaSen If there's a more standard symbol, I'll use it (and edit all the answers).

Serg Lang was a nut. I met him once as a freshman, before classes even started, and he yelled at me about wave/particle duality.

Balarka Sen

9:38 PM

Truly

lolol

bolbteppa

@DanielSank that's incredible, what an honor

DanielSank

@bolbteppa :-)

Semiclassical

I suppose the issue is that you can obtain $\lambda$ in two ways

DanielSank

Look for the purposes of this discussion, $\lambda$ has dimensions of 1/time. No objections.

$\lambda$ is not the mean of the Poisson distribution.

Semiclassical

One is empirical: Let the process run for a long time, count the total number of events, and from that determine the event frequency

bolbteppa

9:40 PM

There's a story online I'm mis-remembering but apparently in a class he once said something and half the class tried to calculate it, and he said 'those that tried to calculate it will become mathematicians' or something

DanielSank

$1/ \lambda$ is the mean of the exponential distribution of times before the next event, however!

Balarka Sen

@DanielSank Perhaps a good idea to point the unfortunate coincidence of notation in the question as a NB?

Anonymous

@BalarkaSen In the physical situation he gave he says $\lambda$ is the "probability per unit time". I'm just saying that that is not a well defined term. So you should go with average number of events in a time interval as wiki says. Or have a look at this post

DanielSank

@BalarkaSen Why? Isn't $\mu$ usually used for mean?

bolbteppa

Another one liner of his is that analysis is just number theory at $\infty$

Semiclassical

9:41 PM

@Blue Here's how I'd define it. Take that same long time run that I said, and break it into a bunch of small time intervals. In each of those, count the number of events per small time interval

Balarka Sen

@DanielSank I think most people are used to $\lambda$ being the parameter of the Poisson distribution (hence the mean).

DanielSank

@BalarkaSen Oye. There aren't enough letters...

Semiclassical

That defines a probability distribution on the number of events per small time interval.

Anonymous

@Semiclassical *and take their average!

Anonymous

That's it

Balarka Sen

9:42 PM

@Blue It's not the average, as DS points out.

Semiclassical

hmm

Balarka Sen

$\lambda \Delta t$ is the mean of the pdf, not $\lambda$

Semiclassical

my example is bad, actually

Balarka Sen

On the interval $[t, t + \Delta t]$

DanielSank

@Blue Yes, you can do it that way.

I'm not convinced it's necessarily better pedagogically, but I do see that it might be more robust a description. Maybe.

I have to think about it when I'm not also drawing mechanical parts :-|

Semiclassical

9:44 PM

lol

Anonymous

@BalarkaSen Now we are back to square 1. In the physical situation we should start with defining lambda as the average number of events in an interval.

Semiclassical

I suspect it should boil down to: Instead of waiting a long time, you can wait a very short time and look for the probability that at least one event happens

Balarka Sen

Oh I mean, sure, why not.

Anonymous

And then show it is equivalent to the $\lambda$ in the Poisson distribution.

Semiclassical

and require that this probability be proportional to the time interval for sufficiently small times.

Balarka Sen

9:45 PM

@Blue Nooo. $\mu = \lambda \Delta t$ in DS's Poisson distribution. Do you realize that?

DanielSank

^ How is everyone not understanding this?

Balarka Sen

In the standard Poisson distribution, $\mu = \lambda$.

Semiclassical

I think I'm understanding it, lol

bolbteppa

@DanielSank there are two derivations of the Poisson distribution in Landau's stat phys, sec 114, one from the Bernoulli distribution (which he also derives) and another from the Gibb's distribution, if that's any help

Anonymous

@BalarkaSen Ah, right. So we show it is equivalent to $\mu$

bolbteppa

9:46 PM

Using gas particles in a box to motivate it

CooperCape

I'm really confused as to why $\Psi(x,0)=e^{-\lambda |x|}$ having $|x|$ instead of $x$ means that $1=2|A|^2\int_0^{\infty}e^{-2\lambda x} dx$ instead of having $1=|A|^2...$

DanielSank

@bolbteppa ~sigh~ I made a post specifically asking to relate the Poisson distribution to a particular physical starting point.

Now everyone wants to offer alternative approaches.

Which is great! But please understand what I actually asked first.

Balarka Sen

^^

Semiclassical

@CooperCape the domain is all real $x$

CooperCape

Right... am I missing something here?

I usually forget the small things and it ruins me.

DanielSank

9:47 PM

@CooperCape It's because if f is even, then $$ 2 \int_0^\infty f(x) dx = \int_{-\infty}^\infty f(x) dx \, .$$

Semiclassical

$\int_{-\infty}^\infty e^{-2\lambda |x|}\,dx=$?

Anonymous

@BalarkaSen Hmm, I was talking about the standard Poisson distribution initially. But okay.

CooperCape

Ohhhh right... thanks both of you :) That actually makes sense now...

DanielSank

@Blue wtf do you mean by "standard"?

Anonymous

So, I guess we're done

DanielSank

9:48 PM

The choice of variables does not make something "standard" or "nonstandard".

Variables are just symbols for heaven's sake.

If you think a conceptual object's standardness is tied to the notation used for that object, you're in for a world of sadness.

Balarka Sen

lol

Anonymous

I could replace variables with elephant emojis. That would just add to the confusion.

Semiclassical

The way I'd put it is that the advantage of DS's way of approaching is that one derives the result from a single assumption: For very short times, the probability of having at least one event is proportional to time.

DanielSank

@Blue Yes, but it would not change the meaning.

Anonymous

Better to stick to widely used variable notations.

DanielSank

9:49 PM

@Blue Yes, usually.

But do you realize that we just wasted half an hour because you refused to read how I was using the symbol?

Semiclassical

"Standard notation" is a fiction, albeit a very convenient one

bolbteppa

@DanielSank I read your post, you can use any derivation to derive it the way you are asking about, seems like the 'time' aspect is throwing you off in your modelling - but since probability per unit time times time is probability, so the time is irrelevant in the derivation until the very end

DanielSank

@bolbteppa I do not understand.

Semiclassical

"unit time times time"

ow

Balarka Sen

lolol

Anonymous

9:51 PM

@DanielSank Well, now you can edit the post so that it becomes less confusing for later visitors. :)

DanielSank

@Blue So what symbol would you like me to use?

and how did you miss the gigantic equation $$\frac{e^{-\lambda T}(\lambda T)^n}{n!} \, ?$$

It's pretty obvious that $\lambda$ has dimensions 1/time there.

Semiclassical

and that's definitely proportional to $T$ when $n=1$ and $T$ is sufficiently small.

(though to follow my statement above I should really talk about the probability of at least one event which is $1-e^{-\lambda T}\approx \lambda T$.)

Anonymous

@DanielSank I'll suggest an edit later, when I get time

DanielSank

@Blue You had 30 minutes to argue but no time to suggest a single symbol?

bolbteppa

In this picture, at the end replace $\overline{N}$ 'average number of particles in the volume $V$ with 'average number of particles you find in a unit time $\lambda$ times time $T$ taken to measure', $\overline{N} = \lambda T$:

DanielSank

9:55 PM

:-P

bolbteppa

DanielSank

@bolbteppa Yeah yeah Floris already wrote that up.

bolbteppa

So what's the problem then

DanielSank

@bolbteppa I have no idea, other than @Blue complaining that "probability per time" isn't a well defined idea.

Balarka Sen

There's no problem, you're just not saying anything new. It's already in the answers.

Semiclassical

9:57 PM

this conversation just feels like people talking past one another

2

bolbteppa

It's still a cool derivation worth repeating :p

Mozibur Ullah

10:26 PM

Conversations are often like that

Sir Cumference

10:51 PM

@Cows Neat, that was my yearbook quote in high school

user400188

@EmilioPisanty thanks for the edit

nbro

Yo yo

dudes

EnlightenedFunky

11:10 PM

Hello

Mozibur Ullah

Hi!

EnlightenedFunky

Is there any physics people that are working on "Instant matter transport"?

Mozibur Ullah

Seriously?

EnlightenedFunky

yeah like sending material from one place to another in an instant

just an idea

ACuriousMind

No, no one is working on that because it would violate causality, i.e. contradict relativity, and hence believed to be impossible

Mozibur Ullah

11:13 PM

Well theres quantum teleportation - it transports state.

But no matter is exchanged.

ACuriousMind

Also, quantum teleportation still needs a classical channel and proceeds therefore at most at the speed of light

EnlightenedFunky

But I have a question then about Bose-Einstein Condensate, doesn't matter behave like a wave?

Mozibur Ullah

Its unlikely to be instantenously for the reasons already point out.

both fermions and bosons have both particle and wave like character.

ACuriousMind

@EnlightenedFunky Quantum objects have features of both classical particles and classical waves, but they really behave like neither.

Mozibur Ullah

The Bose-Einstein condensate occurs because bosons don't obey the Pauli exclusion prinicple. So they can all be in the same state.

True, thats why some wit coined the word - wavicle - but it never really caught on.

At least I never heard any one use it.

EnlightenedFunky

11:17 PM

Ok but you can't increase the speed of the bose-einstein condensate cause you would increase its temperature right?

Mozibur Ullah

That looks like a reasonable assumption.

ACuriousMind

Well, you can certainly accelerate the entire condensate to whatever speed you'd like.

Mozibur Ullah

the other thing about condensates is that they bring quantum effects into the macroscopic arena.

EnlightenedFunky

but would it still exhibit wave properties as when at rest

Mozibur Ullah

I thought he was talking about accelerating the wavicles in the condensate?

EnlightenedFunky

11:19 PM

yeah what Mozibur said

Mozibur Ullah

@curiousone: are condensates why we get super-fluidity?

Actually hoo-haa with extra dimensions in String Theory, why has no-one pointed out that the usual formalism in quantum mechanics has infinite dimensions? I suppose they're not space-time dimensions...

EnlightenedFunky

because if so couldn't we make an instrument that reads that specific wave pattern interfence and turns it to energy wouldn't violate E=mc^2 and create an EM wave, and send it somewhere else and then bring it back to matter from said energy

Mozibur Ullah

Actually, with all the hoo-haa with ...

ooolb

Apr 21 '16 at 20:42, by ACuriousMind

Are you perhaps confusing me with CuriousOne? :P

Mozibur Ullah

I'm not sure whos confusing who with who...

ooolb

11:22 PM

wait

did you just call someone out in chat who doesn't use chat?

Mozibur Ullah

But someones being confusing ... curiouser and curiouser.

ACuriousMind

@MoziburUllah Yes, I think all superfluids are either Bose-Einstein condensates or fermionic condensates like the Cooper pairs of superconductivity, but I'm not entirely sure

ooolb

acm is curiousone confirmed

Mozibur Ullah

I think I typed @curiousone instead of @acuriousmind

@ooolb:

ACuriousMind

@EnlightenedFunky I don't know what that sentence means, sorry.

EnlightenedFunky

11:23 PM

oh ok

0celo7

there goes ACM again

Mozibur Ullah

I think I must be tired (yawn).

ACuriousMind

@MoziburUllah Yeah, you did. But CuriousOne is not active anymore, so it's not a grave confusion ;P

ooolb

@EnlightenedFunky what are you on right now, my dude

0celo7

you killed the competition :o

ooolb

11:24 PM

i want some

Mozibur Ullah

Ok. ;).

Oh no!

EnlightenedFunky

@ooolb Why? I am on nothing... just thinking about something learned

ACuriousMind

@0celo7 That's one possible interpretation ;)

ooolb

@0celo7 you don't have to kill them to get rid of them.

@EnlightenedFunky duuuuuuuuude

im gonna go learn stuff right now

bye

EnlightenedFunky

Couldn't an instrument be made that reads that specific wave pattern interference and turns it to energy wouldn't violate E=mc^2 and create an EM wave, and send it somewhere else and then bring it back to matter from said energy @ACuriousMind better?

0celo7

11:29 PM

@ooolb do you want a more attainable exercise?

ACuriousMind

@EnlightenedFunky I don't know what you mean by "reading" a "specific wave pattern interference" and "turning it into energy". What wave pattern? You can't measure quantum wavefunctions, at least not like you do with actual classical waves. And how would you "turn it into energy"?

ooolb

@0celo7 nah go big or go home m7

im working on it

@EnlightenedFunky is this ironic

EnlightenedFunky

Humm need to learn more quantum mechanics thanks did not think about that

0celo7

                                      @ACuriousMind should I play Civ V?

I dowbloaded it

whoa wtf

ooolb

wat

what is that sorcery

ACuriousMind

11:32 PM

@0celo7 Sure, it's nice

ooolb

I ᴛᴏᴏ, ᴀᴍ ᴄᴏᴏʟ.

ᴅᴏɴ'ᴛ ғᴏʀɢᴇᴛ ᴛᴏ ᴅʀᴏᴘ ᴀ ʟɪᴋᴇ.

ᴀɴᴅ sᴜʙsᴄʀɪʙᴇ ɪғ ʏᴏᴜ ʟɪᴋᴇ ᴡʜᴀᴛ ʏᴏᴜ sᴇᴇ.

ᗩᑕ⋒ᖇᓰᗢᘮᔕᙢᓰﬡᖱ

╭━━━┳━━━╮╱╱╱╱╱╱╱╱╱╱╱╱╱╱╭━╮╭━╮╱╱╱╱╱╭╮
┃╭━╮┃╭━╮┃╱╱╱╱╱╱╱╱╱╱╱╱╱╱┃┃╰╯┃┃╱╱╱╱╱┃┃
┃┃╱┃┃┃╱╰╋╮╭┳━┳┳━━┳╮╭┳━━┫╭╮╭╮┣┳━╮╭━╯┃
┃╰━╯┃┃╱╭┫┃┃┃╭╋┫╭╮┃┃┃┃━━┫┃┃┃┃┣┫╭╮┫╭╮┃
┃╭━╮┃╰━╯┃╰╯┃┃┃┃╰╯┃╰╯┣━━┃┃┃┃┃┃┃┃┃┃╰╯┃
╰╯╱╰┻━━━┻━━┻╯╰┻━━┻━━┻━━┻╯╰╯╰┻┻╯╰┻━━╯

░█▀▀█ ▒█▀▀█ █░░█ █▀▀█ ░▀░ █▀▀█ █░░█ █▀▀ ▒█▀▄▀█ ░▀░ █▀▀▄ █▀▀▄
▒█▄▄█ ▒█░░░ █░░█ █▄▄▀ ▀█▀ █░░█ █░░█ ▀▀█ ▒█▒█▒█ ▀█▀ █░░█ █░░█
▒█░▒█ ▒█▄▄█ ░▀▀▀ ▀░▀▀ ▀▀▀ ▀▀▀▀ ░▀▀▀ ▀▀▀ ▒█░░▒█ ▀▀▀ ▀░░▀ ▀▀▀░

0celo7

that's a but much

ooolb

hmm

was just getting started

0celo7

@ACuriousMind want to play a game with us

ooolb

░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░
░░░█████████████████████████████░░░
░░░██▄██▄██▄██▄██▄██▄██▄██▄██▄██░░░
░░░█████████████████████████████░░░
░░░░░░░░░░░░░░░██▄██░░░░░░░██▄██░░░
░░░░░░░░░░░░░░░█████░░░░░░░█████░░░
░░░░░░░░░░░░░░░██▄██░░░░░░░██▄██░░░
░░░█████████████████████████████░░░
░░░██▄██▄██▄██▄██▄██▄██▄██▄██▄██░░░
░░░█████████████████████████████░░░
░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░
░░░█████████████████████████████░░░
░░░██▄██▄██▄██▄██▄██▄██▄██▄██▄██░░░
░░░█████████████████████████████░░░

it broke

ACuriousMind

11:43 PM

@0celo7 Not today, I need to finish my Shadowrun prep for tomorrow

ooolb

@ACuriousMind notice me :'(

i did that all for you

ACuriousMind

@ooolb But...why?

EnlightenedFunky

lol

bye

0celo7

@ACuriousMind Are there any American adults here right now

ACuriousMind

I have no way of knowing that...why?

0celo7

11:46 PM

United Airlines is being a hecker

they're not beating me senseless, but close to it

ooolb

@ACuriousMind reassurance

0celo7

@ACuriousMind how does having a crazy internet stalker feel?

ACuriousMind

@0celo7 Well, that sucks

@0celo7 I don't know

0celo7

@ACuriousMind well you have one!!!

@ACuriousMind Should I make schnitzel or just sautee the pork

ACuriousMind

Depends. The better the quality, the less inclined I am to make Schnitzel out of it ;P

0celo7

11:52 PM

it's good quality, but I can't get the right flavor on pork without a sauce

ooolb

@0celo7 who? there can only be one. i'll eliminate the competition.

0celo7

jesus dude

you need to fix your priorities

ACuriousMind

^I second that

ooolb

what priorities?

:p

serious question does any of our lives have meaning anymore?

after ACM went to the IT side

i'm still going to learn physics though just to invent an interuniverse travelling machine and travel to the alternate universe where ACM is doing physics.

Transcript for