The h Bar

samjoe

2:03 PM

@ACuriousMind the answer may be incorrect! and I just meant to give him a hint.

ACuriousMind

@samjoe Whether the answer is correct or not does not matter at all for the purposes of the HW policy

samjoe

fine.

ACuriousMind

And whether you meant to give a hint or not is likewise irrelevant - what you did is write down something that could be the actual, full answer to the question.

In any case, we would generally prefer that you don't answer such homework-type question at all. The policy is in place to discourage such questions. Try to find better questions to answer.

0ßelö7

Get rekt

ACuriousMind

In unrelated news, I see 53 items in the close review queue. >3kers, please review!

4

Mike Doonsebury

2:10 PM

Is it a legitimate use of the board to open up a chat room for a personal theory?

0ßelö7

@ACuriousMind does that include me?

physics.stackexchange.com/questions/356086/… can u move this pls

ACuriousMind

@MikeDoonsebury If you have enough reputation to create a room, you can create it for any purpose as long as you abide by the Be Nice policy.

@0ßelö7 Yes, I can, one moment

@0ßelö7 Why would it not?

@SirCumference Unless you're well-rested, I command you to turn back and go to bed now!

Sir Cumference

@ACuriousMind Don't worry, I got 8 hours of sleep. I'm sick XD

0ßelö7

why are you so concerned with his health

ACuriousMind

@0ßelö7 First Law.

0ßelö7

2:16 PM

@ACuriousMind don't talk about fight club?

Sir Cumference

@ACuriousMind A body at rest must stay at rest? So I should sleep?

ACuriousMind

lol

Neither of you has the correct reference

0ßelö7

@SirCumference lmao he wants you to be at rest for infinity, i.e. die

Sir Cumference

> i.e. die

0ßelö7

yes?

Sir Cumference

2:18 PM

Wow, I didn't know what this was, ACM

Anonymous

If the potential function is finite in a region why should the wave function and it's first derivative have to be continuous in that region? (I'm talking in reference to the Kronig-Penney Model)

ACuriousMind

@SirCumference What what was?

As for the "First Law": A hint is that @JohnRennie will probably know what I'm referring to.

4

A: Smoothness constraint of wave function

Here we want to give an easy mathematical bootstrap argument why solutions to the time independent 1D Schrödinger equation (TISE) tend to be rather nice. First formally rewrite the differential form $$-\frac{\hbar^2}{2m} \psi^{\prime\prime}(x) + V(x) \psi(x) ~=~ E \psi(x) \tag{1}$$ into the int...

0ßelö7

what is $\mathcal L^1_\mathrm{loc}$

ACuriousMind

@0ßelö7 Everything relevant is hyperlinked, no?

0ßelö7

@ACuriousMind I think he means $L^1_\mathrm{loc}$

ACuriousMind

2:24 PM

I think he doesn't care about calligraphy that much :P

Anonymous

@ACuriousMind Thanks!

0ßelö7

@ACuriousMind it's an important distinction!

mathcal L could be something else, like a Campanato space

@ACuriousMind do you think I should bother citing the original papers here or just the books I get the material from

Anonymous

As usual I didn't understand many parts of the answer due to the math. But I at least got a basic idea now. :P Mostly looks like Linear Algebra and Analysis though.

0ßelö7

@Blue it is entirely analysis

Secret

[Some time travel comments] Since in the previous paragraph, we have explained how travelling to the future will not necessary result in you to arrive in the future that is resulted as if you have never time travelled (via twin paradox), what is the reason that the past you travelled back, has to be the past you learnt from historical records :?

ACuriousMind

2:27 PM

@0ßelö7 I would probably cite whatever you think is best for people to read if they actually want to follow your references.

0ßelö7

it is kind of important to note that being in 1D is crucial to the argument, @Blue

ACuriousMind

@Secret What previous paragraph? Reminder that this is not your blog.

Anonymous

@0ßelö7 I see. As of now I'm dealing with only 1D SE

0ßelö7

@ACuriousMind Right, good point. The books it is. When I mentioned giving this talk people said they knew the books but never read them

Secret

@ACuriousMind Right, I will put this in the blog instead

0ßelö7

2:28 PM

@Blue it's why PDE is very much harder than ODE

ACuriousMind

@0ßelö7 Ah, yes. I kinda assumed we're in ordinary intro QM territory here where all the potentials are 1d wells

Anonymous

@0ßelö7 I've heard that

0ßelö7

In 1D, if you know $u''$ then you know $u'$ and $u$. But in $\ge$2D, if you know $\Delta u$ then it's HAAAAAAAAAAAAAARD to get info about $u_{ij}$, $u_i$ and $u$.

Anonymous

I'll start learning ODE and PDE during my winter vacation this year. :)

Anonymous

@0ßelö7 Yup, I get that

0ßelö7

2:30 PM

I seriously doubt you'll learn elliptic regularity theory

Balarka Sen

Sounds like we're doing cool stuff

0ßelö7

Schroedinger operators are even worse though, because you want $V$ to live in a really bad space usually

And the domain is typically all of $\Bbb R^n$

@Blue If you want an idea, read the book "Introduction to Spectral Theory" by Hislop and Sigal

Anonymous

@0ßelö7 What are the prerequisites for reading that?

0ßelö7

Basically none, the appendices look very through

I doubt they teach you measure theory though

you'll spend a few weeks reading the appendices if you know none of this material though

Anonymous

@0ßelö7 I see. Thanks for the suggestion. I'll surely look into it during my vacations.

0ßelö7

2:35 PM

Normed vector spaces, $L^p$ spaces and integration, linear operators on Banach spaces, Fourier transform, Sobolev spaces, convolution

@ACuriousMind I need a lemma that takes two slides to state lol

ACuriousMind

@0ßelö7 At that point you should ask yourself whether you really need to tell them exactly what the lemma says

0ßelö7

@ACuriousMind I was actually hoping to give a full proof of the lemma, because it's folklore

you know me, I love proving folklore

Balarka Sen

^ analytical madman

ACuriousMind

@0ßelö7 I'd still maybe put a shorter statement of the lemma into the talk itself and put the full statement on a slide after the last slide that you can pull out if someone asks.

(Being less precise/in-depth than one usually would, but keeping the details in reserve in case it becomes relevant is generally a good strategy for talks, imo)

0ßelö7

I could naturally break it up into 4 sublemmas

but then my talk has like seven lemmas

er, 3 sublemmas

ACuriousMind

2:44 PM

@0ßelö7 Well, I'd omit the explanation of the notation on the slide itself, and since there seems to be two pairs of formulae, I'd just put one of the two and then say that there's another one with suitable substitutions.

0ßelö7

@ACuriousMind Maybe I should break it up, and prove one of them. Then, if I have time, have slides with proofs of the other parts?

ACuriousMind

So that reduces the lemma to two formulae, which should fit on a single slide :P

0ßelö7

Right, I shouldn't have to explain to these guys what a multiderivative is

ACuriousMind

@0ßelö7 Is this a folklore proof where everyone competent in the field will know how it's supposed to go or is there some sort of trick that's not evident to an expert?

Because if it is not the latter, there will probably be no one interested in seeing the proof

0ßelö7

How am I supposed to judge that

ACuriousMind

2:46 PM

@0ßelö7 Ask your advisor?

0ßelö7

Hm, ok

ACuriousMind

I mean, "Hey, I bet you've always wondered how to prove X - here it is" is interesting. "Hey, you know that statement everyone knows how to prove but doesn't bother to write down? Here is the proof written down" significantly less so

0ßelö7

@ACuriousMind Yeah good point. Let me write out the proof carefully and see which parts are actually interesting

@ACuriousMind Well that was my point with Neumann regularity. Everyone thinks it's a routine calculation but you run into subtle issues

ACuriousMind

@0ßelö7 Well, if everyone thinks it routine but it isn't, then that's double interesting!

0ßelö7

@ACuriousMind hence my obsession with Sobolev spaces on manifolds :P

Is $\Bbb R_+^*$ supposed to be $(0,\infty)$?

John Rennie

3:21 PM

@ACuriousMind no physicist shall discuss interpretations of quantum mechanics or by inaction allow interpretations of quantum mechanics to be discussed.

0ßelö7

hmm?

ACuriousMind

@JohnRennie Heh, excellent

0ßelö7

not having a notation index should be a capital offense

0ßelö7

4:12 PM

@ACuriousMind how do I get the half harpoon for restriction?

$\upharpoonright$

nice

@ACuriousMind Is $\{(x,x):x\in X\}$ the diagonal of $X$ or the diagonal of $X\times X$?

just terminology wise

ACuriousMind

I'd say that of $X\times X$

Balarka Sen

the y = x line is the diagonal of R^2 not R

0ßelö7

k

Quantum spaghettification

4:42 PM

Sorry I have a quick question: For questions like this physics.stackexchange.com/questions/356260/… where the accepted answer clearly does not answer the original question what is the best thing to do; downvote, flag or just leave it?

bolbteppa

In a latex thingy, like texmaker, when you press new tab you get a blank file - any way to modify that so it opens a page with a pre-amble and \begin/end{document} already there :p

Leaky Nun

@bolbteppa use TeXstudio and use the wizards :P

0ßelö7

oh thank god

Leaky Nun

@0ßelö7 :O nice

0ßelö7

nothing worse than getting an old ass book and having to decipher the notation

CooperCape

4:57 PM

So this question says express $u^0$ in terms of $u^j$ where $u$ is the four-velocity and I get what $u^0$ and $u^j$ are but I'm a bit confused how to go about this one? I thought maybe using the space-time interval and evaluating for $\frac{dt}{d\tau}$ but it's not workin out for me... :/ Anyone give me a quickie starter please? :p

0ßelö7

is the 4-velocity normalized?

CooperCape

If by that you mean $u\cdot u=-1$ yes? but I may've just seemed liek an idiot by saying that

0ßelö7

Yes, so you have $-(u^0)^2+\mathbf u^2=-1$

CooperCape

Ahhh okay...

so $u^0=\sqrt{1+u^ju_j}$ I guess.. Thanks :D

0ßelö7

5:58 PM

@ACuriousMind what's a more compact way to write $(k+l-n/2-1)\cdots(k+1-n/2)$

I guess it is already pretty compact

might not want to overcomplicate

I feel like it's some combinatorial thing

Balarka Sen

6:19 PM

@0ßelö7 $(k-n/2+\ell-1)!/(k-n/2)!$

It's a permutation coefficient

Anonymous

@BalarkaSen Wouldn't $n$ have to be even for that to be true?

Balarka Sen

Sure, if not you can replace the factorials by the gamma function

Anonymous

@BalarkaSen Okay, yes!

0ßelö7

6:53 PM

@BalarkaSen yeah, but that's not any more compact

ah, but that does help for the convergence proof, thanks

Balarka Sen

kewl

0ßelö7

7:26 PM

@BalarkaSen this is almost done, do you want to look over it

Phase

9:03 PM

Although a physics question, this is still important to chemistry. The delocalized electric field is related to the force (and therefore the repulsive potential) between two electrons. This in turn is what we need to solve the Schrödinger Equation to describe molecules. Short answer: You can calculate the expectation value of the corresponding operator, which comes close to the mentioned superposition. — Feodoran 13 hours ago

:((((

still got closed anyway

0

Q: Delocalised Electron's Electric Field

If we take an electron that's delocalised w.r.t position, how can one evaluate the electric field over some space? Is it some superposition or a sort of field with all the charge at the expectation value of the position?

Would love a full answer

0ßelö7

@ACuriousMind ok it's done

do you want me to resend it?

Phase

9:48 PM

Why do you have to go to the eigenbasis of $M^i$? Why cant you just simply observe that the kronecker which it is equal to is either going to be zero, +1 or -1

ACuriousMind

@Phase How is that observation supposed to relate to the eigenvalues?

Phase

hang on

what if you let i = j, so then you end up with $M^2 + M^2 = 2M^2 = 2\delta ^{ij}I$ so then you get $M = sqrt I$ and get +I or -I with eigenvalues +1 or -1

or am I butchering all of maths by doing this

Because if i != j then you get zero, which doesn't really count as a solution afaik

ACuriousMind

@Phase Well, indeed you get $2M^2 = 2I$, so $M^2 = I$. How does that prove something about the eigenvalues of $M$?

Phase

Because if M = sqrt(I), that means that the eigenvalues of sqrt(I) would be the eigenvalues of M right?

ACuriousMind

What is $\sqrt{I}$?

These are matrices, not numbers

Phase

9:56 PM

Idk, I guess +- I?

ACuriousMind

@Phase Well, did you just define that? That's not a proof!

Phase

i

uhhhhhhhhhhhhhhhhhhhhhhhhhhhh

confuse

0ßelö7

@ACuriousMind So, I take it you don't want the full version?

ACuriousMind

@Phase (I suspect your intuition is correct here, but pretty much everything you'll do to make it into a coherent argument will involve going to the eigenbasis)

Phase

If $M^2$ is in the question, doesn't that mean that $M^a$ is a valid operation as long as $a \in R$?

Idk how to do the nice font

0ßelö7

9:58 PM

@Phase Jesus

Phase

@0ßelö7 i know i know i make your skin crawl with my lack of rigour

ACuriousMind

@0ßelö7 I just looked back at chat and noticed Phase's question, I wasn't purposefully ignoring you - do you want me to look over it? Because I don't think I'll gain much personally from reading the slides.

0ßelö7

@ACuriousMind do you know this already?

ACuriousMind

@Phase Uhhh. You can raise a matrix to an integer power - just multiply it $n$ times with itself, but it is not clear at all what using a non-integer power is supposed to mean

There is such a thing as a square root of a matrix, but I'd expect you to define how to obtain it before using it

Phase

Oh. I guess I just felt that since I was quite an easy thing to imagine a square root of, that I could use that

r i p

ACuriousMind

10:00 PM

@0ßelö7 No, but I won't know it after reading the slides either unless I go read the references, will I?

Phase

Im not entirely sure I understand the question then. If it's quick could you please just write a few lines about how you'd start it?

ACuriousMind

@Phase Well, use the hint: Go to the eigenbasis of $M$, then write out what $M^2 = I$ means in components.

0ßelö7

@ACuriousMind I think you were right, anyone who is good at doing PDE type analysis could prove these lemmas. So in that sense it's quite complete.

Phase

..without imagining the square root of I? just break it into its column space and do it like that?

0ßelö7

The only problem is I did all of the proofs in my notes by hand, but didn't use one of the lemmas

Well, I did, but the lemma is ridiculously general

ACuriousMind

10:02 PM

@Phase Just write down how it looks when you write out the matrices.

0ßelö7

@ACuriousMind I sent it to you and would appreciate general feedback

I fear it's too long, going to read the whole thing out loud now and see

ACuriousMind

@0ßelö7 Okay, will read it soon

0ßelö7

@ACuriousMind Depending on who shows up I might have to review some basic geometry

this is a new seminar so no one knows who will show up

lılostafa

Hurricane hunters aboard a WP-3D aircraft, about to hit the eyewall and penetrate the eye of Hurricane Katrina ^

Balarka Sen

@Phase You could start by writing out what "$\lambda$ is an eigenvalue of $M$" means

lılostafa

10:10 PM

This is fascinating

Phase

@BalarkaSen that it's the image of an normalised-eigenvector in a diagonalised basis?

$Mv - \lambda v = 0$

Balarka Sen

For some $v$, yeah.

Feed that to $M$ again. What do you get?

Phase

$\lambda ^2$?

Balarka Sen

Well, you don't get a number. What's $M(Mv - \lambda v)$?

Phase

0?

the null vector?

Balarka Sen

10:24 PM

One one hand, yes. On the other hand, if you "distribute" the multiplication by the matrix over the expression...?

Phase

$M^2v - M\lambda v$

Balarka Sen

Given the datas about $M$, which is?

Phase

Oh

$Iv - M\lambda v$

$Iv = M\lambda v$

Balarka Sen

$\lambda$ pops out of the right hand side because of linearity

Phase

Sorry, what do you mean "pops out of"?

Balarka Sen

10:26 PM

So you can simplify it further, given $Mv = \lambda v$.

Phase

Oh

$Iv = M^2 v$....?

Balarka Sen

Where did you get $M^2$?

Phase

I replaced the lambda with M

Balarka Sen

I mean $M \lambda v = \lambda M v =$ ... ?

Phase

Oh

$Iv = \lambda ^2 v$?

Balarka Sen

10:28 PM

Yes. What's $Iv$? :P

Phase

v

Balarka Sen

There you go.

Phase

I see, I got confused because I didn't really see what you meant earlier, thanks though

Bernardo Meurer

What up peeps

Phase

Feel pretty shameful for needing this spelled up for me

time to commit an honour suicide

Balarka Sen

10:29 PM

Not before I post my dank meme

Phase

One last question I suppose

Balarka Sen

@lılostafa

Phase

Maybe it's just me having not really done much with Eigenbases but I don't recognise where I "put it in terms of M's eigenbasis". I just wrote it down for some vector v, rather than a space that contains all of the vectors v

is that good enough?

ACuriousMind

@Phase How does the matrix $M$ look in the eigenbasis?

Phase

Oh. Like the identity

ACuriousMind

10:31 PM

@Phase What does that mean?

It's not equal to the identity.

Phase

to cleanse the shame of getting everything wrong

ACuriousMind

Please don't joke like that, confusion is normal when learning things

Phase

RIP, I can remove it if it's offensive

ACuriousMind

I mean, it's not equal to the identity, but it does look, in a certain sense, "like" it. I want you to tell me in what sense exactly.

Phase

I guess I don't really understand the casual language because I'm not familiar with it.

I don't really know what a matrix having a "look" means

Balarka Sen

10:33 PM

Do you know change of basis?

Phase

using the matrix of eigenvalues?

*vectors

JEEEEEEZ

ACuriousMind

@Phase By "looking" in a basis, I mean what its entries in that basis are.

Phase

Well

like +-1 in the diagonal

Balarka Sen

It's not entirely obvious to me what you mean by that but, what I mean is, if you change the basis of your vector space then the matrix changes according to a conjugation formula

ACuriousMind

@Phase Why +-1?

Balarka Sen

10:35 PM

I guess you know that

Phase

Because... $\lambda^2 = 1$ and if you diagonalise it in terms of the eigenvectors then its diagonal with each entry corresponding to a different eigenvector / value

ACuriousMind

@Phase Aha!

So, simply by going to the eigenbasis, the equation $M^2 = I$ tells you $\lambda^2 = 1$ and hence $\lambda = \pm 1$.

That's what the exercise intended you to do.

$\lambda^2 = 1$ is a conclusion here, not something you knew beforehand

The sentence "it's diagonal with each entry corresponding to a eigenvalue" was what I wanted to hear as the answer to "how does it look like?"

Jaime Gallego

I'm bringing a new recruit here lol

ACuriousMind

@JaimeGallego Where is the fresh meat? ;)

Phase

What does it mean though

Jaime Gallego

10:40 PM

ACuriousMind

@Phase What is "it" in that sentence?

Phase

If $\lambda = +-1$, does that mean M has the entries sign flipped? Surely the root of I would be +- I though

by sign flipped i mean like $\lambda _1 = 1$ and $\lambda_2 = -1$

Balarka Sen

was M 2x2

ACuriousMind

@Phase Why are we still talking about the "root" of $I$? Didn't we settle on that not being a good notion here?

Jaime Gallego

I forgot about the points thing lol

Still, he can do Q&A on the site

So that's good

Phase

10:46 PM

@ACuriousMind yeah but I'm just confused

nevermind im dumb

Balarka Sen

It just means M, in a basis of it's eigenvectors, is a diagonal matrix with plus or minus 1 along the diagonals.

Jaime Gallego

@JorgeCalvar Speaketh

Phase

Jesus

How many square roots does $I$ have

Balarka Sen

There are many many matrices which satisfy $M^2 = I$

ACuriousMind

@Phase If by "square root" you mean a matrix that squares to $I$, then infinitely many

Balarka Sen

10:55 PM

Google involutory matrices

If you just take one of those diagonal matrices $M$ with $\pm 1$ along the diagonal entries, take an arbitrary invertible matrix $P$, $P^{-1}MP$ also squares to $I$.

aka "just basechange"

Phase

@JaimeGallego Maybe he had his internet destroyed by the mod who shall not be named

the feared one

ACuriousMind

@JaimeGallego You need 20 rep to be able to talk

Phase

@BalarkaSen I see, are there any cool applications of this in Physics?

Any properties that matrices like those have?

Jaime Gallego

@Phase "Bloody Shog9, Bloody Shog9, Bloody Shog9" ghost jumps out of mirror

2

@ACuriousMind Yeah, I didn't know (before I saw his profile page) whether he had the necessary rep.

ACuriousMind

@JaimeGallego Shows up as 1 rep user to me

Balarka Sen

11:00 PM

@Phase Applications in physics? Not that I know of

They're a neat class of matrices. That's it :P

Ya boi's favorite isometries of $\Bbb R^n$, reflection along a hyperplane, are involutory matrices

Phase

I hear they're basing a lot of the visual effects in the new IT film on the supernatural effects of Shog9 entering a room

They say there's balloons first, then screaming

@BalarkaSen What's a Hyperplane?

Balarka Sen

Forget $n$, take $\Bbb R^2$. Reflect along a line

That's a linear transformation, and the matrix of it is involutory

Because reflecting twice is the same as doing nothing; $A^2 = I$

0ßelö7

@ACuriousMind 50 minutes

perfect

Phase

so any matrix that is its own inverse?

ACuriousMind

@Phase Think about it: By definition, the inverse times the matrix gives $I$. So if $A=A^{-1}$, then $AA^{-1} = A^2 = I$, i.e. $A^2 = I$ is equivalent to being its own inverse.

Phase

11:05 PM

that's quite satisfying

What about if it's like a rotation about a third of 2pi? So that $R^3 = I$

does that have it's own class or is it boring

Balarka Sen

right

yeah these are finite order matrices

not boring

0ßelö7

(all of linear algebra is boring)

Phase

@0ßelö7 mean

Can anyone answer my electron question? I asked it on hbar, then chem.se and I still haven't had a full answer nor resource recommendations

0ßelö7

@ACuriousMind Hmm, the only issue is that I claim that $\int_M 0=\mathrm{vol}\, M$

need to figure out what I did wrong

ACuriousMind

@0ßelö7 Yeah, that doesn't sound right

0ßelö7

11:11 PM

Huh, my main reference has the same typo twice

the recursion relation should start with $u_0(y,y)=1$, not zero

ACuriousMind

@Phase Which one?

Phase

0

Q: Delocalised Electron's Electric Field

If we take an electron that's delocalised w.r.t position, how can one evaluate the electric field over some space? Is it some superposition or a sort of field with all the charge at the expectation value of the position?

quantum-chemistry electrons

0ßelö7

@ACuriousMind I think it's pretty good. I think I'll add in the proof of that lemma just to be safe, but otherwise...?

@ACuriousMind Maybe I should add some words about boundaries or noncompact manifolds?

Sir Cumference

@JaimeGallego Obligatory xkcd

If you actually do this, what really happens is Douglas Hofstadter appears and talks to you for eight hours about strange loops.

Phase

Just reminded me of a time when I was a kid

It was a sleepover and everyone was like 9, and me being an edgelord decided it would be funny to say "candyman" into the mirror in the dark room at night as per the dumb urban legend

a kid cried and i got told off

not the most exciting story but hey ho

Balarka Sen

11:20 PM

lol

i like it

Sir Cumference

@Phase I've heard of bloody mary, but not candyman

Phase

The idea was that some dude would appear behind you and stick a hook in your back or something

Idk, Cornwall is weird

Gonna play some CSGO because my life is beyond redemption. Peace!

Balarka Sen

Hi @Avantgarde

Avantgarde

Oh hey is CSGO on steam?

:O

@BalarkaSen Heylo

Balarka Sen

im finally listening to bish bosch

0ßelö7

11:26 PM

@Avantgarde it's literally made by valve

Balarka Sen

noping the hell out

Avantgarde

lol let me see

@BalarkaSen what?

Balarka Sen

last of the walker trilogy

Avantgarde

Let me check

Balarka Sen

dont lol

listen to The Drift instead

Avantgarde

11:34 PM

Googling Scott Walker gives me an American politician

Balarka Sen

lol

Avantgarde

Google should've known by now what my tastes are, by now

I used 'by now' twice. wow

Balarka Sen

"I used "by now" twice by now"

@Avantgarde youtube.com/watch?v=FbpBxXEPQow

Avantgarde

You like those vocals?

Balarka Sen

very much

Avantgarde

11:39 PM

spooky

Balarka Sen

Tilt and (at least one half of) Drift are just awesome

Interesting how a baroque pop artist went down the line of arcane experimental sonics :p

Avantgarde

Yeah such changes are interesting

Jaime Gallego

@BalarkaSen Forgive my unworthy attempt at memes

Balarka Sen

This is great

I absolutely love the promotional tagline

Avantgarde

JEE, who shrunk the kids whaa?

Balarka Sen

11:45 PM

yup that

@Avantgarde it's a reference to the original poster of Invasion of the Body Snatchers

Jaime Gallego

Honey, I Shrunk the Kids

Honey, I Shrunk the Kids is a 1989 American comic science fiction film. The directorial debut of Joe Johnston and produced by Walt Disney Pictures, it tells the story of an inventor who accidentally shrinks his and his neighbor's kids to a quarter of an inch with his electromagnetic shrinking machine and throws them out into the backyard with the trash, where they must venture into their backyard to return home while fending off insects and other obstacles. Rick Moranis stars as Wayne Szalinski, the inventor who accidentally shrinks his children, Amy (Amy O'Neill) and Nick (Robert Oliveri). Marcia...

Balarka Sen

ah right that's another movie

Avantgarde

Hm, seems like I might've heard this phrase somewhere

Balarka Sen

@Avantgarde One of my favorite tracks from the Tilt (you have heard Farmer In The City, right?): youtube.com/watch?v=wPMsTr3-G3Y

Avantgarde

The vocals don't do it for me. Though whichever style that is, he's doing it pretty well. It's very crisp

Balarka Sen

11:53 PM

True. I can see why you are objecting to the vocals