You know

While area 51 spends billions on making fancy planes

The actual programs for weird spaceships is way underfunded!

IIRC the NASA program for non traditional propulsion methods was under 2 million bucks

And it was pretty wide spanning

then again most avenues were not productive

@DanielSank : here you go, Is there a deep reason for this correspondence? You betchya.

OK I'm off now. Catch you later.

Test $\setminus$

Ah, yes.

Of course, it has to be different from a quotient.

Still, too close to a quotient for my taste

Isn't the set minus supposed to be smaller

Like a - slightly inclined

I have.

Like at a 45 degree angle?

Never seen $\setminus$ @ChrisWhite???

@Danu Wat? Chris suggested that

Chat on mobile sucks

Also hahahaha @DanielSank I'm sure all your questions about decay rates have been cleared up after that response.

Hi guys! So I am really struggling with the concept or wavenumber

Can I walk you through my conceptual though process?

Hi guys! So I am really struggling with the concept of wavenumber

Sure

ok

so first I start with frequency

frequency is the amount of cycles per second

now if I flip that

its the amount of seconds per cycle

Yeah

now i try to apply the same logic to wavelength

and it doesnt work

wavelength is the distance per cycle

but the cycle distance is defined by the wavelength so it will always be 1

Up to a factor it does

the amount of wavelengths per cycle is always one wavelength

@Danu I don't want to delete it because then DanielSank can't see it, but I cannot in good conscience click "Looks OK" or "Skip", really.

and vice versa, the amount of cycles per wavelength is always one cycle

It's only reasonable to delete it @ACuriousMind, IMO

I know that the way I am interpreting this is wrong, so I was wondering what about it is wrong

Nova, wavelength is meters per 2pi radians, then radians per meter is exactly k the wave number

Note the (non essential) factor 2 pi I threw in

@Danu I completely agree. I'll take a screenshot for Daniel, I guess, just in case.

hmm ok ok let me try to figure this out

Nova

so lets say the wavelength is 10 meters, from peak to peak is ten meters

Your problem is this

The length unit is not wavelength it's meter

Cycles per meter is k up to the 2pi which just converts to radians

You see?

um

so 2pi radians is a wave

It's just that you mistakenly took 'wavelength' as the unit of length, but it's 'meter'

2pi radians is one cycle, yes

Deleting his answer...a maelstrom awaits...

Wavelength is meters per cycle, (1/2pi) wave number is cycles per meter

ok ok

wow ok I get it

im so dumb

@Danu Utterly pathetic.

Heyo!

so its a fixed distance, 1 meter, and we are asking how many cycles happen in that 1 meter

@0celo7 It probably won't get deleted because some reviewers think pretty pictures constitute an answer. It even took moderator action to get rid of that cohomology answer.

or how many meters happen in 1 cycle, aka, the wavelength

No wonder I couldn't find it when Sam mentioned it

What was is about

so a wavelength of ten, means 0.1 cycles happen PER meter

Danu you are awesome thank you

Yeah, nice John.

DRIPPING

@JohnDuffield Stop abusing moderator flags. Last warning. Next time it's a suspension.

No fluids in the lab!

Lol did someone get flagged?

Probably me 😊

@0celo7 Just let the nice blue-colored people do their work ;)

Hey I keep them employed

@ACuriousMind The nice blue-colored people have done their work ^

Test. Still alive.

Wow, one hour.

Wonder what he flagged this time

Very well. Where were we?

Homeomorphisms

Hm, nowhere really, it seems.

@0celo7 Ah, yes, but that was a bit ago. New conclusions or questions?

So if a map is not homeomorphic on some set it certainly won't be on a set containing the first set?

@0celo7 Yes

hey guys are there any waves with a lower wave length than gamma waves?

We're gonna do a whole bunch pf covering space theory in topology yayyyy

Yes Nova

It goes up to infinitely small.

(Down to?)

@ChrisWhite what? Im on mobile, aint no stars here

Lie

Srsly

Switch to desktop mode, star, switch back.

Ever seen me spell so badly while not joking?

I don't notice your spelling

@Danu I didn't notice anything unusual :P

Is there a thing like desktop mode?! How?!

Oh hey Im seeing some star now, on the right

But no list and idk how to star stuff

@ACuriousMind Well, that fact is obvious but I don't know how to prove it

@Danu Hit menu, then full site.

Oh hahah @ChrisWhite now I see what youre referring to. No it wasnt me.

Niiiiice

Oh god the zooming is terrible

@0celo7 Well, it's equivalent to "Every restriction of a homeomorphism is a homeomorphism onto its image" by contraposition, no?

Yep.

If you're on a phone it's horrible. Tablets are manageable.

...why are you on mobile around midnight, anyway?

Either your night should be going well enough to not look at your phone, or you should be home :P

Im home. Just dont wanna unpack mah bag

@ACuriousMind I don't know how to prove either

Anyways goodnight all

@0celo7 You're probably overthinking it.

@Danu gn8

Maybe

Hm

Playing Fallout

I found a camera in the bathroom

WHAT WAS HE TAKING A PHOTO OF

@ACuriousMind I need the proofs

@Slereah The graffiti on the walls, obviously

@0celo7 Okay. Let $f :X\to Y$ be a homeomorphism, and let $\bar{f} : U\to f(U)$ be the restriction to a subset $U$ of $X$. Then $\bar{f}$ is bijection (you agreed this is obvious). Let $V\subset U$ be open. Then $\bar{f}(V) = f(V)$, so $\bar{f}(V)$ is open because $f^{-1}$ is continuous. Let $W\subset f(U)$ be open. Then $\bar{f}^{-1}(W) = f^{-1}(W)$ is open because $f$ is continuous. Therefore, $\bar{f}$ is a bicontinuous bijection and hence a homeomorphism.

Is U open

Hm...let's say yes, because otherwise there are some intersections missing, I think.

@ACuriousMind I had that exact same proof

Seemed too obvious

Figured I was missing something.

Called it.

You didn't call anything. You made a lucky guess!!!!

Heyyy

I just read

They probably found a fly

"Fallout 4 has no level cap"

Yesss

I hate level caps so much

@ChrisWhite Wow.

::googles what LIGO is::

Wow @ACuriousMind

Shame on you @ACuriousMind

I'm an engineer and I know what it is

@ChrisWhite LIGO's ground-based, right? Not like, say, LISA.

@Danu lol, did you come back from bed to say that? :D

What, significant wave signals, @ChrisWhite

@ACuriousMind from the dead

Ehh?

Ahh, Big Science at work.

@ChrisWhite no way thats torture

@ChrisWhite That seems a bit wasteful. Also, doesn't that mean the rumors couldn't be trusted, like, at all?

good science is torture.

@ChrisWhite Are the rumors from those top people?

Man! I hope they do find something. But I also thought they wouldnt be in "detection range" yet.

As in not sensitive enough for actual waves to be measured yet

@ChrisWhite Well this is deeply exciting

Yeah sure

But when was something expected? Not yet, right?

@ChrisWhite Is this the kind of thing that will go on the arXiv as soon as confirmed? I don't really know the protocol in astro

if confirmed^

I can see the viXra tittle

God's heartbeat confirmed at LIGO

Photon = Dirac wave on gravitino belt confirmed.

Haha true @ChrisWhite.

Uh SUSY is wrong @Danu

@Danu Seems legit.

My cute particle names say differently

The -inos are cute, but the s- prefix is silly

@ChrisWhite Indeed. I've always wondered how such an instrument would be calibrated though. I mean you have to know what you're "looking at", right?

Sneutrinoino

Who are the top people

Sanity check: the solution book solution here physics.stackexchange.com/questions/218213/… is totally bogus right?

Witten, Some Random Astro Dude and Einstein's corpse?

@0celo7 That... could actually make a good scifi short story\

What could

I'm the most prolific chatter, you have to be more specific. I'm on mobile

oh "gods heartbeat confirmed at LIGO". Something along those lines.

@ChrisWhite So like binary stars that we can observe optically, at the right distance and with the right orbital frequency?

Something like that?

Although do we even have a model for how gravitational waves would be attenuated over distance? I mean inverse-square law and all that, but gravity is non-linear so I would expect matters to be a bit more complicated...

Hmm ok

OK that makes sense

@ChrisWhite First optical... then x-ray & [...]... then neutrino... then gravity (?)

@NeuroFuzzy If the problem is drawn correctly (and I would agree that that is how to interpret the text), yes.

Question: We know that a photon makes a distinct "blip" on an electromagnetic detector. Do we have any idea what the analogue of such a blip would be for a graviton?

Do we even have any guess for how low the intensity has to be to see individual gravitons?

Or is trying to draw an analogy with EM here totally misguided?

I also think that the "blip" scenario, favoured by Feynman, is a bogus argument for the existence of photons

@ACuriousMind physics.stackexchange.com/questions/215173/… is probably relevant for that one.

Anna writes

> Gravitons are to gravitational waves the theoretical analogue of photons for electromagnetic waves.

I have no idea if that's right or not, though.

@MarkMitchison But...we can do those slit experiments with low intensity light where individual blips (actual localized areas of the screen lighting up) accumulate to give the usual interference picture after a long time, no?

Back to bed...

@ACuriousMind Yeah, that's true. But it could be just a bunch of localised electromagnetic wave packets hitting the detectors in different places. Of course this is a bit far-fetched...

But anyway I feel that measuring highly attenuated "classical light", i.e. a laser, is not the best demonstration of photons. It says more about our detectors than the state of the field. But perhaps I am wrong about that.

Well, what's your favourite demonstration?

@MarkMitchison The wave packets aren't localized, are they? They should be spread out on a macroscopic scale (before detection)

@ACuriousMind Photon counting from atomic flourescence seems rather nice.

After you absorb some light, you also detect that there is no longer any light there.

Seems a bit more particle-like than just Poissonian photon counting, where you can randomly get "blips" as close together as your detector resolution will allow

@NeuroFuzzy I was talking about a rather contrived classical explanation for the double-slit experiment. I think people did consider such things in the past, e.g. classical stochastic electrodynamics.

@MarkMitchison Ohh talking about the quality/convincing-ness of demonstrations. Yeah I follow.

lol

@ChrisWhite Yes, yes they do.

@NeuroFuzzy : LIGO is... an interferometer. An earlier interferometer failed to detect something.