The h Bar

@DanielSank Yes.

A limit point of a set is a point such that every neighborhood of the point intersects the set.

@DanielSank Depending on your topology, this will make sense, or not.

For instance, in the space $X=\{a,b\}$ with $\mathcal T=\{X,\emptyset\}$, $b$ is a limit point of $a$.

@ACuriousMind Interesting. The gravitar issues vary by OS.

On my Mac partition, everything is correct. But they're screwed up on Windows.

But I recall Qmechanic being screwed up on OSX.

¯_(ツ)_/¯

3:06 AM

[Philosophical musings] we always said physics models are basically applied maths, so in a sense physics is contained in maths. Then it should be impossible for us to find a physics model that is not also maths, or is it?

Conjecture: if physics is really a subset of maths, then physics must be a mathematical object of sort. Thus a proof of the conjecture (or at least showing it is undecidable) should be possible...?

Background: The above thought is inspired from reading about 1800s gestalt psychology in the theory of art forms book I have been reading

Mr Puta - Green Stuff (Malua Remix) [OFFICIAL]

►

@Obliv hello.

what classes are you taking?

@SirCumference @HDE226868 calling astronomy nerds!

why is this star brightness data measured in electrons/sec

3:24 AM

Oh hey @0celo7

wait why did my picture change :O

magic.

i guess i'm reborn

@0celo7 did u mean me

@Obliv yes

uh physics 3 calc 3 and some shitty generals

yikes

3:26 AM

i don't remember when i left this chat but wow i got lazy these last 2 weeks lol

didn't you get AP credit for generals?

whats that

a typo

oh but you still need a lot of them to graduate

Oh, I just had 3 gen eds left.

And two were because I was lazy and didn't take AP Eng 12

3:27 AM

wtf

How many APs did you take

not many lol just 1 physics 1 calc and a CS

I took 11...

well that's where we differ

you were a super [s]nerd[/s] genius I was not

Super nerd implies studying

I didn't study

Also I'm not a nerd.

I can't code

n00b can't strikethrough

3:29 AM

HALP

~~figure it out yourself~~

ah whatever

god these lab reports are going to take so long D:

@0celo7 so did ur school start up again

WHAT

LOL

what...what are you

3:32 AM

I think you need to get your eyes checked

@Obliv like 2 weeks ago

@Obliv yup

currently working on a project

u were doing some weird chemist stuff a few weeks ago right

something with lasers and cutting metals

For work

oh cool i wish i could work for someone this early on

I work at Starbucks

It's been 5 years since I started listening to hard dance, wow

such good music

3:35 AM

yeah.. right. "ryan we're going to need you to cut these sheets of cerium"

How do you know my name

you're a celebrity

no I'm not you creep

I'll get my girlfriend to beat you up if you get too close

since you have the memory of a goldfish chat.stackexchange.com/…

what'a s goldfish

3:38 AM

a goldfish is something that transforms like a goldfish

what's hard dance @0celo7

like what I linked above

lol wtf is this

DONE WITH HOMEWORK

isn't this hardstyle?

FOR LIKE TWO DAYS

3:42 AM

nice now you can read about GR freak

hardstyle $\subset$ hard dance

seems like it'd be the other way around

no

@Obliv what's a good first order linear ODE

that's not $x=x'$

that's like the only one lol

good in what sense

I take it back

@Obliv that I can solve it and either (a) be disappointed (b) be very happy that I have a legit conjecture

$x'=tx$ isn't good either.

3:47 AM

why not it's just $x = x^t + C$

ezy

...no it's not

no it's not

what is $x=x^t+C$ supposed to be

ok of course my conjecture is wrong

wait why isn't it what I said it was

isn't $t$ just a number

OMG I'M SO DUMB

I haven't done math in so long shit

well this makes my thingie much harder to prove.

Hmm.

Time to get out a GR book.

3:55 AM

@0celo7 whatcha conjecturin'

@Obliv I'm looking at parallel transport on product manifolds

I think there's an elegant way of doing this

But I'm just turning it into an ODE problem

not very nice, but it seems to work

So I need some facts about ODEs

ooooh sounds interesting. what's a product manifold though?

look up Cartesian product

oh i thought there was some other definition of products for manifolds

just a set of ordered pairs between two sets right

well

4:00 AM

Manifold

In mathematics, a manifold is a topological space that locally resembles Euclidean space near each point. More precisely, each point of an n-dimensional manifold has a neighbourhood that is homeomorphic to the Euclidean space of dimension n. One-dimensional manifolds include lines and circles, but not figure eights (because they have crossing points which are not locally homeomorphic to Euclidean 1-space). Two-dimensional manifolds are also called surfaces. Examples include the plane, the sphere, and the torus, which can all be embedded (formed without self-intersections) in three dimensional real...

there is a technical condition on charts

but it's basically just a Cartesian product of charts

yeah if u click the link it scrolls down to it

4:18 AM

@sanya :O how are you yesterday?!!lol

5:18 AM

It would have been much easier if administration procedures have quantum like traits:

I recently have a very annoying scenario where a job application requires filling in two forms A and B

In order to fill in form A, I need to ensure the outcome of form B is sucessfully processed

however, in order to fill in form B, I need form A to be completed first

That is, In order to sucessfully get the job, I need to have $A \text{ and } B$, which means I need to show $B\rightarrow A$, which means I must show $A\rightarrow B$

Classically, the only solutions to this self referential bottleneck is either to forget about the job, or seek an alternate route that can bypass either $A\rightarrow B$ or $B \rightarrow A$

Quantum mechanics would have made this much easier. Given that recent arxiv saying that causality can be in superposition, I could have just assign the states of forms A and B into two qubits a and b respectively, and link them up as a to b or b to a. I then carry out a measurement at the very end. Then the probability distribution obtained by doing this 1000 times will ensure I have $A\rightarrow B$ and $B \rightarrow A$ done without first requiring either one to have done first

, because causality with no well defined order would have allowed me to complete B since I have completed A, but at the same time I have completed B because I already completely A, thus bypassing the order restriction of the bottleneck of this application

::Starting to wonder whether causality with no well defined order will provide a novel solution to the Liar Paradox...::

5:37 AM

(Extension) Maybe I can actually implement this chores I am suffering from into an actual quantum experiment. Suppose I have 3 qubits: A,B and C. A is prepared in some known state $\lvert \psi\rangle$. B and C are initially undetermined. Now wired B and C in a way such that the resulting eigenstate of of B is determined by the eigenstate taken by C, and the eigenstate of C is determined by that of B.

Now place a measurement device M (maybe a polariser, a magnetic field, I am not sure) between A and C, and A and B respectively. Now wired the measurement device in a way such that whether C is being measured or B is being measured first is determined by the state A. Now prepare A in the superposition $\frac{1}{\sqrt{2}}(\lvert 1\rangle + \lvert 0\rangle)$, calculate the final probability of all possibel outcomes of B and C and (B and C).

::Above is an example of a quantum rant!::

DanielSank

6:34 AM

Hey everybody, it's soapbox time:

This is why I don't like turning away questions to engineering sites.

That is all.

6:45 AM

I just use them as distribution, whether it is the Dirac delta or it's distribution derivatives

You can define $\frac{d}{dx}\delta (x)$ via $\int_{\mathbb{R}}\delta (x)dx=-\frac{d}{dx}|_{x=0}$ or via the impulse function $u(x)$ where$\int_{\mathbb{R}}u(x)f(x) dx =\frac{df}{dx}(0)$

Sorry, impulse distribution

Q: Convolution notations and some miscellanous convolution questions

Having said that, it seems I don't understood distributions well enough else the following will not arise

1

Convolution between distributions over the real number line, as it is mentioned in the optics course in my uni and also in wolfram is defined as: $$f(t) * g(t)=\int_{-\infty}^{\infty}f(\tau)g(t-\tau)d\tau$$ For convolutions, there's a nice theorem called Convolution Theorem which states for any...

notation convolution dirac-delta

7:34 AM

damn gravatar craziness

I loved my old one

this one is more funky

user116211

This is too much ;(

Yuggib, you happened to be familiar with distributions in the rigorous way? Maybe you can help me on the above MSE question?

user116211

let me see

first of all, convolutions are only allowed between a distribution and a (smooth/rapidly decaying) function

and of course you can take just partial convolutions, but you have to be careful on what you then do

@yuggib and @0celo7 do you just spend all day on chat?

7:45 AM

@SpaceOtter no, I "work"

me "too" ;)

Are you on apple?

Is there a symbol for partial convolution, or is $*$ conventionally used for convolution of all variables?

@SpaceOtter yes

So I bet you're excellent at swiping between windows when you see someone coming

@Secret if you write $\bigl(f*g(\cdot,y)\bigr)(x)$, it is quite clear that you're taking the convolution only in one variable for $g$ (and all for $f$)

7:48 AM

I see

also, you can write (but I like it less) $f *_1 g$

@SpaceOtter yeah, I'm not bad

yeah, $*_1$ was what I originally have in mind in a similar fashion of partial derivatives

but usually the chat is by far not the worst window that I have open on my desktop while working

I think the bracket one is clearer * write in notebook *

Also regarding your comment about convolution, is a gaussian raidly decaying fast enough to satisfy the convolution criteria (and not blowing up as a result)?

@Secret yes

rapid decrease means that the function, when multiplied by an arbitrary power of the variable or arbitrarily differentiated is still bounded

7:54 AM

ok

@SpaceOtter also, let's say that my work allows me a good amount of freedom in the administration of slacking off times

Oh? Like MMOs XD

ok that leaves one question ,which is that attempt proof of a special case of convolution theorem shown in Q2 of the MSE link. I often heard for physicists who are somewhat aware of the distribution nature of dirac delta that a distribution can never appear outside of an integral (although they might be not mathematically rigorous enough to work in notations of \rangleD,\psi\langle that is typical in the manipulation of distributions)

So I am guessing that I am missing an integral. However since in the convolution I already convolved wrt x and y, I don't know how it can be motivate to at a…

@SpaceOtter well...I don't like them to much

$$e^{-x^2}*\delta(y)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x'^2}\delta(y-x'-y')dx'dy'$$
$$=\int_{-\infty}^{\infty}e^{-x'^2}\delta(x')dx'=e^{-0^2}=1$$
$$\color{red}{\neq e^{-x^2}\delta(y)}$$

Known: Special case of convolution theorem $k(x)*l(y)=k(x)l(y)$

7:59 AM

the convolution with a distribution is calculated differently

also, you can only convolute two functions with an argument in common

8:18 AM

Just btw>>> That's just me stirring s*** by asking christians to explain contradictions that they can't>

8:32 AM

@yuggib, ok I have slightly modified the equation so that the convolution is well defined as it only act on common arguments x and y respectively. However I still encountered an issue.

Define 2D dirac delta, $\delta^2(x,y)=\delta(x)\delta(y)$. Then
$$(0e^{-y^2}+e^{-x^2})*\delta^2(x,y)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(0e^{-y'^2}+e^{-x'^2})\delta(x-x')\delta(y-y')dx'dy'$$
$$=\int_{-\infty}^{\infty}(0e^{-0^2}+e^{-x'^2})\delta(x-x')dx'=0e^{-0^2}+e^{-0^2}=1$$
$$\mathcal{F}(\delta^2(x,y))=1$$

That is, I am now basically convolving wrt x and y for two functions of x and y, but the result is still there. What is that formalism needed to calculate a convolution with a distribution?

your notation is unclear, and to see what a convolution with a distribution is you need to act on a test function

Pardon my physics background...
Let me try again with a test function...

But before we begin, I wan to know whether I am using the correct language in distribution theory here:

Given some test functions $\phi$, $\psi$ we know dirac delta at origin $\delta_0$ is defined by $\langle \delta_0, \psi\rangle =\psi (0)$

Therefore the 2D dirac delta at origin $\delta^{(2)}_0$ is defined by $\langle \delta^{(2)}_0, \phi\rangle =\phi (0,0)$

8:52 AM

@Secret yes

and $\phi$ has to be of rapid decrease

ok (that won't be a problem for our specific case as $e^{-x^2} \in S$ where S is the set of $\psi$ is also rapidly decreasing)

Ok the next step I am not sure...

Now convolution between two distributions $f$ and $g$ are given by

$$(f*g)(x)=\langle f(t),g (x-t) \rangle$$

user3183724

A bit of a long shot, but does anyone know of any papers that look at the Hamiltonian of two two-level systems coupled to some sort of waveguide? If we just look at the two-levels their Hamiltonian would be something like $$\omega_1 \sigma_{z,1}/2 + \omega_2 \sigma_{z,2}/2 + J_{12}\left(\sigma_{+,1}^\dagger\sigma_{-,2} + \sigma_{+,2}^\dagger \sigma_{-,1} \right)$$. When you then bring them on resonance, you get superradiant and subradiant states with respect to the waveguide

I'm just looking for an explicit calculation to show this, which means you have to introduce the continuous modes for the waveguide and do some transformation to get a drive, and then do another transformation to go to symmetric and antisymmetric modes.

@Secret no, there is no convolution between two distributions as I already said

given a distribution $T$, and a rapid decrease function $f$, then $(T*f)$ is the distribution that acts as follows on a test function:

$\langle T*f,\varphi\rangle= \langle T, \tilde{f}*\varphi\rangle$, where $\tilde{f}(x)=f(-x)$

so in fact you actually take the convolution only on rapid decrease functions

I'll let you calculate what $\delta * e^{-(\cdot)^2}$ is... ;-P

9:13 AM

Where can you find your favorited questions?

9:24 AM

@SpaceOtter physics.stackexchange.com/users/128364/…

Ok thank you. I was saving it for later

@SpaceOtter an excellent choice of favourited questions if I may say so :-)

@JohnRennie XD your wiki posts are what explained SR to me essentially. ::round of applause::

It's a long standing hobby horse of mine that using convoluted examples of light clocks is a silly way to teach students SR. Grasping that it's a geometrical theory described by the Minkowski metric is a much better way to understand it.

9:40 AM

Agreed 👏

In that case, $\langle \delta * e^{-(\cdot)^2},\phi\rangle=\langle \delta , e^{-(-\cdot)^2}*\phi\rangle=\langle \delta,e^{-(\cdot)^2}*\phi\rangle=(e^{-(\cdot)^2}*\phi)(0)$ ?

$=\int_{\mathbb{R}}e^{-t^2}\phi(0-t)dt$?

So the effect of the distribution $\delta * e^{-(\cdot)^2}$ is it convolve a given function with a gaussian, and then evaluate the result at x=0?

9:57 AM

@JohnRennie In the Rindler Metric in your post what is $x$ in the coefficient of c^2 dt^2? (the acceleration part in the brackets)

@Secret and that is equivalent to...?

Can't do that on my head, give me a pen worth of time to calculate it on paper

As I have written the Rindler metric the acceleration is along the $x$ axis. The coordinates used are those of the accelerating observer, so the observer's position is at $x=0$ (and $y=z=0$ but we usually ignore the $y$ and $z$ coordinates).

So $x>0$ means positions above the observer i.e. accelerating towards them along the $x$ axis. Likewise $x<0$ means positions below the observer i.e. accelerating away from them along the $x$ axis.

Ok I don't recognise any nonelementary function for that integral other than it is basically $-\langle e^{-(\cdot)^2},\phi\rangle=-\int_{\mathbb{R}}e^{-a^2}\phi(a)da$

Ok makes sense. It's position along the axis, (as usual) facepalm

It hadn't clicked that the direction of acceleration would mean nothing without knowing where the observer was relative to the moving twin

10:09 AM

O wait, if I apply integration by parts I get...

$\int_{\mathbb{R}}e^{-t^2}\phi(-t)dt=[\sqrt{\pi}\phi(-t)]_{\mathbb{R}}-\int_{\ma‌thbb{R}}\sqrt{\pi}\phi'(-t)dt=-\int_{\mathbb{R}}\sqrt{\pi}\phi'(a)da=0$

Still does not look right...

10:51 AM

Qmechanic, what type of theoretical physicists are you, i.e. what field you worked in theoretical physics?

12:32 PM

@Secret he won't answer

Is no one concerned that Slereah has died?

Sir Cumference

@0celo7 That's one way CCD-plus-telescopes can measure brightness

@0celo7 What telescope exactly are you looking at?

More specifically, it measures signals from stars

Basically, there is only one source of signal from a star: the light of the star itself. If the star causes $n$ photons to strike the CCD chip during the exposure, and all of them knock free one electron, then the image should have $n$ electrons. That's the signal.

12:50 PM

@SirCumference Kepler

Sir Cumference

Link to data?

On phone

Sir Cumference

Well it more than likely has a few charge coupled devices.

1:06 PM

Suppose we have a spherical insulator charges with Q inside a spherical conducting shell charges with -2Q. Now why does the conductor creates zero field inside it.

Oh and hello guys

anyone?

@GeorgeSmyridis who said it creates zero field inside it?

My textbook.

It says that the conductor creates zero field inside it.

@GeorgeSmyridis an empty conducting shell yup

@GeorgeSmyridis and also any conductor has no overall electrical field inside it

1:24 PM

@GeorgeSmyridis -2Q is on the outer surface of the conducting shell

@GeorgeSmyridis and it has a thickness which electric field doesn't penetrate it

@GeorgeSmyridis of course there you must have charges on the inner surface of the conducting shell.

So

?

@GeorgeSmyridis so, there you have 3 areas

@GeorgeSmyridis r<b , b<r<c and c<r

@GeorgeSmyridis in the area of b<r<c there is no electrical field.

@GeorgeSmyridis and charges which are placed on the outer surface (r=c) don't have any total effect on the electrical field of the area of r<b

@GeorgeSmyridis got it?

1:43 PM

in the area r<b there is the field that the insulator creates. In this area the conductor creates no field. But i still dont get it. You must have charge in the inner walls. Which should have effect right?

David Z

@2physics Probably overkill to make every message a reply, especially when there's nothing else going on in the room

@0celo7 we can do nothing. He is too deep in the data mining right now

@GeorgeSmyridis of course you do. as long as you have charges on the inner sphere , it attracts some charges of opposite sign on the inner surface of the conducting sphere

@DavidZ I found it useful cuz sometimes I leave the chatroom and using replays by others can notify me if there somebody has responded my msgs and shows where to find it easily also. sorry if it's bothering I'll try not to :)

David Z

Well, it would bother me if I were the one you were pinging. But I guess it's really up to George.

alright, I mean!

lol

David Z

1:57 PM

:-P

hah! today I got published on the best mathematical physics journal B-)

::end of bragging moment::

@Secret I talked to him. He'd dead :(

@yuggib Congrats! Is it that paper about the classical limits of states you talked about a few months ago?

@ACuriousMind nope, another one

@yuggib congrats, what's the name of the journal?

2:00 PM

the one on the classical limit of states in under review, and will be for a while I am afraid

(and I'm not too optimistic it will end up well :( )

@2physics CMP

@GeorgeSmyridis here is another example of that kind

2:16 PM

2:34 PM

@ACuriousMind Lol, first analysis problem set was "terrible" and we now have to write stuff like $\delta(x,N,\epsilon)$ and everything has to come out as $\epsilon$ in the end

@2physics Yeah I get how the charge is distributed but I still don't get the thing about the the filed.

Field

@0celo7 You mean, the others completely borked it and now he tries to teach them to write more stringent proofs?

Keeping in mind what the $\delta$ and $\epsilon$ depend on is a bit difficult at times, and actually writing that out isn't a bad idea. The thing with the "has to come out as $\epsilon$ is still stupid, though

@ACuriousMind I think people were guessing what should come out at the end

like, saying it's $<4\epsilon$ to be safe

lol

@ACuriousMind Thanks for assuming I didn't bork :)

2:46 PM

Well, that's one way to try and trick the TA

In that case, I see a certain value of requiring $\epsilon$ at the end, if only to catch the ones who don't actually understand the inequalities they're using

@yuggib I am not sure what I am supposed to come up with for that integral

3:14 PM

@GeorgeSmyridis I figured it out. there are two approaches to find the E field: 1. you can find the individual fields created by two insulating and conducting spheres in a specific point and then adding them together results in the total value of the E field in that point.

2. you can use the Gauss law formula while both objects(two spheres) are considered present.

in this example, the author has pursued the first approach, therefore he has calculated the E field caused by two objects (spheres) in the areas separately; afterwards he has added them together to find the total value for the E field at the mentioned point.

"separately" means he supposed that there is only a conducting sphere with a net charge of -2Q and it's obvious that , there is no E field inside the conducting sphere. then he supposed it about the insulating sphere which has resulted in the electric field caused by the Q charge of the insulating sphere; and finally he has added these two values (0 and the E field caused by the insulating sphere) together and achieved the total value of the electric field at the points of that area(E2)

that's why it says, the charge on the conducting shell creates zero E field inside the conducting sphere.

3:37 PM

:| don't know how to write the equasions here

Write it in a question section and then send the pic.

René G

@DavidZ Why did you put my question on hold? It's a perfectly reasonable homework question, I posted all the work I had done, etc.

There's a link in the description in the upper right corner of the chat that tells you how to get equations in chat. It's not that complicated.

3:45 PM

@ACuriousMind I know. just don't know how to type eqs using mathjax

@RenéG Showing all your work is not sufficient to make a question on-topic on physics.SE. We expect questions to ask a conceptual physical question, not to ask us to solve an exercise or check someone's work.

@2physics It works just like with LaTeX. If you can't type LaTeX, you should learn it, it is a very valuable skill.

@ACuriousMind yup I also don't know latex. I'll try to learn it soon

3:59 PM

oint\epsilon\vec{E}.\vec{dS}=\sum(Q's\ inside\ S\ +Q's\ on\ S\over2

@ACuriousMind what's the pb with this ? can you help?

\int\epsilon\vec{E}.\vec{dS}=

$\int\epsilon\vec{E}.\vec{dS}=$

yaay

lol

$\oint\epsilon\vec{E}.\vec{dS}=$

$\oint\vec{E}.\vec{dS}=Q_{T}/\epsilon_{0}$

@ACuriousMind I learned it, was not so hard :D thanks

I had forgotten that math mode is indicated by a dollar sign: $. :P

4:18 PM

@SpaceOtter : can you give a link to these JohnRennie wiki posts which explained SR to you?

Ok hang on

what is time This one may be unnecessary but it is sort of one of the three posts

what is time dilation

Twin paradox

@SpaceOtter : so, what is time? In your own words. Take your... time.

There's a saying in physics, usually attributed to Einstein: "you do not really understand something unless you can explain it to your grandmother". If you can't explain something succinctly, you have to ask yourself if you really do understand it. That's why I answer questions. To understand things.

A have a bit of a problem here: if time and space are "equivalent" then why for any given t there is only one point on the worldline, while for a given value of x there might be 0, or 1, or many points of the worldline. — Madan Ivan Aug 15 at 10:43

we have no known instance of detecting any physical phenomenon that is not continous in time

Tsk, look at the time. I have to go I'm afraid.

As far as learning SR is concerned it's a dimension just like the spacial ones (x,y,z) except we don't know why we only goes forward but most evidence seems to point that way despite it's ability to stretch

So when we define an event specifically we give it's time coordinate as well (x,y,z,t)

4:27 PM

@Secret : time and space are not equivalent. Time is a dimension in the sense of measure, but not in the sense of freedom of motion. I can hop forward a metre, but you can't hop forward a second.

Gotta go.

See ya @JohnD

So in SR we treat it just like a spacial dimension only we put it into the same terms hence it's coefficient in the Minkowski metric is $c^2$

@ACuriousMind No TA

@ACuriousMind Yup

@ACuriousMind People were upset because the prof handwaves when writing "obvious" proofs in class, and people just copied his handwaves

For example, he proved uniform convergence preserves continuity for functions $\Bbb R\to\Bbb R$ in class

And on the homework he asked us to prove it for $E\to F$, with $F,E$ Banach

And people just copied his proof for $\Bbb R\to\Bbb R$ and he didn't accept it because it was too sloppy

lol

@0celo7 That is kind of a dick move

Good lecturers should make it clear when something they handwave should be done better in a written-down proof.

For us, our lecturers will have the titles "sketch proof" or somethign simialr to indicate the proof is kinda handwave

@ACuriousMind he said it should be obvious

We were supposed to talk about compactness today but ended up just going over the homework

4:51 PM

I still have no idea what that integral is supposed to mean unless gaussian convolution has its own symbol

what integral

$\int_{\mathbb{R}}e^{-t^2}\phi(-t)dt$

what about it?