CCD

0celo7

Sep 2, 2016 03:19

@SirCumference @HDE226868 calling astronomy nerds!

why is this star brightness data measured in electrons/sec

Obliv

Oh hey @0celo7

wait why did my picture change :O

0celo7

magic.

Obliv

i guess i'm reborn

@0celo7 did u mean me

0celo7

@Obliv yes

Obliv

uh physics 3 calc 3 and some shitty generals

0celo7

Sep 2, 2016 03:26

yikes

Obliv

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

0celo7

didn't you get AP credit for generals?

Obliv

whats that

0celo7

a typo

Obliv

oh but you still need a lot of them to graduate

0celo7

Sep 2, 2016 03:27

Oh, I just had 3 gen eds left.

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

Obliv

wtf

0celo7

How many APs did you take

Obliv

not many lol just 1 physics 1 calc and a CS

0celo7

I took 11...

Obliv

well that's where we differ

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

0celo7

Sep 2, 2016 03:28

Super nerd implies studying

I didn't study

Also I'm not a nerd.

I can't code

n00b can't strikethrough

Obliv

HALP

0celo7

~~figure it out yourself~~

Obliv

ah whatever

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

@0celo7 so did ur school start up again

0celo7

WHAT

Obliv

LOL

0celo7

Sep 2, 2016 03:32

what...what are you

Obliv

I think you need to get your eyes checked

0celo7

@Obliv like 2 weeks ago

@Obliv yup

currently working on a project

Obliv

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

something with lasers and cutting metals

0celo7

For work

Obliv

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

0celo7

Sep 2, 2016 03:34

I work at Starbucks

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

such good music

Obliv

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

0celo7

How do you know my name

Obliv

you're a celebrity

0celo7

no I'm not you creep

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

Obliv

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

0celo7

Sep 2, 2016 03:37

what'a s goldfish

Obliv

a goldfish is something that transforms like a goldfish

what's hard dance @0celo7

0celo7

like what I linked above

Obliv

lol wtf is this

0celo7

DONE WITH HOMEWORK

Obliv

isn't this hardstyle?

0celo7

Sep 2, 2016 03:42

FOR LIKE TWO DAYS

Obliv

nice now you can read about GR freak

0celo7

hardstyle $\subset$ hard dance

Obliv

seems like it'd be the other way around

0celo7

no

@Obliv what's a good first order linear ODE

that's not $x=x'$

that's like the only one lol

Obliv

good in what sense

0celo7

Sep 2, 2016 03:45

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.

Obliv

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

ezy

0celo7

...no it's not

Obliv

no it's not

0celo7

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

ok of course my conjecture is wrong

Obliv

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

0celo7

Sep 2, 2016 03:52

well this makes my thingie much harder to prove.

Hmm.

Time to get out a GR book.

Obliv

@0celo7 whatcha conjecturin'

0celo7

@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

Obliv

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

0celo7

look up Cartesian product

Obliv

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

just a set of ordered pairs between two sets right

0celo7

Sep 2, 2016 04:00

well

Obliv

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...

0celo7

there is a technical condition on charts

but it's basically just a Cartesian product of charts

Obliv

yeah if u click the link it scrolls down to it

2physics

Sep 2, 2016 04:18

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

Secret

Sep 2, 2016 05:18

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...::

Secret

Sep 2, 2016 05:37

(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

Sep 2, 2016 06:34

Hey everybody, it's soapbox time:

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

That is all.

Secret

Sep 2, 2016 06:45

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

Secret

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

1

Q: Convolution notations and some miscellanous convolution questions

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...

yuggib

Sep 2, 2016 07:34

damn gravatar craziness

I loved my old one

this one is more funky

user116211

This is too much ;(

Secret

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

user116211

yuggib

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

Space Otter

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

yuggib

Sep 2, 2016 07:45

@SpaceOtter no, I "work"

Space Otter

me "too" ;)

Are you on apple?

Secret

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

yuggib

@SpaceOtter yes

Space Otter

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

yuggib

@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 $f$)

Secret

Sep 2, 2016 07:48

I see

yuggib

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

@SpaceOtter yeah, I'm not bad

Secret

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

yuggib

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

Secret

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)?

yuggib

@Secret yes

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

Secret

Sep 2, 2016 07:54

ok

yuggib

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

Space Otter

Oh? Like MMOs XD

Secret

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…

yuggib

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

Secret

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

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

yuggib

Sep 2, 2016 07:59

the convolution with a distribution is calculated differently

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

Space Otter

Sep 2, 2016 08:18

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

Secret

Sep 2, 2016 08:32

@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?

yuggib

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

Secret

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 $\langle \delta^{(2)}_0, \phi\rangle =\phi (0,0)$

yuggib

Sep 2, 2016 08:52

@Secret yes

and $\phi$ has to be of rapid decrease

Secret

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.

yuggib

@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

Space Otter

Sep 2, 2016 09:13

Where can you find your favorited questions?

John Rennie

Sep 2, 2016 09:24

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

Space Otter

Ok thank you. I was saving it for later

John Rennie

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

Space Otter

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

John Rennie

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.

Space Otter

Sep 2, 2016 09:40

Agreed 👏

Secret

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?

Space Otter

Sep 2, 2016 09:57

@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)

yuggib

@Secret and that is equivalent to...?

Secret

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

John Rennie

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.

Secret

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

Space Otter

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

Secret

Sep 2, 2016 10:09

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...

Secret

Sep 2, 2016 10:51

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

0celo7

Sep 2, 2016 12:32

@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.

0celo7

Sep 2, 2016 12:50

@SirCumference Kepler

Sir Cumference

Link to data?

0celo7

On phone

Sir Cumference

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

The h Bar

Participants