The h Bar

12:00 AM

@obe i'll let you know where i studied once i graduate. My situation is pretty complicated and talking about it would take a while ;P

user218912

well

user218912

I'm complicating my situation right now :D

Obliv

@obe you're very silly for doing that ;D (would replace silly with something else but we have a nord watching over us)

ACuriousMind

A nord?

Obliv

isn't that what durance (is that who it is? forgot) is?

user218912

12:09 AM

@ACuriousMind help

user218912

should I drop qft...?

user218912

because it's stressing me out to the point where I'm becoming miserable though I still have a 93% in it right now.

ACuriousMind

@Obliv "nords" don't exist in the world he lives in.

@obe I have no idea how I could possibly answer that question.

Obliv

Oh, he looks like a nord or a caveman. or both. man @obe you really need to put things into perspective. maybe go take a quick hike somewhere or something. do some introspection bro

acm may be an a.i., but he is not capable in answering questions like this unless you feed him your whole life story. then you'd have to wait for all the mathematical computation, etc. it'll take a while.

user218912

@ACuriousMind I am doing well but I'm clearly not ready for it because I have to work really hard.

ACuriousMind

12:17 AM

Having to work really hard is not necessarily a sign of not being ready for something.

You should never expect to breeze through something because you think you are "ready for it".

user218912

@ACuriousMind I feel like it's going too fast now.

user218912

is that a bad sign?

user218912

this scattering stuff is killing me.

ACuriousMind

I have no idea how well this is applicable to your situation, but: From my experience, "too fast" is how many students would have characterized the advanced theoretical lectures here. There's usually so much stuff to be done in one semester that the lectures do become rather terse and demanding to follow.

Obliv

@obe is that an intro qft course? like, it can't be all of qft in one semester right?

ACuriousMind

12:29 AM

@Obliv "All" of QFT would take several decades to learn :P

Obliv

yeah I thought so lol.

wait, several decades.. hmm didn't know it was that deep.

ACuriousMind

While it may be "deep", the main issue is that it is broad. Lattice QFT is not high energy scattering QFT is not condensed matter QFT is not non-equilibrium QFT, etc.

While these subfields are all "QFT", after the shared initial steps they diverge wildly in the goals and methods applied.

Also, "several decades" may or may not be hyperbole, depending on your learning speed and what you consider "learning" a field :P

Obliv

but the core concepts of QFT are learnable in much less time, right? Like, although classical EM may have very many sub-fields/areas of study that follow from it, the theory itself isn't that extensive.

ACuriousMind

@Obliv Sure, like with all fields.

Obliv

ok just making sure since I have no idea what QFT is like

akinuri

12:38 AM

I'll be creating a visualization of space distortion in 2D and I was wondering how mass bends the space around it. I mean, does it just bend the space towards its center, or does the mass' rotation also affects this distortion? For example, which distortion in the following image should I be using?

ACuriousMind

@akinuri The spacetime geometry of a rotating mass is given by the Kerr metric (which is different from the non-rotating Schwarzschild metric).

Obliv

@acuriousmind I could use a triple integral to define the volume of a tetrahedron of known dimensions, or I could compute the volume of a function of two vars that obtain the same shape in 3D, right?

ACuriousMind

@Obliv I don't know what your second alternative is supposed to be.

That is, I don't know what it means to "compute the volume of a function of two vars".

Obliv

okay suppose you had a function of the form f(x,y) that has linear relationships to x,y such that it forms a tetrahedron in R^3. Then, you could get the volume of the tetrahedron by integrating w.r.t. x,y for this function, right?

ACuriousMind

I think I know what you mean but you need to be much more precise in what you're saying.

What does it means that a function "forms a tetrahedron in R^3"?

Obliv

12:51 AM

@acuriousmind uhm.. i'm not sure of an example but something of the form f(x,y) = ax + by

it's a contour plot in R^2 but in R^3 it could be a tetrahedron?

ACuriousMind

@Obliv That's a function. If I'm generous I'll grant you that it defines a plane in R^3 if we plot all points $(x,y,f(x.y))$, but you'll have to convince me where the tetrahedron is and why you think integrating $f(x,y)$ gives its volume.

Obliv

in the context of this problem, given 5 vertices of the tetrahedron, finding its volume could be given by $\int \int \int_B dV$ where the bounds of the vertices are placed in their respective integrals. or, $\int \int_B f(x,y)~dA$ where f(x,y) is a function that defines the bounds of the tetrahedron in one of the axis?

ACuriousMind

What is $B$?

What does "placing the bounds of the vertices in their respective integrals" mean?

What does it mean for $f$ to "define the bounds of the tetrahedron in one of the axis"?

Obliv

oh it was 4 vertices*

ACuriousMind

I'm sorry, I'm pretty sure I know what you mean, but if I just take what you write at face value it doesn't make any sense.

I'll also note that you didn't answer my question about where the tetrahedron is and why you think you can compute its volume by integrating $f$.

Obliv

1:03 AM

$B$ would be the bounds of these vertices. So, you're given (0,0,0) (0,0,1) (0,2,0) and (2,2,0) then define bounds as $\int_0^2$ for the $dx$ integral, $\int_0^2$ for dy, etc.

ACuriousMind

That's not how the notation $\int_B$ works!

Obliv

how does it work? i thought it just meant the region that the function exists in

ACuriousMind

If you want to write the integral $\int_0^2 \int_0^2 f(x,y)\mathrm{d}x\mathrm{d}y$ in that notation, you would just write $\int_{[0,2]\times[0,2]} f$.

If you write any $\int_X$, then that means that $X$ is the entire region the function is integrated over, not "the bounds".

Obliv

So it has to be a rectangle of some sort?

ACuriousMind

No.

That was just the easiest example

Obliv

1:08 AM

hmm then I don't get why $B$ is defined by the bounds of the tetrahedron doesn't make sense

it'd be [0,2] x [0,2] x [0,2] in this case, which are the bounds of the tetrahedron

ACuriousMind

You can of course integrate over regions that are not rectangles/cubes, but then you'll not be able to write such neat bounds for $x$ and $y$. For instance, integrating over the top half of a circle would be like $\int_{-1}^{1} \int_0^{\sqrt{1 - x^2}} f(x,y) \mathrm{d}y\mathrm{d}x$.

If we call the top half of the disk $D^+$, then we would simply write the integral as $\int_{D^+} f$ in the $\int_X$-notation.

@Obliv No, $\int_{[0,2]\times[0,2]\times[0,2]}$ would be the integral over the cube centered at $(1,1,1)$ with side length 2.

Obliv

Oh okay. so then $B = [0,2]\times [0,2]\times [0,2]$ which is a cube

ACuriousMind

No, that's not the "bounds" of a cube, that is a cube!

Obliv

LOL

$\int\int\int_B dV$ then would make no sense, as that is the volume of said cube.

ACuriousMind

???

You've completely, utterly, lost me.

Let us go back to the very elementary question: What is $B$?

Obliv

1:15 AM

@ACuriousMind what I wrote above this

ACuriousMind

@Obliv You mean the cube?

Why would we need a cube here? I thought you were asking about some sort of tetrahedron?

Obliv

B is the cube I mentioned, but is only significant for the double integral that is defined by the function. the triple integral isn't the volume of the tetrahedron

ACuriousMind

You cannot integrate a function of two variables over a three-dimensional cube.

Also, you still haven't told me what the function $f$ has to do with the tetrahedron.

Obliv

er, it'd be a square [0,2]x[0,2] in that case.

I'll figure it out, brb.

Sir Cumference

2:58 AM

Guys, great news

I've won the biggest donkey on astronomy award, check the comments

Obliv

3:15 AM

why do you people even answer questions like that @sirC

oh it was put on hold nvm

Sir Cumference

3:54 AM

Woah wtf

What happened to 0celo

How'd he get banned for a month and a half?

Obliv

4:10 AM

I didn't even notice that. lol wtf @sirC

Sir Cumference

@Obliv Yeah nothing in the chat here makes mention of it

Obliv

danu did send 30 msgs to trash so I think something happened b/t him & ocelo7? idk

rip

Sir Cumference

:(

rip in peace

Obliv

@0celo7 I still have my alt if you want it. (jk)

@bernard can u ask 0celo7 if he wants to give his skype to me then can u send it to me through my email?

user218912

who banned 0celo7?

user218912

4:21 AM

I hope ACM does something about it.

user218912

@ACuriousMind are you aware of 0celo7's ban and are you able to do something to lessen it?

Shing

So Ocelo and Dana both got banned?

user218912

what?

Shing

or just Ocelo?

Is he banned in chat?

user218912

that's what it says.

user218912

4:24 AM

chat will suck for the next 3 months. :p

user218912

@SirCumference month and a half? lol it says 89 days.

Sir Cumference

@obe Crud, meant 3 months

user218912

@SirCumference what are you doing?

Sir Cumference

@obe studying

user218912

k

user218912

4:29 AM

I decided to write notes on everything I learn, so everything I know will be in a document.

user218912

that way it will be clear what I do or do not know.

rob

Evening, all

user218912

hi @rob

user218912

are you able to do something about 0celo7's ban? :O

user218912

@ACuriousMind 0celo7 wants to know if you're willing to talk to him on skype about his ban.

rob

4:40 AM

@obe So the policy, which is a good one, is that the mod team doesn't discuss individual suspensions in public.

user218912

@rob I already know the details, I'm asking if you will do something or if you don't care.

user218912

but I'm assuming you can't discuss that either otherwise you would have answered me before. :(

user218912

anyway, please do something!

rob

Mostly I'm a slow typist who chooses his words carefully.

DanielSank

5:07 AM

@ACuriousMind there's a brewing Stack Exchange.

user228700

Hi everyone :-)

user228700

What is job title of the engineers who work at NASA?

rob

@Kaumudi There are many such titles depending on what exactly each person's responsibility is, I think

Why do you ask?

Sir Cumference

5:40 AM

@rob Please do something

user228700

@rob Oh, no reason :-)

user228700

I've a quick question. If the intensity of an electromagnetic wave is given by $I=E/A.t$ where $A$ is the area over which it spreads, then, well, OK, assuming this light falls on a surface that has an area $A$, placed at a distance of $r$ from the source of the light, won't the area in the denominator be $4\pir^2$ and not $A$ itself?

David Z

$E$ is energy, right? In that case, the area depends on how much energy you're taking into account.

If you consider only the energy that falls on the surface, then use the area of the surface.

If you consider all the energy radiated by the source, than use all the area, i.e. $4\pi r^2$.

user228700

@DavidZ What dyou mean..?

user228700

Yes, $E$ is energy.

David Z

5:54 AM

Uh... what do you mean? If that was confusing, maybe there's something I don't understand about the situation you're describing...

user228700

OK, when defining intensity, am I not supposed to take all the energy into account? And then I can just multiply that with the area of the surface to find out how much momentum is being transferred to it, right?

David Z

When you say "all the energy", do you mean all the energy radiated by the source, or all the energy absorbed by the surface?

user228700

I mean the former.

David Z

Ah, well then no, you don't have to take all the energy into account.

user228700

Oh, that's new. Okay, thank you :-)

David Z

5:58 AM

Maybe this will help: intensity is like a density. Just as regular density is mass per unit volume, intensity is the density of power per unit area.

user228700

That does help, thanks!

David Z

If something has a uniform density, then you don't have to take into account all the mass of the object to figure out its density; you only need a little piece. Similarly, if a source has uniform intensity, you don't have to take into account all the energy, you only need a little piece.

@Kaumudi great :-)

By the way, a PSA in response to some earlier posts:

Major moderator actions (such as suspensions) are generally done by consensus of the moderators, and overriding such decisions is always done by consensus (if at all). Asking one of us directly to override something like a suspension is generally pointless.

4

user116211

6:24 AM

WTH is going on ;(

user116211

This is ridiculous Y__Y

user116211

I needed to ask @0celo7 about the third proof of Sylow's Theorem...

user116211

And my luck; he is banned for some sort of brief eternity ;((

user116211

@obe yes; he is the one responsible for the activity of the chat.

user116211

What I shall do for this damn period ;/

user116211

6:31 AM

::Sigh::

Sir Cumference

7:41 AM

The Astronomers' Periodic Table of Elements

That'll drive any chemist nuts

Shing

8:36 AM

Would anyone be kind enough to tell me how much skill one needs to demonstrate it (write the code) in mathematica ?

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