The h Bar

Kenshin

11:00

@ACuriousMind yeah but my question clearly states derive from Newton's laws, not from something else

so I've already demonstrated what level I"m at

AccidentalFourierTransform

Kenshin

lol

@JohnRennie how come the peak wavelength the sun produces (500nm) isn't the same as the peak frequency of light the sun produces (340THz)?

user228700

@JohnRennie I know, and I watched it to check for this door but in all of them, they just walk right in; the door swings open automatically! (Sorry for the latency; had gone to driving class)

user685272

Are you learning to drive a standard transmission?

user228700

Do u mean a manual? If so, yep.

user228700

11:14

I'm almost done, actually; today was my second to last class.

user685272

do you practice on hills?

AccidentalFourierTransform

learning to drive is a waste of time

user228700

Nope. Why? O_O

AccidentalFourierTransform

in a year from now all cars will be controlled by google

user228700

@AccidentalFourierTransform OK why?

user685272

11:16

that's the trickiest part

Kenshin

@AccidentalFourierTransform more lik 10 years

John Rennie

@Kaumudi.H because troll

user228700

:-P

user228700

@JohnRennie I'm doing quite well now! :-D

John Rennie

@Kaumudi.H booked the test?

user685272

11:17

try to stop on a hill and then not roll backward when you start moving

user228700

Nope. That will be done by the driving school...

user228700

@user685272 Uhhh...handbrake?

user685272

no hand brake

user228700

What kind of car doesn't have a hand brake?

user685272

just your feet

they all do have a hand brake

AccidentalFourierTransform

11:18

I dont think you can use the handbrake with your feet

but Im no expert

user228700

@user685272 What izz problem, then?

user685272

this is the ultimate test of your mastery

user228700

@AccidentalFourierTransform I mean, maybe...

AccidentalFourierTransform

the last time I sat on the driver seat was when I took the test

never again

driving is stupid

user685272

the steeper the hill, the harder it is to do. without rolling backward

user228700

11:21

@user685272 But this is plain Physics; on certain hills, this is impossible to do without the handbrake.

John Rennie

@user685272 why would you not just use the handbrake? It's not the ultimate test of mastery, it's being obtuse.

Kenshin

I never need to use the handbrake

user685272

it can be done without the hand brake

user228700

@AccidentalFourierTransform Isn't there a statute of limitations on your driving license?

Kenshin

I find it easier to not use the handbrake to be honest

When you master the hill start without the handbrake, the handbrake just slows you down

user685272

11:22

^

AccidentalFourierTransform

@Kaumudi.H no idea

I dont drive

driving is for poor people

user685272

do you have a driver?

Kenshin

yeah he has a bus driver

AccidentalFourierTransform

no, I have a sister

Kenshin

is she poor

Slereah

11:24

Hail to the bus driver, bus driver, bus driver

Hail to the bus driver, bus driver man

AccidentalFourierTransform

@Kenshin very.

Slereah

He screams and he cusses, he rams other busses

Hail to the bus driver, bus driver man

Kenshin

lol nice peom

user228700

Dammit, I wish I could get through that door...

Slereah

That is a schoolyard rhyme

user228700

11:24

It's probably a glitch in my copy of the game.

Slereah

Made famous in the Simpsons

Busdriver man! The Simpsons

►

Classic Simpsons

Oh god Carter's notation is so bad

user685272

How is your parallel parking? @Kaumudi.H They always test you on that :-)

Slereah

$I^+(S,U)$ is noted as $\underset{U}{\prec} S )$

Have mercy

And I can't seem to be able to avoid Carter's notation because it is used uniquely to define n-degree causal relations

user685272

@JohnRennie In the same way as learning about the obtuse triangle is apart of trigonometry :-)

Slereah

$$\overset{1}{<} S ) = \{ p \in M | (J^+(S) \cap \overline{J^-(p)}) \cup (\overline{J^+(S)} \cap J^-(p)) \neq \varnothing \}$$

I need to come up with a better notation

I'm not even quite sure what that set is supposed to be

Balarka Sen

11:47

@Slereah This is good

first class poetry

very Eliotesque

Here's a bad physics question: Suppose I have a bar of length $\ell$ which has uniformly distributed mass $m$, and it's lying on the ground (so the gravitational force acts uniformly perpendicular to it). Say it's positioned at the $x = 0$ line in the cartesian plane, and I want to move it to the $x = 1$ line. One way to do it is to apply an appropriate amount of force perpendicular to it through the center of mass (aka the midpoint of the bar).

On the other hand, I can apply a certain amount of force at one corner so that it gets a torque and that corner moves to the $x = 1$ line, and then do the same thing with the other corner. (I suppose there is a frictional force involved here to keep the other corner fixed, I dunno).

Is the latter in some way "physically easier" than the former? I just experienced this while moving my bed for cleaning.

user685272

yes

Balarka Sen

So I want to know the rigorous physical reason behind this.

user685272

it involves torque

AccidentalFourierTransform

and a lot of QFT

Slereah

First do you assume the spacetime it happens on to be globally hyperbolic

this is important

Balarka Sen

11:54

sigh

ACuriousMind

Please stop trolling people that come here with genuine questions.

Balarka Sen

I actually enjoyed the troll, and don't think it was harmful, but sure

user685272

the former case has no "lever arm," right?

Balarka Sen

Ah, yeah, that's the thing I suppose

Hm, so what exactly is one minimizing in the latter process? Work/time?

user685272

work = force X perp distance

ACuriousMind

11:59

@BalarkaSen Well, to tell whether it's physically "easier" you'd have to compute the work you put into it in both cases. Assuming you "barely" move the bed (i.e. not with any significant speed) In the first case you work against the frictional force over the distance $d = 1$, in the second case you work against the frictional torque over the two arcs the bed's ends sweep out

apt45

Hi all, does anyone know enough about string theory, holography and AdS/CFT to answer this question? physics.stackexchange.com/questions/343295/…

ACuriousMind

I don't see any "obvious" reason why the result of one of these computations should be lower than the other, though.

Kenshin

yeah I do @apt45

but if you're good at something never do it for free

0celo7

@ACuriousMind how do I know if I'm done in Velen?

ACuriousMind

@Kenshin What did I just say about trolling people who come here with genuine questions?

Kenshin

12:01

I didn't see it I just joined the chat

can u ples repeat it?

AccidentalFourierTransform

are Wilson loops useful at all?

Balarka Sen

@ACuriousMind Am I really doing less (or more) work? It should be the same; I'd think I have to put the same amount of force to do the job in both cases and it moves the same distance.

ACuriousMind

@0celo7 When the only main story quest left points to Novigrad. Although to finish the baron's story you should also do the side quest "Return to Crookbag Bog"

Balarka Sen

I think it's something other than work that gets minimized.

0celo7

@ACuriousMind I did that too.

apt45

12:02

@AccidentalFourierTransform Apparently if you want to prove confinement

ACuriousMind

@BalarkaSen Well, since we're working against friction in this case, the work done can be different although the final positions are the same.

0celo7

The Fiend was pretty tough

Balarka Sen

@ACuriousMind Ahh

Enlightening comment, frictional force is indeed not conservative

ACuriousMind

The work would only be the same if all forces involved were conservative, which is decidedly not the case

AccidentalFourierTransform

@apt45 I guess this is one of the cases physicists abuse the meaning of the word "prove" :-P

user685272

12:03

all 4 feet of the bed are being dragged at once

apt45

@AccidentalFourierTransform ok, up to any proof of the holographic duality ;)

0celo7

@ACuriousMind so how many endings are there?

ACuriousMind

@0celo7 So, what happened to the baron and Anna in your case?

0celo7

@ACuriousMind Baron took his insane wife to the mountains

AccidentalFourierTransform

and she committed adultery

user685272

12:06

:O

ACuriousMind

@0celo7 Ah, well, that's the "nice" ending I guess, for some value of nice.

user685272

that value may be imaginary

perhaps even hypercomplex depending on her condition

0celo7

@ACuriousMind what other endings are there?

ACuriousMind

@AccidentalFourierTransform I've reasoned myself into a corner trying to answer this question. The Ward identity naively applies to the fully resummed propagator. Or...does it? In a free theory, the "fully resummed" propagator is just the ordinary photon propagator $\frac{\eta^\mu\nu}{p^2}$. Contracting that with $p^\mu$ gives $p^\nu/p^2$, which doesn't vanish at all. What's going on?

0celo7

I imagine she dies if you don't kill the tree spirit

Slereah

12:09

@0celo7 Any less ass notation idea for this set?

0celo7

Also the Crones are pretty damn sexy

Slereah

$$\overset{1}{<} S ) = \{ p \in M | (J^+(S) \cap \overline{J^-(p)}) \cup (\overline{J^+(S)} \cap J^-(p)) \neq \varnothing \}$$

0celo7

@ACuriousMind when do I get to kill the Crones?

@Slereah no

On phone, wtf is that

AccidentalFourierTransform

@ACuriousMind you know how the quantum action $\Gamma[\phi]$ works?

its the Legendre transform of $W=\log Z$

and the fully connected correlation functions are given by functional derivatives

ACuriousMind

@0celo7 Yes, but not an easy death. Do you want me to spoil or might you want to wait for a second playthrough in the future?

@0celo7 In due time

Slereah

12:10

@0celo7 it's Carter's n-degree causal relation

ACuriousMind

@AccidentalFourierTransform Which one? The 1PI generating one?

Slereah

It looks very bad

AccidentalFourierTransform

one can "prove" (modulo anomalies) that $\Gamma$ has the same symmetries than the classical action $S$

@ACuriousMind yep

Balarka Sen

wth is 0celo even playing

ACuriousMind

@AccidentalFourierTransform Yeah, I realize one can prove the statement in the path integral approach that way

0celo7

12:11

@Slereah why is there a half open parenthesis

Slereah

$\Gamma$ is nice because you can do pretend GR with it

@0celo7 You'll have to ask Carter about this

AccidentalFourierTransform

that means that all the fully connected functions satisfy the same relations than the standard functions

so if the $n$ point function is transverse

Slereah

He also writes $I^+(S)$ as $\ll S )$

ACuriousMind

I'm less worried about answering the question as such and more worried that the frequent claim that "the fully resummed propagator obeys the ward identity" seems to be false.

AccidentalFourierTransform

its fully connected part is transverse as well

if you let $G_{\mu\nu}=\langle A_\mu A_\nu\rangle$ be the interacting (exact) two-point function

user685272

12:13

Hi @Blue

AccidentalFourierTransform

and $\Delta_{\mu\nu}=\langle A^0_\mu A^0_\nu\rangle$ the free propagator

Anonymous

@Mann If you consider friction to be applied force then the net virtual work won't be zero and that principle (in the screenshot) won't hold.

Anonymous

You need to understand the concept of virtual work first.

AccidentalFourierTransform

then you have $\partial_\mu G^{\mu\nu}=\partial_\mu \Delta^{\mu\nu}$

this is a consequence of Ward-Takahashi

0celo7

god dammit

betrayed again

user685272

12:15

by?

0celo7

AccidentalFourierTransform

and it holds in any gauge ($R_\xi$, Coulomb)

now if you let $\Pi_{\mu\nu}$ be the (amputated) irreducible two point function

Balarka Sen

@0celo7 What in the world

Slereah

I am adrift at sea

AccidentalFourierTransform

you have $G^{\mu\nu}=\Delta^{\mu\nu}+\Delta^{\mu\rho}\Pi_{\rho\sigma}\Delta^{\sigma\nu}+\‌Delta\Pi\Delta\Pi\Delta+\cdots$

Anonymous

12:16

@user685272 Hello

AccidentalFourierTransform

if you use this together with $\partial_\mu G^{\mu\nu}=\partial_\mu \Delta^{\mu\nu}$ you get $\partial_\mu\Pi^{\mu\nu}=0$

Slereah

@0celo7 I think I will have to make the causality section into a part yeah

there is way too much shit

Or several chapters, at least

Maybe causal relations and then causal hierarchies

AccidentalFourierTransform

basically, $\Pi$ is transverse but $G$ is not

but the transverse part of $G$ is "trivial"

0celo7

@ACuriousMind any idea why steam is no longer taking screenshots?

AccidentalFourierTransform

it is equal to the transverse part of the free propagator

the reason is that the transverse part of $A^\mu$, as an operator, is free

$\partial^2(\partial\cdot A)=0$

ACuriousMind

12:19

@AccidentalFourierTransform Yes, that's correct. But the texts claim "the fully resummed propagator obeys the Ward identity", which is false, unless they suddenly take that to mean the general Ward-Takahashi identity.

I guess I'll just have to write a rant that they're simply not explaining well what they're doing

AccidentalFourierTransform

yes, I would say they mean WT instead of just W

there is no consensus about what the W. identity actually is

some books consider W. to be $p\cdot \Pi=0$

some others consider it to be $p\cdot\mathcal M=0$

ACuriousMind

Okay, thanks, I needed to confirm I'm not going (more) insane.

0celo7

@ACuriousMind But I am going insane!

AccidentalFourierTransform

lol

ACuriousMind

@0celo7 I have no idea

user685272

12:22

As they say "crazy like a fox" :P

0celo7

@ACuriousMind steam overlay is shift + tab, yes?

ACuriousMind

@0celo7 yup

0celo7

@ACuriousMind it doesn't seem to work on any game

AccidentalFourierTransform

if $f$ is exact then in general $\int_{\mathbb R^n} f=0$. Under what conditions is the converse true, $\int_{\mathbb R^n} f=0\quad\Rightarrow \quad f=\mathrm d\phi$ (with smooth $f$ but not necessarily of compact support)?

we just need it to decay fast enough?

is that sufficient?

0celo7

that's so wrong it's not even funny

TIL $\int_{\Bbb R^n}1=0$.

AccidentalFourierTransform

12:25

whoops

I meant exact

ACuriousMind

@0celo7 Strange. Have you tried to restart Steam? :P

If yes, check in the options if the "Steam overlay" is deactivated for some reason

0celo7

@ACuriousMind it works in FNV

I just checked

steamcommunity.com/app/292030/discussions/0/133260492055280893

:/

Slereah

"For any timelike submanifold $U$ of $M$, $$]\underset{U}{\succ} \subset ] \gg \subset \underset{U}{\prec}$$"

This is why notation is important

0celo7

are those sexual hieroglyphs?

AccidentalFourierTransform

is that korean?

Slereah

12:33

They are

0celo7

@AccidentalFourierTransform is that first claim true in the case when $f$ is not compactly supported?

AccidentalFourierTransform

well, if it decays fast enough it should be true?

$f=\mathrm d\phi$, $\int_{\mathbb R^n} f=\int_{\partial\mathbb R^n} \phi=0$

:-P

0celo7

$\partial \Bbb R^n$ is nonsense.

You need to do it as a limit of spheres with radius $\to\infty$.

So your $\phi$ needs compact support too.

Slereah

$\partial \Bbb R^n = \varnothing$

Perfectly well defined

Just integrate on the empty set

0celo7

I can imagine some situation where $\phi$ becomes constant at infinity though.

Slereah

12:38

and $\mu [\varnothing] = 0$

So it's 100% true

AccidentalFourierTransform

brb

apt45

Why is AdS/CFT so snubbed in PSE? :'(

0celo7

I've got to get to work

@apt45 because it's not physics.

apt45

@0celo7 :O why do you say so? Maldacena would be very sad If he read your post

Slereah

To hell with Macedonia

What was he thinking

user685272

12:43

symbolically?

Slereah

Yes

This is one of the worst notation I've seen

user685272

is he a mathematician?

Slereah

I'm not sure

Ah, I think I know a good notation for it

Apparently $J^+$ is the degree zero set

So I can probably write it as $J_n^ +$???

0celo7

@apt45 unless you have made verifiable predictions, you're doing witchcraft, not physics

and as a Christian I abhor witchcraft

Slereah

That way you can write it as $$J_n^+ = \{ p \in M | (\overline{J_{n-1}^+(S)} \cap J^-(p)) \cup (J^+(S) \cap \overline{J^-_{n-1}(p)}) \cup ( \bigcup_{r=1}^{n-1} \mathfrak{Whatever}) \neq \varnothing \}$$

0celo7

12:52

flagged for obscenity

Slereah

Apparently with those sets you can generalize the notion of closed timelike curves

I'm not 100% sure but I think globally hyperbolic spacetimes are such that its n-degree causal relations are virtuous

That is, $p \notin J^+_n(p)$ for all $n$

0celo7

@Slereah @ACuriousMind do you use Geforce experience?

Slereah

I do not

0celo7

it's idea of "optimizing" games is to just turn off everything

ACuriousMind

@0celo7 I haven't found it of much use, so I've turned it off.

0celo7

13:00

@ACuriousMind Does it update drivers automatically?

ACuriousMind

@0celo7 Yes

Or, well, it will alert you when new drivers are available, which you can then tell it to install

0celo7

@ACuriousMind Then I'll keep it on. But the optimization does nothing when I can run all games (modulo Ass Creed Unity) on max settings anyway

ACuriousMind

I'm not so obsessed with my drivers being up to date, so I just check them manually once in a while

0celo7

idk why it's telling me to turn off stuff

@ACuriousMind I have no idea what drivers being out of date means

sounds scary

ACuriousMind

Nah (unless there was a really critical bug). New drivers usually just have some more specific optimizations for new games/techniques

This answer now took me a much longer time to write than I care to admit (@AccidentalFourierTransform)

AccidentalFourierTransform

13:05

it seems you assumed $\xi=1$

the analysis holds for general $\xi$

you may want to mention that

ACuriousMind

I think I assumed "I don't care wtf $\xi$ is" ;P

@AccidentalFourierTransform Why are you deleting old, upvoted answers of yours which aren't wrong as far as I can see?

AccidentalFourierTransform

also, you have $(A+A^2+A^3+\cdots)p=0$, where the matrix $A$ is $A=p^{-2}\Pi$. This implies $\frac{Ap}{1-A}=0$, and therefore $Ap=0$, with no need to split $A$ into $f,g$

ACuriousMind

@AccidentalFourierTransform I strongly suspected OP would ask me for a proof of the matrix-valued geometric series if I did that :P

AccidentalFourierTransform

@ACuriousMind they were old, silly and useless answers

Slereah

Ah, there's a decent explanation

If $p \leq_1 q$, it means that $p \in J^+(J^+(q))$

AccidentalFourierTransform

13:12

@ACuriousMind well, please prove that $\sum_{i=1}^\infty (f + gp^2)^i<\infty$

ACuriousMind

@AccidentalFourierTransform Yeah, you're right :D

AccidentalFourierTransform

or delet the answer

Slereah

I think so, anyway

Hence that's why it corresponds to widening the causality slightly

AccidentalFourierTransform

blah blah analytic continuation

0celo7

safe to eat?

AccidentalFourierTransform

13:15

0celo7

I have nothing to eat

thankfully the yogurt is still good

long trips are terrible

Slereah

@0celo7 full of healing penicillin

user685272

what flavour?

Slereah

@user685272 mold

ACuriousMind

@Slereah You're eating mold yogurt? ;P

Slereah

13:18

Isn't that what cheese is, in the end

user685272

@0celo7 not

ACuriousMind

@AccidentalFourierTransform thx, fixed

AccidentalFourierTransform

@ACuriousMind not really :-P

look at the $1+$ term

0celo7

I used eggs last night and didn't check if they were good

ACuriousMind

lol

0celo7

13:20

I didn't shit my brains out so I guess it was all right

ACuriousMind

I can't indices

AccidentalFourierTransform

$P^{\mu\nu}(p) = \frac{\eta^\mu_\lambda}{p^2}1 +...$

👍👍

Slereah

13:33

$$👍^{👍👍}(👍) = \frac{👍^👍_👍}{👍^2}1 +...$$

4

AccidentalFourierTransform

$||😂||=1$

0celo7

13:56

@ACuriousMind why do gross and groB have such different meanings

what is the etymology?

ACuriousMind

@0celo7 I think their etymology is the same - I don't know why the two words took on such different primary meanings in English and German.

But note that the English "gross" retains shades of the German groß, as in "gross mistake", which basically means big mistake

0celo7

@ACuriousMind Mhm.

But things like "gross revenue" have a slightly different meaning still.

user685272

There's a "gross ton" also.

grotesque?

hi @EmilioPisanty

Gross (unit)

In English and related languages, several terms involving the words "great" or "gross" (possibly, from French: grosse thick) relate to numbers involving multiples of exponents of twelve (dozen): A gross refers to a group of 144 items (a dozen dozen or a square dozen). A great gross refers to a group of 1728 items (a dozen gross or a cubic dozen). A small gross or a great hundred refers to a group of 120 items (ten dozen). A gross may be abbreviated as "gr" or "gro". The continued use of these numbers in measurement and counting represents a continuation of the tradition of the duodecimal number...

Emilio Pisanty

14:14

@user685272 boring

FFF system

The furlong/firkin/fortnight (FFF) system is a humorous system of units based on unusual or impractical measurements. The length unit of the system is the furlong, the mass unit is the mass of a firkin of water, and the time unit is the fortnight. Like the SI or metre–kilogram–second systems, there are derived units for velocity, volume, mass and weight, etc. While the FFF system is not used in practice, it has been used as an example in discussions of the relative merits of different systems of units. Some of the FFF units, notably the microfortnight, have been used jokingly in computer science...

user685272

@EmilioPisanty worldcat.org/title/…

can you see that^?

Emilio Pisanty

@user685272 yes

depending on what you mean by "see"

it stays stuck here

so I suspect you might want to classify that under "no"

In other news, I'm happy to see that the helen has not been removed from wikipedia

Helen (unit)

A helen is a humorous unit of measurement based on the concept that Helen of Troy, from the Iliad, had a "face that launched a thousand ships". The helen is thus used to measure quantities of beauty in terms of the theoretical action that could be accomplished by the wielder of such beauty. == Origin == The classic reference to Helen's beauty is Marlowe's lines from the 1592 play The Tragical History of Doctor Faustus, "Was this the face that launched a thousand ships / And burnt the topless towers of Ilium?" In the tradition of humorous pseudounits, then, 1 millihelen is the amount of beauty...

Secret

I want something with $\pi$ Helens

Trollface as universe explodes

user685272

Helen/0 = undefined real beauty

Secret

universe explodes because it is impossible to destroy $\pi$ cities

AccidentalFourierTransform

14:29

well, just redefine $\pi$ to $\pi=3.2$. Easy peasy

Secret

How do you accurately destroy 3.2 cities?

user685272

3 + 1/5 of a city

Secret

$$👍=\iiint \sum_{👍=👍^u👍}\omega_{👍}^{\int^{\int^{👍}d(👍^{e^e})}}+\frac{1}{👍+\frac{👍}{‌0+2👍\frac{3}{4}}}\sqrt{\ln 👍}$$

Emilio Pisanty

@Secret those thumbs up are pretty hard to read

I wonder, though

$$\boldsymbol 👍$$

damn

$$\Large \mathbf{👍}$$

ah, much better

AccidentalFourierTransform

$${\Huge{🍆}}$$

Emilio Pisanty

14:38

\HUGE does not exist

because reasons

I mean, everything LaTeX does is infinitely logical and was baked in by design and if you disagree then you're wrong

so there must be some underlying reason why the largest typeface available is \Huge instead of \HUGE

Secret

NB $\int^{\int^{\int^{\int^\cdots}}}$ is actually valid. Let $F(x)=\int f(x)dx$. Then this nested integral means $F^n(x)=F(F(F(F(\cdots F(x)))))$

Emilio Pisanty

and indeed using \HUGE is just plain-wrong typesetting

Secret

Proof: I proved that in maths chat

0celo7

@EmilioPisanty 🤔

Secret

in Mathematics, Jun 30 at 6:46, by Secret

$$S(n)(x)=\underbrace{\int_0^{\int_0^{\int_0^{\cdots^{\int_0^x x dx} x dx}x dx}x dx}x dx}_{ \text{n times}}$$

Emilio Pisanty

14:40

@0celo7 you hadn't heard?

Secret

in Mathematics, Jun 30 at 6:52, by Secret

$$S(n)(x)=\frac{x^{2n}}{2^{1+2+3+4+\cdots+n}}=\frac{x^{2n}}{2^{\frac{n(n-1)}{2}}‌}$$

0celo7

@EmilioPisanty the evidence seems to point to the opposite

Fake news?

Emilio Pisanty

@0celo7 that depends on where you get your evidence

try tex.se or their chat, for one

or more old-school like latexcommunity

or even TUG

$$🤷‍♂🤦‍♂?$$

apparently not

$$🤷‍🤦?$$

0celo7

🤦🏿‍♂️

user685272

@EmilioPisanty boat?

AccidentalFourierTransform

14:51

143

A: Jokes in the sense of Littlewood: examples?

Let "$\int$" denote $\int_0^x$. We want to find the solution to $$\int f = f-1.$$ We simply "factor out" $f$, getting $1=\left(1-\int\right)f$. Thus, $f=(1-\int)^{-1}1$. Using the geometric series, $$f=\left(1+\int+\iint+\iiint+\cdots\right)1=1+\int_0^x1~dx+\int_0^x\int_0^x1~dx+\cdots$$ Thu...

Slereah

Those aren't jokes, those are just physicist math

user685272

by mathematical physicists

Emilio Pisanty

$\begin{eqnarray} \omega'=\omega \end{eqnarray} \gamma\left(1-\beta\cos\theta\right)$ wft?

Slereah

\begin{eqnarray} \omega'=\omega \end{eqnarray}

No $'s if you use begin and end

user685272

clap clap clap

Emilio Pisanty

14:58

@Slereah no, presumably OP just meant $\begin{eqnarray} \omega'=\omega \end{eqnarray} \gamma\left(1-\beta\cos\theta\right)$

but

why?

just

why?

also, "underling"?

> Jackson justifies this statement underling that the phase of a plane wave is proportional to the number of wave crests that have passed by the observer.

Secret

@AccidentalFourierTransform that's a good one. It's mathematically valid when the inverse kernal exist (I think tha's what it is called?) and indeed used in solving many integrodifferential equations

Fredholm theory

In mathematics, Fredholm theory is a theory of integral equations. In the narrowest sense, Fredholm theory concerns itself with the solution of the Fredholm integral equation. In a broader sense, the abstract structure of Fredholm's theory is given in terms of the spectral theory of Fredholm operators and Fredholm kernels on Hilbert space. The theory is named in honour of Erik Ivar Fredholm. == Overview == The following sections provide a casual sketch of the place of Fredholm theory in the broader context of operator theory and functional analysis. The outline presented here is broad, whereas...

The discrete analogue is the antidifference operator $\Delta^{-1}=\frac{1}{1-e^{D}}$

user685272

@EmilioPisanty "underlining"

Emilio Pisanty

@user685272 doesn't quite fit the bill, though

user685272

perhaps the text has it underlined

Emilio Pisanty

@user685272 it doesn't, I checked ;-)

my guess is OP's English isn't very good

I went ahead and fixed that and other errors

if OP minds, he can roll it back

user685272

15:06

:-D

Slereah

15:54

fucking hell it's hot

Once I buy a flat I'm gonna get an AC

user685272

go to an air conditioned library

Anonymous

Stand in front of your refrigerator with the door open. :P

Anonymous

Or take an ice bath

AccidentalFourierTransform

do the salt and ice challenge all over your body

Transcript for