Mathematics - 2016-06-19 (page 1 of 2)

Leaky Nun

00:10

@Semiclassical How's it now?

Andrew L

Why did you remove it and why is the SE so slow at notifying me

anon

I recall there being a closed form for $\sum |{\rm Aut}(G)|^{-1}$ where $G$ runs over all (isomorphism classes of) finite abelian groups. Can't locate it though. Ideas?

nevermind, apparently Hall proved it equals $\sum |G|^{-1}$ (same $G$s)

Semiclassical

00:31

hi chat

@andrewL because I misread what you wrote

thought you meant taking tan of 0.78 radians, which is nearly pi/4 radians and therefore would have tangent of about 1.

Andrew L

Ahh, I see.

Forever Mozart

01:25

in the evenings I have to spike my coffee with Kahlua to counteract the caffeine

see, I have a method

Akiva Weinberger

01:37

Playing with my Rubik's cube, it occurred to me that $(1~2)$ did not seem to be in $\langle(1~2~3~4),(3~4~5~6)\rangle$, but I couldn't prove it

(Rubik's cube interpretation: One can't switch two cubies by only turning the top and right-side faces.)

Leaky Nun

Well, one can't switch two cubies at all.

anon

"alex, please don't ask what is a stack"

Leaky Nun

"Helium [...] at a standard pressure [...] can never be a solid" :o

Forever Mozart

user174558

@Monad 1. You may not have expressed yourself properly to him 2. He may have misunderstood you 3. A lecturer in mathematics does not mean anything

user174558

01:50

@anon I only know one Alex, and his name is Clark.

Forever Mozart

is 20 references too many for a 10 page paper?

user174558

There is no such thing as too many or too few.

user174558

You should reference what you need to, no more no less.

Akiva Weinberger

@LeakyNun You can if you ignore the edge cubies. (I forgot to mention I was doing this on a 2x2.)

Balarka Sen

random movie scene from something i rewatched recently.

Forever Mozart

01:59

i love the introduction to that movie

user174558

I hate the movie Pi.

user174558

I watched one minute and switched off.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

"Hate" is such a strong word.

Balarka Sen

i like everything about that movie.

user174558

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Hate need not be a strong word. I hate ice cream.

user174558

02:01

My favourite movie is Blue Lagoon: The Awakening.

Forever Mozart

you know the intro scene with the figures moving in the background, and then you suddenly realize they have stopped moving and the music comes in

Balarka Sen

yes

Forever Mozart

i did not understand it but there were some good scenes

user174558

There is a recent movie Midnight Special. It is so terrible.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

@JasperLoy but you still have to put effort into hating ice cream, no?

Forever Mozart

02:03

oh I want to see that, dont spoil please

user174558

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Not much effort.

Balarka Sen

The moving people are Andrei's (the main character's) family. Appears multiple times in dream sequences; I always interpreted it as representing Soviet Russia's suffering.

user174558

I just dreamt Darth Vader was inviting me to a light sabre fight.

user174558

I tried to turn on my light sabre but could not, so used it as an excuse to run away.

Forever Mozart

you know what Freud would say about that

user174558

02:05

Well, to me, dreams are just random collections of thoughts which may not mean anything.

user174558

I learnt not to take my thoughts too seriously.

user174558

Sometimes, I think of all kinds of weird things which I don't mean to do.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Good idea.

Forever Mozart

lots of modern scientists think dreams served an evolutionary purpose

user174558

If your child is worried about his bad thoughts, tell him that they don't mean anything. They are just thoughts, not his real intentions.

user174558

02:07

To me, physics is real science, but the social sciences suffer from bad experiments.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

How are the meds working? @JasperLoy

user174558

To treat humans as subjects in experiements is different from using wooden blocks to show that F=ma.

Forever Mozart

like nightmares are meant to mentally prepare you for certain rare horrible situations

user174558

Every human is different, and statistics on how people react don't mean anything to me except meaningless numbers.

Forever Mozart

like a premonition, to keep you from being eaten by a bigger animal

user174558

02:09

I am not even sure evolution is true. I have not studied it though. I think it is just a theory.

user174558

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Thinking of going off meds by the end of this year.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Hmm...

user174558

Every few days, I hear scientists publishing some study on humans. I wonder how many of these studies are pressure from publish or perish.

user174558

I no longer take these studies seriously, those on human beings that is, because human beings are not blocks of wood for F=ma.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Money talks.

Forever Mozart

02:11

if you cite the chief editor of a journal, will they be more likely to accept your article ?

user174558

Only they know. Every person is different.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Probably

Mike Miller

02:22

If that makes a difference, it's a bad journal.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

corruption is everywhere :p

Forever Mozart

02:39

oh nooooooooooo.

user174558

02:56

My journal is in my drawer, lol.

Forever Mozart

at least its not the garbage can

s.harp

11:46

whats the common notation for real version of $L^2(X)$? (ie equivalence classes of square integrable functions from $X$ to $\mathbb R$ rather than from $X$ to $\mathbb C$.)

s.harp

11:59

After some searching I think $L^2(X;\mathbb R)$ or $L^2(X,\mathbb R)$ is normal, but I have not seen many examples

Owatch

Given: $ax^{2} + bx + c$, is there an easy way to compute the $n^{th}$ terms of $(ax^{2} + bx +c)^{n}$?

Matix

Hi! Can anyone please answer this question?chemistry.stackexchange.com/q/53927/31216

Owatch

$c_{n} = c^{n}$

But the rest I'm unsure.

Tobias Kildetoft

@Matix Why would anyone here be able to answer a chemistry question?

Matix

its actually in a chemistry numerical but deals with math.

its not really chemistry.

Tobias Kildetoft

12:02

@Owatch this is the generalized version of the binomial theorem

Leaky Nun

What's the complexity (lowest) record for bitwise multiplication algorithm in terms of number of bits?

@Owatch Trinomial coefficients.

Matix

how do we find approximate simple whole number ratio?

Owatch

Yes, I seem to remember a bit of that from high school..

Something to do with a pyramid too.

Matix

of 1.098 and 0.277?

Tobias Kildetoft

@LeakyNun As far as I know if $n$ is the greatest number of bits, then it is $nlog(n)$

Leaky Nun

12:03

@TobiasKildetoft And what is the algorithm?

@Matix Euclidean algorithm

Tobias Kildetoft

@LeakyNun No idea. Something involving fast Fourier transforms

Matix

if i divide 0.198 and 0.277 it comes 3.963.....how do i write that as a ratio?

@LeakyNun what do you mean?

Tobias Kildetoft

you did the division incorrectly first of all

Matix

uh i meant 1.098

please help.

Tobias Kildetoft

What numbers are you dividing?

Matix

12:09

1.098 and 0.277

Tobias Kildetoft

and what is wrong with the result?

Matix

i have to find a simple whole number ratio of these two numbers.

Tobias Kildetoft

there is no such, as their ratio is not a whole number

Matix

like 2:3 or something

the approximate ratio

Tobias Kildetoft

how approximate?

Matix

12:12

I dont know about that.

normal approximation maybe

Tobias Kildetoft

then this is not a math question. The correct ratio is 1098/277, possibly with some cancellation

Matix

okay.I ll ask in a chemistry chat or something.

Leaky Nun

@Matix continually take reciprocal of ratio and record decimal part until it is close to an integer

@TobiasKildetoft ok, thanks

Matix

i did not understand @LeakyNun.

Leaky Nun

For example, 3.963 is close to 4

so you can approximate it in one step

for other more complicated ratio like 56:138, you would need to do more stuff

@Matix Are you interested?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

12:20

he's asking in the Chem chat

Leaky Nun

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ whatever

Matix

yes sorry

Leaky Nun

@Matix Do you understand my response?

Akiva Weinberger

Wanna see something ~~stupidly~~ surprisingly hard? math.stackexchange.com/q/1831401/166353

Matix

so the ratio is 4:1?

Leaky Nun

12:22

@Matix Yes. If you're interested I can tell you more.

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ I belong to the chem chat also

Matix

oh and what about 56:132?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:-)

Leaky Nun

@Matix It was 138, but if you want I can do with 132 as well

Matix

how do we do it with 138?

Balarka Sen

@AkivaWeinberger weird

Leaky Nun

12:25

@Matix So we find 138/56, and we find that it is 2.464

this is not close to a whole number

so we remove the 2, and keep 0.464

Do you understand it until now?

Matix

okay i have understood but why take the reciprocal?

Leaky Nun

so, we still want to estimate 0.464, so we take its reciprocal

1/0.464 = 2.154

This is close enough to an integer, so we take it as 2

(If you are not satisfied, you can do more steps until it approaches an integer)

so:
  138
= 56 * 2.464
= 56 * (2 + 0.464)
= 56 * (2 + 1 / 2)
= 56 * (5 / 2)
So 138/56 ~ 5/2

Akiva Weinberger

@BalarkaSen I think it's equivalent to showing that the convex hull of a finite union of rays is closed (at least when all the rays have their vertex at the origin, though I doubt that's needed for it to be true)

Tobias Kildetoft

@AkivaWeinberger Hmm, I would have been tempted to claim that linear maps are closed. Is this not the case?

Matix

i am so bad at this.

thank you

Leaky Nun

12:29

welcome

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

What course are you taking? @Matix

Balarka Sen

@TobiasKildetoft No, it's not true.

Matix

i am only 16.

Leaky Nun

@Matix I just went 17.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

ok

Tobias Kildetoft

12:31

@BalarkaSen I see (I didn't really think that much about it, so there is probably an easy example)

Balarka Sen

@AkivaWeinberger That sounds good to me. Instead of thinking abstractly, I'd start by thinking about the image of each term in $Ax$, for $x_i > 0$. Boundary of those regions are rays passing through origin, and union should be the convex hull. I don't have a good idea how to do this though.

Not union, I meant *intersection.

@Tobias Project the graph of $xy = 1$ in $\Bbb R^2$ to $\Bbb R$ to the x-coordinate?

Akiva Weinberger

@TobiasKildetoft One is given in the question post

Tobias Kildetoft

@AkivaWeinberger Ahh, I only read the initial question. Good to know that my intuition fails as usual :)

Akiva Weinberger

Closed area above graph of $y=e^x$, projected onto $y$-axis

Leaky Nun

Trinomial expansion

In mathematics, a trinomial expansion is the expansion of a power of a sum of three terms into monomials. The expansion is given by ( a + b + c ) n = ∑ i + j + k = n i , j ...

$\Large\displaystyle(a+b+c)^n=\sum_{i,j,k:\ i+j+k=n}\binom{n}{i,j,k}a^ib^jc^k$

Akiva Weinberger

12:38

If $0\notin A(\Delta)$, where $\Delta$ is where the coordinates add to $1$, then it's closed by gerw's proof.

Balarka Sen

@AkivaWeinberger Same idea, but unnecessarily complicated.

But I suppose $e^x$ is a natural choice, as it has a more distinct asymptote along the y-axis than $1/x$. I retract my previous comment.

user 1618033

Some nasty issues with the connectivity here , I wouldn't even dare to start a talk.

Is any of you done with this?

$$\sum _{n=1}^{\infty } \frac{\psi (n) \psi \left(n+\frac{1}{2}\right) \psi \left(n+\frac{1}{3}\right)\cdots\psi \left(n+\frac{1}{m}\right)}{n \left(n+\frac{1}{2}\right) \left(n+\frac{1}{3}\right)\cdots \left(n+\frac{1}{m}\right)}, \ m\ge2$$

Leaky Nun

@user1618033 what is $\psi$?

user 1618033

@LeakyNun a polygamma function (digamma here)

Digamma function

In mathematics, the digamma function is defined as the logarithmic derivative of the gamma function: ψ ( x ) = d d x ln ⁡ ( Γ ( x ) ) = Γ ′ ( x ...

Well, very nasty issues with connectivity. Back later.

Leaky Nun

so $\psi(n)=\dfrac{\gamma'(n)}{\gamma(n)}$

Owatch

12:57

@LeakyNun Yes that is what I wanted, thanks!

SoumyoB

13:38

does anyone here know how to prove the non-existence of a homotopy between 2 paths with same end points when there is a hole in between those 2 paths?

I've been trying to prove it for the past few days but the proof seems really ambiguous...

Semiclassical

That alone isn't going to be enough. Consider a sphere with a puncture at the south pole and a loop around the equator

you can't shrink the loop to a point by moving the loop south, but you can bring it up to the north pole

so the loop is contractible despite the puncture it surrounds.

now, if you treat the plane as having a puncture at infinity (or equivalently, cutting it off somewhere to make a really big disk) then it should be true.

(if that seems like nitpicking, you may be right; i'm not entirely awake right now)

Danu

13:59

When my topology sheet mentions "the surface $K^2$" without further comment, it's probably supposed to be the Klein bottle, right?

Sadams

How many solutions this cauchy equation has y'=x^6+y^4 ,y(2)=-1

Leaky Nun

(ignore me) $y'-y^4=x^6$

$y''-4y^3y'=6x^5$

$y'''-12y^2y'-4y^3y''=30x^4$

Oops, I'm getting nowhere

Danu

lol

Semiclassical

yeah, that's not the kind of ODE you should hope to solve by hand

user243301

This afternoon I will to see the video in YouTube from the official channel of the CIMAT by Cédric Villani, Cédric Villani en Guanajuato, conferencia "El arte vivo de las matemáticas" is in english from the minute 5' (with optional subtittles in spanish). I say it if any user want to see it. Good afternoon.

Leaky Nun

14:05

@Semiclassical yep, found that the hard way

@user243301 "I will to" is a direction translation of "voy a"? "Voy a" is "I am going to", "I will" have no "to" afterwards :)

Semiclassical

ohh, Cedric Villani? neat

Leaky Nun

(sorry for picking grammar)

SoumyoB

14:21

@Semiclassical like I was just imagining a plane like $\mathbb{R^2}$ with a hole at the origin of radius $1$. Now, consider the points $(0,2)$ and $(0,-2)$ and imagine a path, the right half of the circle $x^2+y^2 = 4$ and another path, the left half of the circle. How do I prove that there cannot be a homotopy between these 2 paths?

Sadams

can somebody help with this y'=ycos(x)+xcos(x)-1

trilolil

hello folks

I have some issues with transforming equations from the s domain to the z-domain

example:

$\frac{0.5}{s+j\omega} + \frac{0.5}{s-j\omega}$

SoumyoB

14:38

@Sadams
$\frac{dy}{dx} = (y+x)\cos x -1$
$\Rightarrow \frac{dy}{dx} +1 = (y+x)\cos x$
$\Rightarrow \frac{d(y+x)}{dx} = (y+x)\cos x$
$\Rightarrow \frac{d(y+x)}{y+x} = dx\cos x$
That should do it

Akiva Weinberger

@LeakyNun "I'm going to"

Leaky Nun

@Sadams This is of the form $y'+P(x)y=Q(x)$, which can be solved by integrating factor (although SoumyoB's method is better)

@AkivaWeinberger Well, "I'm going to" is the same as "I am going to" isn't it?

$y'-\cos(x)y=x\cos(x)-1$
$y' e^{-\sin(x)}-\cos(x) y e^{-\sin(x)} = -x\cos(x)\sin(x)+\sin(x)$
$(y e^{-\sin(x)})'=-x\cos(x)\sin(x) + \sin(x)$
$(y e^{-\sin(x)})'=-0.5x\sin(2x) + \sin(x)$
$y e^{-\sin(x)}=-0.125(\sin(2x) - 2x\cos(2x)) - \cos(x) + C$
$y = (-0.125(\sin(2x) - 2x\cos(2x)) - \cos(x) + C) e^{\sin(x)}$

Akiva Weinberger

@LeakyNun Oh, I misread, I thought you thought it was "I will to"

Mike Miller

@Danu Sort of lame notation, but it's the only thing it could be.

user174558

15:49

@leakynun What are you leaking?

Leaky Nun

@JasperLoy guess :p

iwriteonbananas

Hi

Paul Trehiou

hello

Mike Miller

hi

LifeOfPai

Hello

Given the joint density function of the variables $ X,Y$ both variables are given values interval $ [0,1]$.

$$f_{x,y}\begin{cases}
& \frac{2}{3}\text{ if } 0\leq x\leq 1,0\leq y\leq 1,0\leq x+y\leq 1 \\
& \frac{4}{3}\text{ if } 0\leq x\leq 1,0\leq y\leq 1,1\leq x+y\leq 2 \\ \\
& 0\text{ else }
\end{cases}$$

I need to find the joint density function of $f_{x}$.

Someone can help me with this?

What I need to do here?

LifeOfPai

16:13

Someone here and can help with this question? Thank you

Pedro Tamaroff

@LifeOfPai Do you happen to know how the density function $f_X$ and the joint density function $f_{X,Y}$ are related?

LifeOfPai

Yes

Semiclassical

morning chat

Pedro Tamaroff

What's the relation, then?

LifeOfPai

$$f_{x}(a)=\int_{- \infty}^{ \infty} f_{x,y}(a,y)dy$$

By definition

Pedro Tamaroff

16:20

Well, then integrate and you will get $f_X$.

LifeOfPai

But I have 3 part, did not know how to calculate this...

Here I am stuck and can not continue.

Leaky Nun

@LifeOfPai then split the integral into 3 parts

LifeOfPai

This is the question, what will be the domain and what function each part?

Thank you

Leaky Nun

Well, first plot a graph

oeis.org/A079586

My name is here!

Well, my real name.

LifeOfPai

Danu

16:29

@MikeMiller It's even more confusing because the full quote is: "Show that the surfaces $\sum_{g\geq 1}\Bbb R P^2$ and $K^2$ [...]"

Leaky Nun

@LifeOfPai Which part is $2/3$, which part is $4/3$, and which part is $0$?

Mike Miller

Uh, are the reals called $\Bbb B$-something in German?

Oh, ok.

Leaky Nun

Describe in terms of the colour of the region

Danu

I messed up :P

Mike Miller

That sucks

Danu

16:30

Did my TA forget the latter is part of the former?!

LifeOfPai

The bottom (red) is 2/3 and the other 4/3

Leaky Nun

@LifeOfPai not really

the part where the red, orange, and green overlap is 2/3

Mike Miller

What's the full question?

Leaky Nun

where the blue orange green overlap is 4/3

LifeOfPai

Yes that what i mean, sorry

Leaky Nun

16:31

@LifeOfPai Then, imagine the z-direction as coming out of your screen

you would see two triangular prisms

Danu

@MikeMiller In the previous question I (hopefully) showed that a union of two (relatively open) acyclic subsets has trivial cup product in positive degrees. The full question is then: "Show that the surfaces $\sum_{g\geq 1}\Bbb RP^2$ and $K^2$ cannot be written as the union of two open acyclic subsets"

The assertion follows from nontriviality of $\smile$ on $H^1\times H^1$ in $\Bbb Z_2$ coefficients

Leaky Nun

Oh, imagine the density as the z-direction

Mike Miller

sure. I still find the first thing baffling

Leaky Nun

oeis.org/A274292

My name is also here!

(Well, my real name, of course)

Danu

@MikeMiller ... me too :(

LifeOfPai

16:34

@LeakyNun $$\int_{0}^{1}\int_{0}^{x} \frac{2}{3} dydx + \int_{0}^{1}\int_{x}^{1} \frac{4}{3} dydx $$

Mike Miller

I guess it was meant to say the connected sum of $g$ copies of $\Bbb{RP}^2$, where $g$ is fixed and at least 1

Leaky Nun

@LifeOfPai You don't even need integration for this

but it's correct

Semiclassical

@LeakyNun did you happen to look at that paper I linked re: the conducting wire?

Mike Miller

just name that $N_g$ or something lol

Danu

@MikeMiller yeah... but $K^2$ is the case $g=2$... Superfluous at best

Leaky Nun

16:34

@Semiclassical I opened it in my browser and haven't read it, lol

Semiclassical

mmkay

LifeOfPai

@LeakyNun Okay and why not?

what`s your idea?

Danu

@MikeMiller Half of the time we do, but I think our TA copy-pastes exercises from some other source sometimes.

Semiclassical

one basic implication is that both of us were wrong

Leaky Nun

@LifeOfPai You could just treat them as solids and find their volume, that's what integration is

Pedro Tamaroff

16:35

@LifeOfPai Plot the regions and split up your integral.

Leaky Nun

@Semiclassical Well, it'd be great if you could summarize it

Semiclassical

on the one hand, the conducting wire is neutral in the lab frame.

Pedro Tamaroff

It is a piecewise defined integral, if you may.

Mike Miller

lol

Semiclassical

i.e. $n_{eL}=n_{nL}$

Leaky Nun

16:36

the $-$ is not needed

LifeOfPai

You right. Thank you !

Danu

I suspect this because he sometimes forgets to change certain German words, notably "und" (which should be "and")

Semiclassical

eh, depends on whether you're talking about the number density or the charge density.

but fair enough.

Danu

...either that, or he actually typos like that.

Semiclassical

on the other hand, while you can measure the number density of the nuclei in the lab frame (they're at rest), the number density of the electrons is properly measured in the electron frame.

Leaky Nun

16:38

@Semiclassical Where did the extra protons come from?

Semiclassical

hence accounting for Lorentz transformations we conclude $n_{eL}=\gamma n_{eE}$ and $n_{nE}=\gamma n_{nL}$

Pedro Tamaroff

@MikeMiller What's funny?

Leaky Nun

@Semiclassical of accord.

I mean, "agreed"

Semiclassical

and therefore the net charge density in the electron frame is $n_{nE}-n_{eE} = \gamma n_{nL}-\frac{1}{\gamma}n_{eL}=(\gamma -\frac{1}{\gamma})n_{nL}>0$

so yeah, there's a net positive charge in the electron frame.

which is weird to me

LifeOfPai

@LeakyNun that`s my answer:

Leaky Nun

16:42

@Semiclassical why?

Semiclassical

Why is it weird to me?

Leaky Nun

@Semiclassical Yes

Semiclassical

Wrong intuition, I guess.

I mean, it's charge conservation not charge invariance

Leaky Nun

Wait, I thought $\gamma-\frac1\gamma<0$

Mike Miller

i'm in bed crippled by indecision about what to do today

Semiclassical

16:43

$\gamma=1/\sqrt{1-v^2/c^2}$

Leaky Nun

Then isn't $n_{eL}=\frac1\gamma n_{eE}$?

LifeOfPai

@LeakyNun $$f_X(x)=\begin{cases}
& 1/3 \text{ if } 0\leq x\leq X \\
& 2/3 \text{ if } X\leq x\leq 1 \\
& 0 \text{ else}
\end{cases}$$

It`s looks okay ?

Leaky Nun

@LifeOfPai Not really.

Semiclassical

well, $\gamma>1$ for $c>v>0$

Leaky Nun

@Semiclassical well, the electrons are moving, so $n_{eL}<n_{eE}$, right?

LifeOfPai

16:45

@LeakyNun :( So I didnt understend something...

Leaky Nun

@LifeOfPai So it's in terms of $x$, right

Semiclassical

eh, you're right. so i wrote the above relations for the densities backwards

Leaky Nun

so it's actually $\int_{-\infty}^\infty f(x,y) \ \mathrm dy$ right?

Semiclassical

so yeah, should be $(\frac1{\gamma}-\gamma)n_{nL}$ for the net charge density

LifeOfPai

Yes.

Semiclassical

16:47

which is negative, so there's a net negative charge density

main point is just that the wire is charged in the electron's reference frame but not the lab frame.

LifeOfPai

So why it`s dont looks good ? @LeakyNun

Leaky Nun

@LifeOfPai Because it should not be a constant

@LifeOfPai Try writing the dividing point in terms of $x$

LifeOfPai

But $f_{x,y}$ a constant

Leaky Nun

So we are finding the total mass in terms of $x$

which means we look vertically

@Semiclassical guess where the charge went

Charge is conserved

LifeOfPai

@LeakyNun$$\int_{0}^{1}\frac{2}{3} dy, \int_{0}^{1}\frac{4}{3} dy$$

Leaky Nun

16:51

but only under a closed system

Semiclassical

welllll

conservation applies to a given reference frame

to say it's the same in different reference frames would be to say it's invariant

Leaky Nun

@LifeOfPai $f_{\chi}(x)=\int_0^{1-y}\frac23\mathrm dy+\int_{1-y}^1\frac43\mathrm dy$

Semiclassical

and that's a different matter.

Leaky Nun

Alright

charge is still invariant

The positive charge went to itself.

Semiclassical

if you mean charge density---no, it's not. what's invariant is the 4-current.

Four-current

In special and general relativity, the four-current is the four-dimensional analogue of the electric current density, which is used in the geometric context of four-dimensional spacetime, rather than three-dimensional space and time separately. Mathematically it is a four-vector, and is Lorentz covariant. Analogously, it is possible to have any form of "current density", meaning the flow of a quantity per unit time per unit area, see current density for more on this quantity. This article uses the summation convention for indices, see covariance and contravariance of vectors for background on raised...

Leaky Nun

16:53

@Semiclassical I do not understand.

Semiclassical

It's like saying that energy is conserved, so the energy of a system is invariant under Lorentz transformations.

But it's not. What's invariant is $E^2-p^2 c^2$.

Leaky Nun

What is four-current?

@LifeOfPai Do you get it?

Semiclassical

I shouldn't have said the four-current was invariant, btw. Covariant, rather.

LifeOfPai

Yes, Thank you. try to understend

Leaky Nun

The $y$ is a dummy variable.

It represents trying to sum every point on the vertical line we are looking at (a bit inaccurate)

Semiclassical

16:57

anyways, the four-current is defined as $(c\rho,\mathbf{j})$

Leaky Nun

It's like a line transect, if you have studied biology

@Semiclassical More symbols please

Semiclassical

mmkay.

where $\rho$ is the charge density and $\mathbf{j}$ is the current density

Leaky Nun

And what does invariant mean? Its magnitude?

@Semiclassical Sorry, I'll continue tomorrow.

Semiclassical

its magnitude in the sense of $\sqrt{(\rho c)^2-\mathbf{j}^2}$, yeah

LifeOfPai

@LeakyNun I think your domain need to be $$\int_{0}^{1-x}\frac{2}{3} dy + \int_{1-x}^{1}\frac{4}{3} dy$$

Leaky Nun

16:59

bye

Semiclassical

later

Leaky Nun

@LifeOfPai Yes

sorry

so you understand it now

and you can even correct my mistake

bye

LifeOfPai

Yes Thank u!

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$Mathematics$ Mathematics