Scratch what?

What am I scratching?

any sample material that sticks out, at the outer edge of the hole grid

@Loong how big are TEM hole grids?

@KaumudiHarikumar maybe difference is 2d vs 1d conductivity? what textbook are you using? ps saw your extended dialog in here on physics as a profession/ phd, found it interesting.

@JohnRennie youtube.com/watch?v=WvZFD_gRM_M

holy...god

Who are you

@0celo7 question about related to this question: physics.stackexchange.com/questions/279607/… . Lagrangian $L_p(v_\mu)$ is a function of tangent space at a single point $p$- what is the most natural derivative "with respect to" this argument?

It's not the covariant derivative because I'm just considering the lagrangian at a single point...

@NeuroFuzzy Why are you differentiating the Lagrangian at a point? Also, are we on a manifold or just on a vector space?

@0celo7 Well, the Euler-Lagrange equations partly just depend on just that single point. So $v_\mu$ is a vector space but $p$ could be on a manifold.

The guy clearly wants his answer NOT in terms of $L(a,b,c,d)$ (3+1 dimensions, $L(v_\mu)$ could be considered a function of 4 variables if you are a physicist and not a mathematician)

and instead wants it in terms of $L(v)$ with $v$ an abstract one-form...

Do I have to read the question?

What exactly are you confused about

How to naturally/abstractly describe the derivative of $L(v)$ with respect to $v$, with $v$ a one-form.

Note that $\mathrm d \phi=\nabla \phi$ if $\phi$ is scalar.

@NeuroFuzzy Likely not possible.

At least, in all variational stuff I've seen, you just write it in coordinates.

It's pretty obvious what you do then, so it's "natural"

Why is $v$ a 1-form?

Haha alright, that's what I thought but the OP doesn't want that

If $v$ is the velocity, it's a vector.

Oh, no, for fields

@NeuroFuzzy Ok I'll give it a read.

@NeuroFuzzy Oh...that makes things harder.

The space of fields is not a manifold in the usual sense.

@0celo7 oh of course, but the euler lagrange equations just care about $\phi$ and $d\phi$ at a point

erm, the lagrangian does*

Anyways whatever, it is at -1, it isn't worth spending too much time on

He's being difficult on purpose.

He doesn't understand the chain rule?

Oh, no, I see.

He doesn't understand why $\phi$ and $\partial\phi$ are treated as independent.

TBH, I don't know why either off the top of my head. I'll look it up if I were to do some symplectic topology but that's not where I'm taking my research right now.

@0celo7 nono, it's just in the definition of $L$

@NeuroFuzzy Hmm?

When you write down $L=\dot{x}^2-x^2$, everything looks weird. When you write $L(a,b)=b^2-a^2$, then $L:\mathbb{R}^2\to \mathbb{R}$, and you minimize $\int L(x,\dot{x}) dt$, then everything is fine and dandy and rigorous.

@NeuroFuzzy Agreed, but why do we do that?

let's grab Arnold from the physics shelf of shame

Ah.

We don't vary them independently.

But we take the partials independently.

Ohhh yes! That's true. That's a good way to put it.

I need to do some digging

loads PDF list

my precious

Lol. ugh I need to go study for the GREs

this quote is now confusing me.

it's the dedication line of Burke's diff geo

Yaaay hundreds of standard electrodynamics exercises.

Ok, he does things with contact structures

I don't remember anything about that

@NeuroFuzzy I take it back, I do know how to differentiate the Lagrangian abstractly.

Er, I double take it back. I can do it abstactly, but with coordinates. The tangent bundle can be locally trivialized with coordinates $(q,\dot q)$.

oh dear

Then you take derivatives wrt. the bundle coordinates, which is represented by coordinate vector fields on the double tangent bundle.

Ahh double tangent!

It's then clear why a derivative wrt. $\dot q$ makes sense, because $\dot q$ really is just a new variable, it has nothing to do with $q$, until you've imposed the EoM.

It's the EoM that gives you $t\mapsto (q(t),\dot q(t))$.

@NeuroFuzzy Clear now?

@0celo7 yeah. Phrased another way, I could have taken the Lie derivative of $L$ in the direction of a vector on the tangent manifold to the tangent manifold. (& of course it isn't important that it's the Lie derivative, since $L$ is a scalar.)

and of course, $L$ is just a function of independent variables until you plug it into the action and minimize it.

Ahh why are you bringing in the Lie derivative?

Something like $L_vL$ is terrible notation :P

@0celo7 Because just saying "derivative" on abstract things scares me :D

@NeuroFuzzy Huh?

The derivative of a scalar $f$ in the direction of a vector $v$ is just $vf$.

Saying $L_vf$ is waaay overcomplicating things.

@NeuroFuzzy Yeah.

I guess I just wanted to write $L_\mu f$ to have that free index down there...

I don't know. It's clear with components and that's good enough for me.

@NeuroFuzzy Oh god what?

That has no meaning...

Unless you mean $L_{\partial/\partial x^\mu}f$.

But why not directly write $\partial f/\partial x^\mu$?

Do you know what that means on a manifold?

Haha I think once our conversations go on too long we both know the answer and end up nitpicking

Perhaps.

And you know I will end up saying something technically incorrect about abstract index notation if I try to use it for more than half a sentence! :D

I'll stick to my coordinates on physics.stackexchange lol

I like coordinates!

@NeuroFuzzy ...

Abstract index notation?

@0celo7 That's the reason $\nabla_\mu v_\nu$ is so nice, no? We had an argument about this. I took

(oops, hit enter)

I took $\mu$, $\nu$ as fixed integers, representing the covariant derivative with respect to the $\mu$th basis vector.

That's not abstract index notation :P

And you said no, it's a (0,2) tensor

Yeah.

Which it is, but is a headache because in the book I learned it from $\nabla_v$ was defined for some element of the tangent space $v$

So you have to plug in a $v$... being the basis vectors... then you're back to coordinates for a moment.

@NeuroFuzzy No, your book defined it as $\nabla_X$ for $X$ a vector field.

Not just a tangent vector.

It's a nontrivial fact that $\nabla_v$ also makes sense where $v$ is just a tangent vector.

@NeuroFuzzy Basically.

Anyways, have no fear, I'm sure I will be hit over the head with this enough in my differential geometry course next quarter!

@NeuroFuzzy Not likely.

even though I'm sure it will be all flowers butterflies and Riemannian metrics

Lee goes 700 pages without mentioning connections.

Unless you're taking a course on Riemannian geometry specifically, you (probably) won't see them.

Do you know what book the course uses?

@0celo7 Depends on the professor, no, but this is one in 2012. math.ucsd.edu/~benchow/250AFall2012.html Levi-civita connections! Sounds good to me.

@NeuroFuzzy Ah, well one second of looking tells me that's a Riem geo course.

@NeuroFuzzy You won't learn good(™) things like Sard's theorem or cohomology

is there another course you can take?

@0celo7 Not in the math department. I'mgoing to be spending all my time on the algebraic geometry course anyways so it might be worthwhile to just refresh myself on basic diff geo stuff.

oh, why are you taking alg geo?

@0celo7 Because I think I can and a friend (who has taken all the graduate prerequisites) is doing it

I want to take a diff geo course based on Kobyashi-Nomizu

those books are impossible to read on one's own

Hi, everybody.

@DanielSank Hi.

@yuggib Did you say you don't have experience with parabolic PDE?

@yuggib *nonlinear parabolic PDE.

I have some experience with some specific PDE, even of parabolic type; but not so much

@yuggib Do you remember why linear ODEs have solutions everywhere on $\Bbb R$?

Extension of nonlinear pdes to the whole time does not follow the same ideas

I know

I know the proof that ODE's can be solved on some interval, but not that linear ones can be solved everywhere

Use your favorite extension theorem of linear operators

sorry?

Uh, does anyone know why I can't flag posts as very low quality?

The option just kind of disappeared

@KaumudiHarikumar How are you progressing on your goal of reducing world suck?

@0celo7 a linear equation is solved by a linear operator $T $ that satisfies the property $T (t)T (s)=T (t+s) $ for $\lvert t-s\rvert $ small enough right?

Then, you can iterate that process indefinitely splitting any time interval in smaller ones without blowing up (because you're actualli dealing with matrices and reals) apart maybe for an infinite time

@KaumudiHarikumar as I recall harmonic series tend to diverge. We tend to use geometric series as way of calculating quantities, but since harmonic series don't converge they aren't useful for this.

Mathematicians are strange people :-)

@KaumudiHarikumar Very good then.

@KaumudiHarikumar Music.

Hence the name "harmonic".

The musical scale is a geometric series isn't it?

Each note has a frequency that is a multiple of the one below it.

That multiple being the twelth root of two.

Here's a video infographic of what I shared about yesterday

seems to be cursed...whever I enter chat::all the members seem to be dead

like the siren call of life invited them back to the island of world

I think I should be the one to said the above (although nowdays it is less true because as it turns out it is just an annoying timezone issue)

hmm....but not evertime that should happen..even if icomhere midnight

@Xasel It's 09:00 in the UK and around 2 a.m. in the US (depending on the timezone), so the UK members are all at work and the US members are in bed.

Round about 16:00 to 00:00 UTC is the liveliest time. That seems to be the time the UK members are relaxing after work and the US members are looking for something to distract them from their work :-)

I suppose this is the ideal time for the Asian and Far East members, but they don't seem to be as keen on the chat as the rest of us.

hey

There aren't many asian people on the chat on average, and there aren't many australians also, which is why the chat tend to get very quiet during those times where it is afternoon in the southern hemisphere

@KaumudiHarikumar you're the exception :-)

Yes! Annoying cover letter shit is done!, now to continue reading the Noether paper

hey

when doing a mewtonian problem how will i know if i can use the work energy theorem?

have to check that every force is conservative?

try doing a $\oint$ for all your forces about a curve that does nto enclose a singularity, if this vanishes, then you forces are conservative

F is conservative if its an analytic function?

Work energy theorem said that the total KE and PE are conserved, that is, their changes add up to zero i.e. KEf-KEi+PEf-PEi=0

F is conservative if it is curl free

oook

and if i wanna find the total energy?

You sum up the KE and the PE in your system

A newtonian system can only loose energy for example due to friction

it cannot gain?

so i can fasy Initial energy=Final energy-work done by energy

say*

yup, and the sign of your work done will tell you whether you gain or lose energy

whereas for nonconservative forces, energy is always dissipated into heat in the system

work done by force*

Ok so when energy is not conserved i can always take that formula? right ? just subtract the work done by non conservative forces

I have not encountered nonconservative forces calculations in detail, but I suspect you need to first subtract the contributions of the work done by nonconservative forces out of the total energy, before you can use work energy theorem for the system

this is because the work done by nonconservative forces are path dependent thus it depends on how the system get from state A to B

Alternately, you can put the work done by nonconservative forces as part of your system. Then the usual work energy theorem applies as the work doen by nonconservative forces is taken account of by the work term in the theorem

This is how roller coaster problems are often formulated using free body diagrams

@ACuriousMind Sorry for the minor edit-from-review, this time it wasn't intentional. Btw, the question will be probably closed independently from the edit.

@JohnRennie Relating this: I think maybe the constants of a real TOE would be really independent from everything.

To be checked: whether $[\int,\lim]$ form a Lie Algebra

@EmilioPisanty I am really sorry that you find some of the terminology disappointing, please consider, I never use bad words to harm people intentionally.

@Secret On what space

$\int$ maps $L^1$ to $\Bbb C$

What space contains both of those

Oh wait I guess it could be undefinite integrals

Limits do map from functions to reals though

Well lets suppose both lim and integrals resides in real number line i.e. the both does $\mathcal{C}^n(\mathbb{R})\rightarrow\mathbb{R}$

amazon.com/Stochastic-Analysis-Manifolds-Graduate-Mathematics/…

@Slereah ^

@Secret What?

Doesn't the bracket have to be antisymmetric?

@Secret You also have to say on what space the limit is being taken.

For now, assume everything took place in the reals, thus the limit took place in the reals

What?

You're still making no sense.

On what space is it a Lie algebra?

What limit exactly are you taking?

What is the integral?

Riemann, Lebesgue?

What function space are you looking at?

Your question literally has no content.

The limit is of the form $\lim_{(x_1,x_2\cdots x_n)\rightarrow (a_1,a_2\cdots a_n)}$ where $a_i\in \mathbb{R}$. The integral is riemann and the function space is something like $\mathcal{C}^n(\mathbb{R})$ which is always riemannian integrable (which I don't know how to write it because I don't know whether it is $\mathcal{L}^1$

As for the lie algebra, I am not sure. Currently I am just trying to see whether it obeys the lie algebra definitions but it seems as you pointed out, $[\int, \lim]f(\vec{x})$ and $[\lim, \int]f(\vec{x})$ are in general not antisymmetric, thus it seemed to be not a lie algebra

$C^n$ isn't always Riemann integrable.

There are $C^0$ functions that aren't Riemann integrable

@Secret You still haven't said what limit you're taking, and it's not clear to me how you want to apply both taking a limit and then integrating to a single function.

Taking a limit maps a function to a number. Taking an integral maps a function to a number. You can't really do both of those in succession, so what do you mean by that strange commutator notation?

Maybe I have used the commutator notation wrongly, because what I want to investigate is this general expression $\lim_{x\rightarrow b}\int_a^x f(x')dx'-\int_a^x\lim_{x'\rightarrow b} f(x')dx'$ (where all constants are real numbers)and I thought because I saw the integral and limit has the form AB-BA (be treating them as operators), I thought I can just wrote this as $[\int,\lim]f(x)$ as a shorthand.

Why call it a Lie algebra, though

What makes it a Lie algebra rather than a generic type of algebra

Well, maybe you can use that as shorthand, but you have to tell the people you're talking to at least once what you mean by it!

Also, your second term makes no sense - if $x'$ is the integration dummy variable (as indicated by the $\mathrm{d}x'$), then you can't take a limit of it.

you are right... (Argh I seriously have zero intuition in analysis)

Also I always thought [A,B] implies AB-BA whenever it is used. The fact there is a lot of confusion above seemed to suggest I used the [,] notation incorrectly

Anyway, below is the actual question that started the investigation (which I thought I can solve it myself and only need some help by just tackling the above small subquestions)

I am currently reading Page 2 of Noether's paper and I a trying to go from the $\delta f$ expression to the central equation of lagrange. I then attempt to derive that by differentiate both sides. This means I will be differentiating an integral. Now

if the function $f$ is uniformly continous or at least obey the dominant convergence theorem, then I can differentiate it under the integral sign. However, because Noether formulate the functions very generally, I am unable to convince myself that the conditions for differentiation under the integral sign will hold, therefore I tried to explore this problem more generally by 1st principles

That is, I started with this integral $$\frac{d}{dx}\int_a^{x} f(x',u(x'))dx'$$

and then rewrote it in first principles to get

$$\lim_{h\rightarrow 0}\int_a^x\frac{ f(x'+h,u(x'+h))- f(x',u(x'))}{h}dx'$$

Now if differentiation under the integral sign holds, then this is easy to evaluate. Therefore I then tried to investigate the case where it fails by doing the following

@Secret That's equal to $f(x,u(x))$ by the fundamental theorem of calculus.

Why are you trying to "differentiate under the integral sign"?

That's what one has to do if both bounds are functions of $x$ and one wants to differentiate w.r.t. $x$, but here you just have $x$ as the upper bound - the derivative w.r.t. that is just the original function by the fundamental theorem of calculus

@ACuriousMind I am trying to derive $\delta f-\frac{d}{dx}\left(\sum_i \frac{\partial f(x,u(x),\partial_x u(x))}{\partial\frac{du_i(x)}{dx}}\delta u_i(x)\right)=\sum_i \psi_i(x,u(x),\partial_x u(x))\delta u_i(x)$ fromthe given $\int\delta f dx=\int \left(\sum_i\psi_i(x,u(x),\partial_x u(x))\delta u_i \right)dx$ in Noether's paper. If I differentiate by x both sides of the equation involving the integrals, then by chain rule there should be another derivative involve in the integrand summation term

(since that $\sum$ term is a function of x)

you want that limit to fails?

@ManolisLyviakis that question is no longer important, the important one is now highlighted in my reviosu post to ACM and in the next post

@Secret Sorry, you're not making sense. I have no idea what chain rule you're talking about. Noether explicitly tells you that that equation is derived by writing out the boundary terms. You get rid of the integral not by differentiating but by noting that the set you're integrating over is arbitrary, and if $\int_U f(x) = \int_U g(x)$ for any domain of integration $U$, then $f(x) = g(x)$ (except possibly on a set of zero measure).

Because that's the term you get in the "partial integration" Noether also mentions.

I think you might not be ready for reading this paper.

holy fuck

what is this

What does Noether mean by "partial integration" (googling that term so far only give me "integration by parts" but nothing in there suggested it is caused by integration by parts?

@ACuriousMind halp I have quantum mechanical doubts

@ACuriousMind What does $f(X)$ or $f(P)$ mean?

Hey everyone

@KaumudiHarikumar Stefan-Boltzmann law.

^That should be the record for fastest reply ;) @0celo7

you seek the intuiton for it?

@KaumudiHarikumar If you know that law, then look up the law!

Don't ask a question about sweaty people

...

lol

@0celo7 what.

Body, not human body.

@ACuriousMind The Stefan-Boltzmann law question on my thermo final was about sweaty people in a gym :D

@ACuriousMind In general, what is $f(\Omega)$, $\Omega$ an operator, supposed to be?

@KaumudiHarikumar :try to see it like this

Ain't Temepertaure is property of body that is the statistical avergae of kinetic energy

^As far as I am taught

@ACuriousMind For that "partial integration" thing Noether is talking about, is it related to the stoke theorem stuff learned in vector calculus, or is something that I never learnt about at my level?

Greater the temperature the greater is the velocity but see here that with increase in velocity there is increase in collisons among atoms too

it implies there is chnage in acceleration

evethough is avergae out but greater the temperature increases the chances of change in acceleration(collisios)

@ACuriousMind is $f$ supposed to be analytic?

and when charge accelerates, it radiates energy

I need something...

Now can you make connections?

@Secret what line exactly is confusing you

give me the one equality, not a wall of text

Greater the temp --->Greater the chances of drastic chnage is acceleration-->Leading to more emmision of radiation

see this:en.wikipedia.org/wiki/Larmor_formula

@0celo7 Equation no. 4 in Noether's paper, can't seemed to derive that weird $\frac{d}{dx}(\text{big scary expression})$ term

and keyword here is "drastic"

@Secret can you post a screenshot of the paper, I don't have a link

@you questionw hy the radiation increases with temperature

@KaumudiHarikumar : you strill haven't got it?

Link here

@KaumudiHarikumar : Have hilarious joruney with physics(lets hope it's modeled by weierstrass function for fun :P)

huh? is that thing being integrated?

I am not sure, I am not even sure if "partial integration" said by Noether is the usual integration by parts plus stoke theorem commonly used in vector calculus

it is

@0celo7 It's a function of $X$ resp. $P$, obviously :P

Are you asking how functions of operators are defined?

@ACuriousMind Yes.

after seeing the reply of you guys I feel that we should petiton for MathJAx support here on chat too

@0celo7 On the level of a typical QM course, it's just by the power series of $f$.

My prof was like "power series" but that's a bullshit answer.

but then why is there a $\frac{d}{dx}$ thing if it is already being integrated by integration by parts and stokes theorem and where is that $\partial u'$ came from (it does not look like chain rule to me)

@ACuriousMind God dammit I should have taken modern algebra.

@ACuriousMind do you understand Secret's issue

On a rigorous level, you need functional calculus, then you define the new operator by applying that function to the eigenvalues. Basically, you take your self-adjoint $A$ and an eigenbasis $\lvert a\rangle$ and define $f(A)$ by $f(A)\lvert a\rangle = f(a)\lvert a \rangle$. However, to show that this also works for unbounded operators requires math.

I know that all of this is shown in Reed&Simon.

Requires math?

It doesn't require math for bounded operators?

Which Reed & Simon?

@0celo7 Rather much less: For bounded self-adjoint operators, you have the eigenbasis, and defining an operator on an eigenbasis defines the operator. It's just when you don't have an actual eigenbasis (like in the case of position and momentum) that it becomes difficult to rigorously show the existence of the operator. At least I think it's like that

@0celo7 I don't remember, but I think 1 or 2

@ACuriousMind Well clearly $X$ is not a bounded operator.

Yes, I am aware of that :P

And I need to show $[X,f(P)]=\mathrm i\hbar \partial f/\partial P$.

I have a physicist """"proof"""" but it reeks of sewage.

@0celo7 As I said, on the level of a typical QM course you can assume $f$ is analytic and just expand it.

@ACuriousMind But that's wrong >:(

@0celo7: L0l..I thought I was the only person who think that physics...

@0celo7 You can also show it without assuming $f$ is analytic. Go to the momentum representation where you know that $X$ is $\mathrm{i}\hbar\partial_p$. Instead of analyticity you merely use $(f(P)\phi)(p) = f(p)\phi(p)$, which is true even for the most general definition of $f(P)$.

Your "given" is not what you should start from to get the form of $A$.

The calculation that Noether doesn't explicitly do here is the standard derivation of the Euler-Lagrange equations.

@Secret I don't know what $f$ and $A$ are

@ACuriousMind Is that PhD level

@ACuriousMind How to prove that?

@0celo7 $f$ is an arbitrary function, $A$ is an abbreviation for the boundary terms that are usually constrained to vanish after partial integration.

@ACuriousMind $\psi, \delta u$?

@0celo7 Not at all.

@ACuriousMind I dunno, the EL equations are pretty scary.

Function spaces and whatnot.

@0celo7 It's rather straightforward, I'm not writing it out lest you complain that I did your homework again

@ACuriousMind I have a proof that's good enough for the class.

But what you wrote there is $\langle p|f(P)|\psi\rangle=\langle p|f(p)|\psi\rangle$.

Right?

Yes

So we define $f(P)$ by $f(P)|p\rangle=f(p)|p\rangle$?

Yes

Oh, and by completeness that defines it on the whole Hilbert space?

Yes

@ACuriousMind So my only question is what the hell does $\int|x\rangle\langle x|$ mean

(Rigorously, that doesn't work because those $\lvert p\rangle$ don't lie inside the Hilbert space)

@ACuriousMind Ok, that's my second question

third question is why we can work with $|x\rangle$ when that doesn't lie inside the Hilbert space either

Yeah, that's why I said to do this rigorously for unbounded operators is more difficult.

frig

According to the numerous answers on PSE, the EL equation are derived from the action princple, which takes the lagrangian L=T-V. Some derive it form newton laws and the principle of least action. However, these derivation told me nothing on how lagrangians in general are motivated thus it does not help me (except for the classical mech case) on understanding the lagrangian. And then, yesterday I found a PSe answer which talk about lagrangian are motivated by symmetry considerations and

So what do I do?

that answer lead me to the Noether paper

The most important thing is that, I am trying to understand why the lagrangian is T-V, but nearly all answers either start with EL or action princuple, which rely on the definition of the lagrangian

they also tend to ignore the more general case of velocity dependent potentials and I have no idea how more complicated lagrangians are motivated

@Secret I didn't mean the physical motivation for the E-L equations. I mean the actual derivation, the actual (physics) math that is done to arrive at the E-L equations from the principle of extremal action. E.g this is the first hit I get for "derivation of E-L equations", and there's exactly what I'm talking about, in particular the partial integration that appears to confuse you so much.

EL equation derivation is scary

I still don't understand it

@ACuriousMind O cool, nothing in PSE look like this, let me give a read first...

@Secret There is no general principle to "derive" Lagrangians. Lagrangians are guessed.

@Secret You've never seen the derivation of the EL equations on $\Bbb R^n$?

@Secret Are you telling me you're reading some original paper by Noether about calculus of variations without having ever seen its basic application in deriving the E-L equations?

@ACuriousMind That's more ass backwards than anything you've seen from me!

I am literally ZERO BACKGROUND in classical mech because of my tight undergrad that prevent classical mech from slotting into my courses

@Secret What did you do in high school?

@Secret Then...a better idea might be to learn standard classical mechanics first instead of going off to read random papers about it.

We derived the EL equations in my high school physics class

@0celo7 Indeed it is.

@0celo7 After everything else I've heard from you about your physics class, I'm shocked by this revelation. I didn't see the E-L equations till my second semester

@ACuriousMind Now let's compute some sheaf cohomologies

What's a ring again?

@0celo7 Something you put on a finger

...

Whoops.

I meant a chicken, of course.

Chicken ring, yeah.

@ACuriousMind Well, it was just on $\Bbb R^n$

I do not understand the proof on a manifold, although I haven't spent much time trying to

@0celo7 Ah yes, the famous chicken ring.

The problem on a manifold is that the minimizing curve doesn't have to be within one chart

So how the hell do you compute the derivatives

But that's a problem I have with any type of geometric analysis too

@Secret Yes, but physics texts will overlook that

@Secret Do you want the REAL math?

chill

@ACuriousMind Boy I'mma whoop you if you talk back one mo time