The h Bar

Kyle Kanos

15:00

It took a while, but I've finally surpassed Qmechanic for Suggested Edit queue: physics.stackexchange.com/review/suggested-edits/stats!!!

skullpatrol

Congrats!

Johnathon

If a shaft has torques A and B on left side and right respectively, and a distance as x going from left to right, what would the internal torque in the shaft be? Would it be A or B? I'm trying to imagine cutting the shaft and then balancing torque with an internal torque but I'm not sure which side you would consider

0celo7

@Slereah how does one pronounce Henneaux?

Slereah

I'd pronounce it eno

/eno/ in IPA or something

Kyle Kanos

not hen-ee-ax?

or hen-ee-ox?

Slereah

15:05

Nah, -eaux is o in french

and french doesn't really have an /h/ sound anymore

it's just a leftover from the frankish days

Kyle Kanos

well if it's supposed ot be an o sound, why don't they use an o?

Slereah

Why is it knight and not nite

or nit

Kyle Kanos

because it's pronounced kuh-nig-it

skullpatrol

All hail the Monty.

ACuriousMind

@0celo7 Your definition of $w$ is not coordinate-invariant if $u$ is not scalar.

Slereah

15:14

basically it's the usual thing where a word looks normal in the era it's used

but then the pronounciation changes, but not the spelling

ACuriousMind

I mean, sure all that works if you treat $g^{\mu\nu}$ just as an array of functions, but at the end of the day it's still a tensor and picking out particular components from it is still non-sensical unless the resulting expression is properly covariant.

Slereah

that's what you get for not using the bundle

Like a god fearing christian

0celo7

@ACuriousMind it's an array of functions in a chart.

Slereah

But if you change the coordinates, you're not in the same chart

0celo7

and I'm doing all of this in a chart

@Slereah the conservation law you get at the end, $dJ_X=0$, is again covariant

working in coordinates is not a crime

Slereah

15:21

the conservation law is just gonna be the conservation law of $n$ scalar fields tho

0celo7

the conservation law is a PDE, stop trying to insert epistemology

Slereah

Well when you try to flow the metric tensor it will just act on it as something like $g_{\mu\nu}(x) \to g_{\mu\nu}(f(x))$ or something

Which from a PDE perspective is fine I suppose

But that's not the proper transformation law for a tensor

0celo7

it leads to a covariant conservation law, regardless

but somehow it's wrong

unless you can point to a mathematical error in what I have done, then, well, physics is wrong

ACuriousMind

@Slereah The point here is not whether it's the "proper law" or not. The problem is that if 0celo7's derivations were valid, then regardless of what his "non rotation rotation" actually "is", it would still prove a conversation law that's false.

Slereah

Well no

It's just a conservation law on scalar fields

for which it is correct, I think?

ACuriousMind

15:26

@0celo7 Well, that's not true - it's simply mathematically false that the stress-energy is symmetric for all Lagrangians.

It's not physics that's wrong, it's your derivation because it leads to a purely mathematical contradiction. The specific error is rather elusive, though.

0celo7

@ACuriousMind exactly

Slereah

I mean is there a contradiction in it though?

I'm not sure there is, given the hypothesis

If you assume that the fields are scalar fields and $L$ is invariant, I think the stress energy tensor is symmetric

ACuriousMind

@Slereah He isn't assuming the fields are scalar fields. He simply doesn't care whether they are and defines his variation as if they were.

Slereah

Is there much of a difference?

ACuriousMind

@Slereah Yes. Your argument would work if there was something like $\phi_\ast u$ in there and then 0celo7 purposefully replaced it by the scalar expression, but that's not the case. He's just defining $w$ and not choosing to define it in the "obvious" way for a tensor field.

And since indeed the final result is covariant, objecting that this is not coordinate-invariant doesn't work

Slereah

15:35

But if you change coordinates for a collection of scalar fields, the result is also coordinate-invariant

0celo7

the variation is a choice, so I am free to make the choice.

what "should" happen is that it results in no conservation law

but somehow it does here, but an incorrect one

ACuriousMind

@0celo7 I am beginning to think it must be that (2) does not imply (1).

0celo7

@ACuriousMind That was my conclusion before I gave up yesterday

ACuriousMind

(2) is weird, as you correctly observed that holds trivially for all $L$ indepedent of $x$, so it would mean you get all these conserved quantities for basically every theory ever.

Prathyush Poduval

@boides you can check this page out, it's quite good and helpful

0celo7

15:37

yep

CooperCape

Kifflom.

0celo7

@ACuriousMind do you agree that Theorem 1 follows from (1)?

so writing (2) for a solution gives $$L(x,u(x),Du(x))=L(\phi(x),u(x),Du(x))$$

and we agree this is true if $\partial L/\partial x=0$

ACuriousMind

@0celo7 I think something is wrong here in the setup - you act as if $w$ is wholly independent from $y$. But it's not - the flow acting on $x$ already induces a change in the $u$, in the case where the flow is generated by a vector field $X$ this is of course the Lie derivative. Now, you can protest again that you treat $w$ as indepedent from $y$, but you don't - you assume that the whole thingy obeys the E-L equations as we flow along $\tau$,

so you need to respect that the fields need to change as the Lie derivative w.r.t. $X$

0celo7

Hmm, where do I treat $w$ as independent of $y$?

ACuriousMind

In that you claim you're free to pick it in the proof of corollary 2.

When, in fact, it is already determined by you picking $y$ as being generated by $X$

Only if you choose that way (1) is implied by (2)

0celo7

15:47

(1) has no connection between w and y

in (2) I pick a specific w and y

ACuriousMind

@0celo7 Only the implicit one that by saying that everything obeys the E-L equation, the part of $w$ (or rather $m$) that you actually can pick is $m - \mathcal{L}_X$ for a $y$ generated by $X$.

0celo7

No, (1) does not require a solution of the EL equations

ACuriousMind

Now, in Corollary 2, you basically consign yourself to not being allowed to pick that part at all, i.e. the independent variation of the fields is zero. Otherwise, I don't see how (2) would imply (1) with an arbitrary choice of $w$.

0celo7

remember, Noether is about the Lagrangian symmetries, not solutions

one can also write theorem 1 for $u$'s that don't solve the equations

ACuriousMind

@0celo7 Yeah, forget my rambling about the E-L equations, I'm working this out as I go along. So, how is (1) supposed to follow from (2) for an arbitrary choice of $w$ that is not the one induced by the variation $y$?

0celo7

15:51

@ACuriousMind well unless I'm severely wrong here, let's jut interpret (2) as saying there is no explicit $x$ dependence for $L$

ACuriousMind

Okay.

0celo7

now we take some random $u:\Omega\to\Bbb V$ and the associated variation $w$ of $u$ and $y$ of $x$, where $y$ is generated by the flow and $w=u\circ y$

these satisfy the axioms laid out in the first paragraph, agreed?

and now we check that (1) holds

ACuriousMind

How do we check that?

0celo7

OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

there's where the pullback comes in

but a translation is an isometry so you just move the arguments in the integral

wait, rotations are isometries too

yeah, so $\phi(\omega)$ is the isometric image of $\omega$

thus we expect $\int_{\omega}L=\int_{\phi(\omega)}L\circ\phi^{-1}$

evaluated on $w$

now $w\circ\phi^{-1}=u$

so we get $$\int_\omega L(x,w,Dw)=\int_{\phi(\omega)}L(y,u,Du)$$

and then you just move the $y$ to $x$ and you're done

something like that

@ACuriousMind is that at all convincing?

We have $L(x,w,Dw)=L(y,w,Dw)=L(x,u,Du)\circ\phi$, so it follows from the change of variables formula

The issue must be that we really need a pullback and not a composition

ACuriousMind

I feel there's something going on here with you sometimes writing arguments for the second and third component of $L$ and sometimes not.

$L\circ \phi$ is really just $L(\phi(x), u(x),Du(x))$ isn't it?

0celo7

16:03

Well, note that for $u$ fixed, we have a map $F:\Omega\to\Bbb R$ defined by $x\mapsto L(x,u(x),Du(x))$ and I really mean $F\circ\phi$

ACuriousMind

Or, rather, $L\circ \phi$ doesn't really make sense because the domains don't match

Ah, yes.

Slereah

Why do you have $L = L \circ \phi$

0celo7

I don't

Well, depends on what you mean by that

I am NOT claiming that $F=F\circ\phi$

ACuriousMind

So wait, at what point in your derivation do we actually use (2) at all, @0celo7?

0celo7

@ACuriousMind I just want (1) to hold

I figured (2) was reasonable and implies (1)

now I'm not so sure it does

Slereah

16:07

I can't rly contribute much because I'm not quite sure what all the various quantities are

0celo7

so the RHS of (1) is just $$\int_{\phi(\omega)}F(x)\, dx$$

this equals $$\int_{\phi}\phi^*(F(x)\, dx)$$

Anonymous

If I'm asking a question w.r.t to a research paper (which is under a paywall), do I just link to the site with a paywall, or upload it to something like google drive and then share the link to the pdf? I'm talking about the main site. (BTW are such questions allowed on the main site?)

0celo7

assuming $\phi$ is killing, we have $\phi^*dx=dx$

so this is $$\int_\omega \phi^*F(x)\, dx$$

but $F$ is a scalar, so $\phi^*F(x)=F(\phi(x))$

@ACuriousMind Ah. I found the error

Anonymous

@ACuriousMind You around? Could you please have a look at my query above ^

0celo7

$D(w)=D(u\circ\phi)\ne (Du)\circ\phi$

But it is so for translations

thx

Anonymous

16:11

I think linking to sci-hub or something would be unethical...so that option is out

0celo7

inb4 it's true for rotations (I don't think it is)

ACuriousMind

@Blue It's likely that any question that requires the answerers to actually read a specific paper is not a good fit for the site. You can link to the paper however you want, but the essentials necessary for an expert to answer the question should be in the question itself, not in a link

0celo7

@Blue you link to the abstract

ACuriousMind

@0celo7 Heh, well, I didn't do much more than ask you to flesh out how (1) comes about in the end... ;)

Anonymous

@ACuriousMind Oh, got it. Thanks for the advice

0celo7

16:13

@ACuriousMind Can you confirm that this $\ne $ is true for rotations?

It has to be false

Anonymous

I'll try to make the question self-sufficient

Slereah

What is $w$ and $u$

0celo7

rotations are a position dependent transformation so it must be so!

ACuriousMind

@Slereah Names of variables in i.gyazo.com/83e44235d1b8a219eb47ebe2a96c67e1.png

Slereah

I know

What I mean is

What does it correspond to

ACuriousMind

16:14

@0celo7 Yes

Slereah

in non-crazy notation

ACuriousMind

@Slereah The field and the field variation.

Slereah

ah

0celo7

@ACuriousMind other way around

ACuriousMind

I.e. what the usual physicist will write as $\phi$ and $\phi'$

Slereah

16:14

Wait, is it the field variation or the field derivative

ACuriousMind

@0celo7 No, $\delta \phi = m$ :P

0celo7

oh, right

skullpatrol

removed

0celo7

@Slereah $u=\phi$, $w=\phi'$ (variation), $m=\delta\phi$ (infinitesimal variation)

also called a multiplier in math circles

Slereah

what's $Du$ then?

0celo7

16:16

$\partial_\mu\phi$

Slereah

What do you mean by "a variation" then

0celo7

look at the definition...

@ACuriousMind So that was a pretty subtle thing, no?

ACuriousMind

@0celo7 I think it would have been much easier to catch if you had written more detail than "(2) implies (1)" ;)

0celo7

Ah! Eureka! @ACuriousMind Weinberg's Belinfante-Rosenfeld tensor is $T+$ derivatives of $L$ wrt. the derivatives of the field.

So it is the derivative terms that couple to the Lorentz transformations

And the Christodoulou-Noether theorem has corrections involving $\nabla_\mu X^\nu$...

It all makes sense now...

Slereah

and here you were accusing all physics of being wrong

0celo7

16:20

I still stand by that statement

Slereah

I'm still not sure what the problem or solution was but good that it is solved

0celo7

Well, if you don't have kinematic terms there's no spin stuff

Slereah

Probably

$\partial \Lambda A = \Lambda \partial A + A \partial \Lambda$ and all that

0celo7

@Slereah I am slowly making sense of the details of ADM

Slereah

hopefully it will be real

Otherwise we'll have to throw out all those books

Balarka Sen

16:45

@0celo7 You're into PUBG now? :P

Slereah

bbc.com/news/world-europe-42674724

ACuriousMind

> When I see the quality of bread in supermarkets, it is impossible not to get angry. The bread is frozen, some of it comes from Romania or who knows where, nothing is carried out in accordance of the rules of the art.

0celo7

@BalarkaSen why do you say that?

ACuriousMind

Man, the French really hate Romanian bread, eh? :P

Balarka Sen

@0celo7 discord lets me see what game you are playing

0celo7

17:05

@BalarkaSen why is it amusing?

Balarka Sen

I didn't say anything about amusement

skullpatrol

::crickets::

vignette.wikia.nocookie.net/uncyclopedia/images/7/78/…

Semiclassical

17:23

@ACuriousMind bread bigots?

skullpatrol

The upper crust usually are :P

ooolb

17:39

hiiiiiiiiya

skullpatrol

heya

Johnathon

i.imgur.com/6SC0RRa.png for this diagram, would the internal torque at any point x be T_a or T_b?

Anonymous

17:57

@Johnathon Did you have a look at the diagram you posted? :P

Anonymous

Anonymous

I see this ^

Johnathon

whoops my bad

skullpatrol

lol

Anonymous

Find net torque using $(T_{A}x-T_{B}(d-x))\hat{k}$, $\hat{k}$ being the unit vector pointing out of the plane of the paper.

Johnathon

18:04

Sorry I should have made it more clear, T_a and T_b are both torques acting in the same direction

$T_a$ $T_b$ mathjax test

Anonymous

Are they sliding torques (generated due to a couple)? They should be, otherwise a rigid body like that hinged at two points would be fixed. In that case just add up the torques in the same direction.

Johnathon

@Blue It's a shaft sorry, ishould have stated all this at the beginning, my bad

A shaft rotating at constant speed and has uniform diameter

Anonymous

What are the forces generating the torques?

0celo7

@BalarkaSen I just called the Ugandan meme people in PUBG racist @Phase

They called me really mean things

Balarka Sen

looooool

get rekt boi

I don't know how to feel about that meme really. It's so cancerous, but simultaneously too popular

I think the right retaliation is just to come up with a better meme which are not shouted around in game servers by 12 year olds...

Anonymous

18:13

@BalarkaSen There's a paragraph in Nash Sen which I'm having difficulty in understanding. Will you be here for a while?

Balarka Sen

yep

Anonymous

Okaies:

Balarka Sen

which paragraph? I have the book

Johnathon

@Blue I tried to simplify it from this main question: i.imgur.com/dvqCvn8.png what I don't understand is why between 1000 and 2000mm that the internal torque is the same as the torque at D2

Anonymous

@BalarkaSen Page 3. Example 2

Anonymous

18:14

I'm having trouble understanding the deformation of $a_i$ stuff

Anonymous

They're choosing $a_i^{\epsilon}=\epsilon a_i$

Anonymous

$i=0,...,q-1$

Anonymous

Okay so far

Anonymous

But

Anonymous

How is $n_0(a_0^{\epsilon},...,a_{q-1})=n_0(a_0,...,a_{q-1})$ ?

Anonymous

18:16

They just say "evidently"....it's not really evident to me as to what they're doing

Balarka Sen

it's a typo. They mean $n_0(a_0^\epsilon, \cdots, a_{q-1}^\epsilon) = n_0(a_0, \cdots, a_{q-1})$

i.e., the value of the integral doesn't change if you perturb the coefficients

Anonymous

@BalarkaSen Ok. But then how is $n_0(a_0^\epsilon, \cdots, a_{q-1}^\epsilon) = n_0(a_0, \cdots, a_{q-1})$, unless $\epsilon$ is close to $1$ ?

Anonymous

We have to perturb it continuously, no?

Anonymous

They don't mention $\epsilon$ is close to $1$

Balarka Sen

$n_0(a_0, \cdots, a_{q-1})$ is a 1) continuous function 2) takes integer values always.

That forces it to be constant.

Anonymous

18:20

@BalarkaSen Yeah. I got that much. But say $n_0(a_0^3, \cdots, a_{q-1}^3)$ may be $2$ (an integer), while $n_0(a_0, \cdots, a_{q-1})$ is $1$

Balarka Sen

That can't happen, is what I mean by "constant"

$f(z) = n_0(a_0^z, \cdots, a_{q-1}^z)$ is a function which takes integer values for all $z$, and is continuous. If $f(1) = 1$ and $f(3) = 2$, it has to take a non-integer value somewhere in $(1, 3)$ by intermediate value theorem.

I.e,., there must be a $c \in (1, 3)$ such that $1 < f(c) < 2$

Contradiction, because $f(c)$ must be an integer...

Anonymous

Oooo

Anonymous

Understood

Balarka Sen

:)

Anonymous

Thanks !

bolbteppa

18:22

I feel like it's such a defeat to accept you just have to solve Hermite etc using series first and then getting the rest of the craziness like Rodrigues, generating functions etc as the easiest route :(

Balarka Sen

No problem

@Blue $\displaystyle \frac1{2\pi i} \int_C \frac{f'(z)}{f(z)} dz = n_0$ is a beautiful theorem in complex analysis.

It's called the "argument principle". Do you know the geometric interpretation?

Anonymous

@BalarkaSen Yep! I know that proof of that one

Balarka Sen

Nice!

Anonymous

It's $n_0-n_p$ though

Anonymous

In case there are some poles

Balarka Sen

18:29

True, I took a holomorphic function.

Anonymous

Basically we need the fact that $\frac{f'(z)}{f(z)}=\frac{d}{dz}\log f(z)$

Balarka Sen

Oo, careful. $\log$ is a very dodgy function in complex analysis.

Anonymous

Now, if $f(z)$ has multiplicity $m$ at $z=a$

Balarka Sen

That's good for an intuitive picture but you're better off knowing an actual proof!

(In fact that's what my geometric interpretation is)

Anonymous

@BalarkaSen Oh, do you have some other proof?

Anonymous

18:30

I only know this one

Anonymous

If $g(z)$ is holomorphic and non-vanishing over a simply connected set in $\Bbb{C}$

Anonymous

Then $\log g(z)$ is holomorphic

Anonymous

(I'm using the fact that you can then write $f(z)$ as $(z-a)^{m} g(z)$ such that $g(z)$ is holomorphic and non-vanishing)

Balarka Sen

Ah I see your proof will end up being equivalent to mine then.

But you don't need to write log to do this

Anonymous

@BalarkaSen Okay?

Anonymous

18:35

Then how

Balarka Sen

If $f(z) = (z - a)^m g(z)$, $f'(z) = m(z - a)^{m-1} g(z) + (z -a)^m g'(z)$. So $f'(z)/f(z) = m/(z - a) + h(z)$ where $h(z) = g'(z)/g(z)$ is something holomorphic inside $C$.

So that integrates to $2\pi i m$ over $C$

Then add up the multiplicities each of the zeroes contribute

Anonymous

Oh. You're directly finding $f'(z)/f(z)$

Anonymous

That's a good idea

Balarka Sen

Yup

Anonymous

And also there's no problem of zeros in denominator since we're choosing $g(z)$ carefully

Anonymous

18:37

Gotcha

Balarka Sen

Right.

Anonymous

Nice

Balarka Sen

But the geometric interpretation goes something like so:

The integral, as you observed, is $\frac1{2\pi i} \int_C d\log f(z)$

Anonymous

There's a $d/dz$ in there

Balarka Sen

Thanks, edited

What I wrote is a dodgy expression, but let's pretend it's right

Anonymous

18:39

okey

Balarka Sen

log is a multivalued function which changes value by $+2\pi i$ if you loop counterclockwise around $z = 0$, and changes value by $-2\pi i$ if you loop clockwise around $z = 0$

So for each zero $z = z_0$, $f(z_0) = 0$, of multiplicity $m_0$, $C$ loops around that zero of multiplicity $m_0$

That contributes to a $+\frac1{2\pi i} \cdot 2\pi i m_0 = +m_0$ in the integral

Anonymous

Oh, yes. Like for n loops in calculus of residues or contour integration

Anonymous

Understood :)

Balarka Sen

And similarly for each pole $z = z_\infty$, $1/f(z_\infty) = 0$, of multiplicity $m_\infty$, $C$ loops around $\infty$ (clockwise!)

Which contributes similarly to a $-m_\infty$ in the integral

@Blue Right

It's really a rather deep topological fact

Anonymous

@BalarkaSen Yup, cool observation. I was thinking of something like that too :D

Balarka Sen

18:43

Mhm

Anonymous

@BalarkaSen Sounds interesting

Anonymous

I'm at just page 3 and it's already so interesting :P

Balarka Sen

I like the approach of this book

Not hyper-mathematical of course, but neat

vzn

19:04

> My 70y old step dad finally managed to get his paper published. It allegedly proves that spooky action at a distance does not exist and that Bell's theorem is incorrect. He'd be thrilled to get your feedback!

reddit.com/r/Physics/comments/7q3zvw/…

Slereah

your grandfather's probably a crank

Maybe it's Duffield

Anonymous

Duffield is 70yo?

Anonymous

:p

CooperCape

20:12

Allo

Anonymous

@CooperCape ola

Captain Bohemian

The Gauss law has vanishing Poisson bracket with the Hamiltonian constraint.

Anonymous

Reading Gallian and Axler?

CooperCape

Yeaaaahhh

Haven’t really looked much through gallian

Anonymous

Cool. I just got through the first 2 chapters of Gallian

Anonymous

20:14

It's nice

Anonymous

Also reading a topo book

CooperCape

Oooh neat... it’s all group theory at the beginning right?

Anonymous

@CooperCape Yep

CooperCape

I’ll leave topology etc. For when I am better at maths...

(Hah nevverrr)

Anonymous

@CooperCape Before starting topo I suggest mastering at least basic complex analysis, first

CooperCape

20:15

Okay nokay

I got enough maths to do rn anyways

Anonymous

@CooperCape Heh...high school is broring XD

Anonymous

(P.S: Even college is boring)

CooperCape

Depends... on the middle of exams currently eek

Oh... oh no...

College was my getaway ;)

Anonymous

If you will take physics in first year they'll again teach those basic newtonian mech, sound, light, etc along with some programming I guess

Anonymous

And some basic math like multivariable calc and linear algebra (more basic level than Axler)

CooperCape

20:18

Sounds about right (from talking to friends etc.)

I really dislike mechanics tho ;p

Anonymous

The way it is taught in most schools makes it boring

Mithrandir24601

@CooperCape Note: what you think you like and dislike before uni $\neq$ what you actually like and dislike :P

CooperCape

Statics of rigid bodies is duhhhhhhhg

@Mithrandir24601 Good good.... maybe mechanics shall be my ting

Anonymous

It would be nice if they showed more visual simulations and real examples. (Newtonian) Mechanics has some very counter-intuitive results too...at times

CooperCape

Yeah it’s all very rod-like at the mo

Phase

20:26

can someone actually explain to me why the 1st and 3rd laws exist

Surely conservation of momentum and $F = \frac{d}{dt}p$ do it all, no?

CooperCape

Maybe he just wanted a few extra laws to his name.

Having just 1 is a bit basic.

Anonymous

@Phase There are some issues with the definition of inertial frame. Using second law you cannot prove the existence of an inertial frame, which is spoken about in the first law.

Phase

How so though, doesn't $\frac{d}{dt}p \iff a = 0$?

or am I missing something

Anonymous

14

Q: Why is Newton's first law necessary?

Newton's second law says $F=ma$. Now if we put $F=0$ we get $a=0$ which is Newton's first law. So why do we need Newton's first law ? Before asking I did some searching and I got this: Newtons first law is necessary to define inertial reference frame on which the second law can be applied. But...

Anonymous

12

A: Why is Newton's first law necessary?

Newton's first law postulates that there is (at least) one inertial reference frame for every object, in which said object will continue in uniform motion unless acted upon by a force. Newton second law states that, within the inertial reference frame for any object, $F = ma$. Without the first...

Anonymous

20:33

ACM also wrote an answer

Phase

yeah I saw the wild ACM

appearing from the long grass

CooperCape

Was it shiny?

Phase

is that a new thing?

Back in my day Pokémon were just Pokémon

CooperCape

I swear you could always get shiny Pokemon?

They were rare af tho

Phase

really?

CooperCape

20:35

I only played on DS tho was too late to be a game boy child

Phase

i dont remember there being any in 1st generation

CooperCape

Ahh they were gen 2

And up

Phase

back in my day we played Duel Monsters

with large stone tablets

and the loser got banished

Anonymous

There was something called Duel Masters

Anonymous

I forget

Anonymous

20:38

Ah, it was an anime

Phase

sounds like a yugioh rip off

CooperCape

Damn...

I was mario Pokemon and wii sports

Phase

Duel Masters was first serialised 3 years after YGO

that sounds like damn plagiarism to me

CooperCape

Hummmmm

Anonymous

azurilland.com/forums/non-pokemon-forums/entertainment/…

Anonymous

20:42

Best Answer: Duel Masters is a parody of Yu-Gi-Oh.

Phase

I have never seen so much salt from so many permavirgins

CooperCape

Permavirgins....

Now that’s a saying...

0celo7

@Phase I feel like a human!

Phase

@0celo7 you're nearly human!

Just a bit more evolution and you'll be there buddy

0celo7

21:05

> Yield: 11 cups

my god

2.8 L soup

are you kidding me

Thai ginger soup

oh hell yes

DanielSank

21:55

@Blue Wasn't that a YuGiOh! thing?

0celo7

@DanielSank that's duel monsters, the in-universe name for the card game

DanielSank

@0celo7 Oh, right.

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