The h Bar

10:11 PM

I hope that the folks at world building have the sense to to accept straight physics questions. Otherwise we're going to have to police that site from time to time.

in re: worldbuilding.stackexchange.com/questions/47826/…

0celo7

@dmckee Hmm?

Police it?

dmckee

@0celo7 Stop by occasionally and tell them if they are being idiots or not. The physics is not strong on that site.

0celo7

@dmckee Sorry, what?

dmckee

More than a few people think the the intuition they built up by reading a few pop sci book and passing the physics course they had to take in college suffices to extraoplate things almost with out end

0celo7

First, what does "I hope that the folks at world building have the sense to to accept straight physics questions." mean?

dmckee

10:27 PM

Hell ... it means my brain is faster than my fingers.

0celo7

Why are they accepting physics questions?

dmckee

I hope the folks on worldbuilding have the sense to not accept straight physics question.

0celo7

That's better, but I still don't see what's wrong with that

dmckee

@0celo7 They have a physics tag. They often use it in conjunction with [science-based] or [reality-check]. And the answer often contains real bad science.

0celo7

Are you saying real physics questions should get sent to us?

Well, you

dmckee

10:28 PM

I don't care if they send them here or not.

But if they accept them, it's going to be necessary to straighten out the mess occasionally.

Sir Cumference

Would you say that -9 is a perfect square?

0celo7

No

squares are positive

did you not go to school

Sir Cumference

But it's square root is an integer

3i clearly is the square root

0celo7

what is i?

Sir Cumference

sqrt(-1)

0celo7

10:30 PM

clearly not an integer

Sir Cumference

shoot

0celo7

the integers are 0,1,2,3,4,...,negatives

dmckee

See the mess that is the accepted answer (and most of the other answers) on worldbuilding.stackexchange.com/questions/37603/…

0celo7

I do not see i in that list

try again

Sir Cumference

imaginary integers

0celo7

10:31 PM

Ahhh

that title already

Sir Cumference

You're using the set of natural numbers

0celo7

I'm no physicist but I know that's bad

@SirCumference what the hell is that

if you're going to make up definitions, sure

Sir Cumference

i, 2i, 3i, 4i...

0celo7

go for it

Slereah

1/100 the hydrogen mass is pretty fucking heavy for a photon

Sir Cumference

10:31 PM

Dude have you never learned about imaginary numbers

Slereah

That's heavier than an electron

0celo7

@SirCumference I have not seen one of those around

Sir Cumference

@0celo7 Well yeah, it's not a natural number

0celo7

what does i even mean

Slereah

I think with that mass EM would basically be non-existent

Short range interaction

Sir Cumference

10:32 PM

@0celo7 It is defined as $√-1$

0celo7

@Slereah language >:(

Slereah

@SirCumference No it's not

dmckee

It's one thing for a writer to say "Screw the science! I need this for my story." Fine, you need it for your story. But you have to give up on claiming the work is "science based" at that point.

0celo7

@SirCumference nonsense, what is $\sqrt$

Slereah

It's the algebraic closure of $\Bbb R$

Sir Cumference

10:33 PM

@Slereah Wat

Slereah

$\sqrt{-1}$ is a dangerously vague definition

Sir Cumference

@Slereah How is that definition any different? Calling it $\sqrt(-1)$ would make just as much sense, wouldn't it?

@Slereah How?

Slereah

$\sqrt{-1} \times \sqrt{-1} = \sqrt{-1 \times -1} = 1$

$\sqrt{-1} \times \sqrt{-1} = -1$

A risky move

You have to redefine the square root first

Sir Cumference

Fuck it, I never know enough mathematics to speak here

0celo7

...this is basic high school algebra

Sir Cumference

10:35 PM

@0celo7 We never went over abstract algebra, as far as I know

0celo7

@SirCumference $\Bbb C$ is the Clifford algebra of $\Bbb R$, quite simply.

Sir Cumference

@0celo7 Yeah, simple. Thanks.

0celo7

@SirCumference Probably an easier notion than algebraic completeness, tbh.

Sir Cumference

The most complicated math I know is single-variable calculus

0celo7

This is 9th grade algebra.

Sir Cumference

10:37 PM

@0celo7 Where on earth did you learn about that in 9th grade?

Slereah

It's defined as the set of pairs of real $\langle a, b \rangle$ with a specific algebra on it

0celo7

@SirCumference Not about Clifford algebras, but I sure learned to not do what you're doing

dmckee

Now that I'm looking again at that question, the better answers are bubbling up. There is hope after all!

0celo7

It's better to define $i$ to be the object such that $i^2=-1$.

Sir Cumference

@0celo7 What? Misdefining i?

@0celo7 So why does that complicate my original question? "Is -1 a perfect square?"

0celo7

10:38 PM

The only property you ever use is that $i^2=-1$, so this works best.

@SirCumference It's not.

What does a perfect square mean to you?

Sir Cumference

A number whose square root is an integer

What does it mean to you?

0celo7

Perfect squares are defined in $\Bbb N$

Bernard Meurer

@SirCumference A really, really pretty square

0celo7

@BernardMeurer I have a funny picture for you

Bernard Meurer

@0celo7 Is it about Dyson vacuums?

0celo7

10:40 PM

No?

Check your phone

I don't get it, pls explain

Sir Cumference

Goddammit do I really have to go into abstract algebra? I just started going into calc...

0celo7

She doesn't either

Bernard Meurer

@0celo7 What don't you get?

0celo7

The caption

Bernard Meurer

Wtf is this

Boi?

0celo7

10:41 PM

Yeah

Bernard Meurer

I laughed but idk why

0celo7

Same

btw, six months in a week

Bernard Meurer

Dayum, has it really been this long?

0celo7

yup

Bernard Meurer

Well, y'all are basically married now I guess

split and she get's half your fortune

Sir Cumference

10:43 PM

All right, so is it more accurate to define $i$ as the number whose square is -1?

Bernard Meurer

Thankfully half of 0 is still 0 though

0celo7

She's the rich one

Dwight has 7 cars

Bernard Meurer

HAHAHAHA

YOUR FATHER IN LAW'S NAME IS DWIGHT?!

HAHAHAHA

0celo7

Not my father in law...

Bernard Meurer

:v

Dwight Schrute

0celo7

10:45 PM

Also what do you think the d in @dmckee stands for

Bernard Meurer

@0celo7 David

0celo7

oh really, who told you that?

Bernard Meurer

@0celo7 Your mom

l00ser

dmckee

@0celo7 You looked me up at the place I teach. Surely you noticed my name?

0celo7

@dmckee Uh, I did?

Bernard Meurer

10:46 PM

@dmckee Sup David

0celo7

Never call a prof by their first name @BernardMeurer

you little shrew

Bernard Meurer

@0celo7 Ah; Sup Prof. David

dmckee

I thought about changing the user name when I was elected moderator, but we had also just elected David Z, so I thought it would reduce confusion to just do nothing.

'sides, I'm lazy.

Sir Cumference

All right, 0celo, do I have a more accurate definition of i?

Bernard Meurer

@0celo7 Dr. Prof. Mr. McKee

@SirCumference me

dmckee

10:47 PM

@0celo7 My experience was that you can call your research supervisor and others in your group by first names.

But that varies from place to place. Do what the other students do.

Sir Cumference

Can I define $i$ as the number whose square is $-1$?

dmckee

@BernardMeurer Only in Germany, I think. And only if they make my doctorate official.

0celo7

@dmckee Oh, I call my supervisors (I now have two) by their first names

Bernard Meurer

@dmckee The Krauts will never help

0celo7

and when I met with my grad QM prof, he insisted I call him by his first name

But I don't think he understands I'm 18...

Bernard Meurer

10:48 PM

@SirCumference I think so

0celo7

I think he thinks I'm a first year grad student

Not first year undergrad

Bernard Meurer

@0celo7 You're 19 ain't you?

0celo7

@BernardMeurer No

dmckee

@BernardMeurer Apparently there is a process to get made official if you are hired for a university post over there. But I imagine it as being a exceedingly precise snarl of red tape.

Bernard Meurer

@0celo7 Lel, we're the same age now

user218912

10:49 PM

@0celo7 well you do look old.

0celo7

@SirCumference Yes.

Sir Cumference

@0celo7 Cool.

0celo7

pulls out geometric analysis textbook

Ah, spin structures.

Bernard Meurer

@dmckee It's official enough for me babe

Lel, that was weird

Always wanted to use that line

Slereah

us.metamath.org/mpeuni/df-i.html

Proper definition of $i$

0celo7

10:51 PM

@SirCumference The standard basis vector of $\mathrm{C}\ell(\Bbb R)$.

Slereah

It's the standard basis vector of the rank 2 subalgebra

Sir Cumference

Sigh...I need to look into abstract algebra

Slereah

Do you mean

Sir Cumference

Jesus give me a break

Slereah

the AL-JABAR?

0celo7

10:53 PM

Oh, no, it's

erm

the standard basis vector of $\mathrm{C}\ell^1(\Bbb R)$.

Slereah

Not $\mathrm{C}\ell^2(\Bbb R)$?

Sir Cumference

The sad thing is that math is so complex, you can never really get a grasp of everything we've discovered

So I'll always be math deficient

0celo7

@Slereah No, 2 does not exist because $\Bbb R$ is 1-dim.

Slereah

Oh right

0celo7

Although, $\mathrm{Pin}(\Bbb R)=\{-\mathrm i,\mathrm i\}$.

There might be something there of use.

Slereah

10:54 PM

Wait, how does that work then

Sir Cumference

@0celo7 *closing $

Slereah

$i^2 = dx \wedge dx = 0$???

0celo7

sigh

what are you on about

Slereah

Oh wait it's the Clifford product

Hm

What's the Clifford product here

0celo7

@Slereah Do you want me to show you how this works

Slereah

10:56 PM

$i^2 = dx \wedge dx - dx \cdot dx = -1$

I guess?

0celo7

yolo, sure

For any orthonormal basis, $e_i\cdot e_i=-1$, where $\cdot$ is the Clifford product.

Slereah

But then $1 \cdot 1 = -1$

What is the product of two complex numbers

Aren't complex numbers just $\text{Even}(C\ell(\Bbb R^2))$

0celo7

@Slereah Uh

No

It's for a basis of the clifford algebra that that's true

1 is not a basis vector of it...I think

Erm

@Slereah There's something strange

since $\mathrm{C}\ell(\Bbb R)$ is 2-dim there must be two basis vectors

one is 1, which is somehow exempt from that rule

HDE 226868

@SirCumference Isn't it also true that $(-i)^2=-1$?

Slereah

For $z^n = c, c \neq 0$ there is always $n$ solutions

HDE 226868

11:11 PM

Sure, so there's more than one number that satisfies $(\text{number})^2=-1$

Slereah

One more than one, even

$i$ is the unique solution of $z^1 = i$, though

0celo7

11:27 PM

@SirCumference why do you need to know this

@Slereah will you send me Shouten?

you're not using it

Slereah

11:42 PM

Never

Buy your own damn books

0celo7

@Slereah Oh come on why

Slereah

Because property rights

0celo7

Objectivism does not prevent charity

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