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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.
@dmckee Hmm?
Police it?
@0celo7 Stop by occasionally and tell them if they are being idiots or not. The physics is not strong on that site.
@dmckee Sorry, what?
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
First, what does "I hope that the folks at world building have the sense to to accept straight physics questions." mean?
10:27 PM
Hell ... it means my brain is faster than my fingers.
Why are they accepting physics questions?
I hope the folks on worldbuilding have the sense to not accept straight physics question.
That's better, but I still don't see what's wrong with that
@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.
Are you saying real physics questions should get sent to us?
Well, you
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.
Would you say that -9 is a perfect square?
squares are positive
did you not go to school
But it's square root is an integer
3i clearly is the square root
what is i?
10:30 PM
clearly not an integer
the integers are 0,1,2,3,4,...,negatives
See the mess that is the accepted answer (and most of the other answers) on worldbuilding.stackexchange.com/questions/37603/…
I do not see i in that list
try again
imaginary integers
10:31 PM
that title already
You're using the set of natural numbers
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
i, 2i, 3i, 4i...
go for it
1/100 the hydrogen mass is pretty fucking heavy for a photon
10:31 PM
Dude have you never learned about imaginary numbers
That's heavier than an electron
@SirCumference I have not seen one of those around
@0celo7 Well yeah, it's not a natural number
what does i even mean
I think with that mass EM would basically be non-existent
Short range interaction
10:32 PM
@0celo7 It is defined as $√-1$
@Slereah language >:(
@SirCumference No it's not
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.
@SirCumference nonsense, what is $\sqrt$
It's the algebraic closure of $\Bbb R$
10:33 PM
@Slereah Wat
$\sqrt{-1}$ is a dangerously vague definition
@Slereah How is that definition any different? Calling it $\sqrt(-1)$ would make just as much sense, wouldn't it?
@Slereah How?
$\sqrt{-1} \times \sqrt{-1} = \sqrt{-1 \times -1} = 1$
$\sqrt{-1} \times \sqrt{-1} = -1$
A risky move
You have to redefine the square root first
Fuck it, I never know enough mathematics to speak here
...this is basic high school algebra
10:35 PM
@0celo7 We never went over abstract algebra, as far as I know
@SirCumference $\Bbb C$ is the Clifford algebra of $\Bbb R$, quite simply.
@0celo7 Yeah, simple. Thanks.
@SirCumference Probably an easier notion than algebraic completeness, tbh.
The most complicated math I know is single-variable calculus
This is 9th grade algebra.
10:37 PM
@0celo7 Where on earth did you learn about that in 9th grade?
It's defined as the set of pairs of real $\langle a, b \rangle$ with a specific algebra on it
@SirCumference Not about Clifford algebras, but I sure learned to not do what you're doing
Now that I'm looking again at that question, the better answers are bubbling up. There is hope after all!
It's better to define $i$ to be the object such that $i^2=-1$.
@0celo7 What? Misdefining i?
@0celo7 So why does that complicate my original question? "Is -1 a perfect square?"
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?
A number whose square root is an integer
What does it mean to you?
Perfect squares are defined in $\Bbb N$
@SirCumference A really, really pretty square
@BernardMeurer I have a funny picture for you
@0celo7 Is it about Dyson vacuums?
10:40 PM
Check your phone
I don't get it, pls explain
Goddammit do I really have to go into abstract algebra? I just started going into calc...
She doesn't either
@0celo7 What don't you get?
The caption
Wtf is this
10:41 PM
I laughed but idk why
btw, six months in a week
Dayum, has it really been this long?
Well, y'all are basically married now I guess
split and she get's half your fortune
10:43 PM
All right, so is it more accurate to define $i$ as the number whose square is -1?
Thankfully half of 0 is still 0 though
She's the rich one
Dwight has 7 cars
Not my father in law...
Dwight Schrute
10:45 PM
Also what do you think the d in @dmckee stands for
@0celo7 David
oh really, who told you that?
@0celo7 Your mom
@0celo7 You looked me up at the place I teach. Surely you noticed my name?
@dmckee Uh, I did?
10:46 PM
@dmckee Sup David
Never call a prof by their first name @BernardMeurer
you little shrew
@0celo7 Ah; Sup Prof. David
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.
All right, 0celo, do I have a more accurate definition of i?
@0celo7 Dr. Prof. Mr. McKee
@SirCumference me
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.
Can I define $i$ as the number whose square is $-1$?
@BernardMeurer Only in Germany, I think. And only if they make my doctorate official.
@dmckee Oh, I call my supervisors (I now have two) by their first names
@dmckee The Krauts will never help
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...
10:48 PM
@SirCumference I think so
I think he thinks I'm a first year grad student
Not first year undergrad
@0celo7 You're 19 ain't you?
@BernardMeurer No
@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.
@0celo7 Lel, we're the same age now
10:49 PM
@0celo7 well you do look old.
@SirCumference Yes.
@0celo7 Cool.
pulls out geometric analysis textbook
Ah, spin structures.
@dmckee It's official enough for me babe
Lel, that was weird
Always wanted to use that line
Proper definition of $i$
10:51 PM
@SirCumference The standard basis vector of $\mathrm{C}\ell(\Bbb R)$.
It's the standard basis vector of the rank 2 subalgebra
Sigh...I need to look into abstract algebra
Do you mean
Jesus give me a break
10:53 PM
Oh, no, it's
the standard basis vector of $\mathrm{C}\ell^1(\Bbb R)$.
Not $\mathrm{C}\ell^2(\Bbb R)$?
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
@Slereah No, 2 does not exist because $\Bbb R$ is 1-dim.
Oh right
Although, $\mathrm{Pin}(\Bbb R)=\{-\mathrm i,\mathrm i\}$.
There might be something there of use.
10:54 PM
Wait, how does that work then
@0celo7 *closing $
$i^2 = dx \wedge dx = 0$???
what are you on about
Oh wait it's the Clifford product
What's the Clifford product here
@Slereah Do you want me to show you how this works
10:56 PM
$i^2 = dx \wedge dx - dx \cdot dx = -1$
I guess?
yolo, sure
For any orthonormal basis, $e_i\cdot e_i=-1$, where $\cdot$ is the Clifford product.
But then $1 \cdot 1 = -1$
What is the product of two complex numbers
Aren't complex numbers just $\text{Even}(C\ell(\Bbb R^2))$
@Slereah Uh
It's for a basis of the clifford algebra that that's true
1 is not a basis vector of it...I think
@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
@SirCumference Isn't it also true that $(-i)^2=-1$?
For $z^n = c, c \neq 0$ there is always $n$ solutions
11:11 PM
Sure, so there's more than one number that satisfies $(\text{number})^2=-1$
One more than one, even
$i$ is the unique solution of $z^1 = i$, though
11:27 PM
@SirCumference why do you need to know this
@Slereah will you send me Shouten?
you're not using it
11:42 PM
Buy your own damn books
@Slereah Oh come on why
Because property rights
Objectivism does not prevent charity
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