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12:00 AM
I don't think this is the case here, although the effect is close
 
Does she caim to be a mother?
If so, she's up awfully late for that portion of the world doing things not related to kids.
 
Not explicitly
 
Who doesn't know what ass is?
 
It'd be hilarious if this was the case
I'd be willing to entertain the joke :)
 
@0celo7 I think that was an attempt to show how immature we are, given the reply to Danu, but perhaps I've just got an overactive imagination.
 
12:05 AM
@ACuriousMind 4chan is immature. We are not.
 
I agree, though I would find worse words for 4chan.
 
@ACuriousMind Fun, entertaining, etc. are not bad words IMHO.
 
Obviously, those are not the words I had in mind.
But if you find it entertaining, I won't rain on your parade
 
@ACuriousMind I won't complain if it rains, e.g., Benjamins.
Do Germans know what Benjamins are?
 
@0celo7 I'm trying to decipher that and see whether it can enlighten me about the wedge ;)
 
12:07 AM
@Danu I am replying to you. You wondered about a question that I asked @0celo7. Yes, I asked as if I have asked an issue of language, any mature person.
 
@0celo7 Unless their surname is Franklin, no
 
@ACuriousMind Oh lol nothing. It just mentions complex exterior derivatives.
@ACuriousMind C-note
 
@Sofia Ah, yes. I must admit that I found it quite funny that you were not aware of the word 'ass' :)
 
@ACuriousMind Union of 100 singlet simoleons.
 
@0celo7 Don't you see that the "bicomplex" $\Omega^{\ast,\ast}$ there is precisely the $\Lambda^{p,q}$, and its elements are all written with the wedge. I suspect writing the symbol joining one complex to the other as a wedge is convention.
 
12:10 AM
...it's just such a useful word that one sees so much it's hard to imagine someone not coming across it, @Sofia
 
@ACuriousMind I think that $\Omega$ is a section of $\Lambda$, but I get your point.
@ACuriousMind Benjamin = 5 Jacksons
 
(money)^
 
Still no clue?
 
Oh, I spoiled it
 
@Danu Have an internet cookie dollar.
 
12:12 AM
Americanized as f*ck :)
 
@0celo7 Yes, I think you are right, these are the sections, like DeRham is sections of the usual exterior cotangent algebra.
@0celo7 I had no clue till Danu said it
 
@ACuriousMind Have you never been to FreedomLand, Inc.?
 
"I'm out for dead presidents to represent me"
 
Is Franklin on a dollar note, or something?
 
@ACuriousMind Benji is the C-note
 
12:13 AM
@0celo7 Nope, never been across the ocean
 
Jesus, George is on the $1.
 
@0celo7 What the heck is the C-note. A hundred?
 
Abe is the $5.
 
@Danu Shockingly Americanized.
 
12:14 AM
@0celo7 Genuinely a classic.
 
@Danu Forgot all about it, sad to say.
 
The video actually features Jay-Z and his 'friends' playing Monopoly---with real money though.
 
Is there some special reason why when I go to the bathroom, Einstein the cat follows me in and starts rubbing up against my legs?
 
@0celo7 No, all cats do that
 
He clearly wants the D
 
@ACuriousMind A kind spirit... right? ;)
 
Ugh. I am annoyed right now.
 
@Danu Do you mean kindred? I don't get the joke with *kind*^^
 
The AutoReviewScript updated and now it doesn't work
 
Gotta looove "updates" like that
 
1:31 AM
what is the autoreviewscript?
are all reviews autoreviewable
 
It allows you to insert automatically generated text
That way I don't have to manually add in, "Please note that Physics.StackExchange ...."
 
So like comments reminding people not to do things....
 
Right
But it makes it so that I don't have to type it in every time I need to do so
 
@StanShunpike It's the reason Kyle's comments begin to sound robotic when you've read them for the hundredth time ;)
 
Because it is robotic
 
1:33 AM
Yeah, that explains it. I just thought it was because he was boring :-o :p
 
And it's now effing annoying
 
Can you undo the update?
 
I had to go back to the repository (fortunately on GitHub) and install the previous version
 
@StanShunpike lol
Physicists can be boring, but not that boring
 
@KyleKanos did it work?
 
1:38 AM
Yes
And fortunately my replacement text (the common HW one) was still there
 
That's good. I hate extra work for mods. I was really amazed actually yesterday...I was in one of the CS chat rooms for the first time. and I saw people chatting about whether to protect a question. And I didn't realize people actually posted crap answers. That's so stupid. But they were discussing how it was going on and it just baffles me people have the time to post useless stuff.
 
@StanShunpike You don't see discussions about protection here because almost everytime I think oh, that should probably be protected, I find that Qmechanic beat me to it ;)
 
Yeah, some of it is the HNQ effect
Seeing an interesting title (and post) via the Hot Network Question list leads to plain bad answers
I remember there was someone who asked virtually the same question as a HNQ but with one minor tweak. I VTC'd as a duplicate and the guy got a little offended
 
@ACuriousMind QMechanic is like a wizard lol
Mysterious, we don't know where he's from, he seems to know things in advance, and he pulls answers to stuff out of thin air.
 
1:47 AM
@ACuriousMind lolol that's hilarious
 
And I believe he pulls the answers out of his knowledge and books, there are always references ;)
 
It's true. He asked me to provide page numbers in my answer.
@KyleKanos Really? I'm amazed making the HNQ actually makes a difference. I guess though there are a lot of SEs so I don't really appreciate how many people that actually reaches
@ACuriousMind I thought that was homotopy....
 
@StanShunpike I've visited a fair few of the sites due to HNQ
Pretty much any Harry Potter, Marvel/DC, or Tolkien question gets me over to SciFi
 
I think gaming.SE has made a sport of making clickbaity titles like "Is there a downside to killing everyone I meet?"
 
lolol omg that's such a crafty way to get people to click on stuff
 
1:52 AM
@ACuriousMind Or if there is a downside to forcing someone to marry you
 
and ppl do it
you see ads online with titles just like that
and i never click on stuff like that, but some ppl must
 
Yeah, these ads wouldn't exist if there weren't some people who click them
 
Someone's bitter:
0
A: Stark Effect in Hydrogen (solution with Mathematica)

SolarmewWow, thanks for nothing, Physics stack exchange ... I posted this question on Mathematica stack exchange first because I didn't even realize there was a Physics one. And I didn't have my hopes up because it was kind of off topic there even more so than it could've ever possibly been here, but at ...

 
Should I flag this lol?
 
::shrugs:::
@StanShunpike Well, technically, it is an answer to the question since they point out their mistake.
So it's not not an answer.
 
1:54 AM
But could we edit that?
Like if I were a first-timer, I would be put off by that
 
@StanShunpike yes
 
Yes to flag, or yes to edit
?
 
Yes to edit. (You would see the arrow if you weren't on mobile, I guess :P )
 
@StanShunpike Sorry, it links correctly for me. Yes to edit
 
lol mobile, ah the disadvantages. but i'm not complaining, i learn a ton. it works great. i actually only got a smart phone 6 months ago and I'm really loving having internet access wherever i go.
I can go on SE anywhere
So is it true LHC is on again?
I saw Lubos posted something about it yesterday.
 
1:59 AM
Probably, I don't know.
 
lol I take it your not one of the "experimentalists" as Particle Fever calls them
 
@StanShunpike Nope, not really. If they get interesting results, I'll hear of them soon enough.
 
Lol, for me it's sorta like a sporting event. I like to watch all the hype
I love sports.
 
Whether they're currently upgrading or running LHC is not all that interesting, I think
@StanShunpike Heh, go for it, then. I don't like sports :P
 
Not a fan of the Bundesliga?
 
2:05 AM
@StanShunpike Nope. Although, at times, it seems every other German is :D
 
@ACuriousMind Do you not like soccer in general or just German soccer?
@ACuriousMind Your theorist is showing.
 
@0celo7 Soccer in general as a subset of sports in general. I just don't get what's interesting about watching other people do sports.
@0celo7 Yeah, he needs some air at times.
 
Amazing athletes make amazing plays. Have you ever heard of ESPN's sports science segments?
But maybe you won't find it interesting.
@ACuriousMind I'm still not sure what a wedge product is. Any chance you know a convenient way of explaining it?
 
@StanShunpike Yep, my reaction is: meh
@StanShunpike Do you know what a tensor product is?
 
@ACuriousMind ::Unwraps noose rope::
 
2:15 AM
@0celo7 Good thing there's an ocean between us, then.
 
@ACuriousMind ::grumbles::
 
@ACuriousMind not really, but I'm trying to learn about them. Wikipedia defines them using free vector spaces and everyone I talked to said that wasnt needed
Ted Shifrin said I should just use matrices in the beginning but I think if I could learn tensor products it would just be more efficient.
I don't know what an ideal is either.
Tensor products and ideals seem like the key ingredients for defining a wedge product
 
@StanShunpike Understand the tensor product first, then, the wedge is defined out of it. As a slogan, $\wedge$ is the antisymmetrization of $\otimes$.
And, unsurprisingly, I would disagre with the people you talked to, the proper definition of the tensor product is as the free space on a certain basis.
Or as multilinear maps, take your pick
 
Multilinear maps!
I know what those are
Those I am comfortable with
 
Good. The tensors of degree $k$ on a space $V$ are just mulltlinear maps that eat $k$ arguments from $V$, alright?
 
2:21 AM
Sure, that makes sense.
Does that mean they take in both $k$-vectors and $k$-forms?
 
@StanShunpike no forms here. For forms, you need the wedge
A $k$-tensor takes $k$ vectors as arguments
 
Ah, really? I thought it took both. No wonder I am confused.
So a $k$ tensor only accepts $k$ vectors?
 
@StanShunpike Yes. A $k$-tensor takes $k$ different vectors as its arguments, it is a multilinear map $V\times\dots\times V \to \mathbb{R}$
 
Okay, that's very helpful. I have been confusing that point. So no forms yet.
I got confused because schutz says forms behave like vectors
 
@ACuriousMind D: Remember the rug!
 
2:26 AM
@StanShunpike Now, you can take $k$ $1$-tensors $t_i$ and build a $k$-tensor $t$ out of them as $t(v_1,v_2,\dots,v_k) := t_1(v_1)\cdot t_2(v_2)\cdot\dots\cdot t_k(v_k)$.
@0celo7 I'm not doing your weird complex geometry here :P
 
@ACuriousMind What operation is $\cdot$ here?
Is that just dot product?
 
@StanShunpike Multiplication in $\mathbb{R}$
 
OH
I get it
 
Remember, a $1$-tensor that has eaten a vector is just a number
 
okay yeah that makes perfect sense
 
2:30 AM
@StanShunpike: Now, you have to observe that a) the $1$-tensors are just the dual space of $V$, since they're linear maps $V\to\mathbb{R}$, which is the definition of the dual, and so they have a basis $t_i$ and a vector space and that b) $k$-tensors are also a vector space (in the sense that they satisfy the vector space axioms)
 
@ACuriousMind Complex geometry is all mine.
 
@ACuriousMind Yes, I follow.
@0celo7 I think Danu was considering taking a course on that.
 
@StanShunpike Good. Now, let's denote the $2$-tensor I can construct out of pairs of $1$-tensors $t_i,t_j$ as $t_{ij}(v_1,v_2) := (t_i\otimes t_j)(v_1,v_2) := t_i(v_1)\cdot t_j(v_2)$.
One can show that all the possible combinations $t_i\otimes t_j$ form a basis of the space of $2$-tensors.
With me so far?
 
I think so. we started off by constructing $k$-tensors out of $1$-tensors. Now we are saying that this $2$-tensor is simply a construction out of these. Does this mean it forms a vector space?
 
@StanShunpike This $2$-tensor is just a special case of the earlier construction. You agreed that the $k$-tensors abstractly form a vector space. I now claim that the basis of the space of all $2$-tensors is given by these $t_i\otimes t_j$.
similarily, all possible combinations $t_i\otimes t_j\otimes t_k$ are a basis for the space of $3$-tensors, and so on
 
2:38 AM
Makes sense.
Yes, I follow.
So wait, for $g = -t \otimes t + x \otimes x + y \otimes y + z \otimes z$ that works then because we can just add those elements and they form a $2$-tensor?
We can add them because they form a vector space?
 
@StanShunpike Exactly
 
YES!
Awesome :)
 
And every $k$-tensor can be expressed as a sum over the basis $t_{i_1} \otimes \dots \otimes t_{i_k}$ made of the $k$-fold combinations of the basic $1$-tensors.
 
Yes, that makes sense.
That's awesome. I love it.
 
Haha ohhh @0celo7 consider, say, the sequence of partial sums of (1,2,1,3,1,4,1,5,1,...). ie the sequence 1,1+2,1+2+1,1+2+1+3,... then you can have arbitrarily large gaps. Lim sup gap size =infinity. But after every gap comes a gap of size 1! Nothing more to it than that.
 
2:43 AM
@ACuriousMind Is that the basic idea behind the tensor product?
 
@StanShunpike That, uh, basically was it
 
Re a much earlier conversation between you and Danu (?) I can't cite 'cause I'm on my phone.
 
@NeuroFuzzy I see.
 
@ACuriousMind So what is an ideal?
 
@StanShunpike Oh, that would require a bit of ring theory. I'd now just define the wedge by $s\wedge t := s\otimes t - t\otimes s$.
 
2:48 AM
Hmm, amazing. That's awesome. I'm going to have to study the tensor product part to appreciate the wedge part, but that's a really nice way to go about it. Yay! I have a book on exterior algebras coming so this is great. I will hopefully be able to get more out of it now.
@ACuriousMind My uncle gave me a book Topics in Algebra i think it's called
Rings baffle me frankly. I don't know what to do once a group loses it's abelian nature. In fact, I still don't get why rotations aren't abelian. I think that's what I remember someone saying
 
@StanShunpike Then I'm certain you will learn what an ideal is then
 
@StanShunpike You can rotate your body to check this.
 
Try me. How? What's an example?
 
@StanShunpike Rotating something first about the $y$-axis by 90° and then about the $z$-axis by 90° isn't the same as doing it in the reverse order
 
End of the third chapter in Zee IIRC
He has a picture.
 
2:51 AM
Hmm...
Yes, indeed it isn't the same.
 
Abelian groups are boring, they're just sums of copies of $\mathbb{Z}$ and $\mathbb{Z}/n\mathbb{Z}$ :P
 
I think this stuff is fascinating, I just haven't gotten into it yet. I think I need to do some proofs. I haven't yet tried doing the problems in several books.
music was what got me interested in group theory.
Because I always come back to Do.
I always wanted to know
why certain notes are preferred
in certain combinations and if there were ways to describe that through group theory
Anyways, neat stuff
@ACuriousMind When did you first start learning about physics?
 
Well, if you let the mathematician work long enough, they will describe everything ;)
 
lol
I'm curious because some people started very early.
Lubos said he was thinking about this stuff when he was 10. But I grew up in a family that didn't really know much about it.
 
@StanShunpike I'm not sure. I soaked up what my father told me, but my parents are both chemists, so that was only rarely physics. My contact with physics is only through school, I think.
Probably 6th or 7th grade, but I was never someone who thought much about these things. It was good to know that there was knowledge out there, but it could wait.
 
3:02 AM
@ACuriousMind So I'm still struggling to get a grasp on what the current working framework is for understanding blackholes. Is the Schwarzchild metric a tool used for defining blackholes?
I just mean
from what I have heard, GR fails to describe blackholes to some extent.
 
@StanShunpike Oh, don't ask me that, I don't know. My understanding of GR is...superficial
 
@0celo7 Okay, what say you?
@ACuriousMind Does GR interest you? I read somewhere (maybe your profile) that you like gauge theories or something like that...
 
@StanShunpike Correct.
We really haven't solved black holes, because, as @ACuriousMind so elegantly puts it, we use hokus pokus curved spacetime QFT.
 
What do you mean?
 
@StanShunpike We don't really know what happens at the horizon.
 
3:06 AM
Oh, BTW: the baseball game was fun
 
@KyleKanos What are you talking about?
 
There was a surprising number of injuries
 
We don't have any direct evidence of Hawking radiation.
 
@StanShunpike I took my son to his first baseball game
 
@KyleKanos That's awesome! What teams played? My aunt and her family are seeing opening day for cubs cardinals. She is a die-hard cardinals fan
 
3:07 AM
Clemson vs Notre Dame
 
That's cool
 
Eh, Clemson lost
5-1
 
@StanShunpike It conceptually sounds like something that would interest me, but I think I've been deterred by a bad lecture on it, and now I'm so immersed in quantum theories that my interests don't go in that direction anymore.
I mean, it's classical. Ew. ;)
 
@ACuriousMind It's not classical, it's general ;)
 
Lol
@0celo7 But what is a black hole?
That's what I'm confused about. What's our definition of it?
 
3:09 AM
@ACuriousMind It's probably for the best. We don't have any other gauge theory specialists who frequent the chat.
@StanShunpike Roughly, a region of spacetime that is causally disconnected from the rest of spacetime.
 
@StanShunpike Isn't it any object whose escape velocity is $\geq c?$
 
I don't have the precise topological definition memorized.
 
user54412
@StanShunpike The event horizon separates regions of spacetime based on whether there null paths to null infinity
 
^there's an "expert"
:D
 
@ChrisWhite Is what Kyle said captured in that statement then?
 
3:12 AM
@StanShunpike Yes. If there is no path to null infinity inside of the horizon, then light from inside the black hole can't join light from the outside off in the distance.
 
user54412
I'd argue that escape velocity is a misleading way to think about things. In particular, in a Newtonian universe with an arbitrary speed limit of c, you could still escape from a black hole.
 
user54412
After all, with continuous thrust, you can escape at any velocity
 
@0celo7 Oh, I'm probably "advanced", but not a specialist. I'll call myself a specialist when I understand everything in, for example, here.
 
user54412
Moreover, even without continued thrust, launching yourself at less than the escape velocity moves you some distance off the surface, which is decidedly not the case with GR and black holes
 
@ACuriousMind But no one cares about that :P
 
3:14 AM
lol, alright
 
@ACuriousMind I meant a specialist to serve my needs.
Egocentrism and all that.
 
I serve your needs? Now I feel dirty.
 
@ACuriousMind You dissed GR. You should.
 
@ACuriousMind Hey! No one ever told me that!
 
@ChrisWhite Interesting, I never thought about it like that!
@ChrisWhite Quite true. Nice observation.
 
3:18 AM
@NeuroFuzzy It's just the structure theorem for finitely generated modules applied to $\mathbb{Z}$-modules ;)
 
@ACuriousMind is an ideal domain related to ideals?
 
@StanShunpike Yeah, a principal ideal domain is a domain which every ideal is a principal ideal.
...which probably tells you nothing :D
 
Peter piper picked a peck of pickle peppers.
That's about as much as I got out of it
 
Thought so, but I don't have an easy explanation, sorry
 
That's okay, I will do some studying before I ask
I don't like to ask until I feel like I have a chance to get it.
Too much to ask for someone to explain. But if I'm on the cusp, then I ask away.
 
3:24 AM
@ACuriousMind so what the heck do ideals have to do with physics???? I've been learning about them (ideals over commutative rings, specifically)
I actually have a final on them tomorrow morning...
 
My English teacher at Caltech said she was the first person to take Feynman to Huntington Library where they keep a copy of Newton's principia. She said Feynman took the Principa and started reading it. He came to a page where he saw some problem. He thought through the proof and turned the page, only to find Newton had done the same proof.
@ACuriousMind ^
 
@NeuroFuzzy Uh...they're things you can divide out of rings, since they're the kernels of ring morphisms, and it often happens that you want to consider the equalities in some algebra (which is essentially a ring with additional structure) "up to" something, and the way to make that precise is to divide out the ideal these somethings lie in...I don't have a physical explanation of what they are, and I don't know whether there is one.
 
That makes perfect sense to me.
 
Oh. Good :)
 
I guess the problem is then that I don't know where rings pop up in physics
 
3:30 AM
@ACuriousMind What does the phrase "up to" mean in mathematics? I hear it all the time. Like "up to isomorphism" or something like that. The only association I make with it is "up to no good" lol
 
@StanShunpike The prime example is "clock arithmetics", i.e. $\mathbb{Z}/n\mathbb{Z}$. Take some number n, for example 8, and say that two numbers are equal if their difference is a multiple of 8, that is $4 = 44$ "up to $8\mathbb{Z}$".
"up to" means stuff differs only in respects you don't care about, generally
@NeuroFuzzy Yeah, that's a bit difficult to see, and you can live a happy physicists' life without ever knowing what an ideal or a ring is, I think.
You can't live a happy mathematical physicists' life, though ;)
 
@ACuriousMind One could, but I can't. Because that would mean I'd fail my class. :P
I really like what I've seen of group theory so far.
 
Oh, well. You'll survve :)
 
@StanShunpike The precise topological definition of my earlier black hole definition is $B:=M-J^-(\mathcal{J}^+)$ where $J^-$ is the causal past and $\mathcal{J}^+$ is future null infinity.
 
Where is that from?
 
3:40 AM
Straumann.
 
lol that's it I'm getting it
 
@StanShunpike Nah, he doesn't explain what the hell that means.
I just new the page it was on :p
He just cites Hawking, Penrose and Wald I think.
This is equivalent to what Chris White said in words.
 
Oh, I get it.
That's really cool.
Wald has some sections on causality. are they good?
 
@StanShunpike Yes, but demanding.
 
@0celo7 He does put them in the advanced section.
I'm not surprised.
 
3:43 AM
I just had an interesting idea. Since there are points in our universe which appear to be moving away from us at a speed faster than that of light, does our universe satisfy the Newtonian definition of a black hole?
After all, to escape our universe in the Newtonian sense, we would have to go faster than light.
@ACuriousMind Does this sound sufficiently crackpot?
 
@0celo7 I'm not sure. Throw in a bit obscure notation, perhaps?
 
@ACuriousMind Wrap it up in a bow and drop it on 12262's doorstep?
He'll make sure everyone properly defines $c$ first, of course.
One must really understand $c$ before any progress can be made. Everyone is an idiot and obviously does not understand GR.
 
It's the number of meters per meter, of course.
 
Also, you have to define axiomatically the clock you are using.
Otherwise, we're just dumb engineers.
@NeuroFuzzy Go home, childish engineer.
 
3:53 AM
@0celo7 Hey! Some of my best friends are engineers! And, well... OK, they are a bit special.
 
@ACuriousMind I have this urge to \mathrm every post I see now.
 
@0celo7 lol isn't your whole family engineers?
 
@NeuroFuzzy As a future engineer, I have to clarify that I was mocking someone.
There's a troll on this site who wrote a whole page axiomatically defining what a clock is.
 
@0celo7 Just playing along. I think i read some of his posts before too.
 
He said the OP could not know what he was asking because he did not truly know what a clock is.
 
3:55 AM
Does he assume choice?
 
@NeuroFuzzy I have no idea.
 
that was the funniest thing I read all day
oh my god I'll have to use that some time
thanks ;D
 
Welcome.
I have to sleep now.
 
night!
 
@0celo7 Welcome to my world ;)
And good night
I should probably get to sleep, too.
Sun's almost coming up :P
 
3:58 AM
bye!
 
 
3 hours later…
6:52 AM
@ACuriousMind I just went back over the notes I took on tensor products. And then the definition of the wedge product you gave. Here's what I don't understand.
@ACuriousMind the tensor product was defined as $t_i,t_j$ as $t_{ij}(v_1,v_2) := (t_i\otimes t_j)(v_1,v_2) := t_i(v_1)\cdot t_j(v_2)$
But dot product is commutative
And the wedge product definition you gave was
$s\wedge t := s\otimes t - t\otimes s$
So if I let $a$ be the tensors and I write $s = a_i(v_1) \cdot a_j(v_2)$ and $t = a_k(v_3) \cdot a_l(v_4)$, then
Damn sorry let me rewrite that
What I'm trying to say is that
$s$ and $t$ are the tensors so we just write them as a dot product
So that would mean
$s\wedge t = s \cdot t - t \cdot s = 0$
@ACuriousMind there we go, I got it. What I just wrote. And that doesn't make any sense. I must be misunderstanding something conceptually
Forget the part involving $a$ that was a mistake.
 
user54412
7:17 AM
@StanShunpike I think you just need to keep track of all your products and what's acting on what.
 
user54412
start with $s \wedge t = s \otimes t - t \otimes s$
 
user54412
then $(s \wedge t)(u,v) = (s \otimes t)(u,v) - (t \otimes s)(u,v) = s(u) t(v) - t(u) s(v)$
 
user54412
In other words, there's no dot product between $s$ and $t$. Your "dot" product is just the multiplication in the base field, whatever it is. $s(u)$, $t(v)$, etc. are scalars in this field and so can be multiplied in this way.
 
@ChrisWhite but doesn't that still leave me with the result being zero?
 
7:40 AM
In other words, if they are multiplied in the base field, its still $st - ts = 0$
 
8:13 AM
@bolbteppa any thoughts on the above?
 
 
3 hours later…
11:24 AM
@ACuriousMind That he's just so nice to want to solve the problem. Not really a good 'joke'
@ACuriousMind Have you ever done sports at a semi-serious level?
@NeuroFuzzy Theere we go :)
@StanShunpike Rotate a cube, or your phone...
 
11:52 AM
@ACuriousMind I'm not sure one ever can ;)
@0celo7 Oh, you think $c$ is your ally, but you merely adopted $c$. I was born in it, molded by it. I didn’t see Newtonian mechanics until I was already a man.
@StanShunpike Dude, $s(u)\neq s(v)$, and the same holds for $t$
Why would it be zero?
Also hah caught up on chat
Mornings/early afternoon in EU is the perfect time for that---nice 'n' quiet.
 
12:56 PM
@Danu dude, (1) you said dude a lot and (2) dude I didn't see that the inside of the vectors dude flipped dude. So dude no wonder its not zero dude. Dude, I think that like dude that answers the question dude.
Dude where's my car?
 

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