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12:02 PM
Is causality defined in physics?
 
MAFIA: The + cayley table forms a left zero semigroup, which is why it is trivially associative
Since that is a semigroup, perhaps there exists a semigroup of idempotents for it
 
rob
Salutations, all
 
user116211
@rob Welcome, Sir rob.
 
user116211
@Secret You need to ping me ;P
 
It's not interesting enough yet thus I don't want to waste pings
 
user116211
12:12 PM
@Secret There should be.
 
I do not expect there are absorbers (although they are not ruled out when I construct the axioms), therefore if WIP is self consistent, it might be worthwhile to investigate further
Right now I need to check the distributive laws...
 
12:28 PM
1
Q: What is the origin of the term 'degeneracy'?

Leo AlekseyevI've grown accustomed to mentally translating 'degenerate' to 'having equivalent energy', but the origin of the term always puzzled me, particularly in light of the more traditional English usage. Etymology of the word suggests that it is formed from Latin prefix 'de-' and Latin root 'genus'. I...

Off-topic, cf. e.g. this meta post?
 
user116211
@Qmechanic Sounds off-topic.
 
it sounds more like a linguistic question (Etymology) than a physics question
 
user116211
yes.
 
user228700
@Secret: What is the way to identify exceptions to the octet rule (OK, this might be a really dumb question), lest I intriduce coordinate bonds and ruin everything?
 
@DHMO Obvious reasons? I won't inquire since it's your decision, but speaking personnally I can't see any benefit to hiding my own identify - though obviously you're not getting my e-mail address or telephone number :-)
 
12:36 PM
@JohnRennie Thanks for not inquiring
@Kaumudi Wrong chatroom?
which grade are you in? the answer depends on your grade sadly
 
user228700
@DHMO Nah, we discuss everything here :-P
 
user228700
@DHMO I graduated this spring. Not yet in university.
 
Warning, guideline only:
well, to start with, if you have an atom that can form more than 4 bonds, and a lot of halogen atoms surrounding it e.g. $PCl_5$ then it is going to violate octet.

Another example are electron deficient atoms like B, Be. These like to violate octet rule

Metal complexes violate octet like crazy because they obey another rule
 
@Kaumudi ok. in high school you were taught hypervalent molecules, which would not be hypervalent later in your life...
Sugden's hypothesis back in the 20s already explained how those molecules can follow octet
His hypothesis was confirmed in the 50s
 
user228700
@DHMO I may have been taught this (concept) but I've never before come across the term hypervalent molecules. Googling...
 
user228700
12:39 PM
@Secret: OK, thanks. I'm stuck with $ClO_4^-$ and I ruined everything by introducing coordinate bonds ugh :/
 
Also, noble gas havea full octet, thus noble gas compounds must violate octet to form things like $XeF_4$
Usually, it takes rather electronegative atoms to violate octet, but that is not a hard fixed rule
 
user228700
@Secret Hm, OK, this makes sense...
 
user228700
Thanks :-) $ClO_4^-$ is an exception, right..?
 
user228700
I did this: (Yeah, it's been established that I'm dumb)
 
user228700
 
12:44 PM
well, O is pretty electronegative to start with, and Cl only need one electron to compelte octet, thus you have to violate octet to accomodate the other O as what you drawn there
 
yo what is with these temperatures
we're going from 50 to 80 degrees
if I put on long sleves/pants, I'll be hot later
 
user228700
The electrons don't add up :/
 
but it's freezing ouside right now
 
@Kaumudi The bottom O and the left O are both charged... only one of them should be charged
 
user228700
@Secret Yeah :/ When should I introduce coordinate bonds..?
 
12:45 PM
the chlorine should possess 7 electrons
@Kaumudi you shouldn't introduce coordinate bonds when expanding octet
 
user228700
@DHMO What dyou mean by "when expanding octet"?
 
@Kaumudi you are trying to draw the structure of perchlorate by using expanded octet, right
 
@0celo7 don't worry, here our temperature dropped from 30 something C to 11 C at night while it should be spring. The weather is definitely screwed up
 
language
@JohnRennie will ban you
 
user228700
@0celo7 Are u kidding me? U use the f-word all the time!
 
12:48 PM
@Kaumudi of course you can explain its structure by using 3 coordinate bonds instead of an expanded octet
 
@Kaumudi I have a mental disorder, please don't make fun of it
 
but you are trying to apply the expanded-octet hypothesis, right
 
user116211
@Kaumudi He has brain tumour; have some sympathy :(
 
user228700
@DHMO Yeah, that's the thing, I dunno if it'll be an expanded octet beforehand. So I need to ascertain that before starting?
 
@Kaumudi yes.
in the expanded-octet hypothesis, period 2 elements cannot expand octet
 
12:49 PM
2 mins ago, by 0celo7
language
fixed
 
user228700
@MAFIA36790 WHAT?! I didn't know this wtf. @0celo7: That wasn't me making fun of u but dude, I'm so sorry!
 
@Secret yay
 
user228700
@0celo7: Hey, I'm sorry.
 
user116211
@balarka o/
 
'sup
 
user228700
12:51 PM
@DHMO Riight. OK, this is new information.
 
@Kaumudi therefore, nitrate is explained by dative bonds instead
 
user228700
@DHMO Huh? OK, I didn't know that I was supposed to decide whether or not I need to use expanded octet and what not before I start writing the structure!
 
@Kaumudi now you know
 
user228700
@DHMO Yeah, I do, but I'm still not very clear on the whole coordinate bonds thing. So, for elements that can't expand octet, we introduce coordinate bonds?
 
@Kaumudi yes
this is another failure of the expanded-octet hypothesis
i'm clearly biased by calling it a hypothesis
 
user228700
12:55 PM
@DHMO Omg I've been struggling with this my whole life! (:-P) Nobody told me this!
 
@Kaumudi so you just met a person who changed your life :-P
did we just use the same emotion
 
user228700
@DHMO Yeah, the people on this website change my life everyday :-)
 
user228700
@DHMO Yeah..? :-P
 
user228700
OK, thanks! Wow.
 
just think of it as a failed hypothesis: you can do anything with it
 
12:57 PM
MAFIA: Results on the distributive laws are computed: Due to the presence of the + left zero semigroup, the distributive laws (both right and left) trivially holds.

This is in contrast to my expectationo when I construt the axioms, so I will next check whether any of the previosu formulae have suddenly contradict itself due to me making WIP finite
 
user228700
@DHMO Yes, but I gotta use it to solve problems so. Anyway, thanks!
 
@Kaumudi you're welcome
 
If no contradictions found, then perhaps I can continue on the infinite version of WIP, which will be interesting because we can exploit those zero power theorems
 
@Kaumudi I thought you already finished high school
 
user228700
@DHMO I did, but entrance exams.
 
1:00 PM
@Kaumudi alright
 
user228700
I have another quick question, if u don't mind..?
 
ask
 
@Ocelo7 Nice profile picture
 
user228700
When there's a negative charge on a molecule, this negative charge is actually delocalized over the whole molecule, correct?
 
@Kaumudi yes, but in high school it either goes to the most electronegative atom or the center atom
is resonance taught in your high school?
 
user228700
1:03 PM
@DHMO It's funny how u say "in high school" :-P Okay, how to figure out which atom it goes to? (Since the central atom is supposed to be the least electronegative atom, no?)
 
The most electronegative atom tend to have the highest electron density, but at high school level, because we treat electrons classically, it has to be localised at an atom
 
@Kaumudi I thought you said you just finished high school
 
user228700
@DHMO Wut? No, I mean how u said "in high school it either goes to the most electronegative atom or the center atom"
 
@Kaumudi well, both are essentially different systems
the latter is called "formal charge"
@Kaumudi what is funny about that?
 
user228700
@Secret And which atom is that..?
 
1:06 PM
@Kaumudi it depends
ultimately it is another failed theory, so you can do whatever you want
 
user228700
@DHMO Never mind...can you explain that about the latter is formal charge?
 
@Kaumudi the latter system is called formal charge, it assumes that there is no dative bonds
 
usually the most electronegative atom, e.g. two of the O in $SO_4^2-$
 
in fact, it is very easy to convert dative bonds to the formal charge system
 
user228700
@DHMO Which system?
 
1:06 PM
@Kaumudi the formal charge system
 
user228700
@DHMO Yeah, OK...
 
let's take hydronium as an example
it is positively charged
can you draw/find a diagram according to what you learnt?
I just want to know what you learnt
 
user228700
@DHMO U mean its Lewis structure?
 
@Kaumudi yes
 
@Kaumudi do you know any other structure?!
 
user228700
1:09 PM
@Ramanujan How about the shape given by VSEPR theory?
 
@Kaumudi I thought VSEPR adds on the Lewis structure
Whatever, just draw anything
 
user228700
@DHMO Yeah, it does. I was just telling Rananajunan that, oh, never mind. I'll just draw the lewis structure. Drawing...
 
@DHMO vsepr shows repulsion of bonds right?
 
@Ramanujan yes
 
@AlfredCentauri Could I ask a SR question to u?
Is time a scalar or a vector?
 
1:11 PM
@SwapnilDas time is a coordinate
 
Yup, above
Not a quantity?
 
user228700
@DHMO: Brb, sorry...
 
@AlfredCentauri I visualize time as a component of a 4D vector.
 
@SwapnilDas that's correct. Proper time, by the way, is a (Lorentz) scalar.
 
@AlfredCentauri Thanks! Idk what a Lorentz scalar is
 
1:17 PM
@SwapnilDas A Lorentz scalar is invariant under a Lorentz transformation
 
@AlfredCentauri oh, learnt something new. Thanks a lot!
 
@SwapnilDas you're very welcome - gotta go do some GR now...
 
Best of luck! Bye! @AlfredCentauri
 
user116211
@SwapnilDas Time-interval is a vector.
 
MAFIA: Ok all results are computed. It seems we have divided by zero in this system, although in a sense this system is quite trivial. (Graphic abstract in a few sec...)
 
user116211
1:24 PM
All trivial?
 
user228700
I'm back. Sorry about that :/
 
user228700
@DHMO:
 
trivial as in, there are many things that is basically the same blockwise, not that it becomes the trivial structure
 
user116211
okay.
 
You will see it when I present it shortly...
 
1:26 PM
@DHMO also 9 elections.
 
user228700
Hang on, I think I messed up...(Forgive me, I've always sucked at Lewis dot diagrams)
 
user228700
This better:(?)
 
user228700
 
and the charge?
 
user228700
Yeah, I'm not so sure about that :/ Can u help?
 
1:32 PM
@Kaumudi so as I said, there are two systems...
 
user228700
OK..?
 
so I'm not sure which system you want
 
user228700
@DHMO What do u mean..?
 
@Kaumudi I think I should mention both systems then.
the "formal charge" should be easier, so I'll start with that.
it's all about which atom you assign which electron to.
In the formal charge system, the electrons in each bond is shared fairly between the atoms.
 
@MAFIA36790 Oh
 
1:35 PM
could you work out how many electrons belong to each of the four atoms respectively then?
 
user228700
@DHMO Yes. Differences in electronegativity are not taken into account, correct?
 
@Kaumudi correct
 
user228700
@DHMO Uh, u mean work out the formal charge of each atom?
 
@Kaumudi yes
 
user228700
Alright...
 
user228700
1:37 PM
It looks like the F.C of all the H atoms is 0 and that of the O atom is actually +2 :/
 
could you show the step for the O atom?
@Ramanujan very funny
 
@MAFIA36790 Let me know if you find anything contradictory
 
user228700
Well, it has 6 electrons in its isolated state, it's bonded to 4 atoms, all single bonds and it has no lone pairs (going by my diagram, which I don't even know is correct or not), so F.C=6-4-0=+2?
 
user116211
Subsemigroup to what structure?
 
@Kaumudi hint: it is not bonded to 4 atoms
and your diagram is correct
 
user116211
1:40 PM
I mean to what set did you restrict the operation?
 
user228700
@DHMO Wait wait, it does have a lone pair! (According to my diagram) so 6-4-2=0?
 
@Kaumudi did you remembered my words (about formal charge?)
 
@Kaumudi hint: it is not bonded to 4 atoms
 
user228700
@DHMO This. How?
 
@Kaumudi well, for example, it is only bonded to 3 hydrogen atoms
 
user228700
1:41 PM
@DHMO AAARGH, I'm blind!!!
 
@MAFIA36790 The + cayley table is isomorphic to the left zero semigroup of 3 elements, and the distributive law is controlled by it
 
user116211
okay.
 
In mathematics, a null semigroup (also called a zero semigroup) is a semigroup with an absorbing element, called zero, in which the product of any two elements is zero. If every element of the semigroup is a left zero then the semigroup is called a left zero semigroup; a right zero semigroup is defined analogously. According to Clifford and Preston, "In spite of their triviality, these semigroups arise naturally in a number of investigations." == Null semigroup == Let S be a semigroup with zero element 0. Then S is called a null semigroup if for all x and y in S we have xy = 0. === Cayl...
 
user228700
I'm so sorry. There's something wrong with me. Hang on. 6-3-2=1?
 
@Kaumudi correct
 
user116211
1:42 PM
@Secret Yeh, got that.
 
therefore, the formal charge is on the oxygen
 
user228700
@DHMO Which means that...it has one less electron than it should have..?
 
@Kaumudi yes
 
@DHMO isn't O have 0 f.c?
 
@Ramanujan not in hydronium cation
 
user228700
1:43 PM
Alright. So what was ur original point..?
 
@Kaumudi what was your original question..?
 
@DHMO this O has 8 e
 
user228700
@DHMO No, u asked me to draw this at some point...why?
 
@Kaumudi to answer your original question...
@Ramanujan so?
 
user228700
@DHMO My original question was about how to identify which atom has the charge. U said there are two systems...
 
1:45 PM
@Kaumudi yes. so you already learnt the formal charge system, which does not account for the electronegativity difference
 
user228700
@DHMO Right.
 
so should we continue to discuss the system which accounts?
 
@DHMO so it is stable ( without any extra or less elections)
 
user228700
@DHMO Uh, yeah, if that's OK..?
 
@Ramanujan yes, but that is not what formal charge means
@Kaumudi now, which atom should the positive charge be at, according to electronegativity?
 
1:47 PM
@DHMO then what does it mean ( I still thinks in that way) can you improve me
 
user228700
@DHMO Definitely not the oxygen atom, so one of the H atoms.
 
@Kaumudi why "definitely not the oxygen atom"?
 
DHMO then what does it mean ( I still thinks in that way) can you improve me
 
@Ramanujan well, to calculate the formal charge, first you assign each electron to a certain atom according to some rules
 
user228700
@DHMO Well, because it's one of the most electronegative atoms in the periodic table and so, it won't be stable, having a positive charge..?
 
1:48 PM
@Kaumudi yes, but except when bonded with fluorine
now, refer back to your photo
 
user228700
Alright, maybe we need to move this to the CSE chat lest @MAFIA36790 gets pissed again :-P
 
let's say you choose the top hydrogen atom
 
user228700
@DHMO Hm, yes, which is the only element that is more electronegative than oxygen so that makes sense.
 
user228700
@DHMO CSE chatroom?
 
@DHMO can I know rules
 
1:50 PM
@Kaumudi sure
@Ramanujan the rules is that the electrons in each bond are split equally between the two atoms
 
2:11 PM
0
Q: Why isn't my simple electromagnetic coil working?

SoRcE FlarpFor a science fair project I am building a coilgun and so far it has been pretty easy but now I have run into an issue. I have 2 layers of coil wrapped around a straw and when I attach power to the coil and try and put something magnetic inside it it doesn't attract it at all. If you need picture...

Unclear what you're asking, particularly in light of the comment that OP "somehow fixed it by rewinding the coil"?
 
@SwapnilDas yes
 
Anyone good with optics here? @ACuriousMind @ManishEarth?
 
user116211
@AaronAbraham You don't have to ping anyone; just ask. If someone is interested, he/she will respond.
 
Ah well, alright. What about you?
 
user228700
@0celo7: Just. Cannot. Stop:
 
2:20 PM
Huh?
 
You were lamenting about optics the other day (at Chem.SE)
@Kaumudi That sounds....*wrong*....
very
;P
 
It sounds only very slightly wrong by the average wrongness in this room :)
 
user228700
 
user228700
-__- Omg guys. Calm down! ^This is what I meant! @0celo7
 
By the way, @ACuriousMind do you do optics by any chance?
 
2:22 PM
@ACuriousMind I'm at the rocket factory
 
@AaronAbraham I already read your message the first time. What @MAFIA36790 said - unless you know a specific person is uniquely qualified to answer your question, just ask it without pinging anyone - if someone knows the answer and wants to respond, they will.
 
@0celo7 Rolling out V2s ;)
 
I'm getting the hang of it
Running out of ammo though
No one has 5.56
 
@ACuriousMind Will do :)
 
@0celo7 Oh, In FNV. For a moment I thought they would be already letting you construct nuclear rockets :D
 
2:23 PM
That was last semester
 
Would anyone happen to know what causes foam to appear white? Is it total internal ref. or scattering?
 
I was temped to kill the legionnaires at Novak
I got a fr
 
I'm tempted to kill you ;)
 
a grenade launcher
It's my secret weapon for such situation
 
Psst. What game @0celo7 ?
 
2:25 PM
Fallout Casino
 
@AaronAbraham Scattering mostly, as answered here
 
But in that case, why isn't soap water white?
THIS was my doubt...
 
Don't doubt. Question.
 
@0celo7 Oh...sorry, WW2 FPS is my thing...(hence the V2 comment)
 
Because the scattering is at the many surfaces that are in the bubbles. The water has just a single surface where scattering might happen, soapy or not.
 
2:27 PM
@MAFIA36790

More results: The nonassociativity has a special form
\begin{align}
(00)0=1 \neq 0(00)=0\\
(0q)0=q \neq 0(q0)=0\\
(00)q=0 \neq 0(0q)=q\\
(q0)q=q \neq q(0q)=1\\
(0q)q=0 \neq 0(qq)=q\\
(qq)q=0 \neq q(qq)=1\\
\end{align}
These can be wrapped up by the following:
\begin{align}
a(bc)=b^{-1}((ab)c)
\end{align}
 
@ACuriousMind But wouldn't scattering also occur through the bulk of the material?
The same way the sky scatters blue light?
I was under the impression, that scattering was a surface phenomenon.
 
Yes, but much less. The "scattering" here is mostly that the light is partly reflected at every surface, I guess
 
?
Sorry, kinda lost you there..
 
@ACuriousMind Are there any quests you recommend I get?
Can I join the legion?
I want to play skyrim again
 
(P.S @0celo7, Sorry, but I'll be borrowing @ACuriousMind for a bit)
._.
 
2:29 PM
She's mine @AaronAbraham
3
 
No surface is 100% translucent. At each surface of the bubbles, a part of the light is reflected, effective scattering it in random directions after many such events.
 
@0celo7 Most are good
@0celo7 Yes
The main quest choice will be between the NCR, the legion and "I'm doing this on my own", essentially
 
Can I be the boss of he legion
@ACuriousMind does loot scale with difficulty?
 
@0celo7 You can be the boss of my 3rd Panzer unit! ....if you want......
._.
 
2:34 PM
@0celo7 I'll say...yes, but any elaboration might be spoilery
 
( twirls fingers shyly )
 
@0celo7 I don't think so
 
@ACuriousMind remember when you didn't want to spoil saren
@ACuriousMind Damn. I'm addicted.
How did this happen?
It's like that time I said JUST ONE crack pipe
@ACuriousMind How do I know when there are no more quests?
 
2:49 PM
@ACuriousMind Follow up question?
 
@0celo7 Uh. I don't think I ever reached that point
 
Hi docscience :-)
 
Hello
 
@ACuriousMind What causes the scattering at the surfaces?
 
if anyone's interested i suggested docscience ask here about his question:
1
Q: What actually constitutes a 'rip' in the fabric of spacetime?

docscienceScience fiction writers often use the term 'rip' in the fabric of spacetime to provide a means of either time travel or instantaneous travel from one corner of the universe to another. But I've also read and heard bonafide physicists use this term as well, so I suspect there is some credence to i...

 
2:53 PM
Thanks John
 
@docscience: in GR spacetime is a manifold plus a metric.
The manifold dictates the dimensionality and the topology while the metric gives a way of measuring distance.
 
... and I think you pointed out the metric is an input, not an output to the field equations
... so how can one go about ripping it?
 
The manifold topology is an input.
The metric is what you get by solving the Einstein equation.
 
so it is an output?
 
That is we assume a topology, then take the matter/energy distribution and the result is the metric.
 
2:56 PM
... or can be?
 
But a rip would be a change in the topology of the manifold not the metric
 
... which is an input
 
Yes. We assume a manifold right at the start.
And the manifold we normally assume is $\mathcal R^4$ i.e. simply connected 4D spacetime.
 
so then 'ripping' would have to consider another set of mathematics, influences OUTSIDE the field equations?
 
Yes. There is nothing in GR that describes topology change.
You mentioned the quantum foam idea i.e. the idea that at very small distances spacetime is in flux.
And that's a common gee whizz idea on pop sci accounts of cosmology.
 
2:59 PM
Yes, one of the physicists that used the term rip was reffering to very small scales
 

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