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9:14 AM
@Slereah Does GR have ties with condensed matter physics?
a tough question
As far as I know, there currently doesn't exist any model for matter other than perfect fluids in GR
Ah Ic.
@Slereah So did only Einstein work on gr, or have there been any other influential scientists?
@TheStackExchange in second equation what is frame of reference
9:24 AM
Plenty of scientists in GR
@koolman Man
@TheStackExchange man is at rest
9:47 AM
[Division by zero] Caption: The proofs are finally becoming a bit more streamlined. Can we optimise them so we can generate 5 associative structures per 10 mins?
Treating these operations as cosets do really help in speeding things up by exploiting the properties of arrays
why does 0*0 = 1?
Division by zero (Case: involutive 0)
We just set those black numbers, and the rest is controlled by the associative laws
So basically, the scope is: Given an arbitrary finite algebraic structure with some axioms and special elements, what happens when associativity is imposed into the structure and what are the examples (if any)
too hard
4 me
10:09 AM
5 me
10:24 AM
@koolman Man is moving....
See the linked question
Ok, Arrival is not really that good after all...
@Secret What is that? I have a hard time parsing it in this format. Does this have a name?
I don't think it has a name. The workings here is basically applying Light's associativity algorithm by condensing a lot of writing using a similar idea to cosets such that given a set an some element x, xS is the set of all elements in the structure multiplied on the left by x
@Secret Arbitrary structure includes groups, rings, algebras, etc. ?
@G.Bergeron have you read Ted Chiang's short story that it's based on?
10:32 AM
Do you want to?
yes. For the example here, they are any algebraic structure that has a distributive law and two binary operators. Wolfram has a term for such general structures: ringnoids
Oh well, let me know when that maybe firms up into a decision.
10:33 AM
@JohnRennie: I'm sorry! Chemistry took awhile. May we resume our discussion?
So they are generalisation of rings in that you throw away everything except distributivity and associativity
well I don't know what it is at all
Now I'm curious so yes!
@Kaumudi.H Yes. Shall I dig out my picture and repost it?
Some guys in the maths chat said I am basically working on some form of universal algebra
Yes, thanks! ^.^
10:34 AM
where you start with a set, and then impose axioms and see what happens
yes that is how abstract algebra works in general, I was curious to get a sense of what it was exactly
@Secret Is one of the two operation abelian?
Well, for the bit I am currently working on (our of hobby) it is a kind of division by zero algebra using binary operators, which I heavily suspect almost all interesting cases are nonassociative, and throughout the past months I have been trying to formulate proofs to show that.

Nope, I did not assume anything for the binary operators, only closure
@JohnRennie Thanks
10:38 AM
@Kaumudi.H Aha.
Impatient as I am :-)
This is supposed to show the manometer being used in those pictures you posted.
So it's a U tube containing some liquid and the difference in the liquid levels is $h$.
If no pressure is being applied to either end of the tube then the liquid levels should be the same i.e. $h = 0$. Yes?
The above picture you saw is me trying to construct an associative algebraic system that allow the zero element to have a multiplicative inverse. Using arrays to write out Light's associativity test, one can deduce the remaining elements in the multiplicative structure as given by the * cayley table
10:40 AM
Hang on, I will tell u what I understand and then u can tell me if I'm correct..?
@JohnRennie Oh I thought the algebra thing of Secret was based on a short story... O_o
@G.Bergeron :: facepalm :: :-)
After the * cayley table is done, the addition structure can be done easily by imposing distributive law.
Secret's proposed division algebra is actually based on a horror story :-)
10:42 AM
Alright, yes, the liquid should be at the same level if the pressure exertred is the same on both sides. However, when we suck, we are essentially trying to decrease the pressure in our lungs so that the liquid will flow into our...mouth.
Oh, hang on, one second.
@Secret So two binary operations which are both associative and one distributes on the other, right?
@Kaumudi.H yes, you're correct ...
One notable thing about these "zero terms" (products of the zero element that does not equal to zero) algebras is that the + structure is heavily controlled by the zero terms, in that it controls which element dominates/absorbs another with only specifying what additive identity you have in the structure if associativity is to be obeyed

"two binary operations which are both associative and one distributes on the other" yes. So basically an associative ringnoid
@G.Bergeron have you got/do you want that story? Can I delete it from my server now?
10:45 AM
@JohnRennie Done
This is what I'm doing on the side, u see:
Revising while watching TV?
No, not watching T.V, but waiting for the ads.
Oh, for the comedy channel ad ...
Yeah. Still nothing -__- So once every 10 mins or so, I might disappear for 40 seconds, I'm sorry...
10:47 AM
Anyhoo, where did we get to?
So, yes, we try to decrease the pressure in our mouth/lungs so that the liquid will flow into it.
Basically yes. We suck on one end of the tube and as a result the pressure in that end decreases and the liquid rises.
Suppose the hardest we can suck produces a pressure change of $\Delta P$, then the task is to find out how far this makes the liquid rise i.e. what is the height difference $h$.
Right. And when we measure the height difference and plug it into our formula, we are measuring the difference b/w atmospheric pressure and the pressure inside our lungs, yeah..?
$$\Delta P = \rho g h $$
So, that's as far down as we can go while managing to breathe because any farther than that, the pressure inside becomes greater than the pressure outside, yeah?
10:51 AM
Well the liquid in the tube may not be water. From the pictures it looks like mercury.
@Secret Beware of the different usage of the term "oidification". It used to mean degenerate structures, but now it has a more precise meaning in terms of category theory ncatlab.org/nlab/show/Ab-enriched+category
I think it's Cranberry juice :-P
The point is it's measuring the maximum $\Delta P$ our lungs can produce, and this is then equal to $\rho_{water} g h_{max}$ where $h_{max}$ is the maximum depth at which we can suck air through a snorkel.
Yes, okay. That makes sense.
I have a follow-up question.
10:53 AM
You might also want to look at near-ring en.wikipedia.org/wiki/Near-ring
The max. difference b/w the pressure outside and the pressure inside can be like, 0.1-0.2 atm, right?
I don't know offhand, but it's certainly a lot less than 1 atmosphere
Okay, maybe more but there's no way it can be 0.5 atm, yeah?
@G.Bergeron O wow, did not aware that rignoids have a specific meanign in category theory (well I have not read much in category theory yet) By the way, This wolfram link is where I get the notion of rignoid I am familiar with
It's probably Googlable ...
10:55 AM
@Secret I know, it's the older usage
like groupoid that used to refer to magma
The algebraic structure I am working on is weaker than near semirings since they are basically near semirings with annhilation axiom thrown away to allow division by zero
> "Trying to inspire from a very long snorkel may lead to severe cardiac dilatation and failure, as happened to Stigler, who unsuccessfully tried to breathe through a tube at 2-2.5 m"
@Secret Beware, category theory will make you lose your sanity. But it's really cool!
@Kaumudi.H Eek!!
Yeah, definitely b/w 0.1 and 0.2 atm, then since 10m= 1 atm, roughly.
10:58 AM
@Kaumudi.H Yeah I once dived with almost empty tanks... until I started to not be able to suck the air out -_- Its not a snorkel but I'm pretty sure it's the same feeling
@Kaumudi.H Anyway, what were you leading up to asking?
So, okay. Say I have a mug full of...some type of juice on my kitchen floor and I climb onto the loft above my kitchen and there's some type of hose coming out of the glass and into my mouth. And say the loft is around 5 m above my floor.
@G.Bergeron Oof :-/
Dyou think I'd be able to drink the juice?
@Kaumudi.H no
@Kaumudi.H It's the inverted beer funnel
@JohnRennie: Right. But look:
11:02 AM
Backstory of how I start this personal research: Having being taught for so many years that you cannot divide by zero. I then start to wonder whether it is contradictory in all conceivable algebraic structures. I then came across Wheels but even theirs don't allow an inverse of zero. After that, I started to jokingly formulate a conjecture that no nontrivial structures with a multiplicative inverse of zero can exists for all conceivable algebraic structure.
A few weeks later after formulating that conjecture, I started to find some counterexamples of them. Shortly after, with the help of my
@Secret Near-rings don't have it
^ He's about 5 meters above the floor of the class. And he's drinking the juice.
@Kaumudi.H What's happening outside of the frame?
11:04 AM
@Kaumudi.H OK, I haven't seen the experiment, but I have two suggestions:
near rings have + structure forming a group. This is not necessary the case for the strcutures I am constructing as they are in general semigroups under + (and sometimes under * also)
1. He may be using his tongue and cheek muscles to pull in the liquid so the lungs aren't involved. I've never measured it, but I suspect the muscles in the mouth can produce quite a large pressure difference.
@Secret A lot of examples exists. Just take a modular ring $\mathbb{ Z}/n\mathbb{ Z}$ for $n$ not a prime.
Yeah, that makes sense. But he was using his tongue and cheek muscles to suck before as well...
2. with a narrow tube the flow is dominated by the viscosity of the liquid i.e. if you stopped sucking the fluid would fall only slowly.
11:07 AM
@Secret Nope, just a monoid
Riight. And then he starts pulling again...
@Kaumudi.H the volume you can suck in just using your mouth is small. It's OK for pulling in liquids, but you couldn't pull in enough air to breathe that way.
Those have zero divisors, Zero divisors are not inverses of zero. They are nonzero a,b, such that ab=0. They are basically a generalisation of the concept of not allowing division by zero. For example in $\mathbb{Z}/4\mathbb{Z}$ $\frac{2}{2}$ is undefined since $2\cdot 2=0$
I don't think it's a safe assumption that the prof was only using his mouth in the manometer experiment.
@JohnRennie Hm?
11:08 AM
@Secret Ah no ok, they have an identity
Nvm, OK, I understand what u're trying to say.
Yes, that makes sense.
I suppose in principle you could snorkel lower than a metre by pulling in air with your mouth muscles then inhaling that air. But I doubt you could do it fast enough to breathe that way.
Thanks very much _/\ _ Dyou have to go now or dyou have the time for one more question? (I'm really sorry :-/)
@Secret Ok NOW I see what you want. But a question: Isn't removing the annihilation axiom kind of killing what we mean by zero?
More questions are fine. I've finished work and right now I'm killing time before going out to get lunch.
11:10 AM
Okay. This:
I'm hungry!
An there's a post about butter chicken pizza on my right
@G.Bergeron yesterday's lunch :-)
@JohnRennie That's lunch?
11:11 AM
@G.Bergeron Not necessary, a zero element can be either an additive identity, or an absorbing element, or both. If you are working with lattices, a zero element can be said to be the minimal element of the lattice (though I usually don't consider that a zero element)
Four 6oz cheeseburgers with hash browns.
Also the pierogis picture was mmmhh! I haven't eaten those since the last time I was in Poland
killing annihilation (or restricting them to be on sided) does not remove the notion of zero as long there is an additive identity
@JohnRennie: Dyou have a vengeance towards vegetables?
Pierogi aren't that well known in the UK, and they are delicious. Well, they can be delicious.
11:12 AM
@Kaumudi.H Friday's lunch was vegetarian.
@JohnRennie With much butter and fried onions
yesterday, by John Rennie
I've just eaten this:
@G.Bergeron Oooooh, food porn! :-)
@Kaumudi.H yes the pressure is the same according to Pascal's law.
11:14 AM
I really don't get that ^
@Kaumudi.H Think about the gravitational potential
For example, one "trivial" way to divide by zero (element) without being the trivial ring/group/semiring etc. is as follows:
@Kaumudi.H it's a very common conceptual problem, and related questions have been asked many times on the site.
@DanielSank if you want tomorrow ask me about the drunk walk problem, i think i made another method of solving it but im sleepy and lazy so im not gonna work it out tonight
Here, 0 is the two sided additive identity, and has a right inverse of 1
11:16 AM
@JohnRennie Oh..? Okay, what should I be searching for?
@Kaumudi.H I'm honestly not sure what the most intuitive way to explain this is. Maybe it's best to just accept that it's the case.
@Kaumudi.H let me have a look ...
@Kaumudi.H Gravitational potential depends only on the depth
Since it is static, and and the potential is not at a minimum, something else is at play
@Kaumudi.H Why the panic? There are lots of things in physics that we initially have to take on trust and we only fully understand later as we've got more knowledgeable.
11:18 AM
So you expect the pressure to depend only on the depth
Anyway, let me search ...
@JohnRennie :'-( Alright. This makes solving problems a lot harder.
Since pressure is what keeps the water from "falling on itself" under gravity
@JohnRennie Agreed
@G.Bergeron I'm not sure I understand...
Q: Static fluid pressure does *not* depend on the shape, mass, or surface area of the liquid. Huh?

roroI came across this figure that I'm currently unable to accept. I've come up with a thought experiment. Please point out my error: Assume zero atmospheric pressure and no boundary effects of the containers. We are measuring the pressure on the green ball. The red line represents a movable bar...

11:20 AM
@Kaumudi.H What could make an small cube of water move?
Since it doesn't the sum of the forces on the cube is 0
Q: Hydrostatic pressure in a vase at different depths

xeonI am building a water level sensing device on the base of Arduino. The sensor is the eTape by Milontech. The resistance changes almost linearly with depth as the sensor is immersed in liquid (datasheet). The resistance gradient is 150 $\Omega$/inch. The working principle of the sensor is the hyd...

One of the force is given by gravity, a conservative force expressible as a potential
Hmm, these don't have great answers ...
that depends only on depth
11:22 AM
Maybe I'll have a go at explaining myself. Let me draw a picture:
Since the cube of water does not sit at a local minimum of the gravitational potential, something else must retain it
which is the pressure
@G.Bergeron This is a tad difficult for me to understand...
which give a force on the cube equal to the $\Delta$ of the pressure below and above
What's your background?
Graduated high school last spring.
ah ok
I found the same statement a tad weird at that time
11:25 AM
@Kaumudi.H are you happy that $P_a = P_b$?
Anyway, too hungry and John worked me up with the pierogi picture... I'm gonna go get inspired by my refrigerator
What? No, I'm not! Isn't the whole point that I'm not happy with that..?
OK, well pressure is force divided by area. Yes?
And the force is due to the weight of the water. Yes?
11:27 AM
Okay, I know what u're leading up to.
So the pressure is the weight of the water divided by the area of the vessel it's in. Yes?
That while the weight of the column on the right is much larger, that the area is bigger too, yeah?
@Kaumudi.H Correct! :-)
Shall I do the calculation to show this?
What calculation..? :-) I think I get it...
Well if the height of the liquid is $h$ and the area of the vessel is $A$ then the volume of the liquid is $hA$, the mass is therefore $\rho h A$ and the force is $\rho g h A$.
11:30 AM
Hang on tho...
You're about to ask what happens if the vessel is a funny shape?
For the point on the right, we're considering the volume of which column of liquid? The whole thing, right?
I have a Klein bottle shaped mug
Is it a funny shape
11:33 AM
^ Yeah.
@Slereah Pics?
@Kaumudi.H OK, let me draw another diagram ...
@JohnRennie No, no!
Too late :-)
11:34 AM
The question was: "For the point on the right, we're considering the volume of which column of liquid? The whole thing, right?"
:-P U're fast.
nice mug you have there
@Kaumudi.H So far so good?
:-P It's only a diagram. What could be the problem?
OK so $P_a$ is as before the weight of the liquid above $a$ divided by the area of the bit of the pipework above $a$. So $P_a = \rho g h$ as before. Yes?
11:39 AM
And it's not obvious why/how this applies to the pressure $P_b$ at the right side of the pipework.
Uh huh...
But liquid is, well, liquid and that means it flows. Specifically it means that it will flow along a pressure gradient. So if $P_b$ was different to $P_a$ that means water would flow from $a$ to $b$ until the pressures became equal.
Yes. So by that argument, the pressure has to be the same at the same horizontal level...
That does make sense.
11:44 AM
It's far from obvious. Like I said there have been lots of questions about it (though I haven't been great at finding good examples)
Alright, I'll shut up and accept it. I guess I just have trouble figuring out at which points in a system the pressure is the same...esp. when multiple fluids are involved...
In fact I bet you still don't really believe it :-)
Any, um, guidelines?
If your system is in equilbrium, i.e. nothing is flowing, then the pressure at the same horizontal level must be the same.
But you need to be careful to include all sources of pressure.
11:46 AM
For example in our U tube experiment the pressure is the same at the same horizontal level even though the water colum is higher at one side.
And that's because we're sucking on one side. So the air pressure pressing down on the water is lower on the side we are sucking on.
I found the song!!!
@JohnRennie: Sorry :-P One sec...
Ah, OK, I wouldn't have known that. One Republic doesn't feature highly in my list of all time great bands.
@Kaumudi.H Hence my explanation with the pressure as a potential being proportional to the gravitational potential which varies only with height
11:50 AM
@JohnRennie Riight...
Shall I do the calculation for the U tube to show this, or is it obvious?
@JohnRennie No, I don't listen to One Republic either. This one is kinda amazing...now :-)
@JohnRennie Hang on, I'm thinking about this...
@Kaumudi.H At least we no longer have the distraction of the comedy channel :-)
:-P True, but it was on mute anyway.
:-/ Still thinking...
11:56 AM
@JohnRennie You're kidding, right?
Could u explain..? One last time :-|
Well £0.99 is obviously just the starting price, but I bet they sell for under £100. Not that I want one ... much :-)
@Kaumudi.H Explain the U tube expt?
@JohnRennie Yeah, sure, but my "You're kidding, right?" was basically: "Seriously, didn't u just buy another one?!?! :-o"
No, I won't be buying more laptops for a bit.
@JohnRennie Yeah. I think I need a little push...
@JohnRennie :-P Okay, I will believe u.
There is a specific model I want, an Inspiron 7720 SE, but there haven't been any on ebay for a while or at least not at a sensible price.
11:59 AM
Oh, OK...
For clarification: u just want these for...what?
Anyway, back to tubes - let me draw [drum roll] ... A DIAGRAM!!
:-P Sure...
In this diagram we are sucking on the left end of the tube so the liquid has risen there.
lol suck
And my claim is that the pressure at the horizontal dashed line is the same on both sides. Is that OK as a starting point?
Actually this looks good too though at £135 it's a bit expensive.
12:06 PM
Is that Right ... that's a good starting point ?
Yes :-)
Aren't we trying to prove that tho?
Well what's the pressure on the right hand side at the horizontal line?
Atmospheric pressure.
OK, and what's the pressure at the horizontal line at the left hand side?
12:11 PM
What am I saying?
One second.
Oh the correct answer's just gone!
Yeah, it's $P_{suck} + \rho gh$
But I'm only saying that 'cause $P_{suck}$ is smaller than $P_atm$.
@Kaumudi.H Correct.
$P_{suck} < P_{atm}$ because, well, we're sucking.
(Insert joke here)
12:13 PM
Yeah x'D Okay, so there's the calculation.
So we can calculate $h$ because we know the pressures are the same at the same horizontal level: $$P_{suck} + \rho gh = P_{atm}$$
Alright, okay. Now I'm gonna leave u alone to have lunch peacefully. Quick question tho. Do u make lunch or buy from outside?
@Kaumudi.H I rarely buy completely premade meals, but I also frequently don't start completely from scratch.
I see. OK, have a nice lunch! :-) [It's still in B'lore BTW]
@Kaumudi.H I've been trying, and failing, to resist the temptation to check every few minutes :-(
I guess they don't work on Sundays. Hopefully they'll fly it to Chennai overnight.
12:18 PM
:-( Yeah. I decided that nothing will happen till tomorrow.
Yeah, hopefully. Just one more day. I can do this :-P
(They're supposed to have delivered in the next 24 hrs + 12 mins)
Yello :-)
Well like I say I'd guess it will be on a plane overnight. There's still hope.
I'm not so sure about that...
Will it be the end of the world if it arrives on Tuesday? (Silly question, of course it will! :-)
12:20 PM
Like I said, it's only like a 45-minute flight so I see no reason they shouldn't send it tomorrow morning or s'thing...
@JohnRennie When will you be free after lunch ? :)....I needed to discuss some things about this physics.stackexchange.com/questions/298005/… question with you :P It has still received no answers
@JohnRennie Yeah, what kind of a question is that? Of course it will!
Hey, @TheStackExchange: I found the song, u sadist!
Anyway, @JohnRennie: Thanks so so much! :-D For the question and the laptop and everything.
Son of a goat! I've just been downvoted!
@Kaumudi.H Ow..that's bad news...for a sadist :P
But I am feeling better now :P...good song actually!
12:23 PM
@JohnRennie :-O What?! That's a first(?)
@TheStackExchange Bhag ja, before I virtually punch u in the face -__-
@Kaumudi.H no, I get a fair few downvotes. We aren't allowed to speculate about who downvotes because it's anonymous, but I KNOW WHO IT IS :-)
@Kaumudi.H ao ao pakarke toh dikhao :-D
it wasn't me
@JohnRennie :-P I see...
But my head would'nt explode without that song
12:24 PM
@TheStackExchange What's pakarke?
@Kaumudi.H use a "d"
@TheStackExchange It was just, in that moment. Like I said, I'd heard it before.
instead of "r"
12:25 PM
@TheStackExchange Right.
@Kaumudi.H -______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ -
Someone is definitely going to move that into the trash :-zp
I'm home alone with a lot of food....yay!
Anyhoo. Gots to go eat s'thing or my stomach will explode (:-P)
12:27 PM
@TheStackExchange Bhag ja x-(
@TheStackExchange will you come at jee launch pad
@Kaumudi.H KK...get lost :-D
@koolman coming!
@JohnRennie I've a new display picture how do I refresh it here?
12:31 PM
@Mew Wait a few minutes
It will load soon
Within an hour at max
I dont want to wait an hour bro
and how do you know it will be changed within an hour?
@Mew You have to catch a train ?
Or a rocket to Mars?
12:33 PM
I jsut dont' want to see more of Trump
Do you want to see an hour more of Trump?
@Mew Close your eyes, baby!
WTF why ar eyou calling me baby?
@Mew Yeah, he is a great comedian!
12:34 PM
yeah well I want to see Putin
@Mew You are behaving like one...LOL XD...my younger brother is always in a hurry...so it was an analogy
And I am sure that it will load within an hour
Since the script runs every hour
It takes time for the cache to be cleared
12:36 PM
You can request someone with system level access though, to clear the cache immediately
@Mew When did Putin join comedy ?
@JohnRennie who has system levcel access?

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