The h Bar

enumaris

05:44

stupid model won't run...

Secret

06:03

Last night dream has many weird things:

1. the Camel programming language has similar syntax as Python except some awkward $ <comand phrases> : structure. It has the simplicity one liner feature of mathematica's language, object oriented property of Java and other high level languages and its interpreter can be easily embedded as an add-on in browsers and run with minimal time and memory. It's highly objective oriented nature means you can have type variables that are objects from a cloud or even portions of the cloud itself, including basic scripts and functions, making it able to utilise huge cloud computing resources on the f…

2. There is a weird substance that violates the third law of thermodynamics. It has the same temperature and most behaviour as liquid nitrogen, but when it hits the floor, it sometimes turned into a gel like phase

The reason why it violates the third law of thermodynamics is because it cools much faster than the third law will allow for objects of that temperature. For all normal objects such as a hand, a frostbite by a cryogen will take some time (At least a few seconds) to reach the cells at the centre and killing them by hypothermia.

But for this weird "liquid nitrogen", even splashing a little bit of that on the surface of the hand, once the lendenforst effect is gone, the entire bulk of the hand rapidly cools down in milliseconds to the temperature that is between the -197C and the temperature of the hand, meaning that once it successfully wet, your entire hand will behave like a perfect heat conductor and all your cells in the hand dies, and you will need to amputate it

Secret

06:34

Extrapolating, it means either such weird substance can dip beneath absolute zero (because it's rare of cooling does not slow down approaching it), or its cooling curve have cusps, result in abrupt jumps in cooling rate as it approaches absolute zero but never reaching it

Emilio Pisanty

06:56

0

Q: How to compute elliptic integral of the first kind using fortran

I am trying to solve the computational problem using fortran to get the value of Time period based on amplitude. The formula is where value of α ranges from 0 to pi/2 which I can convert degree into radian by using i=1 90 alpha=pi*i/180 end do but what is value of φ here? what should I...

That's a remarkable question

Boy, is OP in for a rude awakening

John Rennie

07:12

@EmilioPisanty I don't understand your answer. The OP's equation contains $\int d\phi$.

Slereah

07:28

Nyello

My dreams are not quite so very deep

Last time I dreamt there was a pamphlet explaining that if you were dreaming about eating human flesh, u probably had nutrient deficiencies

And it listed the nutrients associated with the body parts you ate in your dreams

If you dream about eating eyeballs probably try eating more magnesium mb

Emilio Pisanty

07:41

@JohnRennie not his code

He's asking what phi is

@Slereah you shouldn't eat people

Slereah

that's why you need your magnesium

Emilio Pisanty

The Reluctant Cannibal

►

Slereah

Another dream of mine was that I was in a store to buy a tank to raise oysters

Then I got to the counter and the salesman told me that people would come by to install it

It was then that I realized

That tank was way bigger than I had expected

Fooled by my hubris!

I am not ready for such a gigantic tank of oysters

2

@RyanUnger

I need you to tell me if I am dumb

I am writing a mail to John Earman and I need to know if my question is dumb

Hello,

In your book "Bangs, Crunches, Whimpers and Shrieks", you discuss the notion of causally benign spacetimes (originally from Test fields on compact spacetimes by Yurtsever), and I've been having some troubles understanding some aspects of it.

We have that, given a field φ with equation of motion D(φ) = 0, the open set U is causally regular if there is a smooth extension φ' to M such that D(φ') = 0 and φ' = φ on U. p is causally regular if every neighnourhood Up around p contains a causally regular neighbourhood U'p. M is then benign if every point is causally regular.

Is the solution obvious or am I dum

dacemirror.sci-hub.tw/journal-article/…

if u wish for the paper discussing the topic

PM 2Ring

08:08

Is there a canonical dupe target for standard Twin Paradox questions? Surely there's something better than the one proposed in the comments here. physics.stackexchange.com/questions/485612/…

Slereah

Ask @JohnRennie

He is the twin paradox answerer guy

John Rennie

@PM2Ring here:

81

Q: What is the proper way to explain the twin paradox?

The paradox in the twin paradox is that the situation appears symmetrical so each twin should think the other has aged less, which is of course impossible. There are a thousand explanations out there for why this doesn't happen, but they all end up saying something vague like it's because one tw...

special-relativity time reference-frames relativity

Slereah

SR and GR really are fanciful disciplines

PM 2Ring

@JohnRennie Thanks. That looks familiar. ;) Feel free to wield the hammer.

Slereah

There is absolutely no reasons for the examples to involve spaceships

but it is almost always spaceships

John Rennie

08:15

@Slereah there are few other ways to travel near the speed of light. A large railgun would work but the observer might suffer a rather high acceleration.

Slereah

@JohnRennie You don't need to travel near light speed to do SR

it works at any speed!

Less impressive but as far as calculations go, they're the same

plus spaceships don't go anywhere near lightspeed anyway

PM 2Ring

@Slereah We had a question yesterday involving a relativistic runner.

You must have pretty good running shoes. ;) You'd circumnavigate the planet pretty quickly at that speed. And circular motion at that speed at the radius of the Earth requires about 4.3 billion g centripetal acceleration. BTW, relativistic length contraction only happens in the direction of motion. — PM 2Ring 12 hours ago

Slereah

probably just an ad for sneakers

PM 2Ring

@EmilioPisanty Using standard numerical integration techniques on elliptic integrals of the 1st kind is so old-school. The cool kids use the AGM. ;) And with a little more work, you can do elliptics of the 1st & 2nd kind simultaneously.

Slereah

@PM2Ring I just use the Book

PM 2Ring

08:30

@Secret You can safely dip your fingertips into liquid nitrogen, if you're quick. IME, even though it's about 100° colder, it's safer than handling dry ice, since it doesn't stick to you, and you have that Leidenfrost effect barrier... for about half a second. :)

Slereah

I would still not advise doing it

Don't want your fingers breaking off because your phone rang and you were distracted for a second

John Rennie

@Slereah I've done it. And I still have all my fingers :-)

Slereah

Well I mean sure

I have also handled acids without gloves or goggles

It can go well

But it can also go bad

From a legal standpoint I prefer to not advise it

PM 2Ring

@Slereah That's fair enough. You do have to be careful. But it is safe, if you're quick.

Slereah

It's also mostly fine if acid gets on your skin or if you get a laser in your eye or if you set something on fire

Also I've been electrocuted before, it was fine

Try this at home, kids!

John Rennie

08:35

@Slereah I got electrocuted once and it flipping hurt

Slereah

Well sure, it hurts

But only for a brief time

My old flat had this light which was operated by two different lightswitches

John Rennie

Several seconds in my case. It seemed a lot longer.

Slereah

it was complex to know if it was on or not

so when I had to change the lightbulb, I didn't know if it was on or not

So I got electrocuted a few times

I have also handled explosives drunk

Also fine

John Rennie

In other news, it has stopped raining in Chester after 48 hours of continuous rain.

Apparently this was due to a European weather system that pushed in the UK from the south east.

After Brexit we will be able to stop continental weather systems from entering the UK unless they have a visa.

8

The government assures me of this so it must be true.

Slereah

But I'm pretty sure rain is British weather

So not really an improvement

Emilio Pisanty

08:49

@PM2Ring AGM?

@Slereah that looks outdated, it'll have errors in it

Slereah

I don't think integrals get outdated

they're not made by Apple

PM 2Ring

@EmilioPisanty The arithmetic-geometric mean. See my comment on the question. Gauss discovered this method, but before computers, it was more practical to use a series, since square roots are a PITA when you're doing manual calculation.

If you do a Google search on elliptic integral agm a bunch of helpful articles pop up, but most of them are to PDFs, and it's a pain posting such links, especially on the phone.

The Wikipedia article is a good intro. en.wikipedia.org/wiki/Arithmetic%E2%80%93geometric_mean

Actually, Lagrange discovered the AGM, but Gauss investigated it further.

skullpatrol

09:11

@JohnRennie will snow will have a much harder time getting a visa?

(unless it comes in over the Atlantic)

John Rennie

@skullpatrol the UK generally only gets heavy snow when we get an Arctic air flow pushing in from the north, so sadly Brexit won't affect that :-)

But then England hardly ever gets heavy snow - maybe once every few years. It's a different story in Scotland of course.

skullpatrol

:-)

Emilio Pisanty

09:47

> $e$ is for Euler, one of the most renowned mathematicians of the last millennium. Euler discovered $e$, although what’s more impressive is where he discovered it: in the public writings of Jacob Bernoulli, who actually discovered it.

mathwithbaddrawings.com/2018/02/05/the-abc-book-of-e

@PM2Ring that said, the amount to which I can muster caring about AGMs is pretty nil at the moment...

PM 2Ring

Stigler's law of eponymy

Stigler's law of eponymy, proposed by University of Chicago statistics professor Stephen Stigler in his 1980 publication "Stigler’s law of eponymy", states that no scientific discovery is named after its original discoverer. Examples include Hubble's law which was derived by Georges Lemaître two years before Edwin Hubble, the Pythagorean theorem although it was known to Babylonian mathematicians before the Pythagoreans, and Halley's comet which was observed by astronomers since at least 240 BC. Stigler himself named the sociologist Robert K. Merton as the discoverer of "Stigler's law" to show...

Slereah

10:57

Hm

nlab defines the Nambu-Goto action as $$\int_\Sigma \| d\phi\|^2 $$

Oh wait

I think they're trying to say they're using a different norm from the Polyakov action

But they're not really giving out much details

Because otherwise it wouldn't make sense as $$\| d\phi \| = \sqrt{\gamma^{ab}g_{\mu\nu} \phi^\mu_{,a} \phi^\nu_{,b}}$$

I s'ppose the Nambu-Goto action is p. much just the volume form of the worldvolume

Slereah

11:57

Oh god

Trying to do the Euler-Lagrange equation for a generic Nambu-Goto action

Gonna have to write the determinant in coordinate form D:

This is the worst thing

PM 2Ring

I suppose this OP might be asking about exotic topologies... but maybe not. ;)

Slereah

minuses the question

bolbteppa

57

Q: A life of PhD: is it feasible?

In the next year I will (hopefully successfully) graduate from a PhD programme in pure mathematics. The location is (continental) Western Europe, the topic of the thesis is arithmetic geometry, if it matters. During my PhD experience I have found out the following things being a pure math PhD s...

phd graduate-admissions career-path

my god...

Ryan Unger

@Slereah hi

reading what you wrote

Slereah

Is it dum

I'm pretty sure both Earman and Yurtsever aren't gonna be both wrong but I have no idea why

Ryan Unger

12:07

I think we've had this convo before

what is D?

Slereah

most definately

Since I posted that as a question a few months ago

bolbteppa

Why not latex up the question and insert it as a png in the e-mail

Slereah

D is some differential operator

Usually it's $\Box \phi = 0$

@bolbteppa Might be my age showing but I consider poor form to include bells and whistles in an email

Ryan Unger

"every globally hyperbolic spacetime is causally benign"

Slereah

though I guess I could include a PDF with the question to make it clearer

bolbteppa

12:09

e-mail hasn't existed long enough to allow such etiquette to have been established :p

Ryan Unger

so this definition is ridiculous. Why should $\phi$ even admit a continuous (nevermind satisfying the equation) extension to $M$

Slereah

@RyanUnger That is what I am wondering

Lemme include the actual phrasing of Earman, just in case

pitt.edu/~jearman/Earman_1995BangsCrunches.pdf

p. 180

Ryan Unger

right you're taking the open subset of the torus given by the interior of the square/rectangle?

Slereah

Yeah

I don't think any solution on that subset can be extended to a cylinder

Ryan Unger

so you can show that $\phi$ can be continuously extended to all of $M$ if and only if it is uniformly continuous on $U$

let's assume $U$ bounded otherwise this is completely crazy

now $C^2$ you need even more

do you have the Yurtsever paper

he might mean $\phi\in C^\infty(\overline U)$

Slereah

12:15

aip.scitation.org/doi/10.1063/1.528960

starting on p. 6

Ryan Unger

his definition is fine

note that he talks about the closure as I predicted

it's a theorem of Whitney that what he says is equivalent to what I said

Slereah

Is the example I gave not $C^\infty(\bar{U})$

Ryan Unger

well I've been thinking about that

the thing that's weird there is that your set is such that $\mathrm{int}\overline U\ne U$

you can even just think about a circle minus a point

so you do need uniformly continuous

your function is not even $C^0(\overline U)$

Slereah

Isn't it?

Hm

I mean it's just a cosine on $[-a,a] \times [-b,b]$ for instance

Argh

I am bad at analysis

Ryan Unger

there is a point $x\in \overline U$ such that there are two sequences $x_i,y_i\to x$ in the interior for which $\lim \phi(x_i)\ne \lim \phi(y_i)$ right?

@Slereah sure but the limits are different as you approach -a and +a

so when you identify, bad stuff

$C^0(\overline U)$ means continuous in $U$ and extends continuously to $\partial U$

Slereah

12:23

Well $-a$ and $a$ don't have to be the junction point

The region can be a lot smaller than the torus

Ryan Unger

what

Slereah

I mean let's say the torus is $[-1,1] \times [-1,1]$, and the subset under consideration is $U = [-0.5, 0.5] \times [0.5, 0.5]$

Ryan Unger

then I don't see what the issue is

and that's not open

Slereah

$()$ yes

Isn't the cosine continuous in $\bar{U}$

Ryan Unger

you want to take $M=[-1,1]^2/\sim$ and $U=(-1,1)^2$

Slereah

12:26

But every neighbourhood of every point has to contain a causally regular neighbourhood

So $U$ has to contain a neighbourhood which is CR regular

Ryan Unger

I don't actually see your issue here

the definition seems fine to me

Slereah

Well the definition is fine, I'm just having troubles understanding why it holds in globally hyperbolic manifolds

in this example in particular

Ryan Unger

ohhh

he says that?

how is that globally hyperbolic

are you identifying time?

Slereah

p. 7

"the subset C coincides with M in a globally hyperbolic space-time."

Ryan Unger

page 7?

Slereah

12:28

Well as mentionned previously I just consider a spacelike cylinder in my example

p. 7 of Yurtsever's article

Ryan Unger

my page numbers are in the 3000s

Slereah

3069

(nice)

Ryan Unger

ah sheaf theory

Slereah

Incidentally he also claims that the Minkowski torus is also causally benign for a specific type of identifications

but that is a problem for another day

first to understand why it works for globally hyperbolic spacetimes

I did ask Yurtsever about this paper but he's working with computers now, making real money

Didn't even answer!

Ryan Unger

that's just saying that every point is a cr point right

Slereah

12:31

Yes

So basically every neighbourhood contains a CR neighbourhood

Ryan Unger

this should follow from the usual local existence theory for the Cauchy problem in GH spacetimes

Slereah

Yeah that is Yurtsever's argument

Ryan Unger

ok so if you don't know that, then you need to learn it

Sobolev spaces are good for you

Slereah

Yeah I guess if I look into it mb that will help out understand what's wrong with my example

blah

seems to be a Ringstrom sort of theorem :V

Bloody hell

Ryan Unger

yes you should read Ringstrom's book on the Cauchy problem

Slereah

12:36

Tall order

Hm

About 40 bucks

Reasonable

>Existence of a maximal globally hyperbolic development

>Background from set theory

Is it gonna be Zorn's lemma

Ryan Unger

unfortunately yes

you can remove the need for Zorn though

Slereah

Isn't that theorem concerned with initial data on a whole Cauchy hypersurface, though?

Here it's just an arbitrary open set

Well not totally arbitrary, but not necessarily spanning a Cauchy surface

Oh wait I guess it has sections on local existence

probably should peruse it a bit

Secret

(I was going to comment a couple of posts many hours ago, including some funny ones, but the present atmosphere is too academic serious to it thus I am not doing it to ensure this conversation stay rigorous)

dived into math chat

Ryan Unger

12:58

@Slereah recall that $D^+\cup D^-$ is g.h.

for any open subset of a spacelike hypersurface in a g.h. spacetime

Slereah

True

Plus it's a basis for the topology

So we're fine for every open set

Ryan Unger

yes

that's what you're supposed to do here

Slereah

though currently trying to work out the general formula for the Nambu Goto action

Derivatives of determinants are not pleasant

bolbteppa

What's not general about the usual formula

Slereah

though at least there are identities for it

@bolbteppa not an arbitrary number of dimensions

I mean I assume the formula for that exists somewhere, but most places just do point particles and strings

bolbteppa

13:07

You mean for branes?

Slereah

the $p$-branes yeah

bolbteppa

It's just $S = - T \int d^{n+1} \sigma \sqrt{- \det ( g_{\mu \nu} \partial_{\alpha} X^{\mu} \partial_{\beta} X^{\nu})}$ and can use $\det A = \frac{1}{n!} \varepsilon^{i_1 \ldots i_n} \varepsilon^{j_1 \ldots j_n} A_{i_1 j_1} \ldots A_{i_n j_n}$ to find determinant variations/derivatives easy enough

Slereah

Well

It's not hard

But it is very tedious

plus I'm doing it at work rn :p

So hard to focus too hard on it

bolbteppa

Yeah, recently wrote up the infinitesimal variation of Polyakov properly, showing it's invariant under $\delta h^{\alpha \beta} = \xi^{\gamma} \partial_{\gamma} h^{\alpha \beta} - \partial_{\gamma} \xi^{\beta} h^{\alpha \gamma} - \partial_{\gamma} \xi^{\alpha} h^{\gamma \beta}$ etc looked worse than it was

bolbteppa

13:27

For a point particle we can get from $S = - m \int \sqrt{-\dot{x}^2} d \tau$ to $S = \frac{1}{2} \int e(e^{-2} \dot{x}^2 - m) d \tau$, obviously $e$ looks like the GR metric, and setting $e^2 = g^{\tau \tau}$, $e = \sqrt{-g^{\tau \tau}}$ makes it explicit, but still see a disconnect in calling $e$ an einbein compared to einbeins via $g_{\mu \nu} e_a^{\mu}{} e_b{}^{\nu} = \eta_{ab}$

Slereah

14:16

Also I should probably put it in proper $n$-dimensional form so that I can compute the stress energy tensor

What with the dirac deltas

bolbteppa

\begin{align}
0 &= T_{\alpha \beta} \\
&= - \frac{2}{T} \frac{1}{\sqrt{-h}} \frac{\partial S}{\partial h^{\alpha \beta}(\sigma')} \\
&= - \frac{2}{T} \frac{1}{\sqrt{-h}} \frac{\partial }{\partial h^{\alpha \beta}(\sigma')}[ - \frac{T}{2} \int d^{n+1} \sigma \sqrt{-h} h^{\alpha \beta}(\sigma) g_{\mu \nu} \partial_{\alpha} X^{\mu} \partial_{\beta} X^{\nu} ] \\
&= \frac{1}{\sqrt{-h}} \int d^{n+1} \sigma [- \frac{1}{2} \sqrt{- h} h_{\alpha \beta} h^{\alpha' \beta'} g_{\mu \nu} \partial_{\alpha'} X^{\mu} \partial_{\beta'} X^{\nu} + \sqrt{-h} g_{\mu \nu} \partial_{\alpha} X^{\mu} \partial_{\beta…

That?

Finally making some headway into this octonionsense

Slereah

@bolbteppa I s'ppose

Octonions are apparently the Clifford algebra of $\mathbb{R}^7$ or somesuch

Well, derived from

Probably a tad longish to prove from that tho

Ryan Unger

What does Nambu Goto mean geometrically

bolbteppa

14:34

It's the area of a world-sheet in space-time, just as the SR action of a particle (= 0-brane) is the arc length of a world-line in space-time

Slereah

@RyanUnger it's the volume of the worldvolume

bolbteppa

Trying to see how the exceptional lie algebras are Lie algebras of isometry groups of projective planes involving the complex C's /quaternionic H's/octonionic O's, something something
G_2 Isom(O)
F_4 Isom(O P^2)
E_6 Isom(CxO P^2)
E_7 Isom(HxO P^2)
E_8 Isom(OxO P^2)

Slereah

Just $$\int_\Sigma d\mu[X^*g]$$

For the embedding $X$

bolbteppa

Have to be careful with Baez stuff, but these octonionotes are looking really good

Slereah

Where do the onions come in anyway

bolbteppa

14:49

My thinking is along the lines of: going from $R$ to $C$ via the weird multiplication $(a,b) \cdot (c,d) = (ac+bd,ad-bc)$ can be interated on $C$ to get $H$ and on $H$ to get $O$, with a loss of info at each step (need to read the notes properly), and on going beyond $O$ you lose divisibility (but can keep going to get e.g. 'sedenions'), and without divisibility the notion of inverting is gone which physical/geometrical systems should possess so we stick with these.

Projective geometry seems to be all about isometries of a space, things like parallelism holding from different perspectives of the same situation, so the study of isometry groups of projective planes over the division algebras R, C, H, O is bound to give important symmetry groups, and for some reason the lie algebras of the isometry groups
G_2 Isom(O)
F_4 Isom(O P^2)
E_6 Isom(CxO P^2)
E_7 Isom(HxO P^2)
E_8 Isom(OxO P^2)
are most important

PM 2Ring

Here's an alternate view of the Fano plane. Just imagine there are mirrors on the vertical & horizontal axes through o. ;) gist.github.com/PM2Ring/…

Here's the usual view. gist.github.com/PM2Ring/…

bolbteppa

Why are we using a projective plane P^2 and not a projective line P^1 or something weirder idk etc

PM 2Ring

A finite projective line is pretty boring. :)

bolbteppa

In math.ucr.edu/home/baez/octonions/node13.html he says the Lie groups $SO(n), SU(n), Sp(n)$ "arise naturally as symmetry groups of projective spaces over $R$, $C$, and $H$, respectively. More precisely, they arise as groups of isometries: transformations that preserve a specified Riemannian metric. Let us sketch how this works, as a warmup for the exceptional groups".

PM 2Ring

The Fano plane is just complicated enough to have interesting structure, but not too complicated. So it pops up in a lot of places.

bolbteppa

15:00

In particular $isom(RP^n) \equiv so(n+1)$, $isom(CP^n) \equiv su(n+1)$, $isom(HP^n) \equiv sp(n+1)$, and these give the $A_n,B_,C_n,D_n$ lie algebras, but why we go from $P^n$ to $P^2$ (or no $P^2$ in G_2) in
G_2 Isom(O)
F_4 Isom(O P^2)
E_6 Isom(CxO P^2)
E_7 Isom(HxO P^2)
E_8 Isom(OxO P^2)
idk

Maybe non-associativity of the octionions is why you use P^2 or something...

Oh my god "As we shall explain, the concept of an octonionic projective space $OP^n$ only makes sense for $n \le 2$, due to the nonassociativity of $O$."

This is shocking

So if we're looking for isometry groups of projective spaces which use division algebras as the scalars, and the $A_n, B_n, C_n, D_n$ lie algebras come as tangent spaces to the isometry groups of n-dimensional projective spaces over $R,C$ and $H$ namely $RP^n, CP^n, HP^n$, and the octionions only allow for $OP^0$ (=$O$?), $OP^1$ and $OP^2$ to exist, and the exceptional Lie algebras are tangent spaces of the

$OP^0 = O?$, $O P^2, C \times O \, P^2, H \times O \, P^2, O \times O \, P^2$ isometry groups, where do $C \times O \, P^2, H \times O \, P^2, O \times O \, P^2$ come from

Ryan Unger

15:17

you're doing a Secret

@Slereah you mean $X:\Sigma\to (M,g)$ and you're pulling back the metric to $\Sigma$?

then this is the first variation formula

you can find the formula in Colding and Minicozzi's book, for example

PM 2Ring

Here's the 3D version of the Fano plane, PG(3,2), raytraced using POV-Ray.

in Python on Stack Overflow Chat, Oct 2 '17 at 11:09, by PM 2Ring

bolbteppa

@RyanUnger it's literally just the surface area formula amazing how you would complicate something so simple

@PM2Ring that's pretty cool but it's literally just graphing a mnemonic right no real meaning to it?

Ryan Unger

I'm not complicating anything

PM 2Ring

@bolbteppa It's a mathematical structure. Whether it has any application in physics is another story.

Ryan Unger

first variation of surface area is literally called the first variation formula

go back to school if you don't know that

bolbteppa

15:22

First variation of area formula

In Riemannian geometry, the first variation of area formula relates the mean curvature of a hypersurface to the rate of change of its area as it evolves in the outward normal direction. Let Σ ( t ) {\displaystyle \Sigma (t)} be a smooth family of oriented hypersurfaces in a Riemannian manifold M such that the velocity of each point is given by the outward unit normal at that point. The first variation of area formula is d d t...

Where does this come up in the NG action or it's dynamics etc

PM 2Ring

As for where the Fano plane pops up in maths:

in Python on Stack Overflow Chat, Oct 2 '17 at 12:51, by PM 2Ring

:) If you want to know some of the applications of the Fano plane, take a look at "The Many Names of (7,3,1)" by Ezra Brown, and the follow-up article "Many More Names of (7, 3, 1)".

Ryan Unger

If the NG action is surface area and you want to compute its variation, then that's the formula

Semiclassical

@PM2Ring to be ultimately followed by “somehow even more names of (7,3,1)”

bolbteppa

@PM2Ring is it a mathematical structure or just a mnemonic representing what basis elements multiply into what other basis elements in the octionion multiplication table?

Like just a picture of the 'right-hand rule' for cross products

PM 2Ring

I first got interested in the Fano plane & finite projective geometry via block design, which is pretty useful in designing certain efficient experimental procedures.

bolbteppa

15:30

"Given this, one might naturally guess that the period-8 repetition in the homotopy groups of $O (\infty)$ is in some sense 'caused' by the octonions. As we shall see, this is true. Conversely, Bott periodicity plays a crucial role in the proof that every division algebra over the reals must be of dimension 1, 2, 4, or 8. " madness

PM 2Ring

@bolbteppa I guess you can say it's "just" a mnemonic for the octonion basis elements. Or you can say it's a finite 2D projective geometry. ;) It may seem weird to do geometry in a field with finite sets of points and lines, but there's nothing wrong with it mathematically. And it's good to have multiple ways to conceptualize stuff.

Semiclassical

@PM2Ring i read up on that a while back myself. Amusing to see something seemingly esoteric have very practical implications

bolbteppa

Another nuts comment - the dimensions of the division algebras 1, 2, 4, 8, are the dimensions of the massless Poincare little groups in dimensions 3, 4, 6, 10

PM 2Ring

FWIW, messing around with finite fields is popular in cryptography. The main form of encryption used today, AES, uses arithmetic in the Galois field GF(2^8).

Semiclassical

@PM2Ring it’s a bit like saying that ij=k and it’s cyclic permutations are a mnemonic. You can think of it like such, or you can think of it in terms of 3D rotations

bolbteppa

15:38

The only dimensions in which supersymmetric Yang-Mills can exist...

Semiclassical

It of course makes conceptual sense to think of a group (or some other mathematical structure) as being given by a set of rules/axioms. But for purposes of intuition it helps to have some geometric model which realizes said structure

bolbteppa

@Semiclassical right but how are you going to do that for the octionions with their multiplication table as given here :p

PM 2Ring

There's a classic puzzle from the mid 1800s that can be solved using PG(3, 2). I like how recreational mathematics has sometimes led to the creation of new branches of mathematics.

Kirkman's schoolgirl problem

Kirkman's schoolgirl problem is a problem in combinatorics proposed by Rev. Thomas Penyngton Kirkman in 1850 as Query VI in The Lady's and Gentleman's Diary (pg.48). The problem states: Fifteen young ladies in a school walk out three abreast for seven days in succession: it is required to arrange them daily so that no two shall walk twice abreast. == Solution == If the girls are numbered from 0 to 14, the following arrangement is one solution: A solution to this problem is an example of a Kirkman triple system, which is a Steiner triple system having a parallelism, that is, a partition of the...

bolbteppa

"This is certainly a neat mnemonic, but is there anything deeper lurking behind it? Yes! The Fano plane is the projective plane over the 2-element field $Z_2$. In other words, it consists of lines through the origin in the vector space $Z_2^3$. Since every such line contains a single nonzero element, we can also think of the Fano plane as consisting of the seven nonzero elements of $Z_2^3$."

"Note that planes through the origin of this 3-dimensional vector space give subalgebras of $O$ isomorphic to the quaternions, lines through the origin give subalgebras isomorphic to the complex numbers, and the origin itself gives a subalgebra isomorphic to the real numbers." super weird doing geometry for finite fields...

Semiclassical

too much algebraic geometry for me

Abhas Kumar Sinha

16:13

@PM2Ring why 'Schoolgirl' not for boys 🤔🤔🤔

Another Bad day,

quora.com/Can-you-write-an-answer-with-no-words/answer/…

People still think that Cancer can't be cured

bolbteppa

Projective orthogonal group

In projective geometry and linear algebra, the projective orthogonal group PO is the induced action of the orthogonal group of a quadratic space V = (V,Q) on the associated projective space P(V). Explicitly, the projective orthogonal group is the quotient group PO(V) = O(V)/ZO(V) = O(V)/{±I}where O(V) is the orthogonal group of (V) and ZO(V)={±I} is the subgroup of all orthogonal scalar transformations of V – these consist of the identity and reflection through the origin. These scalars are quotiented out because they act trivially on the projective space and they form the kernel of the action...

There's even a geometrical reason why the Lie algebras $B_n$ and $D_n$ should be different for $SO(2n)$ and $SO(2n+1)$

Abhas Kumar Sinha

That's a Myth, Now in 2019 even 3rd stage Cancer can be treated, that's a complete myth. So my friends who are Cancer Survivors/sufferers don't worry, treatment of Cancer is no big deal, that's completely easy and simple, just few months of pain, then, normal life back!

Very less people are dying due to cancer now then before

ACuriousMind

@Slereah ...why did you not trip the breaker to be sure?

Slereah

16:38

@ACuriousMind did you not read the part about handling explisives drunk

I am no safety namby pamby mama's boy

John Rennie

We experimental physicists have biiiiiiiiig cohones

Sir Cumference

I tried a chat discussion before,& it was most unsatisfactory. My chat buddy agreed that the site badly needed reform & rationalisation. — Michael Walsby 8 hours ago

Ugh great, one of these

ACuriousMind

@JohnRennie Slereah did not strike me as the experimental sort so far :P

John Rennie

Tiny cohones then? :-)

This question has been edited to completely change its meaning. Well, OK, but now I'm sure it's a duplicate of a different question.

I'm sure someone has asked about the effect of time dilation on organisms i.e. whether it really affects age, but I'm damned if I can find it.

VARUN.N RAO

Hey guys, I just wanted to know whether my understanding of QM and Entanglement is reasonably correct.

Please tell me whether these "axioms" completely describe the properties of Spin measurement, Entanglement and do point out if something is not said.

1. If I measure the spin of an electron randomly its 50% up and 50% down
2. Once the measurement done in x direction(and if its up), the electron retains its spin along its axis and how ever many times you measure it along that direction it's always 100% up. And spin in only 1 axis can be known at once, measuring spin in another axis resets…

ACuriousMind

16:50

You can't just say "entangled electrons". There are many entangled states and many of them don't even break Bell's inequality, you need to specify much more carefully how the state is entangled.

VARUN.N RAO

@ACuriousMind "how "? Are there different types of entanglement? I only know entanglement where spin is always opposite. What other types are there?

PM 2Ring

@SirCumference Yeah. What I actually said is:

Semiclassical

1) Depends on how the electron has been prepared. Suppose, for instance, that the source of your electrons was secretly performing a spin measurement along the z-axis prior to you receiving it, and only passing along those which came out spin-up. Then you'd have a 50-50 distribution of spins if you measure along an axis perpendicular to the z-axis, but otherwise there's going to be some bias. (major omission in what I wrote originally)

PM 2Ring

in Discussion between PM 2Ring and Michael Walsby, Jun 1 at 17:45, by PM 2Ring

@MichaelWalsby That stuff I posted is in 6 separate posts. Yes, chat is somewhat limited, and many people have suggested various improvements, to no avail. You can read some info about using chat here.

ACuriousMind

@VARUN.NRAO An entangled state of two or more subsystems is simply one in which you cannot assign definite pure states to each subsystem. The "Bell pairs" that usually feature in Bell-type experiments are just very specific examples of such states.

PM 2Ring

16:53

in Discussion between PM 2Ring and Michael Walsby, Jun 1 at 17:51, by PM 2Ring

OTOH, the chat rooms have some advantages over comments, eg you can reply directly to previous posts. You can display images here, by posting a message just containing the image URL, eg

VARUN.N RAO

@ACuriousMind oh ok. But am I correct in that specific example? I am about to give a small talk so I am verifying.

Semiclassical

"Entangled in the singlet state" is what you'd say for a Bell test

VARUN.N RAO

@Semiclassical Oh yes, I must have been more specific, what I meant by my first point is that if you take random electrons and measure them in random directions and axis, then the probability of you measuring it up and down is 50%. Am I correct in my understanding?

Semiclassical

What you're looking for, in technical terms, is that the initial state of the electron is completely mixed.

VARUN.N RAO

@Semiclassical "singlet state"? Can you please name other types of state so that I can research on them?

Semiclassical

16:58

Any of the three triplet states, for one.

PM 2Ring

Hey @ACuriousMind This guy has reverted your rollback, and deleted the homework tag.

Please do not reverse the edit of the tag again, or the post will be locked and protected from any further edits. The tag - and the "on hold" status - will be removed only if the question is edited to ask about a conceptual question. — ACuriousMind ♦ 2 days ago

Semiclassical

For instance, the (pure) state $\frac{1}{\sqrt{2}}[|\uparrow\downarrow\rangle +|\downarrow\uparrow\rangle]$ is entangled. But it's not the singlet state.

ACuriousMind

@PM2Ring Thanks for the info (perfectly fine case for a custom flag, btw)

Semiclassical

(I'm pretty sure what I just said was true but I'm not 100% so rip)

PM 2Ring

@ACuriousMind Ah, right. :oops: I just remembered about it when I saw you here, so I figured the informal approach would be ok. But next time, I'll use the flag system.

VARUN.N RAO

17:03

@Semiclassical just when I thought I reasonably understood QM

Semiclassical

If you've ever talked about ortho v. parahelium in chemistry, you've run into this

PM 2Ring

@JohnRennie Yeah, he's been editing all his recent questions today, to try to get around the question ban he just earned.

ACuriousMind

@PM2Ring It's generally fine, but e.g. if I had left just after getting the ping I might've forgotten about it and there'd be no trace anywhere

VARUN.N RAO

@Semiclassical I just learnt that in doublet and triplet entanglement there are more electrons involved instead of two, but I think the inverse cosine rule applies there also, there is just more electrons involved. Because the spin of electron is always either up or down

Semiclassical

definitely not. the cosine rule is a consequence of rotational symmetry. if your entangled state isn't rotationally symmetric, I wouldn't expect the cosine rule to hold.

the singlet state has rotational symmetry (because the singlet state has total spin zero) but other entangled pure states won't

VARUN.N RAO

17:09

@Semiclassical oh damnn

@Semiclassical is this the only mistake or are there more??

PM 2Ring

FWIW, in Australia, a singlet is a type of sleeveless shirt, often worn as an undergarment. So I get slightly odd images when thinking of particles in singlet states. :)

VARUN.N RAO

@PM2Ring LOL

Sir Cumference

@PM2Ring Oh so he's referring to you lol

Semiclassical

Honestly, if you just limit yourself to saying "this is for a pair of electrons entangled in the singlet state", you're fine

Sir Cumference

The whole "reform and rationalisation" smells of crackpottery

PM 2Ring

17:13

@SirCumference Yes. I was a bit annoyed that he refused to post in the automatically-created chat room.

Semiclassical

@SirCumference well. the impulse to think that the rules/regs of a particular community may require reworking and attention isn't necessarily a bad one. the presumption that an outsider can see this and that community is entirely blind to it, on the other hand...

PM 2Ring

@SirCumference I don't think he's a crackpot, just a little misinformed, and highly opinionated. And very stubborn.

VARUN.N RAO

@Semiclassical ok so I'm assuming that's the only mistake I have committed.

So my description only concerns singlet entanglement which violates Bell's inequality. And I hope I am reasonably correct on all of my other points.

Thank you Semiclassical and ACuriousMind for taking time to explain this to me

Sir Cumference

@PM2Ring I mean, yeah, that tends to be a feature of them. Certainty that their beliefs are correct and refusal to listen to mainstream explanations

VARUN.N RAO

Good day everyone

Sir Cumference

17:16

Crackpots I mean

PM 2Ring

Eg, -10 & he won't back down, or delete his crap answer: physics.stackexchange.com/a/482821/123208

Nick

Quantum Sound

►

Semiclassical

@VARUN.NRAO As an example of what can go wrong with the state I gave: If both persons measure their particles along the z axis, then they're guaranteed to get opposite results with 100% probability. (That's what you'd expect from the cosine rule---cos^2(theta/2) = 1 when theta=0.)

But if they measure along the x-axis or y-axis, they're guaranteed to get -identical- results.

VARUN.N RAO

@Semiclassical oh interesting. But this entanglement is broken the moment we measure it right?

Semiclassical

For the cosine rule to work, they'd need to get opposite results regardless of what common axis they choose to measure along. This is true for the singlet state (and makes life considerably easier...) but evidently not for this triplet state.

Nick

17:19

@Semiclassical I don't know what you're talking about but I sure do know you're not accounting for time.

PM 2Ring

@SirCumference He might have some crackpot beliefs, but I think his erroneous views are mainly misunderstood or outdated info, almost certainly acquired from popsci sources.

Semiclassical

When I say "If they measure along the x- or y-axis" I mean "had they done that instead of measuring along the z-axis"

And I'm also assuming that I'm making measurements on the entangled pure state I gave earlier, with no kind of detector weirdness

VARUN.N RAO

18:01

@Semiclassical Kind of understood. But the entanglement will be lost on the measurement of one of the electrons in singlet state right? And I hope my other points are correct

Semiclassical

18:20

Right. Note, though, that “unentangled” is not the same as “non-random”

If the first person measures along the z-axis and gets spin-up, then there’s fifty-fifty chance of the second person getting spin-up along the x-axis.

@VARUN.NRAO Forgot to ping the above in reply

Novalium Company

18:34

Hello fellow animals with consciousness.

Slereah

20:24

I don't have a consciousness

I'm a P-zombie

bolbteppa

20:44

"The real numbers are the dependable breadwinner of the family, the complete ordered field we all rely on. The complex numbers are a slightly flashier but still respectable younger brother: not ordered, but algebraically complete. The quaternions, being noncommutative, are the eccentric cousin who is shunned at important family gatherings. But the octonions are the crazy old uncle nobody lets out of the attic: they are nonassociative."

Semiclassical

lool

wonder what that'd make the sedonions etc

(the equivalent of that kid in the sandlot "who got real into the 60s and no one ever heard from him again", maybe)

bolbteppa

haha

bolbteppa

21:03

Can't find why octionionic projective space only makes sense for $n \leq 2$ from skimming :\

"The diagram E6 gives a 78-dimensional compact Lie group that people call E6. It's the isometry group of the bioctonionic projective plane.

The diagram E7 gives a 133-dimensional compact Lie group that people call E7. It's the isometry group of the quateroctonionic projective plane.

The diagram E8 gives a 248-dimensional compact Lie group that people call E8. It's the isometry group of the octooctonionic projective plane. "

**octooctonionic**

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