The h Bar

Semiclassical

00:51

@Blue I played around in Mathematica and confirmed two things.

1) The distribution of vertex counts is well-described by a binomial distribution with probability 1-p and n experiments. (1-p is the survival probability, so it's what you use for the vertices remaining).

For instance, a run of 50000 such prunings on a certain graph gives the following empirical distribution:

the red dots are the empirical frequencies, whereas the blue is the binomial distribution (with n=50 nodes and probability p=.5). so very much on point

2) The distribution of edge counts is not nearly clear-cut. Empirically, you again get a single peak. But if you plot the binomial distribution with n-1 experiments and survival probability (1-p)^2, you get something with the right mean but narrower variance than the empirical distribution:

the fact that the mean value is still accurate means that the result we found earlier is still valid

but the edge distribution isn't actually binomial, so the assumptions we used aren't quite valid

getafix

03:13

I had a quick question about quantum mechanics, for a particle in a linear potential say -c*x, the energy eigen spectrum should be continuous? but we solve the time independent schrodinger equation by transforming it into a airy equation and then the roots of the airy function correspond to energies. I guess this is at odds with my intuition for the energy spectrum to be continous? but then the we normalise the airfunctions to a delta-function like we would for continous case...?

clearly there is a gap in my understanding somewhere

Semiclassical

03:30

the typical place you see the linear potential show up is in the context of wkb

in which case the point is that you're doing an approximation

and the wavefunction you get will only be valid in the vicinity of the turning point, and that's enough

by contrast, the potential away from the turning point need not be linear, and indeed should have a minimum in order to have bound states

in which case you definitely do want quantized energy levels

Avnish Kabaj

@Blue no

Nobody solved my super duper mega ultimate difficult questions

:(

Or maybe it's too difficult for you guys :P

getafix

04:03

@Semiclassical gotcha. so if it is just a linear potential in space, and no boundaries..it is indeed a continuum of energies isn't it?

Semiclassical

i'd say so, yeah

it's a bit weird regardless, though, since it's unbounded below

getafix

yeah..

I can't seem to get good intuition for it, and i can't seem to find a good enough source that discusses the subtleties in detail..

Semiclassical

yeah, and if it's in the context of wkb it's typically a bit annoying

the results you get from wkb are great, but the derivations...yuck

getafix

04:46

yeah i am trying to look at transitions between a|x| and infinite box, with a slanted wall, ...and well in the large L limit (length of the box)..it is just a very shallow linear potential..I can sorta get the solutions to work out, but trying to write reasonable explanations and give me self some solid intuition is hard

Slereah

07:22

morning

Slereah

07:47

Wait

I think I got it

Hm, not sure

But

Maybe you can prove a manifold with two disjoint boundaries has two disjoint collared neighbourhoods from the fact that manifolds are T5 spaces

Anonymous

07:59

@Semiclassical 1. I think that the variance will vary with the nature of the tree 2. I don't think it is logically correct to model the number of edges as a binomial distribution. The reason being that the n-1 experiments aren't exactly independent, unlike the n experiments for the nodes. Consider a tree with a single root node and a million leaf nodes, and compare it with a binary tree with a million nodes for example

Anonymous

Could you run your experiments on a binary tree and check the empirical result once? I have a feeling that for binary trees, our model's variance will exactly match

Anonymous

@AvnishKabaj Meh, it was too easy and boring ;) Google "Markov chain problems"

Avnish Kabaj

08:40

@Blue oh no

Anonymous

@AvnishKabaj The transition matrix method is pretty useful for these kinds of problems. The key is to first represent the whole situation in form of a state diagram

Avnish Kabaj

09:38

Thanks

Waittaminut

You haven't given the answer

You may be wrong for all I know

Anonymous

I am not feeling like doing calculations now :)

Anonymous

Another day

Avnish Kabaj

09:50

You literally have to just multiply 3 fractions

Slereah

10:33

Trying to prove that the geodesic equation is unique when crossing a discontinuity

the horror

Secret

10:48

essential discontinuity = $\infty$ the horror

Anonymous

11:17

@AvnishKabaj Okay, if you insist :P Wolfram gives me $1/3$ if I begin in state B

Anonymous

Matrix $N$

Anonymous

Matrix $B=NR$

Anonymous

Absorbing Markov chain

In the mathematical theory of probability, an absorbing Markov chain is a Markov chain in which every state can reach an absorbing state. An absorbing state is a state that, once entered, cannot be left. Like general Markov chains, there can be continuous-time absorbing Markov chains with an infinite state space. However, this article concentrates on the discrete-time discrete-state-space case. == Formal definition == A Markov chain is an absorbing chain if there is at least one absorbing state and it is possible to go from any state to at least one absorbing state in a finite number of steps.In...

Semiclassical

11:32

@Blue agreed on both counts. But without that assumption I’m not sure how one argues that we expect $(1-p)^2*(n-1)$ edges to survive

As for binary trees, it’s worth a try. I’ll see if I can figure out how to generate them in mma

Slereah

12:08

The Thing - Norwegian folksong "Sámiid Ædnan" HD

►

Anonymous

@Semiclassical I don't think we need the independence assumption since we're talking about the expectation value of the number of edges here. Expectation value of the sum of random variables is equal to the sum of expectation values of the random variables

Anonymous

We need to use the linearity property I think

Anonymous

Let's see it this way:

Anonymous

Assign a random variable to each edge

Anonymous

$X_1,X_2,...,X_{n-1}$

Anonymous

12:13

$X_i$ takes up values $0$ (deleted) or $1$ (not deleted)

Anonymous

$E[X_i] = (1-p)^2.1+(1-(1-p)^2).0$

Anonymous

$\implies E[\sum_i X_i] = (n-1)(1-p)^2$

Anonymous

We don't need independence of the $X_i$'s to use this

Anonymous

They might be dependent. But that will only affect the variance. Not the mean

Avnish Kabaj

@Blue cheater

Anonymous

12:19

@AvnishKabaj Erm?

Anonymous

Wolfram only did the matrix multiplications for me...lol

Semiclassical

@Blue well, that’d be consistent with what I’ve sern

The way for me to test this would be to pick an edge ahead of time, then get statistics on whether said edge survives a pruning

Anonymous

Yeah, or just test on different trees

Anonymous

I'm pretty sure for a binary tree it will follow the binomial model

Anonymous

But I don't know how to prove it

Anonymous

12:27

What's up with the star frenzy

Semiclassical

People love constellations I guess

Anonymous

lol

Semiclassical

I think a key complication is that p(X1,X2) will depend on whether X1,X2 have a common vertex

If they don’t, then i think it factorizes to p(X1)*p(X2)

But not if they have a vertex in common, since this vertex being deleted will affect both

Anonymous

That is true :/

Anonymous

Even for binary trees actually (and if they're not independent they won't follow the binomial distribution and my intuition would be false)

Anonymous

12:36

There might be some other cancelling effect too

Anonymous

Gotta think

Anonymous

Can you think of a case where the edge probabilities are always independent?

Semiclassical

One convenient point is that Xk^n = Xk for any n>0

Anonymous

What is Xk?

Avnish Kabaj

@Blue yes

You weren't able to do it

Na na na na na

:P

Anonymous

12:44

@AvnishKabaj Answer is wrong?

Anonymous

@AvnishKabaj We're talking about a different problem

Semiclassical

An interesting question that comes out of this: Suppose I take (X1+X2+...+Xn)^p and expand that out, then relplace Xk^n -> Xk. I’ll still have some polynomial in these variables, but each term contains at most one factor of Xk. How would I describe the coefficients?

Anonymous

@Semiclassical Multinomial theorem? :P

Semiclassical

For instance one has (X1+X2)^p -> X1+X2+(2^p-2)X1*X2

Anonymous

No idea what we're doing any more

Anonymous

12:53

lol

Semiclassical

Well, suppose you want to compute E[(X1+X2+...+Xn)^p]

You can expand that out, and then use the fact that any power of Xk is just Xk itself since it only takes the values 0,1

So that reduces the problem to finding the expected value of that new polynomial

But I’m not sure about the coeffs

Anonymous

@Semiclassical But why are we raising the sum to the power $p$?

Anonymous

I'm not following the significance of the polynomial

Avnish Kabaj

2 hours ago, by Blue

@AvnishKabaj Okay, if you insist :P Wolfram gives me $1/3$ if I begin in state B

Semiclassical

Because for p=1 it’s the mean number of edges deleted and for p=2 it’s the variance

Avnish Kabaj

12:58

Ooh

Got it

Semiclassical

And more generally it’ll give the higher moments

Anonymous

Oh, I thought p is the probability...lol

Anonymous

Gotcha

Semiclassical

Ah

Wanted to pick something other than n or k

Anonymous

But one point is that $$\text{Var}(X+Y) = E[(X+Y)^2]-E[X+Y]^2$$

Anonymous

13:00

Not what you wrote I guess

Anonymous

But we can get the moments for them, yes

Semiclassical

you're right, I was forgetting that the variance is a central moment

Anonymous

@AvnishKabaj I've no idea what you're saying XD

Semiclassical

the solution to the riddle comes out today, btw, so I'm curious whether this aspect gets acknowledged or not

Avnish Kabaj

Me neither

Semiclassical

13:03

since they only ask about the expected value, they may not touch on it

oh, it's already up

fivethirtyeight.com/features/…

looks like they come to the same answer but with a bit different logic

Anonymous

I don't see them discussing the variance

Semiclassical

if you look at the 4 solutions they link from Tim Black, the first one is basically ours I think

Anonymous

Nice! I'll get back and check them

Anonymous

Gotta rush now

Semiclassical

see ya

Captain Bohemian

13:29

what is the difference between global diffeomorphisms and local diffeomorphisms? are global difeomorphism the coordinate transformations on a trivilized manifold (a manifold which can be covered by a single frame of ccordiantes without singularity)? and local diffeomorphisms are the coordinate transformations on a manifold which needs to be covered by more than one frame of ccordiantes to be free of singularities?

Slereah

13:59

Oh man

Do you know that Thing

Like you're looking for a specific theorem

that may not exist

And you find one paper

that isn't quite what you want

but it's the closest

so you have to read it

"Theorem 3.1. ([55, Theorem 2]) Let $(M, g)$ be a smooth manifold with a $C^{0,1}$-semi-Riemannian metric $g$. Then there exist Filippov solutions of the geodesic equations which are $C^1$-curves."

There's the bastard

Anonymous

Such monstrosity :)

Emilio Pisanty

@Slereah why've you suddenly been saddled with a multi-headed, multi-winged albatross?

2

Slereah

@EmilioPisanty It's the cover of one of the Spivak book

Emilio Pisanty

@Slereah yes, I know

Semiclassical

@blue yeah, looks empirically like the probability of a particular edge surviving is indeed just $(1-p)^2$

Emilio Pisanty

14:05

same question, though

Slereah

It is great

is why

arxiv.org/pdf/1409.1782.pdf

the paper btw

if you wish for something you'll never need to use

Anonymous

@Semiclassical Nice! That makes sense too

Semiclassical

yeah

That still leaves open what the distribution of "# of surviving edges" will be

and I think that's indeed non-universal, with it depending on the degree distribution of your initial tree

Anonymous

BTW Mathematica has this CompleteKaryTree function in built

Semiclassical

nice

I was just creating random trees using this answer: mathematica.stackexchange.com/a/11633/26724

Anonymous

14:10

@Semiclassical Yeah, but the math gets complicated real soon :/ Even for a complete binary tree, I'm not very sure what the edge inter-dependencies would be

Semiclassical

yep

computing the second moment of that variable requires computing E[Xj*Xk] for all distinct pairs of edges j,k

And that, in turn, will depend on whether Xj*Xk have a common vertex

in principle you could do similar computations for higher moments, but then E[X1*X2*X3] will involve a bunch of casework

and so on

I think the variance may just be doable, but anything above that is probably prohibitively tedious

looks like: If X1,X2 have a vertex in common, then P(X1=1,X2=1)=(1-p)^3, P(1,0) = P(0,1) = p(1-p)^2, and P(0,0)=1-P(0,0)-P(1,0)-P(0,1) = p(1+p-p^2)

So E[X1*X2]=(1-p)^3

Anonymous

Well, there are general formulae for the higher moments - skewness, kurtosis and so on (each moment can be recursively calculated using the knowledge of the previous moments). The main task would be to define the joint probability distributions of $X_i,X_j$'s first in a general form

Semiclassical

Another nice thing: X1*X2=0 unless both = 1

So E[X1*X2]=p(X1=1,X2=1)

Which is (1-p)^4 if the edges have no vertices in common and (1-p)^3 if they do

Anonymous

Seems right

Semiclassical

So now it just comes down to how many of the (n-1)(n-2) edge pairs between the n vertices will contain a common vertex

one nice feature of that: $(1-p)^4\leq E[X_1X_2]\leq (1-p)^3$

which implies some upper/lower bounds on the variance of the total # of edges surviving

So while the variance is indeed graph-dependent, it can't vary too drastically between different graphs

so that's neat

Anonymous

14:29

Yep :)

Anonymous

It changes when you allow loops though

Anonymous

But we're talking about trees so it's okay

Semiclassical

yeah

Nobody recognizeable

Can i ask a homework like question in sr here ?

Sir Cumference

If you study graphs in which edges can link more than two nodes, you're more properly called a hyperedgelord.

Semiclassical

14:30

a graph theory phd is the king of the nodes

Anonymous

@SirCumference Man, I'd love to do a graph theory PhD :P

Sir Cumference

@Blue I just wish I could fit in a graph theory class :P

Got a glimpse of it once, it's pretty interesting stuff

Anonymous

Iirc Niel de Beaudrap's PhD was on algebraic graph theory and quantum algorithms

Anonymous

Pretty cool guy. He's on SE

Secret

Quantum graph

In mathematics and physics, a quantum graph is a linear, network-shaped structure of vertices connected by bonds (or edges) with a differential or pseudo-differential operator acting on functions defined on the bonds. Such systems were first studied by Linus Pauling as models of free electrons in organic molecules in the 1930s. They arise in a variety of mathematical contexts, e.g. as model systems in quantum chaos, in the study of waveguides, in photonic crystals and in Anderson localization, or as limit on shrinking thin wires. Quantum graphs have become prominent models in mesoscopic physics...

Novalium Company

14:36

Sup buds. A quick question. Why in special relativity, the inertial reference frame must have no acceleration (constant velocity)? For example in the twin paradox, we consider the frame of reference to be the Earth because it has no acceleration and not the ship which accelerates?

Secret

Geometry in the most minimalistic sense, are time independent relations between primitive objects

Therefore statistics are out

Semiclassical

as in: why can't we imagine that we remain at rest in the spacecraft and the earth moves?

Secret

In other news acceleration = gravity, frame is not inertial

Novalium Company

Meh, I guess it's complicated. @Semiclassical We can image that?

Semiclassical

well, let's make our lives simpler and forget "the earth" for a moment, since that's a big object with a lot of structure

just pick two spaceships. one remains at rest, the other rockets away and then rockets back

Novalium Company

14:45

Alright.

Semiclassical

From the perspective of the second ship, we might say that the first ship is the one that moves away and comes back. So how can there be any asymmetry between the two scenarios?

Novalium Company

I don't know. I was just watching a Professor Dave Explains video.

Slereah

Because that is WRONG

Novalium Company

I guess there is no way to tell?

Slereah

DUN DUN DUUUUN

Semiclassical

14:47

Well, of course it's wrong.

Slereah

People always say you don't need GR for this but I say balderdash

Semiclassical

lol

Slereah

the situation becomes very easy in GR

Novalium Company

So is there an answer how to solve the problem?

Slereah

the first ship is in Minkowski space while the second is in Rindler space!

that is what the switch does

Semiclassical

14:48

I think there's at least one decisive difference between the two scenarios: The person in the first ship never perceives any external force, whereas the second person observes a whole heck of a lot of it as they decelerate and turn around

Slereah

yes

Semiclassical

so while the perceived motions may be equivalent, the two observers have very different experiences otherwise

Slereah

that is the big issue

Gravitational acceleration and other types of accelerations are treated differently

Well, coordinate acceleration here

Semiclassical

right

Novalium Company

youtube.com/watch?v=iIEeSiT3SI4 at 6:14. That's all I know. I have no GR knowledge, yet.

Semiclassical

14:50

I mean, i wouldn't necessarily call this a resolution as such; I don't know GR enough

Novalium Company

I guess that's a too advanced problem and I shouldn't care for now?

Semiclassical

but it at least points to the pertinence of acceleration as the real issue

Sir Cumference

@NovaliumCompany You're better off working up an intuition for classical mechanics before SR

Admittedly though my SR intuition is far better than my classical mechanics intuition

Semiclassical

Of course, this is already implicit in the notion of it being an 'inertial reference frame'

Slereah

just read actual books rather than watching meme videos you tool

Semiclassical

14:52

the moving ship is not an inertial frame, at least not while they're decelerating/accelerating

Novalium Company

@SirCumference Of course, but I'm not jumping into SR seriously. I just wanna know the basic thoughts and concepts. Math is too complicated for 16 y/o who barely understands calculus.

Semiclassical

@NovaliumCompany I think the basic lesson is this: If you want to have results in SR which you can trust without issue, work in an inertial reference frame

Novalium Company

@Semiclassical (thumbs_up)

Semiclassical

otherwise you run into subtleties which you arguably need GR to make real sense of

Slereah

SR is really complicated enough as it is

There's tons of shit they don't really talk about in most SR courses

Semiclassical

14:54

lol

Sir Cumference

@NovaliumCompany Well the thing is SR doesn't deal much with changing functions (e.g. velocities are constant, etc.) so calculus isn't required. GR on the other hand does, and is full of calculus.

Linear algebra is helpful for SR tho

Semiclassical

it's also helpful in QM, but for rather different reasons

Slereah

It's helpful in all of physics

Semiclassical

yeah

it's just that different physics topics use different aspects

Sir Cumference

Although I do think it's a good idea to do basic qualitative reading on higher physics. Had I not known the cool stuff that would be waiting, I'd probably have given up in the early courses

It serves as a battery of motivation in a sense

Plus they kind of expect you know the gist of the subjects beforehand when teaching them.

John Rennie

15:08

@NovaliumCompany this might be interesting:

73

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...

Sir Cumference

@JohnRennie Solution: don't think about it :P

John Rennie

It explains how we handle acceleration in special relativity.

Sir Cumference

Although solving the twins paradox requires calculus @NovaliumCompany

John Rennie

@SirCumference the thing is that the twin paradox only puzzles people because of the piss poor way SR is taught to undergrads. If it was taught properly from the outset no-one would be confused.

Semiclassical

I feel like that answer boils down to: How you understand distance (in the sense of spacetime interval) has to change when you take into account acceleration

Sir Cumference

15:10

@JohnRennie Very true. My only intuition for it at this point is "do the math, you'll see how it works out"

John Rennie

The elapsed time for an observer is the length of their world line. That's all you need to know to understand the twin paradox.

Sir Cumference

Ah right, now I remember. Their worldline is curved and longer right?

Semiclassical

ehh. you also have to be able to understand why your metric should depend on whether you're accelerating or not

Sir Cumference

Been like a year since my SR class

user1414

yup, special relativity is a special case

John Rennie

15:13

@Semiclassical the length of the world line is an invariant. You can use whatever metric is most convenient to calculate it.

user1414

the "piss poor way" SR is taught is also the traditional way :-)

Sir Cumference

^

Semiclassical

sure. and once you declare that the appropriate metric in an inertial reference frame is just the Minkowski metric, then you can work out what the appropriate metric is for a non-inertial frame

Sir Cumference

My course barely mentioned Minkowski diagrams, yet those are the key to understanding it

Semiclassical

but you still need to be able to ground that in the inertial frame

user1414

15:19

@Semiclassical have you TAed any SR courses?

Semiclassical

newp

closest would be Electromagnetism, and it really didn't go into SR at all

(in principle one can talk about SR at the end of an E&M class, but usually there's not time for it in a single semester course)

user1414

in principle SR and GR can be taught back to back :-)

Nobody recognizeable

@JohnRennie can you confirm one solution in a problem of sr here ?

@user1414 yep if ug students hadnt decide to kill themselves then thats the best way to fasten their death. :p

John Rennie

@Nobodyrecognizeable I've just started eating lunch. Ask me tomorrow.

user1414

what's on the menu?

Nobody recognizeable

15:25

@JohnRennie ok. I think got simultaneity preserved as i have to go for dinner as well:-

John Rennie

@user1414 chicken risotto! :-)

user1414

Niiice

:-)

Nobody recognizeable

bbc.com/food/recipes/easy_chicken_and_pea_31687

Yummy

@JohnRennie serves food to us. Lets eat!

user1414

@Nobodyrecognizeable what are you having for dinner?

Anonymous

I'm having boiled rice and veggies :P

user1414

15:32

are you vegetarian?

Nobody recognizeable

@user1414 well i have to go and eat bread @Blue i dont know how to convert roti in english and some vegetable. So goodbye all.

John Rennie

user1414

@Nobodyrecognizeable cya pal

Anonymous

@user1414 Nah, having some acidity issues since yesterday. Also caught a cold due to the weather change. So having light food today

Anonymous

@Nobodyrecognizeable Roti is called Roti in English :)

user1414

15:35

or flatbread

> The word roti is derived from the Sanskrit word रोटिका (rotikā), meaning "bread".

Anonymous

@user1414 Roti is a subset of Flatbread

user1414

indeed

ACuriousMind

I just had burgers.

user1414

McDonalds?

ACuriousMind

And I have hext week off, so that's nice :)

@user1414 Ew, no. There's a little burger joint not far from where I live

John Rennie

15:39

Burgers ... I haven't had burgers for ages. Maybe this weekend.

user1414

Chicken burgers are nice...

Anonymous

I mostly eat the insides of the burger anyway and leave the bread

user1414

that defeats the purpose :P

Anonymous

:P Very few places can make the bread well

Anonymous

Some burgers breads are good

Anonymous

15:43

But most are not

user1414

they're called "buns"

Anonymous

whateva :)

user1414

:P

John Rennie

@user1414 Not in the USA. The word buns means buttocks there.

Anonymous

lolol

John Rennie

15:44

I don't think I'd want to eat a burger between a pair of buttocks.

user1414

bun

Noun: bun (plural buns)

A small bread roll, often sweetened or spiced.
A tight roll of hair worn at the back of the head.
(Ireland) A cupcake.
(slang, Britain) A drunken spree.
(Internet, slang) A newbie.

(3 more not shown…)

Verb: bun (third-person...

ACuriousMind

@JohnRennie Don't google it.

Anonymous

It's getting too visual now

Anonymous

:P

John Rennie

@ACuriousMind an image search for buns does give me pictures of bread. Mind you, that's google.co.uk so it knows I'm English.

user1414

15:47

> 2. A tight roll of hair worn at the back of the head.

or one on each side...

John Rennie

Aha, a Star Wars reference. I always thought Princess Leia had nice buns.

Avnish Kabaj

Yes yes

We have reached 7th grade territory

user1414

you?

Anonymous

@JohnRennie Double entendre?

Anonymous

Okay, don't answer that ^

John Rennie

15:51

@AvnishKabaj 12 to 13 years old? Yes, that's about my mental age :-)

Anonymous

These antacids are cool stuff

Anonymous

Worked faster than I expected....1 hr

user1414

Tums?

Anonymous

Gelusil

Avnish Kabaj

@JohnRennie [insert boys grow old not up meme]

Anonymous

15:56

They are pretty colorful too :D

vzn

lol 7th grade level? great minds think alike :P independent.co.uk/news/world/americas/us-politics/…

danielunderwood

We call them buns over here too

Or at least in my region of the US

enumaris

hmmm

danielunderwood

I've certainly never heard of a bun as a drunken spree though

user1414

sir mixalot does too :P

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