The h Bar

0celo7

16:24

back

ok, let's do this

correct way?

bah

overrated

why is Arnold not here ;_;

@ACuriousMind $g=\phi^*e$

like that?

I'm lost :(

ACuriousMind

@0celo7 What is $g$ and what is $\phi^*$ supposed to be?

0celo7

:(

$g$ is the metric on $V$

ACuriousMind

Inner product, not metric

0celo7

same thing

ACuriousMind

It induces a metric, but you only talk about a metric $g$ where you're on a manifold with a metric tensor

This is just a silly vector space

0celo7

16:27

$\mathbb{R}^n$ is a manifold

ACuriousMind

Anyway, what $\phi^*$ supposed to be and what is $e$

@0celo7 But we are not interested in that structure right now

0celo7

$\phi^*$ is the pullback by $\phi$

and $e$ is the Euclidean metric

but you don't want that

I have no clue what $\phi$ is!

I know it's the vielbein from GR

wait, aren't all vector spaces of dimension $n$ isomorphic

or is that what we're proving

ACuriousMind

@0celo7 That's what we are proving, with the added point that the vector spaces are inner product spaces.

0celo7

oki

ACuriousMind

So, just take the orthonormal basis $e_i$ in $V$ and send it to an orthonormal basis in $\mathbb{E}^n$.

0celo7

16:29

so we know $(e_i,e_j)_V=\delta_{ij}=(\phi(e_i),\phi(e_j))_\mathbb{E}$

ACuriousMind

I.e. define $\phi(e_i)$ to be the vector with $1$ at the $i$th place and $0$ otherwise.

0celo7

GOD DAMN IT

I WAS GETTING THERE

sigh

ACuriousMind

Sorry :(

0celo7

so close

ACuriousMind

But your talk about vielbeins and pullback seemed to me as if you weren't getting there.

0celo7

16:31

I had that $\phi(e_i)$ is an orthonormal basis

I had it

so...what does that have to do with mechanics?

ACuriousMind

Well, the proof that all Galilean spaces are isomorphic goes essentially the same

0celo7

:O

so we have an affine space

ACuriousMind

And knowing that they are all isomorphic, we can be sure that just saying "the setting of mechanics is a Galilean space" fixes the setting uniquely.

0celo7

yeah I get that

ACuriousMind

Okay

0celo7

16:33

but I don't see how they are the same

ACuriousMind

Well, you need to construct an isomorphism again

I.e. take some Galilean space $G$ and find an isomorphism to $\mathbb{R}\times\mathbb{R}^3$.

0celo7

oh, so that's what they actually mean

question: how is space defined for $G$

like how do you have a "time" and "space" direction

how do you separate the two

if two points are in the kernel of $t$, they're in the same "space slice"?

ACuriousMind

@0celo7 Well, $G$ is a 4D affine space with that time map $t : \mathbb{R}^4 \to \mathbb{R}$ on its parallel displacements

@0celo7 Careful, the time map doesn't act on points, it acts on the parallel displacements

0celo7

ok, what exactly does that mean

oh

ACuriousMind

@0celo7 page 4, bottom

0celo7

16:38

parallel displacements are just the vectors connecting points

?

ACuriousMind

@0celo7 Yeah, pretty much

0celo7

got it brah

well

I get that part :)

not how to do the problem, gotta think about that

ok, pick $a,b\in A^4$. If $t(a-b)=0$, these points are simultaneous...so take some other point $c$, for which $t(a-c)=t_0\ne0$. Then solve $t(c-d)=0$ and construct the spatial slice at $t_0$...and so on

in this manner you get spatial slices and times, which gives $\mathbb{R}\times\mathbb{R}^3$

ACuriousMind

Correct idea, I'd say

0celo7

how do I make it formal?

CaptCoonoor

Hello People, Good Day

0celo7

16:46

Good day to you too

ACuriousMind

@0celo7 By giving the isomorphism. You've described how to construct it, but you haven't explicitly stated how to define it

CaptCoonoor

I have a question

0celo7

Hint?

I don't know how to define it explicitly

ACuriousMind

@CaptCoonoor Just ask it, no need to ask for permission

CaptCoonoor

Does light ever reach singularity?

If space time bends to infinity there
Like lim --> infinity

0celo7

16:47

@ACuriousMind Speak for yourself! For my help one must file a competitive application.

ACuriousMind

@0celo7 You've picked $a$ there. Start by defining what you want to send it to

0celo7

@CaptCoonoor I have no idea what that means.

ACuriousMind

21

Q: Does any particle ever reach any singularity inside the black hole?

I am not a professional physicist, so I may say something rubbish in here, but this question has always popped in my mind every time I read or hear anyone speak of particles hitting singularities and "weird things happen". Now to the question at hand, please follow my slow reasoning... As far a...

0celo7

ohhh

CaptCoonoor

Thanks @A

@ACuriousMind

0celo7

16:49

@ACuriousMind Well, make $a$ the origin?

ACuriousMind

@0celo7 Valid choice.

What's the image of $b$, then?

0celo7

Well, I have to draw the vector $b-a$ and assign it time $t_b=t(b-a)$

ACuriousMind

@0celo7 The vector $b-a$ is 4D, and you stated that $t(a-b) = 0$.

0celo7

I did say that

Well, if they're simultaneous the vector is not 4D...

ACuriousMind

The displacement is always a 4D vector, no matter what the points are.

0celo7

16:53

it lies in a 3D subspace

ACuriousMind

@0celo7 Yes, correct. (A single vector even lies in a 1D subspace, always! ;) )

0celo7

but all points $b$ define a 3D subspace

ACuriousMind

Correct, the space $t^{-1}(0)$.

0celo7

yes

ACuriousMind

Now, the Galilean structure includes an inner product on that space, and it comes in handy that we proved all inner product spaces of same dimension are isomorphic

0celo7

16:56

:)

ACuriousMind

So, we can use that to get an isomorphism $\psi : t^{-1}(0) \to \mathbb{R}^3$.

Secret

QUOTE:
Mass is defined asymptotically, yes, but the mass term always comes in the combination Mr, so when r→−∞, where the angular part of the metric goes as r2, you have to replace r by |r| to see what happens. So M goes to −M. Look up the Penrose diagrams in Hawking and Ellis for example. And look up the Kerr metric (too long for these "comments"). If you do AdS/CFT you are doing quantum, and you won't avoid thermalization. Only the classical theory would seem to allow you to get out, God knows in which universe– G. 't Hooft Sep 6 '12 at 16:18

ACuriousMind

What's left for you to go is to get a map $G\to\mathbb{R}\times\mathbb{R}^3$ out of this.

Secret

http://physics.stackexchange.com/questions/211615/how-can-we-determine-the-sizes-of-atoms-and-subatomic-particles
---
No one quotes wikipedia in their paper, and this "paper" looks bogus

0celo7

17:11

@ACuriousMind Makes sense

@ACuriousMind well that's tricky

let me think on it

(busy in lab, so chatting is intermittent)

obe

Practical lab or research lab?

DanielSank

@JohnDuffield The particles within the stress ball all move away from one another in the same way that the confetti particles in a balloon move away from one another if I pop the balloon. What is the data that supports the universe being like a stress ball and not like a popped balloon full of confetti?

@ChrisWhite I believe you are the first person to tell me that this is subtle rather than telling me "because expansion!"

0celo7

17:27

@obe chilling with the grad students in the lab

obe

Cool guy.

Secret

Thought experiment (???)
Is it possible to modify the double slit experiment as follows:
Instead of having a phosphor screen where electrons hit it, it is being replaced by an array of magnets so that for each path the electron will be routed back to the source so that they are fired again at the double slit

Now make some measurements in this system with a detector being placed at one slit (so as to collapse the intereference pattern), and then repeat the measurement again without the detector (so that the intereference pattern can appear again)

user54412

@DanielSank Have you read this paper by Davis and Lineweaver? It's on my list of one of the best papers ever written in astrophysics, precisely because of how many people miss the subtleties of expansion.

3

0celo7

@ChrisWhite I'm gonna be the official undergrad Raman expert

Secret

---
Background of this thought experiment. After reading a bit about matrix mechanics again in wikipedia (and have questions clarified by Acuriousmind). I learnt that the state vector collapses after measurement because it is being mapped to one of the observable's eigenstates (e.g. position or momentum)

For conjugate variables that don't commute this result in the state of the system being mapped to one observable having some well defined eigenstate while the conjugate observable become a superposition of eigenstates of that other observable

0celo7

17:31

I get to do lazar stuff!

user54412

@0celo7 Better to be a ramen expert, but life isn't perfect I suppose...

0celo7

I don't even like Ramen

Secret

I then start to wonder what make the particle gain quantum properties. Because if wavefunction collapse is irreversible as said in the copenhagen interpretation, then an electron that is beign measured in the doubel slit experiment should not be able to produce an interference pattern again if we route it back to the source and fire it again

(But I felt my claim is wrong) therefore I am curious on how they kinda of "reset" themselves so that they can form intereference pattern again?

user54412

@DanielSank In particular, Appendix B of that paper is golden. It catalogs famous smart people getting the whole thing wrong, including Feynman, Weinberg, Peebles, Taylor, Wheeler, and a bunch of textbook authors.

obe

@0celo7 Dr. Octagonapus?

Secret

17:39

Schematics of my thought experiment:

Basically double slit but the phosphor screen is modified so that the detected electrons can be sent back to the source

Need to figure out how to calculate the interaction of the electron with the electric field in the tubing though, cause I felt like the interaction there can possibly screw up what I am trying to probe

Glen The Udderboat

@Secret Is that your head?

Secret

@GlenTheUdderboat it looks nothing like my head as far I can see

ACuriousMind

@GlenTheUdderboat No, it's what is inside of it :P

obe

Depends what you mean by inside.

Glen The Udderboat

@ACuriousMind So, it is a thought-selfie?

ACuriousMind

17:43

lol, probably

Glen The Udderboat

A thoughtie.

Secret

Basically, the reading on matrix mechanics cause me to became curious on
1. whether we can reroute the detected electron to fire at the double slit again
2. If wavefunction collapse is irreversible, then is it possible for this detected electron to produce an interference pattern again in the array of detectors when fired at the double slit?

0celo7

@obe Huh?

obe

?

ACuriousMind

@Secret In principle, yes to both if you find a position detector that doesn't effectively destroy the electron, but why would you be interested in that?

0celo7

17:47

Who is Dr. Octaganopus

obe

Lazors.

sigh...

0celo7

ah, now I remember

obe

lol

0celo7

@ACuriousMind still stuck D:

I'm not sure how to construct the time axis...

ACuriousMind

@0celo7 Observe that we get an isomorphism $\psi_{t_0} : t^{-1}(t_0)\to\mathbb{R}^3$ for any choice of $t_0\in\mathbb{R}$.

0celo7

18:01

@ACuriousMind I have observed this

wait...

what

I have not observed this

Secret

@ACuriousMind
Based on what I understand from reading that brief section on matrix mechanics in wikipedia, a measurement will map the state (e.g. say of an electron) into an observable's eigenstate (e.g. position eigenstate). Now since position and momentum are conjugate variables, it means the momentum state of that electron is a superposition of momentum eigenstates (hence the uncertainty)

Our setup above can presumably be described by some hamiltonian $H$ which has potential energy terms that is due to the electric field used to route the electrons back, and the kinetic energy term will…

obe

Dude I was working on that problem too. @0celo7

No wonder I failed.

ACuriousMind

@0celo7 Tell me when you are certain whether or not you have observed it ;)

0celo7

that's a very quantum question

obe

3 more until 100 profile views.

0celo7

18:04

nope

I have not observed this

$t$ acts on a 4-vector, so how can the image of the inverse be a 3-vector?

ACuriousMind

$t$ is a linear map from a 4D space into a 1D space. By the rank-nullity theorem, the preimages (in particular the kernel) are 3D spaces.

0celo7

but $t_0$ is not in the kernel

ACuriousMind

@0celo7 The preimages of other points are just "shifts" of the kernel

Secret

(I know I am bad at describign it) Give me a sec when I write out a maths version of tha long block of text...

0celo7

@ACuriousMind what...

I thought $t(b-a)$ gives the time separation between two points $a,b$

ACuriousMind

18:10

@0celo7 Take any $x\in\mathbb{R}^4$ with $t(x) = t_0$. Then for every $y\in t^{-1}(t_0)$ you have $t(y-x) = 0$, so $y-x$ is in the kernel. Conversely, for every $c$ in the kernel, you have $t(c+x) = t(c)+t(x) = 0 + t_0 = t_0$, so $c+x\in t^{-1}(t_0)$ this shows that the shift by $x$ gives a bijection between $t^{-1}(0)$ and $t^{-1}(t_0)$.

And that's what I mean with all preimages being "shifts" of the kernel.

Since it's a bijection, every other preimage is also a 3D (affine) subspace.

@Secret Unless the Hamiltonian of that system is the position operator, a position eigenstate will not stay a position eigenstate under time evolution.

What you're trying to do doesn't make sense, sorry.

0celo7

We have the problem where I understand the math but I can't picture what you're talking about.

ACuriousMind

@0celo7 then let's go to a 3D example briefly, because that we can actually picture

0celo7

Ok.

ACuriousMind

Take the projection of $\pi : \mathbb{R}^3\to\mathbb{R}$ onto the $z$-axis

0celo7

ok

ACuriousMind

18:14

The kernel is the $x-y$-plane at $z = 0$

0celo7

yes!

ACuriousMind

The preimage of $1$ is an $x-y$-plane at $z = 1$

0celo7

yes

ACuriousMind

It's the "same plane", just shifted by $1$

0celo7

naturally

ACuriousMind

18:16

So, you have $\pi^{-1}(0)\cong \mathbb{R}^2$ (it's a plane, after all), and also $\pi^{-1}(1)\cong\mathbb{R}^2$

And likewise $\pi^{-1}(z_0)\cong\mathbb{R}^2$ since all these preimages are just planes at $z = z_0$.

0celo7

ok

ACuriousMind

So, I want to do the same one dimension higher, and have isomorphisms $\psi_{t_0} : t^{-1}(t_0)\to\mathbb{R}^3$.

0celo7

waaaaaait

but $t$ gives the time separations!

so in your example above the x-y plane at z=1 is in the kernel

ACuriousMind

No

0celo7

because those vectors have 0 z separation

ACuriousMind

18:22

$t$ is not a map on the Galilean space, it is a map on the displacements

It doesn't really give the time separations, it tells you how much of a given displacement points into the time direction

Only if you apply it to the displacement between two points you get the time separation

But we're not yet worrying about the point in the Galilean space itself, all the maps I've written down so far are just maps on the vector space of displacements

0celo7

but in your example, the vectors in the x-y plane have no component along the z direction

so the projection is 0!

ACuriousMind

No.

0celo7

doesn't matter where along that axis they are

ACuriousMind

The vector is the arrow from the origin to the point

And for the planes with $z\neq 0$, that arrow has a component in $z$-direction

0celo7

ok

ACuriousMind

18:26

that's why I'm currently talking about the displacements, because they're a proper vector space and have an origin.

0celo7

ok

what are we talking about again

ACuriousMind

Well, the ultimate goal is to get an isomorphism $\phi : G\to\mathbb{R}\times\mathbb{R}^3$ for any Galilean space $G$.

0celo7

why is $t^{-1}(t_0)$ 3-dimensional

ok

ACuriousMind

To that end, you took any point $a\in G$ and defined $\phi(a) = 0$.

0celo7

seems to me like you need 4 dimensions for that D:

ACuriousMind

18:29

We're currently tackling the questions how to define the images of other points.

0celo7

I still don't get why that's 3-dimensional

ACuriousMind

@0celo7 But you agreed that the preimages in the 3D case were just planes!

0celo7

well now I disagree because I had the wrong picture

ACuriousMind

Now, in this case, they're all just 3D spaces, in exactly the same way.

@0celo7 You had the wrong picture how the projection is defined, not of what the preimages look like, I'd say.

0celo7

so in the 3-D example

so the vectors go from the origin to the plane

ACuriousMind

18:32

What might be confusing is that the preimages that aren't the kernel are not vector spaces, they're just affine spaces.

0celo7

and we're saying the tips form a plane?

ACuriousMind

@0celo7 Yes, exactly

0celo7

ok but how do you specify the $z$ placement of the plane

ACuriousMind

@0celo7 The plane at $z_0$ is $\pi^{-1}(z_0)$.

I'm not sure if that's your question, though

0celo7

ok ok

so

$t(x)=0$ is a 3-space

ACuriousMind

18:34

You mean, the space of vectors $x$ fulfilling that equation? Yes.

0celo7

and $t(x)=t_0$ is also a 3-space

ACuriousMind

Yes

0celo7

um

wtf is wrong with me

of course it is

well now I get it intuitively

now how to show it mathematically...

ACuriousMind

25 mins ago, by ACuriousMind

@0celo7 Take any $x\in\mathbb{R}^4$ with $t(x) = t_0$. Then for every $y\in t^{-1}(t_0)$ you have $t(y-x) = 0$, so $y-x$ is in the kernel. Conversely, for every $c$ in the kernel, you have $t(c+x) = t(c)+t(x) = 0 + t_0 = t_0$, so $c+x\in t^{-1}(t_0)$ this shows that the shift by $x$ gives a bijection between $t^{-1}(0)$ and $t^{-1}(t_0)$.

I did that already ;)

0celo7

well I didn't understand that

I get $y-x\in \operatorname{ker}t$

and I get $t(c+x)=t_0$

and I get $c+x\in t^{-1}(t_0)$

bolbteppa

18:39

@0celo7 this exercise is a good example of why a proper course in linear algebra would be so useful

0celo7

and then I'm lost

unless

no no still lost

ACuriousMind

@0celo7 Where are you lost?

bolbteppa

@ACuriousMind what was that gauge theory book you linked to before?

ACuriousMind

There's nothing left

What I have written there is the proof that the affine map $t^{-1}(t_0)\to t^{-1}(0), y\mapsto y - x$ is a bijection.

0celo7

"this shows that"

ACuriousMind

18:42

@bolbteppa Since I don't recommend that many books, it probably was Quantization of Gauge Systems by Henneaux/Teitelboim.

But it might have been something else, depending on the context

0celo7

@ACuriousMind :9

what

Slereah

Hey @ACuriousMind

I want to know how to do a QFT derivation in like

a srs manner

Starting from the Lagrangian

Is this process accurate

1) The Lagrangian

0celo7

ok and a bijection can only happen between spaces of equal dimension

alright but why is that a bijection

Slereah

2) Get the conjugate momentum and Hamiltonian

3) Get the EOM

0celo7

@Slereah seriously, can't you wait :/

Slereah

18:44

4) Quanticize all this

It can, yes

ACuriousMind

@0celo7 Because it has the inverse $c\mapsto c+x$.

0celo7

pretty sure Weinberg does that in that order

OHHHH

ok

ACuriousMind

@Slereah You don't need to get the e.o.m., I think. What do you want them for?

0celo7

so where are we now

@ACuriousMind constraints

bolbteppa

Thanks

Secret

18:46

@Acuriousmind

If the electron was not being previously measured, then it will still be in some superposition of position states as the Hamiltonian evolves it and we will expect an interference pattern

But since the above electron has been measured, will it become unable to give an interference pattern since it is being collapsed into the state $|e^-_{(2)}\rangle$, thus it can no longer be changed (since the collapse causes all other position state contributions to go to zero)?

https://en.wikipedia.org/wiki/Wave_function_collapse

ACuriousMind

@0celo7 We've finally concluded that every $t^{-1}(t_0)$ is a 3D affine space.

Slereah

@ACuriousMind I suspected as much, but then how do you get the propagator?

ACuriousMind

@0celo7 I do not count those as "e.o.m.".

0celo7

@ACuriousMind yes, I remain convinced for the time being

Secret

sorry ignore the previous one, there's a copy past problem

ACuriousMind

18:46

@0celo7 Alright

0celo7

people please just let him finish explaining this to me

Secret

@ACuriousMind
Perhaps in maths the above is written like this:
$$\hat{H}_{doubleslitsetup}=T_{e-}(\dot{\mathbf{r}})+V_{routing tube}(\mathbf{r})$$
where $V_{routing tube}(x,y,z)$ is the electric potential within the tube (Not sure how to model this in the simplest way).

Consider an initial state of the electron as a superposition of position states
$$|e^-\rangle=\sum_i c_i|x_i\rangle$$
$$\langle e^-|e^-\rangle=1$$
There exists a detector at a slit such that a position measurement causes the following:

ok much better

0celo7

so what now?

oh for fucks sake

just let him finish

obe

lol

Slereah

If you love @ACuriousMind why don't you marry him

2

0celo7

18:49

@Slereah long hair is a deal breaker

ACuriousMind

@0celo7 Okay, we have the isomorphisms $\psi_{t_0} : t^{-1}(t_0) \to \mathbb{R}^3$. Now define the image of $b\in G$ (not necessarily simultaneous) to be $\phi(b) = (t(b-a),\psi_{t(b-a)}(b-a))$

Slereah

Poor @ACuriousMind

Everyone wants him

ACuriousMind

@Slereah Ah. Well, you do not need the propagator prior to quantizing, I'd say.

Slereah

Hm

How do you construct it afterwards though?

ACuriousMind

@Slereah $\langle \phi(x)\phi(y) \rangle$?

0celo7

18:52

@ACuriousMind yes

that's it?

Slereah

Well yes I suppose, but how do you know what that is

0celo7

well that was easy

ACuriousMind

@0celo7 What's left is to check that this $\phi$ preserves all parts of the Galilean structure.

0celo7

oh fuck me

ACuriousMind

But it's built that way, so it should.

0celo7

18:53

good

how would you do that explicitly

ACuriousMind

@Slereah Show it's the Green's function for the e.o.m., calculate it ;)

Slereah

Hm, I suppose that it is related to the Dyson equation

ACuriousMind

I'm not saying you don't need the propagator, I was just weirded out by it being an extra step in your broad outline

@Slereah Yes, the Schwinger-Dyson equation is precisely the quantum version of the classical e.o.m.

Slereah

Yeah I am just trying to figure out what exactly you need

Because most QFT books are kind of a mess

0celo7

@ACuriousMind yeah I dunno what exactly to check now

Slereah

18:56

there isn't a lot of QFT books with the rigorous derivation from first principles

0celo7

the time function is just the time coordinate

ACuriousMind

@0celo7 Well, first check it's an affine map, i.e. $a-b = \phi(a)-\phi(b)$.

Slereah

there's a lot of going through classical mechanics and using a variety of other formalism

0celo7

@ACuriousMind everything here is linear

right?

Slereah

Schwinger-Dyson equation is pretty easy to do in path integrals but I dunno it in Heisenberg

Let's check zee wikipedia

0celo7

18:57

Zee? Like A. Zee?

ACuriousMind

Then check that it respects the time map, i.e. $t(b) = t'(\phi(b))$ where $t'$ is just the projection onto the first factor in $\mathbb{R}\times\mathbb{R}^3$.

Slereah

Like "the", but frencher

bc I am french

0celo7

@ACuriousMind Well that's trivial, no?

you're not french

Sam is not a French name.

ACuriousMind

Then check that it preserves the simultaneous inner product (this is done because we showed beforehand the isomorphy of inner product spaces and took the $\psi_t$ from that).

@0celo7 Yeah, that's not hard

0celo7

not hard?

literally just read it off

ACuriousMind

18:59

But these are all things you must check after you've written down such a thing

0celo7

we defined it that way

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