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4:24 PM
back
ok, let's do this
correct way?
bah
overrated
why is Arnold not here ;_;
@ACuriousMind $g=\phi^*e$
like that?
I'm lost :(
 
@0celo7 What is $g$ and what is $\phi^*$ supposed to be?
 
:(
$g$ is the metric on $V$
 
Inner product, not metric
 
same thing
 
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
 
4:27 PM
$\mathbb{R}^n$ is a manifold
 
Anyway, what $\phi^*$ supposed to be and what is $e$
@0celo7 But we are not interested in that structure right now
 
$\phi^*$ 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
 
@0celo7 That's what we are proving, with the added point that the vector spaces are inner product spaces.
 
oki
 
So, just take the orthonormal basis $e_i$ in $V$ and send it to an orthonormal basis in $\mathbb{E}^n$.
 
4:29 PM
so we know $(e_i,e_j)_V=\delta_{ij}=(\phi(e_i),\phi(e_j))_\mathbb{E}$
 
I.e. define $\phi(e_i)$ to be the vector with $1$ at the $i$th place and $0$ otherwise.
 
GOD DAMN IT
I WAS GETTING THERE
sigh
 
Sorry :(
 
so close
 
But your talk about vielbeins and pullback seemed to me as if you weren't getting there.
 
4:31 PM
I had that $\phi(e_i)$ is an orthonormal basis
I had it
so...what does that have to do with mechanics?
 
Well, the proof that all Galilean spaces are isomorphic goes essentially the same
 
:O
so we have an affine space
 
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.
 
yeah I get that
 
Okay
 
4:33 PM
but I don't see how they are the same
 
Well, you need to construct an isomorphism again
I.e. take some Galilean space $G$ and find an isomorphism to $\mathbb{R}\times\mathbb{R}^3$.
 
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"?
 
@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
 
ok, what exactly does that mean
oh
 
@0celo7 page 4, bottom
 
4:38 PM
parallel displacements are just the vectors connecting points
?
 
@0celo7 Yeah, pretty much
 
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 $\mathbb{R}\times\mathbb{R}^3$
 
Correct idea, I'd say
 
how do I make it formal?
 
Hello People, Good Day
 
4:46 PM
Good day to you too
 
@0celo7 By giving the isomorphism. You've described how to construct it, but you haven't explicitly stated how to define it
 
I have a question
 
Hint?
I don't know how to define it explicitly
 
@CaptCoonoor Just ask it, no need to ask for permission
 
Does light ever reach singularity?
If space time bends to infinity there
Like lim --> infinity
 
4:47 PM
@ACuriousMind Speak for yourself! For my help one must file a competitive application.
 
@0celo7 You've picked $a$ there. Start by defining what you want to send it to
 
@CaptCoonoor I have no idea what that means.
 
21
Q: Does any particle ever reach any singularity inside the black hole?

user1549I 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...

 
ohhh
 
Thanks @A
@ACuriousMind
 
4:49 PM
@ACuriousMind Well, make $a$ the origin?
 
@0celo7 Valid choice.
What's the image of $b$, then?
 
Well, I have to draw the vector $b-a$ and assign it time $t_b=t(b-a)$
 
@0celo7 The vector $b-a$ is 4D, and you stated that $t(a-b) = 0$.
 
I did say that
Well, if they're simultaneous the vector is not 4D...
 
The displacement is always a 4D vector, no matter what the points are.
 
4:53 PM
it lies in a 3D subspace
 
@0celo7 Yes, correct. (A single vector even lies in a 1D subspace, always! ;) )
 
but all points $b$ define a 3D subspace
 
Correct, the space $t^{-1}(0)$.
 
yes
 
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
 
4:56 PM
:)
 
So, we can use that to get an isomorphism $\psi : t^{-1}(0) \to \mathbb{R}^3$.
 
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
 
What's left for you to go is to get a map $G\to\mathbb{R}\times\mathbb{R}^3$ out of this.
 
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
 
5:11 PM
@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?
 
@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!"
 
5:27 PM
@obe chilling with the grad students in the lab
 
obe
Cool guy.
 
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
 
@ChrisWhite I'm gonna be the official undergrad Raman expert
 
---
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
 
5:31 PM
I get to do lazar stuff!
 
user54412
@0celo7 Better to be a ramen expert, but life isn't perfect I suppose...
 
I don't even like Ramen
 
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?
 
5:39 PM
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
 
@Secret Is that your head?
 
@GlenTheUdderboat it looks nothing like my head as far I can see
 
@GlenTheUdderboat No, it's what is inside of it :P
 
obe
Depends what you mean by inside.
 
@ACuriousMind So, it is a thought-selfie?
 
5:43 PM
lol, probably
 
A thoughtie.
 
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?
 
@obe Huh?
 
obe
?
 
@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?
 
5:47 PM
Who is Dr. Octaganopus
 
obe
sigh...
 
ah, now I remember
 
obe
lol
 
@ACuriousMind still stuck D:
I'm not sure how to construct the time axis...
 
@0celo7 Observe that we get an isomorphism $\psi_{t_0} : t^{-1}(t_0)\to\mathbb{R}^3$ for any choice of $t_0\in\mathbb{R}$.
 
6:01 PM
@ACuriousMind I have observed this
wait...
what
I have not observed this
 
@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.
 
@0celo7 Tell me when you are certain whether or not you have observed it ;)
 
that's a very quantum question
 
obe
3 more until 100 profile views.
 
6:04 PM
nope
I have not observed this
$t$ acts on a 4-vector, so how can the image of the inverse be a 3-vector?
 
$t$ 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.
 
but $t_0$ is not in the kernel
 
@0celo7 The preimages of other points are just "shifts" of the kernel
 
(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...
 
@ACuriousMind what...
I thought $t(b-a)$ gives the time separation between two points $a,b$
 
6:10 PM
@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.
 
We have the problem where I understand the math but I can't picture what you're talking about.
 
@0celo7 then let's go to a 3D example briefly, because that we can actually picture
 
Ok.
 
Take the projection of $\pi : \mathbb{R}^3\to\mathbb{R}$ onto the $z$-axis
 
ok
 
6:14 PM
The kernel is the $x-y$-plane at $z = 0$
 
yes!
 
The preimage of $1$ is an $x-y$-plane at $z = 1$
 
yes
 
It's the "same plane", just shifted by $1$
 
naturally
 
6:16 PM
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 $z = z_0$.
 
ok
 
So, I want to do the same one dimension higher, and have isomorphisms $\psi_{t_0} : t^{-1}(t_0)\to\mathbb{R}^3$.
 
waaaaaait
but $t$ gives the time separations!
so in your example above the x-y plane at z=1 is in the kernel
 
because those vectors have 0 z separation
 
6:22 PM
$t$ 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
 
but in your example, the vectors in the x-y plane have no component along the z direction
so the projection is 0!
 
doesn't matter where along that axis they are
 
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
 
ok
 
6:26 PM
that's why I'm currently talking about the displacements, because they're a proper vector space and have an origin.
 
ok
what are we talking about again
 
Well, the ultimate goal is to get an isomorphism $\phi : G\to\mathbb{R}\times\mathbb{R}^3$ for any Galilean space $G$.
 
why is $t^{-1}(t_0)$ 3-dimensional
ok
 
To that end, you took any point $a\in G$ and defined $\phi(a) = 0$.
 
seems to me like you need 4 dimensions for that D:
 
6:29 PM
We're currently tackling the questions how to define the images of other points.
 
I still don't get why that's 3-dimensional
 
@0celo7 But you agreed that the preimages in the 3D case were just planes!
 
well now I disagree because I had the wrong picture
 
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.
 
so in the 3-D example
so the vectors go from the origin to the plane
 
6:32 PM
What might be confusing is that the preimages that aren't the kernel are not vector spaces, they're just affine spaces.
 
and we're saying the tips form a plane?
 
@0celo7 Yes, exactly
 
ok but how do you specify the $z$ placement of the plane
 
@0celo7 The plane at $z_0$ is $\pi^{-1}(z_0)$.
I'm not sure if that's your question, though
 
ok ok
so
$t(x)=0$ is a 3-space
 
6:34 PM
You mean, the space of vectors $x$ fulfilling that equation? Yes.
 
and $t(x)=t_0$ is also a 3-space
 
um
wtf is wrong with me
of course it is
well now I get it intuitively
now how to show it mathematically...
 
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 ;)
 
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 $c+x\in t^{-1}(t_0)$
 
6:39 PM
@0celo7 this exercise is a good example of why a proper course in linear algebra would be so useful
 
and then I'm lost
unless
no no still lost
 
@0celo7 Where are you lost?
 
@ACuriousMind what was that gauge theory book you linked to before?
 
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.
 
"this shows that"
 
6:42 PM
@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
 
@ACuriousMind :9
what
 
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
 
ok and a bijection can only happen between spaces of equal dimension
alright but why is that a bijection
 
2) Get the conjugate momentum and Hamiltonian
3) Get the EOM
 
@Slereah seriously, can't you wait :/
 
6:44 PM
4) Quanticize all this
It can, yes
 
@0celo7 Because it has the inverse $c\mapsto c+x$.
 
pretty sure Weinberg does that in that order
OHHHH
ok
 
@Slereah You don't need to get the e.o.m., I think. What do you want them for?
 
so where are we now
@ACuriousMind constraints
 
Thanks
 
6:46 PM
@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
 
@0celo7 We've finally concluded that every $t^{-1}(t_0)$ is a 3D affine space.
 
@ACuriousMind I suspected as much, but then how do you get the propagator?
 
@0celo7 I do not count those as "e.o.m.".
 
@ACuriousMind yes, I remain convinced for the time being
 
sorry ignore the previous one, there's a copy past problem
 
6:46 PM
@0celo7 Alright
 
people please just let him finish explaining this to me
 
@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
 
so what now?
oh for fucks sake
just let him finish
 
obe
lol
 
If you love @ACuriousMind why don't you marry him
2
 
6:49 PM
@Slereah long hair is a deal breaker
 
@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 $\phi(b) = (t(b-a),\psi_{t(b-a)}(b-a))$
 
Poor @ACuriousMind
Everyone wants him
 
@Slereah Ah. Well, you do not need the propagator prior to quantizing, I'd say.
 
Hm
How do you construct it afterwards though?
 
@Slereah $\langle \phi(x)\phi(y) \rangle$?
 
6:52 PM
@ACuriousMind yes
that's it?
 
Well yes I suppose, but how do you know what that is
 
well that was easy
 
@0celo7 What's left is to check that this $\phi$ preserves all parts of the Galilean structure.
 
oh fuck me
 
But it's built that way, so it should.
 
6:53 PM
good
how would you do that explicitly
 
@Slereah Show it's the Green's function for the e.o.m., calculate it ;)
 
Hm, I suppose that it is related to the Dyson equation
 
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.
 
Yeah I am just trying to figure out what exactly you need
Because most QFT books are kind of a mess
 
@ACuriousMind yeah I dunno what exactly to check now
 
6:56 PM
there isn't a lot of QFT books with the rigorous derivation from first principles
 
the time function is just the time coordinate
 
@0celo7 Well, first check it's an affine map, i.e. $a-b = \phi(a)-\phi(b)$.
 
there's a lot of going through classical mechanics and using a variety of other formalism
 
@ACuriousMind everything here is linear
right?
 
Schwinger-Dyson equation is pretty easy to do in path integrals but I dunno it in Heisenberg
Let's check zee wikipedia
 
6:57 PM
Zee? Like A. Zee?
 
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 $\mathbb{R}\times\mathbb{R}^3$.
 
Like "the", but frencher
bc I am french
 
@ACuriousMind Well that's trivial, no?
you're not french
Sam is not a French name.
 
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
 
not hard?
literally just read it off
 
6:59 PM
But these are all things you must check after you've written down such a thing
 
we defined it that way
 

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