« first day (2815 days earlier)      last day (2097 days later) » 

danielunderwood
danielunderwood
12:00 AM
someone
don't know if that really counts
 
Emilio Pisanty
Emilio Pisanty
@danielunderwood of course it doesn't
 
danielunderwood
danielunderwood
once
I can't really think of anything that isn't strongly related to one
 
Emilio Pisanty
Emilio Pisanty
of course that doesn't either
@danielunderwood but you'd kinda expect a basic word like that to be usable by beginners, no?
 
danielunderwood
danielunderwood
I expect it to be the first word they learn aside from hello
 
Emilio Pisanty
Emilio Pisanty
(no, of course you wouldn't. the basic words are the ones that get the most use and accumulate the most wear and tear so you expect them to be the most irregular.)
 
Sir Cumference
Sir Cumference
12:14 AM
@EmilioPisanty Lol
 
danielunderwood
danielunderwood
Yeah I guess that makes some sense
 
Sir Cumference
Sir Cumference
Probably my favorite question was this
Which was edited and became this
 
danielunderwood
danielunderwood
12:36 AM
What's up with the revisions, timeline, and clock (join date?) below the name?
In other words, wow
Sometimes I think I don't quite understand something, but wow
Did I say wow?
At least it isn't as bad as searching for cern on youtube
Only do that search if you want to lose more faith in humanity
 
Sir Cumference
Sir Cumference
@danielunderwood A userscript, Stack Overflow Extras
 
danielunderwood
danielunderwood
Oh neato. Looks like there's a whole site for them I didn't even know about stackapps.com
And apparently the nutters just have an issue with cern. I searched black holes and quantums without finding any in the top results
 
 
2 hours later…
danielunderwood
danielunderwood
2:55 AM
Is it possible to do anything with the power of an integral like $(\int f(x) dx)^n$. I was thinking about the earlier conversation about functionals and thought it may be interesting to try to extremize one, but this step seems problematic. I could differentiate $n$ times then go back and integrate $n$ times, but that gets something that I have no idea how to solve and seems a bit fishy anyway.
 
rob
rob
@danielunderwood There's a famous trick for computing the area under a Gaussian that involves squaring the integral and then converting to cylindrical coordinates.
 
danielunderwood
danielunderwood
@rob That's different than the way that's taught in a typical calc class of just changing to polar coordinates, right?
And I think I may have a partial answer since I somehow forgot that you can split an integral of a sum into a sum of integrals. From what I can tell, to extremize $(\int f(x) dx)^2$, you get the condition $\int f(x) dx = 0$...which seems obvious in retrospect
Ahh I think that is the change of coordinates way I was thinking of. I just forgot squaring the integral for some reason
Although $\int f(x) dx$ seems like it may be a trivial solution and there's somehow more structure like Euler-Lagrange. hmmm
 
John Rennie
John Rennie
3:44 AM
@EmilioPisanty Belousov–Zhabotinsky reactions are something all chemistry students will have seen, though it takes effort to get the patterns looking as good as they do in the videos - obviously the videos have been cherry picked.
 
 
1 hour later…
user351417
5:06 AM
@EmilioPisanty I don't see how that relates to the 2 vs 3 -mirror system comparison. But interestingly, I didn't see any discrete consideration of the possibility of asymmetric interference patterns
 
user351417
I guess those would show up within the statistical variations which they mentioned.
 
user351417
 
user351417
Like you'd get pattern a if you have correctly aligned mirrors, but b if the beams aren't collinear.
 
user351417
I can't really figure out whether that'd actually cause an error though... Considering they usually automate the process of counting fringes.
 
Yuzuriha Inori
Yuzuriha Inori
5:46 AM
@BalarkaSen You here at ISI?
 
 
1 hour later…
Slereah
Slereah
7:02 AM
Morning
 
user187604
user187604
7:14 AM
Can someone make me understand this ..
 
Sir Cumference
Sir Cumference
7:41 AM
@user187604 Ok you now understand it
 
user187604
user187604
8:00 AM
@SirCumference bad internet connection makes you understand quickly: p
 
 
2 hours later…
Secret
Secret
9:32 AM
13 hours ago, by Semiclassical
now I want a lojban translation of Parmenide's tautology: "It is necessary both to say and to think that being is; for it is possible that being is, and it is impossible that not-being is"
hmm...
There is Being
But Non-Beings (don't exist)
Sorry, I don't chat. I've made my point. It is up to you to decide if it is helpful or not. Good luck! — safesphere 2 days ago
Also this is plain rude
 
DakkVader
DakkVader
Hey people!
 
apt45
apt45
10:13 AM
Hello. What is the typical size of a hydrogen atom with the electron at the n-th excited level?
 
 
1 hour later…
bolbteppa
bolbteppa
11:41 AM
Alright I think the real problem is understanding classical fields in terms of fibre bundles, and most importantly the transformation law for classical fields in terms of fibre bundles
"One might say that a classical field is a section of an associated fiber bundle over the base space of Minkowski space, with the principal bundle having fibers that are the Lorentz group. Thus, the associated bundle has fibers that transform under the Lorentz group. i.e. transform as representations of the Lorentz group."
 
bolbteppa
bolbteppa
12:18 PM
When you perform a Lorentz transformation on the electromagnetic potential in a Lagrangian, you seem to be transforming the electromagnetic potential by a vector representation of the Lorentz group, and transforming the coordinates the electromagnetic potential is a function of by the Lorentz group directly
i.e. transforming a section of a vector bundle by a group action of a vector representation of the Lorentz group, and transforming the space-time manifold by a group action of the Lorentz group
 
bolbteppa
bolbteppa
12:53 PM
Oh man
"In the fiber bundle formulation, the EM potential is represented not by any particular vector field on a space-time manifold, but by a connection on a principal fiber bundle which has the space-time manifold as its base space. A choice of gauge corresponds to a choice of section for the bundle. To each local section of the bundle there corresponds a unique vector field $A_{\mu}$ on an open set of the base space that represents the potential relative to that section."
 
danielunderwood
danielunderwood
ew math in non-tex
 
bolbteppa
bolbteppa
So a classical spin wave function is a connection on a principal bundle too, as is a KG wave function? :(
 
danielunderwood
danielunderwood
1:10 PM
0
Q: What would happen if the Earth's entire volume was replaced with blueberries?

billybodegaSupposing that the entire Earth was instantaneously replaced with an equal volume of closely packed, but uncompressed blueberries, what would from the perspective of a person on the surface? I.e. would the blueberries float off into space, or would the person sink through to the center of Bluebe...

planets
 
Balarka Sen
Balarka Sen
1:31 PM
@YuzurihaInori I am not.
 
Slereah
Slereah
@bolbteppa Naber
is the cool book
 
bolbteppa
bolbteppa
Deligne is actually using normal language in the intro paragraphs
 
Slereah
Slereah
>using normal language
that is how u recognize the CASUALS
Although Naber does start fairly normal
He derives the Dirac string with Maxwell
 
bolbteppa
bolbteppa
1:52 PM
Basically just trying to 'justify' $\psi_a'(x) = D(\Lambda)_a \, ^b \psi_b(\Lambda^{-1}x)$, it's actually insane, can't seem to find it in Naber
Seems like if you work with Lagrangians you can ignore a load of subtleties and use complicated things without realizing it
 
Slereah
Slereah
Naber is more EM oriented
Not sure what's a good book for like
Spinor bundle shenanigans
@0celo7 would know
 
bolbteppa
bolbteppa
2:12 PM
Seems like you always have a calculus-of-variations action $S[x,\varphi(x),\dots]$ and want to be able to perform coordinate transformations $T(x)$ of $x$, representations of those coordinate transformations $D[T]$ on $\varphi$, and also allow for 'gauge' transformations of $\varphi$, and connections on and/or sections of principal bundles are the 'obvious' way to do this, which is not obvious
$S[T(x),D[T]\varphi(T(x)),\dots] = S[x,\varphi(x),\dots]$
It's nuts how a Lorentz transform of $x$ 'induces' a transformation on $\varphi$ as well as $x$
I mean you can just plug it into a Lagrangian without thinking and ignore this bundle stuff
But in qft they make you face up to it with $\psi'(x) = D(\Lambda) \psi(\Lambda^{-1}x)$ whether you like it or not
That's actually a crazy thing to just define
 
Slereah
Slereah
2:28 PM
I seem to recall some online course that has details on doing spinors properly
i think that one
 
gateprep
gateprep
8
Q: Why are metals malleable and ductile?

mithusengupta123Why are metals malleable and ductile? These two properties seem to be related. Is there a microscopic understanding of these properties possible?

condensed-matter solid-state-physics material-science elasticity metals
I can't understand how they arrived at that slip plane
I am referring to this plane
 
danielunderwood
danielunderwood
Holy text wall that 2nd answer
 
Slereah
Slereah
2:49 PM
good title
 
danielunderwood
danielunderwood
Those UK kids must be crazy smart
 
bolbteppa
bolbteppa
This stuff is incredibly complicated and it's basically just terminology, renaming stuff one already knows :(
 
danielunderwood
danielunderwood
Terminology is scary. Big scary groups like $U(1)$ hah
Well big may not be the best word for that example, but scary for some people I'm sure
 
bolbteppa
bolbteppa
3:10 PM
Nuts
First two pages there
Are basically $\psi'(x) = D(\Lambda) \psi(\Lambda^{-1}x)$ formalized I think
 
Secret
Secret
good title: Well technically the physics is all written, thus i will argue the thing that makes it unreadable to those that are not child prodigies and have absolutely no background in physics and maths will be the maths itself
 
bolbteppa
bolbteppa
$D(\Lambda)$ is a map between fibers, $\psi$ is a section, and $\Lambda^{-1}$ arises because you're considering the action of the Lorentz group on a set of functions/sections, to get $\psi' = \Lambda \cdot \psi = D(\Lambda) \psi \Lambda^{-1}$ or something
 
bolbteppa
bolbteppa
3:45 PM
Consider a group $G$ which acts on a set $X$ as $\cdot : G \times X \to X , (g,x) \mapsto g \cdot x = gx$. If we have a set of functions $F(X)$ defined on $X$, e.g. $f : X \to X, x \mapsto f(x)$, then by defining $(g_1 f)(x) = f(g^{-1} x)$ we see this satisfies $g_2 (g_1 f)(x) = (g_1 f)(g_2^{-1} x) = f(g_1^{-1} g_2^{-1} x) = f[(g_2 g_1)^{-1} x) = (g_2 g_1 f)(x)$ which is the group closure law, and so the action of the group on $X$ has been transferred to the action of $G$ on $F(X)$.
Given a space-time manifold $M$, we attach a vector space to each point $x$ of $M$, referring to $T(M)$ as the set/`bundle' of all these vector spaces attached to $M$. We then define a section $\psi$ as a smooth map from $M$ to $T(M)$ attaching to each point $x$ of $M$ a vector $\psi(x)$ such that $\psi(x)$ is differentiable. If we now consider the action of the Lorentz group $G$ on $M$ such that any element $g$ of $G$ sends every $x$ to $gx$.
We can transfer the action of $G$ on $M$ to the set of sections defined on $M$ by defining $(g \psi)(x) = D(g) \psi(g^{-1} x)$, where $D(g)$ is a map on the bundle $T(M)$ sending the section at $x$ containing $\psi$ to the section at $gx$ containing $g \psi$.
$T(M)$ is kind of a disjoint union of all those vector spaces, and so is just a vector space in a sense, not quite, so $D : G \to T(M)$ is like a 'representation' of $G$ in the tangent bundle, which is not just some simple representation, jesusssss
 
vzn
vzn
4:27 PM
 
Emilio Pisanty
Emilio Pisanty
4:42 PM
 
ACuriousMind
ACuriousMind
@danielunderwood If it's someone new to the chat, just tell them (in a firm but friendly way!), no need to do anything else. If they persist in a disruptive way or you recognize a repeat offender and there doesn't seem to be a room owner around to take care of it, you could raise a custom moderator flag describing the issue, but please don't use the standard rude/offensive flag (unless they're additionally being actually offensive).
@EmilioPisanty Sadly, Newton was wrong about gravity but I still can't fly :'(
 
enumaris
enumaris
do you think if Einstein could meet Newton
he would say
"WRRRRRRONGGGGGGGGGGGGGGGGG"
 
ACuriousMind
ACuriousMind
@enumaris He would say "Falsch!" and Newton would be like "Quid?"
 
enumaris
enumaris
I like my interpretation better
 
bolbteppa
bolbteppa
'Stop talking about alchemy'
 
danielunderwood
danielunderwood
4:54 PM
'What's a Riemann?'
 
enumaris
enumaris
5:07 PM
1
A: Temperature Required to give Particles of Matter Speed comparable to the speed of light

Árpád SzendreiVery good question. As you could figure out, only a macro object would have temperature (we could talk about temperature of a molecule, but that is not this case). You are talking about not the temperature of the particles, but their oscillation speed. And you are talking about how the temperatu...

what the deal with this answer and comments...
 
danielunderwood
danielunderwood
I think he's been reading about the popular subjects and completely missed thermodynamics
That guy almost seems like a bot with the way his questions are formatted
 
ACuriousMind
ACuriousMind
@enumaris I left a comment.
 
enumaris
enumaris
cool beans
 
bolbteppa
bolbteppa
How can $D(\Lambda)$ be a representation if it maps to different fibers/vector spaces :\
 
ACuriousMind
ACuriousMind
@bolbteppa Without a lot more information, that's not a question one can answer :P
 
Secret
Secret
5:20 PM
enumaris and ACuriousmind: He is basically saying that temperature will only affect the motions of molecules, but not the quarks, thus in a sense, there isn't any temperature where it reaches near the speed of light.

However, as you pointed out, quarks and other subatomics don't have a notion of "speed", thus the premise of his answer is wrong
 
enumaris
enumaris
At those temperature there won't be any molecules
just a plasma
 
Secret
Secret
yeah
 
enumaris
enumaris
and you'll also have enough energy to spontaneously create matter-antimatter pairs as well
 
Secret
Secret
Though I must admit, his wording is very weird even under my standards
I need to figuratively bend his comment around a horshoe shaped object to understand what he is saying
 
enumaris
enumaris
 
bolbteppa
bolbteppa
5:23 PM
@ACuriousMind Basically, if $M$ is a spacetime manifold, and $T(M)$ is a vector bundle of Hilbert spaces attached to each point of $M$, and $\psi : M \to T(M)$ is a section, and $G$ is the Lorentz group acting on $M$, we can transfer the action of $G$ on $M$ to $G$ acting on the set of sections on $M$ by defining, for $\Lambda \in G$, $\psi'(x) = (\Lambda \cdot \psi)(x) = [D(\Lambda) \psi \Lambda^{-1}](x) = D(\Lambda) \psi(\Lambda^{-1} x)$, i.e. the transformation law for classical fields.
Here $D(\Lambda)$ is a map sending a section $\psi$ to a section $\psi' = D(\Lambda) \psi \Lambda^{-1}$, I guess it's a representation in the vector space of sections?
 
ACuriousMind
ACuriousMind
@bolbteppa $D$ is just a representation on every fiber vector space.
 
bolbteppa
bolbteppa
Doesn't $D$ send a $\psi$ in a fiber at $x$ to $\psi'$ in a different vector space at $x' = \Lambda x$?
A group rep is defined as something like $D : G \to GL(V)$, this is $D : G \to ?$ involving a ton of different vector spaces I think :(
 
ACuriousMind
ACuriousMind
@bolbteppa Well, the notation is inconsistent and $D$ is overloaded, maybe that is what is confusing here. $D$ is three things: 1. A representation on every fiber vector space individually, let's call it $d_x$. 2. A representation that acts on a collection of fiber element $f_x$ like $d_x(\Lambda) f_x$, let's call it $D$ 3. A representation on the space of sections that acts on a section $\psi$ like $D(\Lambda) \psi \Lambda^{-1}$, let's call it $\mathcal{D}$.
Since all the $d_x$ are isomorphic and we rarely need to call out $D$ specifically, people tend to use the same symbol for all three.
 
bolbteppa
bolbteppa
5:49 PM
hmm
"A representation of a group $G$ is a group action of $G$ on a vector space $V$ by invertible linear maps." So I guess defining $\Lambda \cdot \psi = D(\Lambda) \psi \Lambda^{-1}$ is the group action/representation, the question is ironing it out so it's not overloaded
If $G$ is the Lorentz group and $\mathcal{F}(M;T(M))$ is the function space of sections on $M$, then $\cdot : G \times \mathcal{F}(M;T(M)) \mapsto \mathcal{F}(M;T(M)), (\Lambda,\psi) \mapsto \Lambda \cdot \psi = D(\Lambda) \psi \Lambda^{-1}$
That can't be right, $D(\Lambda)$ is supposed to be a representation, here $D(\Lambda)$ is a map mapping a fiber at e.g. $y$ into a fiber at $\Lambda y$
 
ACuriousMind
ACuriousMind
@bolbteppa No, $D(\Lambda)$ preserves fibers. It is the $\Lambda^{-1}$ on the right side that shifts the fiber.
 
bolbteppa
bolbteppa
6:07 PM
It seems like it should, but if you do it from a bundle pov it actually doesn't seem to, as you can see from the explanation here:
(From the link above)
Or maybe it does
Using their notation, they have $F(f) \circ s \circ f^{-1} = f^* s$ which is the pullback of the section $s$ at $x$
 
ACuriousMind
ACuriousMind
@bolbteppa Yesx, and $F(f)$ is a fiber-preserving diffeomorphism on the bundle, precisely like $D(\Lambda)$ is.
 
enumaris
enumaris
0
Q: Ocean density vs atmosphere density

Árpád SzendreiI understand that the density of the oceans on Earth in on average constant regardless of the depth. It is 1020 kg/m^3 at the surface and 1050 kg/m^3 at deep waters. I understand too that this is not the case with the atmosphere. The density of the atmosphere decreases with height. Though, I ...

quantum-mechanics general-relativity atmospheric-science density
Feels like...this guy is way overthinking things...
using QM and GR to explain the fact that water is incompressible while air is? O____O
 
bolbteppa
bolbteppa
@ACuriousMind So $\psi'(x) = [D(\Lambda) \psi \Lambda^{-1}](x)$ is saying that $\psi'$ in the fiber at $x$ is just $\psi$ in the fiber at $\Lambda^{-1}x$ after being 'rotated' by $D(\Lambda)$, where $D(\Lambda)$ a representation of $\Lambda$ acting in the fiber at $\Lambda^{-1} x$
 
ACuriousMind
ACuriousMind
@bolbteppa Exactly
 
enumaris
enumaris
6:22 PM
wow mickey dees is on door dash...
and it has a 7 dollar delivery fee looool
 
bolbteppa
bolbteppa
Is this some well-known transformation law of fiber bundles, or related to the structure groups or anything
 
ACuriousMind
ACuriousMind
@bolbteppa No. In fact, it is a very special thing that only occurs for "natural" bundles as your text calls them. Note that for a gauge transformation on a gauge bundle, the $\Lambda^{-1}$ is not there (because it wouldn't make sense).
But for natural bundles - those intrinsically arising from the manifold through a functor, not being "tacked onto" it - the map is indeed very natural. You need to associate a bundle map to every $f$, and this is how you do it.
 
bolbteppa
bolbteppa
That's another confusing thing, this rule holds for electromagnetic potentials, where $D(\Lambda)$ is just the 'vector representation' and $\psi = A_{\mu}$, but it's also a section of fiber bundle (apparently only if you fix a gauge for the connection $A_{\mu}$ it then becomes a section which I'm trying to make sense of), the $\Lambda^{-1}$ is there no
 
ACuriousMind
ACuriousMind
@bolbteppa Sure, the rule holds for all fields. Gauge representations (in the narrow sense, i.e. excluding the gauge-y view of e.g. GR) are just "tensored on", they do not interfere with the spacetime structure at all.
But the $\Lambda^{-1}$ is not there for a gauge transformation of a field transforming in a gauge representation. It's just $\psi \mapsto D(A)\psi$ for some element $A$ of the gauge group, and nothing on the left side.
 
bolbteppa
bolbteppa
It sounds like the transformation of a classical field you use in qft, where you'd send e.g. $A_{\mu}(x)$ to $A_{\mu}'(x)$, is a pullback of a section $\psi$ on a vector bundle under a Lorentz transformation, since $\psi' = \Lambda^* \psi = D(\Lambda) \psi \Lambda^{-1}$, and a gauge transformation is some other transformation of sections
 
ACuriousMind
ACuriousMind
6:36 PM
You have to be more specific what you mean by "the transformation"
If you mean a change of coordinates, then yes, a change of coordinates is a pullback.
And a change of coordinates by a Lorentz transformation has precisely the action above.
 
bolbteppa
bolbteppa
It's more complicated than a pullback, usually defined as $(\Lambda^* \psi)(x) = \psi (\Lambda x)$ right
Still need to figure out gauge transformations properly :p
 
ACuriousMind
ACuriousMind
@bolbteppa No, the pullback by $\Lambda$ is precisely $D(\Lambda) \psi \Lambda^{-1}$
That's what the snippet you posted is calling $f^\ast s$, and $f^\ast$ is the pullback.
 
bolbteppa
bolbteppa
Pullback (differential geometry)
Suppose that φ:M→ N is a smooth map between smooth manifolds M and N; then there is an associated linear map from the space of 1-forms on N (the linear space of sections of the cotangent bundle) to the space of 1-forms on M. This linear map is known as the pullback (by φ), and is frequently denoted by φ*. More generally, any covariant tensor field – in particular any differential form – on N may be pulled back to M using φ. When the map φ is a diffeomorphism, then the pullback, together with the pushforward, can be used to transform any tensor field from N to M or vice versa. In particular, if...
The $D(\Lambda)$ seems to mean it's not a pullback as defined there :(
That also seems to ruin calling it a simple group action
 
ACuriousMind
ACuriousMind
@bolbteppa You need to learn to decipher and translate between different terminologies if you want to be nitpicky about these things. $D(\Lambda) \psi \Lambda^{-1}$ is equivalent to the formula that article gives under "Pullback of (covariant) tensor fields".
 
danielunderwood
danielunderwood
@enumaris I think he's trying to learn the "interesting" things but completely missing out on fundamentals
 
bolbteppa
bolbteppa
6:45 PM
 
danielunderwood
danielunderwood
Although I kind of think QM may end up with water being more compressible than classically
 
bolbteppa
bolbteppa
@ACuriousMind I don't see how
 
enumaris
enumaris
@danielunderwood seems like it
 
danielunderwood
danielunderwood
Now what GR ever has to do with compressiblity, I have no idea
 
ACuriousMind
ACuriousMind
@bolbteppa In our case, the map $\varphi$ is $\varphi(x) = \Lambda x$. So $\mathrm{d}\varphi(X) = \Lambda \cdot X$, and that's just the fundamental representation of the Lorentz group on vectors. Once you translate from "multilinear form" to "section of tensor bundle", this becomes a representation $D(\Lambda)$ (the $n$-tensor representation) acting on the tensor-valued field. And $S_{\varphi(x)}$ is just a different notation for $S\circ\varphi^{-1}(x)$.
 
bolbteppa
bolbteppa
7:07 PM
@ACuriousMind $(\Lambda^* \psi)_x(x) = \psi_{\Lambda(x)} [ d(\Lambda x) x] = \psi_{\Lambda(x)} [\Lambda x] = ? $
 
ACuriousMind
ACuriousMind
@bolbteppa You can't write it down as a straightforward equation, because the bijection between multilinear forms and tensors is not a simple equation.
When you try to write it down as an equation, you will end up either with an unreadable mess or something false whose different sides live in different spaces.
 
bolbteppa
bolbteppa
@ACuriousMind alright thanks, will look into it and see if that can be dealt with
 
Sir Cumference
Sir Cumference
manifold through a functor, not being "tacked onto" it - the map is indeed very natural. You need to associate a bundle map to every f
f
, and this is how you do it.

bolbteppa
bolbteppa
That's another confusing thing, this rule holds for electromagnetic potentials, where D(Λ)
D
(
Λ
)
is just the 'vector representation' and ψ=Aμ
ψ
=
A
μ
, but it's also a section of fiber bundle (apparently only if you fix a gauge for the connection Aμ
A
μ
it then becomes a section which I'm trying to make sense of), the Λ−1
 
Secret
Secret
9
Q: Maximum Survivable Time Distortion Factor?

GryphonThis question is somewhat inspired by the number of "time bubble" questions that have been popping up on the site recently. If a real "time bubble" with a large enough time distortion factor existed, entering or leaving the bubble would mean certain death, as parts of the body on the fast side o...

health time-manipulation
Newtonian Temporal Mechanics
 
Semiclassical
Semiclassical
My ignorance of bundle stuff is probably a big part of why I stick with QM over QFT or anything else
There is bundle stuff you can do with QM (eg berry curvature) but it’s not central in the same way
 
ACuriousMind
ACuriousMind
7:21 PM
@SirCumference Uhhh...you okay there?
@Semiclassical A shockingly large amount of QFT people also go without ever really engaging with bundles :P
 
bolbteppa
bolbteppa
If you can't define things properly, a good reason not to :p
@SirCumference ?
 
Semiclassical
Semiclassical
@ACuriousMind heh, true
 
Sir Cumference
Sir Cumference
@ACuriousMind Oh what the hell
 
danielunderwood
danielunderwood
lol
 
Sir Cumference
Sir Cumference
Sigh this has happened before
 
danielunderwood
danielunderwood
7:27 PM
Magical select all + paste?
 
Sir Cumference
Sir Cumference
No idea
@danielunderwood Probably my idiot phone
 
danielunderwood
danielunderwood
Well it provided comic relief I needed lol
 
Sir Cumference
Sir Cumference
Seems to be my life at this point... :P
 
ACuriousMind
ACuriousMind
@SirCumference You should print cards: "Sir Cumference, professional comic relief"
2
 
enumaris
enumaris
7:48 PM
hmmm
 
bolbteppa
bolbteppa
8:10 PM
Bottom of pdf page 15 (book page 3) of this achyra.org/infarinato/files/phdthesis.pdf defines $\varphi^* \xi = T (\varphi) \circ \xi \circ \varphi^{-1}$ i.e. $f^* s = F(f) \circ s \circ f^{-1}$ hmm
as a pushforward not a pullback hmm*
 
Semiclassical
Semiclassical
Elementary QM question about the propagator. Suppose I know K(x,t;x,t’) for all spacetime points for some system
Let me now make a series of measurements of position at times t1,t2,..,tN, obtaining x1,x2,...,xN
 
ACuriousMind
ACuriousMind
@bolbteppa A pushforward would conventionally be written $\phi_\ast$. The mnemonic is that a downstairs star is a foot kicking stuff forwards, and an upstairs star is an arm pulling sfuff backward.
 
Semiclassical
Semiclassical
What’s the relation between the pairwise propagator values ie $K(x_k,t_k;x_{k+1},t_{k+1})$, and the final value ie $K(x_1,t_1;x_N,t_N)$?
 
bolbteppa
bolbteppa
@ACuriousMind that's genius
 
Semiclassical
Semiclassical
I want to say say that the latter is just the product of the former but I’m not sure I’m remembering it right
(Maybe one needs the time between measurements to be infinitesimal?)
 
ACuriousMind
ACuriousMind
8:24 PM
@Semiclassical $K(x_1, t_1; x_N, t_N)$ has nothing to do with your scenario, it is the propagation for a state undisturbed between $t_1$ and $t_N$.
 
bolbteppa
bolbteppa
$K(x_1,t_1;x_N;t_N) = K(x_1;t_1;x_2;t_2)K(x_2,t_2;x_N,t_N) = \dots$ with your one in there somewhere
 
Semiclassical
Semiclassical
Those two answers aren’t quite the same!
 
ACuriousMind
ACuriousMind
What is true is that $\int K(x_i, t_i; x_j t_j)K(x_j,t_j; x_k, t_k) \mathrm{d}x_k = K(x_i, t_i; x_k, t_k)$, i.e. once you no longer care about the intermediate state $x_j$, you get the total probability (density) to go from $x_i$ to $x_k$.
@bolbteppa Yeah, before I heard it I could never remember which star position meant which
 
bolbteppa
bolbteppa
Yeah should include an integral over $x_2,t_2$ in between right
 
Semiclassical
Semiclassical
Right. I think there may still be a claim if the time steps are very small tho
 
ACuriousMind
ACuriousMind
8:30 PM
@Semiclassical The propagator is a manifestation of unitary time evolution. Measurements are not.
 
Semiclassical
Semiclassical
Well, what I have in mind is a bit from Shankar
 
ACuriousMind
ACuriousMind
But, well, the probability density to measure at these step is the product of all propagators, i.e. $\prod_i K(x_i,t_i; x_{i + 1}, t_{i + 1}$. This has nothing to do with $K(x_1,t_1; x_N, t_N)$, however
It becomes $K(x_1,t_1; x_N,t_N)$ once you integrate over all the $x_i$ inbetween, i.e. once you no longer measure.
I guess that's what you were asking - the final amplitude is indeed the product, but the final amplitude is not the same as $K(x_1,t_1; x_N, t_N)$.
 
Semiclassical
Semiclassical
That if you start with a particle localized at x1 at time t (which he interprets to psi(x,0) = delta(x-x1) ) then psi(x,t) = K(x,t,x1,0)
 
ACuriousMind
ACuriousMind
I'm probably getting too worked up about the "ie" part of your question :P
 
Semiclassical
Semiclassical
So the propagator gives the unitary time evolution of a localized state
 
ACuriousMind
ACuriousMind
8:35 PM
Going with the make-believe that $\delta(x)$ is a meaningful state, that is true, yes :P
 
Semiclassical
Semiclassical
hey, if it’s good enough for Shankar...
(The smarter thing would be to consider a narrow Gaussian, of course)
And then something something saddle-point blah blah blah
 
ACuriousMind
ACuriousMind
Yes, ignore my kvetching- where were you going?
 
Semiclassical
Semiclassical
I guess my point has to do with how one creates a state with narrow width in position space
 
enumaris
enumaris
measure the position
 
Semiclassical
Semiclassical
That’s pretty much what I had in mind, yes
(Secretly I have some Bohm stuff on the brain but I see little point in bringing that up)
 
ACuriousMind
ACuriousMind
8:43 PM
::froth forming at the mouth:: Boooooooooooooooohm!
Ahem
 
Semiclassical
Semiclassical
Lol
 
enumaris
enumaris
remember that time I recommended Bohm's book on QM to a QM newbie asking for book recommendations
good times
 
Semiclassical
Semiclassical
Lol
Depends which one really
 
enumaris
enumaris
the one on Bohmian mechanics
 
Semiclassical
Semiclassical
His orthodox textbook is apparently quite good for learning intro QM
 
enumaris
enumaris
8:45 PM
not that one
 
Semiclassical
Semiclassical
His later works, though, are probably not helpful for the newbie :)
I think the error I have is that while any particular sequence of position measurements is a (discretized) Feynman path, there’s an entire space of such paths
I can compute the amplitude along such a path, but to get the amplitude between the endpoints I of course need to sum over all such paths
Which is just the path integral duh
 
ACuriousMind
ACuriousMind
Sounds like you're rediscovering the path integral, yeah :P
 
Semiclassical
Semiclassical
That’s more or less what I had in mind, if I’m honest: the impetus for this was a dBB paper which gave a Bohmian perspective/derivation of the Feynman path integral
I think I get the idea of how that works, but I’m reluctant to look at the details
 
bolbteppa
bolbteppa
9:01 PM
@ACuriousMind So the bottom of page 15 of achyra.org/infarinato/files/phdthesis.pdf writes $f^* s = F(f) \circ s \circ f^{-1}$ (in the notation of the gif above) where $F(f) : T(M) \to T(M)$ is the push-forward, in our case acting as a linear transformation in each fiber not between them, I guess this is what you meant by saying we can't write it down as a formula, we just denote this as e.g. $D(\Lambda)$ in qft?
e.g. we'd have to be able to interpret $D(\Lambda) = e^{-\frac{1}{8} \omega_{\mu \nu}[\gamma^{\mu},\gamma^{\nu}]}$ in the Dirac spinor case as the push-forward of something insane
 
ACuriousMind
ACuriousMind
@bolbteppa I see no expression like $f^\ast s$ there on pg. 15. $F(f)$ is not the pushforward. The pushforward of a vector field is the whole resulting vector field on the target manifold, $F(f)$ itself is just a bundle map. I feel you're trying to hard to distill this into neat formulae - the different formulations are equivalent, but not simply by renaming things.
As I said, $D(\Lambda)$ is an overloaded symbol - sometimes it's the map on a single fiber, sometimes it's the bundle map, sometimes it's a full pushforward or pullback. Since physicists are abysmal at actually defining their symbols, you can't expect to be able to translate it one-to-one into pure mathematical terms of speech.
 
bolbteppa
bolbteppa
@Semiclassical this kind of thing is how the path integral is derived in Zee's qft, e.g. eq. 3 ch. 2 in theory.phys.unm.edu/cahill/523-12/…
 
Semiclassical
Semiclassical
Sounds right
 
ACuriousMind
ACuriousMind
@bolbteppa Now that one is really just the representation on a single vector space, no differential geometry involved.
 
bolbteppa
bolbteppa
@ACuriousMind bottom of pdf page 15 (page 3 of the actual text), last sentence, written as $\varphi_{^*} \xi = T(\varphi) \circ \xi \circ \varphi^{-1}$
 
ACuriousMind
ACuriousMind
9:09 PM
See what I mean? It's impossible to answer the quesiton "What is $D(\Lambda)$?" because physicists tend to use it for a whole host of related actions of the Lorentz group on different, closely related objects
@bolbteppa Yeah, the pushforward is the whole thing, i.e. $\varphi_\ast $, not just $T(\varphi)$.
 
bolbteppa
bolbteppa
Ah sorry, they say $T(\varphi)$ is the tangent map
 
ACuriousMind
ACuriousMind
Indeed. Note that $T( \varphi)$ is more commonly written $\mathrm{d}\phi$.
 
bolbteppa
bolbteppa
@ACuriousMind So the bottom of page 15 of achyra.org/infarinato/files/phdthesis.pdf writes $f_{^*} s = F(f) \circ s \circ f^{-1}$ (in the notation of the gif above) where $F(f) : T(M) \to T(M)$ is the tangent map, in our case acting as a linear transformation in each fiber not between them, I guess this is what you meant by saying we can't write it down as a formula, we just denote this as e.g. $D(\Lambda)$ in qft?
e.g. we'd have to be able to interpret $D(\Lambda) = e^{-\frac{1}{8} \omega_{\mu \nu}[\gamma^{\mu},\gamma^{\nu}]}$ in the Dirac spinor case as the push-forward of something insane
:p
It's written as a nice neat formula here
Representation theory of the Lorentz group
The Lorentz group is a Lie group of symmetries of the spacetime of special relativity. This group can be realized as a collection of matrices, linear transformations, or unitary operators on some Hilbert space; it has a variety of representations. In any relativistically invariant physical theory, these representations must enter in some fashion; physics itself must be made out of them. Indeed, special relativity together with quantum mechanics are the two physical theories that are most thoroughly established, and the conjunction of these two theories is the study of the infinite-dimensional...
 
ACuriousMind
ACuriousMind
@bolbteppa Not only "in our case". A bundle map, by definition, is fiber-preserving. $F(f)$ is always fiber-preserving.
 
Semiclassical
Semiclassical
(the paper i'm referencing is this one, for the fearless: mathreview.uwaterloo.ca/archive/voli/1/oltean.pdf)
 
ACuriousMind
ACuriousMind
9:13 PM
@bolbteppa That article is about representations of the Lorentz group on a single vector space, not about differential geometry and bundles. You're mixing up all the things! :P
The things I said you cannot connect in a neat formula is the notion of the pushforward/pullback of a tensor field conceived of as a multilinear form vs. of a tensor field conceived of as a section of a tensor bundle
 
Semiclassical
Semiclassical
The weirdness (to ramble a bit) is that particle trajectories in dBB always are emphasized as never crossing in configuration space
 
ACuriousMind
ACuriousMind
Go back to the roots, and look at the equivalence of "bilinear forms of vectors" and "matrices". It's a bijection, but it's not "formulaic", it's "conceptual", for lack of better words
 
Semiclassical
Semiclassical
hence if a particle is at a certain point in configuration space at a certain time, then the wavefunction determines what its velocity its
 
ACuriousMind
ACuriousMind
@Semiclassical Question: Are dBB trajectories continuous/differentiable?
Because if they are, they contribute exactly nothing to the path integral. It is a very counterintuitive feature, but the measure of differentiable paths w.r.t. the path integral measure (Wiener measure) is zero.
 
Semiclassical
Semiclassical
I think the story is that the trajectories are always continuous, and 'almost always' differentiable
the principle exception being that they don't have to be differentiable if the particle is localized at a given time
 
ACuriousMind
ACuriousMind
9:18 PM
One of my favourite things to bring out when someone thinks they can understand the path integral by thinking about classical paths: The classical paths are exactly of zero importance in the path integral.
 
Semiclassical
Semiclassical
dBB paths aren't classical tho...
 
ACuriousMind
ACuriousMind
@Semiclassical A "classical path" in this case is anything that is differentiable
 
bolbteppa
bolbteppa
The equivalence of bilinear forms and matrices is just the expansion $B(v,v) = x^i y^j B(e_i,e_j) = x^i y^j B_{ij} = x^T B y$, it's in Lang's linear algebra, the wiki article's $\psi'(x) = D(\Lambda) \psi(\Lambda^{-1} x)$ makes it seem like $\psi'$ is evaluated in a vector space at $x$ while $\psi$ in the vector space at $\Lambda^{-1}x$
 
Semiclassical
Semiclassical
well, this is why I was trying my whole 'make multiple measurements' thing
(with measurements being there mostly to localize the particle at a given time)
in that case, the trajectories you'd get are continuous but generically won't be differentiable at the moment of measurement
 
ACuriousMind
ACuriousMind
@Semiclassical Then you would get the path integral for measurements at every instant of time.
 
Semiclassical
Semiclassical
9:23 PM
Well, for the purposes of calculation you do.
But then, to make sense of the path integral requires such a discretization in the first place...
 
ACuriousMind
ACuriousMind
@Semiclassical No it doesn't, for standard QM, the continuum limit is well-defined, and the path integral measure is the (conditional) Wiener measure
 
Semiclassical
Semiclassical
hmm
 
ACuriousMind
ACuriousMind
It is only in higher dimensions, i.e. QFT, where the continuum limit is generally ill-defined and discretization ("lattice QFT") is required. But not always, see the work of Glimm/Jaffe
 
bolbteppa
bolbteppa
So for a section of a spinor bundle it seems $(\Lambda_{^*} \psi)(x) = [F(\Lambda) \circ \psi \circ \Lambda^{-1}](x) = \psi'(x)$ where $F(\Lambda) = e^{-\frac{1}{8} \omega_{\mu \nu} [\gamma^{\mu},\gamma^{\nu}]}$ is the tangent map acting in the fiber at $\Lambda^{-1} x$, whatever a spinor bundle is :p
 
Semiclassical
Semiclassical
eh, fair enough
I guess my point is that to actually derive such a measure requires thinking about discretizations? But I may be wrong on that
@ACuriousMind Can you give a source which states that precisely? I'm not doubting that there's a valid statement, but I suspect I know how the dBB story will evade it
 
ACuriousMind
ACuriousMind
9:28 PM
@bolbteppa Indeed, that is because it implicitly identifies the tangent/tensor spaces at $x$ and $\Lambda^{-1}x$. And an identification between different fibers is... ::dun dun dun:: ...a connection ("parallel transport"). That's why you need to correct through a "covariant derivative" once you start to care about this - the correction is caused by a non-trivial identification of fibers.
@Semiclassical First chapter of Glimm/Jaffe's book "Quantum mechanics - a functional integral point of view"
 
Semiclassical
Semiclassical
(more or less, I think the point is that the weighting you use for Feynman paths is different than the weighting you use for Bohmian trajectories)
thx
 
ACuriousMind
ACuriousMind
+ every standard mathematical treatment of the Wiener measure, but Glimm/Jaffe is a really condensed place to see it.
 
Semiclassical
Semiclassical
kk
 
bolbteppa
bolbteppa
@ACuriousMind so is $\psi$ more generally a connection rather than a section?
e.g. the whole $\psi' = e^{i\theta} \psi$ thing you make local seems like a connection property
 
ACuriousMind
ACuriousMind
@bolbteppa No, $\psi$ is a section. But if you have a non-trivial connection, the "=", which, remember, means "is identified with", is no longer accurate. The identification between different fibers requires then use of a connection.
You need to transport the tensor at $x$ to the tensor at $\Lambda x$. That's what a connection does, it provides a notion of identification of fibers at different points along paths between those points, parallel transport.
If it sounds like I have difficulty explaining this, it's because it is difficult, and I only grasped it through years of recurring to these issues. A perfect manifestation of von Neumann's adage - "You don't understand mathematics, you get used it to".
 
bolbteppa
bolbteppa
9:35 PM
Yeah it's not easy, it's basically just terminology/concepts and still crazy
 
ACuriousMind
ACuriousMind
Some genius might have written down a lucid and easy explanation somewhere - I am not that genius, nor have I heard of them
 
enumaris
enumaris
question
connection or connexion?
 
bolbteppa
bolbteppa
@Semiclassical I would be surprised if the problem with the continuum limit disappeared for Bohm since it's supposed to be equivalent to the Schrodinger path integral right?
@enumaris show or shew?
 
Semiclassical
Semiclassical
I don't follow. I took ACM's point is that the continuum limit for QM (i.e. the Feynman path integral) really doesn't have any issues
 
bolbteppa
bolbteppa
"shew"
 
Semiclassical
Semiclassical
9:38 PM
so if there's no basic issue with the continuum limit for the Feynman path integral, I don't expect there's one for the Bohmian version of such
 
enumaris
enumaris
I think Bishop and Crittenden call it "connexion"
 
ACuriousMind
ACuriousMind
@enumaris You're free to say whatever you want.
'cti' -> 'xi' is a shift that started in late classical Latin
 
enumaris
enumaris
were those guys Latin?
 
bolbteppa
bolbteppa
Ah, maybe I'm thinking of qft path integrals, I didn't know qm ones were made rigorous if they are
 
ACuriousMind
ACuriousMind
Basically because there's a 'gap' between the 'c' and the 'ti' and rapid speaking hates gaps generally so it gets smoothed out to the gapless 'xi', the sibilant sound at the end of the 'x' goes into the 'i' much more smoothly than the hard dental 't'.
 
Semiclassical
Semiclassical
9:42 PM
yeah, i do tend to say connection like connexion
 
ACuriousMind
ACuriousMind
Besides that, the correlation between English spelling and pronounciation is tenuous at best :P
Though not quite as bad as that of Gaelic
 
Semiclassical
Semiclassical
lol
gotta define the spelling somehow
 
bolbteppa
bolbteppa
'Shew that the fluxion of an ether-based electromagnetic potential must be modified to incorporate a connexion'
 
Semiclassical
Semiclassical
aetheric
 
bolbteppa
bolbteppa
Nice
 
Semiclassical
Semiclassical
9:45 PM
properly I should be using that weird ae
 
ACuriousMind
ACuriousMind
@Semiclassical Actually, the reason for the weak correlation is that English spelling got 'frozen' hundreds of years ago and then some major vowel shifts happened. By contrast, German had a very long period of no unified spelling, and the spelling that got frozen is close the pronounciation because the time that has passed since the freeze has not been long enough for major shifts
 
bolbteppa
bolbteppa
I actually read a seemingly really convincing argument for thinking an aether must exist for the EM field, can't remember it now, something about reference frames
 
Semiclassical
Semiclassical
@ACuriousMind what precipitated said freeze?
 
ACuriousMind
ACuriousMind
@Semiclassical Social convention. I think (but have no actual evidence or sources) that it's related to national identity: The "English" established a firm common identity early on (recall that the Scottish and Welsh, for instance, did not speak English and don't count here), but the Germans remained splintered into a bazillion smaller states until late in the 19th century.
 
Semiclassical
Semiclassical
hmm
I'm surprised there's no specific event that can be correlated with it
e.g. some action by the British government
 
ACuriousMind
ACuriousMind
9:48 PM
@Semiclassical Perhaps there is, I am but an amateur linguist
 
Semiclassical
Semiclassical
 
ACuriousMind
ACuriousMind
I've recently been looking into Gaelic pronounciation and it's interesting how the laws are both rather consistent and absolutely baffling to someone coming from Germanic or Romance languages.
 
Semiclassical
Semiclassical
the first translations/printings of the New Testament in English are mentioned and what I roughly had in mind
 
bolbteppa
bolbteppa
On the one hand it makes a lot of sense to use pushforwards, which are related to changes of variables, and the tangent map part is linear, on the other it seems off/iffy
 
enumaris
enumaris
hmmm
 
enumaris
enumaris
10:45 PM
left with a few issues for the deep speller that seem very non trivial to solve
The good news is, these issues probably wouldn't affect the named entity recognition too much...and it very much does seem that the system does more good than bad heh
 

« first day (2815 days earlier)      last day (2097 days later) »