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00:00 - 21:0021:00 - 00:00

00:03
@ACuriousMind What?
@Slereah exactly :(
What am I gonna do
00:15
you need some gary gum
00:50
Cruz has annihilated the moderators.
Rubio annihilated the moderators and slammed Bush when he tried to jump in.
This is grade A television.
 
5 hours later…
05:29
My most common careless mistake in a maths evaluation is what I called the "3+x3 mistake"

That is, for some reason, there is a high tendency for me to compute 3+3=9 and 3x3=6

I alos made heaps of other careless mistakes of which the only sign that I know there is a mistake is I got 0=0 after I finished the calculations

I have screwed up many exams including my linear algebra one because of that

I had a feeling my frequent mistake that lead to 0=0 might be because I spent too much time literally dividing by zero in abstract algebra, thus my mind got locked up
I end up always do 4 first, unless I know the calculation will be tedious
This is because I otherwise often lose track on what on earth I am computing because all I see in the programs are insert expression, split out expression, only realise later I made a typo
It might be because of my chemistry roots, probably like to get me hands dirty all the time
05:45
My current way to interpret quantum is to treat everything as a superposition, and then only via measurements you get some physical quantities from it. After that, the system continue to do its business as a superposition as it evolve in time

thus I will expect things like probability of finding a particle, the momentum distribution, the energy distribution, angular momentum etc. will continue to fluctuate in time as the system evolves

The problem of this interpretation is that it run into trouble because it is assuming the state vector is an entity as real as electromagnetic fields, whic
There's also something which is kinda a mix between presentism and block time called the growing block view, where the present and past exist, but not the future. The past then grew larger as time progress.

I tend to favour this view because it most closely match our daily experience about time, but I don't think this is compatible with general relativity because the spacetime manifold is a static structure (it does not evolve with time, everything that had happen and will happen is already set to stone by energy momentum in the spacetime)
@Slereah I might want to read that paper in detail, because whenever I saw something like this, the pop science alternative reality concept will quickly pop into my head and I will often confuse the two
 
4 hours later…
09:50
It's not about alternative reality
If anything it was a rather big problem for the existence of CTCs
An initial slice that can develop into a CTC can also develop into another spacetime
(a singular one)
10:07
5
A: Inverse Quantum Operator

Valter MorettiThe operator $X^{-2}$ does exist and is self-adjoint as it follows from standard spectral theory. Its domain is $$D(X^{-2}) := \left\{\psi \in L^2(\mathbb R, dx) \:\left|\: \int_{\mathbb R} x^{-4} |\psi(x)|^2 dx \right.< +\infty\right\}$$ and thereon $$(X^{-2}\psi)(x) := x^{-2}\psi(x)\:.$$ The e...

Ohhh snap
10:53
0
Q: Pointless comments

Matt SThe post I speak of is below. What does the discovery of a pentaquark signify? Whilst OP has hardly been perfect in their articulation of the question, it does to me seem to be clear what they're asking. They want to know about what the discovery of the Pentaquark means for physics. Popular me...

shots fired
Seriously
Who the hell did that
I think it's an excellent comparison, even down to the flags :P
How is American English "simplified" compared to Big Ben
12:34
have you not ever heard Shakespeare?
Shakespeare stole everything from American writers.
@skillpatrol I don't see what a 17th century (or whenever, long ago) poet has to do with anything.
What part of "Traditional" don't you understand? :P
You're telling me British people speak like Shakespeare
They speak traditionally. Nothing more, nothing less pal.
On average, of course :-)
12:50
You're a troll.
Removed, eh? The mods get you?
meh
I think the only time they got me in this room was when I lost my cool with EE ;-)
Oooh ya, once with Sophie :(
But seriously, the "Simplified English" = American English was not an attempted trolling @0celo7
I apologize to the room if it is perceived that way.
13:21
http://physics.stackexchange.com/questions/215317/two-particle-system
Reading this question makes me curious

There's nothing being mentioned that the particles are fermions, thus what prevents them from occupying the same state?
Two particle states for fermions are antisymmetric
If they occupy the same state the wavefunction is 0
I know that, but nothing in that question mentioned they are fermions, they could well be bosons?
Well when you don't care about spin statistics you can just write it as the tensor product of both vectors
Formally bosons should be symmetric and fermions antisymmetric, though
yup, that's why I am also confused about the OP's maths clearly showing that there is some kind of exclusion going on there despite the two particle may not necessary be fermions (because he/she has not state that)
since they are not necessary fermions, we cannot say that is pauli exclusion
so if this effect is real, there must be some more general type of exclusion principle that I have not learnt before in my course
at least that's how I interpret from that question
@Slereah that's me i nthe lab :(
@Secret This is implied somewhere.
13:34
Then it must be had something to do with the tensor product, I am honestly not very famialr with the rigorous treatment of tensor products
Look at how the sign changed, $\lvert 1\rangle\otimes \lvert 0\rangle\to-\lvert0\rangle\otimes\lvert1\rangle$
The question is completely confused about what $\otimes$ denotes and what an equality is.
Also you might have noticed that he got rid of the identical states
Half of these equal signs are not equal signs.
It's antisymmetric, he just doesn't say it explicitely
13:36
second to third is wrong
I think
$\otimes$ is not antisymmetric.
@ACuriousMind right, but that's what OP did
O I see, I thought the sign changed because of some rules in tensor algeba I don't know about

Yeah I noticed the identical states vanished for no reason too
It's just kind of informal notation
@0celo7 Yes, exactly. That calculation makes absolutely no sense.
13:36
calm down
it makes some sense
there are numbers and stuff
@0celo7 is quantum mechanics with tensor products taught at graduate not undergrad years?
It is pop science nonsense, @0celo7
@Slereah It's abuse of notation for no good reason and obscures what is going on
@ACuriousMind aka physics :p
Speaking of
This is again some physicist trying to do math and utterly failing :P
13:37
I should send a mail to Carlip
lol
@PrishChakraborty The mathematics is wrong, so any conclusions you make are also wrong. — 0celo7 7 secs ago
@Slereah why
so Bill Nye is coming to campus today
I think
A complete mess, I would say. — user36790 13 secs ago
Haha based user
when did you guys first learnt about the rigorous treatment of the tensor product and its application in quantum mechanics?
About his whole use of hamiltonians with a non globally hyperbolic spacetime
he must answer for his crimes
crimes? maybe he's just better at this than you
hmm, we have some hypodermic needles in the lab
I could become a heroine addict
That is a sudden turn of events.
13:43
@ACuriousMind I've been considering this for a while
Better take amphetamines, they at least make you better at math
you wanna be a junkie?
@ACuriousMind I have an Adderall prescription, which I stopped taking
felt like cheating in a way
but I guess I need it to not be trash
@skillpatrol junkies can't feel the pain of sucking at math
13:45
true
unless you become a math junkie :P
@0celo7 @skillpatrol @Slereah @ACuriousMind @Danu I mean, what is the minimum stage in uni will one start to learn about the rigorous treatment of $\otimes$ and its applications in quantum mechanics?
@0celo7 Uh, does the time you stopped taking it coincide with the beginning of your perceived decline?
@ACuriousMind : Got my answer, thanks. — Prish Chakraborty 2 mins ago
@ACuriousMind Not sure!
There's no need to involve tensor products really
I didn't take it over the summer, but I didn't try to do any math over the summer, really.
13:47
In the Schrodinger formalism this just corresponds to the product of two wavefunctions
That was just the pre-college relax and play video games and random crap summer.
what did your doctor say?
The doctor can't say anything because my grades are all high As.
There's no measure for trying to read ahead and failing.
still, it is a prescription
Yes, which I take on a voluntary basis.
13:48
for a reason
No, it's there when I need it.
And I don't want to take it because it might show up in drug tests or whatever.
@Secret Unanswerable, since that depends highly on the course structure at your university
@Secret dude I'm a freshman
@0celo7 ...you are being drug tested?
I'm fucking with a spring problem right now
@ACuriousMind If I want to work in the nuclear industry, of course.
My "boss" is getting his on Tuesday for his new job.
13:51
Uh
Okay
prescription drugs are not allowed?
And I'm not having trouble concentrating or whatever. I just don't get things like I used to.
It's very strange, I might even be imagining all of this.
@0celo7 The "of course" is debatable. Mandatory drug tests are forbidden here, for instance.
@ACuriousMind Germany doesn't have a nuclear industry.
yet :P
they took it away after WWII
13:53
@0celo7 We still have some reactors. But the specific industry doesn't matter - it is a general law that only the police can force you to take a drug test.
And most jobs in the nuclear industry (the cool ones) require a clearance or NDA. The former has mandatory testing by law, I'd be surprised if you showed me an example of the latter in an industry involving public safety that doesn't require it.
@ACuriousMind ayy I don't want to work in a reactor
why do people think this
that would be the most boring job ever
@skillpatrol We didn't have nuclear power "after WWII". It was much later.
@0celo7 I did not say that at all.
Ok, but your comment is still irrelevant because I want to work in America.
You should go back to your doctor and talk about this.
And I won't work in Germany unless they get some common sense and rebuild the nuclear industry, I end up doing something like applied math in grad school, or I do something else German friendly.
Drug tests are mandatory for a lot of jobs here, not something that I disagree with, mind you.
13:56
I also didn't want to convince you to come to Germany, I just commented that the "of course" there is not as universal as you might think
Is he a specialist?
If I can't do simple tasks without medication, of course I shouldn't be allowed to work on million dollar experimental equipment.
@ACuriousMind It's an "of course" for America.
@0celo7 I'd say that that depends entirely on the type of medication.
@skillpatrol What is there to talk about? I have good grades and everything is fine. What am I supposed to tell him: "I'm struggling with exercises in books that most people my age would read on 3-4 years."?
I have a perceived decline, there is no actual evidence.
And no, he's not a specialist. Seeing a specialist is a good way to get barred from the industry for life.
14:00
oO How would visiting a doctor get you banned? How would anyone even know you visited one?
@ACuriousMind Medical records.
@0celo7 They are accessible for persons other than your doctors in the US? oO
The government is allowed to know they exist, but not what's in them. So they have to assume you're mentally ill if you go to a psychiatrist.
You have no concept of what a security clearance entails, do you?
How often do you drink alcohol, how often do you have sex, with how many people have you had sex?
I guess my concept of the rights of an individual is a lot stronger :P
No barriers.
The latter two might be military only.
They will ask you if you've had oral sex if you're trying to get a military clearance.
(It's illegal by the UCMJ.)
14:03
How long have you been on this medication?
@0celo7 Does the NSA has a role on this, given how they done all that survellience to all US citizens?
At least it used to be when my parents were in the service, which has been almost a decade.
@0celo7 See, not even in the name of "security clearance" would anyone be allowed to ask those questions to a potential employee here.
I can't make sure those things aren't researched, granted, but at least that would be completely illegal.
@ACuriousMind Right, but you surrender those rights by choosing to get a clearance. So nothing is violated.
They ask your friends and family how you spend your money, if they've seen you taking any pills.
@Secret I don't know.
O well, I guess I should not have asked such questions, but anyway...
14:06
Hm
I wonder how much reading and policies I need to know if I move to US to study or work...?
A lot of the stuff about Pauli Fierz theory talk about the lagrangian
But not the action
@Secret Citizenship.
It's even worse for foreigners.
So the factor of $\sqrt{-g}$ goes unmentionned
@0celo7 It is a violation of the fundamental character of the rights if someone is allowed to make you giving them up a precondition for being employed. The only way anyone should be allowed to make you give up your rights is if you broke a law.
14:07
@ACuriousMind I'm not defending the actions of the employer, but that's nonsense. It's voluntary interaction.
@slereah perhaps you can integrate the lagrangian with $d\tau$ to see if something sensible pops out?
Just like there's no right to be paid $15 an hour. You get paid what you agree upon with the employer.
So if you don't agree to the terms of the clearance, you don't work for them, simple as that. No rights are being violated.
Well no
It's a Lagrangian density
That wouldn't work
no min wage?
min rights
as a person
No, there is no "minimum wage right".
14:10
Hm, even Deser's paper doesn't mention it
And he writes the Hilbert action as $\int d^4x R$
No determinant D:
That scoundrel
I guess the only guy to actually write down the full thing is just Feynman
@Slereah send him to the guillotine, face up :P
@ACuriousMind Again, I don't think they should do it, either, but it's not a violation of rights.
@0celo7 Well, I don't know the laws in the US, but I think that demanding such a large breach of privacy as a precondition for being employed would be illegal here. You may require someone to volunteer all relevant information (e.g. medication they are taking that influences their ability to perform the job), but you may not require them (nor take into consideration) to expose their medical records.
They can't make you expose medical records. They can learn of the existence of them.
And I understand it is illegal, but legality is not morality.
U.S. Gov't definition of "relevant information" = any information that could be used to blackmail you
i.e. hidden prescriptions for e.g. lithium, hard drug use, etc.
They ask about spending to see if you're living beyond your means and might be inclined to take a bribe in exchange for a file.
Snowden certainly made everyone's lives harder
15:09
Why is chat dead
 
2 hours later…
16:41
Why do people with hats look so smug?
The Relativistic Rocket article by John Baez is gone!
Maybe he discovered it contradicts Einstein and The Evidence.
Rockets are pop science constructs
Only found in science fiction
Yup. So is relativity.
Google on "Newtonian action at a distance" for Newton and The Evidence.
Ugh, action at a distance is pop science nonsense
It is done via a fluid filling the ether
The caloric fluid?
17:19
Couldn't tell you
Haven't read a lot of 16th century papers
17:45
lol Springer tech support has no clue why books won't download
18:04
@ACuriousMind I'm confused...geometrically, what is the Hamiltonian? Is it just the thing that generates flows?
@0celo7 Yes.
@ACuriousMind Then I am not confused.
@ACuriousMind So what are the minimum components that one needs to do Hamiltonian mechanics?
i.e. what assumptions/axioms
Since we do Hamiltonian mechanics on the cotangent bundle, the tautological form and symplectic form are not postulated, they are purely geometrical objects intrinsically associated with the cotangent bundle, right?
So all we need is that $X_H$ generates flows and that $\int p\mathrm{d}q-H\mathrm{d}t$ is extremal?
@0celo7 The latter is more abstractly the statement that the action is $\theta(X_H)$ for $\theta$ the tautological one-form, but yes, that's it.
The action is that?
Arnold why did you not tell me this
@ACuriousMind Hmm, I don't see why that's true, you mind explaining?
(that the action is $\theta(X_H)$)
18:28
0
Q: Doubt while computing the animalous dimension of $\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi\bar{\psi}\gamma^{\nu}\partial_{\nu}\psi$

ScardenalliI am following the conventions of http://isites.harvard.edu/fs/docs/icb.topic1146665.files/III-9-RenormalizationGroup.pdf. Consider the QED Lagrangian $$-\frac{1}{4}Z_3F_{\mu\nu}^2+Z_2\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi-Z_2Z_mm\bar{\psi}\psi+Z_eZ_2\sqrt{Z_3}e\bar{\psi}\gamma^{\mu}A_{\mu}\p...

Animalous
I hope there's lions and zebras!
what?
I assume animalous refers to being full of animals
no, you can't read
it says anomalous
oh
I'm retarded
the title is wrong but the text is
why not comment :D
@Slereah do you know why the action is $\theta(X_H)$?
I guess I could just be a buzzkill and correct the title :p
Errr
I think Rovelli talks about it?
Lemme see
@0celo7 The tautological form has locally the form $\theta = p\mathrm{d}q$, while the Hamiltonian vector field is $\frac{\partial H}{\partial p}\partial_q - \frac{\partial H}{\partial q}\partial_p$. Thus, $\theta(X_H) = p\frac{\partial H}{\partial p}$. Plug in a path $(q(t),p(t)$ and integrate it over $t$ and you get $\int p\mathrm{d}q$.
18:34
wait
Sorry, should have said the action is $\int\theta(X_H)[q(t),p(t)]\mathrm{d}t$
is the action principle $\int p\mathrm{d}q$
or with a $-H\mathrm{d}t$
Arnold has both
The action $p dq - H dt$
@ACuriousMind what
what are these square brackets
It's just the canonical transformation of the Lagrangian
18:36
@0celo7 Just taking the arguments. $\theta(X_H)$ is a function $T^*M\to\mathbb{R}$.
ayy what
So you feed it $(q(t),p(t))$ as argument.
Oh, it's bundle coordinates
@ACuriousMind yeah, I don't get the last step
I understand how you got $\theta(X_H)$
@0celo7 $\partial H / \partial p = \dot{q}$ on-shell, then $\dot{q}\mathrm{d}t = \mathrm{d}q$.
18:40
@ACuriousMind rightttttt
was typing that
so I'm confused by what the action really is
is it $\int p\mathrm{d}q-H\mathrm{d}t$ or $\int p\mathrm{d}q$
@0celo7 Any functional of paths which when extremized gives the physical trajectories.
There is not "the" action
I get that
There's a bunch of them which area all more practicable in different situations
but do both of those integrals work?
E.g. the $H\mathrm{d}t$ part is completely superfluous if you fix an energy from the beginning.
18:42
hmm
what do you mean
If you consider only paths with constant $H$ from the beginning, the $H\mathrm{d}t$ part doesn't vary when you vary the paths
ok, really stupoid question time
so it contributes nothing
if H is constant, how do its partial derivatives make sense
do we mean the total time derivative is constant?
@0celo7 Never said they did, I did not fix $H$ to be constant when I took those derivatives
18:44
o.o
what
Also, $H$ is still a function on the entire phase space. Even when you restrict the path to a surface of constant energy, you can still just examine the derivatives of $H$ in directions that point off the surface, i.e. are non-zero
Don't be "confused" thinking you don't understand something. The many different action principles are confusing, and it is difficult to relate them all in a completely rigorous manner.
oki
one last thing
energy is the Noether charge for time translation?
OK I know that
what I don't know it how to explicitly construct the energy as the charge
specifically using the formula on page 88-89 of Arnold
I'm assuming there $I$ is the charge
Well, in the Hamiltonian formalism, it is trivial: $H$ commutes with itself and is hence conserved during time evolution. It generates time evolution. Hence, it is the charge for time evolution.
18:52
I understand that
the page 88 to 89 part is what I don't understand
how to get energy from that, I mean
@0celo7 Uh, just take time evoluition as the $h^s$ and compute away?
$h^t(q)$ is just $q(t)$ for $q(t)$ being a solution of the equation of motion, thus the conserved charge under time translation is $\frac{\partial L}{\partial\dot{q}}\dot{q}$.
uh that's not the total energy
Yeah, should be $-L$ appended.
But Arnold's formulation does not give that.
What, exactly, is a "first integral" in his diction?
a constant of integration I think
It was never really clear or important
1
Q: How do first integrals help you solve differential equations?

JessicaKI am reading about Euler-Lagrange equations and this particular section is a little unclear. Consider the differential equation $$\begin{bmatrix} \dot{x}\\ \dot{y} \end{bmatrix} = \begin{bmatrix} -7x^{\frac{6}{7}}y^{\frac{1}{7}}\\ 7x^{\frac{1}{7}}y^{\frac{6}{7}} \end{bmatrix}\quad (x(0),y(0)) = ...

a first integral defines a level set
19:04
Eureka!
His Noether theorem is not general enough to derive energy as a Noether charge, I think
look at the next page
bottom of 90!
he hid something in a problem I ignored
Yes, exactly. His stated version does not allow for time translations.
@ACuriousMind what is $\tau$ in the hint?
@0celo7 The parameter of a path $(q(\tau),t(\tau))$ in $M_1$.
19:13
how is it related to $t$
Not at all.
then how do we take a derivative
if $t$ and $\tau$ are not related, how is $\mathrm{d}t/\mathrm{d}\tau$ defined?
@0celo7 This is meant on a path, just as $\dot{q} = \frac{\mathrm{d}q}{\mathrm{d}t}$ holds only on paths.
19:16
huh?
how is it defined on a path then
Okay, let's clear something up: When we use coordinates $(q,\dot{q})$ on $TM$, $\dot{q} = \frac{\mathrm{d}q}{\mathrm{d}t}$ does not make sense because what the hell is $t$? Agreed?
right...
$t$ is just the curve parameter
Exactly. In the same way, $\tau$ is the curve parameter for paths in $TM_1$ which has coordinates $(q,t,\dot{q},\dot{t})$.
true, true
ok so I need $I_1$ for time translations
@ACuriousMind So now we get multiple terms for the definition of $I_1$?
because we count $(q,t)$ as coordinates of the base space
I don't understand the question
19:29
what is $I_1$?
@0celo7 The first integral you get by applying Noether's theorem to the Lagrangian system of $L_1$ on $M_1$.
ayy
stop being my TAs
I'm trying to guess the explicit form
@0celo7 huh?
@ACuriousMind ignore that
Okay.
I have to go do something, so expect no reply for ~30 mins.
19:32
don't leave me all alone with the Dutchman
huh, Danu entered and then left
now his picture is faded out, this is strange
Still here
faded picture just means doing stuff in different tabs AFAIK
I thought faded picture means away from chat for a while
20:30
@ACuriousMind Is every symplectic manifold the cotangent space of some configuration space?
Isn't it just in general any 2nD manifold
The sphere is even dimensional and it's not a cotangent bundle of anything
It's not even locally a product
Can we give every 2nd manifold a symplectic form?
I don't think every symplectic manifold is locally a product (and hence it cannot be a bundle)
@0celo7 You need orientability
Because the symplectic form yields a volume form
and the existence of that is equivalent to orientability
So symplectic -> orientable
@0celo7 No, but I think every symplectic manifold is the "quotient" (or reduction) of a cotangent bundle if I'm reading this MO thread correctly
Yeah
20:35
@ACuriousMind Nice link
@ACuriousMind huh
So you just quotient a cotangent bundle by something and it spits out a symplectic manifold?
Well, the cotangent bundle is already a symplectic manifold. There is now the procedure of symplectic reduction (which morally is a quotient in the category of symplectic manifolds, but not quite the usual quotient of manifolds) that produces a lower dimensional symplectic manifold from it. One example would be the reduction of the phase space to the reduced phase space (the constraint surface) in the case of constrained Hamiltonian dynamics.
Spits out another, I should have said.
> category
You know I don't know what that means :P
In another description, it's quotienting out gauge orbits so that the reduction only contains one point from each gauge equivalence class
Sorry, I think symplectic reduction is really not that simple because it is not usually done in most physics contexts I know
@0celo7 Thing that features objects and structure-preserving maps
(= homomorphisms in the appropriate sense)
20:46
I think I knew that
@ACuriousMind When is it used?
@0celo7 Well, it's what one formally does if one eliminates gauge degrees of freedom.
QoGS?
No, QoGS does not talk about formla symplectic reduction at all. Their terminology is quite physicy, and they call it "finding the reduced phase space by solving the constraints". I have two posts on symplectic reduction here and here (in the second part)
your posts depress me
I'll take that as a compliment :P
20:58
...in the sense that I don't even know where to begin to try to understand them
00:00 - 21:0021:00 - 00:00

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