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12:24 AM
Reading ACM's and AFT's QFT tagged answers is basically just as useful as reading a book when learning QFT if not more useful. (definitely more fun)
12:35 AM
@bolbteppa My algorithm is that I learn into the direction for what I want, and if I find one, then check also the math needed for it.
Now I have read this paper in full: Basically, the phase induced in the electronic wavefunction as it interacts with electromagnetic fields in a semi classical manner is the source of the induced emf, thus explaining why the faraday's law and Lorentz force law give the same emf value. In addition, the phase factor that is imposed to ensure a single valued wavefunction in conventional superconductors result in the persistent current,Josephson effect and Messiner effect as the phase arises from a
topological effect in the superconductor
1 hour later…
1:50 AM
@Secret solitons, YAY! =D
2:46 AM
It is nice to see how the phase induced by gauge freedom finally have some sort of physical interpretation, even though not really a direct one. If this result is still reasonably valid up to the QFT level, then the wavefunction may have more physical impact than we thought of.
of course, the result is still consistent with SR. Ultimately, it is Lorentz transformation that give rise to that gauge freedom in the first place
For those interested, let me know if I have misinterpreted this paper
Q: In Feynman functional integrals why do we integrate the action over all time?

zoobySay the definition of a propagator in quantum field theory is: $$G_F(x,y)=\int \phi(x)\phi(y) e^{i S[\phi] } D\phi$$ where S is the action. Why do we integrate the Lagrangian from $t=-\infty$ to $t=+\infty$ instead of from $x_0$ to $y_0$? i.e. $$S[\phi] = \int\limits_{-\infty}^{+\infty} \int\...

Nonlocality is fun
3:33 AM
@PrathyushPoduval I meant that the M-B distribution function is given later on in the book (along with the derivation). Even before they stated the M-B distribution, they used it to derive the average $E$.
Moreover, $<E>$ doesn't evaluate to $\pi k T$: hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/molke.html#c2
4:30 AM
@Blue yeah,you’re right it doesn’t.....in the context the factor doesn’t matter but since it has its own name, it could be something interesting
@Blue usually it’s done the other way around :P, which book are you using?
4:43 AM
@Secret wavefn is REAL™ =D
@PrathyushPoduval This. It does have some topics which Reif misses out.
4:56 AM
@Blue how is it?
@dmckee I was learning from codesdope.com/c-introduction
And I have learn till string topic
Should I restart from that book ?
@JohnRennie What are your thoughts ?
Have a read through the first couple of chapters of K&R and see if you like the style.
^ Best advice.
You may decide you don't like it, in which case stick to the web site.
Can't say I care for the pep-rally style of the link you provided, but that won't prevent it from being a good guide to the language if it settles down.
5:06 AM
Learning a language should be fun. I found it absolutely fascinating and worked through all the problems because they were such fun to do. If you don't find working through K&R enjoyable then don't use it.
You have to be using a pretty contrived definition of 'powerful' to honestly consider c the 'most powerful language'. Its structures map nicely onto a 1970s CPU, and onto the conceptual model of computation that is still widely used, but it lacks a good deal of built-in expressiveness found in other languages.
I still use it for some things, but given that I already know it pretty well it is a good choice for those things. But I use other languages, too, and mostly because they are faster to code in.
Though many of us retain a lot of affection for the old hoss :-)
FORTRAN is the most powerful language because our nukes run on FORTRAN
Ok I will start learning from the book
Teaching FORTRAN to students is awfully close to abuse :-)
5:09 AM
we just get code thrown at us
@JohnRennie Oh I love it for much the same reason that I love m68k assembly: it's got just the right mixture of access to the metal and niceness.
actually, I take that back
we don't usually get code
no examples = wtf
@0ßelö7 I've seen students get FORTRAN books thrown at them. "Make the code work with this new hardware. You'll need this" ::flutter::
To a poor guy who'd never written a line of code in his life, natch.
We gave him a little help over the next few weeks.
I've written plenty, and it makes it worse
I expect FORTRAN to be sensible: it isn't.
^ :-)
5:12 AM
so, what kind of questions should I be prepared for in a talk
"all of them" is not an answer
@0ßelö7 The ones people will ask.
how am I supposed to know which those are?
To be more serious it is helpful to dry run the talk in front of people in your group who are not intimately aware of what you're doing. They will ask some of the right questions.
What is the subject of your talk?
@JohnRennie Overview of the Yamabe problem with mathematical GR in mind
5:14 AM
Another source of possible choices is questions your advisor asked you when you were first explaining a new line of attack to him. Or the ones you asked him when he explained the project to you.
@dmckee This isn't research yet, it's a build-up to it
Are you presenting to specialists in your area, or to a general (but mathematically aware) audience?
@JohnRennie Half and half.
hi chat
The non-specialists will ask about things they don't fully understand but seem interesting. You just need to know your subject well and be interested in their question and they'll be happy.
5:17 AM
I have a question
@JohnRennie I told the organizer I would have a rough outline of my talks by Monday and he said he would help me narrow it down to something manageable. There's actually an incredible amount of stuff I'd need to talk about
The specialists will ask technical questions and sometimes you won't know the answer. In that case return the question i.e. ask the questioner for how they would approach the problem. It isn't an exam viva - it's just a talk.
@KasmirKhaan what's the question?
it is about function compostition
i keep getting it wrong
Sounds more like maths than physics ...
But ask anyway
@JohnRennie The Yamabe problem was interesting because it forced people to care about optimal constants for Sobolev embeddings, but I don't have the time to talk about how one gets the optimal constants
5:19 AM
its is the steps in the proof if the symmetric groups S_dela and S_omega are isomorphic if they have the same order
let theta : delta ---> omega be a bijection
define phi : S_delta ---> S_omega
by phi ( sigma) = theta * sigma * theta ' for all sigma in S_delta
I need to show that this is well defined
the way i think about it is we take an element in the set omega
theta inverse will send it to an elemnt of delta
then sigma of that will be an element in delta
taking theta last will be an element in omega
The proof is truly long.
I guess I should make notes on the rough ideas for all of the proofs that I omit
There's always far more to say than you have time for. Stick to the headlines and if anyone asks about the details tell them you'll be happy to chat after the meeting.
I can post a picture maybe if that helps
I am not an expert on writing using latex
you don't need to be an expert to write simple equations
in any case, you should ask that question in the math chat
or just google it, someone has written the proof down somewhere
5:26 AM
@KasmirKhaan what is the definition of a permutation
it is a bijective map from a set to itself
so you need to show that $\theta\circ \sigma\circ\theta^{-1}$ is a bijection
hint: it is a composition of bijections
if we work on those 3 compositions
first we need an element of delta so thetha inverse can act on it
are you ignoring my hint?
theta inverse will send it to delta
No i just wanted to tell you where i was stuck
let me think about that hint now
5:29 AM
What you are doing is the long way 'round. Think about what I said.
I need to show that that composition is 1-1 and onto
@KasmirKhaan If $f$ and $g$ are bijections, what is $f\circ g$?
and the set here is S_delta
f*g is also a bijection
Ok, so $\theta\circ\sigma\circ\theta^{-1}$ is...
a bijection :)
5:32 AM
aka a permutation
so that's (a)
I tried to think of it piece by piece and i got lost in details
I just need help with understanding the first part =p
Thanks alot for help :)
4 hours later…
9:04 AM
Q: Why are all questions I visit from other sites protected by Qmechanic?

Vladimir FI normally only come when clicking on some questions from other StackExchange sites. Every time the questions are protected by user QMechanic. Is that something like the Community user? I don't see this at other StackExchange sites.

2 hours later…
10:38 AM
Consider two observers, one freely falling and the other stationary at the centre of the Earth. Both have zero proper accelerations but there is nevertheless a relative acceleration. Explain that without GR!
11:31 AM
@JohnRennie That's not what the question is about
I don't disagree that there's something special about gravity, I disagree that this question is the right place to bring that up - the "paradox" the question poses is much more elementary
I think your assumption that you understand what the OP means is a brave one when I suspect that even they don't understand what they mean :-)
2 hours later…
ACM is always the devil ;)
Why are you digging your nose? ^ @JohnRennie :P
@JohnRennie I may or may not have upvoted it based purely on it being hilarious
1:28 PM
@Mithrandir24601 I upvoted first and then read the answer
@Blue I may or may not have also done so :P
@Mithrandir24601 Oh, you did. I know. ;-)
@Blue I can neither confirm nor deny that
I am the observer. I can. :-)
2:01 PM
@ACuriousMind I have a SUGRA
@Blue @Mithrandir24601 you should of course upvote ACM's answer as well, since it's perfectly correct even though it is diabolical :-)
@JohnRennie ACM's answer didn't have a diagram, which, as we all know, is essential to doing physics :)
what is diabolical about it?
2:28 PM
@0ßelö7 that was a (weak) joke ...
2:44 PM
Some random GR stuff:
@JohnRennie trying to turn that into a functional analysis joke
take two neutron stars, and bring them towards collision. Then check the probability of forming a black hole
I wonder if I can simply paste two schwartchild metric together and interpolate the spacetime between
2:56 PM
@0ßelö7 My condolences
@ACuriousMind I am trying to understand this cryptic phrase "AdS is believed to the the ground state of gauged SUGRA"
What does that even mean?
I think that's the viewpoint where you view different manifolds on which you can have a SUGRA as "states", but I never really understood these handwavy claims, either
@Secret no, because the equations are non-linear
Their solutions cannot simply be superimposed
@JohnRennie Well played, sir
@Secret The usual introduction is global.oup.com/academic/product/…
I don't know if it's good
@JohnRennie Jesus, who starred my physicist comment? Please censor
3:05 PM
A: Why are all questions I visit from other sites protected by Qmechanic?

AccidentalFourierTransform10k users can see the statistics of all protected questions here. Here are some screenshots for the rest of users: Let's play a game: some of these users are human, and the rest are AI's. Which is which?

I seem to be the target of a lot of jokey answers today :p
@ACuriousMind We would never dream of joking about you ;)
3:19 PM
Right, @ACuriousMind - I believe I owe you some answers to questions you asked for my AMA...
@ACuriousMind Yes, your taste in death metal is shit, that's why
You need to listen to Cannibal Corpse
@BalarkaSen I'm quoting Lee in my seminar talk
@JohnRennie why are you letting people troll me on the star board
@0ßelö7 I feel uneasy clearing the stars on a comment that has 8 stars and isn't obviously offensive or demeaning.
Whoa! Nine stars :-)
10 now
it's obviously trolling
3:33 PM
I wonder what the record ever number of stars is ...
I am pretty sure I have seen 16 or 17 in the math chat
Not sure what's it here
@0ßelö7 OK, OK, I'll clear the stars on the grounds it's offensive to you.
a lot of people put an active effort in starring that message, you know
@BalarkaSen Be Nice
3:35 PM
I spent 2 seconds of my life in being the 8th starer
But you shouldn't really complain about people starring a post you made in public.
So it's offensive to me that you removed the stars
If I posted saying I'd painted my bottom green then I couldn't complain if people subsequently starred it.
@JohnRennie it was a lie
No, I really have painted my bottom green :-)
3:36 PM
I'm not a physicist and I don't want people thinking so
(I wanted to be able to hide in a field of watermelons)
Alright, this is actual, genuine star-trolling now
@BalarkaSen (Lee has a famous paper on the Yamabe conjecture/positive mass thm)
You realise I'm going to cancel all these stars as soon as everyone looks away. The star board does not need posts about my bottom.
this paper trail is actually impossible to follow
5 mins ago, by John Rennie
But you shouldn't really complain about people starring a post you made in public.
John Rennie confirmed hypocrite
3:41 PM
Excellent, we have effectively sabotaged the star board! :-)
@BalarkaSen Ugh, this is really horrible. Do you know what the Yamabe conjecture is?
Wonderful starry sky, tonight. I should get my binoculars. :)
@0ßelö7 nopee
some serious vaporwave cover art there
3:45 PM
@BalarkaSen In 1960, H. Yamabe thought he could prove the generalized Poincare conjecture by conformally deforming a compact Riemannian manifold and obtaining a metric with constant curvature.
As a first step, he wanted to prove that the conformal class of a given metric always contains a metric with constant scalar curvature.
@BalarkaSen r/surrealmemes, I think you'll like it there.
He thought he had a proof, but in 1968 Trudinger pointed out a serious error in a part of the proof. He repaired 2/3 of it, but the famous $\mu>0$ case was still open.
Then in 1976, using ideas from mathematical quantum mechanics, Aubin gave a simple condition for the conjecture to be solvable. He conjectured that every manifold not conformally equivalent to a sphere satisfied the condition.
@0ßelö7 Ah right I berember this
@ACuriousMind Weak measurements: vague, brief idea: take the system you want to measure and couple it to some sort of quantum device or degree of freedom, that we call a pointer. Weakly interact with the pointer (allowing for perturbation theory to be used) and optionally measure the system, postselecting the outcome states to get the appropriate weak value. Why is this interesting? Aside from a reasonable number of discoveries that arose from this (e.g. Spin Hall Effect of Light),
3:48 PM
it can be used to do interesting things like measure the basis (at the expense of the loosing the info of the state). You can get a complex weak value - e.g. a complex value of position means that both the position and momentum have been changed, on average. Another interesting bit here is that this particular effect can be replicated classically.
This made me question everything I thought I knew about Hamiltonians and led me to the PhD topic I'm doing :) It also seems to provide a link between interpretations. It can actually be used to perform feedback to a qubit. As it shows that averages of observables measured over large ensembles seems to be able to be given rules from classical dynamics [I feel like I didn't describe that bit very well], it's lead me to wonder if this can be used to look at quantum to classical transitions.
@IcEmybReaD beautiful
@BalarkaSen Then in 1984 Schoen gave a proof of Aubin's conjecture, but it in turn assumed a conjecture due to Yau. Schoen claimed Yau had the proof, but never told anyone what it was.
@0ßelö7 kek
So Schoen is credited with the proof, but it's kind of cheating. On the other hand, Lee and Parker gave a different proof that doesn't assume Yau's conjecture.
On the other hand, Schoen's paper is really long, and I think it contains a third proof.
what the fuck? I ignore sakurai for like a year and he turns BLUE?
3:54 PM
@IcEmybReaD Yeah. To honor me and mark the beginning of my QM journey. ;)
@BalarkaSen Schoen and Yau also wrote another proof that's an even slicker version of the Lee proof. Then there's a [personal communication] proof due to Vaugon.
I think Hebey has a complete proof, but his book is obscure and in French.
@IcEmybReaD ARKKK!!
So the conjecture is "solved" if you're willing to read like 6 different papers
@BalarkaSen But the proof is very neat because it uses quantum mechanics and GR
wait a min, it's still the second edition. so what changed other than the cover? (and for what reason?)
@IcEmybReaD probably fixed some typos
3:58 PM
were there lots?
you read it for your course right?
Not many
@BalarkaSen oh god I have rep theory homework
@0ßelö7 thats fun
@BalarkaSen Well, not QM and GR, but is motivated by it
@BalarkaSen If $\mathbf{A}=\nabla \phi$ everywhere in space, where $\phi(x,y,z)$ is a scalar valued function, then what is necessary and sufficient condition on $\phi$ such that $\int_{P_1}^{P_2}\mathbf{A}.d\mathbf{r}$ is path independent. Does $\phi$ being continuous and differentiable everywhere suffice?
4:08 PM
@Blue $\nabla \phi$ needs to make sense, that's all
@BalarkaSen What do you mean by make sense?
This is called the gradient theorem
@Blue Well, $\nabla \phi$ doesn't make sense if $\phi$ isn't differentiable :P
@BalarkaSen Okay. I was asking if differentiablity is enough or we need something stronger than that? Like continuously differentiable?
I guess the proof is just using the multivariable chain rule
4:11 PM
@BalarkaSen Okay, thanks
This is basically fundamental theorem of calculus
"If F is a conservative vector field (also called irrotational, curl-free, or potential), and its components have continuous partial derivatives, the potential of F with respect to a reference point ${\displaystyle \mathbf {r} _{0}}$ is defined in terms of the line integral:"
Scalar potential, simply stated, describes the situation where the difference in the potential energies of an object in two different positions depends only on the positions, not upon the path taken by the object in traveling from one position to the other. It is a scalar field in three-space: a directionless value (scalar) that depends only on its location. A familiar example is potential energy due to gravity. A scalar potential is a fundamental concept in vector analysis and physics (the adjective scalar is frequently omitted if there is no danger of confusion with vector potential). The scalar...
Found this
i don't think the continuity is necessary to define the line integral but prove me wrong
4:16 PM
@BalarkaSen We need it to prove "path-independence" of line integral
We can define the line integral without it for sure.
well if you can tell me where continuity is needed in the proof of gradient theorem you win
Let's take: $f(x,y)=x^2\sin(1/x), x\ne 0, f(0,y)=0$
$f$ being the potential function.
4:23 PM
Wait, I'm trying it out myself before claiming anything
@BalarkaSen The FTC can fail if $f'$ is not $L^1$.
Assuming it it continuous is sufficient.
@0ßelö7 Yup
Oh I see
@Mithrandir24601 Nice, thank you
@BalarkaSen by FTC I mean $\int f'=f$
@0ßelö7 Well, you're basically saying FTC fails if $f'$ is not integrable :P
arent you
4:25 PM
Right, so you need to assume $f\in C^1$
Well, not need, but it's sufficient.
You don't need it
you just need to make sure everything makes sense
Unless you want to get into measure theory, that's a hard thing to do.
@Blue Like I said, if everything makes sense, it holds
C^1 is just a convenience
@BalarkaSen In multiple variables I doubt there's a good condition without GTM. "Integrable along ever path" is a shit condition.
i dont care
4:28 PM
What do you care about?
probably assuming that it's discontinuous on a measure 0 set suffices
well actually that's the Riemann-Lebesgue theorem
@0ßelö7 I care about good, smooth-ass functions
@BalarkaSen a curve is a set of measure zero.
it needs to be discontinuous on a set of $\mathcal H^1$ measure zero :P
ah ok
well frick this shit
why is even Blue bothering about this pedantic thing
isn't everything C^infty to engineers
@BalarkaSen Reason being, $\mathcal H^1$ gives arc length measure on Lipschitz curves.
@BalarkaSen I'm not an engineer! :P
4:31 PM
@0ßelö7 Ah I see
@Blue you're not an analyst either
assume everything is infinitely differentiable and make life easier
@BalarkaSen I interested in everything but good at nothing <---- story of my life.
As an analyst, my advice is to listen to Sir Sen
It's too late for me but you can save yourself
2 minutes of silence in memory of 0celo7
@BalarkaSen Did you understand why continuity of derivatives is needed? Explain! :P
4:35 PM
It's not needed, as we see in the discussion above @Blue
I tried with a circular path and a straight line path for the above example. It gave the same result.
Is he just ignoring me
It's just a convenient assumption
Could someone guide me through that problem? My attempt:
4:36 PM
Which you should always assume
without telling anyone you are assuming it
but assume anyway
I know that the velocity of the centre of mass is constant coz no ext force is acting on the system.
The $v_{com}= \dfrac{v_0}{3}$
Also, there's tension in the string but I am not sure whether it's equal.
Moreover, momentum will be conserved during collision.
I should try with a path that crosses $x=0$ twice perhaps and check
@BalarkaSen I see
@Blue I gave a sufficient condition above
@0ßelö7 Which one?
4:37 PM
$\mathcal H^1(\{x\in \Omega: f\text{ is discontinuous at $x$}\})=0$
@0ßelö7 I don't know what that means
It means that restricted to curves, $f$ should be continuous a.e.
You're asking a very complicated question
Any time you ask a calculus question that involves something discontinuous, the answer is awful.
Let's write a book on differentiable but not C^1 calculus
for freshmen
@0ßelö7 Okay, you could just give me an example where just derivative existing is not sufficient for path-independency of integral. That could serve as motivation for me to pursue it detail further.
4:40 PM
@Blue The point he's bringing up is $\int_a^b f'(t)dt$ need not be $f(b) - f(a)$ if $f'$ is not even integrable over $[a, b]$
In mathematics, Volterra's function, named for Vito Volterra, is a real-valued function V defined on the real line R with the following curious combination of properties: V is differentiable everywhere The derivative V ′ is bounded everywhere The derivative is not Riemann-integrable. == Definition and construction == The function is defined by making use of the Smith–Volterra–Cantor set and "copies" of the function defined by f ( x ) = x 2 sin ⁡ ( ...
@ACuriousMind Is asking that question permissible on the main site?
The standard counterexample.
@Blue Anyway, if $\nabla \phi$ is integrable over any path between $P_1$ and $P_2$, $\int_{P_1}^{P_2} \nabla \phi$ is indeed path-independent
@BalarkaSen What's the condition for integrability of $f'$ over $[a,b]$? Or is the answer to that too complicated?
4:42 PM
@BalarkaSen There's a Lebesgue-measure criterion for $\int f'=f$ to hold. It's not awful.
That's all you need to assume
@Blue Do you know what a null set is?
@Blue Yes, that's precisely what 0celo7 is talking about. You need measure theory to understand that
@0ßelö7 Yes
But I don't know any measure theory. Only high school set theory
4:42 PM
@Blue Ok, tell me
You need to say $f'$ is discontinuous on atmost a set of measure 0
No, not the null set
A null set
@0ßelö7 A set without any elements?
4:43 PM
@0ßelö7 I think no. I don't know the difference between a null set and the null set.
No, $A\subset\Bbb R$ is null if $\forall \epsilon>0$ $\exists$ covering of $A$ by open intervals $I_1,I_2,\dotsc$ such that $\sum_{k\ge1} |I_k|<\epsilon$.
@0ßelö7 Woah. Interesting. I'll read about this later. But thanks for the motivation.
If $f:[a,b]\to\Bbb R$ is bounded, then $\int_a^b f$ exists in the Riemann sense iff $\{x\in [a,b]:f\text{ is discontinuous at $x$}\}$ is a null set.
...why are people starring that?
@0ßelö7 Oh, sort of makes sense. We are using epsilon-delta type of stuff to define the null-set
For every $\epsilon$ we can find a $k$, such that...
4:49 PM
$k$ goes from $1$ to $\infty$
$x \in [a,b]$ such that $f$ is discontinuous should be a null set.
@0ßelö7 Right
$k\in \mathbf{N}$
$A$ being null means that is has zero "length"
4:50 PM
@0ßelö7 Gotcha. That sounds like the beginning of measure theory.
Yes. The proof is so awful it's easier to invent measure theory instead.
(A similar result holds for Riemann integrals in $\Bbb R^n$ and the proof sucks.)
@BalarkaSen I'm stupid, when one does a product of groups are the elements supposed to commute?
5:04 PM
@Abcd Looks very homework-like to me, so no. I'd help you but I'm kinda busy right now
@0ßelö7 (g, 1) and (1, h) commute, yes
How the hell do I compute a socle!?
5:28 PM
@Blue bla bla bla
In the end, "hence proved" :P
@Sid Well, it's not trivial. Epsilon-delta used to drive me crazy. And still does, several times.
@Blue Same here.
Whole of analysis seems to be about epsilon-delta
5:44 PM
@Blue that is offensive
@0ßelö7 At least my analysis course seems to be mostly about that stuff :P
@Blue flagged lol
@IcEmybReaD Fortunately none of the mods are analysts
why do some authors feel like they have to use their own random notation and terminology when they write their books instead of the standard? when the book itself is really well written and readable, that becomes the deal breaker.
If you define $\sum_{n=0}^{\infty} f(x+n) = \sum_{n=0}^{\infty} (e^D)^n f(x) = [1 + e^D + (e^D)^2 + \dots ] f(x) = \dfrac{1}{1 - e^D} f(x)$ then $D^{-1}$ is $D^{-1}f(x) = \int f(x) dx$, but in fact we should define $D^{-1}f(x) = \int_{\infty}^x f(x) dx$ P4, phys.uconn.edu/phys2400/downloads/euler-maclaurin-summation.pdf , any thoughts on the crazy limits?
5:59 PM
That's the same as $-\int_x^\infty$. Not that strange

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