« first day (5085 days earlier)      last day (39 days later) » 
02:00 - 20:0020:00 - 00:00

2:23 AM
Can someone atleast tell me how do I get rid of the $i=\sqrt{-1}$ from the $U \phi U^\dagger=\phi+ [U,\phi]$
The oscillator algebra does not have any $i$. So how can we get rid of $i$?
 
2 hours later…
4:34 AM
@Slereah their discord server is pretty much a ghost town.
5:08 AM
Momentum operator seems to be non-hermitian,so will it's eigenfucnctions,still necessarily span the whole of Hilbert space?
123
123
5:42 AM
Hello Everyone...
6:38 AM
@Arjun if you call something momentum, and it is not an observable that satisfies the postulates of quantum theory, then you need to at least tell us what extension of quantum theory you are using.
@nickbros123 Miao miao more often have to deal with the opposite problem. Some maths student comes to meow with a question whereby they can state and prove all the relevant theorems that miao miao would be using to solve their problem, but couldnt put things together for themselves. And sometimes it is obvious that they are smarter than meow yet they still cant see how to do things themselves. People are just meant to help each other.
@123 because that is just the wrong way to do things. You get what you look for: dead ends.
@qwerty dont even attempt to touch QFT and string theory and whatnot?
@qwerty but if you are the type that wants everything in precise terms from bottom up, then you ought to never be shocked, since the result would be logically obvious...
123
123
@naturallyInconsistent Hello... But i think this way now i am very very comfortable with Newtonian Mechanics and i understand NM this time very well. Just because i read books , asked here any confusions.
@123 ACM just pointed out that whatever you have been doing for 3.5 years is not working. And don't drag me into that.
123
123
I think i am about to end of newtonian mechanics in understanding. Now i am ready to read further LM, HM.
@EE18 This is trivial. $\alpha$ is the apsidal angle that has main term $\pi$; that is a half-orbit; a full orbit has main term $2\pi$
@Obliv nopeu. miao miao just answered
123
123
But it works for me, pls believe me. Now at moment when i read NM books it is very very understandable for me. I just need to understand math and underlying ideas. Nothing else.
7:00 AM
@Obliv Although I have not read Schrödinger's original papers on the topic, I'm quite doubtful of the history that is being presented there. The right-moving waves being $px-Et$ for $(p,E)\in(\mathbb R^+)^2$ is well established in classical physics before quantum theory. But picking $+\mathrm ipx$ and $-\mathrm iEt$, as opposed to $-\mathrm ipx$ and $+\mathrm iEt$, is a convention that didn't matter prior. The wave equations were all homogeneously second order. Only from Schrödinger's
123
123
I think now i understand Newtonian mechanics way better than anyone else. Just because i asked silly questions here, which no one dare to ask or think. I always think about every single detail , which always gave me new knowledge and confidence.
equation onwards did the sign matter, because it mixes a first order time derivative with the second order space derivatives. iirc, it was Dirac who set the conventions in stone.
@123 There is no amount of words that you can give that will make me believe this. You show actual progress, and we will judge accordingly. I expect you will still be stuck on classical mechanics for the next decade.
123
123
@naturallyInconsistent Yes you are right , in front of you i am stuck in newtonian mechanics. But this is not the case.
@ACuriousMind this is totally on you. You keep feeding them...
Note that it is actually pretty interesting how many convention wars there have been in physics, yet nobody in physics thus far has dared to challenge Dirac's choices for representing many things in physics.
123
123
Like since last month i am trying to animate Minkowski Spacetime from lorentz transformation after learning SR. But i am stuck in finding the way how to create new bases.
Even i consult books, youtube, threads, google. No one asked or teach the way i think. Just follow definitions and predefined rules.
7:10 AM
The way you think is just wrong. The standard treatments doing Minkowski diagrams may sometimes be too wordy, sometimes saying too little, but if they are done well, the method to construct all of the things you are talking about, is presented. You are simply insisting upon a catastrophically wrong way to do things and wanting to get somewhere with that.
123
123
My question is that , if i think the wrong way why i succeed at the end of the day.
Wrong things can be used to prove correct things.
123
123
: ) my bad
If you have done any bit of mathematical logic, this is very well known
123
123
You don't know your comments are very very precious and helpful for me.
Which saves my a lot time and mental effort
@naturallyInconsistent You are right. That's why i consult here. Because your knowledge of understanding subject is very vast than i. So even your small comment very very helpful for me.
@naturallyInconsistent No i always try to follow the way at which subject is written. But i never follow criteria without asking questions myself. This way always gave me new knowledge and i can explore the subject more widely.
7:25 AM
gosh I'm sick and now this conversation is annoying me. @123 I was trying hard to give you the benefit of the doubt but you are not behaving sincerely. don't bother saying stuff like "I think now I understand Newtonian mechanics way better than anyone else" when it is a flat out lie, it sounds delusional.
have some humility and introspection
you didn't ask questions that no one else dared ask or think - Ryder rude patiently fielded your questions and linked you to same post you'd seen 3.5 years ago and in August
if things are working out for you, and you understand mechanics so well, why do you need the approval of internet strangers?
@naturallyInconsistent this is a common theme around most math students. math people generally neglect drill math, which I think is a shame. you know banach fixed point theorem, location of roots theorem, and you cant approximate the root of this equation I am giving you? why do you know this then!!! ?
@qwerty hugs
@nickbros123 well, im not talking about drill stuff. I'm talking about linking stuff together. But yes, there are plenty of gaaahhhh just like you mentioned, only that miao miao simply ignored those obvious brainfarts. It is the more intricate, yet also routine physics linkages, that maths people might totally miss.
And those misses are much more interesting to note, especially in the rare cases whereby I've already noted that they can prove all the ingredients
@naturallyInconsistent aww blessed by a cat hug
7:41 AM
there's a lot of that kind of blessing. Went to karaoke until 5am, and now miao miao hath 3 drunkards on mah bed miehehehe
@naturallyInconsistent I guess its in the physics culture to be good at that kind of linkage :P
@naturallyInconsistent you live a wild life lol
@nickbros123 yes, that is one of the big differences between theo phys and maths phys.
@qwerty making up for lost time miehehehehe
it's also nice knowing ppl are safe and sound in myow bed. head pats everywhere. healing maximus
@naturallyInconsistent if I ever make it through QM I'll look at QFT. I did fine the last time I took QM but I dunno if I really understood things deeply. just could do the calculations having finally understood a bit more linear algebra. the dirac book is next up after going through the symmetries stuff
123
123
@qwerty Here i meant to say "dare to ask" such silly question, which i asked. But these are not silly. Also my english is weak
7:49 AM
@qwerty meowth~~
@naturallyInconsistent aw that's sweet!
123
123
Like these days i am thinking again silly question. Like i have read "$mass$" is intrinsic property and "$Force$" is extrinsic property. What about other parameters $velocity$ , $displacement$ , $time$. They are intrinsic or extrinsic or neither.
@naturallyInconsistent I think one can actually spend a great deal of time in just classical physics / philosophy around it, to be fair. If I were the son of an aristrocrat or had a trust fund, Id go down the route in classical physics, what a lot of philosophy peeps around here have gone down ;)
@123 im not interested. I'm disappointed in your behaviour after I stuck up for you yesterday.
123
123
@qwerty No my friend i am always humble. I just blame myself to ask and think silly questions. Pls understand my point of view. Your knowledge is very precious for me.
7:56 AM
humble people don't need to tell other people how humble they are
I'm not going to reply further now. this is clearly not going anywhere
123
123
I just told you dear friend. I always blame myself thinking this way. But what can i do. This way of thinking always helped me
Can we always differentiate every parameters of physics on the basis of 1. Intrinsic and Extrinsic, 2. Absolute and Relative quantity, 3. Cause and Effect, 4. Type of variable (dependent, independent, controlled, constant), 5. Scalar and vector etc...
8:43 AM
@123 how has your way of thinking helped you?
123
123
Because this way of thinking always worked for me.
How do you judge your progress.
@123 one can differentiate wrt any variable of a function
123
123
Like Now i am thinking after yesterday discussion here. Weight is depend on mass $F(m) = mg$ and NGL $F(m) = G \frac{mM}{R^2}$ , if we keep M and R constant. That's why we can equate them. Although way of measuring mass is different.
@ACuriousMind From the point of view of BV-BRST issues, is the difference between a gauge and global symmetry that only gauge symmetries are left on the intersection with the shell?
$\mathrm{Ker}(dS)$ should be all the symmetries of the action right?
123
123
8:49 AM
@RyderRude But velocity is neither intrinsic nor extrinsic it depend on the inertial frame observer. And displacement also neither intrinsic nor extrinsic it is the property of space. And time has no meaning.
I am getting different results properties can have three values intrinsic or extrinsic or neither.
@123 this is not the right way to think about it. e.g. for Coulomb force, ur technique gives F(q1)=kq1q2/r2 and F(m)=ma
we can still equate these two
123
123
Yes
So what is the right way of thinking
the reason we can equate isn't that there's a 'm' in both equtions
123
123
Oookay
i can give the correct version of ur technique
123
123
8:52 AM
Thanks
F(t)=ma(t) and F(t)= k q1q2/( r(t))^2
now u can equate these. The idea is that both are the force on the same particle at the same moment of time $t$, so u equate them
but u r also overthinking it
123
123
Okay. My logic is correct only work for NGL. And by changing logic we can equate coloumbs law . Is this the case
u should use the logic i gave for NGL too
it is the right way to think about it
123
123
Ookay...
Thanks
But in case of weight "$g$" is not a function of time. It is constant. Force change depend on mass
@nickbros123 of course they can. Several hundred years worth of endless arguing...
8:58 AM
g(t)=c is a function of time @123
123
123
@RyderRude Ooh i see.. Ookay
or i should write a(t)=g
@qwerty suddenly they woke up. Now they are safe on the metro~ time for meow meow to wash all the towels~
123
123
Ookay...
May I suggest Newtonian Mechanics by AP French to study from @123
123
123
9:02 AM
@RyderRude We see force of gravity change with mass to keep acceleration constant. Do we use this dependency in force. Or we neglect.
@RyderisnotRude. Yes i have AP French. But the english looks quite hard to me. Definitely i will read it again with more focus.
Good plan :-)
@123 what do u mean
123
123
Thanks. I want to just conclude how dependency works in physics equation.
aww, three Markov processes conversing blissfully~
@naturallyInconsistent ??
123
123
9:06 AM
We know a baseball and cannon ball under free fall have constant acceleration, but both have different masses. It is only possible when force of gravity change with respect to mass.
that is correct
123
123
On the basis of this observation i wrote $F(m) = mg$ and $g$ kept constant near the surface of the earth
yes. Force is a function of mass and the acceleration at the moment
@123 u don't know if a is constant in advance. u get this only after equating ma(t) with the other equation
123
123
@RyderRude Ookay
9:11 AM
ma(t)= Gmn/(r(t))^2. now r(t)^2 is almost R^2 near the surface, so we have ma(t)=Gmn/R^2. Or a(t)=Gn/R^2
123
123
Oookay...
so we have showed that a(t) is independent of t and its constant value of Gn/R^2
123
123
@RyderisnotRude. Thanks
123
123
@RyderRude Ooh, that's how it worked. Nice
9:14 AM
but u r overthinking it
i have to do these weird derivations because u r overthinking things
usually, books would just write ma=Gmn/r^2. then make the approximation, $r~R$, and conclude a=Gn/R^2
123
123
@RyderRude Thanks..
@123 AP French will show you the most efficient way of thinking about Newtonian Mechanics.
123
123
I was thinking this way because of behavior of gravitational force depend on mass. And now i want to understand how dependency in equations work
@RyderisnotRude. I must read it again carefully.
Yes, put a lot of effort into it.
123
123
Indeed i will
9:18 AM
Then use the Cal Tech videos as review.
123
123
I want to understand dependency , because in every branch of physics i saw it. Like L(q(t) , q.(t), t) , H(q(t), p(t))
@RyderisnotRude. I am watching this video
@naturallyInconsistent I saw this example of infinite square well L^2[-a,a] , $\frac{hbar}{i}\frac{\del}{\del x}$ as the corresponding operator for momentum does not seem to work since it is not self-adjoint .So does an operator exist for the momentum in this case?
I'm not familiar with the technical details surrounding Hermitian v.s. self-adjointness that comes with the boundary conditions, so you'd have to ask someone else.
@naturallyInconsistent ohh okk
don't delete that comment! Someone else could have answered that
Lol I'm travelling rn..will repost it after i go to my room
10:16 AM
@Arjun use \partial instead of \del #JustLatexThings
@nickbros123 Sure,thanks :)
10:36 AM
@RyderisnotRude. no.. It is too long
@RyderisnotRude. but i will check it out
11:11 AM
the momentum operator is self adjoint on [a,b] for wavefunctions with period boundary conditions @Arjun
Shankar discusses this in chapter 1
11:48 AM
@Slereah I'm not that familiar with the BV formulation of BRST that I'd feel confident answering that question
I think the gauge symmetries make a point in the configuration space flow along some line, while a regular symmetry takes isolated points to isolated points
not too sure tho
Since apparently critical loci are isolated points if the Hessian isn't degenerate
@Arjun The momentum operator is self-adjoint on the dense subspace of wavefunctions with $\psi(-a) = \psi(a)$ - by their very nature, you can never expect an unbounded operator to be self-adjoint on the entire space, an unbounded everywhere-defined self-adjoint operator does not exist. For the specific case of the particle on an interval, see this answer of mine for a discussion of the domains of definition involved
Note in particular that on $[-a,a]$, $x$ is bounded, so it is everywhere-defined self-adjoint.
@naturallyInconsistent I've been reading arxiv.org/pdf/quant-ph/9907069,see example 2 on page 11-12,he discusses the case where, $\psi(0) = \psi(1)=0$ ,In this case the domain of the adjoint turns out to be bigger than the domain of the p operator.
In the paper above and also in your answer,I quite did not understand how the domain of the adjoint and the adjoint itself have been found.1)Is the adjoint of p in this case unique? 2)How did you arrive at the domain of the adjoint and adjoint itself starting from their definitions as given in the above article?
12:05 PM
@Arjun it's just educated guesswork. E.g. if you try $\psi(-a) = \psi(a) = 0$, you find this is too small, because the adjoint allows a bigger domain, so you make it slightly bigger by just imposing $\psi(-a) = \psi(a)$ without the 0, and you find this "hits the spot" where the domain of the adjoint is the same
this choice of domain is not unique, and in fact may lead to different momentum operators - $p$ is self-adjoint for any choice of boundary conditions $\psi(-a) = \mathrm{e}^{\mathrm{i}\theta}\psi(a)$ for any $\theta\in\mathbb{R}$, see also physics.stackexchange.com/a/577049/50583 by J. Murray and physics.stackexchange.com/a/363410/50583 by Valter Moretti
12:49 PM
@ACuriousMind How the domain of the adjoint itself is found in either cases is not clear to me.From definition it is $D(P^\dagger)$={$\phi\in$ H,such that $\exists\tilde{\phi}\in H$,and $<\phi|P\psi>$=$<\tilde{\phi}|\psi>,\forall\psi\in H$},now this derivation(in the image and also in your answer) shows that all $\phi$ for which $\frac{\hbar}{i}\frac{\partial\phi}{\partial x}$ exists,$\phi$ $\in D(P^\dagger)$.
But this doesen't mean that every $\phi$ that $\in$ $D(P^\dagger)$ has a defined $\frac{\hbar}{i}\frac{\partial\phi}{\partial x}$.I.e this derivation doesen't prove that all elements of $D(P^\dagger)$ are such that ,each one of them has a well defined $\frac{\hbar}{i}\frac{\partial\phi}{\partial x}$,so how do we find the domain of $P^\dagger$?
Henneaux is truly hard to avoid
I guess that the fundamental difference between a gauge symmetry and a global symmetry in those circumstances is that you can make a gauge symmetry arbitrarily small so that the "distance" in configuration space is small, but for a rigid symmetry you just have the same change all over so that you cannot?
Although is it always the case that symmetries split like that
Actually are rigid symmetries even "symmetries" in the sense of being a vector field on the configuration space
ie some vector $v \in T \mathrm{Conf}$ such that $\delta S[v] = 0$
1:13 PM
@Arjun the point of these domains of definition is not that somehow the literal mathematical operation $\partial_x$ would be ill-defined outside of it
in the sense of the weak derivative you have to use for $L^p$ functions, it exists for many (not all) functions, we're implicitly already restricting our domain of definition to the dense subspace of smooth functions
the thing is that an unbounded operator comes with a domain of definition, a rule that says "you can't apply this operator to functions outside of this"
this means that $\partial_x$ with two different domains of definitions is considered to be two different operators in this case
even though you could certainly apply the derivative to all functions regardless of boundary condition
the confusing thing here is that $p = \mathrm{i}\partial_x$ is always equal to its adjoint where their domains of definition match, i.e. self-adjointness here is never broken because we would have $p \neq p^\dagger$, the two operators always act the same on functions they can both act on, it's only broken because the domains are mismatched
@ACuriousMind Okay so to find the elements of $D(P^\dagger)$ we are not looking at the whole of L^p,but in a subspace of L^p in which $\partial_x$ is defined for all elements?
1:28 PM
not really
I'm confused lol
that much is evident, but you have to be a bit more specific about where the confusion is :P
Ideally, you start with an operator $p$ and a domain of definition $D(p)$ that's Hermitian, i.e. $p = p^\dagger$ on $D(p)$.
@ACuriousMind My question is why for all elements of $D(P^\dagger)$, $\partial_x$ should exist
@Arjun if you write out the definition of $D(p^\dagger)$, you end up with $\int (\bar{\phi} p \psi - \overline{p^\dagger \phi}\psi )$, which upon integration by parts, results in something like $\text{boundary terms} + \int \overline{(p-p^\dagger)\phi}\psi = 0$ holding for all $\psi$ for $\phi$ to be in the adjoint. That immediately tells you that, as an operation on functions, $p^\dagger = p$, so functions whose derivative doesn't exist can't be in $D(p^\dagger)$
1:54 PM
@ACuriousMind But you assumed $\phi$'s derivative w.r x exists,when you used integration by parts?Why so?Why does $P\phi$ has to exist?
0
Q: Are global symmetries vectors in configuration space?

SlereahThe generic definition of a symmetry in terms of an action is simply some transformation leaving the action invariant. If we have some configuration space $\mathrm{Conf}$, for instance for a point particle \begin{equation} \mathrm{Conf} = C^\infty(\mathbb{R}, \mathbb{R}^n) \end{equation} and the ...

@ACuriousMind My question is regarding the validity of integration by parts here,since apriori we don't know if p$\phi$ exists
plz halp
@Arjun The adjoint is unique - this derivation shows that if it exists, it has to be the derivative. Again, the proper notion of "derivative" here is the weak derivative, which is defined by this property of integration by parts
@ACuriousMind Wait! So if $\int (\bar{\phi} p \psi$)dx exists it means integration by parts works here?I.e the existance of this integral forces a weak derivative of $\phi$ to exist?
2:04 PM
@Slereah way beyond our paygrades when Slereah is looking for halp.
But I suspect you are correct; although I'd have to first point out that I don't think it is known beforehand how many positions are in the configuration space in the first place. That is, if you were trying to write a computer program to implement all these configuration space action business, it won't be so easy as to state that it is a vector alone.
Some doubts about whether the translation $\phi'$ of a critical field $\phi$ is such that $\delta_{\phi'} S = 0$
@Arjun yes! The weak derivative $\partial_x \phi$ of $\phi$ is by definition the distribution such that $\partial_x \phi [\psi] = \text{boundary terms} -\int \bar{\phi} \partial_x \psi$
@ACuriousMind And it exists if for all $\psi$,\int \bar{\phi} \partial_x \psi$ exists?
2:19 PM
@ACuriousMind Omg! It all makes sense now lol
Tysm!
you're welcome :)
I got stuck cuz until you mentioned I didn't know everyone was talking about the weak derivative the whole time : )
And after understanding what weak derivative meant,things fell into place
2:58 PM
Is there a resource which shows what Electromagnetism looks like in utterly gauge invariant terms?
@SillyGoose uhhh...if you only talk about E and B fields and not the four-potential, i.e. do what every intro to EM text does, everything is already gauge-invariant, isn't it?
the way you are usually taught EM is not as a gauge theory at first
But it’s also not in the usual Lagrange or Hamiltonian formalism
well the reason you introduce the four-potential is because that's more amenable to a Lagrangian description :P
you can either deal in only gauge invariant quantities or you can have a nice covariant Lagrangian formalism, your choice :P
So then i guess i more specifically am wanting to ask how to write down EM in Lagrange or Hamiltonian formalism without introducing gauge potential
@SillyGoose no such formulation of EM is known
I don't think anyone believes it's possible, but I don't know of a proof of impossibility off the top of my head
3:18 PM
in principle of course you could hope to simply take the Hamiltonian action formulation on the fully reduced phase space of standard EM, but the whole reason we have to develop quantization of gauge theories is that obtaining that reduced phase space is typically intractable
3:56 PM
@naturallyInconsistent I'm afraid I still don't follow. The apsidal angle is the angle from the periapsis to the apoapsis. If this is so, and if the question is about how much further along the apoapsis is (in cycle 2 vs cycle 1) relative to a fixed periapsis then haven't we here shown that in cycle 2 the apoapsis is $\pi\epsilon/2$ past the cycle 1 apoapsis?
@EE18 the text literally said perihelion advance and not what you think it is
@ACuriousMind what do you mean by that? Can't you just express the Faraday's tensor in terms of E and B fields and call it done? Those are gauge invariant quantities
@naturallyInconsistent the standard Lagrangian needs to varied w.r.t. to $A$, not $E$ and $B$ to obtain two of the Maxwell equations as its E-L equations
if you vary it w.r.t. to $F$ you don't get the right equations
I thought the problem only comes when you introduce the minimal coupling term $\vec p\to\vec p-q\vec A$
also the other two of the four equations are $\mathrm{d}F = 0$, which are not equations of motion, but simply follow from $F = \mathrm{d}A$; if you don't have $A$, you need to impose $\mathrm{d}F = 0$ as an additional constraint...but imposing such a constraint is equivalent to having a gauge theory
I see
I think we should seriously consider teaching E&M from A upwards, not E and B downwards
4:04 PM
I also discuss this asymmetry between the "electric" and "magnetic" Maxwell equations a bit in this answer on monopoles
@naturallyInconsistent No, I think it's important to first teach the observable part, i.e. explain where Maxwell's equations come from
only after you understand the experimental evidence for these equations I think it's really possible to argue that we'd like an action formulation for them and that this requires us inventing the gauge field
if you start with $A$ it's all rather unmotivated, at least how I imagine it
You'll find that miao miao would never ever present physics decoupled from observations!
then I probably don't understand what you mean by "from A upwards" :P
There is this text by Carver Mead that starts with A field first. The things it starts with would be superconductivity and other similar collective behaviour
But it was very incomplete and nowhere near a good EM textbook
I also think that A field is a much more natural way to treat Brewster's angle
I think one day I might be able to find a way to teach EM starting from A field
also we perhaps shouldn't forget that EM is typically the first introduction to field theory, really (perhaps you did a bit in classical mechanics with vibrations of strings or something but I wouldn't bet on it), introducing both field theory as such and the idea of a field whose value is not observable strikes me as a bit much at once
But today is not that day
4:11 PM
@naturallyInconsistent wald's recent EM text i think aims to do this
however interesting that sounds, the textbook seems to just be in the standard order of topics of EM; but it has a chapter on gauge theory or something
@ACuriousMind I understand your concern but I really don't agree with the basis. The non-observable-ness of A field is similar to how positions are non-observable because choice of origin is left completely free. The affine / torsor issue. It is just that in A field we have an even greater space to be varying by.
@ACuriousMind hm doesn't this suggest that the "mathematics of gauge theory" is actually not a natural way to describe electromagnetism?
@SillyGoose Why would you think so? To me it suggests the exact opposite!
That there is no good action formulation of EM without using gauge theory means gauge theory is the natural way to treat it, no?
Guys would you recommend studying classical field theory from landau lifshitz(as the main text)given one has done most of Griffiths?
@SillyGoose standard order of topics is quite good! There does not need to be revolutions everywhere...
4:15 PM
but to do so, we're required to introduce additionally degrees of freedom that are unobservable and build machinery to remove them at the end of the day; intuitively a natural framework of a physical theory would not require such notions. but i am curious of another opinion
@naturallyInconsistent That's what I've just described in words in my most recent message no? The angle past $\pi$ (which is the aphelion-perihelion angle in cycle 1) which aphelion 2 is at?
@Arjun L&L is good, but why not choose something more modern, using more modern notation, and easier to read?
@naturallyInconsistent what do you recommend for classical field theory(mostly enm)?I have a course next sem titled enm and str,it's based off the first 9 chapters of landau vol 2
@EE18 No, I am specifically telling you that the text defined the apsidal angle $\alpha$ as from perihelion to aphelion half-orbit advance, whereas the next statement in the text was discussion perihelion to perihelion full-orbit perihelion advance.
@SillyGoose Your intuition is just wrong, as evidenced by the immense success of gauge theories describing all of the four fundamental forces.
4:18 PM
@Arjun If it is based off Landau then you're already f-cked so there isn't stuff much nicer. I'd have recommended plenty of other things to read if you weren't cornered like so
@ACuriousMind hm but I think success of a theory and the naturalness of the language it is written in are somewhat disjoint
surely a naturally formulated physical theory is successful, but not vice versa
@SillyGoose Sadly, Heisenberg's insistence that we should only have observable stuff in the theories is a wrong one.
@SillyGoose sure - success is a meaningful and measureable quantity, "naturalness" is not :P
And people vehemently disagree upon what are considered natural
people are naturally inconsistent :p
4:21 PM
I very much reject the idea that nature has to conform to our aesthetics of what theories describing it "should look like"
@naturallyInconsistent XD,you can still do the recommendations,I'll try to read as much this winter break : )
Hm well I guess it seems a strange idea because it seems like gauge theory is like ascribing the identity $1 - 1 = 0$ physical significance (in that such an identity forms the foundations of a physical theory). But really this is a mathematical "trick". So I would not conclude that gauge theory is the theory but rather a heuristic, more accessible version of a theory that could be written entirely in terms of observable quantities.
I don't mean to say the theory as in the end all be all of all physical theories.
I think I am meaning to investigate claims like "gauge fields are necessary to describe the electromagnetic interaction" by my line of questioning
again, in principle there always exists the theory on the reduced phase space, it's just that we can't really compute it that easily
in other contexts you can formulate things in terms of non-local invariant fields, see the work by Strocchi et al. in this answer of mine
4:43 PM
the EM potential has 4 degrees of freedom as a 4-vector but actual degrees of freedom are like 2, which sounds hard to put in a nice object in a Lagrangian :p
too big for a scalar, too small for a vector
@Arjun Miao miao did the whole Feynman lectures (Vol2 is the E&M) and Jackson front part. There were many smaller reads in between, Greiner in particular is wonderful, but I doubt you would have the time.
@naturallyInconsistent That's impressive, did you do all of that while in university?
Feynman lectures was totally before I started uni physics. In particular, Vol 1 only, and that gave meow meow a tremendous head start during year 1, in which I had the time to finish both Vol 2 and 3 and still have social life and all. That died down in year 2 when the maths for theo phys consumed everything.
@naturallyInconsistent I tried reading feynman's lectures but found them a bit hand-wavy,I thought I'll read them after I finish reading physics from other more rigorous books,that way I'll be able to not get stuck and still gain feynman's perspective of the subject
Jackson was during year 2 after finishing Griffiths material.
But that is not anywhere near full understanding; it is only after applying the shit to magnets that shit became holy
@Arjun if you find Feynman handwavy, dont even attempt Susskind
4:55 PM
@naturallyInconsistent I won't lol
I really liked i.e irodov's mechanics and electromagnetism books,loved his attention to detail and clarity,all while not yapping all the way through
lots of ppl say that like irodov
havent touched it but the reputation is glowing
I feel slightly overwhelmed by the amount of stuff i want to study and how fast paced my university seems,I wish I didn't enroll in one.But then one needs a degree and stuff in real world : \
Not to mention the spirit destroying soulless lectures 3 hours a day,it's a massive waste of time
5:34 PM
@Arjun agreed
except analysis, of course !
@Arjun it neednt be soulless. Most of my uni lecturers tried to make them good, and many succeed. Even though miao miao mostly didn't appear, those that do a good job at it, are rightfully rewarded with many students doing well
123
123
@qwerty i have read 5 chapters of AP French with complete understanding. I know all these information priori. I have already worked by hand these things and think and concluded all these results by myself. He didn't encounter my problems of confusion.
@123 Have you tried reading from irodov's mechanics?Whenever i get a doubt while reading a line,he clears it off within the upcoming paragraph lol,I've learnt most of my basic newtonian mechanics from it.I'm currently doing goldstein for lagrangian mechanics and am loving it : )
123
123
5:49 PM
Yes i have more than 100 classical mechanics books also irodov. But didn't read irodov. Now i will read it.
Bruh
123
123
I want to conclude the bases of dependency in equations of mechanics.
@naturallyInconsistent I wish to encounter a teacher like that some day : )
123
123
Like i have seen F(x, t, v) . Some cases force only a function of position alone F(x) like gravity , or time F(t) like EM or velocity F(v) like air drag
Some case two dependencies F(x, t) like gravity and time varying EM.
But it is a property of force itself. How force is changing over time or position or velocity.
@naturallyInconsistent math lectures here are mostly great, but the inherent problem with math I think is that, its not really meant to be taught by a lecturer in front of a blackboard in a 40 hour semester. There would be no room for the "training of the mind to think"
5:55 PM
Im not sure why physics lecturing cant be good there
@naturallyInconsistent I agree with nickbros that math lectures here are good,But am sure nickbros would agree with me that physics lectures kinda suck here :p
I know that...
maybe its just that @arjun and I are suckers for the details
@nickbros123 ...the training is what the weekly exercises are for :P
@naturallyInconsistent I think I will find pure math lectures good even if a donkey teaches it since pure math by definition is axioms->theorems->proofs,no holes of knowledge,but physics on the other hand could get really handwavy and there is room for misinterpretaion of concepts
physics is handwavy in comparison to math, that's not the lecturer's fault; if you want mathematical physics you have to study mathematical physics
@ACuriousMind Agree,but I feel much more content and satified when I study books and think till everything fits for me,this takes time and hence I don't like attending lectures,but some of my teachers are legit bad,I have a prof currently who reads a book a day before and write stuff word by word on the screen,they expect me to attend all their classes and take notes,I can't even do my own studying sitting in the class :'(
@ACuriousMind fair enough, but I do think there is something to be gained in having a go at the results ourselves, before having it be fed to us by the lecturer. I have, though, come to understand that my position is in the minority, and different people might want to gain different things from a course. I personally like to go through a book, use it as a primary source. I really love my current prof cuz he serves as an extra-theorem-inator, as in, he comes in with theorems I havent seen
but I think I gain a lot by having read the material prior
if nothing, atleast it puts things in perspective
@ACuriousMind I still believe(maybe am wrong with this one) that even in theoretical physics one could clearly state the assumptions they've made while deriving something or doing a calculation,I agree it won't be as rigorous as math,but one could still be clear in their steps, and not make me feel like they pulled it out of thin air.
@Arjun well, that's just a bad lecturer then
it's not an intrinsic property of physics lectures that they have to hide their assumptions from you :P
unless by "assumptions" you mean things like "assuming some limit or integral exists", which physics just does all the time, but that's also not specific to lectures :P
6:14 PM
but, playing the devils advocate here a lil bit @arjun, you and my roommate the kind of people who would want a thorough discussion on the gelfand triple :)
@nickbros123 What do you mean? Maybe the structure of "exercises" is different here than what you imagine - I had to do exercise sheets every week and hand them in, and the solutions would be discussed the following week in a dedicated session (led by TAs, not the lecturer)
@ACuriousMind I definitely aim on studying the mathematical axiomatizations of various branches of theoretical physics : ) ,but before that I want to study major branches of physics from canonical physics text-books the physicist way :p
@nickbros123 yeah, that's just not the level of discussion you're going to get from a standard physics course - that doesn't make the lecture bad, it makes your expectations wrong :P
@nickbros123 I would not have ,had I not had the backing of Acm sensei :p
@Slereah I think I wholly agree with Valter's comment - before we are allowed to pull out the Morse lemma, you first need to a) define the topology on $\mathrm{Conf}$ you're using and why b) define the infinite-dimensional operator $\delta$ and c) cite a proof of an appropriate infinite-dimensional version of the Morse lemma
note that e.g. Michor's and Kriegl's global analysis book discusses multiple times that there are several inequivalent smooth structures on $C^\infty(\mathbb{R},\mathbb{R}^n)$
6:18 PM
Yeah after checking I think it may not apply to anything better than a Banach space
which smooth spaces are not
@ACuriousMind yeah, we have a similar structure, weekly / biweekly problem sheets. But they just don't hit the same like, say, proving the heine borel theorem
but I understand
this is kind of unreasonable to many people
But I do wonder because I've seen a few places say that critical points are disconnected, but I think that was the toy model of a finite dimensional configuration space
so idk
Also I would argue that if you started with this kind of Morse argument and just accepted it, you'd hit a roadblock in the other direction because Noether's theorem wants you to have a notion of "continuous global symmetry" on $\mathrm{Conf}$ but that can't happen - if the solutions were discrete points, how is a Noether symmetry supposed to be "continuous"?
either the Morse lemma is invalid in this particular infinite-dimensional setup, or these two viewpoints consider two different topologies on the mapping space
yeah it seemed weird
That's why I hate infinite dimensional spaces
all my homies hate infinite-dimensional spaces
2
unfortunately they seem rather crucial for formulating physics in all sorts of ways :P
6:25 PM
I mean do we need infinitely many things
The harder part is, how to show positively that it is a curve in configuration space
I don't want to have to read Michor 😔
123
123
What does isolated object means? 1. There is no external force 2. External forces are balanced (if external forces balanced it produce stress within the object) 3. Or both
@ACuriousMind @Slereah have you guys studied morse theory?
@nickbros123 not that much tbh
it does pop up occasionally though
@ACuriousMind I'm going through this phase right now XD,some of it seems artificial and indigestible rn,but maybe after a formal course in functional analysis things will ease out xD
@Slereah stll, pretty cool. I just think all my life problems will get solved if I study diff geo/topology
6:32 PM
@123 I consider it to be the first case,no influence from any external body whatsoever,since even if the net force is zero,net torque about some point might not be and total energy might not remain fixed etc
@nickbros123 you will just find new life problems :D
lol
123
123
@Arjun Thanks .. I was saying in second case net force and net torque zero. But it produce stress within the object. Equal and opposite forces at the point of center of mass produce stress. Can it be isolated.
first you'll want to study differential geometry then you'll want to study sheaves, but to study sheaves you have to know category theory. then once you get back to sheaves you'll be interested in algebraic geometry. then you turn into a mathematician
I am en route to becoming a mathematician tbf :)
6:36 PM
@123 I still think no,isolated means no external influences whatsoever(ofc this is an idealization)
123
123
@Arjun Thanks... : )
@123 The force on a particle in Newtonian mechanics ,by a system of other particles is in general a function of its relative position and relative velocity with respect to the particles exerting the force on it. Note also ,depending on the frame you're observing the particle's force field could gain an explicit time dependance,just like the em field example you've given.
Also don't worry too much about what quantites forces depend on, you'll learn about various kinds of forces as you learn more physics : )
123
123
Yes
@SillyGoose this does seem like a good path to go in, though xD. my situation reminds me of Joker in TDK, I just do things, I am just a dog chasing a car, I wouldnt know what to do with one when I caught it :P
@nickbros123 When you finally catch it,maybe start doing physics again? :p
6:46 PM
Thatll have to wait I guess
I need to first convince the rest of my brain to pivot career over to either quant finance or ML
@nickbros123 Lol,maybe talking with the current math graduate students will do the job for you :p
it does the job for the logical part of my brain :)
but thats a small part
If it does the job, for the rest of it flash some pictures of rolls-royces and a villa near the beach :p
maybe it's too much lol
still panders to only the logical part xD
something should "take me off my seat" if u get what I mean
as things stand, whenever someone says: ha, yure doing math, ur gonna end up jobless, my reaction is: oh shit, bad career choice! continues to do math
we're destined to search for our 9th post-doc position with the savings of a 21 yo McDonald's employee when we turn 36
2
6:53 PM
I guess thats the way itll be :)
7:31 PM
What do you get when you cross a mountain climber with a mosquito? Nothing, because you can't cross a scalar with a vector.
@ACuriousMind hey homie
@ACuriousMind I was going through your answer once again and I have the following question,If we restrict the domain of an operator,say using boundary conditions,and if the operator turns out to be self-adjoint on this domain,will it's set of eigenfunctions(including the generalized ones) span the whole of L^2[0,1] or just a subspace of L^2[0,1] in which the vectors also satisfy the boundary conditions?(I'm not sure since in infinite dimensions anything could happen lol)
Can someone refer to me a resource where they evaluate $\langle a_k a_p^\dagger$ for klein gordon field theory. I expect to get BE distribution. I know the corresponding derivation for a single harmonic oscillator, but I cant quite reproduce the steps in field theory.
@Arjun The version of the spectral theorem for unbounded self-adjoint operators holds; if you want to think about this in terms of generalized eigenfunctions spanning the space, you can (I prefer to think in terms of the spectral measure).
@NairitSahoo Why would the vacuum expectation value of the number operator be anything other than 0?
if you find particles in your vacuum it's not a vacuum
@ACuriousMind Oh sorry. I am taking the thermal average, not the VEV: I should have mentioned this
The density matrix is $e^{- \beta H}$
7:46 PM
then you need to consult books on thermal quantum field theory
standard hep-th QFT is at zero temperature and your question makes no sense there
I did consult them. All the books use path integrals. I wanted to know if somebody knows a book which does it the canonical quantization way
the reason they use path integrals is that the operator methods haven't really turned out to be useful in the thermal case :P
via path integrals you can make a connection to ordinary statistical Euclidean FT by Wick rotating and sidestep a lot of issues
Oh... makes sense. Thanks
@ACuriousMind Sorry,but I don't know much about spectral theorem for unbounded self-adjoint operators and also don't know what spectral measure means
Could you give me a physicist's answer to me as of now? :')
02:00 - 20:0020:00 - 00:00

« first day (5085 days earlier)      last day (39 days later) »