i was thinking that perhaps we could identify the wavefunction $|x\rangle$ with the square root of the delta function, $\sqrt {\delta (x-x')}$. then the wavefunction would be square integrable
and it gives a zero probability of finding the particle in any region that doesn't contain $x$
@Obliv Hawking used to quip that asking what's before the big bang is like asking what's north of the north pole - the question does not even make sense within the big bang model.
depends on what exactly you mean by "invariance" and "general coordinate transformations" :P
"Modern" differential geometry tends to prefer writing equations in a coordinate-independent way to begin with, so the notion of being "(in)variant under coordinate transformations" doesn't really make sense
what people often mean by this is that the "form" of the equation is preserved under the things they consider "coordinate transformations", but actually mathematically one would rather formalize this as being (in)variant under diffeomorphisms
Yes...but let's do stuff in oldschool index notation.
The problem is I thought that it doesn't make sense to say equations like $\partial_\nu A^\mu=0$ is "invariant" simply because it is not even a scalar equation...a "quantity" can be invariant...An equation can be atmost covariant.
Hello. I wanted to know if there is a general notion of emergence and algebras in holographic theories
For instance, for AdS/CFT Liu and Leutheusser found that emergence is linked with a type III$_{1}$ algebra, so is this linked with generally all holographic theories?
I am not very familiar with Chandrasekharan, Longo, Pennington and Witten's recent paper for an algebra of observables in static patch de Sitter holography though.
@ACuriousMind I was startled to find out that "Invariance of equations" mean that the quantities which are independent of the state of matter would remain constant under GCT in addition to covariance.
trying to gatekeep the words "invariant" and "covariant" to some narrow rigorous reading will not end well; they have been used for too long and too variously for that :P
@ACuriousMind So it's really another meaning? I was confused because I thought that this is "THE UNIVERSAL" meaning of invariance of an equation. And I couldn't find this meaning elsewhere!
@ACuriousMind it is, however, still useful to attempt to provide clarity and consistency to students by telling them about potential confusions and using words in a gatekept way...
it's not saying "invariance of equations means anything independent of the state of matter is constant" as some sort of general statement, it's saying "we want invariance of equations, and in our case to achieve that we have to make everything constant that doesn't depend on the state of matter"
@ACuriousMind Okay...so the implication goes the other way.
But anyway...is this notion of invariance well known? I mean if I go and say "This equation is invariant"...What would most people think? It is covariant in the usual sense or in this "weird" way the text claims it to be?
Because the words are used by different people in different way I would never assume anything until they write down in formulae what they actually mean :P
Apart from this text, every time I read "this equation is invariant" what they meant was this equation was covariant (sure---invariant under diffeos)...but here I get a whole new different meaning...so I was wondering whether it is well known to all (I understand that the meaning is obviously well known to all...but whether the use of the word is popular)
i think, under this definition, a modification of GR where the dynamics depend on a preferred vector field, would count as a non- co ordinate invariant theory
because this vector field is a constant but is not an invariant quantity, but rather a covariant one
You can just read on in a text without having perfectly understood every single sentence, and they usually become clearer with that additional context; don't stop reading every time you don't immediately and perfectly understand what's going on
No no...I read the rest of the text too (atleast to the end of this topic)...and what they are saying is perfectly consistent with what other people usually says---just the choice of words is...well very different...So I was thinking of asking you guys who have definitely read a lot of sources than me to comment on whether this kind of saying is traditional or not... Btw it is chapter 7 of Gravitation and Spacetime by Ohanian.
A lot of things seem different from the mainstream literature here... E.g. they consider tidal forces as local effects of gravitation...Again the math is consistent with everywhere else but I don't think most of the people would look at tidal force as a "local" measure of gravitation...The catch was in a very different usage of the word "local"
usually, by invariance of an equation, we mean something like Poincaire invariance. under this definition of invariance, the form of the EoM remains invariant under some special co ordinate transformations
@Slereah Hmm...yes and I started comparing with the 2nd meaning "local" meaning at a point...Imagine using this terminology in a discussion of Gaussian normal coordinates!!!
@VaibhavK It's a book actually..."Hans C. Ohanian, Remo Ruffini - Gravitation and Spacetime-Cambridge University Press (2013)"
hey @RyderRude following up on this - chat.stackexchange.com/transcript/message/64490949#64490949. I have drawn it again. ibb.co/fnj077P - and I realize I don't like something there. While I get that the lagrangian $L'$ must contain $y'$, currently on the image, it contains $y' - a$ which is the same thing as $y$ and we kind of get e.o.m there relative to $y$ frame. I think we should write $L'$ relative to $y'$ frame only.
when you were writing to get an expression for L', you wanted to write in terms(relative to y' frame), but what you do is L' = .... - mgy = -mg(y'-a), but L' is not equal to -mgy in the first place(that's my point)
technically, the Lagrangian is a co-ordinate independent function on the manifold. when you choose some co-ordinates, you get functions like L(y, dot y) and L'(y', dot y')
these two are related by a co ordinate transformation.
i consider the motion of the ball relative to their frames. so the person dropping a ball in y' frame, i only consider this ball's motion in y' frame while another person dropping a ball in y frame, i only consider that ball's motion in y frame
@GiorgiLagidze what's happening is that when, (y, dot y) and (y', dot y') are two labels that refer to the same physical situation (the same ball), then we have the equality L(y, dot y)=L'(y', dot y'). u can see that mgy and mg(y'-a) are equal when y'=y+a (as this is when y and y' refer to the same point in the train)
so, given L(y, dot y), you can use this equality to get to L'(y', dot y'). just take the point (y, dot y) and transform it to the other co ordinate system, and use L(y, dot y)=L'(y', dot y')
@GiorgiLagidze it is always a strict equality for co ordinate transformations
mgy = mg(y'-a) holds strictly when y and y' refer to the same point
the reason we have this equality has to do with the fact that the Lagrangian is a co ordinate independent function on abstract points L(p, v), and then functions L1 and L2 we're dealing with are that same function expressed in different co ordinates
@GiorgiLagidze if Landau discusses co ordinate transforms of the Lagrangian, he will explain this much better why L1(y, dot y)=L2(y', dot y')
it has to do with the fact that the action should be invariant, doesnt matter what co ordinate u use. this is so different co ordinates agree on what paths are the physical paths
and also that the Lagrangian and the action are scalars
we have a single function L on asbtract points L(p, v). These L1 and L2 you're dealing with are the composition of that function with the co ordinate choice function
@ACuriousMind Did you ever figure out if a connected space has a chain of overlapping open sets between any two open sets
I am thinking it is maybe showable using the nerve theorem since there should be a cover of the space that's homotopically equivalent to the nerve, and I'm guessing a connected complex would just be equivalent to that
you have two open sets on the manifold. pick any two points in each of them. you can connect them because it's a connected space. all points on the connecting line must have open neighborhoolds because it's a manifold
@Slereah oh. you mean this result in a more general space
J.D.Jackson in his book classical electrodynamics derives macroscopic maxwell's equations by averaging microscopic maxwell's equations ,I wonder if something similar could be done to get the macroscopic energy density from the microscopic one.
In the paper https://arxiv.org/abs/1312.3383 ,the aut...
What is the square root of the Dirac Delta Function? Is it defined for functional integrals? Can it be used to describe quantum wave functions?
\begin{align}
\int_{-\infty}^{\infty}
f(x)\sqrt{\delta(x-a)}dx
\end{align}
it says the square roots always give zero on test functions
also, the eigenstates of momentum are plane waves, whose Fourier transform is the delta function. this means $|x\rangle$ can only be identified with the delta function, instead of the square root
I'm guessing something like the existence of a little nerve complex where the two open sets under consideration are the boundary points of that complex
@naturallyInconsistent 100% confirmed, this guy is from my institute! we were discussing the averaging thing a few weeks back. I dont think the method in the question hes asking, of averaging of energy density, is a suitable method though, there are a lot of information loss when going from micro to macro to viably capture the energy. moreover theres the classic $\langle E^2 \rangle \neq \langle E \rangle^2 $ problem
@Mr.Feynman hi again. i just want to clear up one more thing from yesterday's discussion. in the completeness relation, you're not taking $|\phi (x,y,z)\rangle$ to mean an eigenstate of the specific field operator at $x,y,z$, i.e. $\hat {\phi} (x,y,z)$, right?
@Mr.Feynman because each $|\phi (x,y,z)\rangle$ in the completeness relation is supposed to be a simultaneous eigenstates of all the field operators in $\phi (x,y,z)$, x,y,z $\in R^3$
the state does not correspond to any particular operator
it may be easier to see this point by writing the completenes relation of a lattice field theory in 1+1 spacetime, with a finite volume @Mr.Feynman
the spectrum of one particular field operator is extremely degenerate. we use the simultaneous eigenvectors of all the field operators (i.e. a CSCO) in the completeness relation
just write the completeness relation for a 1+1 lattice field theory with a finite number of field variables
you will see that the completeness relation is really equivalent to the completeness relation written in, say, position basis of multi-particle QM where u use simultaneous eigenstates $|x_1, x_2\rangle$
in a lattice field theory of $n$ real-valued field variables, you will use $|x_1,x_2..... x_n\rangle$ to write your completeness relation, right?
these are just the simultaneous eigenkets of all the field variables
yes, in that sense, the same holds in QFT too. im just saying that $|\phi (x,y,z)\rangle$ in the completeness relation is analogous to $|x_1,x_2....x_n\rangle$
if you want to build these kets as a tensor product of the individual CCR relations, then that's fine (although it would have to be a really weird tensor product for continuum QFT :P)
but in the completeness relation itself, you are using the simultaneous eigenkets. this is also where the functional integration over space-dependent fields comes from
in lattice theory, you have $dx_1 dx_2 ....dx_n$ in ur completeness relation.
in continuum QFT, you have the functional integration
I mean, while for a discrete index $\lvert i\rangle$ you can list each one $\lvert 1\rangle$, $\lvert 2\rangle$, $\lvert 3\rangle$... when we have continuous labels we only write one symbol only, like $\lvert x\rangle$, because that would be an uncountable list
It's really the same thing just that now the label $x$ is going as a subscript of the field but you get the gist from the mesage above
@Mr.Feynman the reason i worry about this notation is that $|\phi (x,y,z)\rangle$ can also make it look like there is one ket corresponding to each x,y,z in R^3. Perhaps we could write it as $|[\phi(x,y,z)]\rangle$ to mean that there is a ket for each function written in square brackets inside the eigenvector
we use square brackets to denote when something depends on an entire function, as opposed to the function's arguments
@Mr.Feynman this notation can also cause this issue. it can mean that $x_i$ is an eigenvalue that is a function of $i$, and that there is one ket for each $i$
when we actually mean that there is one eigenvector corresponding to each function $x(i)$ from $i=1,2,3..N$ to $R$
@ACuriousMind I sometimes catch myself not only failing to understand every sentence, but also that my mind is wondering somewhere else while I am reading 😂 it's like the reading can become a white noise to meditate on or something.
@naturallyInconsistent hi. the multi-particle QM discussion is just for analogy here. the field eigenkets do not have any interpreration in terms of a definite number of particles
@ACuriousMind The worst thing about doing the opposite is that you spend a lot of time on a single sentence and you might or might not suceed but you later realize that the next page explained that very sentence :P
ok :) . imo they are saying that "explanations" are about structural properties of the universe, i.e. they describe the relation of one thing to another (2-place relations). But there is no reason we should deny the existence of 1-place relations (what something is on its own without relation to anything else). it would be unnecessarily restrictive to postulate their non existence
and qualia are 1-place relations. since they are not 2-place, demanding their explanation is a "categorical error" @Amit
yes. but that is also the status quo about qualia that they cant have explanations in terms of physical process. it's based on our intuition about qualia
@Amit I think it's perfectly common to either be tired or have something else on one's mind and then notice you've just read the same damn page 4 times and still no idea what it says. For me it's usually a sign I should go to sleep :P
Sorry Ryder I just dont find it very interesting when it all hangs on intuition, aka mysticism in another jargon. But maybe I didnt dig deep enough, I guess that onus is on me 😂😂
Hi is any one of you willing to clear a particular concept of mine, I usually ask in the problem solving strategies room hut the professor is busy and I sorta need it now.:(
Im sorry I didnt see the info, my question is on a rough inclined plane if a block moves up with some velocity its said that its time of ascent will be less than its time of descent ( no external forces). I understand the reasoning behind this being that the retardation is more than the acceleration but if while going up it reaches 0m/s quicker, how do we know that it has gone all the way to the top, it might just be that it has gone only halfway up, right?
If u had a thin square box of some refractive index that is much larger than the index around it, and let light in through one corner, how does the box light up? Does it just have the single ray going along the diagonal?
Suppose the diagonal had the same index of refraction as the surrounding medium ** forgot to add that
Just trying to better understand fermats principle, like does light ever take 90 degree turns just to circumvent going the long way through an object? hence why i asked before if taking an infinite number of steps thru the diagonal of a square was the same distance travelled as going along the length and then height
if the diagonal has the same index of refraction as the surrounding medium, then for the purposes of light propagation the diagonal and the surrounding medium are the same medium
also how is the diagonal of this box supposed to have a different index of refraction that the rest of the box
I'm confused what physical situation you're imagining here
I thought it does both. I guess if an object is blocking its path, it gets reflected and/or absorbed and it doesnt necessarily care about continuing along its previous trajectory
I was thinking if light could travel from point A to B in a manner where its optimal to go along a straight path, but suboptimal but still better to go along a step-wise path of greater numbers of steps than lower
I am actually most ok with students under time crunch and cannot learn everything that they want, that they are bored with the material currently being taught and cannot wait to start the next topic, and motivation issues. What I have trouble with, however, is complaining when in such combination of situations.
all the fancy stuff like cosmology and quantum theory and whatnot will still be there in a year, or a decade, or whatever, it's not going anywhere, and trying to learn "advanced" topics before the basics can be profoundly confusing because you'll not know the things everyone takes for granted at those levels
what I mean is that regardless of the specific definition of boundedness you're using it should be very simple to see that all operators on finite-dimensional spaces fulfill it
:( I can't see it...it must be something super obvious, maybe. There's only one safe $||A\psi||\le C||\psi||$ for bounded operators...an unbounded operator has no such restriction and an extra condition about the domain... So?
You can't have a dense subspace in a finite dimensional space or what?
@Sanjana Let $\psi = \sum_i c_i \psi_i$ for any orthonormal basis $\psi_i$. Then choose $M$ as the maximum of the norms $\lvert \lvert A \psi_i\rvert \rvert$ (which exists because there are only finitely many $\psi_i$). So $\lvert \lvert A \psi\rvert\rvert \leq M \lvert \lvert \psi\rvert\rvert$.
This also immediately shows you what goes wrong for an unbounded operator in the infinite-dimensional case: There you have infinite bases and the sequence $\lvert \lvert A\psi_i\rvert\rvert$ can diverge to infinity
as for the second question: The only dense subvectorspace of a finite-dimensional vector space is the vector space itself, this is also a nice exercise if you don't see why
Another question--Here user Aleksander Bukva says that there are one dimensional reps of the Heisenberg algebra...What are those for the CCR? Numbers commute so how can there be any 1d rep...are they Grassmannian reps?
@Sanjana Yes - mathematicians love vacuous truths: All finite-dimensional reps are one-dimensional, but this would contradict the commutation relations, so there are no finite-dimensional reps.
The statement "All X are Y" is always true when the set of X is empty
What is the square root of the Dirac Delta Function? Is it defined for functional integrals? Can it be used to describe quantum wave functions?
\begin{align}
\int_{-\infty}^{\infty}
f(x)\sqrt{\delta(x-a)}dx
\end{align}
