7:12 AM
hello
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 $x$

What is the square root of a delta function

@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.

but i dont know how this square root would be defined. maybe it can be defined in the abstract using $\int f(x) (g(x-a))^2 = f(a)$
this g is the square root

Is the notion of invariance of an equation under general coordinate transformations, well known?

@Sanjana yes, the abstract equations on a manifold arent even written in terms of co ordinates

7:20 AM
invariance and not just covariance...of a generic tensorial equation.

@Sanjana It is indeed a well-trodden theory
Originally investigated by Lie himself

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

do you want the equation written in terms of co ordinates to not change its form under general co ordinate transformations?
such equations would have an extreme amount of symmetry. and this is an active transformation symmetry

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.

7:23 AM
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.

@Sanjana you are technically correct but I'm sure you understand what people mean when they call such an equation "invariant" so what's the problem?

@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

If you want some modern version of how differential equations act under group action there's a GR paper on that topic here : arxiv.org/abs/1306.2818
including under diffeomorphisms

@Sanjana I don't know what that means but sure, add another meaning of "invariant" to the list :P

7:30 AM
@Slereah Thanks...I will have a look at it.

"states of matter" is sure a wild introduction to this topic

@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!

pretty sure that's not "the universal meaning", no
I don't even understand the context here - when we're just doing geometry there isn't even such a thing as "matter", so how could this be universal?

Have you ever come across this particular meaning?
I got this from here.

i have no idea why they're using the term "states of matter" here :P

7:37 AM
@Sanjana I don't think you read that right

@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...

@Sanjana under that definition, I think this one is invariant

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"

i think theyre allowing tensor quantities in the equation

@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?

7:41 AM
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)

it's impossible to tell from the small excerpt you posted what the text really means
but given that this reads like an introduction, that's not surprising
if you could perfectly understand everything in the introduction, what would the rest of the text be for :P

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

7:56 AM
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"

they mean "local" as "a tiny neighbourhood" and not "a point"

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

which is a common way of using local
even in math

Poincaire invariance is indeed a stricter kind of invariance than diffeomorphism invariance of equations written in co ordinate independent form

I'm sorry -- Is it possible to point to the paper you are talking about?

8:01 AM
@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)"

It is one of those things you have to get used to I fear

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.

@GiorgiLagidze but L' contains y' and not y.
y'-a can be substituted with y and then you will get L
but that doesnt mean L' contains y. L' is the Lagrangian for the translated frame

why are the two of you still going on about Lagrangians...

@GiorgiLagidze L' only gives u the EoM relative to the y' frame

8:07 AM
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)

i dont know how to explain this one clearly.... It is the issue i talked about yesterday

just tell me this one. do you say that $L' = -mgy = -mg(y'-a)$ ?

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.

@RyderRude i think i got it. no need to go further :P i will answer your question from here - chat.stackexchange.com/transcript/message/64490949#64490949
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
does this sum it up to answer

in some sense, these two functions are equal yes. when you evaluate them on the same "abstract manifold point", you get the same result
but idk if i can call these two functions "the same", as they are defined in different co ordiante systems. they have different signatures

8:15 AM

@GiorgiLagidze yes. and like i said, by initial conditions being the same, we mean y(0), y'(0)=a and dot y(0), dot y'(0)=b
so the initial positions and velocities are the same, but they still refer to different experiments

y(0) =a, y'(0) = a, well, $a$ is relative to each frame. :P

because y=a and y'=a are not the same point
@GiorgiLagidze yes!

seems clear. never mind then, false alarm today :D

ok :)

8:42 AM
@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)

in our case, we don't have strict equality

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')

Is anyone having trouble accessing StackExchange sites?

@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

@Danielillo at least physics.SE works fine

8:49 AM
@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

@Slereah is the connected space a manifold?

9:04 AM
It certainly is
That's of course true here since it is also path connected if connected
But I am wondering in the general case
Nerve theorems seem to be pretty narrow in their applications, though

@Slereah and that's the issue: how general are you dealing with?

i think intuitively this should be true.

is it true for all topological spaces, for a start, I guess

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

a connecting line would imply path connectedness, not connected

9:13 AM
oh
im not operating with the general case then
yes, there are examples of spaces that are connected but not path connected. it says : topologist's sine curve

0

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...

@nickbros123 do we tell him to run away from the abyss?

9:32 AM
22

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

9:53 AM
@naturallyInconsistent omg wait i think i know this guy irl

@Slereah I think this depends on the notion of "chain"?
I feel like what you really want is that the space is path-connected, then you get this chain simply from the neighbourhoods of the points on the path

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
like just a little one dimensional one idk
is that equivalent to path connectedness?

@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

10:15 AM
Here's what I'm vibing to today;(would love to hang out more) but my schedule doesn't permit

10:33 AM
But now I think this is my new vibe

10:46 AM
@nickbros123 tell him to runnnnn

11:02 AM

11:25 AM
@MoreAnonymous it is a genius tune

11:51 AM
@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?
lets say the completeness relation at $t=0$

12:34 PM
@GiorgiLagidze that philosophy room we created is not showing up on "my rooms" anymore

@RyderRude why would I not?

@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

I'm not saying that's wrong. Let me think about it

ok :)

Do you happen to have any reference mentioning this completeness relation?
More specifically any book

12:41 PM
i dont :(
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

@RyderRude Mhh but in that case (assume distinguishable particles for simplicity) you can take individual eigenstates of each position operator
That's how you build $\lvert x_1 x_2... x_N\rangle$
I mean, they are simultaneous eigenstates but they are built from individual eigenstates

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

My point is that by the ETCR your fields commute at different point and like the finite case your basis is built naturally that way
I think it's just a caveat of the continuous notation that you're bringing up right now

@Mr.Feynman yes, i agree with that
i thought maybe there was a misunderstanding, but there isn't. sorry :)
@Mr.Feynman yes, the $|\phi (x,y,z)\rangle$ notation makes it look a bit confusing in this regard, at least to me

12:56 PM
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

yes.
i just thought that notation could cause a misunderstanding that the kets corresponded to the operators at each spacetime point
i just thought that that notation could cause a misunderstanding that the kets corresponded to the operators at each spacetime point

So with a finite number of variables you'd have a similar notation writing $\lvert x_i\rangle$

yes
$i$ is the index here instead of $x,y,z$

@RyderRude No, thank you. Although there wasn't an actual problem, you made me think about something I hadn't thought

great :P

12:59 PM
@RyderRude yes

but isnt that what I was pointing out?

@naturallyInconsistent It's entirely possible
I mean, conversations evolve and sometimes things are lost. I don't remember

@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 $R$

1:18 PM
@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 ðŸ˜‚ 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

@RyderRude let's not play the devil's advocate now :P

@Mr.Feynman that notation can genuinely confuse me lol. maybe it's just me :P
but yes, perhaps it is also standard to use this notation in multi-particle QM

@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

@Amit this is normal
background ablaze

1:31 PM
Normal?? How offensive lol

@Amit i sometimes zone out while reading but i continue reading anyway. i have to go back a few paragraphs

So maybe forcing one self to be focused is violence?
I must understand I must I must I must

@Amit there is putting in effort, and then there is self-abuse. One must strike a balance and not seek extremes, at least not too quickly.

it's torture for sure
some people talk about a thing called the monkey mind
i think it is a concept from monks

@naturallyInconsistent Yes, and since there is no perfect balance I think more often we flip flop between being over and under motivated

1:42 PM
stability achieved

Monkey island

donkey kong has entered the chat

He don't have no time for no monkey business, $\ddot{AA}$ $\ddot{AA}$..

2:00 PM
@GiorgiLagidze we can also use this philosophy room to discuss consciousness. it's become inactive now
there used to be consciousness discussions there a few months ago

There's a philosophy room!! @Mr.Feynman did you join, say something clever and leave yet?? ðŸ˜…ðŸ˜…

yes, and there are a lot of intelligent discussions there on illusionism

Here or in the philosophy SE?

i mean in that room

No but where is it?

2:03 PM
idk it became inactive suddenly. there were big discussions. i contributed a bit too

Ahh so it's on the philosophy SE

do rooms belong to SE sites

You can tell by the sophisticated crème wallpaper

lol

And the big overbearing $\varphi$
Overbearing in a good way

2:08 PM
we were discussing illusionism. i was refuting it mostly
@Amit have you seen this philosophy called structural physicalim : philosophy.stackexchange.com/questions/99656/…

No, can it be called metaphysics?

yes
it is a coherent answer to the hard problem
the post also mentions "the logical contradiction at the heart of illusionism"

So what are they saying can you summarize in 2 sentences? ðŸ˜…ðŸ˜…

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

That's cheating
You used non self explanatory terms lol

2:22 PM
but y :P
@Amit oh. u mean i used jargon

Jargon!!
The forsaken brother of Jörgen

but i cheated with brackets to define them

Ah lemme see

I dont see why it must be true that qualia is 1 place

2:24 PM
yes, but intuitively it seems true

in 2D if u cant go straightly to a point, does it matter if u go 100 small right angle turns towards it or 1 long turn ?

im not saying this is a proved theory. this is just a coherent answer to the hard problem

Intuition is the mother of all.
Lol

Daniel Dennett would however say that qualia can be put into language if you had infinite time. this would.make them multi-place relations

It's coherent but is it self consistent

2:26 PM
yes!

Coherence is just a property of language

it does not unnecessarily deny the existence of something we can observe directly
@Amit lol
yes. i shud say self consistent then

But it doesn't prove quaqualia is self arising
It hangs on that

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

A bit like a religious myth

2:29 PM
i would say it's more unlikely that they can be put into language

So it ends up as mysticism

anyone who claims that has a big onus

Lol
A big what

lol
ðŸ˜‚

@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

2:31 PM
i meant onus to prove. sorry :P

:-) right, ACM, same indicator often for me

@Mr.Feynman Yes, exactly - but I see that quite a lot in this chat :P

not to be confused with the rhyming word

Lost my qualia in Somalia
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 ðŸ˜‚ðŸ˜‚

yes. you will like it if you read more about it. it is def not mysticism to accept 1-place relations
i think i hav given a weird explanation

2:37 PM
Mystics also say you should read more... lol, jk

But non-mystics also say that. now the onus is on u :P
i should say that it is a branch of physicalism. it does not postulate some immaterial substance

Ehehe i'll read more for sure, not saying what ðŸ˜…

this is a good source, I think : plato.stanford.edu/entries/structural-realism (I havent read it yet tho)
im.not yet sure if it's talking about the same philosophy. lemme see
ok i did "find in page : qualia". There was only one hit, but it's indeed the same philosophy. i will also read this full

2:59 PM
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.:(

3:43 PM

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?

sure
how high the block goes is a function of how fast it initially is

Yes thats it I jst wanted to know if it would always reach the highest point on the wedge
Also if i would want to find out the slope of the wedge I would not be able to do so from this data?

from what data?

Like if I were given the initial velocity with which it is projected up, the weight of the block and angle of inclination.

3:57 PM
what's the difference between "angle of inclination" and "slope of the wedge"?

Oh yes tan theta
Ah okay got it, thank you :-)

4:13 PM
@ACuriousMind not everyone can hover above rabbit holes :P

4:25 PM
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

I don't really understand the setup here
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

Oh i was going to build on that. If it simply passes thru as a ray, i was going to add a piece of the refractive index of the box in its path
Then I would put a more "opaque" or like better blocking material if it simply just went through the piece
Also I wouldnt put the piece such that the ray goes perpendicular to the normal line b/t the mediums
I would draw it out but im on mobile atm.

4:40 PM
I don't understand what the presence of the box is supposed to do here at all
but I think the confusion here is that you think light actually propagates as rays?

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

classical light is a wave - you mustn't forget about Huygens' principle
Fermat's principle cannot tell you what happens with obstacles

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

in our modern understanding, it's a statement about the wavefronts from Huygens' principle, really

@Acuriousmind How sure are we that dark matter isnt just a part of ordinary matter? off topic ik

4:46 PM
@Obliv I don't know what this means
if it's "optimal" to go along a straight path, you get a straight path
what's the question

Like the plum pudding model was ditched for the model where there is empty space with positive n neg charges. What if the pudding was dark matter

@Obliv what are your definitions of "ordinary matter" and "dark matter"

Dark matter doesnt interact with light, only gravity
ordinary matter does both..? or maybe ordinary matter only interacts with light and dm is the only gravity part idk

if you don't know what the terms in your question even mean how do you expect me to answer it :P

all i know is the galaxies are spinning too fast ok

4:49 PM
Looking at the sky rn, it seems to be going at a reasonable speed

and what would that have to do with a box and light propagation?

nothing hence "off topic ik"
geometric optics is so boring tho
I just wanna learn new stuff already

then dont do geometric optics and cry at how difficult wave optics is?

I will soon! midterm next fri and i will be done with geo optics i think?

4:52 PM
Anyway, I wanted to learn a touch of chemistry for my geology class but already getting distracted off of the physics involved with the periodic table

you know you are technically not just allowed, but encouraged, to read faster than your lecturer is presenting stuff, ya?

Im not that good
Or motivated

"I just wanna learn new stuff already" and "I'm not that motivated" don't seem like they go together well :P

I mean Im motivated but lack of time is the big factor. Im also motivated by other things like sleep and stuff
Some might even say sleeping well is more optimal in learning than cramming more learning

of course it is
but then the problem isn't that you're not "good" or not "motivated" but simply that you're trying to run before you can walk :P

4:57 PM
i might even say exercise and mental health also play a large role.

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.

I guess I lied then, its not a lack of motivation to learning new material but the lack of motivation to sacrifice other things to do that?
Like i dont get how ppl sit in one spot for 12 hrs a day to learn. I coulda done that as a kid but now not so much

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

@Obliv patience and endurance like that is slowly trained, like a marathon

Just how far away are those things in terms of time needed would you say @ACuriousMind

5:01 PM
but I would not recommend students to train for them; it is very unhealthy

I agree, maybe just for a short while in uni/during ur doctorate but after that u gotta take care of urself so that slows things down

Can unbounded operators act on finite dimensional Hilbert spaces? If so can you give an example from QM?

@Sanjana no; all operators on finite-dimensional spaces are bounded and this should be immediate from the definition of boundedness

@ACuriousMind Umm...but there's no restriction on the dimension of the Hilbert space in the definition, right?

@Sanjana no
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

5:06 PM
If i studied 1000 hrs a year how many years would it take to get to your understanding of physics @acuriousmind

@Obliv I don't think these kinds of questions are useful because people learn very differently

Ok true, but like how many hrs do u think u "put into" physics/math etc? Like 1000/yr is avg 2.7 hrs a day which is doable.
I think in 10000 hrs people are considered to have mastery in something tho that is very vague
ive probably already put in like 4-500 hrs in math/physics so only 9.5 yrs to go potentially

I could not possibly tell you the hours but I went to university for 6 years

:( 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

5:35 PM
Thanks...I just proved it.
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?

(the way to get the direct link is to click the "share" button on the answer)

I was going to ask how you found the number at the end :p

the number at the end is my user number - the "share" button automatically adds it when you're logged in
the second-to-last number is the actual ID of the answer

Every finite dim irrep has to be one dimensional.

So is the catch every finite dim irrep *if it exists* then it is one dimensional.
Doesn't mean that there are finite dim irreps? Like that?
And why don't ** doesn't work sometimes for italics?

markdown doesn't work in multi-line messages
you can either use markdown or have a line break
@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