2:22 AM
@ACuriousMind What is the precise most general condition for an obstruction to have a global chart on some manifold?
I can see that compact manifolds don't have global chart...I mean given some topo information can I compute something and comment "This can't have a global metric on it."?

5 hours later…
7:17 AM
9

A related, but slightly different question appears here: Upper bound on the number of charts needed to cover a topological manifold So, we know that for a topological $n$-manifold there is always a covering by $n+1$ charts, but clearly we can sometimes do better. For instance, with $n$-spheres w...

7:29 AM
@Minsky hi. you are looking for the heat equation :) en.m.wikipedia.org/wiki/Heat_equation
it's not much simpler than the free particle Schrodinger eqn though. there's only so much that can be simplified there without the equation getting trivial. but it's a simpler physical system
@Minsky But this is an analogue of TDSE. the TISE is just the eigenvalue problem, so you can write an identical equation for the heat eqn too

7:55 AM
thanks, i'll see if i can read it @RyderRude. trivial is good for me
free assange !

no, that's not what i meant. it's not trivial :P
18

If we do the wick rotation such that τ = it, then Schrödinger equation, say of a free particle, does have the same form of heat equation. However, it is clear that it admits the wave solution so it is sensible to call it a wave equation. Whether we should treat it as a wave equation or a heat e...

there are also many similar posts if you search

8:12 AM
mainly that paragraph is clear, but not the next one - chat.stackexchange.com/transcript/message/64463527#64463527
I kind of have feeling what ACM is trying to do, but to achieve that, we need to show that the EOM stay the same for L and $L + f\dot q$ which i don't seem to be able to show

yes, thank you; the problem with searching for me, is that most times i don't quite know the names. for example, never seen the heat equation. so, i discuss with gpt, but says random crap, then ask here for some pointers. XD

8:26 AM
@GiorgiLagidze hi. suppose we got Q=dL/dv - f from the original lagrangian and f is non zero. so u modify ur lagrangian to L'=L-fv. now u can compute dL'/dv= dL/dv-f=Q+f-f=Q. so dL'/dv=Q. This implies that dL/dx is 0 on-shell when u use L' instead of L according to the logic in the previous comment
we are trying to modify our lagrangin to L' such that dL'/dv=Q becomes true

I get the part why $Q = \frac{\partial L}{\partial \dot q} -f$. If $f=0$, easy to see why L is independent of x. Okay, say $f$ is non-zero and our transformed lagrangian is $L'=L + f\dot q$.
are we modifying lagrangian in such a way that we account for the fact that eom for $L'$ must be the same as for $L$ ? or we change it in a way that we don't care whether eom are the same ?
$f\dot q$ definitely can't be represented by $\frac{d}{dt} f(q,t)$ , can it ?

ACM said $f\dot{q}$ is a total derivative. but i dont yet see how

exactly :P
The basis of his logic must be that you change your Lagrangian such that transformed lagrangian still yields the same e.o.m and then you put it in $Q$ after that
not only that, this logic only works if lagrangians are different by the $f\dot q$
no worries, better to ask ACM so that we don't go another rabbit hole :P

yes. lets wait for him
@GiorgiLagidze we are choosing the new lagrangian to differ from the old lagrangian by fv

yeah and if you do that, 1) L' and L don't have the same e.o.m 2) the logic of L being independent of x only works for that specific $f\dot q$.

8:40 AM
idk. maybe it's a total derivative somehow. but even then, $f$ depends on the symmetry parameter $a$, and so will $L'$
or maybe $f$ cant depend on $a$ for translations. but it does depend for general symmetries like boosts

try using EL for $L' = L + \dot q f$ and you will realize that it yields the different thing then it would for single L

we get d/dt(dL/dv + f)= dL/dx + v df/dx = dL/dx+ df/dt. and then df/dt cancels from both sides to give the same EL eqn
this is assuming independence of f from v which is true because of yesterday's discussion
ok i got it : fv is the total derivative of $F=\int f dx$, since $f$ is only a function of $x$ @GiorgiLagidze
u can compute dF/dt= $\frac{\partial F}{\partial x}$ dx/dt = fv
ok $f$ also cant depend on the symmetry parameter here because then $Q$ wud too according to Q=dL/dv - f @GiorgiLagidze

9:05 AM
@RyderRude dL/dx + v df/dx = dL/dx+ df/dt how is this correct ?

df/dt= df/dx * dx/dt =df/dx * v. the derivative wrt to x is a partial derivative

hm, you're assuming that f is not a function of t explicitly:) why ?

then the EL would be different too
we r using the EL eqn of time independent lagrangians

that's what I said, why are you assuming that f is not a function of t explicitly ? sure, L' then would yield the different e.o.m. but i guess you just choose such specific $f\dot q$ such that you want E.O.M to stay the same

for time independent lagrangians, f is not explicitly a function of t. f is f(q(t))
@GiorgiLagidze im sure the proof also works for time dependent lagrangians if we use the EL eqn for that in our proof

9:09 AM
I tried it and doesn't work.

$L' = L+f\dot q$
$\frac{\partial L}{\partial q} + \frac{\partial f}{\partial q}\dot q - \frac{d}{dt}(\frac{\partial L}{\partial \dot q} + f) = 0$

en.m.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation im sorry. the EL eqn is the same for time dependent lagrangians

it's the same, but the result that we get from it won't be the same for L' and L

yes. because df/dt = df/dx * v is not valid anymore

9:14 AM
exactly hahaha
i don't know what the proof of ACM actually proves

and the proof of fv being a total derivative also assumes only x dependence of f

hm, better to wait I'd say

@GiorgiLagidze at the very least, the special case proof of time independent lagrangians is also very important
time independent lagrangians show up in all of fundamental physics

yeah, but not the whole proof. thanks ryder. gotta go now and in the evening, i will join again

bye

2 hours later…
11:31 AM
@Sanjana Since a chart by definition is a homeomorphism to $\mathbb{R}^n$ or an open ball in $\mathbb{R}^n$, the existence of a global chart just forces the manifold to be $\mathbb{R}^n$.

Or a subset of which

only open subsets to be precise
but suppose you are given the metric in a chart of a manifold which is not global. can you compute anything with the metric to know if a global chart is possible? @ACuriousMind

@GiorgiLagidze Sorry, part of my argument there disappeared when I pasted it into chat and there's a typo there: For $F$ the anti-derivative of $f$ w.r.t. $q$, since then for not explicitly time-dependent $f$ we have $\frac{\mathrm{d}}{\mathrm{d}t} F = \frac{\partial F}{\partial q}\dot{q}$, so $f\dot{q}$ is the total time derivative of $F$.

@Sanjana as a special case, you can compute the Gaussian curvature everywhere and if it's always positive, you can conclude it's not flat topology
@Sanjana please also see this theorem : en.m.wikipedia.org/wiki/Myers%27s_theorem

12:19 PM
is there any sense in which a Riemannian metric is more natural to describe a manifold than a Lorentzian one?
as in, does curvature of a Riemannian metric have more "visual" interpretation in terms of shape

what does your question mean? What classifies as "more natural"?
One good thing about the Riemannian case is that a Riemannian metric always exists

@Mr.Feynman suppose i gave you a 2D curved lorentzian manifold, embeddable in 3D. then does the the curvature after embedding have an interpretation in terms of shape?
@Mr.Feynman oh
@Mr.Feynman there is also another thing that only the positive definite metric is a metric in the mathematical sense

both are metric in the mathematical sense

I mean the notion of a metric space. only the positive definite metric is the one in that sense

The most general definition of an inner product is that of a symmetric bilinear form over a vector space
@RyderRude In that case we're talking about different usage of the term "metric"

12:28 PM
yes

From a positive definite inner product you may derive a metric(=distance function) in that sense, yes

@Mr.Feynman when we embed a Lorentzian manifold, does the curvature have an interpretation in terms of shape? i mean, does spacetime have an actual shape if we embed it similar to what we see in pop sci GR illustrations?
the pop sci one is still incorrect, to be clear
by shape, i mean a curved surface in the visual sense
but it's 4D, so it cant be "visual". but let's say it was 2D
i think this is correct, because what else would an embedding even do other than give the manifold a curved shape :P
and spacetime is not a metric space

1:12 PM
nvm it has the definition for pseudo riemann manifolds earlier
is it the common view that spacetime has flat topology

1:42 PM
@ACuriousMind Ok...dS$_{d+1}$ space is topologically $\mathbb{R} \times S^d$ (and it indeed does not have a global metric)...but even AdS$_{d+1}$ space is defined via a codimension-1 hyperboloid in the embedding space, then why does it still have a global metric? Is it because we take the universal cover?
@RyderRude Interesting...I was not aware of this. Thanks

@Sanjana what do you mean by "global metric"?
the definition of a (pseudo-)Riemannian manifold $M$ is that the metric $g$ is a function on all of it
i.e. all such manifolds have "global metrics"
so I'm not really following you what you mean when you claim de Sitter space "does not have a global metric" but anti-de Sitter space does

Sorry...I missed the word "static"...why doesn't dS space have a global static metric while AdS space has one?

@Sanjana I think you've omitted a bunch of context here since I don't see what "static" has to do with the rest of this conversation :P

@ACuriousMind Yes, definitely...Sorry. These questions are widely disconnected.

and I don't really know what kind of "why" you're looking for here
when you write dS in static coordinates, you find there's a cosmological horizon

2:02 PM
I understand a "why" is highly subjective...but still I would take this as a consequence instead. Can you provide an alternative "more natural" answer at the level of the embeddings (if it is manifested at that level in any way)?
I mean, both are defined (in one way) as hyperboloids---one sitting and another lying, what privileged one of them to indulge in the pleasure of enjoying a global static metric?

you said "global metric" again :p

Sorry again :) Last time this happened with "weight vectors" and "weights"

well, we usually interpret dS as expanding, right?
but an expanding universe isn't really "static"
so what happens is that this universe is only locally static, since being globally static would violate it expanding (the global timelike Killing vector field would define a notion of time for which the metric was constant in time, i.e. for which the universe was not expanding)
AdS isn't intrinsically expanding, so it doesn't have the same "problem"
I'm sure this handwave is wrong somewhere but if you want the weird technicalities you'll have to wait for @Slereah :P

@ACuriousMind Why do you say this? The geodesic equation in AdS space does look like $\ddot{x}=-\Lambda/3 x$...Doesn't this mean contraction?

no, it's oscillation
that's why I said this was me handwaving :P

2:18 PM
@Sanjana these two theorems are also relevant: en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_theorem

@ACuriousMind So now the question becomes why isn't there a FRLW version of AdS space?
@RyderRude Yeah these are the more famous ones I guess...I knew about these. [Don't know much about AS theorem really though...except what it is talking about]

@Sanjana what do you mean?
FLRW becomes (anti-)de Sitter for $\rho = -p$ constant, i.e. FLRW filled only with dark energy

@ACuriousMind So even that doesn't describe a contracting universe?
How exactly are we characterising expansion here?
The linked answer says it is in the first time derivative of position---why should that be true?
I thought $a(t)$ tells you about the expansion/contraction stuff.

@Sanjana the definition of the Hubble parameter is $v = Hx$
i.e. you're relating velocity, the first derivative of position, to position

So what exactly is the criterion for expansion/contraction?

2:30 PM
"stuff that's farther way is moving faster away/towards us"
there's probably some ambiguity here where different people are using "expanding" and "contracting" in slightly different ways

this could also be due to motion. there has to be a definition in terms of metric

because as you can see in the post I linked, what is true is that in AdS two objects will get closer to each other over time, and something you throw away will return to you
that sounds a lot like what we colloquially mean by "contracting"
so I'd suggest not getting hung up on the words and first trying to be clear over what exactly you actually mean on a technical level

@ACuriousMind Hmm...but, what's your definition of it? I mean in terms of equations if you will...I thought it was $a(t)\ne 1$ for generic $t$, but now it seems this can't be true...because AdS space fulfills this criterion but is still not contracting!

I did say my explanation about dS being expanding was handwaving :P
here's another handwave: if you think about AdS as contracting and dS as expanding, then it should be obvious that only the expanding one will have a horizon (behind which the stuff expanding "away" from the observer will disappear)
but for the contracting universe, there's no equivalent reason we should get a horizon for the static observer

2:46 PM
neither of these handwaves convinces myself fully, though

@ACuriousMind Yeahhhh right...but what's the relation of having a cosmo horizon and having a global static chart? Cosmo horizon is obtained in a non-global static patch of the dS space...globally there's no horizon.

@Sanjana the horizon is the "boundary" you run into when you perform the transformation associated to the time-like Killing vector (i.e. translation in time)
remember that a fully static spacetime would have to have a global time translation isometry
and here we have a time translation isometry, but it can't "leave" the bubble bounded by the horizon

Oh I see...niceeee...
But still can you say anything with regard to the hyperboloids? See...the dS hyperboloid has the time axis as the symmetry axis...but the AdS space doesn't...I have a "feeling" that this has to have some relation with the existence/non-existence of global static metric...But I can't fit some solid (even hand-wavy) reasoning.
Anyway...I must be happy I have the math and two handwaves!

that's another annoying thing: The AdS space people often actually talk about is the cover of the one with periodic time because we don't want our time to be periodic
but it's not always so, so every time someone says "AdS" and then talks about its global properties we first should establish whether time is periodic or not in this context

The decompactified AdS space has the $t \in (-\infty,\infty)$

3:20 PM
Was what you stated only for Riemannian geometry and not Lorentzian?
dS space has a global chart on it but the manifold is not Minkowskian---there are closed spacelike curves...Or does it not cover the poles?

@Sanjana my statement holds for all topological manifolds, regardless of metric
I don't know what you mean by the "global chart" on dS
it's $\mathbb{R}\times S^{n-1}$ and there's no global chart on $S^{n-1}$

@ACuriousMind YES...then how can dS space have a global chart?

it doesn't
which chart do you think is "global"?

but it has a global metric?

I still don't know what you mean by "global metric"
there are no "local" metrics
the metric is a function on the entire manifold by definition, it has nothing to do with charts

3:28 PM
In the wiki [page ](en.wikipedia.org/wiki/De_Sitter_space#Closed_slicing) these are called global coordinates; but how can they be if there is a $S^{n-1}$?
I see the quotation symbols though :p

It is common in physics to use "coordinates" to include spherical coordinates, even though strictly speaking those are not chart coordinates for manifolds

I thought that they are "global" in contrast to Poincare patch(static patch) for example?

@Sanjana they claim the $z_i$ "describe a $S^{n-1}$", but there are n $z_i$ there
so whatever these are, they're not actually a coordinate chart for $S^{n-1}$
because those would have to be $n-1$ coordinates
they're probably meant to be the components of a n-dimensional unit vector lying on the $S^{n-1}$
actually something weird is going on with all the "coordinates" there, they're all $n+1$ coordinates but the article said that the space is $\mathbb{R}\times S^{n-1}$ so it should be only $n$
I think they're all "embedding coordinates"

Yeah they're doing the classic "(A)dS is a hyperquadric in $\mathbb{R}^{n,1}$"

Oh..makes sense now...But I think what Slereah told is also a thing.

I have seen many places one writes $ds^2=-(1+r^2)dt^2+\frac{dr^2}{1+r^2}+r^2d\Omega_{d-1}$ for the $AdS_{d+1}$ and call that "global metric" and the coordinates "global" coordinates...although these don't cover the full space because of the "sphere" part.

3:37 PM
which is usually how people introduce them
they are global in that sense yeah
The same way that spherical coordinates cover the whole sphere
Even though they're not like a map to R^n
But physicists will pretty often say "coordinates" for "some map to a product of Euclidian space and spheres"
or something weirder occasionally

So, once again I was confused by words of Wiki :(

Global metrics(in the way ACM describes it) always exist but global coordinate charts may or may not.

I mean the metric is always global

that's what I keep saying! :P

What you're looking at is just the metric expressed in some coordinate patch

if the metric was not global, then u would have an undefined metric in some regions of the manifold
by "metric is global", we just mean that each point of the manifold is mapped to a metric tensor

4:16 PM
@Relativisticcucumber We all agree that quite often people just arent good at teaching and will be confusing no matter what they do, but people telling you not to fall into rabbit holes is often extremely justified because a lot of stuff in physics is basically accidental and require observation / following the argument / physical insight to understand. If you are always looking for mathematical proofs as to why something is the way it is and no other possibility else,
then it is vastly more likely for you to be stuck exploring theory space that is utterly irrelevant to the physics of our universe, nor learning about what it is we want to talk about. There are quite a lot of people here on Phys.SE who is always mired in rabbit holes and refuse to pluck their heads out of it, and you can witness the disastrous situation they find themselves in but deny is a problem. That is what we are warning against.

4:41 PM
still doing a review on principles of stationary action in my mechanics class but I have a quick question:
why does it say in part b: "The answer is not a simple as replacing the force F by mg."
In part a, I wrote down 3 cases where F is greater than the friction force and $Mg\sin\theta$, less than, and the static case
I can't see how part b wouldn't just be the case where F is greater than the friction &gravity of M

@Obliv It says that because it's true :P
you now have an additional mass in this system that's also moving/being accelerated

oh true
nah i still dont get it lol
the tension of the rope is $F + F_f - Mg\sin\theta$ if we choose down the plane as negative and up the plane as positive
for part a, $F + \mu_kN - Mg\sin\theta$ when the block is moving down the plane

@ACuriousMind Is the notion of a structure group for bundles roughly equivalent to that of the pseudogroup for manifolds

$F - \mu_kN - Mg\sin\theta$ if it's moving up the plane and $|F - Mg\sin\theta| \leq \mu_sN$ for the static case..
and I can't see how you don't just replace $F$ with $mg$ for part b lol

Firstly, you can replace N. Then, for part b, you have to consider how it the string is inextensible and thus the new mass will also accelerate.

4:54 PM
right, i replaced $N = Mg\cos\theta$ in my notes

@Slereah yes, both constitute conditions on the allowed transition functions

Are they related to each other in interesting ways?
I'm not sure I've ever really seen the two discussed together

I think you can exhibit some pseudogroups also as structure groups on the tangent bundle?

The new mass will accelerate for sure, with $F_{net} = ma = mg - Mg\sin\theta \pm F_f$ depending on each case

e.g. when you restrict the transition functions to be isometries, this means their Jacobians make $\mathrm{O}(p,q)$ transition functions on the tangent bundle

4:56 PM
@naturallyInconsistent But how does this change the acceleration on the block M such that $a$ isn't just $\frac{F - Mg\sin\theta \pm F_f}{M}$ like in part a..

could be
I guess the most common application of pseudogroups is G-structures which do involve bundles

@Obliv You kinda have to think of it. One way is to consider the EoM for both masses separately, and then try to stitch them together

Oh I see.. set up EoM for both individually, set Tension on each equal to each other solve for a

I haven't really done much with pseudogroups, though

When I see ACM and Slereah discussing math I'm all like "shhh the adults are talking"

5:02 PM
@Mr.Feynman same

5:17 PM
does sign convention matter as long as ur consistent
like up and down the plane and up and down vertically
I chose + for up the plane, + for down does that work

they do not matter, no
Although I recommend using common one for convenience

5:40 PM
Signs essentially reflect just a choice of ordering

5:54 PM
does this make sense $\int -ky dy = \int -dU(y)$
basically I have a spring hanging from a ceiling and I want a potential energy function so $F = -ky = -\frac{dU}{dy}$
or do I need to solve $\frac{d^2y}{dy^2} = -ky$ first

2 hours later…
7:41 PM
Today I was going thru shankar- fundamentals of physics, mid-high school textbook. It falls under the "all physics in one roof category" that naturallyInconsistent despises, but i thought it was pretty good, wish I had it in my highschool or middle school

do any of ya'll know a good path integral reference? i would just like to know the conceptual basis and the particulars of what it means to "change coordinates" in the context of computing a path integral

@SillyGoose the path integral measure does not rely on a choice of coordinates
it's a measure on paths, not coordinates; what problem are you actually trying to solve this time? ;P
but the classic reference on path integrals is Feynman and Hibbs
if you want the rigorous version, Glimm and Jaffe

@naturallyInconsistent Oh no i do not want to be in a disastrous situation

2 hours later…
9:43 PM
have any of you guys applied to graduate school in the states? i am really struggling with refining my personal statement because i feel there is expected to be this grand reason why one wants to go to graduate school for physics and why one wants to be a physicist, but literally all i have to say is that it interests me and i have no other reason except i like it :PPP
and the prompts are like "why must you go to THIS school to work in THIS physics department" and i am confused about what to say besides i like the topics of those professors. to make matters even worse i like multiple topics so i fear i sound like a flake
the more i look into topics, the fewer i find that i broadly hate :((( i dont even know if theres anything i wouldnt do anymore :,( maybe string theory is a no-go but everything else seems great to me DDDD:
and i dont even know why i like it. how can one really know why they like something? eek merp

@Relativisticcucumber I haven't applied to graduate school like that but I've had to write my share of personal statements like this; stuff like "prof X does interesting stuff and I'm interested in it because of Y" is what they want to hear. That you might also be interested in other topics there isn't relevant in that context, just don't talk about that (omission is not a lie!) and try to come up with why you find that specific stuff fascinating.
It also always feels pretentious/fake to me when I have to come up with grand reasons why I like stuff but you just have to work through the cringe.

9:58 PM
@ACuriousMind but how can one even know why they like something
i have loved science since i was like 4 and i have no idea why
best i can come up with is why i like physics best of all the sciences :((
which involves eliminating every other science

@Relativisticcucumber How would you describe the feeling when you do science you enjoy?

@ACuriousMind illuminating

no, there are no wrong answers here :P
but I can't make a lot out of that single-word description
you have to be a bit more pretentious ;)

vomits
i feel like physics is the only subject that answers my questions in the way that i want them answered
and my main life goal is to answer my questions
but then the next natural question is why physics scratches that itch
and idk math is too much and chem and bio are too little and then here we are
well chem and bio are awesome but like i want to know why in chem and that is physics
bio is just a fun hobby
too tedious

Maybe let me give you example of what I'd write in similar situations. I'm not saying you have to feel the same way, just that this is one way people explain "why" they like things: "I have always enjoyed understanding the world in terms of structures: I enjoyed learning about languages because languages are what structures thought. I enjoyed learning programming because the structures of computer programs are what underlies most of the world around us.
. Theoretical physics - in particular quantum field theory - reveals the structures underlying the most basic building blocks of reality, and so is the natural conclusion of my desire to understand the structure of the things that surround me. The research program of prof X at your institution aligns well with the attempt to understand the most basic structures because <insert some specifics about the research of X here>, and so I am eager to contribute to this research."

10:10 PM
okay i see. so i need to figure out broadly why i like physics but not psychoanalytically

You need to think of this statement as an ad - this isn't about being brutally honest about your psyche, this is about selling other people on you

ah oh no i am bad at that okay i will work on it thank you
i just cringe too hard when i try that
but i must get over it
i like what you wrote. it doesnt make me cringe.

as I said: You just have to work through the cringe :D

okay thank you.

Just write something and then gradually refine it
don't get paralyzed attempting to come up with the perfect statement on the first try

10:13 PM
on a separate note, i am learning the tight binding binding model, and i am confused about the results of something that i worked through. i was going off of tong's notes on band theory and calculated energy levels for the right binding model. it would seem what i have calculated is the energy bands? however, conceptually i dont understand these. [...]
[...] i know that when we have one atom, the solutions to the schrod eq make up orbitals, and i know the tight binding model is exploring what happens when we put neighboring atoms on a lattice, but what do all of these bands look like physically? just a ton and ton of every levels located at every lattice site?
@ACuriousMind i will try :,)

I'm afraid condensed matter theory is not really my forte

i will wait for naturally whom i hope will be inclined to answer hehe
@SillyGoose ymacf

i feel like if i was reading personal statements the only answer I'd consider honest for liking something is that they happened to find out that they like that thing ;_;
there is no proof, only the fact that they willingly do such thing at a deep level

yeah i feel like we all have dispositions and we need to encounter things to see what maximally appeals to those
academia is weird tho bc i dont feel like u do physics for any reason other than you like it

The thing is, the personal statement tells you something about the person who wrote it: What did the person who wrote this consider important, especially in comparison to other applicants? Is the statement specific to the person(s) addressed, or is this generic enough to go to a dozen recipients? etc.
it's not so much that there are "right" and "wrong" answers here, the point is that how you attempt to answer them already reveals a lot

10:38 PM
i don't understand why it says there is quite a bit of algebra. Isn't it just $mg\cos\theta$ that is pushing on $M$
i.e the normal force

@Relativisticcucumber the states are delocalised, Tong's notes discuss this

then just take the horizontal component of it

11:01 PM
yeah I have no idea how $F = Ma = mg\sin\theta$ isn't the answer..

11:19 PM
@fqq i dont understand what this means. so we have eigenstates with energies that form a band, so you mean these states are spread all over the lattice? but that doesnt seem to tell me how i should physically interpret this "energy band" that the solid exhibits?
or am i misunderstanding?