« first day (3998 days earlier)      last day (924 days later) » 
02:00 - 15:0015:00 - 00:00

BannedUser
BannedUser
2:25 AM
Hello for hot air balloon why is $\rho_{in} T_{in}=\rho{out} T{out}$
where rho in is air density inside hot air balloon and rho out is density outside sane for Tenperature T
 
BannedUser
BannedUser
2:39 AM
It was derived from PV=nRT
 
BannedUser
BannedUser
3:35 AM
product of rho and temperature is equal to rho inside and temperature inside and outside
 
John Rennie
John Rennie
4:27 AM
@BannedUser If we take PV = nRT and divide both sides by V we get P = (n/V) RT.
But n/V is the number of moles per unit volume so it is just the density (in moles per cubic metre) so our equation becomes P = ρRT.
Now rearrange this to ρT = P/R. And the pressure is constant, i.e. one atmosphere everywhere, and R is constant so we end up with ρT = constant.
 
antimony
antimony
4:52 AM
> A single photon has no frequency (or wavelength). The wave characteristic of light emerges from a large number of photons.
what ^ that isn't correct is it?
i really don't think that is correct...
(afaik this is not in the context of relativity - where i suppose they could say the actual wavelength is relative to observer etc etc)
since i've seen single photon generation circuits where they bandpass out the pump wavelength etc, well that would be meaningless to apply a bandpass if the generated single photon didn't even have a wavelength
 
John Rennie
John Rennie
If we take the QFT approach then a photon is a mode of the quantum field i.e. an infinite plane wave with a perfectly defined wavelength. So in this case the photon does have a well defined wavelength.
But real photons are not infinitely delocalised and are more reasonably described a wavepacket with some finite length. The localisation restricts the photon to a region of space Δx, so now it has a momentum uncertainty given by the usual Δx Δp = ℏ/2.
But since p = h/λ this means the photon now has a spread of wavelengths rather than a single well defined wavelength. But in most cases it would still be reasonable to say the photon has a wavelength.
 
BannedUser
BannedUser
lord @JohnRennie but why is product of temperature and density of outside and inside the same?
 
antimony
antimony
ahh i see thanks @JohnRennie that is fascinating!
 
John Rennie
John Rennie
@BannedUser Because the pressure is the same inside and outside. A hot air balloon is not sealed. It is open to the atmosphere at the bottom.
 
BannedUser
BannedUser
for hot air balloon is in and air is out
@JohnRennie what about temperature
 
John Rennie
John Rennie
5:03 AM
We worked out ρT = P/R. Yes?
 
BannedUser
BannedUser
yes
 
John Rennie
John Rennie
And the pressure is the same inside and outside the balloon. Yes?
 
BannedUser
BannedUser
yes
Now what about R
 
John Rennie
John Rennie
R is the gas constant.
 
BannedUser
BannedUser
ahhhhhh
I understand it now. Thanks you as always! You saved my day
 
John Rennie
John Rennie
5:05 AM
:-)
 
cOnnectOrTR 12
cOnnectOrTR 12
First we charge a conductor with e then for a small time there is current inside it which soon disappears. This means field inside the conductor is zero now. By gauss law we can show where all the e went. Because the field is zero inside , flux is zero, no charge inside. That means all the charges are now on the surface.
 
antimony
antimony
@JohnRennie this sounds like another way of the reality where the linewidth can never be infinitely small - which i can imagine being the case for a variety of practical reasons. so i suppose this momentum uncertainty is another aspect which contributes to a finite linewidth?
 
cOnnectOrTR 12
cOnnectOrTR 12
@JohnRennie please verify!
 
antimony
antimony
@JohnRennie no hurry with my question, seems you have your hands full already :)
 
John Rennie
John Rennie
@antimony Yes. This the cause of lifetime broadening in spectral lines.
 
antimony
antimony
5:07 AM
ahh lovely thanks :)
 
BannedUser
BannedUser
BTW lord @JohnRennie pressure in is same as outside because " If the pressure inside the balloon is larger than the outside air pressure gas will move from the inside to the outside. When the pressure inside is lower, gas will move into the balloon through the opening. So, at equilibrium, the pressure inside equals the pressure outside." Right?
 
John Rennie
John Rennie
@antimony Emilio has written one of his characteristically excellent answers about this:
10
A: What is the meaning of natural line broadening?

Emilio PisantyDespite the convenient fiction that transitions between quantum states can only happen if the photon energy exactly equals the energy difference between the two states, $$ \hbar \omega = E_2-E_1, $$ if you take that literally then that would imply that even if the photon energy were only off by o...

 
BannedUser
BannedUser
The problem is I don't think pressure will remain same all the time.
 
John Rennie
John Rennie
@BannedUser Why won't the pressure remain the same?
@cOnnectOrTR12 Yes
 
BannedUser
BannedUser
@JohnRennie because if pressure outside and inside is same then Hot air balloon won't go up
because of weight of balloon
I am thinking in terms of force
 
cOnnectOrTR 12
cOnnectOrTR 12
5:12 AM
@JohnRennie That is why we say charges arrange themselves in such a way that outside there is field perpendicular to surface and inside it’s zero.
 
antimony
antimony
wow that is marvellous thank you @JohnRennie!!!
 
John Rennie
John Rennie
@BannedUser Consider an air bubble in water. The pressure of the air inside the bubble is the same as the pressure of the water around it. Yes?
@antimony :-)
 
cOnnectOrTR 12
cOnnectOrTR 12
@JohnRennie right?
 
BannedUser
BannedUser
Hmmm
I will try to draw a diagram
 
John Rennie
John Rennie
@cOnnectOrTR12 It's the reason why the field inside is zero. The reason the field lines are perpendicular to the surface is slightly different.
 
BannedUser
BannedUser
5:22 AM
area will be same and force should be same too since it cancels out or else air bubble will collapse so yes
@JohnRennie
 
John Rennie
John Rennie
OK :-) So the pressure is the same inside and outside the bubble, but the bubble still rises.
 
BannedUser
BannedUser
bouyant
 
John Rennie
John Rennie
But the balloon is basically just a bubble of slightly less dense air in slightly more dense air.
Just like a bubble in water but the density difference is a lot smaller.
So if the bubble in water rises (which it does) our balloon can rise as well (which it does) even though the pressure is the same inside and outside. Yes?
 
Rajorshi Koyal
Rajorshi Koyal
@JohnRennie Sir can we discuss a problem after this discussion?
 
John Rennie
John Rennie
So if we can figure out why the bubble rises this will also tell us why the balloon rises.
@RajorshiKoyal I only answer physics problems I'm afraid.
 
BannedUser
BannedUser
5:27 AM
@JohnRennie yes
 
Rajorshi Koyal
Rajorshi Koyal
Yes but you answered a computer programming question yesterday didn't you?
 
John Rennie
John Rennie
@BannedUser It's because in any fluid the pressure changes with depth according to ΔP = ρgh, where ρ is the density of the fluid. Yes?
 
cOnnectOrTR 12
cOnnectOrTR 12
@JohnRennie what if after the charges reach the surface, that now there is a field inside the conductor? Because there are no charges to develop current so there might be some field inside!
 
John Rennie
John Rennie
It isn't just that the charges reach the surface. Once the charges are at the surface they spread themselves out over the surface to make the field inside zero. That is, the surface charge density will vary over the surface.
It is always possible for the charge density at the surface to vary in such a way that the internal field is zero.
 
cOnnectOrTR 12
cOnnectOrTR 12
That means they are free to move !
@JohnRennie is there a way to show this
 
John Rennie
John Rennie
5:37 AM
They must stay at the surface, but they can move sideways along the surface.
 
BannedUser
BannedUser
sorry I was called outside I hope I haven't wasted your timd
 
John Rennie
John Rennie
@BannedUser Where did we get to ... ?
11 mins ago, by John Rennie
@BannedUser It's because in any fluid the pressure changes with depth according to ΔP = ρgh, where ρ is the density of the fluid. Yes?
 
BannedUser
BannedUser
Yeah this is buoyant force
pressure bottom is larger than combined top and weight of bubble
this implies density of air must be smaller than water
thus it rises up
 
John Rennie
John Rennie
If you consider our hot air balloon, the it is open to the atmosphere at the bottom, so at the bottom of the balloon the pressure is the same inside and outside the balloon.
But at the top of the balloon the pressure inside the balloon is slightly greater than the pressure of the air outside it.
 
BannedUser
BannedUser
ah so this is what you mean
so it will look like a gradient
 
John Rennie
John Rennie
5:46 AM
That's because as we move upwards by the height of the balloon $h$ the pressure inside decreases by $\Delta P_{in} = \rho_{in}gh$ while the pressure outside decreases by $\Delta\rho_{out} = \rho_{out}gh$. Yes?
 
BannedUser
BannedUser
yes correct
so this means top got pressure difference
bottom got equal because of opening
 
John Rennie
John Rennie
Yes :-)
 
BannedUser
BannedUser
Ah stupidly I thought pressure in balloon is constant everywhere forgot to take in account pressure changes with height
@JohnRennie Thank you again boss
 
John Rennie
John Rennie
:-)
 
 
3 hours later…
ACuriousMind
ACuriousMind
8:26 AM
@antimony It may be good to point out here that you almost never have eigenstates of position or momentum (and hence wavelength for photons) for any particles. All "real" particles tend to be such wavepackets as John describes, but you can have situations where the width of the wavepacket is small against the momentum or position resolution we care about and so we pretend they're sharply localized states.
i.e. this isn't special to photons, it's a gap between the idealization we deal with in theory and reality that occurs almost everywhere
 
Slereah
Slereah
8:46 AM
it never happens since they can't be normalized
but then again we also don't have apparatuses to measure $x$ and $p$
there's no infinite precision measurement apparatus
for a start, if the particle isn't in the lab, we can't measure that
so that rules out measuring $\hat{x}$ right away
If we're using like a wire chamber we are measuring their presence within a region as thick as the wire or something
 
cOnnectOrTR 12
cOnnectOrTR 12
9:26 AM
@JohnRennie !
@ACuriousMind !
 
ACuriousMind
ACuriousMind
@cOnnectOrTR12 ?
 
Slereah
Slereah
What is it Lassie
Did someone fall down the well
 
cOnnectOrTR 12
cOnnectOrTR 12
Have you read brief history of time @ACuriousMind
 
ACuriousMind
ACuriousMind
It's been a while, but yes
(man, I read that more than a decade ago...)
 
cOnnectOrTR 12
cOnnectOrTR 12
I have read three chapters seems ok. But are later chapters difficult to understand and require some knowledge of the topics?
 
ACuriousMind
ACuriousMind
9:32 AM
I read it when I was in high school and really didn't have any knowledge of physics beyond that
it's not a textbook
 
cOnnectOrTR 12
cOnnectOrTR 12
@ACuriousMind so did you understood it all roughly? I think maybe it’s good to read some of these books when you are not studying your regular textbooks
 
Slereah
Slereah
I also had the CD-ROM version of it
it was very 90's
 
ACuriousMind
ACuriousMind
sure, I understood it - to the extent that one can "understand" any pop-sci account of physics
it's not an "understanding" that would enable you to solve any cosmology problems or something like that
it's just a general feeling of being a bit better informed than before :P
 
cOnnectOrTR 12
cOnnectOrTR 12
@ACuriousMind ya! When I read from forth chapter every now and then I have lot of questions. It becomes soon boring. How do I approach such books?
Seems like Stephen didn’t care about the viewers and kept on saying whatever he liked
@Slereah I have paperback
@Slereah did you just called me lazy !
 
John Rennie
John Rennie
@cOnnectOrTR12 no
It's a joke that only works if you know about the films about a dog called Lassie.
 
cOnnectOrTR 12
cOnnectOrTR 12
9:42 AM
@JohnRennie why is he calling me a dog
 
John Rennie
John Rennie
He isn't calling you a dog!
When you posted:
16 mins ago, by cOnnectOrTR 12
@JohnRennie !
It looked as if you might be calling for help.
And in the Lassie films Lassie always ended up rescuing someone.
 
cOnnectOrTR 12
cOnnectOrTR 12
@JohnRennie ya! Okay
 
John Rennie
John Rennie
Lassie
Lassie is a fictional female Rough Collie dog, and is featured in a short story by Eric Knight that was later expanded to a full-length novel called Lassie Come-Home. Knight's portrayal of Lassie bears some features in common with another fictional female collie of the same name, featured in the British writer Elizabeth Gaskell's 1859 short story "The Half Brothers". In "The Half Brothers", Lassie is loved only by her young master and guides the adults back to where two boys are lost in a snowstorm.Published in 1940, Knight's novel was filmed by MGM in 1943, as Lassie Come Home with a dog named...
 
bolbteppa
bolbteppa
I didn't bring it up first this time today :D But we can't measure a point particle in classical mechanics with infinite precision either, yet there we don't use this to turn around and say that (classically) particles only exist to within the accuracy we can measure
 
cOnnectOrTR 12
cOnnectOrTR 12
@JohnRennie is it good to read such books! Or it’s of no use!
 
bolbteppa
bolbteppa
9:45 AM
I have written a huge answer to why this claim is just such a big mistake and contradicts just so many QM books here
I even added criticism of the Leader's use of this against LQG at the end
 
John Rennie
John Rennie
@cOnnectOrTR12 Popular science books can be enjoyable. I certainly enjoyed reading A Brief History of Time when it was first published and I didn't know much about quantum mechanics or relativity. But they will not teach you much about physics.
So read popular science books if you enjoy them, but otherwise don't bother.
 
cOnnectOrTR 12
cOnnectOrTR 12
@JohnRennie I read because I think maybe it will broaden up my thinking and views and also what’s going on in the scientific world and about some history !
Sense of a bigger picture!
 
John Rennie
John Rennie
It's a good book, but remember that:
1. it's getting a bit dated now
2. it's Hawking's opinions and not everyone agrees with him
In particular his claims about the Hawking Hartle state for the beginning of the universe are not widely accepted.
 
cOnnectOrTR 12
cOnnectOrTR 12
@JohnRennie can I ask what did you enjoy about it ?
 
bolbteppa
bolbteppa
If you want to see whether you'll like Hawking's book, this is a good documentary to watch
You'll know within the first 10 minutes basically
 
John Rennie
John Rennie
9:51 AM
BBC Radio 6 is playing a lot of Stevie Wonder tracks this morning. I'm bopping along like a teenager (luckily the webcam is off).
 
cOnnectOrTR 12
cOnnectOrTR 12
@JohnRennie what did you enjoy about the book?
 
John Rennie
John Rennie
@cOnnectOrTR12 At the time I read it I knew next to nothing about relativity and I was very curious about it. Hawking's book made me feel like I was beginning to understand it, even though when I actually learned GR I realised that Hawking's book had taught me very little.
 
cOnnectOrTR 12
cOnnectOrTR 12
@JohnRennie what is better? Reading motion in 1D, 2D from a textbook or reading GR from a popular science book?
 
John Rennie
John Rennie
It depends. When I read A Brief History of Time (33 years ago - good grief!) I didn't want to learn GR. I just wanted to be entertained. And I found the book very entertaining.
When I wanted to learn GR I read a textbook on GR.
So if you want to be entertained read a popular science book. If you want to learn physics read a physics textbook.
 
cOnnectOrTR 12
cOnnectOrTR 12
@JohnRennie what is entertainment? When I read I have lots of questions!
 
John Rennie
John Rennie
10:02 AM
Obviously ABHOT doesn't entertain you, so don't read it. Find something else that's fun to do.
It won't teach you any physics so if it isn't entertaining you it isn't doing anything useful.
 
cOnnectOrTR 12
cOnnectOrTR 12
So when you read didn’t you have any questions?
 
John Rennie
John Rennie
It was 33 years ago so I only remember that I enjoyed it, not exactly why I enjoyed it.
 
cOnnectOrTR 12
cOnnectOrTR 12
At what age
 
John Rennie
John Rennie
I read in when it was published, so that would have been in 1988 when I was 27. So I had finished my PhD a couple of years earlier and was working as a colloid scientist.
 
cOnnectOrTR 12
cOnnectOrTR 12
Which college?
 
John Rennie
John Rennie
10:09 AM
I went to Cambridge.
 
cOnnectOrTR 12
cOnnectOrTR 12
:):):)’’
where are you from?
 
ACuriousMind
ACuriousMind
@cOnnectOrTR12 well, if you find it boring, then I'd stop reading it :P
 
John Rennie
John Rennie
That's complicated. My parents are UK nationals but I was born in the Sudan. My father was working there. We moved back to England when I was 8.
 
ACuriousMind
ACuriousMind
in general I don't approach texts with the expectation that they explain everything in the level of detail I need - I'll follow along and make note of what I don't understand and if I'm really interested in it then I'll look it up afterwards elsewhere
 
John Rennie
John Rennie
Little known fact, Michael Atiyah also lived in Sudan when he was young, and in fact in the same city as me (Khartoum). However he had left by the time I was born.
 
ACuriousMind
ACuriousMind
10:14 AM
but in general I'm not particularly bothered by not understanding something immediately - understanding for me is something that grows gradually, from repeated exposure and practice and reading different accounts of it
 
cOnnectOrTR 12
cOnnectOrTR 12
@ACuriousMind doesn’t that means you are stupid?
@JohnRennie :)
 
ACuriousMind
ACuriousMind
@cOnnectOrTR12 I don't think "stupid" is a particularly useful word
it's mostly wielded as an insult by people who do understand something against those who don't, why would I insult myself by thinking of me as stupid?
 
John Rennie
John Rennie
On the other hand I have several times found myself thinking "that was a stupid thing to do", usually in that split second before the pain starts.
 
cOnnectOrTR 12
cOnnectOrTR 12
@ACuriousMind don’t you think if you read something and you get a feeling that it’s not for me the first time that you are not getting it the first time even very slightly then it’s not meant for you
 
John Rennie
John Rennie
Attempting to leap over a parking meter is one such occasion that has stayed in my memory.
 
ACuriousMind
ACuriousMind
10:20 AM
@JohnRennie oh, sure, I have plenty of those - but that was stupid is different from I am stupid ;)
 
John Rennie
John Rennie
:-)
 
cOnnectOrTR 12
cOnnectOrTR 12
Doesn’t reapeated exposure means you are memorising it instead of learning it
@ACuriousMind
 
ACuriousMind
ACuriousMind
@cOnnectOrTR12 No. We were even taught in school to read texts in a step-wise process - first get the general idea, a rough outline, then try to follow more detailed arguments or information. The crucial advantage of textual information is that you can read it as often as you need to, there is no requirement to absorb everything the first time.
@cOnnectOrTR12 If you "memorize" it in the sense of just repeating the text verbatim, then sure. But the point of learning something is that once you have gained understanding, you can explain it in your own words.
 
cOnnectOrTR 12
cOnnectOrTR 12
I didn’t meant to offend you! Stupid in the sense like average not genius
 
ACuriousMind
ACuriousMind
Incidentally, that's why I hate videos as a means of conveying technical information - it's much harder to go back and look something up or re"read" a specific part than it is with a text
 
cOnnectOrTR 12
cOnnectOrTR 12
10:26 AM
And I am not liking brief history of time .
Am I stupid? Why have I not understood a simple concept of charges on a conductor and field inside it is zero in few goes @ACuriousMind
 
ACuriousMind
ACuriousMind
as I said, if you're not liking it, just don't read it unless someone is forcing you to :P
 
cOnnectOrTR 12
cOnnectOrTR 12
Is it because it’s too simple
Does the need to make things complicated makes thing harder and gives a false sense that I am stupid @ACuriousMind ?
 
ACuriousMind
ACuriousMind
@cOnnectOrTR12 As I said, I don't think "stupid" is a useful way to think about this. Perhaps you're missing some prerequisite (e.g. you have often insisted on talking about field lines but several people have told you that's not the most useful way to approach electrodynamics), perhaps the explanations you're reading are not particularly good, perhaps you have other things going on that prevent you from focussing, ...
 
cOnnectOrTR 12
cOnnectOrTR 12
It’s a feeling that it’s good to understand it in terms of field lines
 
ACuriousMind
ACuriousMind
Reasoning rigorously, thinking with the kind of logic that physics is built on and not "everyday intuition", is a skill in itself that is curated by good teachers, but is often neglected. Some people seem to have more natural aptitude than others to think that way, but calling that "stupid" is, again, not useful. It is not a moral failing to not get it.
 
cOnnectOrTR 12
cOnnectOrTR 12
10:40 AM
@ACuriousMind which way ? I don’t get it!
 
ACuriousMind
ACuriousMind
"that way" refers to the "thinking with the kind of logic that physics is built on" from the sentence before that. Sorry, I know my sentences are not always the easiest to read :P
 
cOnnectOrTR 12
cOnnectOrTR 12
Please explain first line
 
ACuriousMind
ACuriousMind
What I mean is that science in general and physics in particular don't use "common sense", they use logic.
But what most people use in everyday life isn't logic, not rigorously, it's just a feeling about "what makes sense" or "intuition"
 
Manas Dogra
Manas Dogra
If I put some moving charged particles in a box, then since the charged particles are moving the current should be non-zero but since it is a "box", there is no change in charge so by i=dq/dt current is zero, which one is correct and why?
 
ACuriousMind
ACuriousMind
this is the largest hurdle I see for people trying to learn elementary physics - the problem is not that the issues are particularly complicated in terms of raw complexity, it's that you have to learn how to think like a scientist in order to begin to grapple with them.
 
Manas Dogra
Manas Dogra
10:44 AM
@ACuriousMind Would you mind if I put this in my whatsapp status? :p
 
ACuriousMind
ACuriousMind
@ManasDogra what are the particles that hit a side of the box doing?
 
cOnnectOrTR 12
cOnnectOrTR 12
@ACuriousMind how far am I from this goal
 
ACuriousMind
ACuriousMind
@ManasDogra no, go nuts :P
@cOnnectOrTR12 I don't know - we don't have mind reading options yet ;)
 
Manas Dogra
Manas Dogra
@ACuriousMind just bouncing back without any loss in energy---no interaction with the box. I mean infinite potential at the boundary.
 
ACuriousMind
ACuriousMind
@ManasDogra well, in that case you have a bunch of tiny currents inside the box but the average current over the box is 0
 
cOnnectOrTR 12
cOnnectOrTR 12
10:49 AM
@ACuriousMind As a scientist how do you see my efforts to understand it in terms of field lines is because I think it’s better? Is it my absolute rigidness or my extra effort from everybody else that will make me a boss of physics?
 
ACuriousMind
ACuriousMind
@cOnnectOrTR12 I think it's an indication you haven't yet realized that physics is a bunch of models and that not every model is appropriate to explain every situation
the "field line model" is neat to understand some things, and not very useful for others
insisting on using only one particular model/visualization is just refusing to engage with the full extent of the actual physics
 
Manas Dogra
Manas Dogra
what if there is a single charged particle?
I mean if there is a single moving electron, then the charge of the electron is not changing so current must be zero but just because it moves it will consist of a current?
 
ACuriousMind
ACuriousMind
you really have no sound logical reasoning to claim that field lines are "better", it's just a feeling - that's exactly the type of unhelpful reasoning I described above
@ManasDogra well, then you have a current given by the motion of the particle and it doesn't really matter that you put it in a box
Current isn't just $I = \partial_t Q$, you can also have current densities $j = \partial_t \rho$, and the charge density here is definitely changing!
 
cOnnectOrTR 12
cOnnectOrTR 12
@ACuriousMind because they are lines in physical sense . I mean literally . Aren’t they?
 
ACuriousMind
ACuriousMind
the simple $I = \partial_t Q$ is only appropriate when you have a bunch of charge $Q$ sitting somewhere and only one path (wire) current can leave it through
@cOnnectOrTR12 I don't know what it means for them to be "lines in physical sense". They are lines we can draw to visualize the underlying vector fields of the electric or magnetic fields.
but all the physical formulae refer not to "field lines", they refer to the mathematical vector fields
 
glS
glS
10:57 AM
@Semiclassical actually, you do know that. You can set bounds on the maximum number of product states whose mixture gives a target separable state. See e.g. physics.stackexchange.com/q/399675/58382
 
ACuriousMind
ACuriousMind
the field lines are a visualization, you can't enter them as input into a formula - they're a qualitative model, not a quantitative one
 
cOnnectOrTR 12
cOnnectOrTR 12
@ACuriousMind then why do we see magnetic field lines around a magnet when we sprinkle iron powder. Are you saying like electric field lines magnetic field lines have no physical existence?
 
ACuriousMind
ACuriousMind
Again, "physical existence" is a much more difficult issue than you're giving it credit for :P
you "see magnetic field lines" with the iron powder because the whole point of why magnetic field lines are a useful concept in the first place is because they tell you in which direction a little magnetic dipole will point when you put it somewhere
 
Manas Dogra
Manas Dogra
@ACuriousMind Oh yes, I was just omitting a term in the continuity equation. Thanks.
These high school texts introduce "current" in the context of electrical circuitry, so I never got to appreciate this in this way.
 
ACuriousMind
ACuriousMind
but even there - if you take finer iron filings in that experiment, you'll "see" more lines that are thinner
 
cOnnectOrTR 12
cOnnectOrTR 12
11:04 AM
When we put potential differences between the flaxseed in oil we see the seeds arrange in curve lines. What are these if not field lines
 
ACuriousMind
ACuriousMind
@cOnnectOrTR12 they are lines of seeds
the "field line" we draw in diagrams is a mathematical abstraction
sometimes we can do experiments - like the ones you know - where the shape of something real follows exactly those abstract lines
whether or not that means that "field lines are real" is not a question physics asks or answers
it means field lines are a useful model to understand these experiments
 
cOnnectOrTR 12
cOnnectOrTR 12
@ACuriousMind which are arranged according to the existing field lines. If there were no lines then how would that arrangement take place
 
ACuriousMind
ACuriousMind
it does not mean field lines are a useful model to understand all experiments
@cOnnectOrTR12 the underlying physical equations (i.e. Maxwell's equations) in terms of the electric and magnetic fields predict this outcome just as well
the field lines are just a much easier way to see what will happen than explicitly solving these equations
but the field lines are just so-called "integral curves" of the vector fields, and the equations tell us e.g. that little dipoles like the iron filings will arrange themselves parallel to these integral curves
i.e. you can start from the fields and derive this without having to reify "field lines" as something real
 
cOnnectOrTR 12
cOnnectOrTR 12
@ACuriousMind reify ?
If you can’t enter it in formula does that make it unreal? We can see it’s pattern experimentally! Why ignore that?
@ACuriousMind how does that make it a concept and not something real.
Doesn’t that mean that we are not that advanced in our maths that we are unable to show field lines, some quantity telling us about field lines
 
ACuriousMind
ACuriousMind
12:02 PM
@cOnnectOrTR12 I'm not saying you should ignore it. I'm saying you shouldn't focus on it to the exclusion of the other ways to think of electrodynamics
 
cOnnectOrTR 12
cOnnectOrTR 12
@ACuriousMind :)
!I need to read more!
 
 
1 hour later…
Nihar Karve
Nihar Karve
1:26 PM
@bolbteppa While I'm not arguing against your answer, I want to point out that Lubos' criticism is justified (if he knows what he's talking about there). Non-separability might not be disagreeable on its own (particularly since the diff-invariant Hilbert space isn't even the physical Hilbert space; one still needs to impose the scalar constraint), but in the formulation of LQG that he refers to, it essentially implies a loss of background independence and incompleteness of the loop basis.
 
bolbteppa
bolbteppa
Yeah I'm just not sure about that, I have read him making general statements about it being a 'disaster' in general, but they also seem to have taken this seriously and it's probably more for the reasons you mentioned
8
A: Quantum Field Theory and Hilbert space dimensionality

Luboš MotlRigged Hilbert spaces have no special relationship to quantum field theory. We can talk about more general elements of "the" Hilbert space such as wave functions looking like distributions. They don't have a finite norm but they're still useful to talk about. Truly physical states that may be re...

Calling such states useful etc but implying they are non-physical and excluded in the 'ultimate' space, and using this same kind of thinking against lqg which I don't think is fair
 
ACuriousMind
ACuriousMind
I think you're confusing the notions of "non-normalizable state" and "separable Hilbert space" here
 
bolbteppa
bolbteppa
Well I'm just saying that the eigenfunctions of a free particle $e^{ikx}$ you get from solving $H \psi = E \psi$ will result in a non-separable basis of some space where we can expand a general wave function in, which the books in my answer do and also physically interpret, that's not technically even a Hilbert space is it
 
ACuriousMind
ACuriousMind
Whether or not we consider plane waves and such "physical states", they are not elements of a Hilbert space. It's not that there's a "separable" space L^2 and then there'd be a "non-separable" Hilbert space that would include also the plane waves. the space the plane waves live in is not a Hilbert space, it's the larger space in the triple of rigged spaces
so the question of whether the Hilbert space of the theory is separable is completely disjoint from the question whether or not the things in the larger space of that rigged triple are "physical" or not
The question "is the Hilbert space separable or not" is equivalent to asking whether or not you can form a countable basis of it, i.e. whether you can for any observable with discrete spectrum find a countable set of eigenvectors so that everything in the Hilbert space can be represented by a convergent series over linear combinations of this countable set
once your Hilbert space isn't separable anymore, many of the usual techniques go out the window, e.g. you can't write $\sum_i c_i v_i$ anymore for an arbitrary vector in the hilbert space and $v_i$ some basis because the basis isn't countable
 
bolbteppa
bolbteppa
1:41 PM
How do you say that for a continuous spectrum
 
ACuriousMind
ACuriousMind
well, the continuous spectra produce "eigenthings" that aren't part of the Hilbert space, so you don't get a basis where you can write a sum, you get something that physicists tend to brazenly write as an integral but which is rather difficult to make mathematically precise
(un?)fortunately, the mathematicians don't seem very interested in making this precise, see mathoverflow.net/q/383952/157071 discussing that this is still a niche subject
 
bolbteppa
bolbteppa
Right, so arguing about separability when talking about a continuous spectrum at the very least is just absolutely not obvious, and not something we can blithely use to write off an entire theory like lqg over without being super careful
 
ACuriousMind
ACuriousMind
well, if you're talking about a continuous spectrum, you're not talking about eigenvectors that lie in a Hilbert space, so the notion of separability doesn't even enter :P
"separability" only makes sense in the context of looking at a Hilbert space
it is a statement about that Hilbert space, not about anything to do with continuous or discrete operators
one problem with non-separable Hilbert spaces as quantum spaces of states is that you would need to go back to the beginning of all algebra you ever did with them and carefully examine each result whether or not it relies on the space being separable
 
bolbteppa
bolbteppa
Right, so if we follow what most of the books in my references do, and actually physically interpret those continuous spectrum eigenfunctions, it means if someone tried to use separability to write this approach off it's simply absurd
 
ACuriousMind
ACuriousMind
yes, because these eigenfunctions don't form a Hilbert space
no one claims they do, everyone is perfectly aware they're non-normalizable
 
bolbteppa
bolbteppa
1:54 PM
So, if we restrict ourselves only to normalizable solutions, the point seems fair, if we admit those non-normalizable wave functions as valid, it's an absurd argument
 
ACuriousMind
ACuriousMind
well, it would be an absurd argument to talk about non-separable Hilbert spaces just because someone does some computation with plane waves
but I'm not sure where you're seeing anyone making that argument or why we're even talking about it :P
the claim about LQG in the answers that apparently set all this off is that it produces a non-separable Hilbert space, not that it uses non-normalized solutions or plane waves or whatever
 
bolbteppa
bolbteppa
Well I basically called the space of eigenfunctions of a free particle a Hilbert space even though it's not really right. Maybe this is a 'separable Rigged Hilbert space' or some analog :p
 
ACuriousMind
ACuriousMind
It's some space associated to the separable Hilbert space $L^2(\mathbb{R}^n)$
e.g. you can take it as the dual of the Schwartz functions inside that, I think
 
bolbteppa
bolbteppa
@ACuriousMind That's a very fair distinction, but I don't think that this is distinction is being made, e.g. here the argument is about conceptual problems arising from treating $\delta$ functions as 'wave functions' and issues related to uncountability, so I think the argument is the same as with the free particle
"This is what you get if you follow the LQG rules directly."
> "You need the uncountable basis of $\sqrt{\delta(x-a)}$ for all values of $a$ to generate the space. The loop quantum gravity Hilbert space is analogous to this example, and it is non-separable"
 
ACuriousMind
ACuriousMind
@bolbteppa I don't even know wtf $\sqrt{\delta(x-x_0)}$ is supposed to be (you can't multiply distributions, so defining a square root for them is not obvious), so Lubos' argument there is pretty strange.
I mean, I kind of understand what he's going for, but I don't think it works
but since I don't know anything about LQG, I'd be careful to conclude anything from that either way :P
 
bolbteppa
bolbteppa
2:10 PM
Lets say he didn't use the square root, it looks like he's still just arguing that using 'generalized eigenfunctions' leads to 'non-separable Hilbert spaces' just by the last quote given above. Lets say we call it a rigged space instead, it looks like he'd be saying that ending up with a rigged space is "analogous" to the lqg case (regardless of what actually is going on, just looking at what he's saying)
 
ACuriousMind
ACuriousMind
I think you're interpreting the analogy much more strongly than intended (I think he's just trying to give an easy example for a non-separable Hilbert space but fails at that), but as I said I don't really know.
In spirit I think he was looking for the example of the Hilbert space of functions that are zero almost everywhere
 
bolbteppa
bolbteppa
 
ACuriousMind
ACuriousMind
the Ashtekar paper that Urs cites seems to agree that the naive fully space of states of LQG is some non-separable entity called the "polymer space"
so I don't think the objection you're trying to construct here is accurate - the LQG people themselves seem well aware that this is a potential problem, and agree this space is non-separable, and indeed construct this polymer space as an uncountable direct sum of separable infinite spaces, which is inseparable
 
bolbteppa
bolbteppa
The first page of that give the sense they are thinking of ending up with a continuous spectrum as being the reason for getting non-seperable Hilbert spaces
 
ACuriousMind
ACuriousMind
they likely disagree over how much of a problem this is, but that's another matter
@bolbteppa they don't talk about a spectrum, they talk about the states being labeled by a continuous parameter. If you have a set of states in your Hilbert space that's labeled by a continuous parameter and are all linearly independent, then you have an inseparable space.
 
Semiclassical
Semiclassical
2:20 PM
@glS ahah. and that $(DD')^2+1$ upper bound squares with what i was seeing in mathematica. (i needed 5 density matrices in the two-qubit case to have more variables than constraints.)
 
ACuriousMind
ACuriousMind
I'm pretty sure Rovelli and Ashtekar would have noticed if the way out here was just "oops, these things don't actually lie inside the Hilbert space so don't worry about the uncountability"
 
Semiclassical
Semiclassical
"maybe my intellectual opponents are careless" is not a reliable strategy
(i mean, it would be against me, but...)
 
ACuriousMind
ACuriousMind
(as an amusing side note: The naming of "continuous spectrum" is a misnomer in the case of non-separable Hilbert spaces: By definition, eigenvalues for which eigenvectors exist inside the Hilbert space belong to the discrete spectrum, no matter if this spectrum is discrete or continuous as a subset of $\mathbb{R}$)
 
Semiclassical
Semiclassical
@ACuriousMind yikes
 
ACuriousMind
ACuriousMind
the naming is of course due to the fact that this corresponds nicely in the case of separable Hilbert spaces and few are crazy enough to do spectral theory in non-separable spaces, but still :P
 
Semiclassical
Semiclassical
2:24 PM
are there any applications where having a non-separable hilbert space is uncontroversial?
 
ACuriousMind
ACuriousMind
I don't think so
 
Semiclassical
Semiclassical
mmkay
 
ACuriousMind
ACuriousMind
but I could be wrong, this is on the very edges of my functional analysis recollection :P
 
Semiclassical
Semiclassical
it's a different issue, but i'm reminded of how..."interesting", the spectral theory of non-self-adjoint operators can be
 
Nihar Karve
Nihar Karve
@Semiclassical yes
 
Slereah
Slereah
2:26 PM
Isn't the Hilbert space of infinitely many particles non-separable
 
Nihar Karve
Nihar Karve
quantum statistical systems in infinite volume
 
ACuriousMind
ACuriousMind
@Slereah well, no, at least not if you mean the space of states where arbitrarily but finitely many particles exist
 
bolbteppa
bolbteppa
> "Quantum space turns out to be described in terms of a basis of .. s-knot states, labeled by discrete quantum numbers. However, the picture is not truly entirely discrete. If the nodes of the s-knots have sufficiently high valence ... [they] are labelled also by certain continuous moduli parameters....
 
Semiclassical
Semiclassical
so something something thermodynamic limit?
 
bolbteppa
bolbteppa
> They spoil the discreetness of the picture and they change the structure of the space of the diffeomorphism-invariant states, $\mathcal{H}_{diff}$.., **making it nonseparable**.

> Nonseparability (absence of a countable basis) is generally regarded as pathological in quantum field theory... the nonseparability of $\mathcal{H}_{diff}$ is not necessarily unacceptable, ... But it is nevertheless disturbing"
 
Slereah
Slereah
2:27 PM
@ACuriousMind I mean infinitely many!
 
bolbteppa
bolbteppa
The assumptions seem to be: a) we expect a discrete spectrum; b) we get a continuous spectrum sometimes; c) this means our Hilbert space is non-separable (where they still call it a Hilbert space).
 
ACuriousMind
ACuriousMind
@bolbteppa no one there has mentioned a spectrum
 
Semiclassical
Semiclassical
as ACM alluded to, continuous vs discrete spectrum isn't a matter of whether the spectrum is a continuous or discrete subset
 
ACuriousMind
ACuriousMind
they're just talking about labels of the states, not about operators
 
bolbteppa
bolbteppa
The continuous moduli parameter point means they get a continuous spectrum and so a non-separable basis, that's the point they are making on the first page if I'm reading it right
 
ACuriousMind
ACuriousMind
2:30 PM
and, as I mentioned above, it would funnily enough be the case that if you took the operator that has each continuously labeled state as an eigenvector, all these labels would be in the discrete part of its spectrum by definition of what a "discrete spectrum" is
 
Semiclassical
Semiclassical
GOTO my last statement
 
ACuriousMind
ACuriousMind
@Slereah then I guess so, but where does that space occur?
is that what the "statistical systems in infinite volume" one is?
 
Slereah
Slereah
I guess!
Homogeneous universe of particles
it's not a wild idea
 
bolbteppa
bolbteppa
@Slereah the first page references the Streater PCT book for a discussion on apparent pathologies
 
Semiclassical
Semiclassical
well
infinite amount of matter is sorta wild
 
Slereah
Slereah
2:31 PM
I mean not that much
 
bolbteppa
bolbteppa
Some of this stuff is just wild to read, e.g. above (1) saying a free scalar field decomposes into an infinite number $i = 1,..,\infty$ of oscillators. It doesn't, it decomposes into a continuously infinite set of ocillators $\int d^3 \mathbf{p}$ oh man...
 
Nihar Karve
Nihar Karve
there's like one active user on the site who actually knows LQG lol
Rovelli himself used to answer here but not anymore
 
Slereah
Slereah
I tried reading Rovelli's book but I sort of start to discombobulate past chapter 6
 
Nihar Karve
Nihar Karve
which one
 
Slereah
Slereah
the Rovelli book
 
bolbteppa
bolbteppa
2:43 PM
which one
 
Semiclassical
Semiclassical
i'd have to be far more comfortable with garden variety QFT before i'd try to read that stuff. and for all my quantum expertise, it's few-body theory that i'm comfortable with
 
Slereah
Slereah
"Quantum gravity"
what other Rovelli book could I be talking about here
 
Nihar Karve
Nihar Karve
Helgoland
 
bolbteppa
bolbteppa
There's another likely way more readable one
 
Semiclassical
Semiclassical
" Covariant Loop Quantum Gravity: An Elementary Introduction to Quantum Gravity and Spinfoam Theory"
 
Slereah
Slereah
2:45 PM
is it worth purchasing?
 
Semiclassical
Semiclassical
that seems like another good candidate in this context :P
 
Slereah
Slereah
By which I mean adding to my giant list of books to purchase bc it is really not a priority
 
bolbteppa
bolbteppa
I'd say so, I'm going to try to read the first few chapters at some stage but it's more part of the general Dirac GR program
 
Slereah
Slereah
I should get DeWitt's book, someday
 
Nihar Karve
Nihar Karve
@Slereah The Global Approach to QFT?
 
bolbteppa
bolbteppa
2:47 PM
@Semiclassical you can set up second quantization as an extension of that, the jump to qft is then pretty direct
 
Semiclassical
Semiclassical
yeah, it's just never really clicked in my brain
i make no claims that this is not a personal failing :P
 
bolbteppa
bolbteppa
Yeah it's absolutely a nightmare to get basic things to click, the usual approach just ignores this and sets it up directly from the commutation relations
Am I wrong or does even the free EM field technically have a continuous spectrum... It looks like they were arguing that having got a discrete '$s$-knot state' spectrum for gravity was an amazing result, but that they also got a continuous spectrum which was a huge problem.
 
Semiclassical
Semiclassical
well. trying to get mathematica to find even one solution to this set of 17 lineaar equations in 21 unknowns exceeds my time on Wolfram Cloud
so...nuts
 
Slereah
Slereah
Doesn't the EM field basically have a discrete spectrum
and is only continuous if you use the generalized basis
 
bolbteppa
bolbteppa
Usually you put it in a finite volume to set it up, but that's just a trick really
 
02:00 - 15:0015:00 - 00:00

« first day (3998 days earlier)      last day (924 days later) »