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12:43 AM
Why is the wave function times its complex conjugate also written as \$|\Psi$|^2 ? I am confused about the purpose of including the absolute value bars.
Is this just by the definition of the inner product in the hilbert space we're concerned with?
 
12:55 AM
nvm got it but i have another question
what makes electrons emit a photon? i have seen this in two cases: 1) in griffiths qm, it says "the result is that a container of hydrogen gives off light (photons)", and when i was reading about Feynman diagrams, i read that electrons approach each other, exchange a photon, and this change in momentum propels the electrons in opposite directions, which can be represented by a Feynman diagram. however, in both cases, it's not clear to me why the electro would emit a photon
 
 
2 hours later…
3:17 AM
How does this guy have 850+ points? physics.stackexchange.com/q/722428/123208 He hasn't posted many questions or answers, and I can't see a bounty on any of his answers. I find it a bit disturbing that someone with a physics PhD is working with a known anti-gravity / free energy crackpot.
 
what does sub quantum mean?
 
3:35 AM
@antimony It's not really a standard term. But it means a level of physics more fundamental than quantum physics.
 
ohh
 
The idea is that there's a simple subquantum layer that explains what we see at the quantum level. Like how the physics of protons, neutrons, electrons and photons explains the chemistry of atoms.
 
but it doesn't necessarily mean quarks and things?
or beyond quarks?
or that is what it means? deeper than quarks?
 
Yes, deeper than quarks. At the quantum level, we have quarks, which appear to be fundamental. But at the subquantum level, the quark is a complex entity. Think of the quark as a picture composed of subquantum pixels.
 
wow
thats a pretty cool idea tbh
but of course, being cool and being correct are not always the same thing haha
are there any experiments which prove these quark pixels?
 
3:46 AM
There's nothing wrong with looking for deeper levels of physics, in principle.
 
right
 
@antimony Right. Physics isn't just about pretty ideas. You need to be able to test those ideas, and see how well they explain what we can observe in the physical world.
Now it can be very difficulf to do effective experiments that explore the world at the quantum level. Generally, the smaller you go, the more energy you need. Which is why we build huge accelerators like the LHC.
 
oh good point
so noone has an accelerator powerful enough to get inside the quark?
yet
 
But we can also do simulated experiments: we create mathematical models, feed them into a computer, crunch the numbers and see if the model corresponds to what our empirical measurements say.
 
oh thats true
i wonder if you could do some of those experiments in space
 
3:57 AM
We've had a lot of success modelling the extreme particle physics that happens deep inside stars. That gives us confidence that our particle physics theories and our astrophysics theories are on track.
 
where there might be some high energy events from natural sources?
ahh yes, right
 
Indeed. Long before we had big accelerators we were trying to figure out the nuclear processes in stars. Obviously, we can't see deep inside stars, but we can get a good idea of what elements they contain from their spectra.
 
oh yes, good point
 
@antimony As far as we know, there isn't anything deeper than quarks. And if there is, it's impossible to get enough energy to see it. Due to color confinement we can't ever isolate a quark. If you try to separate the quarks in a proton it takes so much energy that you end up up producing more quarks & gluons, all bound to each other. We can't see the individual quarks or gluons, but we can see the patterns they make.
 
ohh wow
huh
thats very cool
 
4:11 AM
A jet is a narrow cone of hadrons and other particles produced by the hadronization of a quark or gluon in a particle physics or heavy ion experiment. Particles carrying a color charge, such as quarks, cannot exist in free form because of quantum chromodynamics (QCD) confinement which only allows for colorless states. When an object containing color charge fragments, each fragment carries away some of the color charge. In order to obey confinement, these fragments create other colored objects around them to form colorless objects. The ensemble of these objects is called a jet, since the fragments...
 
incredible
is this line of thinking at all reasonable: if we know approximately the energy which would be required, can we find the equivalent frequency associated with that energy and consider some temporal properties which might occur at that scale?
could it give some hints of the timescales these sub-quark properties might operate at?
 
Yes, that's reasonable. And that gives a timescale on the order of the Planck time. And energy densities similar to what the universe was like within a Planck time of the start of the Big Bang.
Even if you build a galaxy-sized particle accelerator, you probably won't have enough energy to explore the subquantum world... if it exists.
 
4:34 AM
We could do computer simulations of sub-quark stuff, but there are a few problems. It's hard enough to do that at the quark level: it takes huge amounts of computing power, and we have to make various approximations. Those calculations have improved in the last decade or so, due to clever mathematical tricks, but those tricks involve simplifications.
Subquantum theories claim that the subquantum level is simpler, so the calculations should be easier. OTOH, you probably need to perform a whole lot more calculations.
And that can lead to a combinatorial explosion: en.wikipedia.org/wiki/Combinatorial_explosion
 
sorry phonecall bbs
 
No worries.
A similar thing affects string theories. In those theories, a quark isn't a point, it's a tiny string. We can build mathematical models of those strings, and crunch them in a computer, but there are too many possibilities for the string parameters.
So even if some set of parameters lets us perfectly simulate all the known fundamental particles there are zillions of other combinations of parameters that will give the same result, so there's no way to decide which set of parameters is the correct one.
 
 
1 hour later…
5:56 AM
@SillyGoose Do you mean an electron in hydrogen atom? or something like Compton scattering? Note that Feynman diagrams are a pictorial representation of an integral and do not show an actual physical process. The two interaction electrons do not emit photons towards each other.
 
6:38 AM
Deleted
 
7:12 AM
Chad Afternoon
Is it possible to know the Jounal/Conference at which an ArXiv Paper was published?
I just wanted to know if there are any Journal where an ArXiv paper has been accepted or not, otherwise I would have to waste my time reconstructing my whole time that paper and comparing the results (which I think is highly misleading)
 
7:39 AM
@PM2Ring Remember that suggested edits yield +2 rep per edit.
@SillyGoose I rant about the lack of a "why" e.g. here and here
 
When I get to hell I will ask the Devil why he invented the ontology of the quantum field. Weren't the Navier-Stokes equations enough already?
 
8:00 AM
@JohnRennie Why not try to go heaven and ask God ?
 
@Abbas God won't know, it wasn't their invention :-)
 
Hahaha... True! :)
 
@JohnRennie What you mean by invention? What we Humans do is just create our own techniques (like mathematics etc) to understand it. We aren't even sure that what we are using to describe our nature is perfect .
 
I would suggest the possibility that John was, uh, joking? :P
 
@JohnRennie By the way , please answer my question of how to measure closest star .
 
8:04 AM
@Abbas Either you are taking this more seriously than I intended, or you're actually Old Nick in disguise and you're just continuing your master plan to drive all physicists insane.
 
@PM2Ring No, that's not how string theory works - particles are not strings (and, indeed, in some ontology there isn't even really a "string" (cf. here), they are at best "excitations" of strings. I'm not sure what you mean by "too many possibilities" for the string parameters, either.
 
@JohnRennie Sorry , I didn't meant to hurt you :(. You are my hero on Physics SE
 
@Abbas Which question?
Ah, this one?
Aug 5 at 14:40, by Abbas
How do physicists and astronomers find out that alpha centauri is closest to our sun? I mean there are so much stars in the sky so did they calculated distances of each star from sun and after that calculation said that alpha centauri is closest?
Yes, for close stars we can measure their distance using parallax so we know their distances quite accurately.
Measuring parallax is routine these days. It's very unlikely we've missed a star that is closer.
This list covers all known stars, brown dwarfs, and sub-brown dwarfs within 20 light-years (6.13 parsecs) of the Sun. So far, 131 such objects have been found, of which only 22 are bright enough to be visible without a telescope. The visible light needs to reach or exceed the dimmest brightness to be visible to the naked eye from Earth, 6.5 apparent magnitude.Our Solar System, and the other stars/dwarfs listed here, are currently moving within (or near) the Local Interstellar Cloud, roughly 30 light-years (9.2 pc) across. The Local Interstellar Cloud is, in turn, contained inside the Local Bubble...
Wikipedia has a list of all the stars within 20 light years of Earth.
There might be very dim objects that are closer that Alpha Centauri, but would you count them as "stars"?
 
@JohnRennie Yes I am only curious about this . I mean how is it practically possible to measure distances from so many stars and then compare . Wouldn't that take too much time as the sky looks just full of numerous stars
@PM2Ring @JohnRenn See my question is written above this post .
 
It's far easier than you think.
 
8:12 AM
How ?
 
You take a picture of the sky then take another picture 6 months later when the earth has moved by the diameter of its orbit. Then you pass the pictures to a computer and let it compare them.
Stars that are far away will be in the same position in both pictures, while stars that are close to the Earth will have moved due to parallax. So you just look for differences between the two pictures.
It's a really easy measurement. The only problem is if the object is so dim it's hard to see in the pictures.
 
So computers do them .
 
Yes, that's the way it's done these days. In the past it was done by a human looking at the two pictures to try and find differences.
 
Ok, I got it . Thanks for answer
Can you please answer one more
 
you don't really need a computer, you can e.g. just print the two images on transparent background and lay them on top of each other for a simple manual version of what the computer does
 
8:16 AM
Yes, go ahead :-)
 
@ACuriousMind Ah, of course.
 
@ACuriousMind from the text it looks like it's simple to understand but I am so dumb that I can't understand it :(. Please explain what this means
 
parallax is about how much the stars move when the camera moves (in the case of a 6 month distance from one end of Earth's orbit to the other)
 
@ACuriousMind Fair enough. I was over-simplifying. I should've said particles are excitations of strings. I didn't mean to imply that quarks are merely string-shaped pieces of "quark stuff".
 
@Abbas You take the picture and print it on photographic film. Traditionally you use a negative so the dark sky is clear on the film and the stars appear as black dots.
 
8:20 AM
The "too many possibilities" refers to en.wikipedia.org/wiki/String_theory_landscape
 
so when you have the two pictures 6 months apart, you want to determine how much the individual stars moved - so you print the pictures on transparent background and lay them on top of each other and then you can see how much the stars position in one picture is different from the other
@PM2Ring that's a common misunderstanding (that some proponents of ST are guilty of propagating :P): When you compare this landscape to e.g. the landscape of all conceivable QFTs, it is actually smaller, so there isn't really a new problem in ST related to free parameters
the point is that in QFT we don't need to search the landscape - we already have the Standard Model, and it describes our world well
 
I am sorry but I think I am still very dumb to understand that . However I'll refer these chats again in future when I'll be a better learner . Thanks
@JohnRennie please wait . I am just searching the question
 
@Abbas I'll have a look and see if I can find a YouTube video or something similar showing how it's done.
 
@ACuriousMind Right. But in string theory, we don't have an effective way of empirically exploring the landscape, even if we could brute-force enumerate the vast space of possibilities.
 
the free parameters in string theory are much like if I said "oh god, QFT has so many free parameters, how are we ever going to find the standard model" - this is only a problem if you think that we have to arrive at our physics by some sort of random walk through the landscape of all possible theories
 
@ACuriousMind Sure. We can do experiments that tell what the masses of particles are, etc.
 
@PM2Ring what do you mean by "empirically explore"? Each string theory determines a low-energy effective QFT, whose particle content we can of course compare to experiment/the Standard Model
 
@JohnRennie Please see this .physics.stackexchange.com/q/699728/326901. I know it's not a good or conceptual problem but I encountered same problem when I tried this .
 
(almost all ST compactifications yield something very unlike the Standard Model)
 
@JohnRennie Thanks very much,will definitely try
 
8:28 AM
the problem with the size of the string landscape is really only a problem if you think the fine-tuning problem is a problem, and if you think that's a problem, it's exactly as much (if not more) a problem in QFT
there is another problem - that we haven't really found a string theory vacuum whose phenomenology matches exactly what we'd want from the UV completion of the Standard Model, but that's less related to the size of the landscape and more related to the fact that the SM has several features that are hard to generate from most ST vacua
 
@ACuriousMind I thought that multiple compactifications can determine virtually the same effective QFT, which makes it hard to decide between them. Or is that wrong?
 
yes - that's mirror models
 
@JohnRennie This question : physics.stackexchange.com/q/699728/326901
 
but that's just a duality - two a priori different physical theories describing the same actual physics in different terms
there's also dualities among QFTs (e.g. bosonizations, Seiberg duality) and no one worries about these a lot as somehow making it hard to figure out that we should use the Standard Model
 
@Abbas I don't have time to go through right now. Sorry :-(
 
8:39 AM
@JohnRennie It's okay :)
 
 
2 hours later…
fqq
10:56 AM
@ACuriousMind the problem is that many string theorists pushed the idea that string theory solved the fine-tuning problem
 
Terminology question. Consider a satellite on some orbit around Earth. It traces a (great) circle on the surface consisting of points it passes over.

Does this have a name or is it just the great circle traced by the satellite?
 
11:30 AM
@fqq Absolutely, that's why I said "that's a common misunderstanding (that some proponents of ST are guilty of propagating :P)" above ;)
 
 
1 hour later…
12:46 PM
How does one know if a moderator has responded to your flag or decided to ignore it?
 
@MoreAnonymous If you click on the number of helpful flags in your profile you can see an overview over all your flags and whether they were marked as helpful, declined or are still pending.
 
@ACuriousMind where in profile?
Ah found itr
*it
 
 
1 hour later…
2:03 PM
So here's something that's been bothering me
Do the laws of thermodynamics "predict" life cannot be a plague in the universe? I mean if it was then a probable state would be living things ... which is not what we witness when we see what are the probable distributions
 
2:24 PM
@MoreAnonymous most of the universe is not exactly in thermodynamic equilibrium, so how would the "laws of thermodynamics" apply?
 
@ACuriousMind Yea this is another thing that confuses me. How did the universe get itself out of equilibrium?
 
it never started in equilbrium!
 
Doesn't the CMB etc assume equilbrium?
*thermal
 
the point at which it will finally reach it is usually circumscribed as heat death :P
 
@ACuriousMind No wait to get the CMB one assumes a thermal distribution
 
2:27 PM
@MoreAnonymous okay, so: the early, small universe might've been in some sort of equilibrium, it was a hot mess
 
Is that some sort of approximation?
@ACuriousMind How did it get out of equillibirum?
 
then, as it expanded, it cooled
 
I thought that was the most probable state?
But still followed the most probable distribution right?
Temperature went down
 
below a certain temperature, stuff can happen like the formation of atoms, and indeed that happened - then gravity takes over, etc, and you get clumps of matter that aren't in interaction with each other anymore and hence not in meaningful equilibrum
this isn't about thermodynamic laws or probability, it's just that when you cool down a soup of quarks and leptons at some point you'll end up with hydrogen gas
 
@ACuriousMind I think Penrose uses this to argue that the entropy of gravity was not very high
@ACuriousMind But even after phase transition you have a distribution
 
2:31 PM
I'm not sure what you mean
at some point the universe was cold enough that thermal motion wasn't enough anymore to work against gravity
that's the latest point where it stopped being reasonable to apply thermodynamics to the universe as a whole
 
@ACuriousMind what time was this in terms of redshift?
I guess the formation of galaxies time?
 
no idea, I don't have the timeline memorized - whenever macroscopic matter started to form
 
@ACuriousMind So when hydrogen formed?
I would think the hydrogen molecules still would followed the Boltzmann distribution
 
well, just a hot hydrogen gas isn't where thermodynamics breaks down
you need it to be cool enough so that clumps form
that formation of clumps is where the thermodynamic description becomes inappropriate
 
@ACuriousMind I find it strange you can have a system in thermal equilibrium (maximum entropy) and it can take itself out of equilibrium
thats a violation of the 2nd law no?
 
2:35 PM
well, the 2nd law only holds for thermodynamic systems :P
 
@ACuriousMind And isolated systems like our universe?
@ACuriousMind This seems like a cheat btw :P
 
when the universe becomes a bunch of clumps in vacuum, it's not a single thermodynamic system anymore because these clumps don't interact
 
@ACuriousMind I get this heuristically but I do you have any references that back this kind of thinking?
 
I mean, is it not obvious that if I have a box of gas over here and a box of gas over there and there's no connection between them, then they're not the same thermodynamic system?
 
@ACuriousMind They originated from the same box?
I mean I would argue this kind of thinking should apply to the early universe
Where the universe was expanding really quickly
But we know it was in thermal equilibrium then
 
2:39 PM
the point where equilibrium stops being a good assumption is called "freeze out" in cosmology, see e.g. physics.stackexchange.com/q/262592/50583
it happens for different particle species at different points in history
 
@ACuriousMind Yes but even there (I think) that each species had its own thermal distribution
just different temperatures
 
@MoreAnonymous well, sure, for particles like neutrinos
stuff that's subject to gravity clumps together, at least in some places
 
hmmm ... k
@ACuriousMind do you have any thoughts about the noise in LIGO? I mean do you think that's some quantum gravity signature?
 
I have no idea what you're talking about, so no :P
 
I've read people are concerned that its not gaussian
This
I don't get why they think if its not a gaussian its a correlation?
Maybe I'll ask this on stackexchange ...
 
2:47 PM
you expect noise to be Gaußian because of the central limit theorem
something that's not a normal distribution indicates an effect that's not the product of lots of small random error sources, i.e. something that's not noise
 
@ACuriousMind I mean I kinda get that. But is a gaussian invariant?
 
invariant under what?
 
@ACuriousMind frame transfomations
*transformations
 
why would that matter?
the central limit theorem applies in every frame :P
 
@ACuriousMind because there is no privileged frame?
@ACuriousMind Yes but the noise one person measures
should be the same in another frame
 
2:49 PM
independent random variables don't become magically dependent by a frame transformation
 
By that logic the Juttner distribution should be a gaussian too
 
so the central limit theorem predicts that independent random sources of noise will add up to a normal distribution in every frame
it might not the the same gaußian, but noise should still be Gaußian
 
@ACuriousMind The Stosszahlansatz holds in the relativistic case as well and one does not get a gaussian
 
@MoreAnonymous the speeds of particles in a gas of fixed temperature are not independent random variables
 
@ACuriousMind which is why I'm referring to the Stosszahlansatz
 
2:52 PM
okay, no wait: they might be
but the central limit theorem is about the sum of independent random variables
 
@ACuriousMind Wait you concede I'm correct?
I must be dreaming :P
 
you're correct that velocities are independent random variables, yes
you're wrong that this is a contradiction to my CLT argument for noise :P
 
@ACuriousMind So then why isn't the relativisitic gas distribution a gaussian?
 
why should it be? the velocity distribution is not the sum of the individual velocities!
you need to read the CLT properly to apply it - it says that when you sum/superpose a lot of random variables, you get a normal distribution
 
@ACuriousMind But in the galilean case it is?
 
2:55 PM
that's what noise is - there's a lot of uncorrelated sources of small errors, and these all add up to the Gaußian we expect to see as background noise
@MoreAnonymous What Galilean case?
 
@ACuriousMind The maxwell boltzmann distribution
 
MB isn't a normal distribution, either!
 
@ACuriousMind Wait that is a gaussian??
The velocities are normal dsitributed
 
wait, are you talking about the speed or the velocity distribution?
ah, okay
the speed isn't normally distribution because it's cut-off at 0
 
So why doesn't this argument hold in the relativistic case (the Juttner distribution)
?
@ACuriousMind Wow twice a day ... :P
 
3:00 PM
@MoreAnonymous I'm not arguing that MB is due to the central limit theorem!
 
@ACuriousMind But your argument would suggest
it should be?
Atleast as far as I understand
?
 
Why would it? In what way is the MB distribution a sum of random variables?
 
I mean I've heard people argue it
see here for example:
6
Q: Why is velocity normally distributed in a gas, but not energy?

yippy_yayIf one looks at a cubic box of gaseous atoms all initially flying in the same direction at the same speed (but flying at an angle to the walls, so as not to reflect up-and-down against the box walls forever), they will collide with the walls and each other, their previously uniform velocities bec...

 
and you see that Lubos' caveat is explicitly that this doesn't hold in relativistic physics
relativistically, momentum doesn't have a linear relationship with velocity, so you can't argue about collisions like that
that doesn't mean that the central limit theorem doesn't apply to relativity, it just means that relativistically your average momentum/velocity can no longer be viewed as the average of the individual velocities
because the sources of noise add up at a detector, that's just how detectors work :P
 
@ACuriousMind So how does one apply central limit theorem in that case?
@ACuriousMind what is the detector measuring?
I think it depends on that no?
Also in the relativistic gas .. according to you CLT one should get the same distribution independent of the frame. Note: One only gets the Juttner in the rest frame
 
3:17 PM
@MoreAnonymous No, I'm saying you can't apply CLT to the case of relativistic velocities because you can't argue that the momentum after many collision is an average for any type of collision that is not non-relativistic elastic
 
@ACuriousMind wait what?
The Juttner distribution assumes elasticity
 
yes, but not a non-relativistic collision!
the problem is that the velocity/momentum relationship is not linear in rel.
 
@ACuriousMind Isn't $p^\mu = m u^\mu$
?
 
as for the LIGO case: The approximation in which we speak about gravitational waves is one in which these really are waves
i.e. they obey a superposition principle
you could argue that the noise comes from stuff where that approximation isn't valid and so you see non-linear effects of gravity
@MoreAnonymous yes, but the $u$ has a $\gamma$ factor compared to normal velocity in it!
four-velocity isn't velocity
 
@ACuriousMind so are u saying I can use CLT on 4 velocities?
 
3:22 PM
no
 
@ACuriousMind why not?
It's linear now?
 
I'm refusing to say anything about it because I'd have to think much harder about how the derivation of the distributions actually works
 
@ACuriousMind :/
I find it curious you're willing to comment with the same logic is supposed then somehow for LIGO
 
it's not the same logic!
my superposition argument above doesn't have anything to do with how velocities in a gas work
 
@ACuriousMind Okay lemme try to understand
 
3:25 PM
the crux to apply CLT is that you have to have linear superposition of the individual sources of randomness
 
A gravitational wave can be thought as a disturbance in the metric
?
 
I'm saying the usual approximation of far-field gravitational waves has such linear superposition, so the random noise from many sources of such waves will result in a Gaußian noise
the part of this argument you can doubt is that maybe the linear approximation isn't good for whatever's causing the noise at LIGO
 
Okay I get your argument but I'm not convinced ...
 
could be; I don't know anything about gravitational waves specifically, but I'm fairly sure my reasoning here is why people would initially expect noise to be Gaußian
 
I'm gonna pen down my argument on why I think it shouldnt be gaussian
 
3:29 PM
I mean, if you want a qualified answer you should probably just ask on the main site
we have enough GR experts around
 
3:48 PM
@ACuriousMind done
Sorry if it got heated btw
@ACuriousMind Its a bit difficult to tell on the internet the emphasis of "!"
 
no worries, I didn't mean to sound angry or anything
 
@ACuriousMind Coolz
 
4:10 PM
@MoreAnonymous I think the absence of an exclamation point is more likely to sound passive-aggressive in some situations
 
@Feynman_00 really?
 
I guess this is up to personal interpretation of messages
Of course this is not the case in a book, but think of it: would you bother to write an exclamation mark if you were really annoyed?
On the other hand, a full stop is no good :(
 
@Feynman_00 I would!
:P
 
I tend to read messages with exclamation mark as cheery
 
5:02 PM
Hello chad homies
Is it okay to first get the paper at ArXiv, before submitting them for journals?
 
 
2 hours later…
fqq
6:44 PM
It's quite common to submit to the arxiv first and then to a journal if that's what you mean. Not all journals allow it but it's mostly ok in physics
 
I am reading through some sections of Griffiths's Quantum. I was wondering if it is implied that we work with subspaces of Hilbert space when we say a set of eigenfunctions is complete; else, I am left to interpret that no matter what the situation is, your particle can take on any wave function
in other words, a set of eigenfunctions is complete w.r.t. a subspace of Hilbert space, not the whole thing. similar to how (1,0,0) (0,1,0) is a basis which spans a subspace of R3.
 
@SillyGoose I'm not sure what you mean; the statement that eigenfunctions are complete in the context of the spectral theorem is really supposed to mean that: The eigenfunctions of a bounded self-adjoint operator form a basis for the full Hilbert space.
 
For more context, I'm reading the section on definite states of hermitian operators. One of the important mathematical properties of the definite states, i.e., eigenfunctions is that they are "complete" and that "any square-integrable function can be written as a linear combination of them".
oh I see, I guess I do not see why that should be true, but I should read more about it
 
a lot of QM is really much easier if you know a) classical Hamiltonian mechanics and b) finite-dimensional linear algebra
because in the finite-dimensional setting this isn't very hard to prove and there are no caveats
 
In subzero temperatures, would getting wet make you warmer?
 
7:00 PM
Can someone recommend me a GR textbook which focuses on visualizing GR? I'm horrible at spacetime diagrams :/
 
E.g., if you had a glass of super-cooled water, the same temperature as the air, and you poured it all over yourself, would it freezing warm you up?
 
@NoHaxJustRadvylf mostly not; the higher heat conductivity of the water is much worse in most cases than the small benefit of the water staying at 0°C until its frozen
 
So is the dimension of the hermitian operator (represented as a linear transformation) the same dimension as Hilbert space?
 
@SillyGoose what is "the dimension of the hermitian operator"
operators don't have dimensions
 
if an operator is written as an nxn matrix, then i think i mean n as the dimension, which would also be the number of eigenfunctions. if there are the same number of eigenfunctions as the dimension of Hilbert space, i can make better sense of why eigenfunctions form a complete set of functions in Hilbert space (from finite-dimensional linear algebra thinking).
 
7:05 PM
that's not really how it works
if you have a vector space of dimension $n$, then all linear operators on it can be written as nxn matrices, no matter how many eigenvectors they have
 
are you saying then an nxn matrix can have less than n eigenfvectors?
 
yes
it might even have none!
wait, wrong examples
the right example is a 2-by-2 matrix $\begin{pmatrix}2 & 1 \\ 0 & 2 \end{pmatrix}$
this has exactly one eigenvalue, $\lambda = 2$, and only a single eigenvector: $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$
 
so is it then that hermitian matrices have a particular property which guarantees that there are as many eigenvectors as columns (or rows since it is square)?
 
@SillyGoose yes, that's the claim of the spectral theorem: Such operators have "enough eigenvectors" to form a basis of the space
 
@SillyGoose Yes, spectral theorem does
In fact they belong to a special class of operators called "normal" operators
 
7:20 PM
in more "practical" terms, this is equivalent to claiming that there is a basis in which they are diagonal
 
Actually, the eigenvectors of normal operators are orthogonal and hermitian operators have the additional property of having real eigenvalues
 
if you develop the theory of the Jordan normal form, the operators whose eigenvectors form a basis are exactly those whose Jordan normal form is a diagonal matrix
 
I thought you can always form a diagonal matrix from a linear transform (I didn't understand this part of my linear course well). We went over rational canonical form and jordan canonical form
 
the point of the Jordan form is precisely that you can't do that :P
the Jordan form is the closest you can get to diagonalization for arbitary operators
 
oh xD well good im reading through linear again.
 
7:23 PM
My linear algebra course didn't cover Jordan Normal form, dang
 
and the thing that obstructs diagonalization is exactly that there are some eigenvalues that "don't have enough eigenvectors"
 
Is Jordan normal form something I should go back and read?
 
eh
it's kinda neat, but I haven't really needed it anywhere in physics
 
I tend to study on my own things courses don't cover but the book I chose to use that year didn't cover it either
I'll take a look maybe, but now I must focus on differential geometry
By the way @ACuriousMind as much as I agree with you about Hamiltonian mechanics before QM, I think that knowing functional analysis to understand infinite dimensional vector spaces would be even more necessary
 
that's something I'm split on
you could do a lot of QM in finite-dimensional spaces if we started with Stern-Gerlach as the fundamental experiment instead of the double slit
and then, as long as we're explicit at the point where the finite-dimensional algebra breaks down, I think it's necessary to teach people all the physics methods involving $\lvert x\rangle$ and whatnot that don't care about rigor
 
7:44 PM
Hi everyone :) Hope you all are doing alright
 
Hi, :) how are you?
 
I'm much better now! I was somewhat depressed the last 2 weeks :/
 
@ACuriousMind I think that would be ok as long as your aim is to show what is different in the quantum realm but courses also introduce infinite dimensional systems like the harmonic oscillator from the beginning
There is this "scary" feature of QM for which you keep finding out there is nothing special about it
For example, the relation between quantization and the existence of solutions of PDEs
 
8:11 PM
Here is a question: if the components of the metric tensor are constants, then the Christoffel symbols vanish and so does the Riemann tensor. But the commutator of the covariant derivatives still doesnt vanish unless the connection is torsion free. It then seems to me that the Riemann tensor is not a full description of the curvature, since even when it vanishes, there could still be curvature (non-commutation of covariant derivatives). Is my thought correct or am I missing something?
 
8:24 PM
@ShikiRyougi well...it depends on what you mean by "curvature" :P
you've correctly pointed out that the missing piece is torsion
but the Levi-Civita connection, is, by definition, torsion-free, so the Riemann tensor is a complete description of the effects of the LC connection
we call the Riemann tensor curvature because it is directly related to the intuitive notion of a curved surface in the case of manifolds embedded into $\mathbb{R}^n$ and everything induced by the normal LC connection on $\mathbb{R}^n$
there's no torsion by definition in that case, and so the word "curvature" comes from a world where torsion simply doesn't exist
saying that curvature is about covariant derivatives not commuting is not how the word came about/the word comes from the context of the LC connection, which has no torsion by definition
 
8:47 PM
Ahh I see
Yes I'm aware that the LC connection is torsion-free, I was just thinking about it more generally. Suppose we are not working with the LC, how would we define curvature? what would curvature be?
 
@ShikiRyougi I'm not sure what that question means
the tensor we call the curvature tensor still exists in the case of torsion, it's just that there is also torsion
 
I was thinking that our definition curvature should be dependent upon the curvature tensor and torsion; at least, this is how I first understood it. I somewhat identified curvature with the commutator of covariant derivatives. But now I see your point, curvature is described by just the curvature tensor.
 
9:03 PM
@ShikiRyougi maybe this view helps: Curvature is about the shape of manifold when embedded as a submanifold of $\mathbb{R}^n$ and the connection induced as the restriction of the ambient LC connection, torsion is a different property related to choosing a connection that's not the LC connection
 
@ACuriousMind what does ambient mean here?
And thank you, that helped!
 
@ShikiRyougi when you embed $X$ into $Y$, then $Y$ is the "ambient space"
 
Ohh alright
Cool sounding terminology, I love it
 
Ye ye, I just love creative names for stuff like that
 
9:17 PM
There is a search function in this chat, ACM's memory was a big lie :(
 
ambient > 1590s, "surrounding, encircling," from Latin ambientem (nominative ambiens) "a going around," present participle of ambire "to go around, go about," from amb- "around" (from PIE root *ambhi- "around") + ire "go" (from PIE root *ei- "to go"). The notion of "going all around" led to the sense of "encircling, lying all around."
 
@ShikiRyougi Talking about funny terminology, I found out germs are a thing in math
 
@Feynman_00 I never tried to hide that:
Dec 7, 2020 at 18:11, by ACuriousMind
to be fair, one has to remember a specific phrase from the message, but yeah, I'm not remembering the dates of these messages or anything
@Feynman_00 that's the part of math that was invented by farmers
there's stalks growing out of the germs and they form sheaves
 
OMG, those things really exist
 
what part of math invented "viruses" :P
 
9:26 PM
And I thought particle physicists were really creative with "strangeness" or "charm" lol
 
nvm, it probably was cs
viruses was the part of cs invented by biochemists
 
what are you talking about
 
analogy
 
9:42 PM
To the kind souls out there can someone help me with section 1 of this paper?
 
@Feynman_00 XD
@ACuriousMind :o
 
10:51 PM
if i have a physics related question and i have thought it through myself, but I have not tried to look/find the answer or research the question, is it be okay to ask it here?
 
Don't ask about asking, just ask.
I'm actually quoting the description of this room :P
 
11:15 PM
@SillyGoose Seems like a moral dilemma. Usually, I first search for the answer and if I still don't understand I will ask - but it really depends on the question. You can sometimes tell that a question probably hasn't an answer somewhere easily accessible.
 

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