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North Læraðr
North Læraðr
12:07 AM
@JohnRennie Uh, that is NOT what I got
 
 
2 hours later…
More Anonymous
More Anonymous
2:02 AM
@ACuriousMind the idea that entropy is observer dependent seems spooky
To me
 
Terran Swett
Terran Swett
2:57 AM
I'm reading about quantum mechanics again. I hope nobody minds if I give a bit of a monologue punctuated with the occasional vague question :D
I looked at the Schrödinger equation. It looks like that's the core of it (or, at least, the core of one formulation of it). That's the equation that specifies how the state of the system changes over time. Right?
The simplest case, perhaps, is a single non-relativistic particle in 1-dinensional space with constant potential.
My understanding is that in that case, the Schrödinger equation comes down to the following: the time derivative of the wavefunction is i times a real constant times the second derivative of the wavefunction. Right?
That's a wave equation, but it's not the wave equation I'm used to.
I meant to write "second spatial derivative" for "second derivative" above.
Anyway, wave equations. The "standard" wave equation states that the second time derivative is a positive real constant times the second spatial derivative. The result is purely real waves with constant velocity—no dispersion.
The Schrödinger equation, on the other hand, produces waves with both a real and an imaginary component, 90 degrees out of phase, and where the phase velocity of the wave is inversely proportional to the wavelength, which in turn means that the frequency is proportional to the square of the wavelength.
Right?
 
John Rennie
John Rennie
3:52 AM
@NorthLæraðr what did you get?
 
John Rennie
John Rennie
4:08 AM
@TerranSwett we have a tendency to write the solution for a free particle as a complex exponential because it's mathematically convenient, but of course we can do exactly the same for the "standard" wave equation.
And yes the waves we get for a free particle have different phase and group velocities.
 
 
1 hour later…
Terran Swett
Terran Swett
5:16 AM
@JohnRennie Yeah, looks like the phase velocity is half the group velocity, right?
 
John Rennie
John Rennie
You should get the product of the group and phase velocities equals c²
 
Terran Swett
Terran Swett
For the non-relativistic equation?
That equation doesn't even have c in it.
 
John Rennie
John Rennie
Good question. I'd have to think about it and it's 6 a.m. here so I'm not fully awake yet.
 
Terran Swett
Terran Swett
By the way, how can the standard wave equation be rewritten so that waves look like complex exponentials instead of real sine waves?
My first thought was du/dt = i du/dx, but that won't work.
 
Slereah
Slereah
the velocity of quantum waves is determined by their momentum
 
Terran Swett
Terran Swett
5:25 AM
Wikipedia seems to say that for a particle in 1-d space, if it's reasonably well-localized in both position and momentum, the particle will travel at twice its own phase velocity. en.wikipedia.org/wiki/…
It will, of course, spread out as it does so.
 
John Rennie
John Rennie
@TerranSwett $y = e^{i(\omega t + kx)}$ is a solution of the standard wave equation. Yes?
 
Terran Swett
Terran Swett
Oh right, I think it is.
But something like $y = (\sin t) (\sin x)$ is also a solution of the standard wave equation, but not the Schrödinger equation.
But, of course, that sinusoid is just the sum of two complex exponentials with opposite frequencies. In the standard wave equation, the two complex exponentials move in opposite directions... I think... do they?
No, it's the sum of four complex exponential waves: two with positive frequency, two with negative frequency, two traveling forwards and two traveling backwards.
That solution is impossible in the Schrödinger equation because the sign of the frequency determines which direction the wave travels.
 
 
4 hours later…
Hammad
Hammad
9:51 AM
Someone could shed a light on physics informed machine learning in summary?
This link is related to this topic phys.org/news/…
 
ACuriousMind
ACuriousMind
10:10 AM
@MoreAnonymous You don't need black holes for that, see physics.stackexchange.com/q/315104/50583
The concept of thermodynamics is inherently about the observer lacking information about the full microstate of the system - entropy measures that lack of information.
 
Hammad
Hammad
@ACuriousMind Thermodynamics in three lines!
 
More Anonymous
More Anonymous
10:27 AM
@ACuriousMind Yes I've seen that before but I wasn't convinced :/
I feel like there are $2$ situations when probability gets involved: 1. when we don't know (ignorance) and 2. The likelihood of in how many
times a certain event happens
I thought entropy was a statement of the latter
 
ACuriousMind
ACuriousMind
the point is that in classical mechanics, there is on a fundamental level no such thing as "likelihood". The classical world is fully mechanistic, and probabilities only arise from us being uncertain about the true microstate and being incapable of solving the equations of motions for complex systems in such a way that we could compute future microstates.
so in thermodynamics we gesture at some other parameters that define a macrostate - i.e. an ensemble of microstates - and then we can talk about how likely (in a frequentist sense) it is that some particular event on the level of microstates happens among all states in this ensemble
 
More Anonymous
More Anonymous
@ACuriousMind In classical mechanics there is also likelihood for example how frequent will a gas molecule collide. Now I do agree if we know the initial conditions we can calculate the collisions. But this does not negate my point. As an example consider the prime numbers obviously a number is prime or isnt yet we have Prime number theorem
and Probabilistic Number Theory
 
ACuriousMind
ACuriousMind
@MoreAnonymous If we could solve the underlying e.o.m. for every gas particle, we would not need to talk about this frequency, but could in fact determine when each particle will collide with the next
the moment we reduce ourselves to only talking about the average frequency, we are erasing information about the particular microstate
 
More Anonymous
More Anonymous
@ACuriousMind Indeed just as if we knew if every number was prime or composite
 
ACuriousMind
ACuriousMind
and the frequency is only well-defined w.r.t. to some ensemble of states - that's what the macrostate is
so in order to have the frequency well-defined we must erase the information about the particular microstate
you could talk about frequencies of collision following from one particular microstate but that's not what thermodynamics does - entropy is a property of the macrostate
this is obvious from its very definition - it counts the number of microstates in a macrostate!
i.e. the number of fully-defined states that could be "hidden" under the incomplete information that is the macrostate
 
More Anonymous
More Anonymous
10:41 AM
I'm still uncertain how this enables you to distinguish between the source of the probabilities I spoke about?
 
ACuriousMind
ACuriousMind
they are the same
 
More Anonymous
More Anonymous
I mean I can could equally say all macrostates are equally likely (no mention of any observer here) in thermal equillibrium
 
ACuriousMind
ACuriousMind
that makes no sense
 
More Anonymous
More Anonymous
@ACuriousMind Probability due to ignorance = Probability due to likelihood?
 
ACuriousMind
ACuriousMind
A macrostate is defined by some values such as temperature and volume. In thermal equilibrium you have one particular set of these parameters, i.e. a single macrostate
the thing is that observers can differ by what they consider characteristic of the macrostate, i.e. they differ in the possible microstates they assign to the system
that's what the answer to the question I linked tries to explain via the example with the colorblind observer
The thing about the notion of "macrostate" is that it's 'sized' exactly such that an observer can determine that a system in is one specific macrostate - which is really nothing more than a specific collection of microstates about which the observer is uncertain.
normally thermodynamics doesn't talk about this much because in most situations the observers don't meaningfully differ in their ability to determine the macrostate
 
More Anonymous
More Anonymous
10:49 AM
@ACuriousMind But in the blackhole example they do?
Any other examples?
 
ACuriousMind
ACuriousMind
the three classical ensembles of thermodynamics are a straightforward example of this
in some of these, the observer is able to determine a fixed total number of particles, in others they're not
there is no objective property of a system "being in" one of these examples - but for different systems different ensembles are useful because in practice our ability to observe them differs
 
More Anonymous
More Anonymous
But all observers would agree the particles are not fixed right?
(making it objective again)?
 
ACuriousMind
ACuriousMind
Not necessarily. For a crude analogy, consider humans on earth: We die and are born constantly, but on some scales you can track individual births and deaths and on other scales you need to treat this as some average rate of births/deaths
all observers agree the number of humans fluctuates, but they don't agree on whether we need to treat the actual number of humans as a property of the state we can precisely determine or as something whose precise value is unknown.
for the former, the number of humans is part of the macrostate - all possible microstates of "humans on earth" in their models have the same number of humans - while for the latter, the exact number is an unknowable part of the microstate and the macrostate just has something like the rate of births/deaths
this is like the grand canonical ensemble, where the number of particles is not known, compared to the other two
it's not about whether or not a particular property of a system changes or not - it's about whether or not it is useful for your modelling purposes to treat it as part of the microstate or as part of the macrostate
 
More Anonymous
More Anonymous
Is the E.T Jaynes mainstream physics or does it make some physicists uncomfortable too?
 
ACuriousMind
ACuriousMind
I'd bet it makes some physicists uncomfortable :P
 
More Anonymous
More Anonymous
11:00 AM
I think I'll need time to assimilate this point of view.
 
ACuriousMind
ACuriousMind
but QM and the measurement problem also make many uncomfortable, that doesn't make QM non-mainstream :P
 
More Anonymous
More Anonymous
I agree ... But it does make the resolutions non-mainstream :P
 
ACuriousMind
ACuriousMind
The discomfort mainly arises because this is a clear example in classical physics where it becomes evident that physics is really just modelling reality rather than always describing intrinsic features of it, but we've grown up with thinking about things like "temperature" as intrinsic properties of systems.
 
More Anonymous
More Anonymous
I believe in my childhood fairy tales and will not yield them easily :P
 
 
3 hours later…
JingleBells
JingleBells
1:36 PM
How and from where do you guys think AGI will most likely happen? Will it happen gradually with many research papers from many scientists or do you think some Russian kid will unlock the formula in his basement?
 
Charlie
Charlie
How are you drawing the line between AGI and non-AGI? I wouldn't say there's a clear threshold at which point AI becomes AGI
 
JingleBells
JingleBells
By AGI I mean an algorithm that can generalize across many domains. Meaning, if I built an algorithm that has learned to play tennis, then I can sit down with the algorithm and teach it to drive a car by only introducing it to new data but no manual modification of it's internal structure.
Singularity... that kinda stuff
maybe it surpasses our knowledge of many domains and invents new stuff...
 
Charlie
Charlie
My answer still depends on what you mean by "many domains", that's still ambiguous. That being said my guess would be it will be a gradual process as it has been so far (although breakthroughs happen)
 
JingleBells
JingleBells
1:51 PM
We, humans, we can learn almost anything. Any task, solve problems, play games, play sports... so many things. And we haven't found a single algorithm that can generalize to do all the things we're capable of. That algorithm is what I'm referring to AGI.
 
skullpatrol
skullpatrol
we took a looong time to evolve
a Russian kid in his basement probably can't do it alone, these days
if, that's what you meant
 
JingleBells
JingleBells
sure, I agree that AGI will probably happen thanks to the efforts of many researchers from all around the world. I think that each paper and each discovery takes us closer and closer. There, of course, will be a point in time when a group of researchers or a single person will combine all the knowledge, papers, get an idea, try it out, and summon AGI.
 
skullpatrol
skullpatrol
along with each paper and each discovery comes more and more complex analysis which fewer and fewer people can understand
 
JingleBells
JingleBells
Not necessarily. No one said AGI will be complicated.
 
skullpatrol
skullpatrol
Nobody said it would be simple
 
JingleBells
JingleBells
2:07 PM
Sure, but your sentence was pretty certain, and not probable.
 
skullpatrol
skullpatrol
it was just a general observation
 
JingleBells
JingleBells
Oh, sure. I thought it was in the context of AGI. In general, as more papers come, things get layered up and become more complicated, etc... but AGI might require few papers and they might be simple.
 
Azmuth
Azmuth
2:29 PM
@JingleBells AGI models Already Exists
But, each of them have their own problems...
I'm pretty sure, in next 50 years better models will come with less flaws...
Check this room then:
Some movie suggestions?
I'm bored now :(
 
skullpatrol
skullpatrol
Raging bull
 
Azmuth
Azmuth
@skullpatrol Plot?
 
skullpatrol
skullpatrol
Boxing
 
Azmuth
Azmuth
means a basic story?
 
skullpatrol
skullpatrol
Classic rise and fall of a champion
 
Azmuth
Azmuth
2:39 PM
wow!
have u watched 500 days of summer?
(emotional complex)
 
skullpatrol
skullpatrol
nope
 
Azmuth
Azmuth
I'm downloading that one.
 
skullpatrol
skullpatrol
coolio
 
Azmuth
Azmuth
Raging bull seems cool!
IMDB 8.2!!!
WOW
 
JingleBells
JingleBells
2:56 PM
@Azmuth Depends on what you mean by AGI, but the way most people understand AGI, no, it does not completely exist today.
 
Azmuth
Azmuth
@JingleBells What's your definition to AGI?
 
 
1 hour later…
John Rennie
John Rennie
4:16 PM
This has (deservedly) won this years Ignoble prize for physics
 
ACuriousMind
ACuriousMind
> Earthworms are non-regulated animals, and therefore this research did not require the approval of our Institutional Animal Ethics Committee. However, the worms were treated as humane as practical and afterwards they were placed into a worm farm where they fully recovered.
I'm glad the worms are okay
 
Slereah
Slereah
4:56 PM
Mine aren't
Working on wormholes, I have to use plenty of worms
 
Faded Giant
Faded Giant
Worms?
user image
4
 
ZeroTheHero
ZeroTheHero
5:16 PM
@ACuriousMind Allo.. regarding that very recent case of spam... it's not deleted yet but it is "fair" to edit out the link?
I mean: it's rare to see such a blatant case of spam.
 
Faded Giant
Faded Giant
@ZeroTheHero Please don't edit spam posts. Just flag them.
 
ACuriousMind
ACuriousMind
5:33 PM
@ZeroTheHero What spam? :) Don't edit spam posts, it might cause the spam flag to be declined (e.g. if the moderator reviewing it doesn't realize the post was edited) and confuses the Charcoal tools
 
JingleBells
JingleBells
can we make an ai that looks at an image/3d model of a skull or skeleton and builds the flesh?
I mean, does something like this exist?
An ML model guesses what a person looked like based on his/her skull
 
Charlie
Charlie
Is skull shape enough to determine that much about the face beside just shape?
 
Faded Giant
Faded Giant
Forensic facial reconstruction
Forensic facial reconstruction (or forensic facial approximation) is the process of recreating the face of an individual (whose identity is often not known) from their skeletal remains through an amalgamation of artistry, anthropology, osteology, and anatomy. It is easily the most subjective—as well as one of the most controversial—techniques in the field of forensic anthropology. Despite this controversy, facial reconstruction has proved successful frequently enough that research and methodological developments continue to be advanced. In addition to remains involved in criminal investigations...
 
JingleBells
JingleBells
@Charlie Donno, but I've seen face reconstructions for a long time... I suppose you need other data but... you get the basic idea.
 
Charlie
Charlie
@FadedGiant oh surprisingly accurate fair enough
 
Yashas
Yashas
6:06 PM
How do I find the maxima of $\frac{|\lambda|^{2n}}{n!}$?
where n is an integer
I approximated n! to the gamma function since the maxima wouldn't change. I found the derivative and simplified but ended up with an equation which has n in upper limit of a summation.
Context: Harmonic oscillators and coherent states where I have expressed $|\lambda\rangle$ (a coherent state where $\lambda$ denotes the label from $\mathbb{N} \cup \{0\}$) as an infinite summation of $f(n)|n\rangle$ terms where $f(n)$ is $e^{-\frac{|\lambda|^2}{2}} \frac{\lambda^n}{\sqrt{n!}}$. I want to find the maxima of $f(n)$.
 
ACuriousMind
ACuriousMind
@Yashas Why do you want to find its maximum?
 
ZeroTheHero
ZeroTheHero
6:32 PM
@ACuriousMind right. Good point. Thanks for the tip.
Indeed the “Community” user deleted the post shortly after I asked my question.
 
Yashas
Yashas
@ACuriousMind I was curious. It looked interesting because it was a linear combination of eigenfunctions and I need to get $\lambda$ as the answer.
 
ACuriousMind
ACuriousMind
@Yashas Alright, I just didn't see a physical motivation to look for it, hence my asking. Try looking at that expression for $n+1$, and express it in terms of it for $n$ times a factor, i.e. $\frac{\lambda^{2(n+1)}}{(n+1)!} = k(n) \frac{\lambda^{2n}}{n!}$. The maximum is where $k(n)$ becomes smaller than 1.
 
North Læraðr
North Læraðr
7:31 PM
@JohnRennie -1.50 * 10^11
 
 
4 hours later…
Stupid question inc
Stupid question inc
11:31 PM
Do you have any advice on good physics courses for self-learning?
please tag me if you know
 

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