« first day (4523 days earlier)      last day (700 days later) » 
00:00 - 18:0018:00 - 00:00

18:20
If only there wasn't a real nature out there physics could be beautiful lol
That buggy thing called experimental reality
@Amit I just take the manifold as having ontological existence. It's so clean to think about :P
Yeah, but is 4 dimensions really a manifold? Maybe it's a fewfold?
Lol
But i also accept that my thinking is wrong. Nature doesnt know shit about manifolds
I wonder if GR can be formulated with a Riemannian (++++) metric
Some people suspect that spacetime is ++++ at the microscopic scales. Has to be a niche idea tho
I dont think GR can be formulated with ++++ tho
18:29
Interesting. But then I wouldn't want to do it cheaply like, just set the $t$ values of 4-vectors to be negative to make up for that and preserve lorentz invariance. That's cheating lol
U mean imaginary t?
No, i did not mean that
I read people suspect actual ++++ at microscopic scales, no cheating
Its apparently becuz euclidean qft path integral is nicer
mathematically
After all what does lorentz invariance says, that if a massive thingy chases a light beam, no matter who's measuring how close he gets, they all get the same result right
The Euclidean stuff is even crazier, as i found today :P
It says massless particles remain at rest wrt every frame of reference
But with these small'ish guys can we really measure this
Euclidean particles apparently hav each a speed limit, depending on their mass
18:34
I don't know much about QFT. I only remember daggers and integrals
@Amit i think it's straight up wrong. Our universe's massless particles are photons. They r so not at rest
What is straight up wrong?
That massless particles remain at rest in our universe
Ah, it's not clear what it would mean even... there's also the HUP...
Yeah
Well, in the classical theory, they remain at rest
In the quantum theory, it implies they must hav a well defined momentum of 0
Which wud mean infinite uncertainty in position, yeah
18:37
I always get a feeling that a lot of QM was developed still with a vague hope that "okay it's all crazy now, but if we make it just a bit more crazy so that we can calculate better, it'll all become clear physically..."
Kind of like accepting having to pass a dark tunnel to reach the light lol
Just a bit more
They embraced all of the craziness
Im surprised they figured it out so fast tho
Yes because it may be true that we will get a better picture just by making our calculations more accurate. On the other hand, at what point do we say that the model has become so mathematical that we have no idea how to even ask about the physics of what it models
But then again, QM doesnt hav a single founder. It took a whole army
No doubt it took a lot of good people
Something like the Feynman rules say, I think everyone kind of admits that what's going on physically is not what the diagrams represent. Yet, it's clear that something is going on that these diagrams are modelling, and we don't exactly know what that is
Otherwise it wouldn't work as well
Those diagrams really dont mean shit :P
18:42
lol, I was trying to put it more mildly
U can draw any successive matrix multiplications like that
It shows up even in classical field theory :P
Sorry, i mean tensor products, not matrix
@RyderRude 2->2 scattering is a 4-point function.
@ACuriousMind I thought u meant that the 2-point function was the coulomb potential
how would that follow from what I said?
Idk...
Lol
@ACuriousMind What happened is that I didn't know the exact derivation of Coulomb's law using scattering. But I knew that the 2-p function was the Coulomb potential
18:47
@RyderRude In classical field theory probabilities are not the result though right
@Amit yeah. The result is the field solution to the interacting field theory
Those diagrams just indicate tensor products
But the tensors we have are of infinite dimension here
In GR, each index runs from 1-4
And Penrose already covered that case with diagrams ^_^
lol it's a testament to how unlikable tensors are I guess, every time they pop out in any dimension, someone finds a diagrammatic way to avoid them
No one even calls them tensors in this context
But i guess, they cud b called that, mathematically
I think very formal people of the quantum realm are somewhat aware of tensor spaces
lol
@RyderRude Feynman diagrams have nothing to do with tensors
you're perhaps thinking of Penrose diagrams, which are entirely different creatures
18:52
No, i really meant feynman diagrams
But by tensor, i dont mean the GR definition of tensor :whatever transforms likena tensor is a tensor
Mathematically Feynman diagrammatics are a way to organize "perturbation" series
I mean each term represents matrix multiplication
I don't know what that is supposed to mean
I mean the Green's function $G(x1, x2) $ is the matrix here
that's not a matrix, it's a function
18:54
Yeah, but it can b thought of as a linear map from two infinite dimensional vectors to numbers
I don't know why you'd do that in this context
https://arxiv.org/abs/2207.06135
Someone thought of a way to relate the two at least. But I don't know nearly enough about FD to understand this
I have an example using Einstein notation
The calculation involved in Feynman diagrams is something like : G(x, x1) G(x1, x2) G(x2, y). Here, see that x1 and x2 are repeating indices. This means they r integrated over @ACuriousMind
Do you mean the scattering matrix
18:58
oh, you just mean that when you have to diagrams with external legs and you connect the legs you integrate over the variable associated with the legs in the same way you sum over the indices of two tensors when you contract them
Exactly!
It is not a tensor in the GR sense. It does not transform like a tensor or anything
A tensor never breaks a contract
I think it's the wrong way around to think of this as a "tensor" property, it's just how convolution of functions works
It is a tensor in the lineae map sense
I've seen some people use the Einstein notation for it
18:59
tensor contraction is a special case of this convolution
$$f_a g^a = \int f(x) g(x) dx$$
Probably convolution makes more sense here, yeah
Yeah I think to really be a tensor you need the multilinearity thingy first
or whatever it was
or, well, inner product/pairing/Whatever, it's not exactly convolution
19:00
I was using the word tensor in the general sense : As just a linear map @ACuriousMind
@RyderRude sure but nothing about the functions here is linear
"linear map" is the general word for linear maps
An inner product is a multilinear map, like the metric tensor is basically right
Lol
I swear I saw a post about this
I mean, the Green's function (2-point function) specifically is a linear operator since it's the kernel for solving a linear differential equation
but the diagrammatics involve vertices, too, which aren't linear in this sense
19:19
@ACuriousMind I guess we can say that only Einstein notation is useful here, despite them not having a linear map interpretation in this context.
Schwartz's book uses this notation for Feynman diagram terms
In the case of QFT Feynman diagrams, the Green's functions dont have a linear map interpretation
In the case of classical field theory feynman diagrams, the Green's functions have a linear map interpretation
19:32
Hello. I need to know how to calculate SPL change from a sound source to the listening position at a distance.
Please help me find this out it's so important and urgent! thank you.
0
Q: Can a question be deleted or closed if it exists on another stackexchange site?

AureliusCan a question be closed,deleted, or called a duplicate if it has already been asked with or without answers on another stackexchange site? Since many topics of Physics and Biology tend to intersect with Chemistry, is it possible that a question might be called a duplicate for being asked on anot...

@Amit Thanks! already saw this one but it's not saying how SPL changes from one source to one listener. does it?
well if you know the SPL at the source then you can substitute 0 for the distance of the known SPL right?
I think it has to be an inverse square law either way
ah no sorry, for pressure it's just inverse no square
The point is that I wanna find the SPL at the source
19:42
yuh so multiply by the distance
I think.
and the distances are like r1/r2 so I cannot use 0 in this equation
cause it would give me 0 or infinity ! which is nonsense in rl.
so don't get too close to the speaker or you will hear infinite volume lol
well yeah this is when you have a listener in the middle and two speakers and you want to calculate spl for the second spekaer
this is an idealized situation you see, don't go to the singularity of the sound
i mean what exactly is meant by finding the SPL at the source, if it's a point source it doesn't make sense it will be infinite by definition
20:02
finding spl at the listening position, not the source
@amit I already know the sound pressure the speaker is producing, but I wanna see, in the distance the listener has to the speaker, how the spl changes, how would it be at the listening position.
yeah but I'm saying you can't take any position to be r=0
we don't do it ever, even for gravity we can't calculate that
well, whether that's an "even" or not lol, we can't
i mean at r=0 you can't define the value
the fact that you have r1/r2 over there means that you have to say what is the ratio of the distances....
@Amit On the PDF I have, the information consists of a "SPL at the membrane of speaker" and also "Virtual point Distance". Does this mean that the SPL at the membrane is measured at Virtual point distance?
I don't know what is "virtual point distance". However, clearly the membrane is some distnace away from the source, so it's not r=0 anymore...
Anyway just find out what is the distance between the membrane and the true source and you'll have your $r_1$
Thank you!
Welcome
20:27
0
Q: Is there a way to search for down voted or negative questions to answer?

AureliusQuestions with negative votes are generally not displayed at the top and sometimes hidden. Is there a way to specifically look up for negative voted or inactive questions on the site? The questions with negative votes are generally not shown but sometimes some of those questions might actually be...

 
2 hours later…
22:36
Every now and again you find simple masterpiece level posts
20
A: Is there a high level reason why the inverse square law of gravitation yields periodic orbits without precession?

Francois ZieglerThere are: Bertrand’s theorem, which says that the isotropic oscillator and Kepler potentials are the only analytic radial ones all of whose nonrectilinear bounded orbits are closed. (Recommendation: Albouy - Lectures on the two-body problem. But he says: “The proof of this theorem provides ver...

23:23
Does anyone know if there exists a quick way to check if a finite dimensional matrix has degenerate eigenvalues?
there is this theorem, kayleigh something?
Ideally it would be a more abstract way to reason than defining the matrix explicitly and computing the eigenvalues :P
cayley hamilton
that tells you can tell it from the characteristic polynomial
maybe I'm botching it up, that's more or less what I remember
actually seems like you don't need cayley hamilton, just to inspect the polynomial
@SillyGoose I'm not sure what you mean by more abstract
So M is the matrix I am trying to determine if can have a degenerate spectrum, and so I am trying to see if there is a way to check this while avoiding explicitly defining the matrix :P
like maybe there is a list of properties like: If your matrix has X characteristic, its spectrum is degenerate
but it's crazy you are raising something to the powuh of a matrix??? lol kidding
degenerate means like if any eigenvalue repeats?
23:33
yes
what do the eigenvalues represent?
energy levels?
wait actually i think i have misunderstood something... i might be looking for something else
the eigenvalues here are just computationally important because they let us compute the minimum of a function defined in terms of M
but H is the Hamiltonian right? it should have something to do with energy?
do the exponentials multiply both parts of the tensor product? then they would cancel? something is weird about that
the functio nof interest is 3.28, which is the linear entropy (2nd order von neumann entropy) of a time evolved, environment-traced-out density matrix
what is this stat mech?
23:39
well i think it is easiest to think of the exponential as its taylor series. so the action of the exponential (to the power of hte hamiltonian) on the density matrix amounts to considering the action of the hamiltonian on the density matrix which in general is not trivial
interesting but i have no idea..
i see also a pauli matrix thrown in for the fun?
i think it is more on the side of quantum but perhaps nearingg the intersection of quantum and stat mech
well the pauli matrix comes from a nice way to write the initial system state we are dealing with i believe
Is this a QFT book?
I hope someone else can help, that's beyond me
no no this is someone's thesis that i am working off of i know not qft :P
ah, cool ok
23:47
but i think i got the wrong idea of what to ask :). i think my question should have been in general can matrices of the same dimension have the same maximum eigenvalue, which I think is quite plausible to be true
what is a maximum eigenvalue
the largest eigenvalue in its spectrum
ahh that's a physics thing
i mean in linear algebra you do have a spectral radius, but since you constrain apparently two matrices to have the same one it must be due to physical reasons right
but this project is quite cool :D. it is about trying to factor some Hilbert space into system and environment based on quasi-classical criteria. So for each factorization, we assign a score by computing the linear entropy (computationally nice way to measure the likelihood of a state to entangle) of some candidate quasi-classical state(s). Then we search for the space with the smallest score.
sounds nice
23:52
well there is no constrain per se, but having two matrices with the same maximum eigenvalue in my context corresponds to two factorizations being equally good
@SillyGoose for such questions, just remember that the diagonal matrix with diagonal entries $\lambda_i$ trivially has eigenvalues $\lambda_i$. So there are infinitely many operators/matrices with the same highest eigenvalue - their diagonalisations just differ in the lower $\lambda_i$
and so i was just trying to reason that i should be getting many or multiple factorizations being equally good (which computationally i am)
hm i see okay that is sufficient for me
in fact, there are infinitely many operators with the exact same eigenvalues!
just take the diagonal matrix $D = \mathrm{diag}(\lambda_1,\dots,\lambda_n)$ and any unitary $U$ then $UDU^{-1}$ has the same eigenvalues
since there are infinitely many unitaries there are infinitely many operators with the same eigenvalues
well now i've got to figure out how to classify these minima :P.
what minima
23:57
well actually i know how to classify them, but i think implementing it into code will be a bit difficult because there are infinitely many
well so I have a 4-dimensional Hilbert space factored into the tensor product of 2 2-dimensional spaces. I am starting with an initial candidate pointer state which is [[1,0],[0,0]] \otimes (maximally mixed state), where the first tensor factor is the system.
I also have a Hamiltonian which is the 1D ising model hamiltonian. Everything is already put into the tensor product factorization such that this initial candidate pointer state when time evolved has no entanglement entropy.
00:00 - 18:0018:00 - 00:00

« first day (4523 days earlier)      last day (700 days later) »