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12:15 AM
physics needs to be kept pure
it's like how people complain about bad grammar becoming more standard
 
 
6 hours later…
6:09 AM
morning
 
 
1 hour later…
7:25 AM
Hm
How to prove geometric optics
Apparently the rigorous process involves the Luneberg-Kline asymptotic expansion
But it seems long and tedious
 
7:46 AM
"Every mathematician knows it is impossible to understand an elementary course in thermodynamics."
:D
 
8:06 AM
"According to Gibbs, the geometrical structure of thermodynamics is described by a contact manifold, equipped with the contact form, whose zeroes define the laws of thermodynamics $$dE = T dS - P DV$$
the horror
"Substances are Legendre submanifolds of the Gibbs manifold"
"Legendre singularities and the icosahedron"
Help
This article keeps getting worse
 
8:54 AM
Least useful article
 
9:06 AM
@DanielSank he said, unassumingly?
Official congratulations to @DanielSank for his role in the (now-released) Quantum Supremacy paper by the Google team!
15
 
9:35 AM
@SirCumference ha! It's not 'bad grammar' becoming standard, it's people's understanding of linguistics being horribly outdated
@EmilioPisanty and hooray! Quantum Computers can officially do things! :D
 
hip hip hooray!
 
10:22 AM
English grammar is nothing but bad grammar of previous incarnations of the language
 
Geometric optics is the theory of electromagnetic waves which are such that their amplitude and direction of the solution of the wave equation remains constant over any local region, so that the solution reduces to a plane waves locally, the condition for this to hold is that the wavelength goes to zero so that the wave can always be considered plane over any local region.
The mathematical way to say this is that asolution $f = A e^{i\psi}$ of the wave function is such that $\psi = \psi_0 + \frac{\partial \psi}{\partial \mathbf{r}} \cdot d \mathbf{r} + \frac{\partial \psi}{\partial t} dt + ... = a_0 + \mathbf{k} \cdot d \mathbf{r} - \omega dt = a_0 - k_{\mu} dx^{\mu}$ i.e. the phase is a plane wave to first order, and this dominates, so that the first derivative $k_{\mu}$ of the phase $\psi$ satisfies $k_{\mu} k^{\mu} = 0$, which is the Eikonal equation.
You can also get this by plugging $f$ into the wave equation $\partial_{\mu} \partial^{\mu} f = 0$ and neglecting all terms except the $k_{\mu} k^{\mu}$ term which again gives the Eikonal equation.
 
10:41 AM
wave equation* not wave function gah
What is rigor going to add to this
 
10:59 AM
The hard part is showing that $E \approx e^{i\psi}$ is a decent approximation, though
 
why is $|\Psi|^2$ proportional to probabilities?
 
@Ultradark That is how the theory was constructed
$|\Psi|^2$ obeys all the rules of a probability distribution
Therefore it can be interpreted as such
 
can you normalize a pdf with a non-constant factor?
 
I mean if it's not constant it's not really a factor
 
can you normalize a pdf with a function? (I mean)
why is it always a "normalizing constant"
 
11:08 AM
Because the Hilbert space is a vector space
There's no concept of multiplying two wavefunctions together
Also if you have to normalize it with another function, why not use the result as your wavefunction to start with
what's the point of the first non-normalized wavefunction
 
i was just asking in general, but I guess that makes sense physically
 
The point of the Hilbert space is that it's the minimal mathematical structure to get the results of QM
it's a vector space because of the superposition principle
It has an inner product to give you a probability distribution
 
do you know what's going on here?
It looks like the points at infinity are being "blown up" to get from the LHS to the RHS
 
what does the text that accompanies the diagrams suggest?
 
I'm gathering that there's some conformal map that sends minkowski space to ADS
What does 3 dimensional Minkowski space conformally compactified to a cube look like?
 
11:26 AM
sounds like an exercise left to the reader :P
 
Do you know the answer?
 
nope, sorry
 
There's no pictures of this, I'm kinda surprised
they are all 2d
 
If you want conformal compactification to a cube just use some basic $(0,1)$ to $\mathbb{R}$ diffeomorphism
Like $\arctan(x)$
 
in a two 2d compactification there are 4 points at infinity
am I right to say there will be 8 points at infinity in a 3d compactification?
 
11:44 AM
Why 8?
 
because the cube has 8 corners
 
Oh
 
$f(t,\mathbf{r}) = A(t,\mathbf{r})e^{i\psi(t,\mathbf{r})}$ is just a complex number representation of the solution of the wave equation $\partial_{\mu} \partial^{\mu} f = 0$ so there's no approximation (for the potential $A_{\mu}$ not $E$)
 
Isn't it the $n-1$ polytope rather than just the corners?
 
The approximation comes from assuming $A$ and $\psi$ have certain properties
0
Q: Spin J $A_J(s,t) = - \frac{g^2(-s)^J}{t-M^2}$ Amplitude?

bolbteppaIn GSW equation (1.1.2) they define the scattering amplitude for a spin $J$ particle at high energies as $$A_J(s,t) = - \frac{g^2(-s)^J}{t-M^2}$$ mentioning it is an asymptotic approximation to a formula involving Legendre polynomials, and in this stackexchange post it is also defined, and in Wil...

No votes for the origin of string theory :(
 
11:54 AM
@Slereah It very well could be, why do you think that?
 
Dunno
vague intuition
 
I don't think it's that because let's say 3d Minkowski is situated inside a regular octahedron instead
 
Well I mean, usually the higher dimensional conformal compactification aren't done as a polygon, really
Usually you compactify it in spherical coordinates
So that the "canonical" diagram is just the usual diagram $\times S^{n-2}$
Let's see if there's a standard compactification of $\mathbb{R}^2$ as a square
 
can you give me an example of such a canonical diagram ? (for $\times S^{n-2}$)
 
Well the canonical one is the Minkowski metric in spherical coordinates
$$ds^2 = -dt^2 + dr^2 + r^2 d\Omega^2$$
Switch to null radial coordinates
$$ds^2 = -dudv + (\frac{v - u}{2})^2d\Omega^2$$
Then use the conformal factor $[(1+u^2)(1+v^2)]^{-1}$
$$ds^2 = -\frac{dudv}{(1+u^2)(1+v^2)} + \frac{(v - u)^2}{2(1+u^2)(1+v^2)}d\Omega^2$$
Do a coordinate change with $u = \tan(p)$, $v = \tan(q)$
and then $T = (p+q) / 2$ and $R = (q-p) / 2$
You end up with $$ds^2 = -dT^2 + dR^2 + \sin^{n-2}(R) d\Omega$$
Or something like that
It's actually more $\mathbb{R} \times S^3$ if you consider the maximally extended spacetime, but the conformal compactification of Minkowski itself is just $\mathbb{R}^2 \times S^2$
That's the fairly standard conformal compactification that you see on every diagram of Minkowski space
probably not unique, though, but I'm not sure there are others that are as palatable
 
12:12 PM
I don't understand why maximally extended spacetime would be $\Bbb R\times S^3$
 
@Ultradark just look at the metric
$dr^2 + \sin^{2}(r) d\Omega$ is the metric of a $3$-sphere
 
if that's the case then wouldn't it be impossible to visualise in 3d?
 
Certainly, though 1) this is in 4 dimensions 2) this is why you suppress the angular degrees of freedom on diagrams
in 2+1 dimensions it would be $dr^2 + \sin(r) d\theta$ which is... also probably a sphere, but not the standard metric for one?
So you have $\mathbb{R} \times S^2$
 
perfect
 
I'm not sure there are other decent compactification of Minkowski space that have properties as nice as the Penrose one, though
The huge benefit of such compactification is that the radial part has the same causal structure as Minkowski space
so it's extremely easy to get the causal structure at a glance
 
12:24 PM
okay. you could map the version inside the sphere to the cube because they are homeomorphic
 
Errrrr
I mean I suppose, yes?
Although that's the surface of a cube
And not a cube itself
This isn't $\mathbb{R}^3 \to (0,1)^3$
 
yeah, so starting with $S^2$ with the causal structure of maximally extended Minkowski spacetime (supressing angular degrees of freedom) in the bulk of $S^2$
 
Also if you map the sphere to a cube you're gonna lose the radial ~ Minkowski structure
 
are you sure about that?
 
No, but I would guess it to be likely
 
12:40 PM
doesn't the 2d maximally extended compactified spacetime have the radial structure?
 
Yes, but it's a very simple spacetime
the metric is just $$ds^2 = -dT^2 + dR^2$$
Fairly easy to see it's just Minkowski spacetime
 
@Slereah thanks for the help with this
 
1:29 PM
Does anyone remember when math was simple
Before we had manifolds we had shapes
And the word "number" used to be meaningful
All right I gotta wake up
 
@SirCumference where do you put the cut off line
Remember in Ancient Greece when they murdered a guy because he proved that the square root wasn't a rational number
 
Probably when something as basic as addition leads to groups and other things
 
I mean really I think people underestimate how nasty old timey math used to be
 
@Slereah Er, no I actually don't
 
Simpler math also meant much more complicated proofs for some theorems
 
1:32 PM
That's pretty insane
 
Hippasus of Metapontum (; Greek: Ἵππασος ὁ Μεταποντῖνος, Híppasos; c. 530 - c. 450 BC), was a Pythagorean philosopher. Little is known about his life or his beliefs, but he is sometimes credited with the discovery of the existence of irrational numbers. The discovery of irrational numbers is said to have been shocking to the Pythagoreans, and Hippasus is supposed to have drowned at sea, apparently as a punishment from the gods for divulging this. However, the few ancient sources which describe this story either do not mention Hippasus by name (e.g. Pappus) or alternatively tell that Hippasus drowned...
"It is related to Hippasus that he was a Pythagorean, and that, owing to his being the first to publish and describe the sphere from the twelve pentagons, he perished at sea for his impiety, but he received credit for the discovery, though really it all belonged to HIM (for in this way they refer to Pythagoras, and they do not call him by his name)."
 
A punishment from the gods for discovering irrational numbers
That's a sentence I didn't think I'd read
 
The Pythagoreans were big on the universe being math
Irrationals didn't sit well with them
 
I understand constructible numbers
Those can be irrational
 
How do you feel about algebraic numbers
 
1:44 PM
I don’t like polynomials
 
1 is a polynomial solution
 
How about elliptic functions
 
Those are not numbers
 
What is a number tho
I don't think it actually has a definition
 
Well see above
each number set is constructed differently
and each of them can be constructed in a variety of ways
I think you have the ~number object~ in category theory if you want to define numbers independently of their constructions
IIRC natural number objects are basically just the Peano axioms
 
1:58 PM
0 := { }, 1 = { { } }, 2 = { { }, { { } } }, ... (I think)
 
@Mithrandir24601 depends on what you mean by "do things", right? quantum computers have been factoring numbers, in the lab, since at least 2001. Or did you mean useful things?
=P
 
@bolbteppa IIRC 0 is $\{ \varnothing \}$ in Russell's axioms
but it is $\varnothing$ in the von Neumann and the other construction
 
Lets keep it within the realm of reality and not n-reality
 
@EmilioPisanty I mean ' things that aren't trivial on a desktop pc' :P
 
Right maybe it's 0 = {{ }}
 
1:59 PM
Well no, what you're describing is the von Neumann one, I think
The Russell axioms are much worse
$1$ is defined as the equivalence of every set of cardinality $1$
 
Although then again Russell has hierarchic sets so maybe it's not that bad
 
0 = { }, 1 = {0} = { { } } , 2 = {0,1} = { {}, {{}} }, ...
 
1 is like $$1 = \{ A | \forall x \in A, \not\exists y, x \neq y \wedge y \in A \}$$
 
How you define a mass in space, because there is no inertia in space, because of no force.
 
2:02 PM
So $1 \in 2$
and $0 \in 1$ (take that $0 = 1$ examples like $x = y \mapsto 0 = 1$!)
 
Yeah that is the benefit of the von Neumann axioms
 
@bolbteppa $1!!\neq 0$
 
since you have $a < b \sim a \in b$
 
Hello, can anyone tell me why we can't use reduced mass in pulley system?
I let's say we have a pulley, on one end of the string we have one object of mass m_1, and the other end of mass m_2.
Then the tension in string comes out to be \frac{2m_1m_2}{m_1+m_2}, but lets say we take reduced mass, and consider the other object stationary.
 
@SirCumference of course they do
 
2:06 PM
In that case, we have tension equal to half of actual tension.
 
Christodoulou starts his physics classes with the definition of numbers because it’s so important
And physicists don’t know these things
 
We learned about the numbers
Back in kindergarten
1
2
possibly others
 
Using apples and slices of pizza
 
Those are graded algebras @bolbteppa
Different matter
 
This must be definitely because it's in respect to mass m_1. So when taking with respect to ground, what happens such that acceleration becomes double?
I tried to find out the answer but I couldn't understand.
I thought it should be half of what reduced mass gives, but it's actually double.
I asked my teacher but he said that he'll teach properly with center of mass, but I want to know the answer.
 
2:16 PM
I refuse to believe that is a word
 
Why
 
Very silly
 
It does exactly what you think it does
 
Unlikely, since I don't know what a sheaf is
I don't know how to make something more sheaf-like than previously
 
Well that article is a bit incomprehensible
But check Wells’ book
 
2:19 PM
finding a good category theory ressource ain't easy
 
@yuvrajsingh Why do you think forces do not exist in space?
 
Though I'm thinking of getting that spivak book
 
Sheaves of rings/Abelian groups are not hard to understand
 
I hear it's alright?
I vaguely recall that it is basically mapping objects to open sets
or vice versa
since that's how AQFT works
 
No real results in any of this stuff, just more words to define things you already know
 
2:26 PM
Implying physics contains real results...
 
QED laughs at you
 
QED is old physics
Hard to argue with that
 
@RyanUnger i don't think i've ever heard the word "number" come up in any of my undergrad math courses
Aside from the usual phrases like "natural numbers"
 
Numbers only arise as combinatorial or normalization factors
 
In principle I agree with @bolbteppa but I’d rather defend number theorists than agree with a physicist
This pains me
 
2:31 PM
number theory is a meme
 
Still need to read Riemann's paper properly
 
A body in space, is like floating, I does not have frame of reference.
 
@yuvrajsingh hi
 
@SirCumference what does this mean?
 
@RyanUnger your respect for set theorists equates to my respect for number theorists :P
Jul 3 at 15:51, by Ryan Unger
set theory is a meme
 
2:40 PM
Wow what an awful take
 
2:55 PM
Number theory doesn't really help anyone
And it's not a beautiful field either
It's probably the only math that I can't manage to get any interest in
 
@SirCumference this is such an awful take
Jesus
 
@EmilioPisanty Thanks!
 
Isn't number theory used in cryptography
 
I finally found someone with worse math tastes than me
Huzzah
 
@Slereah That's like the only cited application
but even as a pure math field, it's just not beautiful
like really how often has anyone said "holy crap that makes so much sense" regarding a number theory theorem
 
3:10 PM
@SirCumference i mean it's more applications than GR
 
yeah but GR is cool
ok i'm explaining this badly but still
 
Do you even know what number theorists do
 
they study $\mathbb{Z}$ and its properties right?
 
physics.stackexchange.com/questions/509647/… Ever notice how the less clear the question is, the more aggressively the OP's will respond to people questioning them?
 
OK i actually need to finish my analysis hw
 
3:23 PM
@RyanUnger do they find new numbers
2
 
@PM2Ring where's the python room?
I have some questions but they're probably too basic for main SO
 
@Slereah yep.
Finding large numbers is a hard problem
If you want to do it correctly
 
Any python question can be asked here :
 
3:51 PM
@RyanUnger i dunno, in terms of enjoyment I'd rank math fields as analysis > logic/set theory > algebra > things i don't like > number theory
like if there's some really cool aspect about studying primes and other aspects of the integers, i'm just missing it
 
quite lucky that it's rarely used in physics
Imagine if all particle masses had to be prime numbers
 
@Slereah i think we can say "quite lucky that's it's rarely used in [any subject]" aside from cryptography :P
 
Too bad cryptography is so important now!
 
4:06 PM
to be fair the fact that people study set theory as a career means it's gotta have some very interesting aspect i haven't gotten to, so i ought to keep an open mind
 
Like 100 years ago almost nobody had any reason to give a shit about cryptography
Unless you were a SPY or something
And now we use it all the time
And even then I think old timey cryptography used very little number theory
it was mostly just substitution and transposition
It was usually enough in the pre-computer era
Earliest prime-based cryptography is 1874 apparently
 
@SirCumference Meant to say number theory, tired
@Slereah it's weird to think we're living in a post-computer era when it only really began a few decades ago
like pre-computer era was 99.99999999999% of human history and now suddenly everything is different
i dunno i'm sleepy
 
well I mean a lot of things changed in recent human history, to be fair
Also like 90% of human history was just hunting animals and making stylish loincloths
so obviously yes things have changed a lot
 
@Slereah we still make stylish loincloths, so that's some continuity there
 
nah they don't make them like they used to
 
4:36 PM
does anyone know where dimensional analysis is explained a bit "rigorously"? I know the computations are not hard, but I don't really understand from which principles we work. How can we tell from the Hamiltonian (without resorting to classical physics) that energy has dimensions mass*lenghte^2*time^-2?
Say we have a potential $V_0\delta(x)$. Then our Hamiltonian looks like $p^2/2m+V_0\delta(x)$. So.. how do we decide which are going to be our "base quantities”? I mean, I guess since $p=-i\hbar\partial/\partial x$ (in one dimension) we know the dimensions of $p$ if we know those of $\hbar$ and $x$. We can set the dimension of $x$ to be length, and we can include mass as well (since $m$ is in the Hamiltonian anyways)
However, from $\Delta E\Delta t\geq\hbar/2$, it would seem that we should also set $\hbar$ to something, but that’s not the case, since we know that $E=1/2mv^2$, so $E$ is fixed, and therefore $\hbar$. But.. how can I derive that, I guess?
in this expression of the expectation value of the kinetic energy, we still have something in terms of $\hbar$
I just wish it was explained somewhere consicely and rigorously, because whenever I have to do dimensional analysis, I just resort to random classical formulas to get something that I can recognise, but I kind of want to have an "algorithm" or an explanation (from first principles/rules) from which I can derive things if I've forgotten how it works
 
I mean there isn't a whole lot of math in dimensional analysis
 
I know that
 
The most math you can use for dimensional analysis is graded algebras
 
the math isn't the problem
the problem (for me) is identifying which things we are going to set as fundamental quantities/dimensions I guess, based purely from the equation we're working with
 
It's usually based on the context really?
The same equations can have different dimensions
 
4:45 PM
well, that's why I gave an example hamiltonian right
$p^2/(2m)+V_0\delta(x)$
 
Well it all depends on how you define everything
 
that's my question exactly:x
I don't know what the thought process behind that is
 
Usually a momentum is $\sim M \cdot L/T$
 
well....
could we work with the operator definition from QM?
I kind of don't want to resort to classical physics:x
 
Well at some point of your theory, you have to relate the operators you're using with real life measurements
 
4:47 PM
the reason for that is that in QM, momentum is given in terms of $\hbar$
 
And that's where you can relate the two
I'm sure you can have two operators which are basically identical as far as mathematical properties go
 
and I kind of don't want to pick random formulas, like $E=\hbar f$ to know what to do with $\hbar$
 
But are different as far as units go
What you need is some basic idea of what you're measuring
 
well say we want to give(/measure) the ground state energy
then dimensional analysis can give an indication what it will be
but I don't really know what to do with $\hbar$
 
ie if you have an observable $\hat{A}$, which gives rise to the expectation value $\langle \hat{A} \rangle$, then $\hat{A}$ will have the dimension of that measurement
Well, not quite
since the wavefunction has a dimension, too
That one isn't too hard to work out
 
4:49 PM
okay, so our observable here is $\hat H$, for the sake of a context
@Slereah uh:x
 
@ShaVuklia why can't you resort to classical physics when there has to be a classical limit for things like energy
 
$$\langle \Psi, \Psi \rangle = \int \Psi^*(x) \Psi(x) d^nx 1$$
 
ah, right okay
 
Therefore $$[\Psi^2] = [L^n]^{-1}$$
 
@bolbteppa I don't want to use random classical formulas. I feel like it's easy enough to get it from commutation relations and the hamiltonian itself
@Slereah right that makes sense
but does it really matter for the ground state energy? we have $H\psi=E\psi$ anyways
 
4:51 PM
But I mean dimensions are an arbitrary label in the end
 
Where does the Schrodinger equation even come from without considering a classical limit
 
so the dimensions of $\psi$ kind of don't matter right
 
It's what meaning we apply to them that is important
Well it does!
Since $[\Psi] = [L^{-3/2}]$
 
uh:x
right okay
but how does that help us to determine the ground state energy?
 
Well I guess Dirac has a way but ....
 
4:52 PM
@bolbteppa well I'm kind of taking that as a starting point
(I want to start somewhere, and not just do random things, because I'm afraid I might otherwise end up in a circular reasoning)
 
$\hbar$ has units of $[E][T]$ and so $i \hbar \partial_t \psi = \hat{H} \psi$ shows $\hat{H}$ has units of energy
 
As I said really
What you want is just to relate the operators with the measurements
Otherwise it's all arbitrary
 
@Slereah but I don't understand how that process goes:(
 
The measurement of position is $$\langle \hat{x} \rangle$$
 
@ShaVuklia there's only one way you can trust you wont commit circular reasoning in my mind
The Course of Theoretical Physics is a ten-volume series of books covering theoretical physics that was initiated by Lev Landau and written in collaboration with his student Evgeny Lifshitz starting in the late 1930s. It is said that Landau composed much of the series in his head while in an NKVD prison in 1938-1939. However, almost all of the actual writing of the early volumes was done by Lifshitz, giving rise to the witticism, "not a word of Landau and not a thought of Lifshitz". The first eight volumes were finished in the 1950s, written in Russian and translated into English in the late 1950s...
 
4:54 PM
Therefore $x$ has the unit of length
 
@bolbteppa I guess this is fair
 
@ShaVuklia It's one of those rarely brough up thing in physics
The linking of the mathematical formalism to the actual experiments
The rules of correspondance, as they are called
 
@RyanUnger ask your new cool friends what they think of those books
 
@RyanUnger Ask your cool new friends to scan that thesis
 
@Slereah alumni library privileges give you a way to get it?
 
4:57 PM
Evening.
 
Who knows, but I tried and I really can't get it
Short of stealing it
 
I think the only thing that confuses me now, is what to do with $\hbar$
 
What's the word for the thing people vote in? Like, the word poll is only for elections. What's the word for any topic?
 
we can say choose time and length and mass as your fundamental dimensions, but what do we do with $\hbar$?
 
4:58 PM
From $\psi \approx e^{iS/\hbar}$ we see $\hbar$ has units of energy times time so that the argument of the exponential is dimensionless since the action has units of energy times time
 
yea, I'm convinced between the relation of $\hbar$ and energy and time
 
via $S = \int [T - V] dt $
 
but I feel like we have one degree of freedom left
 
:52262680 So if I vote whether Justin Bieber to wear his pants up or down, the form will be called an election, not a poll or something else?
 
4:59 PM
If $\hat H=i\hbar\partial/\partial t$, then we either have to choose energy or $\hbar$ as a fundamental quantity, no?
 
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