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12:21 AM
@JMac I think it's just because of the dog pictures
The pictures really help understand the physics too
 
 
3 hours later…
3:48 AM
Which identities are to be used to show the equivalence between the two expressions of $v_{av}$?
 
y'all see the new way eigenvectors can now be derived from eigenvalues...how has something like that not been discovered in 150 years yo...
 
@enumaris chat.stackexchange.com/transcript/message/52564016#52564016 Apparently in Tao's blog someone actually pointed out an unpublished manuscript from 2014 that showed the same arxiv.org/abs/1401.4580
 
4:33 AM
even if it's the same, that's still 2014
this shoulda been found like 1926 or 1880 or something wtf?
maybe 1950-1980's latest?
 
 
3 hours later…
7:33 AM
> Parke said. “I’m expecting someday to get an email from somebody that says, ‘If you look at this obscure paper by [the 19th-century mathematician] Cauchy, in the third appendix in a footnote, it’s there.’”
 
 
2 hours later…
9:20 AM
whoa
the 2010s will come to an end in a month
soon enough we'll be hearing kids say "i was born in the 20s"
 
10:02 AM
After the government prints money, where do they go? To the central bank/s and to borrowers?
 
 
2 hours later…
11:53 AM
@NovaliumCompany Country-dependent. For example, in the USA, not the government prints money, but a private company, in private hands. In my home country, the government prints the cash, and it can go in the first round either to the central bank or to private banks.
@NovaliumCompany I am not sure, how does it go. Afaik it does not go to the central bank. The important thing (who has how many cash by whom) happens electronically, and printed cash moves only if it is needed.
 
@SirCumference Why aren't kids born in 2000 called millenials
 
12:10 PM
hello
I have trick question
for example the equation of rigid rotor is
$$\psi=ke^{i \phi}\sin \theta$$
then is it possible to find out l by inspection
m is 1 over here
or do I need to use the energy operator
 
 
2 hours later…
2:17 PM
@NovaliumCompany You live in bolivia? How's your new Prez..? Heard a lot of unrest there.
@NovaliumCompany Inflation is not good, but money devaluing is good, an unethical technique called, Currency Manipulation to increase the value of goods in the country by devaluing the money, currently seen in larger terms in the trade war
@NovaliumCompany Forex reserve market.
@AvnishKabaj Are both $\theta$ and $\phi$ variables?
 
2:56 PM
If I ask you out on a date, would your answer be the same as your answer to this question?
 
 
1 hour later…
4:03 PM
wtf
 
 
2 hours later…
6:30 PM
How is the integral that represents a wave packet , that is $\int_{-\infty}^{\infty} A(k) e^{i(kx-\omega t)}dk$ equivalent, or equal, to the sum of different sinusoidal waves with different wave numbers and amplitude, that is $A_1 e^{i(k_1x-\omega_1 t)}+A_2 e^{i(k_2x-\omega_2 t)}+...$?
If the integral is a Riemann integral, one can write this as something like $\sum_{j=1}^{n} A(k_j) e^{i(k_jx-\omega_j t)}$, but what would one take the limit of and where is the $\Delta k_j$ term in the sum that becomes $dk$?
And how does one interpret the infinite limits in the integral?
 
@schn There's nothing particularly "physicsy" here, it's just math: Both your expressions are Fourier transforms (+ the $\mathrm{e}^{\mathrm{i}\omega t}$ term from the Schrödinger equation). In one case you're Fourier transforming the function on the reals $A : \mathbb{R}\to\mathbb{R}, k\mapsto A(k)$, in the other a discrete function $A:\mathbb{Z}\to\mathbb{R}, j\mapsto A(k_j)$.
 
@ACuriousMind Is it possible to write $\int_{-\infty}^{\infty} A(k) e^{i(kx-\omega t)}dk$ out as a sum?
 
No
Integrals are not sums, they are generalizations of sums.
 
It could be a limit of a sum, no?
 
Depending on your definition of the integral (Riemann, Lebesgue,...), it is, yes.
 
6:41 PM
What kind of integral is this then? (Riemann, Lebesgue,...)
 
The physicist typically doesn't care much, but the Lebesgue integral plays well with the Fourier transform in general. But this is a mathematical detail usually irrelevant to the physics.
The pragmatic answer is that it's just an integral, use whatever definition of it you're most comfortable with.
(If you are not comfortable with any, it may be best to first learn the math before diving into the physics)
 
@ACuriousMind wha n-no
rigor needs to be a thing
otherwise the model becomes less useful
 
@ACuriousMind It is just unclear how $\int_{-\infty}^{\infty} A(k) e^{i(kx-\omega t)}dk$ is equal to $A_1 e^{i(k_1x-\omega_1 t)}+A_2 e^{i(k_2x-\omega_2 t)}+...$, when, if we treat the integral as the limit of a Riemann sum, it can't be written as such.
 
@schn Who claims they're "equal"?
 
though for most QM purposes you deal with "nice" (Riemann integrable) functions
 
6:47 PM
@SirCumference I'm all for rigor in the right places, but physics is full of integrals that aren't well-defined in any approach!
 
@ACuriousMind i know, i see them and i think what's the point
like if we wanna ditch rigor then we're turning into philosophy
 
@ACuriousMind Isn't $\int_{-\infty}^{\infty} A(k) e^{i(kx-\omega t)}dk$ representing the superposing of many sinusoidal wave functions?
 
@schn Yes - uncountably many, which is why you can't write it as a sum (sums are usually only countable).
 
less rigor necessarily means less accuracy and more handwaviness
@schn well that's the motivation yes
heuristically they're weighted by a $dk$
 
@SirCumference Unlike philosophy, we still have experiment to ground us.
 
6:50 PM
@ACuriousMind experiment can't verify theory if the theory is vaguely defined :P
 
People don't use "ill-defined" path integrals because of their philosophical pleasantness, but because they work.
 
well i guess it could if i don't think too much on what the theory is saying, but eh
kind of defeats the purpose
 
From mathematical breakthroughs like the Donaldson invariant to countless QFT predictions in HEP and CMT, they've shown that - regardless of how rigorous they may be - they often get the right result anyway.
 
as an example, if i don't define my terminology then i can more or less handwavily say my experimental results match up with my theoretical predictions
less rigor necessarily implies our theory makes vaguer assertions
 
@SirCumference I'm not interesting in debating the virtue of "rigor" in the abstract.
As a matter of fact, QFT is not fully rigorous, but is the most accurate theory of the universe we have. It does not make "vague assertions", it computes the values of experimentally accessible quantities to unmatched precision!
 
6:54 PM
welp i guess it's just me but when the logic and theory is removed from physics in favor of ad hoc arguments, the interesting part of the subject is destroyed
 
hello
 
@ACuriousMind Mathematically speaking, and in the case of the Riemann integral, one only writes the integral notation if the sum has a finite limit as the mesh size approaches zero, correct? Hence shouldn't there be a mathematical expression that comes before one writes $\int_{-\infty}^{\infty} A(k) e^{i(kx-\omega t)}dk$?
 
shouldn't we put a force greater than gravity's to actually move the object?
so shouldn't the work be greater than that?
 
@ACuriousMind If I assert "this integral equals x" without stating what type of integral I'm using, it's by definition a vague assertion :P
yeah you can get away with it but it can still use some more precision
 
@schn Sure, it's a limit of Riemann sums, cf. Wikipedia
The integral is defined to be that limit, so that we don't have to write it out every time.
 
6:57 PM
@ACuriousMind Correct, so there should be a Riemann sum associated with it, right?
 
@schn that riemann sum would be the fourier series iirc
 
@SirCumference But a physical theory does not make assertions about the values of random integrals! It makes assertions about the behaviour of the natural world - the path integrals are tools to arrive at the assertions, but not part of the assertions themselves.
@schn I don't understand the question, I'm afraid.
 
@ACuriousMind Physical theories define many concepts mathematically. There's probably a good number of observables or functions or etc. that are defined in terms of integration
 
A rigorous physical theory is a beautiful thing, but the fact of the matter is that the rigor is almost always developed in hindsight, not while the theory is made.
 
Integration was more or less an example, but in a theory built with mathematics, precision is pretty important
 
7:01 PM
@SirCumference How would the Fourier series representing the Riemann sum associated with $\int_{-\infty}^{\infty} A(k) e^{i(kx-\omega t)}dk$ look like?
 
Newtonian mechanics wasn't rigorous when it was invented - the tools to do rigorous analysis didn't even exist yet! QM wasn't rigorous when it was invented, neither was GR (Einstein famously said that he didn't understand GR anymore after the mathematicians invaded it, after all!) and neither is QFT.
 
@schn here are my notes i wrote a couple of years ago on the topic
they may not be the best
 
Look, I like rigor! But I do not believe that it is good to pretend that mathematically rigorous physics is always superior to more traditional physics done with a good measure of intuition. Sometimes, rigor is necessary to avoid us going astray - but sometimes it just slows us down. There's is an art in choosing the right place to be rigorous.
 
@ACuriousMind yeah, but now that the tools do exist, it seems illogical to ignore them when we can make our theory more precise
Mathematics is built on logic. It doesn't require you to throw away intuition :P
In fact practically every concept is a formalization of an intuitive notion
Nothing's defined for the sake of it
 
@SirCumference Do you actually know what rigorous QM looks like? I understand very well e.g. why introductory physics courses do not do rigorous QM in infinite-dimensional spaces. The functional analytic machinery necessary for that would itself consume at least one semester - without doing any physics!
 
7:07 PM
@ACuriousMind No my QM course was crap in rigor
I get your point from a teaching perspective, but at the end of the day there ought to be a correct answer to whether we're using a Riemann or Lebesgue integral in a part of our theory (as an example)
 
@SirCumference @ACuriousMind Thanks for your time. Will look through the notes.
 
@SirCumference I agree - but that doesn't mean I have to insist on pointing that out every time I talk about it. You started this discussion when I answered a question (arguably a "teaching moment"!) and said it often didn't matter.
 
@ACuriousMind Well I didn't realize you were specifically using it for a teaching moment, I was caught up in the statement "The physicist typically doesn't care much...this is a mathematical detail usually irrelevant to the physics." :P
 
Note the "typically" and "usually" - I didn't say it never mattered, or that we should never care about it.
 
7:27 PM
I was born to be alive.
 
8:05 PM
Who here exists?
 
Just give me an intuition how to approach Q4 part a
I know it is based on rule of mixtures
 
 
1 hour later…
9:18 PM
Who's impartial to a little Fierz'ing here and there...
 

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