> 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.’”
@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.
@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?
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 $A:\mathbb{Z}\to\mathbb{R}, j\mapsto A(k_j)$.
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 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.
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!
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
@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 $\int_{-\infty}^{\infty} A(k) e^{i(kx-\omega t)}dk$?
@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.
@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.
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.
@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!
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 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
