5 ms·
Thanks Professor Hossenfelder. You mentioned that the amateur physicists don't know enough math. Well, some of us amateur physicists have concluded that the ac
by graycat 5y ago
Thanks Professor Hossenfelder.
You mentioned that the amateur physicists don't know enough math. Well, some of us amateur physicists have concluded that the academic physicists don't know enough math.
E.g., when I was trying to be a ugrad physics major, the prof in the E&M course tried to prove the divergence theorem. His proof was, in one word, junk. Sadly as a ungrad math major, the math department didn't prove it either. By now, sure, I've gone through fairly careful treatments including the Jacobian, ..., exterior algebra of differential forms (texts by Fleming, Rudin, Apostol, ...). Even some of those treatments are a bit fishy in places and incomplete: Physicists keep using old versions of differentials, and the math books got rid of those 50+ years ago; clean ups are not too difficult but need to be done by either physics or math.
The Riemann integral over the whole real line is fishy (smells like old fish). The improper integral approach of integrating from -a to a and then taking the limit as a goes to infinity is also fishy. There is a good clean up -- the Lebesgue theory.
Back to the divergence theorem, physics wants to have a 2 dimensional measure they write as dA and a three dimensional measure they write as dV, but there is a gap in their argument: They are necessarily working with what measure theory calls product measure and in particular Fubini's theorem, and there needs to be some orthogonality that is not made clear, in either math or physics.
I'm tired of screaming myself to a sore throat over the claims that the wave functions of quantum mechanics form a Hilbert space. Hilbert space is a math idea, and in math departments we know quite well what a Hilbert space is, a complete inner product space. The "complete" part is in the sense of Cauchy, that every Cauchy convergent sequence converges. Well, the wave functions fail that completeness assumption.
You mentioned that you can tell backgrounds just from use of language: Right, so can math people! So, when an MIT physics guy teaches quantum mechanics and says that "the wave functions are differentiable and also continuous" we know his background is in physics, not math -- EVERY differentiable function is continuous.
It goes on this way: Net, to us math people, it is the physics people who are short on the math.
Then beyond the math, some of us amateurs have some other issues with physics:
Physics in quantum mechanics keeps talking about superposition and linearity. Us math people know in fine detail just what linearity is -- it's central to linear algebra, Banach space, Hilbert space, integration, differentiation, and more. I have to question if the physics people have a clear understanding of linearity. Then there is superposition -- tough even to make a mathematically credible guess just what physics means by superposition.
The last time I was reading from Feynman's Lectures on Physics I recall, before I wanted to get my money back, that he claimed that a particle of unknown position had probability distribution uniform over all of space. WRONG! It's a trivial argument, maybe even at the middle school level, to show that no such distribution can exist.
Then setting aside the math and going for something like actual physics, a particle goes through a beam splitter such as the one in the Mach-Zehnder interferometer. On one path from the beam splitter there is a detector. We send in one particle. The beam splitter yields the wave function in two physically separated parts on two very different paths; we believe this from our experience with the full Mach-Zehnder interferometer.
This trial we get a signal from the detector. Okay, but what happens to the other part of the wave function? The quantum mechanics lectures and texts don't say.
Professor Hossenfelder, some of us amateur physics people are actually math people (e.g., with a Ph.D. degree from a world famous research university and have published peer-reviewed original research in math with theorems and proofs) and look at physics people as very short on math and also sometimes even their physics. Uh, for $50 an hour, I will tutor you in linearity, Lebesgue theory, Hilbert space, and probability theory (based on sigma algebras).
- ncmncm 5y agoGood one. But if you really do want to get the maths onto firm footing, maybe you should be working to encode it all under Lean.
- Koshkin 5y ago> WRONG! All caps. You have earned 5 points! [0] [0] https://math.ucr.edu/home/baez/crackpot.html https://math.ucr.edu/home/baez/crackpot.html
- graycat 5y agoSince I am 100% correct about a statement of Feynman, I have in particular shown that the crackpot theory is wrong, a crackpot in itself!
- hither_shores 5y ago> Back to the divergence theorem, physics wants to have a 2 dimensional measure they write as dA and a three dimensional measure they write as dV, but there is a gap in their argument: They are necessarily working with what measure theory calls product measure and in particular Fubini's theorem, and there needs to be some orthogonality that is not made clear, in either math or physics. Those are differential forms, defined explicitly in terms of submanifolds of euclidean space. Turning them into measures on the corresponding manifolds requires picking an arbitrary "unit speed" parameterization (or equivalent data) to absorb the equally arbitrary (from the intrinsic perspective) choice of embedding. Sticking with differential forms and letting smoothness assumptions handle all the measure-theoretic difficulties for you is the principled choice and also the easy choice, so it should come as no surprise when physicists choose it. > Physics in quantum mechanics keeps talking about superposition and linearity. Us math people know in fine detail just what linearity is -- it's central to linear algebra, Banach space, Hilbert space, integration, differentiation, and more. I have to question if the physics people have a clear understanding of linearity. Then there is superposition -- tough even to make a mathematically credible guess just what physics means by superposition. obeys the superposition principle = has dynamics given by a linear operator. Or sometimes a bounded linear operator, continuous linear operator, etc. depending on context. Physics is no different from math in this respect. > I'm tired of screaming myself to a sore throat over the claims that the wave functions of quantum mechanics form a Hilbert space. Hilbert space is a math idea, and in math departments we know quite well what a Hilbert space is, a complete inner product space. The "complete" part is in the sense of Cauchy, that every Cauchy convergent sequence converges. Well, the wave functions fail that completeness assumption. State spaces in finite dimensional QM are projective Hilbert spaces, in a perfectly literal sense, with all the completeness properties you expect. You're right that intro QM mischaracterizes it though: the position (wlog momentum) "basis" is actually defined with respect to a rigged Hilbert space structure Psi \subset H \subset Psi^*. This is also the appropriate approach for the infinite dimensional case ... and a completely inappropriate approach for teaching freshmen. > The last time I was reading from Feynman's Lectures on Physics I recall, before I wanted to get my money back, that he claimed that a particle of unknown position had probability distribution uniform over all of space. WRONG! It's a trivial argument, maybe even at the middle school level, to show that no such distribution can exist. Again, he's teaching freshmen: don't confuse lies-to-children with genuine misunderstanding. Physicists are well aware that plane waves are not actual states.