@Черенки I guess you should start reading relativity seriously. Your question requires a very long answer. BTW, I don't think I can help much either (because I haven't read much beyond Lorentz transforms)
Is this legal to post my physics question here for verification only but not for answer?
I just want to make sure that my question is on-topic, not off-topic, and fits the rules, like it's not too broad and etc.
I have already read all the "How to ask" sections on the help center and then in ph...
@Черенки Start with Shankar's lectures. I found those very helpful. For book recommendations you can check the corresponding MIT OCW syllabus page for SR.
@0celóñe7 No idea what "relatively open" is supposed to mean in that context. Generally I'd consider "for $Y\subset Z$, $X\subset Y$ is relatively open if $X = U\cap Y$ for some open set $U\subset Z$", but that doesn't seem to be the case here
@AccidentalFourierTransform I do not have 20 reputation points to talk in the chat unfortunately :( So I am writing here. OK, fine, I do not care if what I described really is a classical field theory or not. What I am asking is whether "A classical theory plus a vacuum state plus creation and annihilation operators plus Born's rule" can be defined as QFT such that other results of QFT essentially are derived from this definition. That's all I ask really. If this question is a misunderstanding, then I will ask separate questions, dividing them into chunks. — Lucia Guzheim24 hours ago
@Черенки I don't know what level you're at, but a couple of years ago, one of my friends worked on this for people transitioning from A level to undergrad physics
@ACuriousMind The 'I' evolves with time, so the 'I' at one moment is not the same 'I' as the next, hence the 'I' at the moment that was typed will never exist again ;)
@Mithrandir24601 Continuity of identity is a tricky question. If the I at one moment is not the same as the I at the next, then what does the phrase "the I" even denote?
@Mithrandir24601 Ah, then we end up at the question why we would consider $I_1(t_1)$ and $I_1(t_2)$ "the same" but markedly different from $I_2(t_2)$, where $I_i$ are the identity functions associated to different brains.
@Mithrandir24601 Ah, "the same" was a poor choice of phrasing. We generally feel that all the $I_1(t)$ are closer "related" than $I_2$. If you don't hold that view. If you disavow that view, then you're off the hook.
The other day someone from the WSJ asked me to explain why quantum mechanics is weird.
I went off on a rather well-reasoned and carefully articulated rant about how it's not weird and that people only say that to sound cool.
I even went so far as to say that Feynman's famous statement that {anyone who thinks they understand quantum mechanics is wrong} is destructive, stupid, and should not be repeated.
@ACuriousMind it's continuous in one direction (time), but discrete in another (different brains) - as such, some $I_j\left( t_1\right)$s are closer to $I_i\left( t_1\right)$ than $I_i\left( t_2\right)$
Apparently the movie player wasn't working. My dad noticed that the plane was full of physicists and commented to Feynman that you'd think on a plane full of physicists you'd have someone who could fix a video player.
@DanielSank I never understood why he made these strange claims, given that there's that famous "magnets" interview where he pretty much says that he can't explain the fundamental theory of magnetism without the other person knowing a bunch of prerequisites/math.
When I make a matrix of eigenvectors to change the basis of the matrix, does the non-commutative nature of matrices make how I make the matrix of eigenvectors important? As in, one column before the other or vice versa