@ACuriousMind Yeah, I have to read about the Moyal bracket in detail, but I did read your answer. It seemed to say more that "dequantization o quantization = identity" than what "dequantization" should be.

There should be something interesting going on here, but I will say it after I have given more thought.

@bolbteppa Haha, this is smart. I think once you make this rigorous, you're saying any that the commutator of a pair of operators which are functional calculus of $x$ and $p$ has a formal Taylor expansion in $[x, p]$, which in the limit $\hbar \to 0$ goes to the Poisson bracket. OK, but you see why this is an artifact of the formal properties that the commutator and bracket share, and doesn't really explain the classical limit, yes?

Anyway, I come here with a bad question. Does the norm of the 4-acceleration have a nice formula I can just use?

$\sqrt{a^\mu a_\mu}$? :P

Lol

I'm not certain what you're looking for

I was hoping to not having to manipulate, this is cheat sheet stuff, I wasn't asking a trick question.

I get $A = (c\gamma\gamma', a\gamma^2 + v\gamma\gamma')$ I think

That has a terrifying norm so I don't want to manipulate

There must be someone who's written down a better expression in terms of just $x, x' =v, x'' = a$

@BalarkaSen the usual trick is to go to the frame where $a$ and $v$ are parallel

in that case some of the terms cancel and you're just left with $\gamma^3 \lvert a\rvert$

oh wait, you can't "go to that frame"

you just have to be lucky that this is the case :P

yeah seems useful anyway

but if $v$ and $a$ are randomly positioned, I don't think you have any chance at getting a "nice" expression

Yeah, doesn't seem nice

@Yashas are you here

Man, he isnt pingable

is there anyone here that can answer some extrememy basic machine learning doubts

@Slereah (sorry if its an unnecessary ping) but pls ping me if you are available for some help

@BalarkaSen not sure what you mean by 'doesn't really explain the classical limit', if one wants to know why $\hbar \to 0$ is taken as being the definition of the classical limit that can be explained too, but once you accept it then it's obvious that some equation like $\frac{d}{dt} \hat{f}(t) = \frac{i}{\hbar} [\hat{H}, \hat{f}]$ which describes a system with 'quantum turned on' is going to have to work when it's turned off which is taken as the $\hbar \to 0$ limit

$a^{\mu}$ only has a complicated formula because you've taken a covariant quantity and re-expressed it in three-dimensional form, it's $a^{\mu} = \frac{d^2x^{\mu}}{ds^2}$ in terms of a world-line arclength parametrization

@bolbteppa I know but that doesn't help in computations. Of course, $s = c \tau$, and if you unwind that in coordinates you'll get my expression.

Yeah it's a mess, makes electromagnetism a lot messier

@bolbteppa I mean to say that your computation is sort of tautological if you think about it.

It's clever, but

The reason you get the Poisson bracket when you expand $[f(x, p), g(x, p)]$ out in Taylor series on $[x, p] = i \hbar$ and take limit as $\hbar \to 0$ is because of the formal properties of the commutator as a Lie bracket

I will admit that your computation can be made 100% rigorous. It's just that after doing that it feels tautologically true, just because for both the Poisson bracket and the $i\hbar$-scaled commutator, $(x_i, x_j) = (p_i, p_j) = 0, (x_i, p_j) = \delta_{ij}$

I know I am not being precise about what kind of explanation I want. I can say more to explain what I want.

I feel like I can get away with not being precise since this is a physics chat :)

In Cantor's diagonal argument they say that s is not within the enumeration list $s_1,s_2,...s_n,...$ but how can they know that it isn't there?...It isn't a part of $s_1,s_2,...,s_n$ by construction but how can it be NOT one of the $s_{n+1},s_{n+2},...$?

By construction the $n$'th entry of $s_n$ is always different from the $n$'th entry of $s$ in the enumeration, and since the original set is countable it's elements can be listed, but we're listing them all and it's never in the list we're making, how can it be there if you never write it down

@bolbteppa The original set is countable but infinite, so we are not really writing all the elements...How do we know that it isn't there in the part which we haven't yet written down, i.e., after the the $n$'th entry of $s_n$?

What is meant by an abstract noun?

Could someone help me out?

@ManasDogra wait it's not countable sorry, this is proving that, if you begin assuming it's countable this shows by contradiction it isn't as the wiki states, the point is you can't count i.e. enumerate all it's elements

@bolbteppa When you said it's countable I assumed that you are talking of the preliminary assumption that it is countable(after which a contradiction is shown).That's why I agreed with you upon that it is countable.

yeah that's what I meant, you begin assuming it's a countable set, then because by the definition of the set of elements we're working with, we can also always choose a $s_n$ to have it's $n$'th element different from some $s$, but this can be iterated indefinitely to construct a listing in bijection with $\mathbb{N}$, however $s$ is never in this list, a contradiction if you assume the original set is countable

But what I don't understand is say take $s_1,s_2,...,s_n$ Now we can make a s by "complementing the diagonal values". It isn't one of the $s_i$s for i=1,2,...,n(by construction) What guarantees that it even isn't one of the $s_j$s for j=n+1,n+2,...

@bolbteppa I understood that in the sense of bijections with the set of natural numbers, but not the diagonal argument...although they are equivalent!

The point is your procedure generates a listing of elements of the set that can be put in bijection with $\mathbb{N}$, if you are able to stop at some point with a finite amount of them left then yeah the $s$ would have to be one of the remaining ones, but there isn't a finite amount of them left so you can always generate a list of elements which can be labelled by $\mathbb{N}$ which is what it means to count it's elements

@bolbteppa Give me a bit of time to understand this.

@bolbteppa "but there isn't a finite amount of them left" so why can't I conclude that my constructed s CAN BE in the set of what's already left(precisely infinite of them)?

You can and it is, but the original of elements is uncountable

If my s is in the set then why is it uncountable? I mean, we have merely made a s which isn't one of the first n elements of the enumerated list but not a s which isn't in the rest of the list...So why is it uncountable when I can always make a "big" list which contains the specially constructed s which wasn't a part of the first n elements of the previously enumerated list?

If possible explain without invoking the bijection with Natural numbers thing which is equivalent to but logically independent of the diagonal argument(is it? not sure)

Because the construction goes on forever and at each stage the new element we count is such that it disagrees with $s$, it means that $s$ is never in the list of elements and never will be, so the original set can't be counted because it has elements that can't be placed in the listing we're constructing, but we assumed that every element of the set could be put in a listing, this makes no sense unless the assumption is flawed to begin with

@bolbteppa Finally understood it. Thanks a lot

If you have a finite double well potential, can you write the solutions to the Schroedinger equation as superposition of two different square well potential wavefunctions?

You are allowed to rescale the masses

Hello World of Physiks..

It seems to me that physicists love to symmetrize products of non-commuting factors. Why is this often allowed or needed?

do you have a specific example in mind

I guess, is the answer very situation dependent? I have operators that don't commute when I introduce a magnetic field by minimal coupling. My book says that the solution is to write the products as a symmetric part and an antisymmetric part.

Hi, this question was asked before but I didn't actually get what I was looking for so I thought I'd post the question here instead.

Say person A just created a particle (high probability of one-particle state), is the probability of a very far away detector getting triggered (probability of finding a particle outside of its light cone) non-zero according to QFT?