@0celo7 Well, in the 1d case you're constructing the Lebesgue measure by declaring intervals $[a,b]$ to have measure $b-a$ basically. A countable set clearly has inner measure 0 then since you can't fit any non-trivial intervals inside it.
(I may not have used "inner measure" in its formal sense here, I hope you know what I'm trying to say)
@ACuriousMind you did use inner measure correctly btw
inner Lebesgue measure is generated by putting $(a,b)$ inside of things
@ACuriousMind Recall that one can define "measure zero" without measure theory. I want to see if one can prove that [0,1] does not have measure zero without mentioning measure theory.
@ACuriousMind Ah. One needs compactness. It's not true that $[0,1]\cap \Bbb Q$ is compact.
@BernardoMeurer What I have to show is that if $I$ is a closed interval, and $I_1,I_2,\dotsc$ form a cover, then $\sum_{j\ge 1}|I_j|\ge |I|$. That doesn't seem crazy.
@0celo7 If you're afraid of dying, go the ER. If you're whinging about your unmentionables for amusement value, stop. We can't help you here. As @ACuriousMind says, let's change the subject.
@ACuriousMind For brevity, let $A_q$ denote $A\cap\Bbb Q$, where $A\subset\Bbb R$. We say that $A\subset \Bbb Q$ has $\Bbb Q$-measure zero if for $\epsilon>0$ in $\Bbb Q$, there is an open covering of $A$ by rational intervals $I_j$ such that $\sum_{j\ge 1}|I_j|<\epsilon$. Does this seem reasonable?
Lift to $\Bbb R$; we obtain a covering of $(0,1)$ by intervals. Using standard techniques, we can show that the remaining sets have at least volume $1$. But this is unchanged in $\Bbb Q$, so we have $\sum_{i\ge 1}|I_i|\ge 1$.
"That time when you poop so much in night that it empties your stomach really bad and ur all tired so you lie in bed and the empty stomach seems like the best thing in the world and combined with the AC and the fact that you washed yourself with water and are feeling that good freezing chill thing makes it the best of the bestest moments of exsitence.." - Cactus
@MartianCactus Trying it is nice. Proving it mathematically is also nice, in case your plane mirror isn't in an infinitely-large room. Try it this way:
Figure out, for a concave curved mirror, what the condition is for the image to be inverted vs. upright. That you can test, using the curved mirror you probably have in your house.
Then flatten your curved mirror by making the radius of curvature arbitrarily large, and see whether it's still possible to produce the inverted image.
For a bonus you can also show that a convex mirror always makes upright virtual images of real objects, and flatten that one, too.
idk yet how to do tiff with mirrors in real life, I only know how to fraw ray diagrams and stuff as we haven't yet been taken to the lab(next grade which will be starting in a month will do that)
@MartianCactus Not your first language, no worries at all.
@0celo7 I never climbed the trees around Perkins much because I wasn't on that side of the hill. I used to spend a lot of time in the one by the Nielsen upper-level entrance, and in a few of the super-climbable ones on the north side of the hill, below Ayres.
@0celo7 You used to have to jump to get to the lowest branch, but not far. But it could be that the limb I used has been pruned since the last time I was there.
@MartianCactus For reflections, some of the rays may cross at places other than the images.
That's part of the reason that some people teach lenses before mirrors: there's a "goes into" side and a "goes out of" side.
If you do the three principal rays (goes in parallel, goes out parallel, goes through optical center), then all three should meet in exactly one place.
Well two points: they all start from the object, and all (perhaps virtually) pass through the image.
@bolbteppa Let's run through it for $[0,1]$. Cover $[0,1]$ by open intervals $\{I_j\}_{j\ge 1}$. Since $[0,1]$ is compact, $I_1,\dotsc, I_N$ covers. Agreed?
For $I=(a,b)$, write $|I|=b-a$. We want to show that $\sum_{i=1}^N|I_j|\ge 1$.
This is pretty much clear, because the union of $I_j$'s contains some $(-\epsilon,1+\epsilon)$ since $[0,1]$ is also connected
So $\sum_{j\ge 1}|I_j|\ge 1$, so you can't make it $<\epsilon$.
The cover $\{I_j\}$ does cover $[0,1] \cap \mathbb{Q}$, but so what? The measure of the cover of $[0,1] \cap \mathbb{Q}$ is different to the measure of $[0,1] \cap \mathbb{Q}$.
anecdote of the day: Enrico Fermi did not usually take notes, but during the 1948 Pocono conference he took voluminous notes during Julian Schwinger’s lecture. When he got back to Chicago, he assembled a group consisting of two professors, Edward Teller and Gregory Wentzel, and four graduate students, Geoff Chew, Murph Goldberger, Marshall Rosenbluth, and Chen-Ning Yang (all to become major figures later).
The group met in Fermi’s office several times a week, a couple of hours each time, to try to figure out what Schwinger had done. After 6 weeks, everyone was exhausted. Then someone asked, “Didn’t Feynman also speak?” The three professors, who had attended the conference, said yes. But when pressed, not Fermi, nor Teller, nor Wentzel could recall what Feynman had said. All they remembered was his strange notation: p with a funny slash through it.
What I remember is that if you have a countable collection of real numbers, such as the rationals in $[0,1]$, then you can surround each of the points by open intervals of length $\varepsilon/2^n, n \geq 1$, so that the measure of the set is less than or equal to $\sum_{n=1}^{\infty} \frac{\varepsilon}{2^n} = \varepsilon$, since $ \varepsilon$ is arbitrary, your set is of measure zero, if you try to cover $[0,1]$ in this way $\varepsilon$ is not arbitrary.
@Runlikehell Weinberg is an excellent place to find new bits and insights if you already know QFT and can decipher his idiosyncratic notation. It's a terrible place if you want to learn it.
@0celo7 can you expand your argument about QFT being nonsense? I dislike it mainly cause i'm having an hard time transitioning from a bachelor in astrophysics to a master in theoretical physics, but me having an hard time is not enough to justify my frustration
@Runlikehell I think many people are having a hard time with QFT because it's not as settled as other subjects. Most of the stuff they learned before QFT is something we've known for a while to be basically "done", several iterations of books and papers went over refining those subjects and nowadays they can be taught in a more-or-less easily digestible manner by a competent lecturer
But QFT is not - we're still figuring it out, really. Don't let the fact that there are "textbooks" fool you, there are so many different approaches to QFT and everyone claims to have the right one but no one really has figured out how it "should" be done
@Runlikehell My frustration with QFT has little to do with it being difficult or poorly explains, it's that (interacting QFT in 4 dimensions) it is not well-defined.
How you get from normal QM to the number formalism and creation/annihilation operators and wave functions as linear combinations of operators is sincerely mental unless you read that chapter
It's not this neat thing of what a "physical theory" is in classical mechanics or hydrodynamics or whatever, it's a huge mess of different approaches and explanations and arguments and tools that somehow all fit together but how exactly is a matter of spirited debate.
If I have the Lagrangian of a system, derive the equations of motion, and then solve the coupled differential equations, what is left to do in order to find the normal modes and vibrational frequencies? The scenario is a double well with one particle in each. Both particles are connected by a spring. I could probably work through the details but I just need someone to point me in the correct general direction.
@0celo7 More than not well defined it looks like a mess to me so far. It is all formal manipulation, so it looks more abstract and less physical than anything i've studied so far, at the same time this manipulations are sometimes sick and they make it depart from mathematics too
@Runlikehell roughly, if you apply Heisenberg ("there is no concept of the path of a particle", ch. 1 sec. 1) to more than one identical particle, you get what Landau calls 'the principle of indistinguishability', and so you immediately see the notion of coordinates $x_1, x_2$ of multi-particle wave functions $\psi(x_1,x_2)$ etc... lose the relevance they once had
you need new less redundant variables, such as the number of particles in a given stationary state, for example. Hence the number formalism. To get it, you note that a Hamiltonian which acts on single particle wave functions at a time merely changes the stationary state the wave function was in, so QM becomes a theory of probability distributions for the numbers of particles in stationary states, and you can derive the explicit formalism basically from this calculation (if you push it)
If you follow that procedure through a little bit, you will explicitly derive the commutation relations for fields that QFT books postulate as axioms, and you can get both Fermi/Bose commutation relations directly, so you can see where QFT came from
