The h Bar

19:01

I dont think such a rule exists

ACuriousMind

@Sanjana My argument above shows why $\lvert \sin(x)\rvert$ is purely even. The half-sine wave is just $\frac{1}{2}(\lvert \sin(x) \rvert + \sin(x))$, so its Fourier series differs from the $\lvert \sin(x)\rvert$ only by an order 1 sine term and a factor of 2 in the cosine terms.

naturallyInconsistent

ahhhh

Sanjana

Noticed that...but was looking for a generic rule. But okay, good enough.
Can you tell me something about the $|x|$?

ACuriousMind

what about it?

$\lvert x\rvert$ is not a periodic function, it doesn't have a unique Fourier series

Sanjana

Oh sorry, I forgot to mention I meant consider $|x|$ in the range $[-\pi,\pi]$ and then periodically continue it. I mean consider the function to be defined on $S^1$.
Then it has a Fourier series expansion like $$|x|=\frac{\pi}{2}-\frac{1}{\pi} \sum_{n \in \text{even no.s}} \frac{\cos(nx)}{n^2}$$

Same question: Why only evens? Is it answerable in the same way as above?

@ACuriousMind Using this?

naturallyInconsistent

19:13

Maybe consider the integral / derivatives. |x| is really a sawtooth wave, and its derivative is the square wave.

Obliv

19:23

ACuriousMind

@Sanjana 1. Your Fourier series is wrong, it should contain purely odd terms. 2. Since $\lvert x\rvert$ is even, it just comes from the Fourier coefficients of $x$ on $[0,\pi]$, and it so happens that that Fourier series is purely odd.

Obliv

Does the last statement imply there would be a diffraction pattern without a slit? That feels wrong

I guess what does "If one dispenses with the slit in Figure 4 and merely assumes an
original beam of constant irradiance across a finite width b, all our results follow in the same way." mean

ACuriousMind

@Obliv "That feels wrong" is not a physical argument :P

Sanjana

@ACuriousMind 1. Oh crap...I mistyped it. 2. Why does the coefficients coming from that of $x$ on $[0,\pi]$ imply that the $n \in \text{odd no.s}$?

Obliv

well, is a laser a collimated beam?

idk what even is a collimated beam

do they naturally occur lol (i believe they are basically plane waves)

ACuriousMind

19:30

@Sanjana $f(x) = x$ is odd, and the Fourier series of real and odd functions have only odd coefficients

There's really just a bunch of simple rules here, you just need to be a bit creative in combining them

Sanjana

@ACuriousMind I thought $f(x)$ is odd implies $b_n=0$ and leaves the $n$s of $a_n$ unrestricted?

ACuriousMind

ah, sorry, I mixed up the two notions of "odd"

sine/cosine series are annoying :P

Sanjana

@ACuriousMind True :(

@ACuriousMind But thanks for this...This was really helpful

Jagerber48

@Obliv I think this is getting at the idea that beams diffract whether or not there are obstructions in the way. But I don't really like talking about "diffraction" as a concept at all. Waves follow the wave equation, always. When a wave encounters obstructions it has certain non-particle like behaviors, but when the wave is in free space it ALSO has non-particle like behaviors. You can call these both diffraction if you like.

In practice there is no such thing as a collimated beams.

ACuriousMind

@Sanjana Here's the argument for the $\lvert x\rvert$: If we have $f(x) + f(x+L/2) = c$ for some constant $c$, then we have $a_n + (-1)^n a_n = c_n$, where $c_n$ are the Fourier coefficients for $c$, which are non-zero only for odd $n$. So this shows $a_n = 0$ for $n$ even.

Jagerber48

19:40

Laser beams are Gaussian beams and no Gaussian beam has a fixed radius as a function of its propagation distance. However, if the distance between your observation point at the waist of the Gaussian beam is much less than the Rayleigh range for the beam then it will be approximately "collimated" (meaning the waist isn't changing as a function of z)

Plane waves have a fixed radius (infinite) as a function of propagation distance but since they're spread over all space they require infinite energy. There are also things called Bessel beams which have a fixed profile as a function of propagation distance, but it turns out that these beams also require infinite energy.

Obliv

"Even highly collimated laser beams are subject to beam spreading as
they propagate, due to diffraction. It is a fundamental consequence of the
wave nature of light that beams of finite transverse extent must spread as
they propagate."

Sanjana

@ACuriousMind Ah! I almost thought about that

Jagerber48

The transverse profile of any finite-energy beam will change as a function of propagation distance.

Obliv

is there some deeper reasoning behind this? @Jagerber48

Sanjana

Here was my argument. I can generalise this by adding a constant which you did for the $|x|$ case. Thanks

Jagerber48

19:42

@Obliv The deepest reasoning that I can give you is that it is a consequence of the wave equation that governs the propagation of light. In the paraxial case this equation is the paraxial wave equation.

Next comment won't be more satisfying, but it may be interesting: Incidentally I think the paraxial wave equation is the same as the Schrodinger equation. So the reason a Gaussian beam spreads as it propagates is the same reason a Gaussian wavefunction spreads as a function of time.

Obliv

what's the paraxial wave equation? The gaussian beam one?

Jagerber48

Check out this book: onlinelibrary.wiley.com/doi/book/10.1002/0471213748

Obliv

thank you i will look into it

Jagerber48

I'm not sure I'll get the progression right. We have electric and magnetic fields, these are vector-valued fields at each point in space. Maxwell's equations tell us how the electric and magnetic fields behave. In particular, we see that there are certain solutions of Maxwells equations (radiation) which solve the vector Helmholtz equation. This is a vector differential equation

I think you can separate the vector Helmholtz equation into scalar Helmholtz equations which are scalar differential equations. But then, if you have a solution where the field is field is not varying rapidly on the length scale of a single wavelength then you the field approximately solves the paraxial Helmholtz equation, or paraxial wave equation.

Helmholtz equation

In mathematics, the Helmholtz equation is the eigenvalue problem for the Laplace operator. It corresponds to the linear partial differential equation where ∇2 is the Laplace operator, k2 is the eigenvalue, and f is the (eigen)function. When the equation is applied to waves, k is known as the wave number. The Helmholtz equation has a variety of applications in physics and other sciences, including the wave equation, the diffusion equation, and the Schrödinger equation for a free particle. == Motivation and uses == The Helmholtz equation often arises in the study of physical problems invol...

When you do "beam optics" in the lab you are in a situation 99% of the time where you can use the paraxial wave equation to describe your laser beams.

ACuriousMind

at Obliv's level I'd not overkill this with actually solving wave equations: The reason a finite beam will inevitably spread is just Huygens' principle, and for a point at the boundary of the beam the waves emitted from it cannot be reached (and hence not be cancelled) in the direction orthogonal to the beam axis

Jagerber48

19:50

For example, the Gaussian beam solution comes out of paraxial wave equation

@ACuriousMind Yeah that's probably better than what I said. I've always wanted to see (or do for myself) a derivation of Huygen's principle from the Helmholtz equation.

I think it's just right there in the Green's function or whatever. I just haven't ever thought hard about it.

I guess with that explanation I wonder why some beams can converge before they diverge though. I think from the heat equation I have intuition about how heat can spread. But the wave equation (and Huygen's principle) allows interference to happen which allows beams to diverge, converge, and interfere.

Obliv

i just know the general form $\nabla^2 \psi = \frac{1}{v^2}\frac{\partial^2 \psi}{\partial t^2}$

but i'm more than happy to have explanations that are beyond my level. it helps organize ideas in my head

haven't heard of scalar differential equations before

Jagerber48

@Obliv Yeah, that's the scalar wave equation. From that you can get the Helmholtz equation then the paraxial Helmholtz equation. I'd need to pull that book out, but I think you can see without too much trouble that beams should not stay constant as a function of propagation distance from the paraxial equation, so it could be a good place to go to see if it will help. See e.g. en.wikipedia.org/wiki/Fresnel_diffraction which that Wiki page says is an exact solution to the Paraxial eqn.

Obliv

I see it says it's the solution to the Helmholtz equation, but I don't see paraxial equation

Jagerber48

en.wikipedia.org/wiki/Helmholtz_equation#Paraxial_approximation I was just quoting/paraphrasing the bottom of this section of this page

Obliv

yeah that's not easy to follow but i get the general idea

Jagerber48

20:06

On the spinor stuff, here's the new story I want to tell. We have a manifold. That manifold has a tangent bundle from which we can construct tensor bundles. Now say we have a group that acts on the manifold. That group has representations. The tangent/tensor bundles on the manifold allow us to "attach" these representations to the manifold as some type of bundle (something something principal bundle).

The easy way to attach vectors to spacetime (like the electric field) would be to say it is a type of tangent bundle. But if we want to be tricky, we could say that vectors fields are spin-1 representation of SO(3) (or Spin(3)) that we attach to the manifold

We could then say that (Pauli) spinors are spin-1/2 representations of Spin(3) that we attach to the manifold in the same way

This gives us a parallel treatment between vectors and spinors...

ACuriousMind

@Jagerber48 No, it doesn't, because the correct notion of "attaching" a bundle to the manifold in the way tangent vectors are attached is through the soldering form whose definition requires the tangent bundle to be defined independently

the point is that the tangent bundle/frame bundle of a manifold is really special

and yes, spinors are attached by looking at the orthonormal frame bundle and looking for a lift to a Spin(n) bundle, the spinors live then in associated bundles to that Spin(n) bundle, but there are topological obstructions for this to exist as I've already mentioned - every manifold has tangent vectors, not every manifold has spinors

Jagerber48

@ACuriousMind I'm not sure this goes against my story? In my story I did say the tangent bundle is defined first, and we use that to attach what I'm calling "representation" bundles to the manifold. Then I'm saying that vector fields that we encounter in physics can be considered a spin-1 representations that we attach as "representation" bundles

Rather than taking the obvious strategy of directly constructing them as a type of tangent bundle

ACuriousMind

I find that rather convoluted

Jagerber48

It was inspired by AccidentalFourierTranform's answer to my question: physics.stackexchange.com/a/793366/128186

I find it convoluted too, but if it's coherent it's towards a more satisfactory explanation for me I think.

The story I'm telling is my formalization of when physicists say "an X is something that transforms like an X". It means there is a spacetime manifold, a group acts on that manifold, and we can "attach" spaces that also transform under that group (e.g. vector or tensors spaces or spinor spaces), but in different ways.

ACuriousMind

@Jagerber48 This is only correct when the group in question is a spacetime symmetry; for internal symmetries there is no group action on the manifold itself

Jagerber48

20:19

Ah that's a good point.

But wait.. here's my problem.

I want to be able to apply the same group action to the manifold, to attached vector spaces, and to attached spinor spaces all at once.

Because when people say something like "if you perform a rotation then ... spinors behave a certain way" or something.

That has to mean you're rotating the manifold/coordinates on the manifold and spinors transform in some corresponding way. Like.. if the manifold isn't transforming I can't understand how the mathematical transformation could be seen as physically meaningful

ACuriousMind

@Jagerber48 that's why the frame bundle is special: The Jacobians of diffeomorphisms (transformations of the manifold) give you an element of the frame bundle at every point, which then gives you an element of the Spin(n) bundle as the covering of the frame bundle, and this then acts on all the things in a consistent manner

naturallyInconsistent

Well, you can have a gauge transformation that the internal space is symmetric around, but isn't tied to the manifold's transformations

Jagerber48

@ACuriousMind Ok.. so I think my story is still correct I just needed to be more precise? The story is: I have a manifold, I have some groups, I use those groups to attach representations of the groups to the manifold. I now apply some group to the manifold, this... gives me something about the frame bundle... which I then use to transform the attached vectors/tensors/spinors?

@naturallyInconsistent This is getting way out of my wheelhouse, but aren't gauge transformations SPECIFICALLY without physical consequence?

naturallyInconsistent

Yes, they are specifically without physical consequence, but we describe electroweak and fundamental strong interactions with U(1)xSU(2) and SU(3) respectively, and those are not transforming when you do a spacetime transform.

ACuriousMind

@Jagerber48 I mean you don't have random groups, they have to be related to the frame bundle (e.g. SO(p,q) is just the fiber of the orthonormal frame bundle) in order for diffeomorphisms to induce transformations of the vectors in that way

and you get the spinors by lifting the covering map from Spin(p,q) to SO(p,q) to a map of a Spin(p,q)-bundle covering the SO(p,q) bundle (which can be obstructed)

Jagerber48

20:31

@ACuriousMind Ok, but that's kind of the thing no one says that I've been missing! The story I've heard about spinors is like... we have rotations and vectors, and those are fine. But there's this weird group called Spin(3). We can make representations of that and define spinor fields. But my question is like: Why? Let me just pick some other crazy group that transforms however I like and make fields of that.

But the key (to helping with my difficulty) is that there is a relationship between transformations on the manifold and transformations on the attached spaces.

ACuriousMind

@Jagerber48 oh I think in that case your confusion lies in an entirely different place:

4 hours ago, by ACuriousMind

and the whole underlying reason spinors appear in quantum theories but not classical theories is representation-theoretic, too: All the proper linear representations of SO(n) arise as tensor products of vectors, so you don't need anything except vectors to build all the classical objects, but in the quantum theory, we have to admit projective representations, which includes spinors

^that is the reason we need to look at universal covers and their representations to begin with

Jagerber48

What should I research to learn about why we have to admit projective representations in quantum theory

ACuriousMind

for starters, look at this Q&A of mine :P

John Zimmerman

Why don't physicists study $M^{n,k}$ (minkowski space in arbitrary dimensions of space and time)

Jagerber48

This is interesting, and I think I want to understand it as further motivation. But I think the thing that has bothered me about "an X is something that transforms like X" is "what are the limits on how we can define things in this manner". And for that I think I just needed a better understanding of how, when we attach things to the manifold, there is still a relationship between transformations on the manifold and transformations of what we attach.

That is, I can attach whatever representations I like to the manifold so long as I can coherently relate transformations on the manifold to transformations on the representations I attach. I think this is related to something that had been discussed earlier about how you can't attach spinors to all manifold because they don't all permit the right structure.

ACuriousMind

20:42

well that's just because there are topological obstructions to there being a Spin(p,q)-bundle covering the frame bundle, c. en.wikipedia.org/wiki/…

@JohnZimmerman because more than one time dimension messes up the evolution equations; there's no well-posed Cauchy problems in more than one time dimension

and fundamentally physics is about predicting "the future" from "the present"

John Zimmerman

@ACuriousMind I don't understand why I can't just specify Cauchy data on all the time variables

ACuriousMind

@JohnZimmerman In more than one time dimension, you always have CTCs even in flat Minkowski space

John Zimmerman

Critical theory catastrophe's?

closed timelike curves

oh okay

looking at this purely mathematically is this studied do you think?

ACuriousMind

is what studied?

John Zimmerman

I'm going to think about this more before I say anything else

but the CTC in Minkowski space makes sense to me

I mean that what you said about the CTC in Minkowski space for multiple time variables makes sense

Obliv

21:34

what is meant by "missing from the pattern"?

is it not still in the pattern but the irradiance at that point is 0?

oh I guess I can see why it's called missing. You'd expect a bright spot somewhere in either pattern but it's not there due to the interference of both

@ACuriousmind how did you keep up with your classes in undergrad/grad school were you just always ahead because you liked learning in your spare time?

It sucks because I'm enjoying learning thru my textbook but i don't think it's at the correct pace (cramming before the final) but during the semester i had a million and one reasons to get distracted

what is an angular breadth?

ACuriousMind

22:00

@Obliv I'm afraid I actually mostly learned by listening to the lectures :P

Obliv

waaat. So you didn't have to sift through textbooks or online notes to understand concepts as much?

i guess if the professor does a good enough job it's not necessary ;O

i kinda like the feeling of intimately engaging with a textbook it's like reading a novel except the story is derivation of mathematical models governing physical systems

i kinda regret talking so badly of this textbook at the beginning of the semester. it might seem dense but it's all more or less self contained if you care enough to learn

Obliv

22:26

eh nvm i won't complain. it's worded fine

user587860

22:52

I am depressed again :(

user587860

Is it really impossible for a web programming AA student to apply to US universities to study Physics? Bolbteppa provided valuable insights about this matter, albet it was a very negative answer. But I'm just unable to control my thoughts that I am worthless and a failure

Obliv

whats AA student

user587860

AA student means a student who's enrolled in a 2-year college

Obliv

are you located within the US

user587860

No, I am in Turkey

user587860

22:55

I have portraits of Newton, Einstein, and Witten on the rooms of my wall and I am crying everyday by looking at them. Because I want to do physics like them

Obliv

I have a cousin in turkey, he's gonna study engineering in japan. I don't think it's impossible by any means.

I'm also turkish (was born there and came to america at age 4)

user587860

Wait a second, seriously?

user587860

That's fantastic

Obliv

It depends on your goals and financial situation. If you can afford the out of state tuition costs, or moving to the US and waiting until you have in state tuition status then you can attend most state universities

user587860

Yeah, that's the problem. Universities in the US, except the top ones, do not provide need-based financial aid for international students

Obliv

22:57

at my community college before, there were tons of international students. The idea is you apply to a 4 year after you get your associates. You can apply for scholarships, grants, etc even as a foreign student

I'd definitely try to speak to financial aid specialists at schools of interest. Visiting their website & talking to people is the best way. Helps if your english is solid

Another issue I see is that once people finish their studies, they need to update their immigration visa so they can continue to stay. that part i'm not too familiar with

user587860

@Obliv What about top places like MIT?

Obliv

@Supersymmetry this is a semi-recent state legislation that came out and helps people like myself to afford school. If you can figure out what it takes to be an NJ resident, you could potentially qualify. (Your citizenship status doesn't matter)

I have no idea, I'm only familiar with NJ schools. You can check out their website and talk to people at the relevant offices

user587860

Okay, I do not want to put laughably difficult goals for myself, especially given the situation I am unfortunately in. But I also do not want to give up and try my best to change something in my life

Obliv

You can do it if you really want to. I think the hardest part is the immigration stuff, but if you ever meet an obstacle try to get more info by talking to people. Often times we give up even though things are possibly within reach..

Although, there are probably plenty of good physics programs in europe and outside of the US. I guess applying to schools comes first, then thinking about making the arrangements

user587860

Yeah, what matters is to get admitted first. If there's no admission, there's no need to worry about residency permit and so on.

Obliv

23:16

but what I'm saying is you can be admitted to basically every 2 year institution, assuming you can afford out of state tuition. If not, you can just find a place to live for X amount of time to get in state status, then apply to said universities @Supersymmetry

like the thing I linked makes community college & state university free if you can gain residential status (and make under 65k a year, otherwise there will be a small cost that scales with your income)

but the issue is getting the visa to be allowed to stay in the US without going to school or having a job that sponsors your visa.

if you could get the student visa by applying to a community college, you'd qualify for residency status in maybe a year and then the rest of school would be free.

user587860

@Obliv Yes, but what I do not understand is, if you're granted an admission to a top college in the US, then they explicitly state that there's no issue with residency permit unless you've studied more than two and half a semester at the college from which you're applying to the university. Then, you're no longer eligible for Bachelor programme.

user587860

Like, if such an issue could happen, how would other international admits attend the places they got admitted to?

Obliv

I have no idea what that means.. you're saying they don't accept transfers?

user587860

Apologies for the confusion. No, I am suggesting that if, for instance, Harvard admits an international student in X country, who's enrolled at a university for no more than two and half a semester...

user587860

Then they're obviously aware of the possible issues with VISA residency permit, or whatever it is called. So why would they offer an applicant admission who cannot attend there for such reasons? As far as I know, they do not even ask questions in their application form, related to residency permit.

user587860

23:32

I'm just trying to suggest that getting admitted to a school in the US can be used to get residency permit

Obliv

(sorry for lack of proper accents) Ben demek istedim: New jerseya gelebilsen student visala or baska visala, bir kac ay/yil sonra "NJ resident" oluyorsun ve o sekilde bu "garden state guarantee" ve community college bedava oluyor (or cok indirim veriyorlar <$100k kazansan). Sen sadece yuksek universitylere gecmek istiyorsan directly, private ve state school degisik policyler oluyor. For example: princeton does not accept community college transfers (and maybe all international tranfers idk)

yes if you get admitted and can get a student visa, then after some time you get state residency status. Then, maybe it will be cheaper. (I'm speculating, maybe based on your visa status the discount doesn't count) but you should ask financial aid people at schools

user587860

@Obliv Your Turkish is fine :) Evet, anlıyorum.

user587860

But given that it's extremely competitive to get an admission, especially for someone in my situation, is there a path that you can suggest I follow? To at least give myself the best chance

Obliv

ilk once dusunmen lazim budu: Para biriktirmek istiyorsan, belki en zor girmeyi okularinla baslamicaksin. Yani, scholarship vermeseler, hepsini kendin oduceksin. Belki daha rahat olur community collegea gecsen ve sonradan transfer yapmayi. Ama visa situation bilmiyorum nasil oluyor. Sana kac yil visa vericekler ve community college icin veriyormula arastirsan

en iyi okulara gecmek zorunda degilsin, mence. Physicsni seviyorsan ve cok okusan, state university de calisabilirsin belki. Burda bir kac tane orenciler biliyorum ve onlar senin kadar bilmiyorlar mence.

ve onlar PhD yapiyorlar

user587860

PhD yapanlar benim kadar bilmiyor mu diyorsun?

Obliv

23:42

seninle cok konusmadim ama string theory hakkinda konusdugun icin I think so

there is a wide range of students here in the US. People at the lower end of the spectrum like myself barely understand diff. eq. and then there are people who are learning diff. geometry in middleschool/high school and both ends of the spectrum attend top schools and mid schools

user587860

Ben de diff geo yaptim high schoolda

user587860

@Obliv Riemannian geometry

user587860

Kendime self-teach ettim bu stufflari hep

Obliv

adamin ismini soylemek istemiyorum ama ben lise bitirdim saatda bu websiteni buldum ve o burda hep diff geo riemannian geometry sorular ve problemler hakkinda konusuyordu. Ben hic bilmedim cocuklar bu seyler oreniyordu. Senin gibi kendi orenmis middleschool/high school da

ve bir normal university gidiyordu engineering icin ve herkes soyluyordu senin potential cok buyukdu, math veya physics yapsan daha cok basarlicaksin. He is at princeton now doing phd..

sizin gibi insanlari cok merak ediyorum

this is good practice for my turkish lol it's so bad

i'd be so screwed if i went back (i keep thinking in both karachay and turkish)

user587860

@Obliv No, your Turkish is really great, and we can continue discussing in this language if you prefer.

user587860

23:56

Evet, benim gibi daha early study etmis insanlar var mutlaka. Ama ben de maalesef cesitli durumlardan dolayi, istedigim bolume (fizige) ve universiteye giremedim. Bu da beni cok depresif yapiyo

user587860

Finansal durumum da cok iyi degil

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