The h Bar

Mr. Feynman

15:27

@RyderRude Well, at least the war between you two is funny

Obliv

15:52

Why is the classical wave eq. in the form $v^2\frac{\partial^2 y}{\partial x^2}=\frac{\partial^2 y}{\partial t^2}$ space & time invariant, but $\frac{\partial y}{\partial t} = -v\frac{\partial y}{\partial x}$ isn't? Is it because the signs are different

that right side eq. is used in deriving the left side form so I don't get why they don't have the same properties

Obliv

16:06

nvm found an answer

naturallyInconsistent

16:33

@nickbros123 there are very nice treatments of them

Slereah

16:54

ncatlab.org/schreiber/files/dcct170811.pdf

Looks like the Schreiber masterpiece is going along fine

Why does he have an abstract, it's a thousand pages

is that supposed to be a paper

Silly Goose

17:10

When talking about anyons in QFT do we lose access to algebras?

ACuriousMind

@SillyGoose what

Slereah

yeah a cop comes by and take away all the algebras

He is saying that EM is a $(\mathbb{R} / \mathbb{Z})$-principal connection

Silly Goose

Slereah

Pompous

Too good to say a circle

Silly Goose

so the paper there talks about defining "anyonic comm relations" by basically picking up some prescribed phase if you switch two creation operators (for example). at least this is to my understanding

ACuriousMind

17:20

yes, anyons are defined by the phase they pick up under exchange

I don't see what this has to do with "losing access to algebras", nor what that even means :P

Slereah

it is still very much an algebra

Algebra-er, if anything

Silly Goose

so i mean this looks like you have some vector space $V(\mathbb{F})$ and some elements $A, B \in V$ and you define a map $\chi: V \times V \to V$ by $\chi(A, B) = [A, B] + (1 - e^{i\varphi(A, B)})BA$

but i mean this is not bilinear in general

hence, this map would not define an algebra in general

ACuriousMind

why would you look at that weird map or demand that it's bilinear

Silly Goose

that weird map is exactly the condition that a phase is picked up after exchange

ACuriousMind

a map is not a condition

Silly Goose

17:23

a vector space with a bilinear map (satisfying other conditions) becomes an algebra. this occurs in the usual case of bosons (to my understanding).

ACuriousMind

I'm afraid I'm not following you at all

We have the algebra of c/a operators, which is generated by the c/a operators $a,a^\dagger$. In the bosonic case, $aa^\dagger = a^\dagger a$, in the fermionic case $aa^\dagger = -a^\dagger a$. In the anyonic case, $aa^\dagger = \mathrm{e}^{\mathrm{i}\phi}a^\dagger a $ for some $\phi \neq 1, \phi \neq -1$.

Silly Goose

what is the map that turns the set of c/a operators into an algebra

ACuriousMind

I don't understand the question

Slereah

Braiding map

errr

That's the thing that turns it into anyons

The map that turns it to an algebra is just the composition of maps?

ACuriousMind

it's an algebra, there is some abstract multiplication operation between them; if you want to be needlessly formal about it we're taking the free algebra generated by two elements and quotienting out any of the respective commutation relations above to obtain the bosonic/fermionic/anyonic c/a algebra

Silly Goose

17:27

i mean any mathematical structure is with respect to the things defining said structure. if i have a topological space, it is a set together with a topology. it is the combination which makes the structure a topological space. in the case of an algebra, the underlying vector space is more clear, but it is unclear what bilinear map is being placed atop the underlying vector space to make it an algebra

Slereah

$a$ and $a^\dagger$ are linear maps on a Hilbert space, you can compose them

The "product" is map composition

this is still true for the anyonic case

Silly Goose

but i mean this is not what we look for representations of

so maybe i am quite confused. in QM for instance, we have algebras of operators defined by commutation relations: e.g. $[\sigma_i, \sigma_j] = 2i\epsilon_{ijk}\sigma_k$.

Then we look for representations of this algebra

not necessarily of the algebra defined by the vector space of pauli matrices together with (say) matrix multiplication as a way to product them

i thought analogously, the algebra we are interested in is not this "ambient algebra" defined by producting c/a operators, but the algebra defined by their characteristic commutation relations.

ACuriousMind

again, the formal way to say how such relations define an algebra is to take the free algebra and quotient the relations you want to hold out of it

Silly Goose

okay so my question is how to understand that quotienting the relation for anyons still results in an algebra again

ACuriousMind

it's just a relation

you can quotient out arbitrary algebraic relations out of the free algebra and get an algebra in which that relation holds

there's nothing special about commutation or anti-commutation relations in this respect

Slereah

17:39

they don't even have to have any fancy relation under exchange

ACuriousMind

you are perhaps confusing this with Lie algebras where of course it is special that we can't actually take products of the elements themselves but only have the Lie bracket, but here we are only considering an algebra of operators that does not need to be a Lie algebra

Obliv

17:55

What's wrong with wave equations of the form $f(x^2-vt)$ or $f(x^2-vt^2)$, In classical mechanics, frequency isn't the energy of the wave, so this shouldn't violate energy conservation right

Slereah

If you plug it in the wave equation you get $$x^2 f'' + f' - \ddot{f}$$

Obliv

Do we use $f(x-vt)$ because it leads to the 2nd order diff. form of the classical wave eq. which is symmetric in space/time so that we could describe light, and particles with schrodinger's equation? Since to get non-rel. one-dim. time-independent schrodinger's equation we input DeBroglie wavelength into a classical wave, take two time derivatives to get momentum^2 then input that into the classical energy conservation law.

Slereah

not really something very wave equation-like

We mostly care about waves because of the wave equation

Obliv

@Slereah in theory, you could model certain things with that eq. without breaking any laws tho pre-schrodinger/qm right

Slereah

Since it is the evolution for a lot of interesting systems

You could have weirder equations with propagating solutions, of course

like something like that idk

Telegrapher's equations

The telegrapher's equations (or just telegraph equations) are a set of two coupled, linear equations that predict the voltage and current distributions on a linear electrical transmission line. The equations are important because they allow transmission lines to be analyzed using circuit theory. The equations and their solutions are applicable from 0 Hz (i.e. direct current) to frequencies at which the transmission line structure can support higher order non-TEM modes.: 282–286 The equations can be expressed in both the time domain and the frequency domain. In the time domain the independe...

It's still propagating but not quite like the wave equation

or this

Korteweg–De Vries equation

In mathematics, the Korteweg–De Vries (KdV) equation is a partial differential equation (PDE) which serves as a mathematical model of waves on shallow water surfaces. It is particularly notable as the prototypical example of an integrable PDE and exhibits many of the expected behaviors for an integrable PDE, such as a large number of explicit solutions, in particular soliton solutions, and an infinite number of conserved quantities, despite the nonlinearity which typically renders PDEs intractable. The KdV can be solved by the inverse scattering method (ISM). In fact, Gardner, Greene, Kruskal and...

woosh

A lot of modelling of real macroscopic "waves" tends to not obey the wave equation

Obliv

18:04

If I wanted to derive wave equations of the most general form in 3 spatial, 1 temporal, dimensions, what would I need? Before ever postulating physical constraints like energy conservation etc

Slereah

because there are messy things

@Obliv what do you call a "wave"

Obliv

something with spatial and temporal periods

Slereah

I mean the solutions I posted don't even have periods

Obliv

like, $f:\mathbb{R}^4\to\mathbb{R}^4$ defined by $f(x,y,z,t)=...$ something

oh

Slereah

Do you just mean "periodic functions"

Because that's just the theory of Fourier series

Obliv

18:06

Well, I don't think I want only periodic functions I want to understand in the most general sense, I suppose. So diff. equations that aren't spatially and temporally symmetric?

would be that most general form? otherwise they'd necessarily be periodic?

Slereah

So just functions that aren't homogeneous?

Obliv

something that propogates with time I guess, so not even waves

Slereah

that sounds pretty wide as a class :p

You'd be better off just classifying functions that are homogeneous

Obliv

Eh I forgot what homogeneous means. I think when you have only derivatives and no constants in the diff eq

That changes a lot I presume?

Yeah I will just study some general forms of diff eqs and solutions so I can get a better idea of the physics of classical/qm wave functions

like what the forms signify

PM 2Ring

@Obliv Here are some cute wave things from Greg Egan. gregegan.net/APPLETS/07/07.html & gregegan.net/APPLETS/20/20.html

nickbros123

18:17

@naturallyInconsistent Im sure there are. But I thought I would do it after getting a decent grounding on analysis and algebra, perhaps from Loomis and Sternberg, or even Hubbard and Hubbard. Also, I was making a joke :)

user430580

18:28

You should all study gravity, I heard that is a field with potential.

Obliv

18:46

Not sure if I'm doing this right.

I said: We have $\psi_1(0) = Ae^{k(0)}=A$, $\psi_1'(0)=kAe^{k(0)}=kA$, $\psi_2(0) = B(0)^2+C(0)+D=D$, $\psi_2'(0)=2B(0)+C=C$. Since we require $\psi'$ continuous at finite potential discontinuities, we require $\psi_1'(0)=\psi_2'(0) \implies kA = C$. Since $\psi$ must be everywhere continuous, we have $\psi_1(0)=\psi_2(0) \implies A = D$. So we can rewrite $\psi_2(x) = Bx^2+kAx+A = Ae^{kx}$ thus $B = A(e^{kx}-k+1)$

to normalize the total solution, we just take $\int\limits_{-\infty}^{\infty}\psi^2dx = 1$ and plug in $\psi$ for each discontinuity?

No, I'd do it just for $\psi_1$ to get $A$ in terms of $k$ and then plug that in for the 2nd psi, 3rd psi is 0 so doesn't matter

Obliv

19:01

I can't tell if this is just an error, or what, but they're writing the intervals for the wave function as closed even though it's unbounded on both sides

like, $x\in [-\infty,\infty]$

Mr. Feynman

19:43

@Obliv physics books :P

Relativisticcucumber

@ACuriousMind struggles. hmmm. because $\nabla ^2(\frac{1}{r}) = \frac{1}{r^2}\frac{\partial}{\partial r}(-\frac{r^2}{r^2}) = -4\pi \delta(r)$ and here im not allowed to "reduce" r's. so why can i reduce r's in other places?? clearly i dont understand.

ACuriousMind

@Relativisticcucumber The first equality is simply not true at $r=0$

Relativisticcucumber

@ACuriousMind oh no how can that be

ACuriousMind

the rule you applied there is for differentiable functions, but $\frac{1}{r}$ is not a differentiable function at $r=0$

Relativisticcucumber

thus the confusion begins. ok i see so the issue is not reducing r's. its differentiating nondifferentiable things?

ACuriousMind

19:50

and everywhere for $r\neq 0$ it's true

Relativisticcucumber

but $\frac{r^2}{r} = r$ also doesnt make sense at $r=0$

ACuriousMind

@Relativisticcucumber yes, the problem is that the notion of derivative used so that $\nabla^2(\frac{1}{r}) = -4\pi\delta(r)$ is true is not the ordinary notion of derivative, but the physicists don't tell you that because they're afraid of math :P

@Relativisticcucumber that's because that's an equality of functions, and neither side is defined at $r=0$. In contrast, $\Delta \frac{1}{r} = -4\pi\delta$ is an equality of distributions, which properly mathematically doesn't even make sense to evaluate at single points because the r.h.s. is not a function

Relativisticcucumber

@ACuriousMind i am still rather confused. because what i did was try to take the laplacian of a scalar potential so how is it even correct to just throw a distribution in here

this is one of those problems where i get the right answer yet i have zero understanding of the material behind the problem

sad

ACuriousMind

to be honest I think most people who use the equality $\Delta\frac{1}{r} = -4\pi\delta$ in this context don't have any more understanding than you do, either

mathematically, this is a distributional derivative

now, you might ask why it's okay to just say that if we in physics always pretend to work with functions, but the fact is that if you really think about what a density - such as the r.h.s. in the EM Poisson equation $\Delta \phi = \rho$ - is, it's a distribution because the point value of a density at a point is actually irrelevant physically, what matters is its integrals over volumes - in distributional terms, the application of the distribution to the indicator functions of those volumes.

Relativisticcucumber

the thing im sus about is that to solve the problem i do $\nabla \frac{1}{r}$. this seems to be saying the laplacian of $\frac{1}{r}$ is problematic but the gradient is not??

ACuriousMind

20:05

it's also sus, the problem is that you want to have thing this be "valid" in some sense at $r=0$, but in the sense of ordinary functions that derivative does not exist at $r=0$ because the function isn't even defined there

the typical physics approach is just to compute stuff and show it works without explaining how this derivative can mathematically even exist; but it's not as if physicists ever wonder about stuff like limits actually existing in any other place so at least it's consistent :P

Relativisticcucumber

20:18

god

physics pains me

Claudio

you're gonna face this problem if you read about central potentials problems in Cohen-T, they have something about this in the appendix, but the answers on MSE are much better imo

Or even better, you might wanna study distribution theory

and then try to fully understand this answer. I haven't tried yet but it's on my list of things to do in the future

@Relativisticcucumber lol

Relativisticcucumber

this is less bad than my condensed matter prof today. he did a massive cringe

i need to find the expression but it was canceling a term and a sum of that term

maybe it was correct somehow. i need to explore the notes.

it was this expression:

canceled the c on first term with c's in second term

sobs how can i make sense of this

Claudio

20:36

lol Im in pain for you :|

good luck tho

Vincent Thacker

21:26

@Relativisticcucumber Distributions like the dirac delta are defined by how they act in integrals

So the proof of the equality of distributions is integrating over the volume containing the origin

user430580

23:17

A cop stopped Heisenberg when he was driving one day and said "Sir, do you realize that you were doing exactly 90 miles an hour?". Heisenberg replies "Oh no, now I have no idea where I was".

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