Is it a misnomer when an electrostatics problem specifies a conductor to have $\rho$ distributed uniformly throughout

shouldn't it just be $\sigma$ spread across the surface area

\sigma for surface charge

oh right

and you're more-or-less right i'd say---i'd say it's just "charge" which is uniformly distributed, and for a surface that's a uniform sigma

hmm okay. So the picture should still be charge on the surface. It might just be a way for the question to not "give away" that fact.

right

or shorthand for it

sometimes we get so used to saying it the quick way that we forget that it's not obvious

Even in the classical picture, it's not implied that $\mathbf{E}=0$ in a conductor btw

well, you do have to say something about electrostatic equilibrium

it has to be assumed since it's not impossible for charge to exist along the axes/points of symmetry for the conductor

i.e., a single point charge in the center of a sphere, or a continuous line of charge along the cylindrical axis of symmetry for a cylindrical conductor

true

there's probably some semantics about what the right 'definition' is

the formal definition i'd probably fall back on is "a conductor is what you get if you start with a dielectric material and take $\epsilon\to \infty$."

@SillyGoose

oh i think the implication was that it wasn't a conductor but an insulator btw

also that is adorable @Relativisticcucumber

@RyderRude the domain of the functor is the group. It isn't the category of groups

Any group is a category with a single object

@Obliv in reality it is very logical in electrostatics. The definition of conductor is a material where charge is free to move, at equilibrium (which is were electrostatics works) no charge should be moving and for that to happen you need $\vec{E} = 0$ inside the conductor or else the charge would be free to move

For the level of the discourse it is extremely clear that it is some kind of misuderstanding. It should be an insulator. But when you learn further and need to model what the electrons inside a metal is doing, then we would treat all the freed electrons in a statistical thermodynamics model, separate from all the ions. The first approximation is then to smear the ions out into a uniform charge density background. Later, when you learn crystalline physics, then you can upgrade to crystalline

ionic background

@Relativisticcucumber amazing

XD sidney coleman just lighting up a cigarette at the beginning of lecture: youtube.com/watch?v=mnlHLd3UtYA

@alam Hi :-)

Hi:)

Can I move this to the problem solving room? We prefer the chat here to be about general principles rather than specific problems.

Ok

→ 5 messages moved to Problem Solving Strategies

is this an appropriate writing down of the Hamiltonian of an ideal scattering experiment?

i remark that I do not explicitly define what the Hamiltonian is between the asymptotic behavior and the bounds of the finite interaction interval $\mathcal{I}$

@alam hi

@VincentThacker so the map between objects is just $F(SO(2))=R^2$, right? I'm using $SO(2)$ to denote the one object of the group

i understand the map between morphisms.. the map between objects is weird becuz ive never seen a map of this form

i am also wondering if this is an appropriate interpretation of the $S$ matirx

in QFT we have these dirac deltas pop up that also encode conservation of (insert quantity) to my understanding

bleb well i think sakurai is saying that $1 - blah$ is the amplitude for no scattering and $blah$ is the amplitude for scattering

@SillyGoose yes

i don't understand how this quantity can be a distribution? at what point did we turn it into a distribution... it should just be an inner product i would have thought and so a complex number

but it is expected to be a distribution becuz it shud b delta in the absence of scattering

i should clarify that the incoming and outgoing state at this point is assumed to be a particle in a length $L$ box for now, but i guess i see your point once we take the $L \to \infty$ limit

okay i guess that makes sense. if the incoming and outgoing waves are plane waves then surely the $S$-matrix is not a discretely indexed matrix (in the momentum basis)

okay so it is really an artifact of the idealized set up

and then when we move into a real picture maybe we integrate twice: once over $k$ and once over $k'$ to get our packets

but the momentum space is discrete for qft in a box

i guess it depends on how u r setting things up

hm well i do have momentum space as being discrete at this point

so i have the kronecker delta indexed by $k$, and the dirac delta just appears from evaluating the integral in (72)

oh

but i think what you said makes sense bc ultimately i will take $L \to \infty$

that is how sakurai sets it up

so the $S$-matrix here is really like an $S$-matrix density or something

@RyderRude Yes

Is there any modern physics book which is not just plug the numbers and get the answer( I have books like Arthur beiser and paul tipler but they just throw formulas and in example number that have to be plugged and get the answer)

@SillyGoose the integral $\int dt'\mathrm{e}^{i\omega_{kk'}t'}$ is not convergent in the ordinary sense

It's the integral representation of the Dirac delta

More formally, you would understand this by learning about Fourier transforms of distributions