The h Bar

12:47 AM

To paraphrase Wilde, to lose one giant is unfortunate, to lose four is downright careless.

5:47 AM

@Slereah You were so right

Even the guy who gave the correct theory gave a couple of theories which didn't work out...See this-- inis.iaea.org/collection/NCLCollectionStore/_Public/46/035/…

The Nobel laureate confesses his own mistakes :)

Slereah

7:03 AM

I'm always amazed that physics exist really

Imagine looking at some spots on some photo film exposed to a rock and try to deduce that quantum mechanics exists

More Anonymous

8:29 AM

@Slereah Its the achievement of many great minds!

B. Brekke

I have some creation and annihilation operators and want to apply Wick's theorem. These are conventional second quantization operators in the Heisenberg picture. Do they have to be time-ordered or is this only the case for field operators?

Slereah

There is a standard convention for ladder operators to be time ordered, yes

which corresponds to the time ordering of the field operators

B. Brekke

Okay thanks, that sounds consistent

Slereah

Famously if you do not time order them, there will be a divergent term due to a Dirac function

Due to the commutator $$[a^\dagger, a] \approx \delta$$

evaluated at the same point this causes a divergence

This is equivalent to the Hamiltonian having an infinite sum of a constant term in less abstract methods

Physics Meta

8:52 AM

0

Q: Would original posts meant to share knowledge instead of ask questions be on topic on the site?

For a long time the way I tended to see the Schrodinger Equation written, some of the notation was like a foreign language to me, and so I couldn't work out how to actually use it to model anything, but then I saw it all the terms that were written out in what was a mathematical foreign language ...

Slereah

9:23 AM

"Marino has done a little research, and finds that Modi asked to join several monasteries, all of which rejected him because he didn't have a college degree"

Can't even join a cult without a degree these days

Slereah

10:03 AM

"It is a fun exercise to show that $f$ does not lie in any $1$-parameter subgroup of $\text{Diff}(S^1)$."

It's not a fun exercise at all!

Manas Dogra

I have seen there are some symmetries in the equation of motion which isn't present in the lagrangian...where do these extra symmetries come from?
Like galilean invariance in free particle EOM but no such invariance in the lagrangian..

bolbteppa

Harmonic oscillator eom $m \ddot{x} = - k x$ has a symmetry of the eom that is not a symmetry of the Lagrangian $L = \frac{1}{2} m \dot{x}^2 - \frac{1}{2} k x^2$, can you guess it

ACuriousMind

@ManasDogra Galilean invariance of the free particle is actually a (quasi-)symmetry of the action!

bolbteppa

$m \to 1/k$ and $k \to 1 / m$

ACuriousMind

@bolbteppa that's not a symmetry, symmetries act on the dynamical variables

but in the same spirit there is the symmetry of the e.o.m. $x\mapsto \lambda x$ which is not a symmetry of the action

Manas Dogra

10:16 AM

@bolbteppa scale transformations?

@ACuriousMind oh u said it :(

ACuriousMind

::rummages:: Qmechanic discusses various types of symmetries here

the symmetries of the e.o.m. don't really have any deeper meaning (though they can be useful in solving htem)

it's the quasi-symmetries of the action that are relevant for Noether's theorem etc.

Manas Dogra

But where from they are coming?

bolbteppa

It's not a symmetry of the Hamiltonian $H = \frac{p^2}{2m} + \frac{1}{2} k x^2$ either, but in the Hamiltonian formalism we can do canonical transformations, and $x \to p$, $p \to - x$ is a canonical transformation, so the Hamiltonian's are canonically equivalent and the physics is the same, which is what a symmetry encodes, just because it's different to a Lie symmetry of the coordinates and velocities doesn't mean it's not a symmetry, this is the idea behind T-duality symmetries

ACuriousMind

@bolbteppa that is an entirely different notion of "symmetry" - which is why we usually call these "dualities"!

@ManasDogra I don't understand the question

where does any symmetry "come from"?

Manas Dogra

I mean..we have a lagrangian and it does not have a particular symmetry..and the moment I put it in euler lagrange equation we get some extra symmetries..

bolbteppa

10:23 AM

@ACuriousMind it's still a symmetry, and refered to in the literature as a symmetry, in this example it's still a symmetry of the equations of motion that is not a symmetry of the Lagrangian which is what was asked for

Manas Dogra

Manipulating some equations without adding extra content should not give extra symmetries right?

ACuriousMind

@ManasDogra Why not? The functions in the E-L equations are combinations of derivatives of the Lagrangian, and a function and its derivatives can have different symmetries!

(just look at any polynomial $x^n$)

bolbteppa

@ManasDogra what about this: arxiv.org/abs/1705.08446

ACuriousMind

the point is that a symmetry of the e.o.m. and a symmetry of the action simply are two different things with different meanings

if you just call both "symmetry" it sure is mysterious why they're not the same, but that's a language problem, not a conceptual problem

bolbteppa

There's some good examples in there of a free particle (eq. 5, 6, 19, 20), I'm not sure what this means in general

Manas Dogra

10:41 AM

@ACuriousMind Ohhh okaayy...so it's the E-L who is giving these extra wings

@bolbteppa thanks I will have a look

@ACuriousMind about the charge density of a delta function in a moving frame which we discussed a few days earlier...since in the moving frame delta(x) is the charge density..should $\int q\delta(x)dV'$ give us the charge in the moving frame? (I think not cz the answer is not coming out to be q which should due to Lorentz invariance of the charge).

ACuriousMind

@ManasDogra What's the problem? Your charge density is $q\delta(x)$ in every frame, so regardless of frame you have $\int q\delta(x) = q$ as the total charge

Manas Dogra

The problem is the volume element is dV' the primed frame which is $dV/ \gamma$
What to do about the extra gamma?

ACuriousMind

whatever $\gamma$ you get from the $\delta$ has to cancel against the one you get from the $\mathrm{d}V$

Manas Dogra

@ACuriousMind Yes but there isn't one gamma with the $\delta$ because it is invariant in all frames as you said.

ACuriousMind

ah, I get the problem

@ManasDogra remember that $\rho\mapsto \gamma\rho$, right? This integral is not just an integral over a delta function, it's an integral over $\rho$

Manas Dogra

10:54 AM

@ACuriousMind Yes $\rho' dV'$ should give us q.

ACuriousMind

the thing is, when we say "the charge density is $q\delta(x)$ in every frame", we mean that the integral for total charge looks like $\int q\delta(x)\mathrm{d}V$ in every frame

i.e. this is just the same thing you already know - the transformation of the $\delta$ cancels against the transformation of the $\mathrm{d}V$

Ishwaran

Hello. I am having this doubt for a long time, when we draw image formation for various lenses and mirrors, we are taking 2 to 3 light rays passing through focus, parellel to incident ray and another one hitting the pole of the mirror, why we are taking 3 light rays like this ?. Are we checking Common case scenario ?

ACuriousMind

densities sort of live with an implied $\mathrm{d}V$ behind them, very similar to the frequent confusion about how the spectral densities in Planck's law look different for frequency and wavelength because there's an implied $\mathrm{d}\lambda$ or $\mathrm{d}\nu$ "behind" them

so when we say the charge density is $q\delta(x)$ everywhere, that already includes the transformation of the $\mathrm{d}V$, you don't need to transform it a second time

Manas Dogra

Tell me one thing if I have a function f(x)...and if I want to look at it from a boosted frame, I will see f'(x') right? not f(x') or f'(x), right?

@ACuriousMind Hmm I get it now...thing is our professor said that the answer should be delta(x)/gamma...he simply transformed the coordinates only and didn't care about the transformation of the density...saying that the transformation of coordinates is enough :)

Slereah

11:28 AM

If you do a coordinate transform then a value of the function at a point $x'$ will be $f'(x')$

$f'(x')$ is related to $f(x)$ via the transformation

Nihar Karve

in what sense are color charges and the electric charge related? The former does not appear in a gauge transformation of the quark field: $U(x)=\exp(i\lambda_a(x)T_a)$ whereas the latter does: $U(x)=\exp(-ine\lambda(x))$

Slereah

in a more abstract sense you have that the actual function is $f(p)$ with $p$ a spacetime point, and then you can express this in any coordinate system via $f(\phi^{-1}(x))$

Manas Dogra

@Slereah So if rho(x) is the charge density in unprimed frame...in the primed frame it will be rho'(x') right?

Slereah

Therefore you have that $$f(p) = f(\phi_1^{-1}(x)) = f(\phi_2^{-1}(x'))$$

ACuriousMind

@NiharKarve by "color charges" you mean stuff like "green", "blue" or "antiblue-red"?

Slereah

11:32 AM

that's what the relation between the two is

For compactness people usually get rid of the coordinate map

so $f \circ \phi_1 = f$ and $f \circ\phi_2 = f'$

Fairly obviously since they are homeomorphism you can change one into the other by $x = (\phi_2 \circ \phi_1^{-1})(x')$

That's the coordinate transformation

Nihar Karve

@ACuriousMind yes

ACuriousMind

@NiharKarve ah, then color "charge" is actually a misnomer - a better analogue to "electric charge" in the world of the strong force is the representation (fundamental, adjoint, etc.) of SU(3) something transforms in - since the electric charge of something classifies the U(1) representation it transforms in

but someone called the individual possible states in a given SU(3) rep "color charges", and the name stuck

Slereah

Also in a simpler way isn't it just a case of shoveling the constants in the function

$\lambda^a(x)$ is a somewhat arbitrary vector, you can put some "charges" in it if you want

Although... argh

I forget

ACuriousMind

@Slereah but the color charges aren't constants like $q$ for the electric charge!

Slereah

I don't do much QCD I'm afraid!

I guess it's hard when you don't have free quarks

ACuriousMind

11:39 AM

they're just labels "red blue green" for a possible basis of the fundamental rep, and eight labels like "red-antiblue" for a possible basis of the adjoint rep

Slereah

Aren't the red-green-blue of constant "norm" so to speak?

ACuriousMind

what norm?

Slereah

Not sure

Haven't done much QCD myself

ACuriousMind

I mean, sure this is a unitary rep so there's some norm

Slereah

Also isn't there a coupling constant for QCD?

Nihar Karve

11:41 AM

@ACuriousMind yes, that's what I was thinking. Is there a way to label all representations uniquely (Casimirs?)

ACuriousMind

but, again, color charge in the usual sense is not a number

Slereah

I know it's not a "constant" but still

Nihar Karve

yes

ACuriousMind

@NiharKarve the usual classification of SU(N) reps is via Young diagrams

Slereah

Bad news

They're very bad for SU(3)

Nihar Karve

11:42 AM

really?

Slereah

Well they are worse than SU(2) certainly

ACuriousMind

of course there's some Casimirs behind that but people tend to favour the diagrams and I trust this is because the Casimirs just aren't as neat as the spin number for su(2)

Nihar Karve

fair enough

Slereah

I'm sure SU(26) is even worse, but that's not often encountered

The diagrams give you the reps but not the Clebsch-Gordan coefficients, which is unfortunate

(That's what I had to find out last time I had to work with them)

ACuriousMind

I try not to get into situations where I need any actual values of CG coefficients, that's always a pain :P

Slereah

11:44 AM

I had to compute meson cross sections unfortunately :p

make little pions collide

ACuriousMind

poor pions

Nihar Karve

@Slereah i don't think they're really that much difficult provided you don't get into representations with lots of boxes

Slereah

Don't feel too bad for them they mostly get out untouched

Although I'm not 100% sure

I have suspiscions that my probabilities were wrong

Didn't add up to 1 IIRC, which is usually a bad sign

But I think $\pi + \pi \to \pi + \pi$ was pretty likely for all pions

The reps are classified by isospin and strangeness in the case of mesons IIRC, if that helps

You get those pretty diagrams

What's the Noether charge of QCD?

isn't there an equivalent?

6

A: What is the Noether charge associated with the the color $SU(3)$ symmetry of QCD?

The $\mathrm{SU}(3)$ gauge symmetry is a local symmetry, and therefore it is not Noether's first, but Noether's second theorem that applies to it, which does not yield conserved quantities. For $\mathrm{U}(1)$ gauge symmetries like the electromagnetic symmetry, there is also a global $\mathrm{U}...

nevermind @ACuriousMind already answered

Can't even measure the number of Gell-Manns of a quark

What role does the coupling constant play there though?

Nihar Karve

12:01 PM

in what?

Slereah

You could have some SU(3) associated field with one coupling constant and another one with a different coupling constant

analogue to electric charges

but maybe more analogous to the permittivity of free space

Nihar Karve

[deleted]

Slereah

$T_a$ is just a basis for $su(3)$ though

$g$ is an actual parameter of the theory

Nihar Karve

what is a basis of an Lie algebra

Slereah

Generators

ACuriousMind

12:06 PM

@NiharKarve just...a basis. It's a vector space, after all

bolbteppa

The basis of the underlying vector space

ACuriousMind

people often call that "a set of generators" or whatever, but it's really just a basis :P

Nihar Karve

what I meant was that the choice of $T_a$ determines the representation, which is why $T_a$ is just like the $n$ of electric charge

Slereah

I mean yeah but that doesn't influence the magnitude of the interaction

Nihar Karve

but it does influence the way the interaction works, no?

Slereah

12:08 PM

I mean sure

quarks and mesons and baryons behave differently

and they're different reps of $SU(3)$

ACuriousMind

@Slereah we have to be careful about what we mean by "electric charge" actually

Slereah

yeah I guess it's tricky to compare them directly here

ACuriousMind

I was thinking of "multiple of fundamental charge", but there's of course also a coupling constant of the field to something with that fundamental charge

the analogue to the latter is obviously just the QCD coupling constants

the analogue to the former is the representation

Slereah

If different flavors had different coupling constants you could say they have a different color charges

arxiv.org/pdf/hep-th/0607203.pdf

Classical QCD looks pretty ugly still

Ishwaran

Can anyone help with my question ?

Nihar Karve

12:16 PM

so I've still managed to confuse myself here: why isn't there a 1-dimensional "internal" electric space in the same way that there is a 3-dimensional internal space for fundamental quarks?

Slereah

What do you mean

$U(1)$ is 1-dimensional

it acts on the 1-dimensional space $\mathbb{R}$

Well $\mathbb{C}$

If you have a complex scalar field, the $U(1)$ symmetry will mean that the configurations of the field related by a phase are identical

Nihar Karve

I think I got it - all members of the vector space $\mathbb C^*$ are gauge-equivalent to each other, unlike in the case of $SU(3)$ on $\mathrm{GL}(3,\mathbb C)$

yeah there you go

ACuriousMind

@NiharKarve there is an internal 1d space, you just don't see it!

@NiharKarve no, this is not true - the global U(1) symmetry is not gauge

Slereah

The same is true of $SU(3)$ except it is a bit more complicated since most vector spaces won't admit a non-trivial rep of it

So obviously you're gonna see $\mathbb{C}^3$ more often

But most fields you can work a $U(1)$ symmetry on since it's just phase-related

So it's a bit less obvious

ACuriousMind

The thing is that you see the 3d internal space because it gives you an index on the field - if you have a field transforming in a lorentz rep $V$ and in internal space $I$, then the total space is $V\otimes I$

but if $I$ is 1d, then $V\otimes I = V$

Slereah

12:26 PM

You can put an index on $U(1)$ gauge stuff of course

ACuriousMind

and so you just get the electromagnetic symmetry acting as a phase on the original $V$

Slereah

But the simplest basis for it is $1$

It's not the most exciting

Well, if you want to make it look better

$$T_0 = \begin{pmatrix} 1\\ \end{pmatrix}$$

If you're working with a scalar field, it will apply itself to the vector space $U(1) \times \mathbb{C}$, so essentially $e^{i\alpha(x)} \phi(x)$

But of course that's the same function space

Nihar Karve

@ACuriousMind got it, thanks

Slereah

it does mean that you can't have a charged real scalar field though

bolbteppa

1

A: What exactly is the color charge in QCD?

Yes, color charge has nothing to do with "colors". The naming is indeed arbitrary; we could have called them yellow, blue, and cyan, or A, B, and C. What's important is that there are 3 of them. As to your first question: the term "charge" carries a slightly different meaning in the context of ...

It's a good question that's probably very trivial, why isn't there an "$e$"-type thing in the exponential $U(x)=\exp(i\lambda_a(x)T_a)$ like there is in $U(x)=\exp(-ine\lambda(x))$, the end of that is giving some of an answer but I don't see why

Nihar Karve

12:39 PM

well you can just rescale your parameters $\lambda_a$ or just set $g=1$, nothing too deep

bolbteppa

I'll say Griffiths does write one in

Slereah

@NiharKarve What if you have more than one coupling constant though

Nihar Karve

@bolbteppa in those conventions it would indeed be $\exp(-ig\lambda^a(x)T^a)$

Slereah

It's hard to think about this properly because I guess I am imagining two little classical quarks in a scattering scenario

with a little gluon field around them

But I'm guessing that's probably not a great metaphore for QCD

maybe you can rescale two different coupling constants without issues

Nihar Karve

...what's stopping one from transforming as $\exp(-ig_1\lambda^a(x)T^a_{\mathcal{R}_1})$ and the other as $\exp(-ig_2\lambda^a(x)T^a_{\mathcal{R}_2})$?

Slereah

12:44 PM

Dunno, I guess maybe if they don't include the coupling constant in some formalisms maybe it's not important?

But maybe it's just because all flavors have the same coupling in QCD

but uuuuh

bolbteppa

Color charge is just not the same as electric charge, I think that's the point, that there is a coupling in the exponential which quantifies how the gauge fields couple to the quarks is like qed

Slereah

What about

Another SU(3) theory

There's SU(3) theories that aren't QCD

Maybe they have varying coupling

although idk if SU(3) sigma models have like a connection as QCD does, so maybe that's not a great comparison

Nihar Karve

@Slereah classical Yang-Mills is evil, you end up thinking that there'll always be a nice 1/r^2 force which is attractive in the singlet channel and repulsive in the adjoint

then of course you get hit with confinement

Slereah

Usually the sigma models are boring $U^\dagger U$ terms

Actually I guess they're even global SU(3) symmetries those theories

Maybe they do have charges

bolbteppa

Another good related post on this by he who shall not be named

35

A: Mathematically, what is color charge?

Color charge is the representation of the SU(3) gauge group. The representation theory of SU(3) is described below: The basic representation is called the "3" or the fundamental, or defining, representation. It is a triplet of complex numbers $V^i$, which transform under a 3 by 3 SU(3) matrix by...

'The tensorial method is never taught for some reason, but it is the quickest way to do Clebsch-Gordon decompositions in real life.'

Slereah

12:55 PM

The tensorial method was hard to find and the only place I could find it used birdtrack notation

It was not a fun summer

Fortunately in modern times wikipedia just has a page on SU(3) Clebsch Gordan

bolbteppa

Yeah it's very hard to find much on this, there's a chapter on it in this book

1

Q: Good alternatives to Georgi's Lie Algebras in Particle Physics?

I just started reading Lie Algebras in Particle Physics by Howard Georgi and I'm finding it frustratingly fuzzy. What I mean is that he often makes non-trivial statements without proof, is imprecise with definitions, and so on. However, what other books I can find seem to swing a bit too far the ...

particle-physics resource-recommendations group-theory representation-theory

Manas Dogra

1:09 PM

For fermions we can get the intrinsic parity by seeing that the parity operator is $\gamma^0$...what about bosons?

In particular why is the intrinsic parity of most gauge bosons -1?

Slereah

isn't the parity operator for gauge bosons just gonna be whatever operator flips the vector

Nihar Karve

I believe there is a slick argument where you couple the massless vector boson to a complex scalar

then $A_\mu$ has to transform the same way as $\partial_\mu$ for the Lagrangian to be invariant, making the intrinsic parity -1

bolbteppa

You'd have to study it properly, e.g. one has to consider 'orbital' parity and 'internal' parity, and why does parity act that way for spinors, there's apparently two slightly different ways to define parity on spinors too because $P^2 = \pm 1$ are apparently both consistent

ACuriousMind

@ManasDogra What do you mean? Intrinsic parity is a convention, there's no "operator" for it.

e.g. we have no clue what the "intrinsic parity" of leptons really is, since lepton conservation means it always cancels out of all reactions we could use to try and figure it out

you have to fix some intrinsic parity by convention and then work your way through the rest of your model

Manas Dogra

@ACuriousMind The parity operator for dirac spinors is $\gamma^0$.

So the fact that the parity of photon is -1 is just a convention once I fix the parity of free fermions to be -1?

ACuriousMind

1:27 PM

the underlying problem is that the effect of intrinsic parity is just a phase factor

but global phases in QM are irrelevant

all that matters is relative phases, and all you can meaningfully say is "A has opposite parity to B"

whether you implement that by A having parity +1 and B having parity -1 or vice versa doesn't really matter

Manas Dogra

But that's not what I am asking here

@ACuriousMind Yes yes I get that...I am fixing the parity of Dirac fermions to be +1 and antifermions to be -1. Does this help me somehow to tell that the intrinsic parity of Photons to be -1?

if yes how..if not how..

For example, Mark Thomson's book on particle physics says that it can be shown from QFT that vector bosons have parity -1. I am simply asking, how so?

Nihar Karve

@ManasDogra isn't it simply enforced by the $e\bar\psi\gamma_\mu A^\mu\psi$ coupling term

bolbteppa

Parity (physics)

In quantum mechanics, a parity transformation (also called parity inversion) is the flip in the sign of one spatial coordinate. In three dimensions, it can also refer to the simultaneous flip in the sign of all three spatial coordinates (a point reflection): P : ( x y z...

Manas Dogra

@NiharKarve Yeah I think it has to be but I don't see directly how.

bolbteppa

Basically a photon is a state, it is described by a creation operator acting on a vacuum, where the creation operator came from the photon field, in the Coulomb gauge it's obviously just a usual vector field so the vector field will change by a sign under parity, thus it's Fourier modes (and the momenta they depend on) will also change under parity, so it has odd 'intrinsic parity' as the wiki states

Nihar Karve

1:39 PM

@ManasDogra $\bar\psi\gamma_\mu\psi$ overall has parity -1. The coupling term above must be invariant under parity, i.e. a parity of +1. Since you contract with the photon field, $A_\mu$must have parity -1 since $-1\times-1=+1$

Manas Dogra

@bolbteppa As I understand it, to impose parity invariance while quantizing we get a condition that the annihilation operator is odd under parity and that tells us the parity of photon is -1. hmm.

@bolbteppa Yes yes..Thank you so much

@NiharKarve Hmm this clears things up for me..Thank you so much.

RaMathuzen

3:00 PM

Is there any article in the physics world dealing with a particle-(like a point object) that emits a magnetic field in all directions? ( for example, like a point charge emits an Electric field)?

John Rennie

@RaMathuzen Yes, it is called a magnetic monopole

RaMathuzen

Ah I forgot that

John Rennie

However no magnetic monopole has ever been observed and for various theoretical reasons we believe they do not exist.

RaMathuzen

@JohnRennie Ok I just asked because I was thinking about a thought experiment that is how will be the motion of a charged particle in presence of field due to a monopole

Manas Dogra

I think apart from that...one can also construct real world current densities which emits such a magnetic field..
Fields don't uniquely determine the sources afterall..

Slereah

3:10 PM

I mean magnetic monopoles may just be rare

ACuriousMind

@ManasDogra no

Slereah

IIRC they are invariant and if they were present during the early universe they may just have been scattered by inflation

ACuriousMind

A monopole field has $\nabla \cdot B \propto \rho_\text{mag}$, where $\rho_\text{mag}$ is the magnetic charge density, but real-world magnetic fields have $\nabla \cdot B = 0$ always

Slereah

there are objects which sort of act like magnetic monopoles in condensed matter physics IIRC?

I vaguely remember something like that

also you can just have EM fields stemming from no charge at all if you are creative

I'm not sure if you can have magnetic fields that way though

I'd have to check Wheeler's stuff on the topic

Magnetic monopole

In particle physics, a magnetic monopole is a hypothetical elementary particle that is an isolated magnet with only one magnetic pole (a north pole without a south pole or vice versa). A magnetic monopole would have a net "magnetic charge". Modern interest in the concept stems from particle theories, notably the grand unified and superstring theories, which predict their existence.Magnetism in bar magnets and electromagnets is not caused by magnetic monopoles, and indeed, there is no known experimental or observational evidence that magnetic monopoles exist. Some condensed matter systems contain...

Manas Dogra

Oh yes...forgot that exists :(

Was thinking about spherically inverting a compact region of constant potential...but that trick works only for electrostatic and gravitational potentials only.

cOnnectOrTR 12

3:25 PM

Hi @JohnRennie

John Rennie

@cOnnectOrTR12 Hi :-)

cOnnectOrTR 12

When we find the E due to a charged conductor (eg a thick plate)it comes as E=sigma/e0. It has been derived by taking a Gaussian surface one end inside the conductor and other outside. We take charges on one side and not on other side. But when we talk about field due to thin sheet we draw Gaussian surface cylinder with both ends on both sides. Why have we ignored the charges on other side of the thick conductor. Does it not spread on both surfaces in case of thick conductor?

Slereah

Gauss theorem holds true no matter what shape you're using

John Rennie

It depends on what exactly we mean by "area charge density"

Slereah

It may be more or less easy to compute depending on what you pick of course

John Rennie

3:29 PM

Suppose we take 1 m² of the plate (of any thickness) then the area charge density is the total charge in that square metre including both sides of plate. OK so far?

cOnnectOrTR 12

Okay

Including both sides ?

John Rennie

Yes. And assuming the plate is symmetrical the charge density will be the same on both surfaces, so the surface charge density on each surface will be half the area charge density.

The problem is that people tend to use area charge density and surface charge density interchangeably for a plate often without carefully defining what they mean.

cOnnectOrTR 12

You mean area charge density is the total charge in total surface area

Of plate including all 6 sides

Or two opposite surfaces only

John Rennie

@cOnnectOrTR12 we normally take the plate to be infinite, or at least so big that we can ignore the edges. So we only need consider the two faces of the plate.

cOnnectOrTR 12

Ok

So

Why choose different Gaussian closed surfaces one being inside the conductor and other through the surface of thin plate

John Rennie

3:41 PM

Personal preference I guess. Either method will give the same result.

cOnnectOrTR 12

So in the case of thick conductor I can take a Gaussian surface all the way through the conductor and out on both sides

John Rennie

Yes

cOnnectOrTR 12

But that will make E= sigma/2e0 instead of sigma/e0. And the field inside the conductor is zero too

I guess it has been taken inside the surface because there is no field inside the conductor due to surface charge on one side

Unlike the case of thin sheet

The charges there make field on both sides

John Rennie

If you use a surface that goes through the sheet then to find the total charge inside the surface you have to use the area charge density. Yes?

cOnnectOrTR 12

Ok

Go on

Physics Meta

3:54 PM

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Q: What's the criteria for a comment "no longer needed" flag?

Reading the answer by SuperCiocia to this question Should we flag comments like "Thank you"?, I learned about the comment flag for "no longer needed". I thought this was interesting because, from time to time, I stumble on dead-end questions that are being kept alive from Roomba solely based on h...

discussion comments flagging

John Rennie

But if you put one side of your Gaussian surface inside the plate then you need to use only the surface charge density, because your surface only encloses one surface.

And the surface charge density is half the area charge density, so the end result is going to come out the same.

cOnnectOrTR 12

Yes in that case 2 will cancel out

But when we take a sheet then we don’t take 2sigmaA.

Why?

Because it’s the same charge on both sides. One very thin line of charge producing fields on both sides and

Penetrating both sides of the Gaussian surface

John Rennie

We normally take σ to mean the area charge density of the plate, so the surface charge density is σ/2.

cOnnectOrTR 12

In resnik surface charge density is defined as sigma

Is 21:26 good

cOnnectOrTR 12

4:15 PM

@JohnRennie am I right!

Iram Haque

5:14 PM

there is a part of my lecture in intro to QM which I am struggling to understand. It may be out of context so I understand if it is not a question possible to answer:

What I don't understand is why one electron can be in spin up and the other in spin down

Slereah

Because those are two different states

different sets of quantum numbers

Iram Haque

Right but what I dont understand is how spin up and spin down are possible spin states

Slereah

Ground state of hydrogen $s$ has two possible states for example, $(0,0,-1/2)$ and $(0,0,+1/2)$

Did you see how spin worked?

Iram Haque

Okay so what I am confused is when I read the addition of angular momenta section in griffiths

Here he mentions that for two spin 1/2 particles these are the possible spin states

there is no state which is |x,y> = |up,down>

only linear combinations of |up,down> and |down,up>

Which boils my question how we can say that one particle is in spin up and one is in spin down, when there is no state which has that configuration?

Slereah

That screenshot you posted is the case for what happens with two particles of spin $1/2$

There are four possible solutions, since you have $2 \times 2$ possibilities

One of them is spin $0$, the rest is spin $1$

what the first image is about is for one particle

Either it's $|\uparrow\rangle$ or $|\downarrow\rangle$

Iram Haque

5:22 PM

holyy

that makes so much sense

thank you

Yes

so much

No problem

7:50 PM

hey

just wondering about something. are Cauchy surfaces related to Lorentz surfaces?

and can you "deform" a Cauchy surface and/or a Lorentz surface in the sense of continuously randomizing the geometry of the surface (as is done in random geometry), using SLE curves/higher dim. analogues

Slereah

By Lorentz surface do you mean a $1+1$ dimensional manifold

If so no, they are very much not the same thing

A Lorentz surface is just a surface (as in 2 dimensional) with a Lorentz metric on it

Cauchy surface is a surface where you put your boundary data on a manifold

and typically in physics those are Riemannian submanifolds and not necessarily 2 dimensional

geocalc33

8:08 PM

@Slereah okay thank you

im trying to understand Cauchy surfaces as they relate to foliations

as well. from what I understand Cauchy surfaces can foliate spacetime?

Slereah

Well, it is to be hoped, yes

But not all of them can

geocalc33

so if a certain Cauchy surface foliates a spactime

Slereah

Well a Cauchy (hyper)surface doesn't foliate spacetime, you need a collection of them

geocalc33

yeah

if you have that collection

they are assumed to be smooth right?

but not necessarily analytic?

Slereah

8:25 PM

Things are usually smooth in physics certainly

it needs to be spacelike so it needs to be $C^1$ at least

geocalc33

so assuming it's analytic. do physicists ever deform the geometry (sort of like a diffeomorphism) but one that doesn't preserve smoothness?

keywords for the deformation: random surfaces, random geometry, SLE curve

I'm guessing nobody does that lol

Slereah

Nothing I know anyway

Also GR doesn't usually assume analytic

geocalc33

just sorta exploring this idea recreationally

Slereah

Analytic isn't used much in physics since people use compact supports a lot

geocalc33

oh okay

so I am interpreting an analytic foliation of Cauchy "paths" (family of 1d manifolds) of a flat 1+1 spacetime, and trying to turn a dial (smooth parameter) to evolve the Cauchy paths to be more and more random, but still remain bounded in a semi compact space

in theory this will deform the geometry in the bounded compact space will force the Cauchy paths into factal curves

then assuming we know the locations of some points initially, then it's possible to have different probability measures for generating the coordinate values of other points

Slereah

9:20 PM

Not really my area of expertise I'm afraid

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