3:58 AM
Hello Guys..

4 hours later…
8:18 AM
Hello @JohnRennie Sir.
Pls see the question I don't understand the situation of this question. About axis, cylinder position, wall position diagram is not very good.
Also how flat beam is located with the cylinder and wall.

8:49 AM
Understand the solution based on figure. But don't understand the situation of question.

2 hours later…
11:08 AM
Morning

3 hours later…
2:03 PM
The authors of the book calls the group transformations in the internal space as gauge transformations. I don't usually see the term "gauge transformations" referred to this.

1 hour later…
3:20 PM
It's a common enough term
Also "vertical transformations"

3:34 PM
This guy has been repeatedly spamming us with his personal theory.

Invite him to the chat, sir. Perhaps we could reason with him.

@JohnRennie just flag if you think it doesn't answer the question, but note that posting your own work while disclosing that it's yours does not meet the SE definition of spam (undisclosed and/or persistent irrelevant advertising, in particular for commercial gain)

I did. I am just expressing my dismay here :-)

@JohnRennie well, you flagged as spam, which this is not - someone posting an answer to a question saying "I've written a paper about this, here's the abstract" is potentially a valid answer to a question here

hello
How is everyone here?

3:44 PM
hi

Hope everything is going safe for you all

hello :-)

@RewCie fine thanks, how are you?

Not good actually
was alone in the room and was having a lot of mood swings, so came here to see what everyone is doing
Anything new for you all in 2021?
What is everyone thinking? 2021 will go better or what?

3:46 PM
@Slereah do you mean people also use vertical transformations to refer to the group transformations in the internal space?

@RewCie hope for the best; but, be prepared for the worst

It may be imaginable considering the term horizontal distribution.

@Bohemianrelativist yes, that's what he means (the terms comes from vertical and horizontal bundles)

@user85795 Ah yes, Life used to be easier when I was a kid.
Then things messed up

a gauge theory has a principal bundle involved, and the gauge transformations are transformations on the fibers of this bundle, i.e. vertical transformations

3:50 PM
@RewCie We all have to grow up, physically at least :-)

Yes :-) That's a good thing.... everyone has to.... :-)

mentally, we have a choice

@JohnRennie that's rather unfortunate, considering he has tenure at the reputed TIFR in India

Yeah that's a bit shocking to see

What!!?
Isn't that a fake account?

3:55 PM
@RewCie possibly, but whoever owns the account seems to be advocating the same theory that he has put forward on his university page

@NiharKarve Shocked to know that David Tong was too at same uni...
Probably he's marketing his papers

if you're shocked at tenured older physicists proposing, uh, seemingly ill-considered new amazing theories I would suggest you haven't been paying attention, it's almost part of their lifecycle :P

haha okay :-)

There are much worse possibilities
in the lifecycles

@user85795 unless the event involved is the destruction of the universe, that's always trivially true

3:59 PM
haha lol :-)
XD
Does 2021 seems to get better than 2020?

13 messages deleted

:56848340 Okay, my apologies

Now I see the term _metric spacetime_ in the book __Teleparallel Gravity__ - 'Curvature and torsion are tensorial properties of Lorentz connections. This becomes evident if we note that many different connections can be defined on the very same _metric spacetime_. Of course, when restricted to the specific case of General Relativity, where only the zero-torsion spin connection is present, universality
of gravitation allows its curvature to be interpreted—together with the metric—as
part of the spacetime definition, and one can then talk about “spacetime curvature”.'

@Bohemianrelativist this was worth reading on this topic too

@Bohemianrelativist since I do not know how your books defines "metric spacetime", I cannot say
but I suspect they use it as a synonym for (pseudo-)Riemannian manifold, not for metric space

4:06 PM

@user85795 What about dropping this topic was unclear to you?

oh okay, I know him....

Lest we forget

@JohnRennie Do you know how much time will it take my phone to complete factory reset?
I want to format and renew everything...

@RewCie it usually only takes a minute or two.

4:09 PM
Do you know how to get them back?
to phone, post factory reset?

@RewCie it's not something I've ever done as I keep all my photos on my PC not on the phone or Google Photos.

I tried USB cable to copy DCIM folder to Laptop, for 1.8 GB of folder, the copy was running @ 4 kb/s
Would take almost 5-6 hours to complete
I wanted to get back... so....

@RewCie Can't you copy over wireless? That's what i do.

@JohnRennie nope. My Bluetooth driver on linux has some issues
I was reading phonetics (RP)... That's how they were pronounced. p - प, b - ब, t - ट, d - ड, k - क, g - ग .....

4:26 PM
What does (RP) stand for? @RewCie

@RewCie wifi not Bluetooth

4:42 PM
@Bohemianrelativist "Vertical" refers to the fact that the transformations happen in the fiber
ie if a point $a$ moves to $b$, you have $\pi(a) = \pi(b)$
They remain at the same vertical position on the manifold

If anyone is interested there is a long interview with Freeman Dyson on YouTube here. It is broken up info short sections and that makes it easy to pick out the bits that interest you.
3
He has some interesting stories about meeting the great names in the history of QM.

@JohnRennie How? I don't know how to use Wifi to Transfer files to Linux..

@RewCie I've only ever transferred files to Windows, but I'm sure Googling would find you articles on how to do it in Linux.

Okay... :-)

@RewCie I would say that "t" is emphatically not "ट" in RP English

4:52 PM
you can't say what 't' is in RP English to begin with because English does not use phonetic spelling :P
cf. that old joke about writing fish as ghoti

I'm actually struggling to find a single RP English word that includes the "ट" sound, which is why I said that.

Wiki says that letter stands for a retroflex sound that doesn't exist in English

@ACuriousMind I've seen the one with "potato" as "GHOUGHPHTHEIGHTTEEAU"
Actually there are better ones than that

@JohnRennie I don't think any PhD would agree with #95 in that series.

@NiharKarve Don't blame us! If the rest of you hadn't spent the last few millennia invading us and bringing your languages with you English wouldn't be such a mess.

5:04 PM
lol

@user85795 I enjoyed my PhD immensely. It was then, and remains still, the best three years of my life.
But I hear stories, especially from America, of students finding it a traumatic experience.

@JohnRennie the terrible non-phonetic spelling is actually much more a result of the orthography being fixed hundreds of years back and never really having been changed after that
it was phonetic - but only back when they started writing things down, not now

accent is a sticky topic
It's too easy to not see the forest for the trees.

5:23 PM
Is there a physical meaning of a representation (or moving between representations)?
I'm reading something that says that $Y^{(l)}_m$ are l-dimensional representations of the rotation group, which I take to mean that states $| lm \rangle$ are in different representations, while I always thought that you pick a representation and stick with it for a given physical object

@DanielUnderwood the physical meaning of a representation depends on what it is a representation of
in your case, the spherical harmonics for fixel $\ell$ are a representation of the orbital angular momentum algebra - if this is conserved/constant in your specific case, you only need one of these representations, but if you have to consider different angular momenta, you obviously need several of these representations

So you're saying the physical meaning is back in the group rather than a given representation?
And is this something that's specific for quantum mechanics? You usually stick to a vector representation classically, right (or at least stick to what you started with)?

@DanielUnderwood well, classically representations - if they appear - appear in a different way because the space of states is not a vector space there
or rather, the observables are not usually linear operators on it, so everything works quite differently classically - the reason representations appear "so much" in QM is because "observables are linear operators on a Hilbert space" means that you have algebras of observables - like angular momentum - that are being represented on the state space

5:40 PM
@JohnRennie have you ever considered going back and doing another PhD?

if you want to talk about something similar classically you'd probably talk about Hamiltonian actions and moment maps on phase space, but that usually seen as a bit of an advanced/exotic topic since most of mechanics does just fine without talking about it

hmmm I think I've mixed things up by seeing groups and trying to apply the same notion to both symmetry of an action and writing out quantum states, while really they're entirely different concepts

@user85795 I think doing a PhD requires a singleminded approach and focus that is only easy when you're young. I doubt I could find the concentration to do another PhD now.

@DanielUnderwood note that a representation of a group is a much more specific concept than that of an action of a group
both classical and quantum mechanics have actions of the observables on their spaces of states - but only in quantum mechanics does this action necessarily take the form of a linear/projective representation on a vector space

Sorry, I meant action as in the integral of the Lagrangian rather than the action of a group element -- like you're trying to make your action Lorentz invariant or something...though I don't know that I've ever actually thought about that in the context of a group
I think my line of thought is Symmetry -> there has to be a group -> you have to be working in a representation and I've somehow conflated that with the representation of a quantum state

5:50 PM
@DanielUnderwood sure, the symmetries are groups/algebras that act on the variables in the action, i.e. the state space
(or on paths/histories through state space, if you want to be precise)
but that state space is e.g. in the Hamiltonian formulation a symplectic manifold, not a vector space, and so the action of the symmetry groups is not as a linear representation
i.e. I'm saying your second step "group -> you have to work in a representation" isn't necessarily true - it is only true if the group is acting linearly on a vector space

6:04 PM
Maybe that's the bit that I'm missing. Everything that I've seen from the group side seems to work in linear representations

Linear representations are the most common since they are the ones acting on vector spaces
which are your fields and such

7:02 PM
what is the dimension of a hilbert space containing Dirac spinors?

isnt it a fock space?

In qft yes, but if the Dirac spinor is seen as solution to the Dirac equation interpreted as a wave equation?

@john The wavefunction of a particle with spin space $H_s$ lives in the infinite-dimensional $L^2(\mathbb{R}^3)\otimes H_s$, i.e. just the space of square-integrable functions valued in the spinor space
(note that for a spinless particle that transforms in the trivial 1d spin representation, this reduces to the usual $L^2(\mathbb{R}^3)$ for wavefunctions without spin)

7:24 PM
@ACuriousMind a 4-components spinor is similar to a 4-vector. Then shouldn’t the spinor be an element of a 4-dimensional space?

@john sure, for a spinor $H_s$ is 4-dimensional

The full wavefunction is infinite dimensional, the spin state itself lives in a 4-dimensional complex vector space @john

but the solutions to the Dirac equation are not 4-component spinors, they are functions valued in 4-component spinors
since you specified "seen as solution to the Dirac equation" I assumed you were talking about spinor-valued wavefunctions, not just a spinor

Easy to check since any solution of the Dirac equation is also a solution of the KG equation :p
It has to be somewhat functiony
So the Hilbert space has the same cardinality
even though it is "four times" larger
It has a basis $4 \aleph_0$ instead of $\aleph_0$

@ACuriousMind actually never heard of “functions valued in 4-component spinors”, I was referring to spinors, my bad

7:57 PM
It's a spinor field, if you prefer

Oh ok yes, definitely talking about just spinors
Anyway, thank you guys

Reading through some lecture notes in electrodynamics:
"Electromagnetic waves are fundamental solutions of Maxwell’s equations in the absence of charges and currents."
Why absence of charges and currents?

Just pass them through the equations
You can see they correspond to a charge and current of 0
and vice versa, if you consider the equations with zero charge and current, after some manipulations, you end up on the wave equation

I see, so if the charge and current density is 0, then you just get wave equations?

8:12 PM
yes

Thanks.

It's a standard exercise to show it

Should do it.
@Slereah I was thinking, one has four equations, wouldn't it be enough with just two to get the wave equations for $\mathbf E$ and $\mathbf B$ respectively? :)

Yes, that's how you do it
Take the two equations of E, put them together, and that's the wave equation for E

I see.

8:16 PM
There's some vector calculus manipulation to do

@Slereah Do you know if this is done in Griffith's or in Jackson's book?

It's a standard calculation
It's in all EM books

@Slereah Section 9.2 in Griffith if I'm not mistaken.
Follows from applying the curl to Ampere's and Faraday's law.

it's in chapter 6 of Jackson

8:32 PM
Nice, thanks.

9:24 PM
@schn it takes about two lines in the covariant formulation: up to signs and coefficients Maxwell's equations involving sources are $\partial_{\mu} F^{\mu \nu} = j^{\nu}$ and $j^{\nu} = 0$ if the charge and current density is zero, so $\partial_{\mu} F^{\mu \nu} = 0$, and in the Lorentz gauge $\partial_{\mu} A^{\mu} = 0$ this reduces to $\partial^2 A_{\mu} = 0$ which is the four-dimensional wave equation.
The $E$ and $B$ fields defined in terms of this don't change the form of the solution, still plane waves

@bolbteppa Interesting. On a related note, do you know from which definition the second equaility follows in...
Is this simply the definition of the real part of a complex wave?

Indeed applying $\partial^2$ to e.g. $E_i = \partial_0 A_i - \partial_i A_0$ shows $E$ and similarly $B$ satisfies the wave equation
I think it follows from expanding $B = \nabla \times \mathbf{A}$ for a plane wave $\mathbf{A}$, one can show it can be expressed in terms of $\mathbf{E}$

I see. Makes sense, so the amplitude with its complex conjugate is its length squared. However, why does the curl disappear?

It doesn't dissappear, note (for a plane wave) up to factors that $\mathbf{E} \approx \mathbf{A}$ and $\mathbf{B} = \nabla \times \mathbf{A}$ for a plane wave reduces to $\mathbf{k} \times \mathbf{A}$ up to factors, so up to those factors one has $\mathbf{E} \times \mathbf{B} = \mathbf{E} \times (\mathbf{k} \times \mathbf{A})$ which is a triple product so using the BAC-CAB identity and $\nabla \cdot \mathbf{E} = \mathbf{k} \cdot \mathbf{A} = 0$ it should reduce as above

9:44 PM
@bolbteppa How come $\nabla \cdot \mathbf{E} = \mathbf{k} \cdot \mathbf{A}$?

Yeah I should say for a monochromatic plane wave, in general it's that for a sum of different modes

?
Im considering using it as a starting point to learn group and representation theory
and how they use in physics

@bolbteppa Why does $\mathbf k$ substitute $\nabla$?

I'm thinking of Coulomb gauge now right, in the Coulomb gauge $\mathbf{E} = - \dot{\mathbf{A}}$ with $\mathbf{A} = \mathbf{A}_{\mathbf{k}} e^{i \mathbf{k} \cdot \mathbf{r}}$ (monochromatic) so that $\nabla \cdot \mathbf{E} = \nabla \cdot \dot{\mathbf{A}} = \mathbf{k} \cdot \dot{\mathbf{A}} = 0$, Lorentz gauge I'd need to think but in your formula I think it's using Coulomb gauge I'd need to check the context

I think Woit's notes are generally considered good

9:57 PM
@MoreAnonymous you read the criticism of it and it barely talking about probability

they seem very comprehensive

@bolbteppa I mean Im familiar enough with the Born rule

Weyl wrote a book on group theory and qm, probably a safer choice

Prefer the newer writing styles more

From the amazon reviews: 'As an alternative view, CN Yang has said about this book that "Almost every theoretical physicist born before 1935 has a copy...but very few read it. Most are not accustomed to Weyl's concentration on the structural aspects of physics and feel uncomfortable with his emphasis on concepts. The book was just too abstract for most physicists."'

10:44 PM
Woit also seems to have the most recent course following his book on youtube youtube.com/channel/UCdSEKN94Jo1xjUHVyfEfIjg/playlists

@DanielUnderwood thanks

In Griffith's electrodynamics book, chapter 10.1.3, when the scalar potential $V$ is presented using the Coulomb gauge, he states that $\mathbf E$ will only change after sufficient time has elapsed for the "news" to arrive. How does this follow from $\mathbf E$ written in terms of potentials? It has the partial derivative of $\mathbf A$ with respect to time in it, is he referring to that term?
Kindly ping me if you'd like to share some insight.

11:11 PM
@schn The equation for the electromagnetic fields is a wave equation in vacuum for a propagation speed of $c$, i.e. the speed of light.
or you can see this more explicitly in Jefimenko's equations, which express the general solution to Maxwell's equations in terms of retarded fields/potentials at distant places at an earlier time (earlier by precisely the time something with speed $c$ would need to get from there to the point under consideration

@ACuriousMind Thanks for the insight.
Is there a disadvantage with the Coulomb gauge?
Is it "incorrect" in any way?

no gauge is incorrect, that's the whole point of gauges - each one is equally as "correct" as the other

Right :)
However, $V$ is zero far away...is this also the case in other gauges?

nothing forbids you to choose a gauge where it's not
however if it is zero in one gauge, then it will be pure gauge, i.e. a total derivative, in all others due to how the gauge transformations act

Thanks for the replies.
@ACuriousMind On a related note, in the Coulomb gauge, the equation for $\mathbf A$ reads
What is the solution to this equation?
For $V$ one has
with solution

11:35 PM
@schn see Wiki

11:55 PM
@ACuriousMind In the Lorenz gauge, does $V$ also vanish far away?

I'm afraid I don't have the various properties of the gauges in EM memorized :P Why does it matter?