The h Bar

12:01 AM

@123 You can imagine the origin of the coordinate system at B, and the diagram in the 2nd picture has just drawn an (x,z) plane at $y = - l$ to highlight the similarities with the A case

123

12:16 AM

@bolbteppa Thanks. for the confirmation , the same results i have concluded by thinking origin at B.

Is there any physical meaning of magnitude and direction of angular momentum? Because different origin give different magnitude and direction of angular momentum and torque.

naturallyInconsistent

2:53 AM

@Obliv miehehehe

Silly Goose

3:38 AM

if i parameterize the unitary matrices with the generalized gellmann matrices, is the parameterization periodic?

i think the answer is no

hmm

Silly Goose

3:51 AM

amazing

Silly Goose

6:18 AM

consider the classical electromagnetic field strength tensor $F^{\mu\nu}$

I am wondering how to think about getting the right sign when defining the electric field components

is it like: $F^{0\nu} = F^{0i} = \partial^0 A^i - \partial^i A^0 = \partial_0 A^i + \partial_i A^0 = -E^i$? where I am using $(+, -, -, -)$ signature. where the final equality is our usual definition of the electric field. I think this is wrong because the indices get messed up in the penultimate equality...

Silly Goose

8:04 AM

if I am computing $\lvert E \lvert^2$ in tensor notation, is this $E^i E^i$? I think this should be the case. As opposed to $E_i E^i$.

naturallyInconsistent

@SillyGoose you usually do not compute $|E|^2$ in tensor notation

Slereah

The relevant metric tensor is the spatial part of the metric tensor

So in flat space $\delta_{ij} E^i E^j$

Silly Goose

okay i see

bleb I am getting that the electromagnetic energy density is negative...

Slereah

With some appropriate tensor manipulation you can find out that $F^{\mu\nu} F_{\mu\nu}$ is the same as $E^2 - B^2$

By using identities related to the definitions of the Levi-Civita tensor and projections

Silly Goose

right i get that it is equal to $2(E^2 - B^2)$

naturallyInconsistent

8:12 AM

When trying to figure out the direction in which certain particles with certain momenta are moving, it is necessary to be using the contravariant components $p^\mu=(+|E|,p^i)$ instead of the covariant components. The Lorentz invariant phase is $k_\mu x^\mu=k_t x^t+k_i x^i=\eta^{tt}(k^t x^t-k^i x^i)$; the momentum is $\hat p_\mu=-i\nabla_\mu$. Between these definitions you can figure out the derivative part of the Faraday tensor. However, I do not know how you are defining the 4-potential $A$

Silly Goose

8:28 AM

man chasing down negative signs ;_;

naturallyInconsistent

9:15 AM

Do you actually need to chase down the negative signs? Is it not acceptable to just put in the minus signs to make them correct as you need them?

Relativisticcucumber

@SillyGoose lmfao what is this

naturallyInconsistent

There are some bits that are internal constraints and so you have to get them correct. However, there are also quite a lot of bits that are conventions, and people can choose different conventions.

123

Hello Everyone...

Is there any physical meaning of magnitude and direction of angular momentum and torque?

Mr. Feynman

10:32 AM

@naturallyInconsistent NO

I guess I would be happier if I did but I can't

123

11:38 AM

Pls see highlighted step, i don't understand how they put $\rho_{j} = R_{\perp}$

DIRAC1930

@Slereah I remember you posting ages ago a picture of your bookshelf

Slereah

12:13 PM

What about it

John Zimmerman

12:27 PM

is there a name for a continuum of mutually intersecting gedoesically compactified AdS spaces? I'm developing a model of something

John Zimmerman

12:38 PM

DIRAC1930

Got any new additions?

Slereah

Sure, I buy a new one once in a while

DIRAC1930

I wish textbooks werent so expensive

John Zimmerman

me too

but there are some nice resources online

for free

DIRAC1930

12:55 PM

Come to think of it, Statistical Mechanics is the only area in physics that is most likely complete in its fundamentals

lucabtz

@DIRAC1930 what does this mean

Mr. Feynman

Stat mech should be called Stat mess

@lucabtz yooo

lucabtz

@Mr.Feynman hey la

DIRAC1930

Well there is still progress to be made in all other fields but stat mech is largely (in it's equilibrium form) is largely complete in its foundations

lucabtz

@DIRAC1930 I mean quantum statistical mechanics is as complete as complete you consider quantum mechanics

Same for classical

Also classical mechanics is pretty complete too. Idk understand what you mean

DIRAC1930

1:01 PM

I meant between GR, QM, QFT and Stat Mech

lucabtz

@DIRAC1930 I don't think stat mech can be more complete than qm by definition

It literally depends on qm so

Mr. Feynman

@lucabtz are you on the other side now?

DIRAC1930

If you have states, and a Hamitlonian, all you need is the fudamental posulate of statistcal mechanics and that's it

Even if QM changes slightly I doubt stat mech will change

Mr. Feynman

Before I get it: isn't GSW a bit outdated?

I heard it doesn't even mention CFT

lucabtz

@Mr.Feynman what side

@DIRAC1930 it's a postulate of qm that states are element of Hilbert spaces and so on

DIRAC1930

1:06 PM

@lucabtz So you believe this will change?

Didn't think so

lucabtz

@DIRAC1930 I didn't say that

I'm just saying that quantum statistical mechanics is as complete as quantum mechanics is

It's valid when quantum mechanics is valid AND you have a statistically important number of particles

Mr. Feynman

@lucabtz mhhh

Oh god don't tell me you forgot

Mar 13 at 15:25, by lucabtz

omg im graduating on 27th march

DIRAC1930

You don't even need QM... You can do classical statistical mechancs....and nothing changes in it's fundamental composition

lucabtz

1:42 PM

@Mr.Feynman HAHAHAH

No I remembered

Mr. Feynman

Even if you did it would have been a fun story to tell your grandchildren

Arjun

1:59 PM

If $\hat A$ is a linear operator and if $\forall$ functions f,$\int_{-\infty}^\infty \hat Afdx=0$ ,does it mean $\hat A$ =$\hat0$ ?( where ,$\hat 0f=0 $, $\forall f$)

DIRAC1930

2:33 PM

@Arjun It might be useful to give some context behind your question i.e. the answer you get might be different if you are doing pure mathematics compared with physics

Ryder Rude

@123 theyve used $\sum m_j \rho _j =M R_{\perp}$. vector $\rho _j$ carries the $x$ and $y$ components of the position vector of mass $m_j$, while vector $R_{\perp}$ carries the $x$ and $y$ components of the center of mass.

@123 this follows from the definition of center of mass : $R_{\perp}=\frac{1}{M} \sum m_j \rho _j$

DIRAC1930

2:51 PM

@JohnZimmerman This stuff looks completely insane lol

Ryder Rude

@JohnZimmerman hi. r u making ur own theory

DIRAC1930

3:12 PM

@RyderRude What are you interested in/currently working on?

Ryder Rude

3:25 PM

@DIRAC1930 ive been learning geometry stuff...

@DIRAC1930 u?

DIRAC1930

I've decided to reunderstand statistical mechanics so that I can understand Fermi liquid theory

So I am going through L&L 5

and then when I have that in control, trying to understand L&L 9 (the chapter that motivates Fermi Liquid Theory)

Then go through the non-rel QFT calculations in L&L 9

Ryder Rude

3:41 PM

great

DIRAC1930

Come along for the ride

Ryder Rude

not me :P.. i have trouble committing

plus ive been trying to read some other stuff

@DIRAC1930 good luck tho :)

DIRAC1930

@RyderRude Okay, you are forgiven :)

Arjun

4:03 PM

@DIRAC1930 In this lecture(youtube.com/…) at about 25 mins allan adams uses this property $\int_{-\infty}^\infty \hat Afdx=0\implies \hat A$ =$\hat0$ to find Adjoint of the operator A

He says if $\forall$ functions f & g,$\int_{-\infty}^\infty g*(\hat A-C^\dagger)fdx=0$ then $\hat A$ =$C^\dagger$

Here g* is the complex conjugate of the function g and $A^\dagger$ means the adjoint of the operator $\hat A$

DIRAC1930

4:31 PM

@Arjun This seems like a reasonable thing to write

ACuriousMind

@Arjun Note that $\int g(A-C^\dagger)f$ is the inner product between $g$ and $(A-C^\dagger)f$. A vector whose inner product with all other vectors is zero is zero, so this means $(A-C^\dagger)f = 0$ for all $f$. But an operator that yields zero on all vectors is the zero operator, so $A-C^\dagger = 0$

Arjun

@ACuriousMind thanks and can we apply linear algebra results to functions directly?(I'm totally new to this stuff) Like why the above inner product rule works for functions just like how it works for vectors is not intuitive to me

John Zimmerman

5:10 PM

@RyderRude @DIRAC1930 yes I am making my own theory and it's not as wild as you might think

arxiv.org/abs/hep-th/0302067

Sean carroll and monica guica published that a while ago.

but I had not found it until a month ago. I had been thinking about a related theory years prior

ACuriousMind

@Arjun this is "infinite-dimensional linear algebra" aka functional analysis - formally the vector space here is the Hilbert space of square-Integrable functions $L^2(\mathbb{R})$

See en.wikipedia.org/wiki/Lp_space

DIRAC1930

If you type this in Mathematica
ResourceData["Books in Stephen Wolfram's Library"]
You can see which books Wolfram has in his personal library

It is quite interesting

Mr. Feynman

8:34 PM

@ACuriousMind since you said that you went through GSW: do you think it is somehow outdated (I read in a comment that in 1995 Superstring theory had another revolution)?

ACuriousMind

@Mr.Feynman did you read my full statement about it? :P

Mar 15 at 21:00, by ACuriousMind

@DIRAC1930 it's been a while since I read it; since I have almost no memory of it (as opposed to the dislike I still carry for BBS) I guess it was okay :P

can't tell you how outdated something is that I don't really remember

Mr. Feynman

I did and I forgot :P

> Another two volume set. It is now over 20 years old and takes a slightly old-fashioned
route through the subject, ** with no explicit mention of conformal field theory **. How-
ever, it does contain much good material and the explanations are uniformly excellent.
Volume one is most relevant for these lectures.

That's the scariest part in Tong's "review"

Oh, basically DIRAC1930 asked the same question a couple of days ago, sorry for that

Silly Goose

what does "flat" mean in flat torus?

this is the context

ACuriousMind

@SillyGoose that it's flat

Silly Goose

well my naive interpretation is two concentric circles :P that doesn't seem like a torus

ACuriousMind

8:48 PM

well, look at what the text is about: it's talking about a metric/measure/volume form

what does "flat" mean to you in that context?

when we say "Euclidean space is flat", we don't mean that all Euclidean spaces fit into a plane (as you seem to have attempted with the two concentric circle), do we?

Silly Goose

Hm well so then by Euclidean space is flat do we mean that Euclidean space has no "curvature" but I don't know what curvature is

@Relativisticcucumber my working code XD

lucabtz

@Mr.Feynman today I will tell you about that time I forgot I was graduating

Mr. Feynman

Oh no grandpa is back at it with his zoomer stories

lucabtz

@Mr.Feynman lol

@SillyGoose you can see the torus as a quotient of a hypercube identifying the opposite faces

That is also intuitively flat

ACuriousMind

@SillyGoose in that case the simplest statement to make here is that the torus they're considering is the one you get as a quotient of flat space, i.e. $\mathbb{R}^2/\mathbb{Z}^2$

and the metric on the torus is the one inherited from $\mathbb{R}^2$ through the quotient, not the one you would induce on the usual donut embedded in $\mathbb{R}^3$

Silly Goose

9:03 PM

hm so in this particular case our construction seems pretty sensitive to the underlying topological space. In particular homeomorphic underlying top spaces won’t give us our construction of interest. Is this only because we’re trying to inherit structure from a particular metric, which itself induces the underlying topological space? Im not sure if that makes sense :P. I am trying to understand what initial data our construction is sensitive to

ACuriousMind

what construction do you mean, and in what sense do you think it's "sensitive to the underlying topological space"?

Silly Goose

I mean the construction of the fubini study measure

well i am confused about what would be the difference between considering the “flat torus” vs the “usual donut embedding” for the purpose of constructing FS measure

ACuriousMind

@SillyGoose the metric on it

and so also the volume form/measure on it

the statement "flat torus" is just the statement that your $\mathrm{d}\Omega_n$ contains no functions of $\nu_i$

it's just a constant top degree form

that's one way to say something is flat: there are coordinates in which its volume form is constant

Silly Goose

Hm so the “underlying space” here is a topological space and a metric? Where the metric is not necessarily a metric that induces the topology on the underlying space?

123

@RyderRude Thanks may be i understand that. From diagram if i take infinitesimal xy-plane of where $\rho_{j}$ is in diagram i will get CoM_1 of that plane by $R_{xy1} = \frac{\sum\rho_{1j}x_{1j}}{M_1}$, then i will move $\rho_{j}$ slight forward/backward then again find the CoM_2 of that plane. If find CoM of all that planes we will get the final result of $R_{\perp}$

ACuriousMind

9:08 PM

if you don't know enough differential geometry to know what curvature is, this is either irrelevant for the rest of the text and you can ignore it or you should perhaps learn some Riemannian geometry first :P

Silly Goose

For now I’m just trying to understand enough to do some concrete computations with the fubini study measure :P

i would like to compute FS-average of a function of pure states from a $2^n$ dimension system, for instance

In an ideal world i would indeed just learn differential geometry right now:P

well maybe i can get acquainted with the basic concepts still…

ACuriousMind

@SillyGoose In Riemannian geometry you can prove that given an underlying topological space $X$, for all Riemannian manifolds $(X,g)$, $g$ induces the same topology as $X$ originally has (By certain continuity requirements on $g$ that are part of the definition of a Riemannian manifold)

so, no: the metric always induces the topology of the underlying space, it's just that you can have many different metrics inducing the same topology that still have different curvatures

Silly Goose

Oh okay

ACuriousMind

if curvature was equivalent to the mere topological data, what would be the point in the additional structure :P

@SillyGoose ...but you only need the formula for $\mathrm{\Omega}_n$ for that

you don't need to understand anything else about it :P

Silly Goose

Bleb okay i should just proceed and see if i can work with it:P

DIRAC1930

9:51 PM

Something fishy must have been going on in Dublin in 1942

Wonder if this had anything to do with the UK equivelant of the Manhattan project

ACuriousMind

what's supposed to be fishy about the photo?

it's just a bunch of back-then contemporary physicists at a prestigious institute

DIRAC1930

Dirac, Schrodinger and Heitler (the guy who essentially wrote the quintessential 1930s/early 40s QFT book)

Dirac didn't usually work in Dublin

ACuriousMind

...people can be at places where they don't usually work?

DIRAC1930

The above is in 1942

I wonder if it had anything to do with the Tube Alloys project

Dirac was known to have done some calculations for it

Obliv

if something fishy is going to happen somewhere, I doubt they'd want a group photo celebrating the occasion?

but I'm guessing you're being facetious and don't really mean fishy, rather you're interested in what was going on there at the time.

ACuriousMind

10:01 PM

@DIRAC1930 So? Not all academics were spending 100% of their time on the war effort

Schrödinger was literally the director of the DIAS in Dublin at the time

DIRAC1930

He was the only member than Heitler joined working on the strong interactions

ACuriousMind

"hm, must be something fishy when the famous director of the local institute is photographed with other famous people"

of course Schrödinger would have invited people like Dirac to come there

Obliv

Nah, they're definitely hiding something. The elites have a time machine, i saw it in an anime once.

DIRAC1930

Good one.....

Obliv

how many of those people can you guys name? I recognize some but can only name a few

who is the 2nd from the right on the first row

ACuriousMind

10:07 PM

heitler was assistant professor at DIAS at that time, too

two of your three "suspicious people" just worked there

DIRAC1930

@Obliv Ask ACuriousMind, he knows Schrodinger personally judging by this comment "of course Schrödinger would have invited people like Dirac to come there"

ACuriousMind

> Heitler remained at Bristol eight years, until 1941, when he became a professor at the Dublin Institute for Advanced Studies, which was arranged there by Erwin Schrödinger, Director of the School for Theoretical Physics. He has been described as the "unsung hero of DIAS in the 1940s".

(from Wiki)

Obliv

A perfect alibi if you ask me. Like working at a bank and getting paid weekly, accruing wealth overtime in discretion. You're robbing them without them even knowing.

DIRAC1930

Until Schrodinger met ACuriousMind and personally told him "of course I'm going to invite people like Dirac to come there"

ACuriousMind

from the institute itself:

> A major attempt to combat the isolation of the School was made in 1942 by holding the first colloquium which lasted from 16th. to 29th. July. The speakers from abroad were P. A. M. Dirac, who delivered five lectures on Quantum Electrodynamics, and A. S. Eddington, who gave the same number of lectures on Unification of Relativity Theory and Quantum Theory. These two sets of lectures were published as Numbers 1 and 2 of the Communications of the Dublin Institute for Advanced Studies, Series A.

So Schrödinger and Heitler were already there, Dirac was invited to lecture on QED. What's fishy about that?

Obliv

10:09 PM

jeeez, i didn't even recognize schrodinger lol. I've only seen earlier photos :\

shrug

DIRAC1930

Good one...

Obliv

is that to me... or..

DIRAC1930

Well both of you literally took an innocent comment and completely twisted it for reasons I have no idea why except to be deliberatly argumentative

ACuriousMind

@DIRAC1930 What do you mean "twisted"? You claimed there was something "fishy" about the photo from the colloquium and it might have had something to do with secret war projects, I'm providing explanations for why this might really just have been an ordinary meeting of physicists

what's your point in alleging something like that if you don't like people arguing about it?

Obliv

Excuse me, I make dumb remarks all the time @Dirac1930

I like to poke fun, don't mind me.

DIRAC1930

10:26 PM

For anyone actually interested there seems to be a dissertation written on Dirac's involvement in the Tube Alloys project here diginole.lib.fsu.edu/islandora/object/fsu:770703/datastream/PDF/… . On pg 37 there is the first letter he recieved in 1942. It would be interesting to know who the group mentioned in that letter is and where exactly Dirac did his work

"These included
German theoretical physicist and later atomic spy Klaus Fuchs,59 British experimental physicist
Patrick Maynard Stuart Blackett,60 and Michael Polyani, a British-Hungarian polymathematician who made contributions in several fields of study."

ACuriousMind

> Four universities provided the locations where the experiments were taking place. The laboratory at the University of Birmingham was responsible for all the theoretical work, such as what size of critical mass was needed for an explosion. It was run by Peierls, with the help of fellow German refugee scientist Klaus Fuchs.

(again, from Wiki)

the "group" was most likely the MAUD committee or its successors

> The division of the MAUD Committee at Cambridge was jointly led by Bragg and Cockcroft.[37] It included Bretscher, Feather, Halban, Kowarski, Herbert Freundlich and Nicholas Kemmer. Paul Dirac assisted as a consultant, although he was not formally part of the team.

(Wiki)

DIRAC1930

Page 17 to 18 has details about Birmingham here royalsocietypublishing.org/doi/pdf/10.1098/rsbm.1986.0006

Obliv

I feel like Wiki & looking at some random persons thesis isn't really all that productive. If you are serious about uncovering some "fishy" business you should put your detective hat on and find primary sources of information. (I didn't look at the citations for either, but I'd start there)

DIRAC1930

And it just mentions "During the war there was little opportunity for foreign travel, but
Dirac paid several visits to the Dublin Institute for Advanced Study,
where Schrodinger was his host."

I think fishy means something different where I live

It just means 'slightly weird'

ACuriousMind

to me it's more like a colloquial term for "suspicious"

DIRAC1930

10:37 PM

Page 20 has an interesting comment

In M ichaelm as term
1945 H arish-C handra arrived from India, w here he had worked w ith
H. J. Bhabha at Bombay. He venerated D irac but becam e persuaded by
his experience at C am bridge that he was not suited to theoretical physics.
As to his reason for abandoning physics, he m entioned a conversation
with D irac in which he said that he had discovered a lack of rigour in
D irac’s work on the Lorentz group. D irac replied ‘ I am not interested
in proofs but only in w hat nature does.’

Obliv

weird also depends on the context of what is normal, which only you really know.

DIRAC1930

Well even a new school starting in 1 year and having that caliber of guests is already very weird

ACuriousMind

@DIRAC1930 Ireland was "neutral" during the war and until 1948 still technically a subject of the British crown, so this is about the most benign form of "foreign" travel there was at that time

Obliv

(forgive me) maybe so, but not all island visits are benign.

DIRAC1930

Theres also Eddington in the front row I believe

ACuriousMind

10:40 PM

@DIRAC1930 that's the second "speaker from abroad" mentioned in my excerpt from the institute history above

apparently Dirac and Eddington were the highlights of that particular event

DIRAC1930

It turns out not all those people in that picture are phycists

That preist is in a different subject

Oh actually he's a mathematician

He does many things

" the School of Theoretical Physics and the School of Celtic Studies" this is a very weird combination

Obliv

Who is the 1st person in the 1st row?

DIRAC1930

Someone who looks like they don't want to be there

ACuriousMind

@Obliv A mathematical physicist named Sheila Tinney, again, cf. Wiki

DIRAC1930

Sheila Tinney

Sheila Christina Tinney (née Power; 15 January 1918 – 27 March 2010) was an Irish mathematical physicist. Her 1941 PhD from the University of Edinburgh, completed under the supervision of Max Born in just two years, is believed to make her the first Irish-born and -raised woman to receive a doctorate in the mathematical sciences. == Life == Sheila Christina Power was the fourth of six children born in Galway city to Michael Power [a.k.a. Mícheál de Paor, originally from rural Kilkenny, Chair of Mathematics at University College Galway (UCG) from 1912 to 1955] and Christina Cunniffe (who died in...

Obliv

10:45 PM

who is this guy

ACuriousMind

@DIRAC1930 a bit less weird if you consider that the institute was Irish and founded explicitly as an Irish prestige project

DIRAC1930

That guy clearly wants to be there

ACuriousMind

(also at my uni there was a party each semester that combined the natural sciences and...theology :P)

@Obliv it's Peng Huanwu

Obliv

I bet german uni parties get wild.

ACuriousMind

literally the only guy not in the front row that anyone seems to bother naming

Obliv

10:48 PM

wait wasn't china like on the opposite team

how'd this dude get in

DIRAC1930

Btw they changed buildings

I think this is the original building buildingsofireland.ie/building-images-iiif/niah/images/…

Obliv

WAIT yeah why did they only name him lmao

I guess for people like me that are curious about him. I'm the target audience I guess.

ACuriousMind

@Obliv nope, "China" didn't really participate in WWII except by getting invaded by the Japanese (which were enemies of the Allies!) and having a lot of civil war until Mao won

so by "enemy of my enemy is my friend", it makes sense they didn't consider a Chinese physicist a threat

DIRAC1930

I'm just surprised this happened all in the space of 1 year

It seems like such a good idea

To have these very small, purely research universities

I suppose the IAS in Princeton is kind of like this but larger scale

Obliv

I live like 2 hrs from princeton. Want me to do some sleuthing for you

ACuriousMind

10:55 PM

this was literally supposed to be the Irish version of the IAS

and I guess for its first few years it was

lucabtz

@DIRAC1930 it's because that's how research has worked for ages. There are plenty of conferences where you see big names together today as well and it makes sense there were at the time too

Obliv

what are the "motivations" of such institutes? I'm just imagining some kind of justice league

ACuriousMind

what

Obliv

I don't know, something that isn't tied to a single nation lol

DIRAC1930

@Obliv Well the IAS in Princeton became a safe haven for Jewish scientists during WW2 i think

ACuriousMind

10:58 PM

I mean, of course they're usually going to present as international but of course the government funding them is hoping to reap the prestige for its own nation :P

DIRAC1930

@lucabtz Universities don't just pop up in 1 year and have the likes of Schrodinger, Dirac, Eddington, Heitler all together. It is a unique circumstance and a special case that is interesting

The new building looks horrible compared to the old one

I reckon if they kept the old one, it would still be regarded as prestigious

Obliv

Regardless, it's been ~80 years since then so I doubt any of that matters now aside from historical significance. There are way more "fishy" things going on today.. probably.

DIRAC1930

That's true

Wow, I just realised that I havent learned anything really that's newer than 60/70 years old

Obliv

Physicists that get too involved with political affairs probably don't have long careers anyway is my guess.

must be nice. @DIRAC1930 I'm still in the 1800s

the end of this semester i'll finally do blackbody radiation though and then QM after

DIRAC1930

I wish I could go back to before I learned rel QFT and just have done non-rel QFT instead

Obliv

11:08 PM

But I'm not trying to be a theoretician so it doesn't really matter

Wait aren't you an english teacher?

ACuriousMind

@Obliv Germany had a physicist has chancellor for 16 years; she's still alive

DIRAC1930

What are you trying to be?

ACuriousMind

@Obliv no that's Sanjana

123

In the 2nd paragraph they take radius of circle $\ell cos\alpha$ which is wrong it should be $\ell sin\alpha$ and after few lines they said $|r| = \ell$.

Pls correct me. Where i am doing mistake

DIRAC1930

I'm just a failed condensed matter theorist and a failed fundamental theoretical physicist

Obliv

11:12 PM

@ACuriousMind Hah, well that's probably quite rare. My depiction of physicists are that they're pretty bad with politics/social issues.

Silly Goose

I am confused about whether the expression for electromagnetic potential energy is gauge invariant or not

Obliv

@123 I'm guessing the diagram is drawn incorrectly, just pretend $\alpha$ is the angle $\frac{\pi}{2} - \alpha$ in the drawing?

Silly Goose

also about the aharnov-bohm effect at a superficial level: so can we interpret this effect to be a consequence of the fact that we assumed two electromagnetic potentials to be gauge equivalent because under "usual circumstances" they are. when in reality some nontrivial mathematical result actually implies that they are not gauge equivalent?

ACuriousMind

@SillyGoose what do you mean by "the [...] electromagnetic potential energy"?

123

@Obliv Oh.. I see. Thanks. Let me think this way

ACuriousMind

11:19 PM

outside of electrostatics there is no "potential energy" in the usual sense there

Silly Goose

well let me ask what i was originally looking into

ACuriousMind

@SillyGoose No.

The A-B effect for two gauge-equivalent potentials is the same

Silly Goose

so i wanted a canonical example of when the hamiltonian cannot be interpreted as total energy. i also wanted this example to not be a dissipative system.

i thought the hamiltonian corresponding to the electromagnetism lagrangian was such an example. however, i am not so sure after reading a physicsstack post, which i will try to find again

ACuriousMind

@SillyGoose then you should've read this answer of mine :P

@SillyGoose "the Hamiltonian" for gauge theories is a bit of a subtle issue, because it depends on whether you mean the naive or the "extended" Hamiltonian, among other things

remember my mention of the Hamiltonian formulation of gauge theories being a constrained theory? that's what you're running into, and most elementary texts don't really treat the constrained formalism properly

Obliv

PSE has grown considerably since I first joined. It's comforting to know if I ever have a physics related question there's a good chance it's been asked on main.

And a decent chance ACM has a really nice answer to it

Idk how to feel about questions like this. Makes it feel like quora

inb4 acm says to do your daily review queues then

Silly Goose

11:34 PM

where would be the most precise (and in modern notation) exposition of the aharnov-bohm effect? I don't mean for it to have to talk in the differential geometric language, but for it still to be precise as in to not make incorrect statements. would this just be one of the original papers of the time?

ACuriousMind

most certainly not :P

but it's really not all that complicated - the integral around the loop depends only on the flux through the solenoid, which is gauge invariant

but any "modern" exposition should really talk in terms of geometric concepts like holonomy of the connection

Silly Goose

is the magnetic field inside the solenoid time-varying?

I am trying to connect this to adiabatic evolution and the notion of geometric phase from there

ACuriousMind

not usually

the current through the solenoid is usually constant

Silly Goose

I am confusing in what spaces each operation is taking place. For instance, to my understanding, on wikipedia, e.g., we are talking about the situation in which a charged particle starts in an eigenstate and ends in the same eigenstate as it moves around some solenoid in physical space, so this is a "loop" is state space.

but i am not so sure what would even cause this sort of adibatic evolution of the charged particle

ACuriousMind

@SillyGoose it's just any time-varying dynamics where the Hamiltonian changes slowly enough for the adiabatic assumption to be valid

the A-B effect is mathematically the same as a geometric phase, but it's not about adiabatic evolution, you just shoot stuff in paths around the solenoid

Obliv

11:44 PM

what is adiabatic evolution? $Q = 0$? :D

Silly Goose

so we have a solenoid and its terms in the hamiltonian. then we impose another term in the hamiltonian that vanishes on and inside the solenoid?

ACuriousMind

in this context it's the assumption that an initial eigenstate of the Hamiltonian evolves such that it remains an eigenstate of the time-varying Hamiltonian at all times

@SillyGoose I don't know what you mean

Silly Goose

@Obliv it is the time evolution of a quantum system under certain assumptions. namely, that the hamiltonian changes much slower than energy eigenstates do.

@ACuriousMind i am trying to understand what the total hamiltonian of our charged particle + solenoid system is

ACuriousMind

I don't know how that's relevant

you're not considering the solenoid as part of your system in the A-B effect, it's a background field

the particles aren't acting on the solenoid, it's just humming away and the particles are moving around it

2

Silly Goose

okay i see so then I think i mean to ask the total hamiltonian of the charged particle

ACuriousMind

11:48 PM

that's just the usual Hamiltonian of a particle in an EM field

Silly Goose

well i guess I am confused what it means to shoot this particle in physical space in a path from one side of the solenoid to the other

123

@Obliv Thanks it worked

Obliv

Nice

Silly Goose

@Obliv this is what i gather to be the formal definition of adiabatic evolution. in particular, eq (17). this reduces to what ACM says when the evolved state is an energy eigenstate of the initial hamiltonian $H(0)$

Obliv

you lost me at "adiabatic regime"

and bra-ket notation

123

11:53 PM

@Obliv They said wrong sentence in the question part. Rod skewed at an angle $\alpha$ ,

Obliv

and eigenstates

ACuriousMind

@SillyGoose see physics.stackexchange.com/a/100305/50583

Silly Goose

an adiabatic regime is a formalization of the sentence "a time interval in which the hamiltonian varies much slower than the states of the system". eigenstates of the hamiltonian are states describing the system when the system has a definite energy. so if i measure an ensemble of my system prepared in such a state, I would receive with 100% probability an energy value $E$

@ACuriousMind oh okay this might answer all my questions. thank you

Obliv

is $n;t$ an index or something?

Silly Goose

@Obliv yes $n$ corresponds to having an energy $E_n$ and $t$ is the continuous parameter that we call time.

this is one way to define an adiabatic regime

but again it is just a formalization of a conceptually simple idea

Obliv

11:58 PM

story of 20th century+ physics and math

Silly Goose

The left-hand side of is a definition for the characteristic time of the Hamiltonian to appreciably change. Likewise, the right-hand side is a definition for the characteristic time for the energy eigenstates to appreciable change. The approximation for the right-hand side is the inner product one obtains when the Hamiltonian is time-independent.

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