The h Bar

user54412

00:01

But can a hooker satisfy @BernardMeurer as well as I can?

Joe Stavitsky

I've been at this for over 12 hours. This onedrive.live.com/… is my calculation for this problem onedrive.live.com/…

Could someone please tell me where I went wrong?

0celo7

00:16

@ChrisWhite I needeth thy help

Bernard Meurer

@ChrisWhite Nope

0celo7

@ChrisWhite Does Springer let you order the MyCopy of this?

@ChrisWhite It's a different kind of satisfaction.

Bernard Meurer

@ChrisWhite Would you be a cookie and help me out with a small piece of code when you have the time? I'm working on a multithreaded ackermann calculator

0celo7

@ChrisWhite Please...

It takes five seconds

3075

00:31

@0celo7 did you log off?

0celo7

@ChrisWhite If you don't want to check, please tell me.

A simple "no" is a thousand times better than nothing.

user54412

@0celo7 don't think I have MyCopy

0celo7

@ChrisWhite Are you on Princeton wifi?

user54412

@0celo7 often, though often that doesn't count for much

0celo7

@ChrisWhite I mean right now

user54412

00:44

stalker...

0celo7

Oh please

user54412

@BernardMeurer I can take a look.

user54412

@BernardMeurer Have you ever written an hdf5 file?

0celo7

@ChrisWhite If you see the MyCopy button for other books but not that one, it answers my question.

like rd.springer.com/book/10.1007/978-1-4614-3612-6

user54412

::connects to network::

user54412

00:48

I see MyCopy for both

0celo7

What the shit, librarian

user54412

25 dollars each

0celo7

@ChrisWhite Are you willing to order the Riemannian geometry one for me? I can send you the money however.

user54412

That's... desperate :p

user54412

I actually don't think there's any reasonable way to get money to me.

0celo7

00:52

@ChrisWhite Snapchat :P

You can send snaps with money :D

user54412

o.O

user54412

I've been told you only send snaps of GR books with cats on them

0celo7

@ChrisWhite Also of the campus, the people I stalk, and non-GR books.

@ChrisWhite I'm assuming you use some kind of bank

3075

01:12

paypal?

Bernard Meurer

01:53

@ChrisWhite A what file?

Also I'm willing to give you my hand in marriage over basic valgrind instructions

@ChrisWhite I'll put the code in a GitHub repo and send you the link tomorrow, do you have any preferred mean of communication?

dmckee

02:45

Hmmm ... One sentence answer with sophisticated vocabulary and non-trivial sentence structure on the same general topic, but not actually addressing the question. Copy sentence; paste into google between quotation marks; find original source on the first try; leave comment about plagiarism; delete. Next!

Secret

02:59

It is said that the reason why there is spontaneous emission but not absorption is because the emitted photon has higher entropy

This PSE then talks about how both photon absorption and emission result in entropy production
http://physics.stackexchange.com/questions/119872/do-individual-photons-increase-entropy-as-they-travel

In some sense it is clear on how an absorbed photon will result in the energy to be distributed into the multiple degenerate states of the object as internal energy, hence increasing entropy in the process. But if that's the case, then by symmetry, we should also exp…

3. what are the concepts needed in order to write/derive a mathematical expression that describe the entropy production in an absorption and an emission process (assuming I am dealign with a two level system with some degeneracy for simplicity). Do I need to consider about photon gas?

user116211

> Quantum mechanics is not probabilistic but deterministic and uncertain.

user116211

I'm quite confused with it.

user116211

I know unitary time-evolution is deterministic....

user116211

but what about the wavefunction-collapse? This is not at all deterministic?

user116211

And what is meant by uncertain? I'm really confused....

Secret

03:12

Decoherence as far I know is random.
And then you have the uncertainty principle where it actually make a difference which observable you want to measure first given two conjugate operators A and B

user116211

@Secret Yes, this is what commutativity of operators means....

user116211

but what is the difference between uncertainty and randomness?

Secret

So in a sense, despite we know the probability amplitude evolves deterministically (hence the probability also evolves deterministically) under the hamiltonian, the decoherence caused by a measurement randomly project the system into one of its eigenstates

So in a sense, the uncertainty is not due to us not able to keep track of all variables (because there aren't any, unless you consider nonlocal ones such as in Bohm mechanics), but it is something fundamental

As for the details, I am still reading that non commutative probability theory that Acuriousmind referred me last night, thus I a…

it makes a nice comparison to chaos, where it is also deterministic but uncertain. Except it is not random because the uncertainty is due to us having too many variables to track

Given the EXACT same starting conditions, you get the same evolution for a chaotic system
Differ by a little, and the evolution will started to diverge exponentially, and you get something like a fractal when you plot the phase diagram

(PS I am not that good at chaos theory, thus if you see any misconception, please check)

user116211

@Secret: Thanks; I'd catch you later.

Secret

03:43

(unrelated notes, because it might be dead wrong) Until a change of understanding occurs when I finally finish reading about the non communtative probabilities and von neumman measurements, my current way to interpret quantum mechanics is as follows:
1. States are real things and they can superimpose
2. States evolve deterministically under time translation
3. States can become correlated in which case what is previously considered as two states is now just one state

Basically, treating as if the maths has a physical interpretation. So far it does help me to grasp some concepts of entangle…

The only thing that is not from our intuition that I need to accept for the above interpretation is that randomness exists, and states are undetermined until they are measured (i.e. realism)

vzn

via reddit/ physics

@Secret reminds me of bose-einstein condensate. its "electron gas" in a sense.

some of your questions remind me of experiments profiled in "the quantum challenge" by greenstein/ zajonc, a really great book for this line of study.

0celo7

@Secret Do you know what an antisymmetric isomorphism is?

vzn

the atom as an obj/ entity that "dissipates energy" wrt entropy and photon emission/ absorption is a compelling picture and think there is some (major?) support for it.

Secret

I only know group isomorphisms, thus anything else I am not sure other than wild guess from the words

"some kind of map that preserve structure between two mathematical objects except with a flip in sign?"

user54412

@BernardMeurer my school email works

Secret

03:57

turns out I am not very far, look to the section "relation preserving isomorphism"
https://en.wikipedia.org/wiki/Isomorphism

Bernard Meurer

@ChrisWhite cjwhite[at]princeton.edu correct?

user54412

yep

Bernard Meurer

Alrighty, I'll hit it to you tomorrow, thanks for offering me a hand :)

user116211

@BernardMeurer: o/

Bernard Meurer

@MAFIA36790 Hey there mate

user116211

04:00

So, all settled? Are you tension-free now? @bernard

Bernard Meurer

@MAFIA36790 Not entirely, but much better

Thanks for asking :)

user116211

:D keep updating.

Bernard Meurer

@MAFIA36790 Will do!

Secret

My current understanding of entanglement is as follows:
1. There are two states A and B, and for each the possible eigenstates (outcomes) are + and -
2. Some time in the past, A and B interacted. They are no correlated and must be described by some new state $C=A \otimes B$. In particular, the possible outcomes are correlated in such as way: ++ or --
3. We then physically separate the two states far away, and then make a measurement
4. Decoherence occurs and randomly, one of the outcomes out of the possibilities is obtained

Secret

above is subjected to change when I finished reading those two resources referred to me

user116211

04:09

3

Q: "Randomness" versus "uncertainty"

Highly rated PhysicsSE contributor @CuriousOne regularly makes the following claim about quantum mechanics (e.g. here): There is no randomness in quantum mechanics, there is only uncertainty. I want to know what this is supposed to mean.

user116211

@Secret: ^^^

Secret

One prediction from my above interpretation is as follows:

suppose I have a pair of entangled photons A and B at earth and another pair of entangled photons C and D at andromeda, then given that they have not been close at some time in the past (close as defined that they are only timeline separated), when we calculate the evolution of these two systems, the state should be able to be decomposed into states (AB) and (CD) until at least 2.5 million years later, where they might no longer be able to be decomposed

(above To be checked with the actual maths)

typo: timelike, not timeline (stupid autocorrect!)

my interpretation will predict that if the above is not true, then either there is a nonlocal interaction between the two pairs of entangled states, or there is something wrong in my understandning of quantum mechanics

0celo7

04:45

@ACuriousMind So I came across this term "antisymmetric isomorphism" in my travels. A googling brought me to this wiki page. Now I need this on a vector space. So let $f:V\to V$ be an isomorphism. Then it is antisymmetric if $f(x)=y$ and $f(y)=x$ implies $x=y$?

@ACuriousMind Furthermore, is it true that if $V$ is odd-dimensional, there is no such antisymmetric isomorphism?

Jester Tran

05:04

Suppose (X,d) is a metric space and Y is a subset of X. What is the difference between the following statements:
1. Y is closed.
2. Y is closed in X?

Slereah

07:53

A set is always closed with respect to another set

(and a topology)

Slereah

08:20

"Just to see that we have is right, let us look at one particular representation (which we will simply pull out of a hat)"

Damn you Peskin

Secret

https://en.wikipedia.org/wiki/Magnetic_potential#Magnetic_vector_potential
I wonder if a magnetic vector potential can be interpreted this way (assuming magnetic fields are real and not just bookeeping device (as concluded by the previous discussion with danielsank)):

Since all magnetic field lines (except real monopoles) must be closed into a loop, the magnetic vector potential measures the strength of the magnetic field (in terms of how large the microscopic circulation (curl) is) and also how a given loop of magnetic field orient in space (the direction of the magnetic vector potential)

Slereah

Ugh

Computing algebras for the Lorentz group

So bloody long

Secret

The above interpretation has a nice (insert suitable word) that the usual analogy of scalar potential being like a hilly landscape can be recovered, if we simply generalise the notion of height into the notion of how fast something is whirling

Secret

As for what a magnetic field is (other than a bookkeeping device), I will just stick to the usual definition that it is simply one component of the electromagnetic field tensor for now

i.e. a consequence of relativity (electric field in a moving frame)

Secret

08:37

Analogy aside, it has to be stressed that the main reason of introducing potentials is because of choosing a gauge

and I don't think there is an easy way to explain to the public why it is intuitive or natural without going too abstract

Slereah

Secret

physicists, where a difference within an order of magnitude means nothing to them

this comic also seemed to accidentally demonstrate homeomorphism, I think (I don't remember if triangles are homeomorphic to circles given that the former has corners)

Slereah

Why would corners matter for a homeomorphism

Secret

08:56

Because I don't remember, my memory is failing -_-

ACuriousMind

09:35

@0celo7 Have you "tested" this definition on some simple examples? :P

ACuriousMind

10:06

Sigh...and there goes my last close vote for the day

John Rennie

11:12

@barrycarter: if you're still interested in relativistic acceleration, someone pointed out an article on Susskind's web site that has the clearest explanation I've seen:

lecture-notes.co.uk/susskind/special-relativity/lecture-6/…

Alex

Hi all

Slereah

Hello

Slereah

11:55

physics.umd.edu/courses/Phys624/agashe/F10/solutions/HW4.pdf

Holy buns

The proof for Dirac matrices and the Lorentz algebra is long as shit

Forget that noise

I'll do it later

0celo7

@ACuriousMind I'm sure you'll cite the identity as a counterexample?

ACuriousMind

@0celo7 To your question whether or not such things exist on odd-dimensional spaces? Yes.

0celo7

Let's say "besides the identity".

ACuriousMind

More generally, every idempotent endomorphism is "antisymmetric" in that sense.

ramsay

hello,
one quick question in case of light, when we plot it on spacetime diagram with some observers moving and some stationary why can't we turn spacetime diagram but rather we just squash it? (i mean lorrentz contract it)

Slereah

12:06

because a boost isn't a rotation

if it was, you'd get some pretty peculiar results

ramsay

yup i agree we will get different speed of light , but is this valid only when light is in the diagram i mean can we change the orientation of axis so to make our worldlines straight or we have to use trick by gicen lorrentz ONLY?

is my quesiton clear?

skill patrol

no

ramsay

OK, wait then!

to make green worldline straight why can't i just turn my axes but why i have top use lorrentz trick to get green worldline straight?

now am i clear?

please ping me if you get answer.
thank you!
bye :-D

0celo7

12:22

@ACuriousMind Yeah, I figured that one out.

Secret

@ramsay
The constancy of the speed of light in flat spacetime is a postulate in special relativity. It basically ties the space and time dimensions together in a specific way (i.e. the minkowski metric $ds^2=-(dt)^2+(dx)^2+(dy)^2+(dz)^2$)

Even if you don't draw the light ray, it is always "there" in a sense that every event has to make a light cone, and nothing can cut across it under a transformation

If one can simply made the green line straight by rotating the piece of paper, then you can end up cases where a timelike worldline suddenly become spacelike, which is not allowed because th…

Put it in another way, attempt to switch frames of reference in spacetime by just rotating the spacetime diagram will violate causality, and no physicists want that (and there are no experiemntal evidence of that ether)

0celo7

@ACuriousMind And I take it there's no rule against odd dimensional spaces having idempotent self maps?

@ACuriousMind I have an isomorphism $A:E\to E$, where $E$ is odd-dimensional. But I'm supposed to show that $A$ is "antisymmetric" and therefore conclude $E$ is even-dimensional, a contradiction.

> However, the rank of a skew-symmetric map is always even

PROOF

@ACuriousMind Wait, do they simply mean the matrix of the transformation is antisymmetric?

@ACuriousMind Does "antisymmetric isomorphism" in the sense of the Wiki article $\Leftrightarrow$ the matrix is antisymmetric?

Secret

12:45

https://www.sciencedaily.com/releases/2016/04/160412160346.htm
Now we are getting something interesting

Perhaps we only have a maximum of 400ms for each interval awaring what we are doing

using this finding as a guess, it seems that our perception of the flow of time is our brain stitching discrete pieces of information together

But even if that is true, what is responsible for that order of stimulus being received by the brain?
and why do the brain move through the order of sequence at all, given our best model of spacetime is a static entity

We all knew that the speed of light is the ultimate limit that information can transfer, so we might as well simplify the above scenario as how the brain received e.g. red light after green light after blue light

Secret

That is, the subjective experience that time flows (or that motion is perceived) boils down to what does the stepping through the values of the proper time parameter?

If there is no such stepping occur, there should be no change at all being perceived (because in terms of the spacetime diagram, we stay put on an event in the wordline, no matter how subsequent events is labelled with an increasing sequence of numbers $\tau$)

0celo7

@ACuriousMind Ugh, they mean that the tensor $v\mapsto A(v)$ is an antisymmetric tensor :/

Secret

or is it the mere fact that there is an ordering to the events exists is enough to define motion and change?

0celo7

13:01

@ACuriousMind Why do we write sequences without labeling the arrows

short exact sequences, etc.

@ACuriousMind Did you once quote this theorem? "A compact homogeneous space with Ric􏰆=0 is flat. Inparticular, any Ricci flat homogeneous space is flat"?

Secret

On another note, something even more interesting from emergence

sciencedaily.com/releases/2016/04/160412105320.htm

ramsay

@Secret you are, were always wonderful :-D
Thank you

ACuriousMind

@0celo7 How am I supposed to know what "they" mean? I don't even know who "they" are!

@0celo7 No.

@0celo7 what?

ramsay

ACuriousMind, are you here?

ACuriousMind

@ramsay Yes?

ramsay

13:11

:-D nothing i just wanted to know whether you have blocked me or not

skill patrol

wassup with all the permanently pinned messages?

they're blocking all the good stuff...

ramsay

13:33

@Secret what is $ds^2$ and all that stuff in $d's$ ?, i know this is super stupid question.

yuggib

sigh...

skill patrol

why the sigh?

yuggib

once the title of my papers was like "Mode Regularization for N = 1, 2 SUSY Sigma Model"

(well just once but it counts)

Slereah

that's a very titley title

yuggib

now I am preparing something entitled "Commutative limit points of nets of non-commutative measures on Weyl algebras"

Slereah

13:35

It could have been generated by Snarxiv

If you need more title ideas

snarxiv.org

4

"A Hypersurface Defect During Inflation and Monopoles"

"Some General Frameworks as Vortices in Heterotic Strings Deformed by Quasi-primary Operators"

"Conformal Gauge Theories Compactified on N Copies of C^n"

skill patrol

13:46

@yuggib i still don't understand the reason for the sigh?

Slereah

Oh my god

The source code of snarxiv

:D

abstract ::=
<asentence>. <bsentence>. <csentence>. <dsentence>.
| <asentence>. <adverb>, <asentence>. <bsentence>. <csentence>. <dsentence>.
| <asentence>. <bsentence>. <csentence>. <dsentence>. <closing>.
| <asentence>. <adverb>, <asentence>. <bsentence>. <csentence>. <dsentence>. <closing>.
| <statement>. <csentence>. <csentence>. <dsentence>.
| <statement>. <adverb>, <asentence>. <csentence>. <csentence>. <dsentence>.
| <statement>. <adverb>, <asentence>. <csentence>. <csentence>. <dsentence>. <closing>.

Here's a simple model for your abstracts, fellows

Secret

it is basically how the metric tensor for minkowski space is written. $ds^2 $is an invariant called the squared (spacetime) interval

Think it is kinda like a pythagorus theorem for flat spacetime

AccidentalFourierTransform

@DarthPlagueis man, I'm telling you: I didn't downvote, I was trying to help

ramsay

Oh, i see :-D , thank you Secret

AccidentalFourierTransform

@DarthPlagueis you can go and downvote all my posts, its fine

yuggib

13:54

@skillpatrol I am getting too mathematical

AccidentalFourierTransform

@DarthPlagueis but dont call me an asshole when its you whos being immature

yuggib

and not in the "cool mathematics" sense...in the "excessively technical mathematics" sense

skill patrol

@yuggib ok, i get it now

Slereah

You should write a fun paper

To relax

yuggib

@Slereah fun for whom?

Slereah

13:58

The Reader

yuggib

for me, even the one I'm writing now is fun :-P

Slereah

and yourself, hopefully

Well then don't complain you NERD

skill patrol

@AccidentalFourierTransform the user will not see your pings unless they have been in the room

yuggib

@AccidentalFourierTransform OfACat

Slereah

rimshot

0celo7

14:02

@ACuriousMind I'm pretty sure you have used it

@ACuriousMind well there's this short exact sequence $0\to \mathfrak{iso}_p\to\mathfrak{iso}\to \mathfrak{t}_p\to 0$ but I have no clue what each $\to$ is!

skill patrol

@0celo7 what do you think of my complaint about the star board being over taken by mods pinning messages?

0celo7

@skillpatrol it occurred to me

skill patrol

like half the board

0celo7

Mods do what mods want to do

I see no use in fighting them

skill patrol

mod=god

0celo7

14:06

pretty much

@DavidZ doesn't even tell me why he bans me any more

the first few times he at least sent an email

skill patrol

did you ask?

last time, i mean

0celo7

No

It doesn't matter anyway

He can always find a new reason to ban me 😪

skill patrol

hey, one of the pinned messages disappeared

ACuriousMind

@0celo7 Yes, I might have used it, I don't remember

0celo7

@ACuriousMind I saw it randomly while researching killing fields, and thought I remembered you using it

skill patrol

14:10

thanks, whoever did it

:-)

ACuriousMind

@0celo7 Then that's either bad style or you simply don't have the necessary knowledge - usually, unlabeled arrows are either the obvious ones or they have been constructed beforehand.

Slereah

Hm

ACuriousMind

However, in short exact sequences, the arrows are almost always obvious: The first one is the inclusion, and the second one is taking the quotient by what has been included.

Slereah

How to show that $[\gamma^0, \gamma^i] \gamma^0 = - \gamma^0 [\gamma^0, \gamma^i] $

ACuriousMind

@skillpatrol They automatically become unpinned after 14 days or so

0celo7

14:13

@ACuriousMind taking the quotient?

skill patrol

@ACuriousMind oh yeah, i forgot

Slereah

Oh wait

ACuriousMind

@0celo7 Well, I assumed you are in a category that has quotients :P

Slereah

I guess it's easy to show that the time part and spatial part of gamma anticommutes

Since $\{\gamma^0, \gamma^i\} = 0$

0celo7

@ACuriousMind wtf are you talking about

ACuriousMind

14:14

@0celo7 Do you not know what a quotient is?

0celo7

Maybe

What are we quotienting?

ACuriousMind

E.g. of a group by a normal subgroup, or of a ring by an ideal

0celo7

Yes, I know how to do that

ACuriousMind

@0celo7 That I cannot know because you didn't tell me what kind of sequence that was.

0celo7

Vector spaces

ACuriousMind

14:16

You know how to take a quotient of vector spaces.

They're just abelian groups.

0celo7

But what the hell quotient am I taking

Slereah

Guys

Is there a real difference between the usual Dirac lagrangian

$\bar \psi (i\gamma^\mu \partial_\mu - m) \psi$

And the "correct" one

ACuriousMind

@0celo7 Uh...that of $\mathfrak{iso}$ by $\mathfrak{iso}_p$, whatever that is?

Slereah

$\frac{i}{2}(\bar \psi \gamma^\mu \partial_\mu\psi + \psi \gamma^\mu \partial_\mu\bar \psi) - m\bar\psi \psi)$

ACuriousMind

For vector spaces, every s.e.s. $0\to A \to B\to C \to 0$ is one where $B\cong A\oplus C$ in such a fashion that the first map is the inclusion of $A$ and the second map is the projection onto $C$

Emilio Pisanty

14:23

0

Q: Abstract areas of mathematics in physics

I know the most important ideas of quantum mechanics relies on Hilbert spaces and Group theory. Similiarly, the most important ideas of general relativity relies on manifolds, tensor calculus, and differential geometry. Moreover, complex analysis and differential equations are used in physics. I ...

mathematical-physics mathematics

↑ I'm gonna go get some popcorn

0celo7

@ACuriousMind iso is the Lie algebra of the isometry group and iso_p is the set of killing vectors at p which happen to vanish

ACuriousMind

@Slereah $\psi\gamma\bar\psi$ doesn't make sense. $\bar\psi$ is a row vector, $\psi$ is a column vector, $\gamma$ doesn't naturally act on those in this order.

@0celo7 So what is your question?

Slereah

Well you know what I mean

0celo7

@ACuriousMind I'll work out the proof and let you know.

ACuriousMind

@Slereah No, I don't, else I would have offered a guess

0celo7

14:25

I might not have one, I was just lamenting the fact that these books make you hunt for the arrow definitions.

Slereah

$\frac{i}{2}(\bar \psi \gamma^\mu \partial_\mu\psi + \partial_\mu\bar \psi \gamma^\mu \psi) - m\bar\psi \psi)$

IF YOU PREFER

M'lady

AccidentalFourierTransform

@skillpatrol crap... whatever, thank you!

@yuggib Meow! ;-)

skill patrol

@AccidentalFourierTransform np

0celo7

Or is t_p the zero killing vectors. Oh well, in class now

ACuriousMind

@Slereah Ah...well, just integrate by parts, no?

Slereah

14:27

Could be!

ACuriousMind

I.e. the actions are equal up to a boundary term, hence they have the same e.o.m.

Slereah

Sounds about right

Although

WHAT IF THE BOUNDARY TERM IS IMPORTANT

How do you even deal with boundary terms if the space is compact

aaaaah

ACuriousMind

@Slereah There are no boundary terms for integrals over compact spaces

Slereah

Exactly

How does integration by part work out then

ACuriousMind

It works, the boundary term is just always zero!

Slereah

14:29

that is good to know

ACuriousMind

...and now I hope that's actually true :D

0celo7

@ACuriousMind I have wondered this

ACuriousMind

But yes, it should be

0celo7

Is a compact manifold with boundary always boundaryless?

We always "use" this

ACuriousMind

@0celo7 Of course not, closed intervals are not boundaryless

Bass

14:30

@Slereah Aren't they just the conjugate forms of each other? There's the Dirac lagrangian $\mathcal{L}_\rightarrow = \bar\psi(i\gamma^\mu\partial_\mu-m)\psi$, and there's its conjugate form $\mathcal{L}_\leftarrow=(\mathcal{L}_\rightarrow)^\dagger$ where the partial derivatives act on $\bar\psi$ instead of on $\psi$. Since they lead to the same EOMs, you can mix them like $\mathcal L=(1/2)(\mathcal L_\leftarrow+\mathcal L_\rightarrow)$.

ACuriousMind

On a manifold with boundary you of course have boundary terms

0celo7

I'm confused

Slereah

No, one is symmetric and the other isn't

0celo7

How the hell does Stokes theorem work anyway

Slereah

But I think it's an integration by parts business, yeah

Bass

14:32

@Slereah Isn't the symmetric one just the sum of the two non-symmetric forms?

Slereah

No

The derivative applies to the spinor and its conjugate in that one

Oh yes I suppose that's what you meant

yuggib

@EmilioPisanty I don't think many people will answer...

Bass

@Slereah Yep, there's the "asymmetric one" $\mathcal L_\rightarrow=\bar\psi(i\gamma^\mu\partial_\mu-m)\psi$, and there's its conjugate $\mathcal L_\leftarrow=\bar\psi(i\gamma^\mu\overset{\leftarrow}{\partial_\mu}+m)\psi = -(\mathcal L_\leftarrow)^\dagger$.

ACuriousMind

@yuggib Perhaps someone will do p-adic quantum mechanics ;P

Slereah

@ACuriousMind : It's been done before, I think

ACuriousMind

14:36

@Slereah I know

yuggib

@ACuriousMind ahahah luckily the guy asked for mainstream physics

Secret

> No, you cannot think of the vector potential as the "momentum per unit charge stored" because the vector potential is not unique. I can add any gradient of a scalar function to it and the result is still a valid vector potential. – ACuriousMind 4 hours ago

Bass

@Slereah So there's no need for partial integration.

yuggib

let's not forget there's also adelic QM

Secret

@acuriousmind when we tried to fix the gauge freedom for the scalar potential, we often sort of end up fixing some zero reference point for the scalar potential

Is an analogous thing happened for the vector potential when we fix the gauge freedom by fixing the grad(V) term, that is the grad(V) is sort of like a zero value for the vector potential except it is a vector quantity?

ACuriousMind

14:40

@Secret Yes, but it destroys the simple interpretation of "choosing a zero reference point" as $\nabla \phi$ is an arbitrary curl-free vector field.

Secret

make sense

ACuriousMind

The freedom here is a true local gauge freedom, and corresponds to the $A$ overcounting the actual d.o.f.

0celo7

@Slereah Kobayashi Nomizu is in New York

At least the first volume is

Slereah

Good thing we're not on 9/11 then

0celo7

Jesus

Oh, it's in NJ now.

I swear to god if it's in Chinese

Slereah

14:49

Secret

This is chinese (simplified chinese to be exact)

Jiminion

Dumb question: Is SUSY tied to string theory, or are they separate and separable things?

Secret

the two character word above is time and the two character word to the right is space

@Jiminion If my memory serves, a lot of versions of string theory predict SUSY particles, thus if SUSY is out, string theory won't live long

0celo7

@Slereah it's not a relativity book

yuggib

@Jiminion I think they are separable

Obliv

14:51

does "has an inverse" imply an inverse function or simply an inverse (strictly speaking)?

yuggib

@Obliv depends from the context

Secret

An inverse is anything that when operated on an element in a group gives an identity element

Obliv

what if there are more than 1 identity elements?

like $f(a_1) = b$ and $f(a_2) = b$ the inverse $g(b) = a_1$ and $g(b) = a_2$ so if you say $b$ has an inverse

doesn't really imply a function does it?

David Z

Isn't there a proof that a group's identity element is unique?

Slereah

yes

yuggib

14:53

@DavidZ of course

Secret

but I think obliv is talking about inverses and identity elements in general, in which case we might be in semigroup or mooned territory

yuggib

I like mooned territory

Secret

monoid, stupid autocorrect!

Slereah

Hm

Peskin says that $\sigma^2 \vec \sigma^* = - \vec \sigma\sigma^2 $ is an identity

But I can't prove it at all

Isn't $\sigma^2$ just $4 I$

And $\sigma$ hermitian

I guess $\sigma_2^* = -\sigma_2$, if we go by representation, but it's the only one

Secret

https://en.wikipedia.org/wiki/Inverse_element
Just like (S,*) can have several left identities or several right identities, it is possible for an element to have several left inverses or several right inverses (but note that their definition above uses a two-sided identity e). It can even have several left inverses and several right inverses.

If the operation * is associative then if an element has both a left inverse and a right inverse, they are equal. In other words, in a monoid (an associative unital magma) every element has at most one inverse (as defined in this section). In a monoid…

Thus you cannot have more than one two sided inverse unless your structure is non associative (which is a pain to work with because you need all those brackets)

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