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00:36
does anyone know of a (shorter) introduction to gauge theories? as in, more like an introductory note of a length like 20 pages or so
@ClaudioMenchinelli yes
@SillyGoose what do you expect to learn in 20 pages :P
a very broad picture of the structures involved :P at least a naming of what mathematical things correspond to what physical things
also: What kind of "gauge theories"? Arbitrary ones? Yang-Mills theories? Literally just electromagnetism? Lagrangian formulation? Hamiltonian formulation?
(but not necessarily definitions of the mathematical things)
maybe just electromagnetism or the simplest case that still shows off the characteristics of a general gauge theory
I mean, the core characteristic is just that the theory is symmetric under a space-time dependent symmetry $A\mapsto A + \mathrm{d}f$ for any function $f$
you don't even need a single page for that :P
00:44
what is $A$?
the 4-potential?
@SillyGoose I mean you could start here
A gauge theory is a type of theory in physics. The word gauge means a measurement, a thickness, an in-between distance (as in railroad tracks), or a resulting number of units per certain parameter (a number of loops in an inch of fabric or a number of lead balls in a pound of ammunition). Modern theories describe physical forces in terms of fields, e.g., the electromagnetic field, the gravitational field, and fields that describe forces between the elementary particles. A general feature of these field theories is that the fundamental fields cannot be directly measured; however, some associated...
@ACuriousMind what is meant by the theory is symmetric?
At least as far as I've been introduced to them, it's largely about finding potentials that yield equivalent forces/fields/etc.
@SillyGoose just the ordinary sense of symmetry? Like, the action is invariant under this transformation
00:46
Think about how shifting your potential by a constant doesn't have any effect on the force
There are generally other ways you could change a potential and still end up with the same physical results
Different potentials that yield identical physical results (regardless of which potential you pick) are said to be gauge invariant among each other
Typically the choice of potential is just a matter of mathematical convenience. E.g. sometimes if we have a given electric field, it'll be useful to choose a potential that satisfies certain conditions (e.g. regarding its derivative)
@ACuriousMind should i have in mind quasi-symmetries as described by QM here: physics.stackexchange.com/questions/11313/…? Instead of leaving the action entirely invariant: $\delta S = 0$?
@SillyGoose you should always be prepared for "symmetry" to actually mean that notion of quasi-symmetry
but that's not special to gauge theories at all
@SirCumference hm i think i get this notion, i think i am wanting to gain some familiarity with the terminology/structures of gauge theory
the problem is that there really isn't a lot of elementary examples of gauge theories
@SillyGoose As usual I would start with a simpler intuition/familiarity and build up from there
00:51
or, rather, there isn't a lot of non-artificial elementary examples
really usually the only gauge theory you'll see is electromagnetism until we get to Yang-Mills theories to describe QCD
In e.g. electromagnetism, there are plenty of potentials that all yield the same electromagnetic fields, so it's often a matter of using the most mathematically convenient one (i.e. the most preferable "gauge")
and electromagnetism is a really poor example of the general "structures" of gauge theories for various reasons
cripes
Tbh for now just keep it simple and know that it's a matter of swapping between potentials and still getting the same end result
There's a much deeper discussion on the relationships between gauge invariant potentials, but at the moment I'd say go through the Introduction wikipedia page
e.g. you'll find out about how you can identify "convenient" potentials by making some mathematical restrictions (for instance, requiring $\nabla \cdot A = 0$)
as well as whether certain restrictions uniquely identify a potential or not
> In field theories, different configurations of the unobservable fields can result in identical observable quantities. A transformation from one such field configuration to another is called a gauge transformation; the lack of change in the measurable quantities, despite the field being transformed, is a property called gauge invariance.
This line is pretty much the general idea
there's already one problem: Because physics often only encounters gauge theories in the context of EM and Yang-Mills theories, people tend to talk about it as field theories, but there are definitions of gauge theories that apply to finite-dimensional (i.e. non-field-theoretic) systems just as well
yet understanding the formulation of gauge theories in this general sense will take a lot of effort to connect to the way physics usually talks about them
 
3 hours later…
03:49
You guys know anything about the standard model?
03:59
What are we supposed to do with a question like this?
04:20
I rly like EnM cos there are no particles in the fundamental equations of Maxwell, you just need $\rho(\vec r, t)$ $\vec J(\vec r, t)$ and you're basically done in principle. Everything's a fluid !1!1!1
consider the irreps of $sl(2; \mathbb{C})$ labeled by positive integers $m \in \mathbb{N}$.
is the fact that irreps labeled by some fixed $m$ are isomorphic representations a consequence of Shur's lemma?
I am not seeing how this proof I am following (from B.C. Hall) proves that same $m$ irreps of $sl(2;\mathbb{C})$ are isomorphic
it seems like i could use shur's lemma given that i construct an intertwining map between the irreps $\pi_m$ and $\pi_{m'}$ where $m = m'$ that is nonzero
oh wait
nvm I don't get it
05:19
@nickbros123 until your fluid is made of point charges...
05:31
@naturallyInconsistent ignores and proceeds to average the density and work with smooth fluids anyway
 
2 hours later…
07:28
@thesmartwaterbear yes
08:09
@SillyGoose I mean, if you have an intertwining map, that is the isomorphism, isn't it?
08:21
well i haven't come up with an explicit intertwining map, and the proof in Hall does not mention Shur's lemma. So I feel I am missing a more direct way of proving the statement
y is an equivalence class defined using $f^{-1} gf$? in linear algebra, this means a change of basis. what does this mean in english in the abstract?
we have access to the above explicit constraints on the irreps of sl(2;C)
@SillyGoose The map is just mapping the eigenvectors $u_i$ of one representation to the eigenvectors $u'_i$ of the other
you just need to verify this is an intertwiner if it's not obvious to you
(also it's Schur not "Shur")
 
1 hour later…
09:51
all abelian grps r simple
but not all simple grps r abelian
this means that the derived subgroup of a non-abelian grp can be identity?
@RyderRude no they aren't
trivially, $\mathbb{Z}/n\mathbb{Z}$ is not simple for $n$ not prime
i havnt learned quotients
ok so derived subgroup is a recipe to make normal subgroups
it's just the cyclic group of order $n$ (also written $\mathbb{Z}_n$)
09:54
so all simple grps must have identity as the derived subgroup
@ACuriousMind oh
this means that non abliean grps can hav identity as the derived subgroup?
i will write the logic clearly : 1. derived subgroups are always normal subgroups 2. simple groups have no normal subgroups. 3. the derived subgroup of a simple group must be identity
is this correct
which point is incorrect
it is suposd to be logical implication
your 2nd point is not true
but it's their definition : no non trivial normal subgroups
but you didn't say "non-trivial"
and the identity is not the only trivial normal subgroup
10:00
oooh
so that's y 3 doesnt hold
is the other option the entire group as the subgroup?
is this trivial
and normal
so 3 gets modified to : the derived subgroup of a simple is group is either the entire group or identity
so simple grps need not be abliean
only the latter case is abelian
thanksss
10:26
i feel there's no many definitions in grp.theory
math used to be grounded before this
i dont like abstract things as much anymore
11:06
what is the physical meaning of $f^{-1}gf$?
it is suposd to give equivalence. like how x rotations are the same kind of thing as z rotations
Just because something is mathematically called an "equivalence class" doesn't mean it has to have some sort of physical meaning of everything in that class being equivalent. An abstract group doesn't have any "physical" meaning at all.
there's many different equivalence relations on groups, conjugacy is just one of them
so this is just one kind of equivalence
in what sense does it give an equivalent element
I don't understand the question
$g\sim f^{-1}gf$ is an equivalence relation
why is this called an equivalence relation
purely by the mathematical definition of what an equivalence relation is
11:13
oh
they're called that because you can quotient any set by any equivalence relation and get a smaller set in which elements that were "equivalent" in the original set are now equal
the idea being that an "equivalence" generalizes the notion of equality
ooh so we r identifying equivalent elements in the original set as a single element and forming a new grp?
is this related to quotient grp?
oh so this is a general set theoretic idea
not just grp theory
but there are many different equivalence relations on a group (any set, really), it just turns out that conjugacy is a useful tool in group theory, e.g. each conjugacy class of a finite group corresponds to an irrep of the group
11:19
ok any relation which satisfies those three criteria defines equivalence of the elements in a set
and this f^{-1}gf does the three criteria
@ACuriousMind oh
so this is y this is useful. otherwise, i can construct arbitrary relations that satisfy the three criteria
so this quotient thing is a general idea
related to sets
Also: Conjugacy should be familiar to you from linear algebra - there we call two conjugate matrices "similar", and you should know that similar matrices share a lot of properties
i identified this with a change of basis there, which gives the same matrices in a different representation
i havent heard of "similar" matrices. but i know change of basis
ooh but these matrices have the same eigenvectors and eigenvalues
u can interpret it as an active transform instead of a change of basis
@ACuriousMind the book mentioned one thing..the equivalent elements have the same cyclic structure
but im not sure how this is useful... i think ill know in rep theory
this notion was useful in constructing quotient grp
so it has uses i guess
a change of basis pretty intuitive to me
but $f^{-1}gf$... idk how to interpret it
maybe i shud think of these as matrices as all finite grps are matrix grps
but one thing is that, in a change of basis, we r allowed to use arbitrary bases $B$
@RyderRude "change of basis" is indeed the intuition here - two matrices are similar if and only if they are the expression for the same linear operator in different bases
in this group notion, two elements are equivalent only if another element relates them. so there may exist a change of basis between the matrices and they may still not be equivalent in this group notion of equivalence
this is y i hav trouble relating this to a change of basis
@RyderRude Why would that be?
for any change of basis, there is an invertible matrix $B$ (its columns are the new basis vectors in terms of the old basis vectors) that relates matrices $M$ in the original basis to matrices $BMB^{-1}$ in the new basis
11:34
lets say i have permutation matrixes and two non equivalent permutations and there still exists a change of basis transform between them
yes but what if B is not part of the group?
so even if B exists, the matrices r not equivalent
sure, if you restrict to matrix groups that are smaller than $\mathrm{GL}(V)$, then you're excluding some changes of basis
that's why conjugacy generalizes the notion of similar matrices
yeaah
so it's just equivalency under chnage of bases but restricted
it's like a group acts on itself using two notions
one is gG
and the other is g^{-1} G g
in QM, we never use the gG notion of a group acting on itself
the heisenberg picture uses g^{-1} G g
You have to think about the latter in the sense of matrices: When a group $G$ acts on a set $X$, then you can do a "change of basis" by $g\in G$ by $x\mapsto gx$. Then the action of any other $h\in G$ on $X$ goes to $x\mapsto g^{-1}hgx$ by the same logic as for actual basis changes in linear algebra
discussed on math.SE e.g. here
thankss
 
3 hours later…
14:37
Why is the normal to a stationary hypersurface proportional to $\partial_\mu r$?
Shouldn't it be proportional to a timelike killing vector?
14:54
@Slereah basically I'm trying to understand why the $g^{rr}=0$ is a sufficient condition for an apparent horizon in Kerr spacetime. Of course that condition means that a hypersurface normal to $\partial_\mu r$ is lightlike but I'm confused about the statement below
The null condition is obvious but what about the "stationary surface" condition?
Am I possibly misinterpreting the meaning of "stationary" here? One thing is that a spacetime is stationary (there is an asymptotically timelike Killing vector field) but a hypersurface?
15:11
Wait, I guess that for hypersurfaces "stationary" means that the Raychaudhuri expansion is zero
 
3 hours later…
18:12
happy new year
bit late but have been busy
@Mr.Feynman what book is this?
18:28
@ekardnam_ Eric Poisson: A relativist's toolkit
2024 turned me into a baby Slereah
@Mr.Feynman that's why it looked familiar then
Is there some villain backstory or something?
19:20
wait no
you don't need to use Schur's lemma here at all
I'm not sure why you're so fixated on that idea :P
From the picture you posted, you've already shown that every irrep is spanned by $m+1$ vectors $u_0,\dots, u_m$ with the given property. The claim that two representations of the same $m$ are isomorphic is just that $u_i\mapsto u'_i$ (where $u'_0,\dots,u'_m$ spans the second representation) is an intertwiner
hm okay i guess i can see it by direct computation
but i still feel like i am missing the point since it seems like this result should be easy to see :P
I find it easy to see :P
you already showed essentially what the structure of each irrep is here, and it's only dependent on $m$
if mapping the two $u_i, u_i'$ did not produce an intertwiner, then that would mean there's some additional structure not captured by the relations/actions on the $u_i$ you wrote down that can be violated by the intertwiner while still preserving those relations
but the $u_i$ span the space, and the action of every Lie algebra element is specified
it's the same logic as "all vector spaces of the same dimension are isomorphic" - "all representations with the same labels are isomorphic", you just have take care that the labels (here:$m$) actually specify everything
19:50
hmi think i am getting caught up because i know that the dimensions of the representation spaces being equal does not tell you in general if the representations are equivalent. and so i am trying to see why in this special case the aforementioned is true
20:00
because you showed that the representation of dimension $m+1$ always has this structure labeled by $m$
and conversely, you can construct a representation of dimension $m+1$ just from this data
so how could there be two different representations of dimension $m+1$ if they both have this structure of the $u_m$?
okay i think then what i am confused about is what it means for two representations to be equivalent
i can see the intertwiner definition, but i don't get what structures are being preserved i guess
because for, say, vector space isomorphisms, it is clear that what you are preserving is the axioms of vector space and nothing else
it is not so clear to me what in particular is being preserved: the axioms of representations?
@SillyGoose Think of vector space isomorphisms as changes of basis. The intertwiner condition says that for the change of basis given by the intertwiner, the matrices that represent the group in the first representation are transformed by this change of basis into the matrices in the second representation
i.e. this is just the statement that two representations that are the same except that you chose different bases for the underlying vector space are equivalent
 
2 hours later…
21:50
depending on the gas model, ideal or otherwise, do the laws of motion vary much?
It's said that the molecules in an ideal gas obey Newton's laws of motion (under the constraints of what is an ideal gas)
Curious how different the laws of motion become with different systems, or is it always governed by some foundational mechanics like QM or something
@Obliv Well I guess it's referring to the fact that Newton's laws don't hold in some nonclassical situations
Namely the third law not holding in special relativity (see this discussion from Taylor's book "Classical Mechanics")
So I guess for a relativistic gas that law wouldn't be applicable
22:15
nvm i get it. noble gases are monatomic lol
@SirCumference I'm taking EM & thermo this semester and I think my EM class should cover SR more in depth. Going to learn a lot hopefully
My only real goal is to be able to have a solid foundation for 20th century physics. like loretnz transforms & SR should be very familiar or easy to derive idk
23:02
Hi people!
I have a question. In hadronic collisions at CERN for example, many times we shift from the laboratory frame to boosted frames along the beam axis, why we do that?
is it true that any set $S$ is equal to the union of its elements (singletons)?
@SillyGoose what problems do you have (dis)proving this statement?
@imbAF ...because it's convenient for some computations? Presumably you're boosting with the velocity of the particles of the beam, i.e. in into the rest frame of the beam, right?
hm i guess i don't but i'm not sure if my reasoning is quite right...
I mean, you can boost at different velocities other then the velocity of the hadronic center of mass
But I guess so
Let $S$ be a set. For each element $s \in S$ construct the singleton $\{s\}$. Then, by definition $\bigcup_{s \in S} \{s\} = \{s \in S \lvert s \in S \} = S$.
i dont know i feel like for very basic set stuff, there may be more "philosophical problems" lurking or something :P
23:19
I mean if you wanted to properly "prove" this statement you'd first have to be careful about what your formal definition of the union actually is etc.
e.g. you have $x\in A\cup B\iff x\in A \vee x\in B$ and then proving such set-theoretic statements is manipulating a bunch of tedious but obvious logical equivalences
Consider some collection of sets $\{S_x\}$, each of which is contained in some universe $U$. Define $\bigcup_{x \in X} S_x = \{s \in U \ \lvert \ s \in S_x \text{ for any } x \in X\}$
hm well maybe i shall just assert this as a fact :P
math.SE has a bunch of proofs here

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