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 $f$
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...
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: $\delta S = 0$?
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")
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.
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
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
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
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
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
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
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
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
@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
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
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
@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?
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
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
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
@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
@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
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?
@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?
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
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