@SillyGoose that's not a feeling, that's because QM deals in vector spaces as state spaces exclusively, while classical mechanics only does so accidentally
Separately, suppose i have the EoM of the proca lagrangian density. This EoM is not gauge invariant w.r.t. adding an arbitrary divergence to the field. Is this manifested in the fact that the “Lorentz gauge” is fixed for the proca EoM?
i.e. the proca EoM is only equivalent to the Maxwell EoM in the m->0 limit if the Lorentz gauge is chosen for the maxwell EoM
Based on what you've said above you'll do fine, everybody struggles with these and you just go with it and when it clicks it will have been worth it, try the 'shorter course' and dip into the main volumes if/when you need or for more info, you'll almost certainly learn more this way
I am a bit confused though because if i decompose the proca eom into the klein gordan part and what i just wrote above, isn’t it not unique the ways we can recombine these two conditions into a single equation
so the proca eom is $\partial_\mu Z^{\mu\nu} + m^2 Z^\nu = 0$, which is equivalent to $\square Z^\mu + \partial^\mu \partial_\nu Z^\nu + m^2Z^\mu = 0$, which implies (upon contracting with a partial derivative) $\partial_\nu Z^\nu = 0$.
This allows us to write that the proca EoM implies that $\square Z^\mu + m^2 Z^\mu = 0$ given $\partial_\mu Z^\mu = 0$
it seems like this final statement is "equivalent" to the original proca EoM
but I can recombine these two conditions in more than one way, not just back into the original proca EoM
@SillyGoose What do you mean it seems like that? You just said that $A\iff B$ so for every $B\implies C$, we also have $A\implies C$. It's literally impossible for one element of an equivalent pair to imply "more" than the other
I managed to do it: I expressed a $2 \times 2$ matrix with a homogeneous polynomial $ax+by+c xy+d$. The coefficients are more or less simple: $d$ is the "1st" element of the matrix $a_{11}$, $c$ is the trace-"antitrace" (Antitrace is the sum of anti-diagonal elements) i.e. $a_{11}+a_{22}-a_{12}-a_{21}$. Finally, $a$ is -difference between the elements on the first column (i.e. $-(a_{11}-a_{21})$) and $b$ is -difference between the elements on the first row (i.e. $-(a_{11}-a_{12})$).
One curious property which I cannot explain "intuitively" is $x \to x^m$ or $y \to y^n$ for $n \in \mathbb{N}$or both is keeping the coefficients same i.e. the same matrix can be described by $ax^n+bx^m+c x^l y^p+d$ for any $n,m,l,p \in \mathbb{N}$!
I mean that there are essentially (at least) two very different paths to "mathematize" gauge theories - either via the Lagrangian formalism, principal bundles etc. or via the Hamiltonian formalism, the theory of constraints, etc.
but in either case, classical EM is a very boring theory where a lot of the abstract gauge theory isn't really needed - the gauge group is Abelian - so I'm not sure what you hope to gain from this emphasis
@SillyGoose There is a little known fact about classical EM i.e. it has quasisoliton solutions too: the Hopfion solution. This is described in Nastase's classical field theory text which deals with a lot of classical field theory stuff including EM. I am not saying that it is what you are looking for, though.
@SillyGoose I think you missed the million of times that ACM was talking about how he hated that framing. People worked out the various QFT, and then worked backwards to argue that they have certain local gauge symmetries. They did not take theories that have global gauge symmetries and then made them local. Or imposed any gauge invariance. And all that jazz.
@SillyGoose Look at it like this: You don't have to "fix" how the field transforms. There's an infinite family of possible gauge transformations that act on $A$ as $A\mapsto A+\mathrm{d}\alpha$ and on $\psi$ as $\psi\mapsto \mathrm{e}^{\mathrm{i}e\alpha}\psi$, parametrized by the charge $e$. Exactly one of these families will be a symmetry of the action, namely the one whose $e$ matches the $e$ in the interaction term between $\psi$ and $A$.
in particular if you omit the interaction term, then $e=0$ is a symmetry and the field remains uncharged
that this transformation is a symmetry of the action is just a fact, it does not depend on any choices
you don't need to fix a priori how anything transforms under "the gauge symmetry"
you just look at the action without any preconceived notions and notice that that transformation is a symmetry
just like you notice any other symmetry: You don't look at the Kepler problem and declare "a priori" how stuff has to transform under the hidden SO(4) symmetry associated with the LRL vector, you just happen to stumble upon this conserved quantity and then figure out the symmetry associated to it - all physical symmetries are "in the action" the moment you write the action down, there are no additional inputs needed
@SillyGoose by Noether's second theorem, gauge symmetries are associated with identities between the EL equations of the fields, and ultimately gauge symmetries mean that your generalized coordinates/fields are not truly independent. In the Hamiltonian formalism that manifests directly as first-class constraints.
Under mild assumptions, the existence of gauge symmetries is equivalent to such constraints, meaning if your theory is unconstrained after Legendre transform, it cannot have any gauge symmetries
in my view above there's a lot of potential symmetries, but in fact it's very rare any of them turns out to be an actual symmetry
I'm designing a sensor which will measure scattered light. There will be a sensing beam, and a feedback beam. The sensing beam will carry in the order of 1 uW (micro). The feedback beam will carry in the order of 10 mW (milli).
I’m thinking about crossing two of the beams, because space is at premium. There’s no fundamental reason for crossing them.
The Monster Group is the largest "sporadic simple group" with order:
808,017,424,794,512,875,886,459,904,961,710,757,005,754,368,000,000,000
I read that if you compactify 26D bosonic string theory on the 24D torus given by the Leech Lattice, then you end up with a vertex algebra with Monster Gr...
@ACuriousMind so i guess i am wondering why we are specializing to transformations of the form you wrote down. I could do something crazy like $A^\mu \mapsto A^\mu + B^\mu(x)$
@ACuriousMind Okay so we really are considering every possible action on the fields—it turns out that the potential actual gauge transformations are of the form you wrote
I remember an insightful conversation with ACM on the same topic. "Gauge symmetry arises by making global symmetries local" is an extremely widespread misconception (mis-motivation?) with origins in the wording in the original Yang Mills papers from the '50s.
this is actually easier than finding global symmetries because there's no algorithm to determine all conserved quantities of a system, but the passage to the Hamiltonian formalism does have an algorithm to generate all constraints (but unfortunately there the devil is in the details of proving equivalence between first-class constraints and gauge symmetries instead)
although I agree with ACM that the gauge principle is... well, a principle and there is no a priori reason to "gauge" a global symmetry (that is, to construct a Lagrangian having the "local version" of that global symmetry), it works fine nonetheless
@Mr.Feynman is this true in every case? It is sort of surprising i guess if so because then a constant group is a symmetry of the action iff a parameterized set of corresponding groups is a symmetry of the action
I don't really understand the question about the "general mathematical concept" - we're looking at transformations of coordinate/field space that leave the action invariant
In other words: saying that you should promote symmetries to local for some gibberish reason is not ok, but saying that it's the right "recipe", stating it as a principle is ok
I wouldn't interpret it further. The gauge principle just is.
i guess so what is the mathematical structure of a “gauge group”. Like $e^{-ief(x)}$ is like $U(1)$ but as a function of the function $f(x)$ so like $U(1) \times F$
I think here it's about being careful in the way you state it. Like the global symmetries of complex scalar field lagrangians or Dirac Lagrangian or the QCD Lagrangian
Slereah is just alluding to the fact that when your spacetime isn't $\mathbb{R}^4$, you might not have gauge transformations (or gauge fields, for that matter) that are functions on the entire spacetime
that's where we get into the bundle stuff again :P
and in fact that's what mathematicians mostly worry about when they say they're doing "gauge theory" - their gauge theory is in the end the theory of principal bundles over arbitrary manifolds
is a gauge transformation a local coordinate transformation? Like it is an assignment of a coordinate transformation to (possibly not) each point of spacerime
@SillyGoose ill-defined question: Like with global symmetries, you can either talk about having "two symmetry groups $G$ and $H$" or you can talk about having "one symmetry group $G\times H$"
@SillyGoose no, it's not a coordinate transformation
that function $\alpha$ then acts (usually) on every field $\phi_i$ in some representation $V_i,\rho_i$ as $\phi_i(x)\mapsto \rho_i(\alpha(x))\phi_i(x)$
yes, that's right, and when someone says "coordinates" without qualification they 99% of the time mean the spacetime coordinates, not the fields as coordinates for field space
so when we say that the standard model is associated with U(1) x SU(2) x SU(3) are we saying that we have three types of fields that each transform under reps of the aforementioned three groups? These will end up being the “gauge bosons” that mediate interactions?
@DIRAC1930 This was the holding magnet of a sensor of the structure-borne sound monitoring system. It was attached to the outside of the reactor pressure vessel.
note that the photon is associated to the EM 4-potential $A$, but $A$ does not really transform in a representation of U(1): $A\mapsto A + \mathrm{d}\alpha$ is not of the form of action via a representation above, my "(usually)" is untrue exactly for the gauge fields that correspond to the gauge bosons
the gauge fields are special because they're actually the local components of a connection form; the one other place in physics where you encounter this kind of transformation behaviour is for the Christoffel symbols that transform linearly under the Jacobian but + an extra term, that extra term is analogous to this $\mathrm{d}\alpha$
and in fact for non-Abelian gauge groups you then have $A\mapsto gAg^ {-1} + \mathrm{d}\alpha$ for $g = \mathrm{e}^{\mathrm{i}\alpha}$
but you'll learn all this if you just go the normal way and learn about QCD and electroweak theory after learning about QED
i am confused about why for an abelian gauge group the conjugation by group elements on the field disappears, which looks like the group element commuting with the image of the field. But i didn’t think the image of the field lived in the lie algebra in any sense
there isn't really a truly separate "weak" mediator field because of how the symmetry breaking works :P
the $\mathrm{U}(1)$ of electromagnetism is not the factor in the $\mathrm{U}(1)\times\mathrm{SU}(2)$, that U(1) is for hypercharge, not electric charge
that wasn't your question, but it's one more reason to maybe just sit down and learn the stuff systematically instead of worrying about not understanding very brief descriptions of how it works now ahead of time
@Mr.Feynman I remember this seminar by Vafa on the string landscape. He was saying that with quantum gravity there are some restrictions on which symmetry can be global. Honestly I don't remember the details though
@ACuriousMind instantons t
Yay
I really find annoying I had no course in instantons in my master
The Monster Group is the largest "sporadic simple group" with order:
808,017,424,794,512,875,886,459,904,961,710,757,005,754,368,000,000,000
I read that if you compactify 26D bosonic string theory on the 24D torus given by the Leech Lattice, then you end up with a vertex algebra with Monster Gr...
