46

The problem In case you were not aware of this, finding a proof for confinement is one of the Millenium Problems by the Clay Mathematics Institute. You can find the (detailed) answer to your question in the official problem description by Arthur Jaffe and Edward Witten. In short: proving confinement is essentially equivalent to showing that a quantum Yang-...


26

Most of the forces induced by a point particle follows the 1/r^2 rule No, it's the forces mediated by point particles with no mass and charge that follow the the 1/r^2 rule. then why does strong force don't obey it? The inverse square law is a consequence of the particles having no mass and/or charge. Such particles have long/infinite lifetimes and can ...


20

A theory is typically described by a Lagrangian, and varying this gives us the equations of motion of the system. The symmetries you describe are symmetries of the Lagrangian i.e. they are transformations that leave the Lagrangian unchanged. It would be nice to think that the Lagrangians that describe our leading theories of physics were derived in some ...


19

A theory is usually denominated a 'gauge theory' if all the interactions in that theory are introduced by promoting global symmetries to gauge symmetries. Note that a gauge theory is a gauge invariant theory, but a gauge invariant theory doesn't has to be a gauge theory (for example, the Standard Model is gauge invariant, but it's not a gauge theory since ...


18

As Lubos Motl and twistor59 explain, a necessary condition for unitarity is that the Yang Mills (YM) gauge group $G$ with corresponding Lie algebra $g$ should be real and have a positive (semi)definite associative/invariant bilinear form $\kappa: g\times g \to \mathbb{R}$, cf. the kinetic part of the Yang Mills action. The bilinear form $\kappa$ is often ...


18

If we write the field strength in terms of "electric" and "magnetic" fields $\vec{E}$ and $\vec{B}$, the relevant expression can be written as $$\text{Tr}F_{\mu\nu}\tilde{F}^{\mu\nu}=4\,\text{Tr}\vec{E}\cdot\vec{B}.$$. Under parity transformations, $\vec{E}\rightarrow-\vec{E}$ and $\vec{B}\rightarrow\vec{B}$, while under charge conjugation, $\vec{E}\...


15

Free nonabelian gauge theory is the limit of zero coupling. So it is true, in a sense, that free $SU(N_c)$ gauge theory is a theory of $N_c^2 -1$ free "photons." There's a crucial subtlety, though: it's a gauge theory, so we only consider gauge-invariant states to be physical. Thus, already in the free theory, there is a "Gauss law" constraint that renders ...


14

In any supersymmetric theory you can choose the gauge coupling to be the coefficient of $W_\alpha^2$ in the superpotential. This gauge coupling runs only at one-loop, which is a fundamental consequence of the non-renormalization theorems. The other possible running coefficients are the kinetic terms, $Z(\mu)QQ^\dagger$. These generally get renormalized to ...


14

It's not a sufficient explanation. There are asymptotically free theories which are not strongly coupled in the IR. The rate at which the coupling gets strong is important. In QCD, it seems to get strong very quickly near the confinement scale, so that beyond a certain scale, you only see hadrons. It is not really understood how this works. The ...


14

You've got things slightly backwards. In constructive QFT, one almost always starts in Euclidean spacetime -- where it is 'easy' to define the path integral -- and then analytically continues to get correlation functions on Minkowski spacetime. (The Osterwalder-Schrader Theorem tells you when this analytic continuation 'works', meaning when the resulting ...


13

The possibility of spontaneous Lorentz symmetry violation due to the infrared problem of the Dirac-Maxwell equation was conjectured a long time ago by Frohlich, Morchio and Strocchi, in references [1,2] mentioned in the given Balachandran and Vaidya article. In perturbative QED, we usually assume that the scattering states are free eigenstates of the number ...


13

The usual action for Yang-Mills theory is, using differential forms $$S = \int \operatorname{tr} (F \wedge \star F)$$ where $\star$ is the Hodge dual. Now note that the integral of a differential form is always defined with respect to an orientation, and the Hodge dual is also defined with respect to an orientation. Parity is reversing the orientation, which ...


12

In Yang-Mills, the gauge connection plays the role of a potential and the curvature form plays the role of a "field strength". In GR, the metric tensor plays the role of a potential, and the connection plays the role of a field strength. This is why, in particular, the gravitational force is not a real force, as the connection is not gauge-covariant. Of ...


11

Wu and Yang (1968) found a static solution to the sourceless SU(2) Yang-Mills equations, (please, see the following two relatively recent articles containing a rather detailed description of the solution: Marinho, Oliveira, Carlson, Frederico and Ngome The solution constitutes of a generalization of the Abelian Dirac monopole. The vector potential is given ...


11

The original papers by Gerard 't Hooft himself are quite readable. On the Phase Transition Towards Permanent Quark Confinement A Property of Electric and Magnetic Flux in Nonabelian Gauge Theories Topology of the Gauge Condition and New Confinement Phases in Nonabelian Gauge Theories Whenever I open these papers, I'm always awestruck.


11

I recommend that you to read the chapter 15.2 in "The Quantum Theory of Fields" Volume 2 by Steven Weinberg, he answers precisely your question. Here a short summary In a gauge theory with algebra generators satisfying $$ [t_\alpha,t_\beta]=iC^\gamma_{\alpha\beta}t_\gamma $$ it can be checked that the field strength tensor $F^\beta_{\mu\nu}$ transforms as ...


11

It seems the resolution to OP's question lies in the difference between the Levi-Civita symbol, which is not a tensor and whose values are only $0$ and $\pm 1$; and the Levi-Civita tensor, whose definition differs from the Levi-Civita symbol by a factor of $\sqrt{|\det(g_{\mu\nu})|}$.


11

I) Vanishing field-strength $F=0$ does not imply that the gauge potential $A$ is pure gauge. It only holds locally. There could be global obstructions. In fact, topological obstructions could happen even if the gauge group $G$ is Abelian. II) Let us sketched the proof of the local statement in a sufficiently small neighborhood $\Omega\subseteq M$ of a point ...


11

Bundles and compactified spacetime A gauge theory cannot be looked at purely locally, it has inherently global features on cannot see locally. The proper mathematical formalization of a Yang-Mills gauge theory is that the gauge field $A$ is a connection on a principal bundle $P\to M$ over spacetime $M$. However, in practice, it turns out that physicists don'...


10

Yes, one traditional alternative to the path integral formalism is the operator formalism. For QED with abelian gauge group, the old quantization formulation is the Gupta-Bleuler formulation. For QCD/Yang-Mills theory with non-abelian gauge group, the Gupta-Bleuler formulation is replaced by the BRST formulation. The BRST formulation exists in at least 3 ...


9

From the beginning of the wikipedia page on Yang-Mills theory (have you read it?): "Yang–Mills theory is a gauge theory based on the SU(N) group ... ... In early 1954, Chen Ning Yang and Robert Mills extended the concept of gauge theory for abelian groups, e.g. quantum electrodynamics, to nonabelian groups to provide ... ... This prompted a significant ...


9

Since this question is still open and therefore not definitely answerable at present, I save the valuable discussion of the topic in the comments as an answer such that it does not get lost: This is just an accident of 10 dimensions--- there is too much supersymmetry to have a full SUSY superspace. It's a very good question, but research level, if you ...


9

It's because you want the kinetic part of the Yang Mills action $$ \int Tr({\bf{F^2}}) dV$$ to be positive definite. To guarantee this, the Lie algebra inner product you're using (Killing form) needs to be positive definite. This is guaranteed if the gauge group is compact and semi-simple. (I'm not sure if it's only if G is compact and semi simple though. ...


9

First impressions based on a quick read of the preprint: I'm out of my depth on this! I couldn't tell you if their derivation is correct, but assuming that it is: They don't treat real QCD. They study SU(2) YM without quarks. The authors claim they can do real QCD and get the same result, but this is not demonstrated in the paper (they defer this to a later ...


9

Gravity isn't Yang-Mills theory in the narrow sense – well, except for equivalences such as AdS/CFT or Matrix theory that imply that a quantum gravitational theory is fully equivalent to a gauge theory living in a different space (e.g. in AdS/CFT, on the boundary of the AdS space). However, gravity is a gauge theory in the broader sense because it's ...


9

OK, blathering it is. The key thing to understand in Yang-Mills theory (whether or not you're trying to be mathematically rigorous) is that gauge symmetry is not a physical symmetry. (Its job is to keep track of the overcounting in a redundant description.) Classically, this means that the observables -- the numerical quantities which we can measure -- ...


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