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5

First, the gauge invariance means that the solutions $A_\mu(x^\alpha)$ are not unique. For every solution, the gauge transformations of it are solutions, too. That may be a problem because sometimes we want to have specific values of $A_\mu(x^\alpha)$ that answer a physical question. Second, we sometimes gauge fix because the equations simplify. For ...


4

The problem is that they have too many solutions if the gauge is not fixed. Imagine you have some initial values, and want to solve it on the computer. Then you have to solve the equations for the next time step given the values for the previous one. But you have to compute the values for, say, four variables but have only three equations. You somehow ...


3

The first answer to such a question must always be: A gauge symmetry has no "physical" meaning, it is an artifact of our choice for the coordinates/fields with which we describe the system (cf. Gauge symmetry is not a symmetry?, What is the importance of vector potential not being unique?, "Quantization of gauge systems" by Henneaux and Teitelboim). Any ...


2

1) Note that the non-relativistic gluon model of glueballs has even less justification than the non-relativistic quark model of baryons and mesons. This is because the model messes up the spin assignments: Massless gluons have only two spin states, but non-relativistic gluons have three. 2) The color quantum numbers are trivial: The product $$ [8]\times ...


2

It depends on what you mean by vacuum. If you mean a field configuration that has $F=0$, then the gauge potential is locally pure gauge (cf. this answer by Qmechanic), so there can only be global obstructions to the gauge equivalence class of the $A$ corresponding to $F=0$ being $A=0$ everywhere. On $\mathbb{R}^{1,3}$, there is no such obstruction. If ...


2

$$ F^2 \to F^2 + \alpha_i f^{ijk} F_{j\mu\nu} F_k^{\mu\nu} $$ $f^{ijk}$ is completely antisymmetric in all its indices and is being contracted with something that is symmetric in $jk$. Thus, the action is invariant. We can see that $f^{ijk}$ is totally antisymmetric as follows $$ [T^i , T^j] = f^{ij}{}_k T^k \implies \text{tr} \left( [ T^i , T^j ] T^k ...


2

1) Universal covering groups are groups with the property of being simply connected. Each algebra has a unique covering group. The other groups, $\{G\}$, associated to the same algebra can be obtained from the covering group in the following way $$G=\frac{\tilde G}{Ker(\rho)},$$ where $Ker(\rho)$ is the kernel of the group homomorphism $\rho:\tilde ...


2

The solution that you wrote in your last (not numbered) equation is not a basis of a Hilbert space of sections because the phase factor: $(-1)^n e^{2i\pi A}$ depends on $n$. The phase factor should not depend on $n$. Please see your (correct) equation (2) defining the boundary conditions, in which the phase factor does not depend on $n$. Thus there is no ...


2

The field stregth tensor of a Yang-Mills theory is defined as $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu+ie\left[A_\mu,A_\nu\right].$$ In general, the gauge field is in the adjoint representation of the gauge group (we normally say it takes value in the algebra) so it is written as $$A_\mu=A_\mu^aT_a,\quad a=1,2,\ldots dim(G),$$ where $dim(G)$ ...


2

This is a very broad question, so there are many ways to answer it. Here is one interpretation. A principal distinction between gauge symmetries and global symmetries is that gauge symmetries lead to long-range interactions between charged particles; the gauge symmetry demands the existence of a massless field which can propagate over arbitrarily long ...


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This is just very sloppy language on the paper's part. As you say, gauge bosons are very real and their existence has physically measurable consequences (otherwise, why would we ever waste time talking about them?). (By the way "photons and electrons" are not good examples of non-gauge particles, because photons are also gauge bosons :) .) The paper just ...


2

When Maxwell formulated his equations he did so using quaternions, which BTW is a more elegant formalism, and Heaviside formulated them as we normally read them. Our standard vector forms of the Maxwell equations are more convenient for electrical engineering. These equations linearly add the electric and magnetic fields. This linear property is a signature ...


1

Quote from "Lectures on D-branes, Constantin P. Bachas" Now within type-II perturbation theory there are no such elementary RR sources. Indeed, if a closed-string state were a source for a RR (p + 1)-form,then the trilinear coupling $$< closed| C_{(p+1)} |closed >$$ would not vanish. This is impossible because the coupling involves an ...


1

If the Fourier transform of $f(x)$ is $\tilde{f}(k)$, then the Fourier transform of $df/dx$ is $ik\tilde{f}(k)$. Proof: $$\frac{df}{dx} = \frac{d}{dx} \int \frac{dk}{2\pi} \tilde{f}(k) e^{ikx} = \int \frac{dk}{2\pi} \left[ik\tilde{f}(k) \right] e^{ikx}.$$ This explains the momentum factors, so we've reduced the task to showing $$\Gamma(x, y, z) \sim ...


1

They're not different at all! Seiberg has used the term electric-magnetic duality in the title of his famous paper, too. His duality is an electric-magnetic duality because the duality relates two descriptions and objects that are electrically charged under the gauge group on one side (e.g. quarks and gluons) are mapped to solitons (forms of magnetic ...


1

The mathematical framework that I am familiar with for abelian p-form gauge theory (the one promoted by Freed, Moore and others) is that of Cheeger-Simons differential forms. In this framework, the space of topologically trivial p-form gauge fields over a manifold $X$ (the analogue of 1-form gauge fields on the trivial $U(1)$ bundle) are identified with the ...



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