# Tag Info

0

A first-class constraint eliminates 2 degrees of freedom because it, one the one hand, relates the $p_i$ and the $q^i$ with an equation, and on the other hand generates a one-parameter group of gauge transformations on the constraint surface, where all states lying in the same orbit have to be physically identified. So you lose one d.o.f. because of the ...

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As far as I know, gauginos couple to bilinear combinations of particles and their partners (sparticles). E.g. as photino is neutral, it couples to neutral combinations like electrons and anti-selectrons. Gluinos couple to quark-anti-squark pairs with accordingly chosen colour. Considering MSSM here, and of course, at energies where SUSY is unbroken.

2

I believe this is a valid gauge condition. You can use the hint given by AccidentalFourierTransform in a comment and for arbitrary $A^\mu$ find such $\Omega$ that $(A_\mu+\Omega_{,\mu})(A^\mu+\Omega^{,\mu})=0$, as the latter equation can be resolved with respect to $\Omega_{,0}$, so a Cauchy problem can be posed. Of course, the usual mathematical issues of ...

4

No, the gauge group of electromagnetism is still $\mathrm{U }(1)$. This is because there is a difference between the gauge group $G$ of a physical theory on a manifold $M$ and the group of gauge transformations $\mathcal{G}$ consisting of local functions $M\to G$ and the algebra of infinitesimal gauge transformations consisting of local functions ...

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OP is basically asking (v4) the following. How the equation $$\int\! \mathrm{d}^4x~ \left(i{\cal A}(x) + i\sum_r J_r(x) \langle F^r\left[x,\Phi\right] \rangle_J\right) ~=~0, \tag{4.5}$$ or equivalently, $$\int\! \mathrm{d}^4x~ \left(i{\cal A}(x) + i\sum_r J_r(x) F^r\left[x,\frac{\delta}{i \delta J}\right]\right)Z[J] ~=~0, \tag{4.5'}$$ ...

4

The derivation in the given reference indeed seems confused and inconsistent. The crucial error seems to me that $$S[\phi + \epsilon\delta\phi] = S[\phi]$$ is just not true for an infinitesimal symmetry. The definition of a symmetry is that $S[\phi']=S[\phi]$ (modulo boundary terms) for the finite transformation $\phi\mapsto \phi'$. Writing this ...

0

This is an example of lie-algebra valued 1-forms. Actually you may write explicitly, $A = A_{\mu} ^a T^a dx^{\mu}$. Since the generators also anti-commute so we get the result. And for the same reason sometimes you will find expressions like $[A,A]$ in literature for your term.

6

That's because you are forgetting that $A$ has a Yang-Mills index. You better write this in components, which reads $\epsilon^{\mu\nu\rho} g_{IJ} \Big( A^I_\mu \partial_\nu A_\rho^J + \frac{1}{3} f^J{}_{KL} A^I_\mu A^K_\nu A^L_\rho \Big)$ This component notation also answers your second question.

1

A massless vector is the gauge field $A_\mu$ and the massless symmetric traceless tensor is the metric perturbation $h_{\mu\nu}$. They can couple only to conserved currents, namely the $U(1)$ vector current $j_\mu$ and the stress tensor $T_{\mu\nu}$ respectively. The requirement for this arises from gauge invariance. Coupling a gauge field to a current ...

2

I think what they mean is FI-term is not gauge invariant under the full gauge symmetry of the theory, but under this remaining gauge freedom after WZ gauge, which is $U(1)$.

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In a relativistic field theory, a conserved quantity according to Noether's theorem is given by a conserved current (density) $J^\mu$, i.e. $\partial_\mu J^\mu=0$. Hence, the contradiction you suggest does not really exist. The symmetry corresponding to conservation of electric charge is indeed the global part of the $U(1)$ gauge symmetry.

4

Yes, you missed the whole point of gauge invariance which clearly has to be a symmetry – the Hamiltonian has to be and is invariant under these transformations. In the case of a global symmetry, a Hamiltonian may afford "not to be symmetric" under it. But once a group of transformations is said to be a gauge symmetry, the failure of $H$ to be invariant would ...

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I used to count dofs in terms of spinor representation. There are two spinor representation of SO(3,1) (1/2,0) and (0,1/2) denoted by dot and no-dot indices. Vector (spin 1) representation in spinor indices is (1/2,0) X (0,1/2) = (1/2,1/2) which has 4 dofs. If the theory is massless, the theory is gauge invariant, there is one gauge dof, which can be ...

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