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3

Let us here assume that the classical theory is given by a Hamiltonian (as opposed to a Lagrangian) formulation, so that we have a Poisson bracket $\{\cdot,\cdot\}_{PB}$ (and so that we can discuss whether or not the generators form a Poisson algebra or not). A generic theory will have constraints (and corresponding Lagrange multipliers). The constraints ...


2

There are two $SU(3)$ symmetries that are often discussed with regards to QCD. There is a gauge symmetry which corresponds to color charge which is mediated by the gluon and there is an approximate global flavor symmetry which acts on the flavors of the quarks (turns an up into down quark for example). All stable hadrons are color singlets and thus don't ...


2

Let $M$ be a manifold, let $V$ be a finite-dimensional vector space, and let $\Omega$ be a sample space (in the sense of probability). For each point $p\in M$, let $X(p):\Omega\to V$ be a random variable. A mapping $T_V:V\to V$ is called a target space transformation and a mapping $T_M:M\to M$ is called a space transformation (or spacetime transformation ...


3

The correct electroweak gauge group is $SU(2)_L \times U(1)_Y$ where $Y$ denotes the weak hypercharge. After the Higgs field spontaneously breaks this exact symmetry, third generator of $SU(2)_L$ (weak isospin) and weak hypercharge combine to give the remaining unbroken $U(1)_{em}$. Gauge bosons and fermions fall under different representations of this ...


3

Actually we have the following Lie algebra isomorphism $$u(1)\oplus su(2)\cong u(2),$$ and there exists the following Lie group isomorphism $$[U(1)\times SU(2)]/\mathbb{Z}_2 ~\cong~ U(2).$$ In other words, there is a two-to-one map between $U(1)\times SU(2)$ and $U(2)$. So in that sense the Glashow-Salam-Weinberg $U(1)\times SU(2)$ model already contains ...


8

Nice question! The short answer is that the group is not $SU(2)\times U(1)$, it is $SU(2)_L \times U(1)_{em}$. In other words the two groups act on different standard model particles differently. For example the left handed neutrino does interact weakly and so transforms under the $SU(2)_L$, but is electrically neutral so it doesn't transform under the ...


0

The symmetry of the Cauchy stress tensor is the result of applying conservation of angular momentum to an infinitesimal material element. This derivation assumes that there are no body moments. If there are body moments, the only way that angular momentum can be conserved is if the stress tensor is asymmetric. This is apparently important in the analysis of ...


1

Symmetry of that tensor directly arises from the standard relation of momenta of forces and temporal derivative of angular momentum. So, when assuming standard hypotheses of Newtonian Physics, the stress tensor necessarily turns out to be symmetric. Failure of symmetry is actually equivalent to the failure of the usual interplay of momenta of forces and ...


0

I don't see why there could't be asymmetries for r>c, considering the fact that the charge distribution is not symmetric. So what is the argument for this symmetry? Look at the figure below, In the initial stage, there is an electric field within the conductor( electric field created by the the charge $q$ ) which lasts for an $infinitesimal$ $period$ ...


3

There are 3 actions of the Galilean group on the free particle: On the configuration space, on the phase space and on the quantum state space (wave functions). The Galilean Lie algebra is faithfully realized on the configuration space by means of vector fields, but its lifted action on Poisson algebra of functions on the phase apace and on the wave functions ...


2

Consider a theory of fields $\phi:M\to T$ where $M$ is a manifold, and $T$ is a set. In physics, $T$ is often either a vector space or a manifold. We call $M$ the domain of the theory, and we call $T$ the target space. of the theory. We call a function from $M$ to $T$ a field configuration, and the set of all field configurations is denoted $\mathcal F$. ...


0

Comments to the question (v2): It seems that OP assumes that $\alpha$ is independent of $x$, i.e., OP considers a global quasisymmetry $$\delta A_{\mu}=\alpha \dot{A}_{\mu}.$$ The corresponding conserved quantity is energy, cf. Example 1 on the Wikipedia page for Noether's (first) theorem.


2

We may approach the problem via differential forms, or ordinary tensor calculus: Differential Forms: The field strength tensor $F$ is a differential form given by the exterior derivative of the 1-form $A$, i.e. $F=\mathrm{d}A$ which in components is $\partial_{[\mu}A_{\nu]}$. To add a total derivative to the form $A$ is equivalent to adding the exterior ...


3

You are right concerning the parity transformation, it implies the degeneracy of all states with finite momentum. The effect of the translation symmetry does not imply more than what is known from the conservation of momentum, as both operators are closely related. Indeed, the translation operator is given by $$\hat T(a)=e^{i \hat P a}$$ (up to a sign, ...


4

The central extensions are classified by the second cohomology group: http://en.wikipedia.org/wiki/Group_extension . If this group is trivial then each central extension is semidirect (and hence in some sense trivial). In particular, this is the case for the Poincare group but not for the Galilei group. However, if you want to take a nonrelativistic limit ...


2

In quantum field theory, when manipulating the path integral, we naively assume the measures (or strictly speaking the product of the measures and integrand) are invariant under the gauge transformations. In a fundamental paper, Fujikawa demonstrated the flaw in this assumption (in certain cases), and how to rigorously compute the analogue of a Jacobian ...


1

Suppose the potential on the outside metal surface is $V_s$. Then the potential outside the sphere is given by the solution of Laplace's equation on that domain. However, it's easy to show that the potential $$V(\mathbf{r})=V_s\frac{c}{|\mathbf{r}|}$$ satisfies Laplace's equation on the domain $|\mathbf{r}|\geq c$. Since solutions to Laplace's equation are ...


4

The multiplication by a phase of the wave function commutes with the action of the Galilean group. It is always possible to add a generator, commuting with a Lie algebra generators, to form a Larger Lie algebra. In this case, the larger Lie algebra is called a central extension of the former. The origin of the name is that the added generator (or ...



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