# Tag Info

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It is the celebrated spin connection on the tangent space, gauging Lorentz rotations so you can take Lorentz covariant derivatives on spinors---you would not be able to do Supergravity without it. As you see, however, $\omega_\mu^{ab}$ is a composite gauge field, that is, it is is an elaborate function of Vierbeine (or Vielbeine) and their derivatives, ...

4

Hint: Eq. (6) in its current form (v4) is meaningless since the lhs. depends on $x$, while the rhs. is integrated over $x$. The functional derivative $$\tag{A}\frac{\delta F}{\delta u(x^{\prime})}~\stackrel{(B)}{=}~\frac{\delta u(x)}{\delta u(x^{\prime})}~=~\delta(x\!-\!x^{\prime})$$ in eq. (6) for the functional $$\tag{B} F[u]~:=~u(x)~=~\int \! ... 3 There most definitely is, and your text should have used it in defining the unitary gauge more conventionally: the SU(2) group element parameterization of physics, that is the rotation matrix for spinors R. Absorb v into the definition of σ, where it belongs and from where it can re-emerge at will.$$R=\exp (i\theta ~\hat{n}\cdot\vec{\sigma})=I \cos \theta ...

3

Going from action to EOMs is simple: it is just (functional) differentiation. Going the other way from EOMs to the action is hard: It is (functional) integration, and sometimes impossible! OP is now essentially asking: Can we integrate one more time? Well, not the action itself. But if we replace the EOMs and the Lagrangian $L$ with their dynamical ...

3

Going the way stated in the question's title is easy: The Euler-Lagrange condition is, inherently, a condition on the action -- the statement is that the classical path is the path for which the action takes a minimum value for the path. Since this is a statement about the value of the action, and the action is Lorentz-invariant, then this minimum value is ...

2

The key is: Landau theory doesn't assume the order parameter is small. All it assumes is that the free energy is analytic in the order parameter. One then usually expands this free energy up to some order (which is possibly by definition of 'analytic'). It is key to realize that expanding a function in a variable to some order does not mean this variable has ...

2

Consider a map $$S \ni\phi \mapsto F[\phi] \in \mathbb R$$ defined on a class $S$ of smooth functions $\phi$ defined on the compact set $\Omega \subset \mathbb R^n$ obtained by taking the closure of an open set with regular boundary $\partial \Omega$. Thus the map $F$ associates a real number $F[\phi]$ to each function $\phi\in S$. We say that the ...

2

The standard way to put in a temperature is to go to imaginary time (euclidean space) and impose periodic/anti-periodic boundary conditions on bosonic/fermionic fields $$\phi(x,\tau)=\phi(x,\tau+\beta) , \;\;\;\; \psi(x,\tau)=-\psi(x,\tau+\beta),$$ where $\beta=1/T$. This ensures that the path integral represents the partition function $Z=Tr[\exp(-\beta ... 2 Well, not really. We COULD write hamiltonian as square root - if we know, what is a square root of an operator. Of course we have simple approximation: $$\sqrt{1+x}=1+\frac x2-\frac{x^2}{8}+O(x^3)$$ Using this we could write your hamiltonian as: $$\mathcal H=mc^2\sqrt{1+\frac{p^2}{m^2c^2}}=mc^2+\frac{p^2}{2m}+O(p^4).$$ The problem is that this form of ... 2 We should probably start by pointing out that no Weyl fermion has ever been observed. The recent observations are of quasiparticles that behave like Weyl fermions. Speaking rather loosely (and at the risk of upsetting the QFT experts hereabouts) a Dirac fermion can be viewed as a sum of two Weyl fermions, and the observations are of paired quasiparticles ... 2 Yes. You are correct. A non-relativistic theory would be invariant under the Galilean group. Lorentz invariance (specifically, invariance under Lorentz boosts) is what defines a relativistic theory. 2 I cannot quite vouch for exhaustive panoramas, but the crucial point is that GL(N), SU(N) matrices are representable in a nonhermitean basis discovered by Sylvester in 1882, the clock and shift matrices which he called nonions for N=3 (long before the Gell-Mann basis!), sedenions, etc. Their braiding relations, and maximal grading, and hence commutators, ... 2 I) Assuming that the variational problem for the action$S=\int \! d^nx~{\cal L}\$ is well-posed (with appropriate boundary conditions), the field-theoretic Euler-Lagrange (EL) equations read in general $$\tag{1} 0~\approx~\frac{\delta S}{\delta \phi^{\alpha}} ~=~\frac{\partial {\cal L}}{\partial \phi^{\alpha}} -\sum_{\mu} \frac{d}{dx^{\mu}} \frac{\partial ... 1 And regarding why it's called a "free" theory, it's not specific to a momentum-space formulation. It's "free" because the Lagrangian is quadratic in the fields, and therefore the equations of motion (what you get from plugging the Lagrangian into the Euler-Lagrange equation) are linear in the fields. Therefore you can superpose different classical ... 1 A general advice: Before trying to understand Hamiltonian field theory, make sure you understand Lagrangian field theory. Before trying to understand Lagrangian field theory, make sure you understand Lagrangian point mechanics. In Lagrangian point mechanics, the functional derivative of the action is$$\tag{1} \frac{\delta S}{\delta q(t)} ...

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