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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

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 ...

9

Your question is not specific to inflation, and really applies to any case where a bosonic quantum field behaves semiclassically due to macroscopically large occupation numbers. One very simple example of this is the Stark effect in quantum mechanics, where a Hyrodgen atom is placed in a uniform electric field. The atom is treated as a quantum mechanical ...

4

Actually, the metric variational definition for the stress-energy tensor (due to Hilbert, as remarked by Qmechanic) is an universal improvement procedure for the canonical stress-energy tensor (and hence not always concides with the latter), in a sense which will be made precise below. Such a procedure is necessary because the canonical stress-energy tensor, ...

2

Well, you cannot take any ol' matter theory in flat Minkowski space and stick in a curved metric tensor $g_{\mu\nu}$ in the matter action as you like, if that's what you're implying. The caveat is that the resulting matter action $S_{\rm m}[\Phi, g]$ should be a general relativistic diffeomorphism-invariant functional. Then the Hilbert stress-energy-momentum ...

0

I question your premise: in classical field theory point singularities are not necessary, let alone inevitable. The most simple and relevant example is the gravitational potential of the Earth: $$\phi(r)\propto\begin{cases} r^2 & r<R_\oplus\\ \frac{1}{r} & r>R_\oplus \end{cases}$$ which is well-behaved and finite everywhere ($\phi\in\mathscr ... 0 Symmetry, stability and dimension analysis. You can consider a scalar field theory, for instance. A dynamical action for such a theory must be$S = \int d^4 x \, \, \partial_\mu \phi \partial^\mu \phi $because i. Lorentz symmetry indicates that all the indices must be properly contracted ii. The field equations must not exceed second ... 0 The point is that one should distinguish between a total spacetime derivative $$\frac{d}{dx^{\mu}}~=~ \frac{\partial }{\partial x^{\mu}}+ \phi_{,\mu} \frac{\partial }{\partial \phi}+ \phi_{,\mu\nu} \frac{\partial }{\partial \phi_{,\nu}}+\ldots \tag{1}$$ (where ellipsis denotes contributions in case of higher space-time derivatives), and an explicit ... 2 There are two kinds of derivatives we should differentiate: $$\frac{\mathrm d\mathcal L}{\mathrm dx}=\lim_{h\to 0}\frac{1}{h}\big[\mathcal L(\phi(x+h),\phi'(x+h),x+h)-\mathcal L(\phi(x),\phi'(x),x)\big]\tag{1}$$ and $$\frac{\partial\mathcal L}{\partial x}=\lim_{h\to 0}\frac{1}{h}\big[\mathcal L(\phi(x),\phi'(x),x+h)-\mathcal ... 2 Yes, there is systematic way called the Noether procedure. Simply you write down all possible 2-derivative fermionic terms with arbitrary coefficients and vary the action using the SUSY transformation rules. Then, you fix the coefficients to obtain the invariance up to a total derivative. When you have 4-derivatives, there are two cases: a. Off-Shell ... 0 For the coupling of p-brane Gauge fields to B 2-forms and U(1) Gauge fields check the Chern-Simons effective action for Dp branes. For coupling to fermions check some SUGRA actions, for instance the D=11 SUGRA has the graviton, the gravitino and 3-form Gauge field. Coupling to scalars is omnipresent in dimensional reductions of SUGRA actions. 2 This is a simple application of the chain rule$$ \frac{\partial}{\partial x^\mu} \phi \big( \Lambda_\nu{}^\mu x^\nu \big) = \frac{\partial}{\partial x^\mu} \big( \Lambda_\rho{}^\nu x^\rho \big) \left[ \frac{ \partial }{ \partial y^\nu } \phi (y) \right]_{y^\mu = \Lambda_\nu{}^\mu x^\nu} = \Lambda_\mu{}^\nu \left[ \frac{ \partial }{ \partial y^\nu } ... 2 My 2 cents on it is that in QM (be it "standard" QM or QFT) one describes only the state of a particle. Having said that, the most general state for a single particle is indeed a wave packet. Now, if you localise certainly a particle at some point in time, then later on it will be associated with a spreading wave packet because of Heisenberg indeterminacy ... 6 A particle is not a wavepacket. And there are no particle states for interacting theories. We define particle states in QFT by expanding the free field into its Fourier modes and using these modes as creation/annihilation operators for particle states - the mode of momentum$p$creates the particle state$\lvert p\rangle$with momentum$p$. The Hilbert ... 1 No, not every force field you can imagine is conservative. A force field is conservative if its line integral is path-independent, that is to say, $$\int_C \vec F \cdot d \vec x = U(a)-U(b)$$ For every curve C with end points$a,b$.$U$is some scalar function. If the force field is the gradient of a scalar field, this is automatically verified: ... 6 Theorem: let$L$be a homogeneous function of degree$k$; then the on-shell lagrangian is a total derivative. Proof: according to the Euler's homogeneous function theorem, $$k\ L(q,\dot q)=q\frac{\partial L}{\partial q}+\dot q\frac{\partial L}{\partial \dot q}\tag{1}$$ On the other hand, because of the Euler-Lagrange equations, $$(1)=q\frac{\partial ... 0 In the equation provided for r, there is a mistake. The equation is got by equating the centripetal force acting on the charge with the magnetic force on the charge since the circular motion is provided by the magnetic field. So,$$r=\frac{mv}{qB}$$the v is the velocity of charge, not the voltage. To calculate the velocity, you are provided with ... 1 There are two different values represented by v in this problem: lowercase v is the velocity of the electron uppercase V is the voltage that accelerates the electron In both the force equation F=qvB and the radius equation r = mv/qB, v refers to the velocity of the electron. I believe this is where your mistake was, since you said that the ... 1 It is, a priori, completely correct to add both primary and secondary constraints to the Hamiltonian density by Lagrange multipliers. What is not correct is how you determined the equations of motion: There is no "F^{i0}" in the Hamiltonian theory! It is called \pi^i there and it is not dependent on \partial_i A_0, it is an independent canonical ... 1 A general advice: Before trying to understand Hamiltonian field theory, make sure you understand Lagrangian field theory. Before trying to understand Lagrangian field theory, make sure you understand Lagrangian point mechanics. In Lagrangian point mechanics, the functional derivative of the action is$$\tag{1} \frac{\delta S}{\delta q(t)} ... 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 ...

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