New answers tagged

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OP is right. In general, to make the action principle work, we only need to impose boundary conditions (BCs) at the boundary of spacetime. Hence we shouldn't impose eq. (6) in the interior of spacetime. In other words, OP should preferably redo their above analysis using the action rather than the Lagrangian.


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Let's consider a simpler problem: the vibrations of a string. We will describe it with the following Lagrangian density $$\mathscr{L} = \frac{1}{2}\partial_a\phi\partial^a\phi\,.$$ The equation of motion is $$0 = \partial_t^2\phi - \partial_x^2\phi\,.$$ The solution of this equation is any function of the form $$\phi = f(t + x) + g(t - x)\,.$$ We must now ...


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First, for fermion component notation in the Martin's textbook. Forget your notations for a while, and start from the beginning. For Weyl spinors, let me replace the dagger (h.c.) with the bar to avoid clutter (which is quite a common practice). This bar (or dagger) always accompanies the dotted indices, upper or lower, while undotted indices are always ...


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There are two facts to be distinguished. Electrons are what we loosely call particles, so they only ever occur in discrete numbers. Millikan demonstrated the discreteness of charge. Secondly, localised states in general have discrete energies. Examples of these are atomic and molecular states. Free electrons have continuous electrons. So called free ...


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I'll address what I understand to be your question, namely Where is the discreteness of the number of excitations of QFT coming from, even in free propagation that apparently does not involve a potential and the corresponding compactness associated with discreteness? (This is the magic of Fock space often referred to as "second quantization", an ...


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In normal quantum mechanics, we consider an individual particle, turn its momentum and position into vector operators $\hat{P}_i$ and $\hat{X}_i$, and enforce the canonical commutation relations $[\hat{X}_i,\hat{P}_j]=i\hbar\delta_{ij}$. In Quantum Field Theory, we want to apply the laws of quantum mechanics to the field itself, and not to particles. A field ...


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Let us work with the semi-disk $D$ with radial time flowing from the origin. Let us make the time $\tau=1$ be the contour boundary of the semi-disk while $\tau=0$ be the origin. Also, for simplicity let us neglect $b$ and $c$ for a while. An arbitrary first-quantized state of the string at $\tau=1$ is given by a functional $\Psi(X|_{\tau=1}(\sigma))$. A ...


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TL;DR: JoshuaTS' answer is exactly right: (Minus) the 3rd term ${\cal V}_3=\frac{1}{8\pi G}(\nabla\Phi)^2$ is the energy density of the gravitational field. In total OP's Lagrangian density contains 3 terms: ${\cal L}={\cal T}_1-{\cal V}_2-{\cal V}_3$. Here ${\cal T}_1$ is a kinetic term for matter, while ${\cal V}_2=\rho\Phi$ is an interaction/source term ...


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The last term represents the energy that is carried in the gravitational field itself. If you are familiar with electrostatics, this would be the equivalent of the statement that the energy stored in an electric field is $\frac{1}{2}\epsilon|\mathbf{E}|^2=\frac{1}{2}\epsilon(-\nabla V)^2$, where $V$ is the electric potential (the electric equivalent to $\phi$...


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The magnetic field originating from a bar magnet is continuously decreasing when moving away from the bar magnet. It does not have "lines" of stronger field strength. The lines that are forming are a consequence of the magnetic fields generated by the iron fillings themselves. An iron particle possesses a magnetic moment that aligns with the ...


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As far as I can see, your formula is correct. Let's transform this formula slightly. From $[\alpha_k,\alpha_{-k}] = 0$, it follows $$ \varphi_k \equiv \frac{v_k}{u_k} = \frac{v_{-k}}{u_{-k}}. $$ Last equality together with $|u_k|^2 - |v_k|^2 = 1$ and $|u_{-k}|^2-|v_{-k}|^2 = 1$ leads to equalities $$ u_k = \frac{e^{i\gamma_k}}{\sqrt{1-|\varphi_k|^2}}, \ v_k =...


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Maxwell's equations describe a massless vector (spin-1) U1-gauge field (the photon-field). Other particles have different properties (spin, mass, coupling to other fields), and have different equations of motion.


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Maxwell's equations can be written as a massless wave equation and this is a special case of Einstein's energy-momentum relation in wave form. The general case is the Klein-Gordon equation, which is satisfied by any free non-interacting field.


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Equations of motion (EOM) are typically the equations that determine the time-evolution of the system. E.g. in Newtonian mechanics, Newton's 2nd law is the EOM. (One should avoid referring to the kinematic $suvat$-equations as EOM to avoid confusion.) For Lagrangian systems, the Euler-Lagrange (EL) equations are referred to as EOM, even in case of EL ...


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The practical (although probably not the most rigorous) definition comes from looking at the derivatives in the equation. Equations of motion describe the evolution of the position of particles, so the independent variable is time (or a parameter like proper time) and the dependent variables are spatial position (or spacetime event locations). You'll see a ...


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Field equations tell you how fields change in spacetime, whereas equations of motion tell you how arbitrary physical objects move in spacetime. In other words, field equations are equations of motion for a field. Normally the term EOM is used in classical mechanics to denote the motion of a single body or system of bodies (though it certainly is used in ...


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There is no "precise distinction" between these terms. The "field equations" are just important equations of a field theory, which may or may not be the equations of motion for that theory. And even which equations are "equations of motion" is not unique! In the Lagrangian formulation of electromagnetism coupled to a charged ...


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So, to complete the half solution in the answer one need only notice that multiplying the $\frac{\partial S}{\partial\phi^i}$ the expression for $F^{ij}$ in terms of $F^{ia}$ and $F^{ijk}$, one is left with $$\frac{\partial S}{\partial\phi^i}F^{ij}=-\frac{\partial S}{\partial\phi^i}F^{ia}R^j_a.$$ However, recall that this trivial gauge transformation was ...


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I have done more direct approach to this task in first orfder in $\epsilon$. Let's start with dirct calculation of propogator: $$ \langle \phi(p) \phi(-p) \rangle = \frac{1}{p^2+m^2} - \frac{4 \cdot 3 \cdot g}{(p^2 + m^2)^2} \int^\Lambda \frac{d^{4-\epsilon}k}{(2\pi)^{4-\epsilon}} \frac{1}{k^2+m^2} + \dots $$ $$ \int^\Lambda \frac{d^{d}k}{(2\pi)^d} \frac{...


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FWIW, Dirichlet boundary conditions (BCs) are not the only type of BCs. There are also e.g. Neumann or Robin BCs. The relevant BCs depends on the physical system at hand. This is true even if no variational formulation exists. In the context of a variational problem, say a stationary action formulation, if the Lagrangian depends on spacetime derivatives of ...


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3/2 spin particles are studied in the framework of Supergravity. So textbooks and lectures on Supergravity will certainly contain a study on 3/2-fermions since the supersymmetric partner(s) of the graviton, the gravitino(s) are 3/2-spin particles. On the other hand, as such particles are not part of the Standard Model (or variants of it) and Supergravity is ...


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If not, why not? Because we have not yet observed a fundamental particle with $3/2$ spin. If yes, how would one build such a theory? The Rarita–Schwinger equation.


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The language is loose. If you press an expert to specify exactly which piece of the principal-bundle picture is called "the gauge field," you might get different answers on different days of the week. One answer is... yes, all of it. Sometimes we might call the $c_i$ the components of the gauge field.$^\dagger$ The idea is that the generators $T^...


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In this answer we will focus on the cubic term, which seems to be OP's main question. The trilinear form $$t\equiv\langle\cdot,[\cdot,\cdot]\rangle: \mathfrak{g}\times \mathfrak{g}\times\mathfrak{g}\to \mathbb{C}\tag{A}$$ is totally antisymmetric, because the bilinear form $\langle\cdot,\cdot\rangle$ is invariant/associative. Consider fields ${\bf e}$ that ...


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$\newcommand{\rto}{\overset{\scriptscriptstyle r\to\infty}{\longrightarrow}} \newcommand{\v}[1]{\boldsymbol{#1}} \newcommand{\t}{\tau} \newcommand{\pd}{\partial} \newcommand{\demeqq}{\overset{!}{=}}$ One should make a distinction between time-dependent and time-independent gauge transformations. I will denote $x=(\t,\v x)$. What is described in the book is ...


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The action should be invariant under the action of gauge group, therefore, you have to construct something, that when considering the action of gauge group remains unchanged. Assume that the field stregth tensor $F$ transform under some representation of $G$: $$ F \Rightarrow U^{i_1 \ldots i_n}_{j_1 \ldots j_m} F $$ To make this more clear, let us list ...


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Revised answer: The first one was wrong even if three people liked it! The "Lagrangian" derived in the cited paper is first order in the time derivative of $c_i$ and so the "Lagrangian" is really that of the Hamiltonian action principle. This starts from the action functional $$ S[p,q] = \int \{\sum_i p_i\dot q_i -H(p,q)\}dt $$ whose ...


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Some books wish to have a simple-looking commutator for the annihilationa and creation operators: $$ [\hat a_p,\hat a^\dagger_q]= (2\pi)^3 \delta^3(p-q) $$ while other prefer the normalization $$ [\hat a_p,\hat a^\dagger_q]= (2\pi)^3 2E_p\delta^3(p-q), $$ which is is a Lorentz covariant normalization in that states created by $a^\dagger_p$ have a higher ...


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The Hamiltonian formulation explicitily breaks Lorentz by choosing a particular direction in spacetime, namely $t$. More generally, one can foliate spacetime with spacetime surfaces $\Sigma_t$, where $t$ denotes a coordinate along a congruence, which can be thought as a time coordinate. This construction is used in Hamiltonian formulation of GR, which is ...


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Here is a very simple examples of collective modes: the normal modes of a mass-spring chain. This is pretty close to phonon modes in a solid state system. But the same principle applies to any other situation where many parts of a system are coupled and oscillate at the same frequency. https://www.youtube.com/watch?v=WE-HB8DBFU4


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QFT (quantum field theory) quantizes the classical fields, which are regarded as operators. A free real scalar field $\phi (x)$ is described as a mode expansion, which in the Schroedinger picture shows $$ \phi (\vec x) = \int \frac{d^3p}{(2 \pi)^3} \frac{1}{\sqrt{2 E_p}} \left(a (\vec p) e^{i \vec p \cdot \vec x} + a^\dagger (\vec p) e^{-i \vec p \cdot \vec ...


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In the written Lagrangian is not in Gaussian units. Gaussian still keeps the c. If it is in 'natural units', then e is dimensionless as are velocities. Also, mass, energy, momentum, frequency, inverse time, inverse length all have the same units. The units for each are chosen to be whatever is convenient, but should be the same for each object in a ...


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I had the same question a few months back. I decided that I want to look into the History of Physics on how the concept of Field came into Physics. I can list you some references which can embark you on the same journey I had gone through. Good Luck. Just go through the same order as it is listed below: Ernan McMullin (2002). "The Origins of the Field ...


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Advanced Topics in QFT (Shifman) talks about U(1) anomaly in 1+1, Aspects of Symmetry (Coleman) have topics on Higgs 1+1, and probably something about it in David Tong lectures.


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From the point of view of finite temperature QFT there will emerge some difference. Consider the grand canonical partition function: $$ Z = \text{Tr} \ e^{-\beta (H - \mu N)} $$ Where $H$ is the Hamiltonian, and $\mu, N$ are the chemical potential and particle number, correspondingly. The Green's function, defined as : $$ G_{\beta, \mu} (t, x, y) = Z^{-1} \...


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Sometimes, rather than applying the Euler-Lagrange equations, one can use the functional derivative method directly to the action, namely, we have the equality, $$\frac{\mathrm d}{\mathrm d \epsilon} S[\phi + \epsilon f] \Bigg\vert_{\epsilon=0} = \int\frac{\delta S}{\delta \phi}(x) f(x) \,\mathrm dx.$$ Then after applying appropriate boundary conditions and ...


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For what it's worth, the correct field-theoretic Euler-Lagrange (EL) equation reads in general $$ 0~\approx~\frac{\delta S}{\delta\phi} ~=~\frac{\partial {\cal L}}{\partial\phi} -\sum_{\mu} \frac{d}{dx^{\mu}} \frac{\partial {\cal L}}{\partial (\partial_{\mu}\phi)} + \sum_{\mu\leq \nu} \frac{d}{dx^{\mu}} \frac{d}{dx^{\nu}} \frac{\partial {\cal L}}{\partial (\...


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Varying the Lagrangian, \begin{align} \delta {\cal L} &= \delta \phi \Box \phi + \phi \Box \delta \phi \\ &= \delta \phi \Box \phi - \partial_\mu \phi \partial^\mu \delta \phi + \partial_\mu ( \phi \partial^\mu \delta \phi ) \\ &= 2 \delta \phi \Box \phi + \partial_\mu ( \phi \partial^\mu \delta \phi - \partial^\mu \phi \delta \phi ) \\ \...


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You are correct that the usual quantum textbook derivation of the translation action $$ e^{a\partial_x}f(x)=f(x+a) $$ is totally bogus since we cannot assume analyticity. A route that applies to much larger class of functions is via Fourier transforms. Assume that we can write $$ f(x)= \int_{-\infty}^{\infty} \tilde f(k) e^{ikx} dk $$ then we can proceed ...


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I am trying too to understand how this works, and I think the following argument from Mandl and Shaw "Quantum field theory" could help you, even if it probably won't answer completely your qestion as I find it still an open problem for myself. Anyway at least in the case of electromagnetism it worked well for me as there is a classical theory to ...


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Gauge symmetries aren't symmetries; they are redundancies in our description. We don't ever need to work with a theory written in a gauge invariant form; you could always choose a gauge and call this "the Lagrangian". Gauge symmetry amounts to a bookkeeping device to track different equivalent representations of the theory. This is useful because ...


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I propose that the local, that is, gauge symmetry has no physical meaning. This is based on the fact that classical electromagnetism can be better formulated without gauge invariance. See my peer reviewed paper at https://arxiv.org/abs/physics/0106078.


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Besides the fact that gauge theory works, the reason we promote global symmetries to local ones is because we value locality. It would be strange if to realize a symmetry, we need to perform some change everywhere in the universe. Relaxing this constraint leads to gauge theory.


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You are not really inducing a local symmetry as you are requiring it. If you have a Lagrangian $\mathcal{L}$ of the type $(\partial^\mu \phi)^\dagger\partial_\mu \phi + \phi^\dagger \phi$, you trivially have a global $U(1)$ symmetry, meaning you can shift the field $\phi$ by a constant phase $\varphi$ across all of space without changing $\mathcal{L}$: $\phi ...


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I think the local character comes from analyzing known phenomena. A widely studied theory like Electromagnetism has a gauge invariance in the classical formulation. We see therefore that changing gauge and having the same theory might persist going quantum, having it in the quantum regime assures us that it will persist classically. Then, we see that in ...


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Here is one approach: Let us for simplicity only consider static (i.e. time-independent) configurations. The time-dependent case is left to the reader. The stationary solutions are the 2 ground states, the kink and the antikink. (The 2 latter have a moduli parameter.) The 2 ground states are obviously locally stable. That the kink and the antikink are ...


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