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First, terminology: You are not "determining the gauge group", what you are doing in gauge fixing is determining a smooth choice of (hopefully only) one representant of an equivalence class of field configurations called the gauge orbit. Geometrically, you are seeking a section which intersects each gauge orbit exactly once. The problem of finding a gauge ...

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The red lines represent equipotential lines, which have the additional property of having an electric field of zero. The electric field at a point on the red line can be defined as $$\mathbf E_{red}=\sum_{i=1}^{4}\mathbf E_i$$ $$\mathbf E_i=\frac{1}{4\pi\varepsilon_0}\frac{q_i}{r_i^2}\cdot{\mathbf{\hat{r}_{0i}}}$$ Let's take the case where all charges ...

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The Dirac equation is more restrictive than the Klein-Gordon equation. For every solution to the Dirac equation, its components will be a solution of the Klein-Gordon equation, but the converse isn't true: if you form a spinor whose components are solutions of the Klein-Gordon equation, it might not solve the Dirac equation. If we start with the ...

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Two different Lagrangians give different canonical momentum. If two different Lagrangians differ by a surface term then they differ by a total divergence. And they thus yield the same actions hence have the same equations of motion. When you do integration by parts you produce a surface term (the difference between the two). Imagine subtraction the two ...

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I am tempted to give an experimentalist's point of view, though of course I agree with Lubos and Savanah's answers. Consider a disk of radius R, then circumference is 2πR. Now, make this disk rotate at velocity of the order of c(speed of light). Here it is evident that you are defining a center of mass system for a disk and a person/machinery that will ...

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What makes the case of the Higgs field different from that of other particles is that the Higgs field in the vacuum has a nonzero expectation value. So, if the electromagnetic field is in its lowest energy state then that means that the field strength will be zero on average (there are still quantum fluctuations, but on average it is zero). But for the Higgs ...

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In quantum field theory it is actually not obvious how many fields there are since fields can have components. If we have two fields $A$ and $B$, we can consider them to be merely components of the same field. Or reversely, if $A_1$ and $A_2$ are components of a field we can relabel them $A$ and $B$. However, it rubs physicists the wrong way to split fields ...

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Every particle has a corresponding field that permeates all of space in the same way the Higgs has a field that does so. The spin up electron. The spin down electron. The spin up positron. The spin down positron. The up quarks (all three colors and both spins). The down quark (all three colors and both spins). Same for the charm, strange, top and bottom. ...

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GR can be recast into an equivalent but conceptually quite different form, using teleparallel gravity. This approach introduces the Weitzenboeck connection, which has no curvature, but has torsion. The presence of torsion indicates that gravity is not geometrized. Recall that in GR, we can always choose a locally inertial coordinate system such that the ...

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I wrote this answer assuming you don't know much GR The analogue of the Electromagnetic field in General Relativity is the metric tensor $g_{\mu\nu}$. The value of the metric is determined by the mass energy distribution via Einsteins equation, which is analogous to Maxwell's Equations. In the equation below $G_{\mu\nu}$ is a tensor which depends on ...

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The Gauge Theory of Gravity (GTG) by Lasenby, Doran and Gull has a background spacetime with fields on it. It is basically derived from the same physical principles but as a background theory. It ends up not being the same theory, for instance it doesn't have the same isotropic solutions, and I think it does not allow time travel and such (unlike General ...

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It's simple actually if you write it out, Note that both the first identity and the second identity give you 16 scalar results. The 16 results of second identity are the same as the 16 results from the first identity but multiplied with values from the spinors. These extra multiplication factors are the same independent for any of the summed $\sigma^\mu$. ...

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To someone moving on the edge of the disk, the disk is length contracted in the direction of travel. So to them, it is not a circle, no problems with geometry. Euclidean geometry only holds for an inertial plane of simultaneity. By the end of this answer you should know when and how to use Euclidean geometry and what it means (or doesn't). And a collection ...

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First, terminology: Symmetry groups are not "defined on domains". Symmetry groups exist in the abstract, and they are then represented on certain spaces. If we have a spacetime manifold $\mathcal{M}$, then the fields are functions $$f : \mathcal{M} \to V$$ where $V$ is some vector space upon which a representation $\rho : \mathrm{SO}(1,3)\to ... 2 This is just relativity of simultaneity again. A similar thing happens if you have a bunch of spaceships in a line that fire their thrusters at a fixed time. Different observers will disagree about whether they fired at the same time and will disagree about the spacing. Always in a consistent way. So I'd like to address the concept of geometry by not having ... 6 No, it doesn't violate the rules of geometry, it violates the rules of Euclidean geometry. Simple conclusion: for an observer fixed to a disk rotating uniformly relative to an inertial frame, the spatial geometry is non-Euclidean; in particular, the ratio of a circle's circumference to its diameter depends both on the circle's diameter and center position. ... 6 What is wrong is the idea that one can actually make the disk rotate; and it will remain perfectly rigid. In reality, what this correct argument shows is that relativity doesn't admit the existence of any perfectly rigid bodies. This is a perfectly basic, settled, and indisputable textbook material that every mature physicist knows. The first sentence of ... 2 Here we assume that OP's question asks about$\phi^4$-theory in 1+1D, where the lagrangian density reads $$\tag{1} {\cal L}~=~\frac{1}{2}\dot\phi^2 -{\cal U}, \qquad {\cal U}~:=~ \frac{1}{2} \phi^{\prime 2} + {\cal V},\qquad \phi \in C^1(\mathbb{R}^2),$$ where the$\phi^4$-potential density $$\tag{2} {\cal V}(\phi)~\propto~(\phi^2-v^2)^2~ \geq~ 0$$ ... 0 The energy density of the state$\pm v$is going to be something like$\propto μ^4$, if you are using the basic$\varphi^4$theory. While the energy of the domain wall is finite, the energy of the vacuum state is not, and so the transition to the vacuum state iver all space will be infinite. 1 [I somewhat haphazardly pieced this answer together, so I'm not absolutely certain the conclusion is correct.] Cayley's theorem is useless here, because the group isomorphism it produces is not required to preserve any kind of topology on the groups, in particular not notions of continuity or differentiability. On the infinite symmetric group$S_\infty$on ... 2 Formally, the meaning you assign is just the usual meaning of the derivative. $$\partial_\mu \psi(x^\nu) = \lim_{h \to 0} \frac{\psi(x^\nu + h\delta^\nu_\mu) - \psi(x^\nu)}{h}$$ You can indeed compute it componentwise, because you can subtract two spinors, as in the equation above, just by subtracting their components. The object you get has sixteen ... 0 Basically your confusion is caused by a bias about what a vector is. Everyone agrees that you can add two vector and get another. Everyone agrees you can scale a vector and get another vector. Sometimes we square a vector and get a scalar, but some people say that is "merely" an abuse of notation. But that is just a special case of ... 2 You have a few different questions here, so let's try to go through them one by one. When we make the chiral symmetry local, have we introduced a gauge symmetry, or some analogue of a gauge symmetry? When you make the chiral symmetry local you introduce a gauge symmetry. The terms "gauge symmetry" and "local symmetry" are two different ways of saying the ... 1 Hints: Then potential term$\frac{1}{2}(\nabla\phi)^2$is semipositive definite and is only zero for a$x$-independent configuration$\phi$. If one completes the square of the potential $$V(\phi)~=~\frac{\lambda}{4}\phi^4-\frac{\mu^2}{2}\phi^2~=~ \frac{\lambda}{4} \left(\phi^2-\frac{\mu^2}{\lambda}\right)^2-\frac{\mu^4}{4\lambda},$$ then it becomes clear ... 1 With a Lagrangian like:$\mathcal{L} = \partial_\mu \phi^\dagger \, \partial^\mu \phi - V(\phi) = \mathring{\phi^\dagger} \mathring{\phi} + \partial_i \phi^\dagger \, \partial^i\phi - V(\phi) $, the Hamiltonian is: \begin{equation*} \mathcal{H} = \frac{\partial \mathcal{L}}{\partial \mathring{\phi}} \mathring{\phi} + \mathring{\phi^\dagger} ... 0 Step back and ask how you know whether a Lagrangian,$L=L(Q_i,\dot Q_i),$is correct. At the classical level the only answer is whether it gives the correct equations of motion as the Euler-Lagrange equations: $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot Q_i}\right)= \frac{\partial L}{\partial Q_i}.$$ Which is unchanged if you replace ... 1 Conventions do not change physics. If they would, we would not call them conventions. When studying Lagrangian mechanics, you may have noticed that you can multiply a lagrangian by any constant, and receive the same dynamics. Thus, we often (Or always) choose the constant such that the term$(\partial_0\phi)^2\$ appears with a positive sign. (And often with ...

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No, the Lagrangian density is different: $$\mathcal{L} = \pm \frac{1}{2} \partial^{\mu} \phi \partial_{\mu} \phi.$$ The Hamiltonian density is actually the same in both conventions. However, this has no physical meaning. The choice of the signature is purely conventional.

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