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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 ...


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. ...


3

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 ...


3

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 ...


3

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 ...


2

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 ...


2

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 ...


2

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 ...


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 ...


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 ...


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 ...


2

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.


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 $$\tag{2} {\cal V}(\phi)~\propto~(\phi^2-v^2)^2~ \geq~ 0$$ ...


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 ...


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} ...


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

[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 ...


1

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 ...


1

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 ...


1

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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