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2

The meaning of $\sigma_{ij}$ is force in direction $j$ applied in a surface whose normal is in the direction $i$. Therefore $\sigma_{xx}$ is an x-directed force applied in a surface whose normal is in the x direction, which we interpret as pressure. When $i\neq j$ we call it shear, but the idea is the same. This drawing from Wikipedia might be helpful:


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We do if there is a color trace. The term $D_{\mu}D_{\nu}F^{\mu\nu}=\frac{1}{2}[D_{\mu},D_{\nu}]F^{\mu\nu}$ is proportional to $F_{\mu\nu} F^{\mu\nu}$.


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As the more upvoted answer said, if there are color indices then the covariant derivative doesn't commute with itself and the expression you wrote makes sense. If not, symmetry arguments about the symmetric nature of the derivative and the connection and the anti-symmetric nature of the curvature tensor are enough to reason like I did below that the ...


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I think you should not take the Peskin and Schröder quote too seriously. They are likely just using the Fourier relationship "short distance <-> high momenta" and the idea that there are propagators $\langle \phi^2 \rangle$ (which are the "fluctuation"/variance of $\phi$, see this post) associated to the Feynman lines of virtual particles, so "small ...


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The state of the universe is not homogeneous and isotropic, but the laws of physics are. For example, the speed of light propagation is the same in all directions, and the mass of the electron is not a function of position. Writing down a Lagrangian requires an assumption about the laws of physics (or more precisely, an assumption about the dynamics). There ...


0

We all know that, for a test particle (classical) in a gravitational field, the motion is only determined by the geodesic lines Actually, that isn't quite right. The geodesic lines don't actually exist in any objective sense. They're abstract things that are used to model particle motion, but's it's wrong to think that light curves because it follows a ...


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Why do you say it is not a soliton solution? I did not verify your answer but assuming its true, then it seems like a soliton to me. Your vacuum consists of $\phi=0,\pm 1$. This solution obviously interpolates between the two vacua $\phi(+\infty)\rightarrow +1$ and $\phi(-\infty)\rightarrow -1$, and is therefore topologically stable. Moreover the region ...


3

It was pointed out by @Peter Anderson in the comment that you forgot the transformation of the derivative, which in infinitesimal form should read $$\delta \partial_n = - g^{lm} \Lambda_{mn}\partial_l$$ which comes from the Lorentz transformation $$\partial_n \to g^{lm}(L^{-1})_{mn} \partial_l$$ (the metric is there to keep the indices in agreement with OP's ...


1

I suspect you have misunderstood what is meant by the term field in relation to the Higgs boson. You say: an electron creates a radial electric field but in no way it can interact with the field it created and you are quite correct that the electron creates an electrostatic field around it. However in this context the term field refers to a quantum ...


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Take an Abelian example. The $U(1)$ gauge invariant kinetic term of the photon is given by $$\mathscr{L}=-\frac{1}{4}F_{\mu \nu}F^{\mu \nu}$$ Where $$F_{\mu \nu} = \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}.$$ That is $\mathscr{L}$ is invariant under the transformation: $A_{\mu}(x) \rightarrow A_{\mu}(x)-\partial_{\mu} \eta(x)$ for any $\eta$ and $x$. If ...


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Mass comes from things other than the Higgs field--the latter is just the main contributor. What gives the Higgs boson its mass is still up for debate--for a more detailed discussion, see the following post: How does the Higgs Boson gain mass itself?


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Actually you can formulate Gauss's law for the gravitational field as well: $$\oint_S \vec g \cdot d\vec A=-4\pi G M, $$ where on the left you have the gravitational flux through a closed surface and $M$ is the mass inside the volume. $G$ is the gravitational constant. When you call this quantity on the left $\Phi_G$ and write the mass as an integral of ...


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In analogy with Gauss's law for the electric field $\nabla\cdot \vec{E}=\rho/\epsilon_0$, the flux of the gravitational field through a closed surface is proportional to the mass contained inside the surface. There is an approximation to General Relativity called Gravitoelectromagnetism (see Wikipedia page of this name. It's relationship with Newton's law ...



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