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1

Going to the conformal gauge is nothing but using coordinates in which the metric is diagonal (in euclidean space this is called isothermal coordinates ). Therefore in order to show that the Polyakov action is Weyl-invariant without using the conformal gauge, it is sufficient to show that the action does not dependent on the coordinate choice at all. ...


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What is the exact meaning of homogeneity in cosmology? Conifold and Milad have adequately explained the distinction between homogeneous and isotropic, so I'll answer on a different tack: See the Einstein digital papers where he said 'empty' space in its physical relation is neither homogeneous nor isotropic. A gravitational field is a place where space is ...


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Here is a short summary inspired by Barbara Ryden: Homogeneity: No preferred location Isotropic: No preferred direction And here are some examples to clarify things: Example of homogeneous but not isotropic: A forest, it looks the same no matter where you are, but trees make the vertical direction distinct. Example of Isotropic but not homogeneous: When ...


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Homogeneity in cosmology means uniformity from point to point, not only in composition or content, but in geometry as well. An empty space with a singularity is still non-homogeneous. Isotropy at every point does imply homogeneity, but we are not in a position to observe the universe from every point. Mathematically, isotropy at any two distinct points ...


1

No, you can not write this transformation as $c_{i\uparrow}^\dagger|0\rangle=c_{i\downarrow}^\dagger|0\rangle$, because $c_{i\uparrow}^\dagger|0\rangle$ and $c_{i\downarrow}^\dagger|0\rangle$ are two orthogonal quantum states: they can not be equal. The transformation $c_{i\uparrow}^\dagger|0\rangle\to c_{i\uparrow}|0\rangle$ you start with is also wrong, ...


1

That a Hamiltonian preserves a symmetry means $$ [H, C] = 0 \Rightarrow CHC^\dagger = CHC^{-1} = H$$ For a unitary symmetry operator $C$ (or anti-unitary if it is time reversal). The Hamiltonian of a crystalline condensed matter system written in terms of the Bloch matrix is: $$ H = \sum_{\vec k} \psi^\dagger(\vec k) H(\vec k) \psi(\vec k) $$ Where ...


5

Lee Smolin doesn't mean that the most fundemental physical theory can have no symmetry. What he means is that symmetry shouldn't be the guiding principle in discerning fundemental physical theories. While symmetry is mathematically useful, it doesn't provide a sufficient reason to accept a theory, this goes back to Leibniz's principle of sufficient reason. ...


3

This is a heuristic explanation of Witten's statement, without going into the subtleties of axiomatic quantum field theory issues, such as vacuum polarization or renormalization. A particle is characterized by a definite momentum plus possible other quantum numbers. Thus, one particle states are by definition states with a definite eigenvalues of the ...


0

It is the total $H = \sum_k H_k$ invariant under time reversal


2

I think the original source of this claim is the famous unpublished paper of Luescher and Mack. Everyone's citing it. It is more rigorous mathematically and general (they don't assume parity) than Di Francesco. It starts on pages 1-2 of the manuscript. The proof below is basically the same proof, just with added details and a little bit different notation. ...


2

There's no real symmetry argument that can explain why the $z$ component is zero. This is because it doesn't have to be: there can always be a uniform magnetic field added to the existing field, without affecting Maxwell's equations. They are all differential equations in the fields, like the following: $$\nabla\times\mathbf{B} = 0$$ which is Ampere's ...


1

There's an argument I think briefly mentioned in Aristotles Physics; where he argues that an object travels in a straight line since to veer would require a cause; he actually says for what reason would it move up or down or to the left? In another sense it's an argument from symmetry; actually this is related to Newtons First Law where cause is interpreted ...


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Your own source says Let it be given that the light travels from point A to point B. The whole argument is prefaced with the restriction that we have light emitted from A and received at B. In your example, you want light to start at A and go around B, which is an entirely different thing altogether and has nothing to do with Leibniz. Note also that ...


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We can draw a circle using a line segment as a radius, but one of those pints that defines the line segments will be in the center. We cannot trace the circle and go from the center at the same time.


2

The mass terms for the $ \sigma $ and $ \vec{ \pi } $ fields are, \begin{equation} m _\sigma \sigma \sigma + m _\pi \vec{ \pi } \cdot \vec{ \pi } \end{equation} You have two terms that are going to turn into each other under a symmetry transformation. Thus they need to have the same coefficient in order to remain invariant under the symmetry (feel free ...


0

No discrete symmetries are not infinitesimal and the derivation of Noether's theorem requires an infinitesimal symmetry to work as such.. Noether theorem is only valid for infinitesimal symmetries. However if you have a discrete symmetry which can expressed as an exponential form of some infinitesimal symmetry then you do have a corresponding conserved ...



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