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Two contradictory derivations of Killing equation

As an overall comment, I stress that conservation of $Q$ is valid for the Killing vector $\xi$ if the considered curve is a geodesic. Let us come to the issue. First of all, generally speaking, the ...
Valter Moretti's user avatar
2 votes

Two contradictory derivations of Killing equation

Both approaches are fine. In the first approach, the analysis is done at the coordinate/component level of the equations. Simply asking the question how does $Q$ very with $\tau$ if we write ...
TimRias's user avatar
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1 vote

Why Consider Only Triplet States for Spin in $2$-Electron Systems?

This is slightly strange. In any case what matters is that the fermionic states should be fully antisymmetric w/r to permutations of the particles, and that means spin and spatial degrees of freedom. ...
ZeroTheHero's user avatar
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Does quasi-symmetry preserve the solution of the equation of motion?

Boundary condition is a part of the very definition of your field theory, classical or quantum. When e.g Peskin & Shroeder leaves out the surface term, "all fields and derivatives vanish at ...
T.P. Ho's user avatar
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Does all symmetry breaking have corresponding unitary group?

In general you can have broken symmetry groups different from $U(n)$. For example the quantum Ising model has a discrete symmetry (enacted by $P=\prod_j \sigma^z_j$) that breaks spontaneously in the ...
lcv's user avatar
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Does all symmetry breaking have corresponding unitary group?

As in this SE post, you can have the Lagrangian with $SO(N)$ symmetry $${\mathcal{L}} = \frac{1}{2}(\partial_\mu \Phi)^T (\partial^\mu \Phi) - \left(\frac{1}{2}\mu^2 \Phi^T \Phi + \frac{1}{4}\lambda (...
Gabriel Ybarra Marcaida's user avatar
1 vote

Two contradictory derivations of Killing equation

I think the problem lies in the notation. I guess Tong is treating the components $\xi_\mu$ as scalars for which we have $\frac{\mathrm{d}}{\mathrm{d}\tau}=u^\alpha\partial_\alpha$. And the ...
Silas's user avatar
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Parity transformation of the $\pi^{0}\rightarrow\gamma\gamma$ process

... how do I apply P? Spacelike components of a vector/tensor reverse their components. But the Levi-Civita symbol is a pseudotensor, so its three space like reversals amount to changing the ...
Cosmas Zachos's user avatar
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Understanding Symmetries and Invariances in Electrostatic Fields

The concept of symmetry plays an important role in all branches of Physics, and electrostatics is a good playground for developing an intuition on how symmetries works. A little disclaimer: I do not ...
Giovanbattista Favorito's user avatar
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Charge conjugation and Transition form factors

Perhaps the conceptual confusion here regards the difference between gauge symmetries and global symmetries. Gauge 'symmetries' are really just redundancies---different descriptions of entirely the ...
SethK's user avatar
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2 votes
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How does inserting an operator in the path integral change the equation of motion?

The missing conceptual point to appreciate is that when you are calculating a correlation function it can be useful to interpret the same expression in different ways. We wish to calculate the ...
SethK's user avatar
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How does inserting an operator in the path integral change the equation of motion?

Yes, the Wilson line (2.42) is now part of the action in the path integral. Yes, the symmetry defect operator (SDO) (2.41) is now part of the action in the path integral. References: T.D. Brennan &...
Qmechanic's user avatar
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2 votes
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Using particle-hole symmetry of the Hubbard model to study the model at different densities

$\newcommand{\dag}{\dagger}$I realized it's because there are two spins that must be accounted for. Under the particle-hole transformation, we have \begin{align*} N = \sum_{i, \sigma} a^\dag_{i, \...
zeroknowledgeprover's user avatar
1 vote

Is the FRW metric, based on spatial homogeneity and isotropy, rotationally and translationally invariant? If so, how?

If what you're confused about is why the metric seems to depend on the coordinates and thus might not be translationally or rotationally invariant, look at the Minkowski metric (flat spacetime) in ...
controlgroup's user avatar
2 votes
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Is the FRW metric, based on spatial homogeneity and isotropy, rotationally and translationally invariant? If so, how?

Roughly speaking, in addition to Einstein equations, the (spatial) FLRW metric is constructed by assuming that, at fixed time, in the Riemannian manifold defining space: (a) metric properties are ...
Valter Moretti's user avatar
3 votes

Designing a thought experiment on Noether's Theorem

Experiments on crystals are translationally variant because the crystal structure is only the same up to translations that reproduce the same structure, in such cases there is "crystal momentum&...
mike1994's user avatar
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Probabilistic behavior of quantum mechanics

If you have two identically prepared systems, say two copies of the state $ \alpha \vert \uparrow \rangle + \beta \vert \downarrow\rangle$, with $\vert \alpha\vert^2+\vert\beta\vert^2=1$, you might ...
ZeroTheHero's user avatar
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5 votes

Have all the symmetries of the standard model of particle physics been found?

My answer is also ‘No’ but by direct construction: over the past few years, various research groups have understood new symmetries of the Standard Model. Over the past decade, field theorists have ...
SethK's user avatar
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Checks of anomaly cancellation

A lot of the information you gave me is beyond my knowledge, not being a science professional myself or even a student. But I like reading physics textbooks from time to time and the chapter on ...
Fern's user avatar
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2 votes
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Doubts in circuit analysis

Maybe I am loosing some detail here, but assuming that all resistances are equal, and looking at the symmetry of the circuit, all dpps and currents should be the same in module if you flip the circuit ...
ebenezer's user avatar
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Checks of anomaly cancellation

... if 𝐺 is a global symmetry of the classical Lagrangian, then one has to check 𝐺×𝐻² anomalies, where 𝐻 is one of the SM gauge groups. On may check them for academic purposes, not consistency, ...
Cosmas Zachos's user avatar
1 vote

Variation in the context of symmetries

The Einstein-Hilbert action is $$ S = \frac{1}{2\kappa} \int R \sqrt{-g} d^4x$$ with $\kappa$ the dimensionful constants. If you set $R=0$ at the outset, you have the action $S = 0$, which is a ...
Brick's user avatar
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Variation in the context of symmetries

It's hard to say for sure, since you don't remember the details, but usually we don't assume anything about the solutions to the equations of motion when deriving them (by computing the variation of ...
Bairrao's user avatar
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Deriving the Minkowski Metric from homogeneity of space-time and the isotropy of space

The Wikipedia page (at least as it stands when I read it today) says that "it follows" from the assumptions stated in the question that the spacetime interval between two events 1 and 2 is $$...
Brick's user avatar
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1 vote

Killing tensor in the Kerr metric

The Killing equation is a overdetermined system of PDEs of finite type, as a result, there is a algorithm to compute all Killing tensors for a given metric. See for example arXiv 1704.02074.
liyiontheway's user avatar
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Behavior of the Electric- and Magnetic-field under time reversal and parity

For the parity of B, the arguments can be confusing because there are at least two ways of thinking about the question of mirror symmetry and parity. I like the following. Suppose that you observe an ...
benjamin nachumi's user avatar
0 votes

Misconceptions in spontaneous symmetry breaking

For completeness let us write down the potential $$V(\phi)=-\mu^2\phi^2+\lambda \phi^4$$ In what follows we will use the term "invariant", which means: "does not change". The ...
Maik H.'s user avatar
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Relationship Between Ground State Wavefunctions' Amplitudes Under Discrete Symmetry Operations

I think that you are correct in what you have. The authors of your paper have defined the action of $\hat T$ on the states $|\sigma\rangle$ but it is not clear what they mean by $\psi(\hat T \sigma)$ ...
mike stone's user avatar
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2 votes
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Derivation of Noether Current in Condensed Matter Field Theory by Altland and Simons

A&S assumes that the action has a strict$^1$ (rather than a quasi-)symmetry, and so that the bare Noether current (1.43) without improvement terms suffice. -- $^1$ NB: It is implicitly assumed ...
Qmechanic's user avatar
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