# Tag Info

3

To add to David Hammen's answer on the question: When numerically integrating this, together with Euler's equation of rotation, is there a way to ensure that the determinant of $R$ remains equal to one (otherwise $\vec{x}(t)$ will also be scaled)? Method 1 Dumb But Effective Naïve Multiplication Whilst you are getting up to speed with more ...

3

Do I need to use the angular velocity vector in the rotating or inertial reference frame for this? Yes. You can do it either way. I start with the expression that relates the time derivative of a vector quantity $\boldsymbol u$ in the inertial and rotating frames: $$\left(\frac {d\boldsymbol u}{dt}\right)_I = \left(\frac {d\boldsymbol u}{dt}\right)_R ... 1 We have a number m that shows up in a purely mathematical context and then we interpret it as having a physical meaning. The question is: What is the intuition that connects the two? The intuition connecting the two is essentially identical with the reason why Wigner connects the purely physical notion of an elementary particle with the purely ... 1 The 4 generators of SU(5) are not all "equivalent". In general, the generators of the group/algebra satisfy a defining equation of the form$$[T^i,T^j]=f^{ijk} T^k$$so depending on the structure constants f^{ijk} ,it is possible for example that$$[T^1,T^2]\neq[T^2,T^3]\quad\text{etc,}$$so it is important which generators are broken. In terms of ... 1 Since you have not provided a direct reference, it's hard to be completely sure (and particularly to pin down the details), but there's really only one general idea that this can refer to. In general, any arbitrary isometry S of euclidean space has the form$$ \mathbf x\mapsto R\mathbf x+\mathbf t, $$where \mathbf t is an arbitrary vector and ... 1 I think I can see where this is going, but I don't understand your notation so I will use different, but I hope that this will make sense.... The point here is that if translation is a symmetry operation (call it t) and rotation is a symmetry operation (call it C_n - where n is the number of C_n operations required for a full turn - i.e. an n-fold ... 1 Here we will for simplicity just consider an arbitrary finite-dimensional complex^1 semisimple Lie algebra \mathfrak{g}. I) One may show that the CSAs are precisely the maximal toral Lie subalgebras of \mathfrak{g}. In particular CSAs are abelian. Also the Killing form \kappa:\mathfrak{g}\times \mathfrak{g}\to \mathbb{C} (which is ... 1 As Wikipedia explains, E_7 refers to several, closely related real and complex Lie groups and Lie algebras. All the various E_7 Lie groups (algebras) are Lie subgroups (subalgebras) of the complex Lie group E_7 (algebra e_7), respectively. The latter has complex dimension 133 and rank 7. Specifically, E_{7(7)}\equiv E_{7(+7)}\equiv E_{7,7} ... 2 \renewcommand{ket}[1]{|#1\rangle} The basic logical connection here is$$\text{symmetry} \rightarrow \text{degeneracy} \rightarrow \text{avoided crossing} \rightarrow \text{band gap} \, .$$\textrm{symmetry}\rightarrow \textrm{degeneracy} Consider an operator S and let T(t) = \exp[-i H t / \hbar] be the time evolution operator. If$$ [ T(t), S] = 0 ...

1

If somebodys interested in the solution I just solved it myself :P ﻿﻿﻿ \begin{split} t^a t^b \otimes t^a t^b &= \frac{1}{16} [\lambda^a \lambda^b \otimes \lambda^a \lambda^b] \\ &= \frac{1}{16} \left[\frac{2}{N_c}\delta^{ab} \mathbb{1} + d^{abc} \lambda_c + ...

2

If there is only one band maximum in the BZ, this point is one of the high-symmetry points of the BZ. However, there can be cases where there are many points which are a band maximum and they are not at one of the high-symmetry points of the BZ. These points however are all connected by a symmetry operation. An example of a system with band minima away ...

2

I do not agree with the answer given by @ACuriousMind. @Scardenalli has asked for a compact Ricci-flat Riemannian manifold $M$ having as isometry group $U(1)\times SU(2)\times SU(3)$. This does not imply that $M$ must be a symmetric Riemannian manifold. However, the answer to @Scardenalli's question is still no, and it follows from a classical result in ...

1

Saying that $\textrm{SU}(2)$ describes internal symmetries as the isospin is somewhat incorrect. Rather, it is the gauge group describing the weak interaction. In quantum field theory the equations of motion for the fields and the particles involved therein are described by means of a Lagrangian which is supposed to be invariant under some gauge group of ...

1

Since you start with a group $G$ which is a symmetry of the theory (ie of Lagrangian) there will be some generators for $G$. Call them $T_1,\ldots,T_N$. In the original Lagrangian, (before the Higgs gets a vev) you can do a transformation with any of the $T_i$'s and the Lagrangian will not change. After the field gets a vev, you can try the same and you will ...

0

There is a real Lie group $\tilde{Diff}(S^1)$ which is a $U(1)$ central extension of the real Lie group $Diff(S^1)$, and the Virasoro algebra is the Lie algebra of this Lie group. The central extension $\tilde{Diff}(S^1)$ can be realized geometrically in two ways. The first is via a Hilbert space embedding (as in the book of Pressley-Segal), and the second ...

Top 50 recent answers are included