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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 ...
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 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} ... 3 My answer to my own question here may be helpful. No, there is no anti-isospin as there is no anti-spin, because SU(2) has no complex representations. In contrast, SU(3) has complex representations and therefore the conjugate charge is different from normal charge, which means in the case of SU(3) color A complex representation R, is a group ... 2 General comment to the question (v3): Non-abelian YM [such as, e.g., YM with gauge group SU(2) or SU(3)] has besides quartic gauge boson interactions also cubic gauge boson interactions, while abelian YM (aka. QED) has neither. This is because the Feynman-rules for the cubic (quartic) gauge boson vertices are linear (quadratic) in the Lie algebra ... 5 Let us start with U(1) electromagnetism and see why it does not have such interactions. The field strength tensor is given by F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu, and the relevant part of the QED Lagrangian is proportional to F_{\mu\nu}F^{\mu\nu}. This means that the Lagrangian has only terms that are at most quadratic in the gauge field ... 4 LHS: (\cos\phi+i\sigma_z\sin\phi)\sigma_y(\cos\phi-i\sigma_z\sin\phi)=\cos^2\phi\sigma_y-i\cos\phi\sin\phi[\sigma_y,\sigma_z]+\sin^2\phi \sigma_z\sigma_y\sigma_z=(\cos^2\phi-\sin^2\phi)\sigma_y+2\cos\phi\sin\phi\sigma_x=\cos 2\phi \sigma_y+\sigma_x\sin2\phi=(\cos 2\phi+i\sin 2\phi\sigma_z)\sigma_y=e^{2i\phi\sigma_z}\sigma_y 4 Use fact that the matrices i\,\sigma form the Lie algebra \mathfrak{su}(2) and then the adjoint representation "braiding relationship" that:$$\exp(\phi\,\mathrm{ad}(Z))\,Y = \mathrm{Ad}(e^{\phi\,Z})\,Y\tag{1} and here $Z$ and $Y$ are $3\times 1$ column matrices that stand for superpositions of basis matrices in the Lie algebra in question. ...