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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

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. ...



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