New answers tagged

1

Power of Legendre polynomials: $$\frac{1}{\| \mathbf{x-y} \|}=\frac{1}{\sqrt{r^{2}-2rr'\cos \theta+r'^{2}}}= \frac{1}{r} \sum_{n=0}^{\infty} \left( \frac{r'}{r} \right)^{n} P_{n} (\cos \theta)$$ where $\displaystyle \, \cos \theta = \frac{\mathbf{x} \cdot \mathbf{y}}{\| \mathbf{y} \| \| \mathbf{x} \|} \,$ and $\displaystyle \, \begin{pmatrix} \mathbf{r'} ...


1

When you work in finite dimension, all these questions are basically trivial (this adresses mainly your first two questions). There is only one topology on any finite-dimensional vector space that makes it into a separated locally convex topological space. This is most likely the only topology you will ever see on any such space. And of course this is the ...


1

A bit of Theoretical Physics in QCD I think this is best answered from a physics perspective. Let $\mathcal{L}_{QCD}$ be a QCD Lagrangian that is invariant under the gauge transformation, $$\psi(x) \to e^{-i\alpha}\psi(x)$$ If all masses are equal in $\mathcal{L}_{QCD}$ then there is also an invariance, $$ \psi(x) \to e^{-i\alpha^aT_a}\psi(x) $$ where ...


0

You can check for yourself that in $G^3$ that is the Pauli algebra with basis $ e_1, e_2, e_3$ $ (1+ e_1) (1 - e_1) =0 $ . The claim is that there are zero divisors, not it is impossible in principle to divide two elements.


5

Not every element of a Clifford algebra has the form $v$ for $v$ a vector. In general an element of a Clifford algebra is a sum of products of such things, and most of these aren't invertible. Explicitly, let $\text{Cliff}(p, q)$ be the Clifford algebra over $\mathbb{R}$ generated by $p$ generators $e_1, \dots e_p$ satisfying $e_i^2 = -1$ and $q$ ...



Top 50 recent answers are included