Gerben
Reputation
643
Top tag
Next privilege 1,000 Rep.
Create tags
 Jan19 awarded Yearling Jan19 awarded Yearling Jun8 awarded Constituent Jun8 awarded Caucus Jan19 awarded Yearling Nov13 comment Anti-curl operator @Max: that doesn't change much, but you should be careful about what you mean by a 'closed surface'. If $D$ is a manifold with boundary $S$, then $$\oint_S \vec{B}\cdot\vec{ds}= \int_D \nabla \cdot \vec{B}$$ (through Stokes' theorem), so that doesn't help you much. Nov13 comment Anti-curl operator @Max: the 'de Rham cohomology' is a general way of studing whether, given some space $M$, the implication $df = 0 \rightarrow f = d a$ for some $a$ holds. Here, $d$ is some generalized derivative operator - the notion includes grad, div and curl. In particular, the curl corresponds to the first cohomology group, so the mathematician's way of stating your question is 'is $H^1(M)$ trivial'? For $\mathbb{R}^n$ it's true, but for spaces like $\mathbb{S}^1$ or $\mathbb{R}^2 - \{p\}$ for some point $p$, it's not. Nov13 comment Anti-curl operator Indeed, you'll need to impose some conditions at infinity. I guess that $\nabla \times \mathbf{B}(\mathbf{r}) = \mathcal{O}(1/r^2)$ suffices, but someone competent should check this. Of course, there are also smoothness conditions on $\mathbf{B}$; $\mathcal{C}^2$ should suffice, I reckon. Nov13 comment Anti-curl operator @Max: simple connectedness is crucial for this idea. If the domain on which $\mathbf{B}$ is defined isn't simply connected, then you don't have the implication $\nabla \cdot \mathbf{B} = 0 \Rightarrow \mathbf{B} = \nabla \times \mathbf{A}$ for some $\mathbf{A}.$ Nov12 revised Anti-curl operator added 80 characters in body Nov12 answered Anti-curl operator Nov3 comment How do you solve the Initial value probelm $dp/dt = 10p(1-p), p(0)=0.1$? @Gardel: for your proposed solution $P(t)$, we have $P(0) = 1/(1-9) = -1/8 \neq 0.1$. Check your signs! Oct31 comment If $K_i=\operatorname{ker}(T^i)$ and $\exists r\in\mathbb{N}:K_r=K_{r+1}$ then prove $K_r=K_{r+i}$ $\forall i\geq 1$ The part $K_r \subset K_{r+i}$ should be easy. Here's a hint for the converse: let $v \in K_{r+2}.$ Then $T^{r+2}(v) = 0 \rightarrow T(v) \in K_{r+1} = K_r$ (by hypothesis) $\rightarrow v \in K_{r+1} \rightarrow K_{r+2} \subset K_{r+1} \rightarrow K_{r+2} = K_r.$ Oct28 comment an homeomorphism from the plane to the disc In terms of complex numbers, you can write this as $z \mapsto [|z|/(1 + |z|)] \times z.$ Oct22 comment Induction proof concerning a sum of binomial coefficients: $\sum_{j=m}^n\binom{j}{m}=\binom{n+1}{m+1}$ @Alex: when you go to the URL FoolForMath provided, it shows a picture with the above identity represented on Pascal's triangle, on which the relevant numbers form a 'hockey stick'. Oct22 comment Checking a calculation: Simultaneous equations @craig: if this is a tunneling problem in quantum mechanics, $g(x)$ should probably be a function of $\kappa x$, where $\kappa \neq k$. Oct21 comment Casimir operator of a Lie Group .. how can i calculate ??' No! The algebra of translation operators is Abelian, i.e. everything commutes. You can only find a Casimir invariant if an algebra is semisimple; in practice, this means that $g_{ab}$ is non-degenerate - which isn't the case for the translation operator algebra. The main idea is that the Casimir is a special ('distinguished') element of the algebra (technically: the universal enveloping algebra). For $\mathfrak{so}(3)$, $L^2$ is (up to a scale factor) the only nontrivial element that commutes with all $L_i$. For the algebra of the $P_i$, any product of $P_i$ would work... Oct21 comment Casimir operator of a Lie Group .. how can i calculate ??' By the way: I made a mistake. I believe it should be $\epsilon_{akl} \epsilon_{blk} = -2 \delta_{ab}.$ However, this comes down to multiplying a Casimir by -2, which isn't a big deal (can you see why?). And yes, that formula can be used as the definition of $g_{ab}$ (it's entirely equivalent to the mathematicians' definition of the Killing form). Oct21 comment Casimir operator of a Lie Group .. how can i calculate ??' For the first example: since $[P_i,P_j] = 0,$ all structure constants are zero, so the Killing form $g_{ab}$ vanishes. Since $H := g_{ab} X^a X^b$, $H = 0$ in this case. In the second case, you need slightly more work: $C_{ab}^d = \epsilon_{abd},$ so $g_{ab} = C_{ak}^l C_{bl}^k = \epsilon_{akl} \epsilon_{blk} = \delta_{ab},$ which means that $H = \delta_{ab} X^a X^b = L_x^2 + L_y^2 + L_z^2.$ Oct21 comment Is there any relation between the principal eigenvector of the original matrix and its inverse? There isn't some direct relationship between the principal e.v. of $A$ and the principal e.v. of $A^{-1}$, no.