User soulphysics - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:36:09Z http://mathoverflow.net/feeds/user/7671 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68680/counterexamples-in-pde/118327#118327 Answer by soulphysics for Counterexamples in PDE soulphysics 2013-01-08T01:42:57Z 2013-01-08T01:42:57Z <p>A particularly simple example is Norton's dome, with height given as a function of radial distance on the surface of the dome by</p> <p>$h = \frac{2}{3g}r^{3/2}$</p> <p>where $g$ is the gravitational constant near the surface of the earth. The dome has a curvature singularity at the apex. And, if we model a mass at the ($r=0$) apex of this dome with zero velocity, we find that Newton's equation does not have a unique solution; the mass can "fall" at any arbitrary time $t$ for no reason at all.</p> <p>Norton's paper about the dome: <a href="http://philsci-archive.pitt.edu/2943/" rel="nofollow">http://philsci-archive.pitt.edu/2943/</a></p> <p>A helpful reply: <a href="http://philsci-archive.pitt.edu/3195/" rel="nofollow">http://philsci-archive.pitt.edu/3195/</a></p> http://mathoverflow.net/questions/112236/complete-collections-of-hamiltonian-flows-on-symplectic-manifolds "Complete" collections of Hamiltonian flows on symplectic manifolds soulphysics 2012-11-13T00:31:03Z 2012-11-13T00:31:03Z <p>Question for Math Overflow</p> <p>Let $M$ be a smooth $2n$-dimensional manifold and $\Omega$ a symplectic form, and let $(q,p)=(q_1,...,q_n,p_1,...,p_n)$ be a local (Darboux) coordinate system on $M$. Let $Q_i:M\rightarrow\mathbb{R}$ be the projection function $Q_i(q,p)=q_i$, and similarly let $P_i(q,p)=p_i$. These projection functions have the following property: if a function $f:M\rightarrow\mathbb{R}$ Poisson commutes with all of them, in that <code>$\{Q_i,f \}=\{P_i,f\}=0$</code> for each $i=1,...,n$, then it must be a constant function, $f(q,p)=k\in\mathbb{R}$.</p> <p>I am interested in when the generators of a collection of Hamiltonian flows will have this property. Let $\varphi^{s_i}_\sigma$ and $\varphi^{r_i}_\rho$ be the local Hamiltonian flows generated by a set of functions $s_i$ and $r_i$, for $i=1,...,n$. (That is, $\varphi^{s_i}_\sigma$ is a group of symplectic transformations parametrized by $\sigma$, which threads the vector field on $M$ generated by $s_i$, and similarly for $\varphi^{r_i}_\rho$.) If the set of $2n$ functions $s_i$, $r_i$ satisfy the property above (that <code>$\{s_i,f \}=\{r_i,f\}=0$</code> for all $i$ only if $f$ is constant), then I'll call this collection of flows <em>complete</em>. (This is different, of course, from the notion of a "complete Hamiltonian flow.")</p> <p>When is a collection of Hamiltonian flows complete in this way? Has anyone heard of or worked on an interesting criterion?</p> <p>For example, I thought it might be interesting if one of the groups were known to represent 'translations' of some interesting quantity. One way to have this would be to have a maximal set of Poisson commuting functions $q_1, ..., q_n$ such that for all $i,j\in\mathbb{R}$ and for all $\sigma\in\mathbb{R}$,</p> <p>$q_i\circ\varphi^{s_j}_\sigma = q_i + \delta_{ij}\sigma$</p> <p>$q_i\circ\varphi^{r_j}_\rho = q_i$</p> <p>(where $\delta_{ij}=1$ if $i=j$ and $0$ otherwise). Is there something beyond this that might make $\varphi^{s_i}_\sigma$, $\varphi^{r_j}_\rho$ a complete collection in the sense above? I'd be curious to hear your thoughts.</p> http://mathoverflow.net/questions/98534/when-do-commuting-hamiltonian-flows-have-commuting-generators When do commuting Hamiltonian flows have commuting generators? soulphysics 2012-06-01T02:45:59Z 2012-06-04T17:01:43Z <p>Let $(P,\Omega)$ be a symplectic manifold, and let $[\cdot,\cdot]$ be the natural Poisson bracket. Let $\varphi^h(a)$ be the Hamiltonian flow generated by the smooth function $h:P\rightarrow\mathbb{R}$, and let $\varphi^g(b)$ be the Hamiltonian flow generated by the smooth function $g:P\rightarrow\mathbb{R}$.</p> <p>Suppose the two flows commute, $\varphi^g(b)\varphi^h(a) = \varphi^h(a)\varphi^g(b)$. Are there interesting circumstances under which it follows that the generators commute as well, $[g,h]=0$?</p> <p>I know that the flows commuting implies that $[g,h]=k$ for some constant $k$; a reference for the proof of this can be found, for example, in Arnold's (1989) <em>Mathematical Methods of classical mechanics</em>, pg. 218 Cor. 9. It is also easy to show that if $[g,h]=0$, then $\varphi^g(b)\varphi^h(a) = \varphi^h(a)\varphi^g(b)$. But under what circumstances does the converse hold?</p> <p>It's interesting that in quantum mechanics, the analogous relations hold in both directions. That is, if $G$ and $H$ are linear self-adjoint operators on a Hilbert space, and if $e^{ibG}$ and $e^{iaH}$ are the continuous one-parameter unitary groups they generate, then $e^{ibG}e^{iaH}=e^{iaH}e^{ibG}$ if and only if $[H,G]=0$ (where now $[\cdot,\cdot]$ is the commutator bracket).</p> http://mathoverflow.net/questions/78573/classical-analogue-of-the-stone-von-neumann-theorem Classical analogue of the Stone-von Neumann Theorem? soulphysics 2011-10-19T12:40:36Z 2011-10-26T22:25:04Z <p>Let $U_s$, $V_t$ be a pair of continuous $n$-parameter groups ($n &lt; \infty$) of unitary operators on a complex Hilbert space $\mathcal{H}$. The <a href="http://en.wikipedia.org/wiki/Stone%E2%80%93von_Neumann_theorem" rel="nofollow">Stone-von Neumann Theorem</a> establishes that any such pair forming an irreducible representation of the Weyl relations,</p> <p>$U_sV_t = e^{is\cdot t}V_tU_s$</p> <p>is unitarily equivalent to the Schrödinger representation, and hence that all such representations are unitarily equivalent. (Note: the Weyl relations in this context are equivalent to the canonical commutation relations (CCRs) $[Q,P]\psi=i\psi$ for all $\psi$ in the common dense domain of $Q$ and $P$, where $Q$ and $P$ are the generators of $V$ and $U$.)</p> <p><strong>Question:</strong> Is there a known analogue of this result in the context of classical Hamiltonian mechanics?</p> <p>I don't know of a classical analogue of the Weyl relations. But there is a classical analogue of the CCRs, which is the Poisson bracket $\{q,p\}=1$. So, here's how I imagine a classical analogue of the Stone-von Neumann theorem might look (just a rough attempt, really!).</p> <p>Let $\mathcal{M}$ be a smooth $2n$-dimensional manifold and $\omega$ a symplectic form on $\mathcal{M}$. Let $\xi = (q,p)$ be any global coordinate system on $\mathcal{M}$, and let $Q:\mathcal{M}\rightarrow\mathbb{R}$ and $P:\mathcal{M}\rightarrow\mathbb{R}$ be the projections onto $q$ and $p$, respectively. Then (conjecture): all such pairs ($Q$, $P)$ satisfying,</p> <p>$\{Q,P\}=1$</p> <p>where $\{\cdot,\cdot\}$ is the Poisson bracket associated with $(\mathcal{M}, \omega)$, are related by a single canonical transformation.</p> <p>Does this seem like a reasonable way to formulate the classical analogue? Is the status of this conjecture obvious? Your thoughts are appreciated!</p> http://mathoverflow.net/questions/54925/do-unitary-bijections-act-invariantly-on-irreducible-representations Do unitary bijections act invariantly on irreducible representations? soulphysics 2011-02-09T20:49:03Z 2011-02-09T22:47:55Z <p>Let $\mathcal{A}$ be a $C^*$ algebra. Let $(\pi, \mathcal{H})$ be a faithful, irreducible, unitary, Hilbert space representation of $\mathcal{A}$; i.e., $\pi:\mathcal{A}\rightarrow\mathcal{B}(\mathcal{H})$ is an injective *-homomorphism from $\mathcal{A}$ into a subset of the bounded operators $\mathcal{B}(\mathcal{H})$ on a Hilbert space $\mathcal{H}$, such that the resulting representation is irreducible. (A representation is irreducible if whenever $\pi(\mathcal{A})$ acts invariantly on a subspace $\mathcal{H}^\prime \subseteq \mathcal{H}$, then $\mathcal{H}^\prime = \mathbf{0}$ or $\mathcal{H}^\prime = \mathcal{H}$.)</p> <p>Let $U \in \mathcal{B}(\mathcal{H})$ be any unitary ($U^* = U^{-1}$) bijection on $\mathcal{H}$, not necessarily in $\pi(\mathcal{A})$.</p> <p>Does it follow that $U\pi(\mathcal{A})U^{*} = \pi(\mathcal{A})$?</p> http://mathoverflow.net/questions/50288/is-expra-expar-for-real-r-and-a-in-a-banach-space Is exp(rA) = (exp(A))^r for real r and A in a Banach space? soulphysics 2010-12-24T15:29:40Z 2010-12-25T00:22:47Z <p>Is $e^{(rA)} = (e^{A})^r$ when $r \in \mathbb{R}$ and $A$ is an element of a Banach algebra?</p> <p>Clearly if $n$ is an integer, then</p> <p>$e^{(nA)} = e^{A+A \cdots +A} = e^{A}e^{A}\cdots e^{A} = (e^{A})^n$,</p> <p>where the second equality follows from the Baker-Hausdorff lemma and the fact that [A,A]=0. On the other hand, I think the equality is not generally true when $r \in \mathbb{C}$. But what about the reals?</p> <p>Many thanks for your thoughts!</p> http://mathoverflow.net/questions/40666/what-is-the-outer-automorphism-group-of-sun What is the outer automorphism group of SU(n)? soulphysics 2010-09-30T19:43:49Z 2010-10-03T01:15:43Z <p>All the automorphisms of $SU(2)$ seem to be inner, which would mean that $\mathrm{Out}$ $SU(2)$ is trivial. Is that correct? Is this true in general $SU(n)$? I can't quite see -- any thoughts would be helpful.</p> http://mathoverflow.net/questions/30269/mathematics-in-nature/32472#32472 Answer by soulphysics for mathematics in nature soulphysics 2010-07-19T11:50:39Z 2010-07-19T11:56:38Z <p>A few quick examples:</p> <ul> <li>Breeding bunnies $\rightarrow$ <strong>the Fibonacci sequence</strong></li> <li>Projectile/planetary motion $\rightarrow$ <strong>conic sections</strong></li> <li>Natural springs and sinks $\rightarrow$ <strong>Gauss' law</strong></li> <li>Galilean relativity + constant speed of light $\rightarrow$ <strong>non-Euclidean geometry</strong></li> <li>Relativistic gravity $\rightarrow$ <strong>intrinsic curvature</strong></li> <li>Invariances of Maxwell equations $\rightarrow$ <strong>Conformal transformations</strong></li> <li>Observables in quantum mechanics $\rightarrow$ <strong>Lie groups (a la Wigner)</strong></li> </ul> <p>EDIT:</p> <ul> <li>Symmetries of the electron $\rightarrow$ <strong>Quaternions</strong></li> </ul> http://mathoverflow.net/questions/32169/why-are-they-called-isothermal-coordinates/32171#32171 Answer by soulphysics for Why are they called isothermal coordinates? soulphysics 2010-07-16T13:57:42Z 2010-07-16T14:03:03Z <p>According to <a href="http://books.google.com/books?id=TGw98Z6Cv-EC&amp;lpg=PA518&amp;dq=isothermal%2520coordinates&amp;pg=PA518#v=onepage&amp;q=%2522isothermal%2522%2520is%2520due%2520to%2520Lame&amp;f=false" rel="nofollow">Gray, Abbena and Salamon</a>, that's the name given to such coordinate systems by Gabriel Lamé in his 1833 study of heat transfer. The reason is, if you've got a thermally isolated surface of constant heat conduction, the constant coordinate lines are isotherms iff the coordinates are isothermal.</p> http://mathoverflow.net/questions/32164/how-big-is-the-center-of-an-orthogonal-group How big is the center of an orthogonal group? soulphysics 2010-07-16T12:56:07Z 2010-07-16T12:56:07Z <p>How big is the center of an arbitrary orthogonal group $O(m,n)$?</p> <p>In the special case of the "circle group" $O(2)$, it's clear that $|\zeta O(2)|$ = 1. In the case of $O(3)$, it seems clear that the center has two elements $\zeta O(3) = \lbrace 1, -1 \rbrace$. I can see this by visualizing a sphere in an arbitrary $(i, j, k)$ basis, and observing that both the identity and the "complete" reversal $(i, j, k) \mapsto (-i, -j, -k)$ commute with everthing.</p> <p>But I'd like a simple way to see how the situation changes for more general orthogonal groups like the (inhomogeneous) Lorentz group $O(3,1)$.</p> http://mathoverflow.net/questions/98534/when-do-commuting-hamiltonian-flows-have-commuting-generators/98541#98541 Comment by soulphysics soulphysics 2012-06-03T20:22:59Z 2012-06-03T20:22:59Z &gt; There they commute iff... $[G,H]$ is a constant multiple of the identity. I understand $\mathbb{P}\mathcal{H}$ to contain the equivalence classes of vectors related by a multiplicative constant. But then, isn't calling $[G,H]$ a &quot;constant multiple of the identity&quot; the same as saying $[G,H]=0$? It seems to me that on the projective space $\mathbb{P}\mathcal{H}$, it is also the case that $[e^{ibG}, e^{iaH}]=0$ iff $[G,H]=0$. What am I missing? [P.S.: Thanks for the Roels and Weinstein reference, that is very helpful.] http://mathoverflow.net/questions/98534/when-do-commuting-hamiltonian-flows-have-commuting-generators/98543#98543 Comment by soulphysics soulphysics 2012-06-02T22:25:39Z 2012-06-02T22:25:39Z As it happens I am exceedingly lowbrow, touch&#233;! ;) Just in case, would you happen to have a reference for the Souriau approach? http://mathoverflow.net/questions/50288/is-expra-expar-for-real-r-and-a-in-a-banach-space/50314#50314 Comment by soulphysics soulphysics 2010-12-26T17:21:30Z 2010-12-26T17:21:30Z Oh dear -- you're of course correct, the claim is not generally true. Thanks! http://mathoverflow.net/questions/50288/is-expra-expar-for-real-r-and-a-in-a-banach-space Comment by soulphysics soulphysics 2010-12-24T16:44:29Z 2010-12-24T16:44:29Z @fedja -- One way is the Taylor expansion: $A^r = \Sigma^\infty_{n=0}\frac{r^n \log^n(A)}{n!},$ where the logarithms are again defined by their Taylor expansions. @arsmath -- That's a way to define the operator $e^X$ (where $X$ is in the Banach algebra), but what's needed is $X^r$. http://mathoverflow.net/questions/40666/what-is-the-outer-automorphism-group-of-sun/40668#40668 Comment by soulphysics soulphysics 2010-09-30T20:06:52Z 2010-09-30T20:06:52Z Excellent, I can see that now. Thanks very much! And now it seems that complex conjugation sends the fundamental representation of $SU(2)$ to itself. So, am I correct to think that $\mathrm{Out}$ $SU(2)$ is trivial? http://mathoverflow.net/questions/30269/mathematics-in-nature/32472#32472 Comment by soulphysics soulphysics 2010-07-21T12:12:34Z 2010-07-21T12:12:34Z @Victor -- obviously. But the question asks for natural phenomena that &quot;perfectly illustrate&quot; a mathematical concept, not &quot;historically inspired&quot;. Sheesh. http://mathoverflow.net/questions/30269/mathematics-in-nature/32472#32472 Comment by soulphysics soulphysics 2010-07-19T12:00:27Z 2010-07-19T12:00:27Z Hey, it's an idealization, ok? And anyway, if it was good enough for Fibonacci himself... ;) http://mathoverflow.net/questions/32164/how-big-is-the-center-of-an-orthogonal-group Comment by soulphysics soulphysics 2010-07-16T15:29:56Z 2010-07-16T15:29:56Z Thanks Robin -- I've convinced myself now. Although the Lorentz group has four interesting discrete transformations, only I and -I are central. http://mathoverflow.net/questions/32164/how-big-is-the-center-of-an-orthogonal-group Comment by soulphysics soulphysics 2010-07-16T14:06:01Z 2010-07-16T14:06:01Z Robin -- could you elaborate? I seem to recall from somewhere that the Lorentz group has a 4-element center... http://mathoverflow.net/questions/32164/how-big-is-the-center-of-an-orthogonal-group Comment by soulphysics soulphysics 2010-07-16T13:20:00Z 2010-07-16T13:20:00Z Thanks -- I was thinking about $O$ over the real number field, for which what I said is correct. But as you suggest, the question gets more interesting for $O$ over arbitrary fields!