User lizao li - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:46:56Z http://mathoverflow.net/feeds/user/16464 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109070/applications-of-hodge-de-rham-laplacian-on-p-forms-p-neq-0-n-in-physics-or-en Applications of Hodge-de Rham Laplacian on p-forms ($p\neq 0,n$) in physics or engineering Lizao Li 2012-10-07T15:53:39Z 2012-10-15T00:28:30Z <p>The Hodge-de Rham Laplacian $L=(d+d^*)^2$, where $d$ is the boundary operator of the de Rham complex, is well-known in the math community. Recently, I tried very hard to search for examples of its use in physics and engineering applications (for giving talks or future teaching). More specifically, I want to find applications of the Hodge-de Rham Laplacian on scalar(real or complex)-valued differential p-forms on $\Omega\in\mathbb{R}^n$ with natural boundary conditions and $p\neq 0,n$, where not only the cohomology but also the solution itself is of interest (thinking about PDEs).</p> <p>To my surprise, I can only find one such example: in $\Omega\in \mathbb{R}^3$, the 1-form and 2-form Laplacians are the better-known vector Laplacian used in the Maxwell equations. </p> <p>Note that "natural boundary" condition is very important. It means the boundary condition comes from the need to perform integration by parts. For example in coordinates, one of the homogeneous natural boundary conditions for 2-Laplacian in $\mathbb{R}^3$ is $u\times n =0$ and $(\nabla\times u)\cdot n =0$. But $u=0$ on $\partial \Omega$ is not "natural" here. That's why:</p> <ol> <li>Stokes or Navier-Stokes does not count, because the vector Laplacian there is really the strain opeartor on div-free fields (the boundary condition is $u=0$ on $\partial \Omega$). </li> <li>For the similar reason about the boundary condition, the "vector Laplacian" given by a system of $m$ scalar Laplacians (or its corresponding heat or wave) equation does not qualify.</li> </ol> <p>It is somewhat unbelievable that the Maxwell equations is the only example. Are there any more applications?</p> http://mathoverflow.net/questions/106847/geometric-derivation-of-the-einsteins-field-equation-from-the-hilbert-action Geometric derivation of the Einsteinâ€™s field equation from the Hilbert action. Lizao Li 2012-09-10T21:33:08Z 2012-09-10T21:33:08Z <p>It is well-known that the equation for stationary solutions of the Einstein-Hilbert functional is given by the Einstein field equation (for a statement, see <a href="http://mathoverflow.net/questions/106786/coordinate-free-derivation-of-the-einsteins-field-equation-from-the-hilbert-acti" rel="nofollow">previous question</a>). The standard derivation of this is through <a href="http://en.wikipedia.org/wiki/Fundamental_theorem_of_Riemannian_geometry#TheKoszulformula" rel="nofollow">Koszul's formulae</a> either in coordinates (for example <a href="http://en.wikipedia.org/wiki/Einstein%25E2%2580%2593Hilbert_action" rel="nofollow">wikipedia</a>), or in abstract index notation (for example, in Wald's General Relativty), or in coordinate-free notation (for example, as pointed out by <a href="http://mathoverflow.net/questions/106786/coordinate-free-derivation-of-the-einsteins-field-equation-from-the-hilbert-acti" rel="nofollow">Thomas Richard</a> in Besse - Einstein manifold). This approach is mainly algebraic by using the definition in terms of Koszul's formulae and then calculus in various notations. Essentially the derivation is a direct calculation without the need to even mention the manifold. </p> <p>I am wondering if there is a way to derive/interprete the statement refered to at the beginning using an alternative method which is more geometric, ie. using parallel transport or alike. The criterion for "geometric" being (a) a direct reference to the manifold is necessary; or (b) a picture, at least in principle a mental picture, can be drawn to at least carry the main idea of the derivation (of course, pictures of formulae don't count).</p> http://mathoverflow.net/questions/106786/coordinate-free-derivation-of-the-einsteins-field-equation-from-the-hilbert-acti Coordinate-free derivation of the Einstein's field equation from the Hilbert action. Lizao Li 2012-09-10T07:38:37Z 2012-09-10T09:04:36Z <p>It is well-known that the equation for stationary solutions of the Einstein-Hilbert functional (without matter and cosmological constant, which is irrelevant here):<br> $$S = \int_M R \mu_g,$$ is given by the Einstein's field equation: $$Ric -\frac{1}{2}R g = 0,$$ where $\mu_g$ is the canonical volume form given by the metric $g$, $Ric$ is the Ricci curvature and $R$ is the Ricci scalar.</p> <p>The standard derivation of the above statement seems to be a not so hard but not so pleasant direct calculation, either in coordinates or abstract indices, expanding everything in terms of the Christoffel symbol and eventually in terms of $g$ and then calculus.</p> <p>My questions is: is there a more geometric and coordinate-free way to derive this? </p> http://mathoverflow.net/questions/59186/regge-calculus-questions-of-consistency-resolved/105363#105363 Answer by Lizao Li for Regge calculus: Questions of consistency resolved? Lizao Li 2012-08-23T23:57:33Z 2012-09-10T07:18:14Z <p>The consistency is proved by Cheeger, M\"uller and Schrader in 1984, "On the Curvature of Piecewise Flat Sapces". Roughly speaking, given a smooth Riemannian manifold with a smooth metric, there exists a sequence of triangulation, on which Regge's definition converges to the smooth curvature as a measure. </p> <p>At the linearized level, there is also a recent paper on the consistency: Christiansen 2011, "On The Linearization of Regge Calculus". One of the theorems in the paper is that we have consistency between linearized Regge and linearized Einstein equation as well.</p> <p>That said, when you talk about the convergence of a numerical algorithm, it depends on a lot of other things as well, such as your formulation of the Einstein's equation. You will also need some form of stability to ensure convergence. Those questions remain to be solved (hopefully in my thesis:-).</p> http://mathoverflow.net/questions/116363/numerical-linear-algebra-how-to-compute-btc1b-efficiently Comment by Lizao Li Lizao Li 2013-02-04T05:03:44Z 2013-02-04T05:03:44Z I would suggest first check what is wrong with CG. If for the bigger systems, it takes the similar number of iterations to convergence compared to the smaller system, then I don't know anything you can do about it. But if the bigger system takes a lot more iterations, you should look for additoinal structures of the system and find a good preconditioner. http://mathoverflow.net/questions/109070/applications-of-hodge-de-rham-laplacian-on-p-forms-p-neq-0-n-in-physics-or-en/109263#109263 Comment by Lizao Li Lizao Li 2012-10-11T20:17:17Z 2012-10-11T20:17:17Z Thank you very much. Sounds very interesting. Your mathematical description definitely helped a lot for me to understand it. http://mathoverflow.net/questions/109070/applications-of-hodge-de-rham-laplacian-on-p-forms-p-neq-0-n-in-physics-or-en/109099#109099 Comment by Lizao Li Lizao Li 2012-10-08T22:57:48Z 2012-10-08T22:57:48Z Indeed many Gauge theories can be formulated as Laplace or Dirac operators on principal-bundle-valued differential forms. Here I am thinking about more down-to-earth scalar valued ones as just motivations for talks and so on... Thanks all the same. http://mathoverflow.net/questions/109070/applications-of-hodge-de-rham-laplacian-on-p-forms-p-neq-0-n-in-physics-or-en/109073#109073 Comment by Lizao Li Lizao Li 2012-10-08T22:51:31Z 2012-10-08T22:51:31Z Very nice. Thanks. http://mathoverflow.net/questions/109070/applications-of-hodge-de-rham-laplacian-on-p-forms-p-neq-0-n-in-physics-or-en Comment by Lizao Li Lizao Li 2012-10-08T22:47:55Z 2012-10-08T22:47:55Z @Yang, actually I excluded the 0- and 3-forms because they are just the usual scalar Laplacians with different boundary conditions. There are way too many examples of them that I don't need to try hard to find plenty of motivation. http://mathoverflow.net/questions/106847/geometric-derivation-of-the-einsteins-field-equation-from-the-hilbert-action Comment by Lizao Li Lizao Li 2012-09-11T15:50:11Z 2012-09-11T15:50:11Z saying that the problem here can be in any way as elementary as that. But I was hoping there was more to the pretty formulae than calculus alone. http://mathoverflow.net/questions/106847/geometric-derivation-of-the-einsteins-field-equation-from-the-hilbert-action Comment by Lizao Li Lizao Li 2012-09-11T15:48:25Z 2012-09-11T15:48:25Z @Willie Wong - Thank you for the comment. I call it can be done without &quot;direct reference to the manifold&quot; because you can define the Riemann tensor by the Koszul formula and prove the equivalence as a calculus exercise. For example, you want to find the Fermat point in a triangle. Yes you can solve it through &quot;variational approach&quot; by calculus. Or you can do by geomtry. There is a huge difference aesthetically. The relevance of the example here is that the Ricci and scalar curvatures too can be characterized by a lot of other things other than their direct formulae alone. Of course I am not http://mathoverflow.net/questions/106847/geometric-derivation-of-the-einsteins-field-equation-from-the-hilbert-action Comment by Lizao Li Lizao Li 2012-09-11T15:18:07Z 2012-09-11T15:18:07Z But I don't really see how things fit in here. Also, it seems even in MTW - Gravitation they just give a direct derivation. http://mathoverflow.net/questions/106847/geometric-derivation-of-the-einsteins-field-equation-from-the-hilbert-action Comment by Lizao Li Lizao Li 2012-09-11T15:17:26Z 2012-09-11T15:17:26Z Thank you for the comments. @Deane Yang - I am not that optimistic about it as well. But the situation for the geodesic equation is somewhat different. The equation itself has the Christoffel symbol in it. So I don't think there is much you can do about it. The meaning of the various curvatures appeared here are beyond mere calculation tools. Hilbert action is the total curvature (integration of the 2nd Lipschitz-Killing curvature), which at least for simple objects have a very clear interpretation other than the formulae itself. The Einstein tensor has geometric characterizations too. http://mathoverflow.net/questions/106786/coordinate-free-derivation-of-the-einsteins-field-equation-from-the-hilbert-acti/106792#106792 Comment by Lizao Li Lizao Li 2012-09-10T20:54:58Z 2012-09-10T20:54:58Z Thank you very much. That is indeed a coordinate-free calculation. Still, it is just the standard derivation in the coordinate-free notation. So I didn't formulate my problem well as I was hoping for something more geometrical rather than algebra, say in terms of parallel transport or something of that kind. Thanks all the same.