User qhliu - MathOverflow

Answer by QHLIU for Curvature dependence of the Laplacian operator acting on a n-1 dimensional compact submanifold in the n-dimensional Euclidian space

2013-04-16T01:28:47Z

Dear Rbega, Thank you! For a two dimensional surface, I can prove a much simpler relation by direct computations: \begin{equation*} 1/\sqrt{g}\partial _{\mu }g^{\mu \nu }\sqrt{g}\partial _ {\nu }\mathbf{R}=2 \mathbf{\mathbf{H}_{R}}\text{.} \end{equation*}

Now, Let me calculate explicitly in general according to your formula, with use of the Einstein summation convention. Since we deal with a $n-1$ dimensional surface \begin{equation*} {\mathbf{R}} = ({ X_ {1},X_ {2},...,X_ {n} } )= X_ {j}\mathbf{i}_{Xj} \end{equation*} with $\mathbf{i}_ {X_{j}}$ denoting the unit normal along $j$-th Cartesian coordinate, we would have $ \Delta _ {\mathbf{R}^{n}} $ $ \mathbf{R} $ $=\partial _ {X_{i}} $ $ \partial _ {X_{i}} \mathbf{R} =0, $ and \begin{equation*} \nabla _ {\mathbf{R}^{n}} \mathbf{R} \end{equation*} \begin{equation*} \equiv (\mathbf{i} _ {X_{i}} \partial _ {X_{i}}) (X_{j}\mathbf{i} _ {X_{j}}) \end{equation*} \begin{equation*} = \mathbf{i} _ {X_{i}} \delta _ {ij} \mathbf{i} _ {X_{j}}, \end{equation*} and so \begin{equation*} \mathbf{H}_{R}\mathbf{\cdot }\nabla _{\mathbf{R}^{n}}\mathbf{R=\mathbf{H} _{R}.} \end{equation*} If I\ am correct, please tell me what is $f(\mathbf{n},\mathbf{n})$ in our problem \begin{equation*} \mathbf{R}=X_{j}\mathbf{i}_{Xj}\text{,} \end{equation*} and what is $\nabla _{\mathbf{R}^{n}}^{2}$? In physics, we usually use two forms: one is the \begin{equation*} 1/\sqrt{g}\partial _{\mu }g^{\mu \nu }\sqrt{g}\partial _{\nu } \end{equation*} on the surface, and another is $ \partial _ {X_{i}} \partial _ {X_{i}}$ in the $n$ dimensional Euclidean space, what is the $\nabla _{\mathbf{R} ^{n}}^{2}$?

Curvature dependence of the Laplacian operator acting on a n-1 dimensional compact submanifold in the n-dimensional Euclidian space

2013-04-15T14:03:36Z

Possibly a simple question in differential geometry (maybe not accurate but understandable in mathematical terms): Given an compact surface $ \mathbf {R} $ in $n$ Euclidean space parameterized by $n-1$ variables $ (x_1,x_2,...,x_{n-1}) $ in the following:

$ \mathbf {R} $={ $ X_1,X_2,X_3,...,X_n$ }, ($ X_i=X_i(x_1,x_2,...,x_{n-1}$ ) is the $i$-th Cartesian coordinate)

Then, what is the result of Laplacian operator $∇^2=(1/(\sqrt{g})\partial_{μ}g^{μυ}\sqrt{g} \partial_{υ} $ acting on the $ \mathbf {R} $ as $∇^2 \mathbf {R}$ ? I think that it should be a result that purely depends on the extrinsic curvatures, and also a geometric invariant. Please offer me the result together with a reference which is accessible to a physicist. Thanks.

what is parameterization of the Trefoil knot surface in R³?

2012-03-17T05:06:18Z

what is parameterization e.g., (x(u,v),y(u,v),z(u,v)), of the Trefoil knot surface in R³ whose cross section of the surface can be circular, or in general elliptic?

Thanks!

Are these operators defined on 2D surfaces self-adjoint?

2010-09-04T09:11:54Z

My research group finds/proposes a fundamental operator in quantum mechanics, the Cartesian momentum as I called (I think for mathematician the ref. 2007 is sufficient). However, I do not know whether it is self-adjoint or not (we are all physicists). If a mathematician can give a definite answer to it for even simple surfaces such as cylindrical and spherical, he has then a nice paper.

The standard representation of the curved smooth surface $M$ embedded\ in $ R^{3}$ is,

$\mathbf{r}(\xi ,\zeta )\mathbf{=}\left( x(\xi ,\zeta ),y(\xi ,\zeta ),z(\xi ,\zeta )\right)$.

The covariant derivatives of $\mathbf{r}$ are $\mathbf{r}_{\mu }=\partial \mathbf{r}/ \partial x^{\mu }$ .

The contravariant derivatives

$\mathbf{r}^{\mu }\equiv g^{\mu \upsilon }\mathbf{r}_{\upsilon }$

is the generalized inverse of the covariant ones $\mathbf{r}_{\mu }$.

The unit normal vector at point $(\xi ,\zeta )$ is $\mathbf{n=r}^{\xi } \times \mathbf{r}^{\zeta }/ \sqrt{g}$.

The Hermitian Cartesian momentum $\mathbf{p}$ takes a compact form,

$\mathbf{p=}-i\hbar (\mathbf{r}^{\mu }\partial _{\mu }+H\mathbf{n),}$

where $H$ is the mean curvature of the surface. When the motion is constraint-free or in a flat plane, i.e., when $H=0$, the constraint induced terms $H\mathbf{n}$ vanish. Then the Cartesian momentum operator reproduces its usual form as, $\mathbf{p=}-i\hbar \nabla $.

For a particle moves on the surface of a sphere of radius $r$, $ x=r\sin \theta \cos \varphi ,\text{ }y=r\sin \theta \sin \varphi ,\text{ }z=r\cos \theta$,

the hermitian operators for Cartesian momenta $p_{i}$ are respectively,

$p_{x} =-\frac{i\hbar }{r}(\cos \theta \cos \varphi \frac{\partial }{\partial \theta }-\frac{\sin \varphi }{\sin \theta }\frac{\partial }{\partial \varphi }-\sin \theta \cos \varphi ), $

$p_{y} =-\frac{i\hbar }{r}(\cos \theta \sin \varphi \frac{\partial }{\partial \theta }+\frac{\cos \varphi }{\sin \theta }\frac{\partial }{\partial \varphi }-\sin \theta \sin \varphi ), $

$p_{z} =\frac{i\hbar }{r}(\sin \theta \frac{\partial }{\partial \theta }+\cos \theta ).$

On the spherical surface, the complete set of the spherical harmonics defines the Hilbert space.

Refs.

2003, Liu Q H and Liu T G, Int. Quantum Hamiltonian for the Rigid Rotator, J. Theoret. Phys. 42(2003)2877.

2004, Liu Q H, Hou J X, Xiao Y P and Li L X, Quantum Motion on 2D Surface of Nonspherical Topology, Int. J. Theoret. Phys. 43(2004)1011.

2005, Xiao Y P, Lai M M, Hou J X, Chen X W and Liu Q H, A Secondary Operator Ordering Problem for a Charged Rigid Planar Rotator in Uniform Magnetic Field, Comm. Theoret. Phys. 44(2005)49.

2006a, Lai M M, Wang X, Xiao Y P and Liu Q H, Gauge Transformation and Constraint Induced Operator Ordering for Charged Rigid Planar Rotator in Uniform Magnetic Field, Comm. Theoret. Phys. 46(2006) 843.

2006b, Wang X, Xiao Y P, Liu T G, Lai M M and Rao, Quantum Motion on 2D Surfaces of Spherical Topology, Int. J. Theoret. Phys. 45(2006)2509.

2006c, Liu Q H, Universality of Operator Ordering in Kinetic Energy Operator for Particles Moving on two Dimensional Surfaces, Int. J. Theoret. Phys. 45(2006)2167.

2007, Liu Q H., Tong C L., Lai M M., Constraint-induced mean curvature dependence of Cartesian momentum operators J. Phys. A 40(2007)4161.

2010, Zhu X M, Xu M and Liu Q H, Wave packets on spherical surface viewed from expectation values of Cartesian variables, Int. J. Geom. Meth. Mod. Phys., 7(2010)411-423.

An asymptotic expression for the solution to the squares problem suggested by statistical mechanics

2010-09-11T10:52:09Z

The $s$ squares problem is to count the number $r_s (n)$ of integer solutions $(x_1,x_2,...,x_s)$ of the Diophantine equation $x_{1}^{2}+x_{2}^{2}+...+x_{s}^{2}=n$ in which changing the sign or order of the $x_i$ ’s gives distinct solutions. If looking into the statistical mechanics for classical ideal gas in 3D, we meet with the same thing with $s=3N$, $N$ is the number of particles. But now the $3N$ squares problem is to count the number of the microstates in the so-called microscope ensemble. The following asymptotic expression of $r_{3N}(n)$ is experimentally validated, so it is physically proved:

$r_{3N}(n)\approx \frac{{\pi}^{3N/2}}{\Gamma (3N/2)} {{n}^{3N/2-1}}$, in thermodynamic limit $n/N=const.$ and $n \to \infty$ .

My question is: How to give an estimate of the error, and does anyone know such a formula in mathematical literature?

Ref.

S.C.Miline, New infinite families of exact sums of squares formulas, Jacobi elliptic functions and Ramanujan’s tau function, Proc. Natl. Acad. Sci. USA, 1996, 93:15004-15008, and references cited therein.

complete estimates of the error for a well-known asymptotic expression of partition p(n,m)

2010-09-01T09:43:21Z

Let $p(n,m)$ be the number of partitions of an integer $n$ into integers $\le m$, we have a well-known asymptotic expression:

For a fixed $m$ and $n\to\infty$, $$p(n,m)=\frac{n^{m-1}}{m!(m-1)!} (1+O(1/n)) $$

My question is: why the error $O(1/n)$ is independent of $m$? Or how can it be extended for $m$ growing slowly with $n$? Please help me to find the answer or the references. Thanks.

Answer by QHLIU for complete estimates of the error for a well-known asymptotic expression of partition p(n,m)

2010-09-01T10:38:43Z

Thank Robin Chapman very much for editing.

There is a nice asymptotic expression for partition $q(n,M)$ that denotes the number of partitions of $n$ with $M$ parts all distinct: As $n\to\infty$,

$$ q(n,M)\approx \frac{(n-1)!}{M!(M-1)!(n-M)!}\left( 1+O\left( \frac{M^{3}}{n} \right) \right)$$

Isn't there no similar asymptotic expression for partition $p(n,m)$?

Comment by QHLIU

2013-04-16T03:07:50Z

my clarification sees in the form of answer below.

Comment by QHLIU

2013-04-15T15:39:07Z

Thank you for your answer. But in my question, the Laplacian(-Beltrami) operator takes a definite form, corresponding to your $\Delta$. Then what does the difference between $\Delta$ and $\Delta _ {\mathbb {R}^n}$? It appears a compact form of the result for the definite $\mathbf{R}=\{X_{1},X_{2},...,X_{n}\}$.

Comment by QHLIU

2012-03-18T00:26:23Z

@Jyrki Lahtonen Trefoil knot surface in R³ seems quite inevitable in physics world, e.g. in the string theory, electromagnetism, fluid membrane, condensed matter physics, nanostructure, ..., cf. Faddeev model (hopfion.com/faddeev.html), Linked and knotted beams of light (nature.com/nphys/journal/v4/n9/abs/nphys1056.html), etc. What I am interested if quantum constrained motion on the surface possibly results in some novel consequence, cf. Geometric momentum: The proper momentum for a free particle on a two-dimensional sphere (pra.aps.org/abstract/PRA/v84/i4/e042101).

Comment by QHLIU

2012-03-17T13:52:52Z

Thank you very much!

Comment by QHLIU

2010-09-12T03:22:33Z

@ Martin Gisser: iopscience.iop.org/1751-8121/40/15/007 or arxiv.org/abs/0910.0769

Comment by QHLIU

2010-09-11T13:55:58Z

fedja: I admit that physical way of reasoning usually loses the mathematical rigor, but it is supported by the experiments. Besides, physical reasoning enriches the mathematical studies, such as the Fourier series, distribution theory, etc. All primitive physical results are then mathematically proved correct; no exception is so far identified. Why isn’t it wrong? The approximate solution to the squares problem is easily attainable by treating the lattice (hyper-)spherical surface as a smooth one, and the number of the solutions is then nothing but the area of the surface area.

Comment by QHLIU

2010-09-11T12:20:11Z

Robin Chapman: Thanks a lot. Some English and presentation problems are fixed.

Comment by QHLIU

2010-09-11T11:56:18Z

Robin Chapman: yes, it is a mathematical problem that has not been proved yet. In physics, such a formula is experimentally verifed so physically correct. A ref: P. K. Pathria, Statistical Mechanics, 2nd ed. (1997) p.17

Comment by QHLIU

2010-09-11T11:42:00Z

Robin Chapman: "Experimentally proved" amounts to "mathematically proved"? No. In mathematics, we have to know estimates of the error, and it is a mathematical problem.

Comment by QHLIU

2010-09-11T11:37:23Z

In statistical mechanics for classical ideal gas, the so-called thermodynamic limit must be imposed. So, $n$ and $N$ must be linearly dependent and $n to \infty$. However, I do not know if the asymptotic expression holds without the thermodynamic limit, nor the error of the expression

Comment by QHLIU

2010-09-11T11:23:59Z

To Robin Chapman: true, it is an asymptotic expression in limit $N\to\infty$. In statistical mechanics, it is also a famous problem. If there is such a mathematical literature, I am sure no physicist has been aware of its existence. It's hard to accept a fact that physicists in this circle have all overlooked such a fundamental result.

Comment by QHLIU

2010-09-11T11:15:06Z

To Gjergji Zaimi: not inside a sphere, but on the surface of it.

Comment by QHLIU

2010-09-11T05:18:34Z

Yang's comment is absolutely true!

Comment by QHLIU

2010-09-04T08:20:54Z

Please be noted that in G. Szekeres' two papers in 1951 and 1953, the asymptotic formulae for $p(n,m)$ valid only under strong condition that $m$ is related to $n^2$. In physics, no asymptotic formula not useful unless $m \alpha n$ ($\alpha >0$ and as $m$ is large).