7,645 reputation
2832
bio website supermath.info
location Virgina USA
age 36
visits member for 2 years, 1 month
seen yesterday

I'm interested in mathematics which is used to frame physical theory. I guess that means I'm at least a little interested in just about anything.

Currently I'm trying to understand Cartan's Method of moving frames as it applies to various classification questions of low-dimensional geometry.

I also have interest in superanalysis and certain problems of hypercomplex analysis.

More generally, I'm just looking for interested students who want to exceed the status-quo of undergraduate mathematics. Ideally, our interests overlap.


Jul
22
comment How to take limit *along a path*
The point of calculating path limits is not to confirm the value of a limit exists. Rather, the point of path limits in practice is to confirm the limit cannot exist.
Jul
22
reviewed Approve suggested edit on How to use the Mehler kernel to get the solution of the Quantum harmonic oscillator with a given initial condition
Jul
22
reviewed Reject suggested edit on A system of nonlinear equations with sine and cosine functions
Jul
21
reviewed Reject suggested edit on I am searching for an unusual real-valued function.
Jul
21
answered The best balance in studying Mathematics?
Jul
21
comment What is the meaning of the symbol $\nabla^k$?
@IlikeSerena sometimes a minus is included for the Laplacian. There is not a universal convention on this point.
Jul
20
comment Should I read about Manifolds or Algebraic Topology?
Sounds like you already have some background in topology so I'd recommend reading the manifold text. Incidentally, that is a great book with a lot of insight.
Jul
20
comment $\dfrac{\partial}{\partial x}\left(\int_{g(x)}^{h(x)}f(y)\, dy \right)= f(h(x))h'(x)-f(g(x))g'(x)$
it seems $\frac{d}{dx}$ would be a reasonable notation here as the differentiated function is only a function of $x$ (not $y$)
Jul
20
comment Is it normal that a pure math student doesn't know vector analysis?
@math.n00b for the $\nabla (\vec{A} \cdot \vec{B})$ calculation, I recommend starting with the cross product terms then using the double $\epsilon$-identity on each term. A direct assault would seem to require you changing $\delta$-'s back into $\epsilon$ which requires a bit more imagination.
Jul
20
comment Open problems in Federer's Geometric Measure Theory
I'll put a bounty on your question here once it's possible.
Jul
20
comment Is it normal that a pure math student doesn't know vector analysis?
Nearly perfect, however, in the last line we should have $\partial_i (\nabla \cdot \vec{A})-\nabla^2 A_i = (\nabla(\nabla \cdot \vec{A})- \nabla^2 \vec{A})_i$. Which, is what you want as you calculated the $i$-th component of $\nabla \times (\nabla \times \vec{A})$. Nice work!
Jul
20
comment Is it normal that a pure math student doesn't know vector analysis?
$\vec{A} \cdot \nabla = A_x \partial_x+A_y \partial_y+A_z \partial_z$. The trap is that we are so accustomed to using $\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$ for vectors that we are tempted to suppose it applies to operators, but, it doesn't. The notation $\nabla \cdot \vec{A}$ and $\vec{A} \cdot \nabla$ are quite different.
Jul
20
comment Is it normal that a pure math student doesn't know vector analysis?
@math.n00b there is a marked difference between the first two terms in the identity in your post. For instance, we could just as well write $(\nabla \cdot \vec{A} )\vec{B}$ or $\vec{B}(\nabla \cdot \vec{A} )$ either way, we're just scalar multiplying the divergence of $\vec{A}$ on $\vec{B}$. The term $\vec{A} \cdot \nabla$ is a differential operator. It is no longer the case we can exchange the order of the last two terms as, for instance, $(\vec{A} \cdot \nabla) \vec{B} \neq \vec{B}(\vec{A} \cdot \nabla)$. On the lhs is a vector, on the rhs an operator. More to the point ...
Jul
19
comment Is it normal that a pure math student doesn't know vector analysis?
@Semiclassical incidentally, you might like physicsoverflow.org I haven't said much of use there yet, but it seems like a good community to support.
Jul
19
comment Is it normal that a pure math student doesn't know vector analysis?
@Semiclassical well, at first dummies are a bit unsettling (hence their absence in my calculation here intended for math audience) but, after a few hundred pages of calculation and omitting a few thousand $\sum$ symbols, you get numb to the abuse. In truth, it is very freeing, in combination with multi-index notation, you can say so much with so little. That said, I am not versed in en.wikipedia.org/wiki/Penrose_graphical_notation . Interesting.
Jul
19
answered Is it normal that a pure math student doesn't know vector analysis?
Jul
19
comment Is it normal that a pure math student doesn't know vector analysis?
For special relativity, it's hard to beat the mathematical simplicity and clarity of Resnicks text. amazon.com/Introduction-Special-Relativity-Robert-Resnick/dp/… . Also, if you want to look at calculus III written by someone who cares about $\nabla$ and curvelinear coordinates you might look at supermath.info/OldCalculusIIIresources.html see pages 360-425 or for my recent edit with a bit less of topics supermath.info/CalculusIIIf2014.pdf
Jul
18
comment Open problems in Federer's Geometric Measure Theory
I wonder if you might get better input from math overflow on this one.
Jul
18
comment Partial derivative or something else?
@Leucippus has a good point, in Einstein notation using $i=1,2,3$ whereas $\mu = 0, 1, 2, 3$ with metric $\eta = \text{diag}(-1,1,1,1)$ we have $3$-D Laplacian $\triangle^2 = \partial_i\partial^i$ or the D'Alembertian $\Box^2=\partial^{\mu}\partial_{\mu}$ which, in a sense, is the $4$-D Laplacian for special relativity. I joke that $\circ^2$ should be the Laplacian for an infinite dimensional space. I have seen a pentagon used for $5$-dimensions in a talk by en.wikipedia.org/wiki/Peter_van_Nieuwenhuizen . Also, see en.wikipedia.org/wiki/D%27Alembert_operator
Jul
18
comment Conceptual Question on different representations of Hyperplanes, Higher Standpoint, Coordinate-free
Of course, in the context of a vector space where we can order the basis however we like then it is possible to solve for the last. In my answer, the last variable is trivially a function of the first $n-1$ variables. That sort of result is also available for abstract manifolds see the last paragraph in the picture in math.stackexchange.com/q/345542/36530. It seems wikipedia is lacking a decent article on this... surprising.