Recent content by cliowa

Thanks for that clarification. When you say do you really mean that the process of thought is that 1) when you apply an electric field to a medium bound charges (dipole moments) will be "created" (locally) and 2) you then construct a polarization field that has divergence equal to the density...

Hey all, I'm studying laser-matter interactions and was wondering: Is there any physical meaning to a non-vanishing polarization field with non-trivial constitutive relation but vanishing divergence? (By non-trivial I mean the constitutive equation does not stipulate that the polarization and...

I agree completely. What puzzles me is rather the way the electrostatic potential can be "acquired" through contact. Example: A bird that first sat on some tree (which has the same potential as the earth) and then flies off to a power line, where he (suddenly?) "acquires" the potential of the...

Thinking in terms of electric circuits and electrostatic potential I understand how an electric current arises as manifestation of a difference in potential. How does this work at a more microscopic level? I.e. how does an electron know what potential it's environment is at? E.g.: If I...

Attention: This is completely wrong in full generality! By the very definition of a Banach space, this is only true in a Banach space. This is misleading. "The other direction" simply means you want to prove that the real numbers are a Banach space. Even if you forget about the construction...

Think of a space of functions: Say you fix some interval, look at the continuous functions...

Notice that the function f in \sum_{n=0}^{\infty} 2^{n}f(2^{n}) need not be well-defined for all arguments in the real numbers, but only for the natural numbers (including the zero). What you have is a positive monotone decreasing sequence (which is denoted here by f(n) but could just as well be...

Are you given any special topology on the sets X,Y? You could start by assuming that f(x) is not a limit point of f(A), i.e. there exists a neighborhood U of f(x) in Y, such that...

You don't even need Lagrange for that. Say you have a cyclic group G of order m, then for any a in G the order of a is at most n, right? Do you know what to do next? regards...Cliowa

So it should be clear to you that in your case (where n=12) you don't have a group with respect to the usual mod operation. If you consider an additive group {0,...,n-1}, where 0 is the identity, taking mod n works. Notice that the crucial point is that for 2 numbers to be identified mod n can...

Your mod-calculations are a bit odd. When you talk about taking {1,...,11} with multiplication mod 12, you identify all mupltiples of 12 with 1, etc, i.e. you identify 24 and 1, 25 and 2, 26 and 3, 27 and ...

Do you know what the cotangent bundle is? One can view the cotangent bundle as the tangent bundle with the tangent spaces replaced by their dual spaces. So an element of a cotangent space acts on the tangent space, i.e. you feed it a tangent vector and out comes a number. That's precisely what...

Say your function f is in C^k. Let f_n denote the n-th Fourier coefficient. Then one can prove that n^k f_n\rightarrow 0 as n\rightarrow \infty. Is that enough for your problem?

I guess the only resolution would be to have the product of two distributions depending on two different variables, a double integral kind of thing. But physicist manage to suffocate every clue about the real meaning of what they're doing using their fancy integral notation anyway.

Thanks, DeadWolfe, that's exactly what I'm saying: You need to use the definition of y, the level set.