Recent content by krome

http://arxiv.org/abs/1308.2676 p.33

That's really cool. I know nothing about coding so I can't help you. I'm just wondering if you've compared your result with the known exact result via the image charge method. It would be cool to see the quality of this lattice method.

As far as I understand, such a charge distribution (a) always exists; and (b) is unique. This is a boundary value problem for Poisson's equation and existence and uniqueness of solutions (given completely specified boundary conditions) is a theorem.

I believe what you have here is tensor notation but with the tensor sign left out to reduce clutter. By the tensor A \otimes B of two matrices A,B, we mean at every entry of A, insert a copy of B multiplied by that entry in A. For example, i'll work out the first term in that Hamiltonian. I'll...

\langle N \rangle is the Fermi-Dirac distribution, which is derived on that wikipedia page. So, you can perform the derivative yourself and verify the second equality. The first equality can be derived as follows. First, \displaystyle \langle ( \Delta N )^2 \rangle = \langle (N - \langle N...

I'm assuming you know what to do without the absorbing wall (i.e. how to derive the appropriate diffusion equation and show that the solution is Gaussian of some particular width in the continuum limit etc. etc. etc.) The diffusion equation is linear and has a unique solution given complete and...

You are correct that the angle between Q and P is not the same as the angle between Q' and P. However, the length of Q is also not the same as the length of Q'. The cross product depends on both of these things. It might not be immediately obvious why these two changes exactly compensate for one...

"Linear" in "Linear Algebra" means "closed under addition". In physics a "linear system" is one that satisfies the superposition principle, which is just the physics way of saying closed under addition. This means that if S_1 and S_2 are two possible states of the system (i.e. two possible...

The 100 in sin(100t) has units of radians per second. When you multiply 100\ \text{rad/s} with t seconds, you get an angle in radians. Then you can take the sine of that. The sine of something that has units of seconds is meaningless. This is the reason why you should NEVER EVER EVER solve...

Your solution and the solution in your textbook are the same.

Good lord! I must be going mad or selectively blind. I swear when I read this last night the second term did not have a factor of 2! :confused:

I may be misunderstanding something, but I think the second term (with the double sum in i and j) should be multiplied by 2. Either that or the sum in j should be over j \neq i rather than j<i. Anyway, you are correct to say that the blue term needs to be expanded further. Just try writing...

L is a linear map, which means L(aA+bB) = aL(A) + bL(B).

Let X_n=\mathbb{R} and X= \prod_{n \in \mathbb{N}} X_n = \mathbb{R}^{\mathbb{N}}. Suppose U_{n \in \mathbb{N}} is a countable neighborhood basis for the point \vec{0} = (0,0, \ldots ) \in X. See if you can think of an open neighborhood of \vec{0} that does not contain anyone of the U_n's...

The solution says IF aA+bB=0, THEN a=b=0. That is what it means for the vectors A and B to be linearly independent. Vectors in a basis must be linearly independent.