Recent content by scoobmx

Does this operator (in 3D): ε_{ijk}∇_k = \begin{pmatrix} 0 & \frac{\partial}{\partial z} & -\frac{\partial}{\partial y}\\ -\frac{\partial}{\partial z} & 0 & \frac{\partial}{\partial x}\\ \frac{\partial}{\partial y} & -\frac{\partial}{\partial x} & 0 \end{pmatrix} have a formal name and a more...

http://en.wikipedia.org/wiki/Acousto-optic_modulator

Electrons are negatively charged and so move against electric field lines naturally. It's the opposite of the positive test charge.

It is.

Electrons moving from -ve to +ve are not moving against the field. They are negatively charged and so are moving with the field.

I don't think the size of the orbit is the only factor. The period also has to be in there somewhere I believe.

http://en.wikipedia.org/wiki/Quantum_decoherence

Only the normal component is continuous.. so B is parallel to surface at the boundary.

The AMOUNT of photons relates to the intensity. The amplitude gets normalized so that the probability of finding it somewhere is 1. So the amplitude itself means nothing but its phase and distribution does.

The physical principle is the acceleration is toward the gravitation body, which if you place at the origin means it accelerates everything around it straight toward itself, which is the radius direction.

Ah, my mistake, I just read the problem again to find it's a metal sphere. You're going to want to add a dipole-like term into the superposition solution. This is because a metal sphere screening a uniform external field will move its surface charge to produce dipole-like fields. The...

The vector has implied dimensionality. Zero scalar means 0 in 1 dimension. Zero vector means 0 in all components of the vector.

Can you quote this 'general solution'? I suggested superposition because the external electric field doesn't affect the charge distribution of the sphere (at least I assumed it was all bound charges) so in the electric field point of view it's just the sum of the 2 parts. You do not need to...

http://en.wikipedia.org/wiki/Magnetohydrodynamic_drive

It has to do with the density of states of the D band. The energy spacing between states is small enough that the cost of energy of moving some antiparallel electrons up to higher energy parallel states (because the lower energy states are already occupied) is met by the energy gained from...