Recent content by lys04

Why not Would this be a valid example? Suppose instead of the X basis was ##|+ \rangle = \frac{\sqrt{2}}{\sqrt{10}} |0 \rangle + \frac{\sqrt{8}}{\sqrt{10}} |1 \rangle ## And Alice sends out a 0 encoded in the X basis i.e a ##| + \rangle## and the eve tries to intercept by making a measurement...

In the BB84 protocol, we have two basis, the Z basis which consists of |0> and |1> which represents bit values of 0 and 1 respectively in the Z basis. From this we construct another basis, the X basis which consists of ##|+> = \frac{1}{\sqrt(2)} |0> + \frac{1}{\sqrt(2)} |1>## and ##|-> =...

Yeah and also only the perpendicular component contribute to the magnetic dipole moment which kinda surprises me since the parallel component is the bit that's moving? And also this annoys me but what do I do with the dl that's perpendicular to the direction of v? it seems like that's being...

So then ##dV'=dA dl' = \frac{dA dl}{\gamma} = \frac{dV}{\gamma}##? This is dV' when the component of dl is parallel to the direction of v. How do I deal with the component that's perpendicular? It doesn't change right? (But is the perpendicular component really needed anyways?) So then if I plug...

Sorry for the late reply. Hm so by length contraction for the component of dl parallel to the direction of v it gets contracted? dl’=dl/gamma?

Yeah I thought about that, I’m not sure how to do it tho since the direction is not specified but instead in some arbitrary direction.

A charge distribution stationary in its own frame S’ has a static charge density ##\rho ’##, a total charge of 0 and a net electric dipole ##\vec{p'}##. An observer in frame S sees the charge distribution moving with constant velocity ##\vec{v}=c \vec{\beta}## What is the magnetic dipole moment...

From my understanding, The electric field, ##\vec{E}## in both of these equations are referring to the same electric field, the electric field produced by free charges in vacuum? ##\vec{D}## refers to the electric displacement field density INSIDE the dielectric and $$\epsilon_0 \vec{E}$$ is...

If the charge distribution has non zero total charge then the dipole moment depends on where you measure it. But it doesn’t matter if the charge distribution has 0 net charge

Ah yeah I can see how that's possible. But given a non-zero divergence, would the dipole moment be given by ##\vec{p}(\vec{x_0}) =\int_{surface} (\vec{x’}-\vec{x_0}) \sigma (\vec{x’}) dA’ + \int_{volume} (\vec{x’}-\vec{x_0}) \rho (\vec{x’}) dV’ ##? No, this is not a homework problem.

^^

Yeah, I'm not sure how to do that since in the case where the width was infinite by symmetry the fields below and above would only have horizontal components and the vertical ones would all cancel out.

Here's what I'm thinking: Since the width is L and the current is flowing in positive z direction, there is a surface current density of $$\vec{K}=\frac{I}{L} \vec{z}$$ Find the vector potential due to one infinitely long wire in the z direction Add a lot of them together to form a finite width...

Oh and by the Uniqueness theorem, the placement of image charges are unique too right?