711 reputation
114
bio website none
location None
age 23
visits member for 1 year, 9 months
seen Aug 27 at 21:36

NTS


Apr
4
asked Why does the index of refraction change the direction of light
Mar
29
comment cgs Gauss' system of units
@ChrisWhite Neither did I, my teacher uses it for dipole and he writes it like that in his notes (they're computer written so it's not bad calligraphy). All optics book I've seen use just $p$.
Mar
29
comment cgs Gauss' system of units
@LubošMotl Thanks, that's both relieving and it sucks. I will try to get used to cgs Gauss' system as I am seeing it is actually used.
Mar
29
comment cgs Gauss' system of units
@MarkEichenlaub Ok, generally, I understand where both systems come from, but I still have not clear how to change between systems. The only thing I could do is to go to the beginning and start deriving in both systems to see the differences.
Mar
29
revised cgs Gauss' system of units
added 345 characters in body
Mar
29
asked cgs Gauss' system of units
Mar
25
comment Why is $\vec j\cdot \vec e$ the joule dissipation?
And so what is its physical interpretation microscopically. What's that product at those scales, how must I see that?
Mar
25
answered So gravity turns things round
Mar
25
asked Why is $\vec j\cdot \vec e$ the joule dissipation?
Mar
21
accepted Fast question about Lagrangian
Mar
19
comment Monochromatic wave
Ok, thank you for the answer.
Mar
19
accepted Monochromatic wave
Mar
18
asked Monochromatic wave
Mar
18
accepted Circular polarisation
Mar
18
comment Circular polarisation
Ok, thanks for your help.
Mar
17
comment Circular polarisation
Thank you. And, in this particular case, the conclusion of that is that for $E$ and $B$ to be a solutions of Maxwell's equations, that $k$ must have only $z$ component, because $E_0$ and $B_0$ are in the $x-y$ plane, and they're perpendicular. The fact that $\vec{B} = \vec{n}\times\vec{E}$ means that if $B$ makes circles, $E$ must make the too, right? So if they're asking me if those can be waves in the empty space, I should say that only if $k$ has only $z$ component, because otherwise they wouldn't be a solution to Maxwell's equation.?
Mar
17
asked Circular polarisation
Mar
17
answered Why molecular forces do not obey inverse square law?
Mar
5
accepted Dealing with experimental data
Mar
5
comment Dealing with experimental data
Ok, that's good enough, I will accept this answer and follow your advice (it's totally not worth it). I will simplify things.