/sci/ - Science & Math

its time for those 30k after graduation electrical engineers to show their worth

Phyisicsfag here. Can any of you recommend me some textbooks that explain various applications of electromagnetic waves.

>>8034917
Woops, I meant
[math]\vec{E}=\vec{A}*e^{i(\vec{k}\cdot{r}-\omega t)}[/math]

>>8034908
Sure, the electric field of light in a vacuum is given by:
[math]\vec{E}=\vec{A}*e^{i\(vec{k}\cdot{r}-\omega t)}[/math]
with [math]\vec{k}[/math] orthogonal to[math]\vec{A}[/math]. If, however the relative speed between you and the light wave was zero, then [math]\omega=k*c=0[/math] and we end with the problem stated, but a static, sinusoidal electric field hardly strikes me as improbable thought experiment.
What would this current density indicate though?

>>8034893
But interestingly it holds up the continuity equation:
[math]\nabla\cdot\vec{J}=i\frac{t}{\mu_0}(\vec{k}\cdot\vec{k}\times(\vec{k}\times\vec{A}))*e^{i\vec{k}\cdot\vec{r}}=0[/math]
so it's pretty weird.

>>8034893
How would one create an electric field propagating through space with a constant speed? If it had a constant speed, it would be a light wave. I stumbled upon this while considering the implications of the speed of light being equal to 0 in all systems, rather than what it is now.

>>8034883
So basically what I said earlier.

>>8034864
But we have no charge:
[math]\nabla\cdot\vec{E}=i\vec{k}\cdot\vec{E}=0[/math]as the k-vector is orthogonal to the field direction (k orthogonal to A).
So we have no charge, but we have a current? Explain if you will/can.

Right, so noting that
[math]\vec{k} is orthogonal to \vec{A}[/math] and [math]\frac{\partial e^{i\vec{k}\cdot\vec{r}}}{\partial x_j}=ik_j e^{i\vec{k}\cdot\vec{r}}[/math]
we get
[math] \nabla \times \vec{E} = i\vec{k}\times\vec{E}=-\frac{\partial \vec{B}}{\partial t}[/math]
[math]\vec{B}=-i\vec{k}\times\vec{E}*t[/math]
[math]\nabla\times\vec{B}=\mu_0\vec{J}[/math] since the time derivative of the electric field is zero. Now, we get:
[math] \vec{J}=\frac{t}{\mu_0}(\vec{k}\times(\vec{k}\times\vec{E})[/math]
[math]=\frac{t}{\mu_0}(\vec{k}\times(\vec{k}\times\vec{A}))*e^{i\vec{k}\cdot\vec{r}}[/math]

>>8034768
int(f(t)*dr)=int(f(t)*dr/dt*dt)
It's just the chain rule.

>>8034730
Wow, yeah, excuse me for assuming that the area of a rectangle is equal to the product of its two orthogonal sides.
No it doesn't require stronger considerations of the function. Again, we start with an approximation, then, since an infinite area can have any fucking shape we choose, we choose the simpler one.

>>8034764
And how would I do that?

>>8034752
Are you mocking me?
Did you try to actually do the math?

testing maths: [math]e^x[math]

>>8034560
But K is orthogonal to A, giving a non-zero cross product.

>>8034629
the area under a specific piece of curve within x and dx can if f(x+dx)-f(x) is approximately 0 be given by f(x)dx.
Hence, to get the area under the curve we take the sum of all these minor contributions:
sum(f(x)*dx) from x=a to x=b.
Letting dx-> 0 gives the exact area and WOOPS, that's the definition of an integral.

>>8031436
Thermodynamics motherfucker do you speak it?
It's literally statistical mechanics.

Does LateX code work on this board?
$\nabla \mu_0^2 $

Does LateX code work on this board?
$\nabla $

Hi /sci/, I was wondering what would happen if you plugged in a static sinusoidal electric field into Maxwell's equations, and I got something strange.
If we let A be a vector carrying magnitude and direction of the E-field, and k being the wave vector for the field and let K be orthogonal to A, we get:
E=Aexp(i*k*r), we can substitute the spatial derivation (usually nabla, but I'll just call it D because nabla is hard to make)for ik (still a vector), and get: D*E=0
DxE=-dB/dt=i(kxA)*exp(ik*r)
B=-i(kxA)exp(ikr)*t
DxB=µJ+(c^-2)*dE/dt=µJ=[kx(kxA)]*t*exp(ikr)

J=t/µ*[kx(kxA)]exp(ikr)

We have a current density although we have no charge. The continuity equation is still satisfied: D*J=-d(density)/dt=k*[kx(kxA)]*t/µ*exp(ikr)=0

So we have no charge but we have a current density. What does that mean?
Pic sorta related, mfw.

>>5053429
I already answered
>>5053401

>>5053414
that sounds fucking awesome. I wish I had something like that too ;__;

>>5053404
doesn't look like highschool physics to me, unless american highschool do harder physics than ours (italy), I'm sure first time I saw a request for a derivative in a problem was at uni

still working on it, but at least I think I got it.

Write the hamiltonian for an interacting electron

H= 1/2m (p- e/c A)^2 + eV

you can get A and V from maxwell's equations from E and B

from here, you can use Hamilton's equations dH/dq = -dp/dt, and p(t) integrating that.
Then change into polar coordinates, and get dH/d(angle) = - dL/dt (being L the momentum associated with the angular variable).

The force at t=0 should be simply eE(0) + v(0)xB(0)

I'll get back to you after I've done the calculations