Hi Veeky Forums, 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.
William Perry
DxE = 0
Benjamin Myers
But K is orthogonal to A, giving a non-zero cross product.
Nathan Sanchez
Fine, DxE = [math] \epsilon [/math]
Ian Johnson
testing maths: [math]e^x[math]
Jose Martinez
Please use TeX for fucks sake
Gavin Fisher
Are you mocking me? Did you try to actually do the math?
James Perry
And how would I do that?
Samuel Brooks
[math] \nabla \cdot B = 0 [\math] But instead of a \ you use a /
Connor Stewart
Thanks bruh. [math]/nabla [/math]
Adam Jones
It means that's the current distribution you'd need to get that E-field.
Thomas Diaz
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]
Adam Nelson
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.
Lucas Myers
>We have a current density although we have no charge maybe it simply means that such an electrical field is impossible to obtain without charges. How the fuck would you create such an electric field anyway.
Jack Carter
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.
Jackson Richardson
>How would one create an electric field propagating through space with a constant speed?
by placing a charge anywhere
light already has constant speed, I don't get it. Can you elaborate?
Elijah Hughes
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.
Dominic Ross
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?
Cameron Scott
Woops, I meant [math]\vec{E}=\vec{A}*e^{i(\vec{k}\cdot{r}-\omega t)}[/math]
Gabriel Carter
You just explained why it wouldn't work though. The electric field of light *doesn't* have zero frequency, and you can't boost to a frame where it does. Maxwell's equations have Lorentz symmetry built into them. All this is telling you is that this is not a field configuration.
Jace Rodriguez
**not a VALID configuration.
Ayden Rivera
If you have a static electric field, the curl of E is ALWAYS ZERO. Otherwise it's not a valid E field.
Ryder Smith
[eqn]\nabla \times (\vec A ~e^{i \vec k \cdot \vec r} )= i \vec k \times (\vec A ~e^{i \vec k \cdot \vec r})[/eqn]
Are you sure this is correct?
Justin Green
I can't really follow your math, but of course you can have no charge and a current density. Indeed, a current carrying wire is modeled as having no net charge.
Lucas Thomas
A field can't be sinusoidal *and* static. Static = DC.
Charles Sanders
Basically, you're ignoring d/dt E when you can't.
Jonathan Martinez
it can. It doesn't have to vary in time, it can just be sinusoidal through space.