Hi Veeky Forums...

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.

DxE = 0

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

Fine, DxE = [math] \epsilon [/math]

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

Please use TeX for fucks sake

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

And how would I do that?

[math] \nabla \cdot B = 0 [\math]
But instead of a \ you use a /

Thanks bruh.
[math]/nabla [/math]

It means that's the current distribution you'd need to get that E-field.

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]

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.

>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.

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.

>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?

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.

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?

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

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.

**not a VALID configuration.

If you have a static electric field, the curl of E is ALWAYS ZERO.
Otherwise it's not a valid E field.

[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?

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.

A field can't be sinusoidal *and* static. Static = DC.

Basically, you're ignoring d/dt E when you can't.

it can. It doesn't have to vary in time, it can just be sinusoidal through space.