I I don't get Poisson's equation

[MaestroBach]

So I suddenly realized that I don't understand something about Poisson's equation, and more specifically, how it relates to the laplacian, which says ∇^2Φ = 0. I've read that poisson's equation, when looking in regions of no charge, reduce to laplacian, and yeah that makes sense, but why is Poisson's equation a thing at all in the first place?

What I mean by that is, can't I rewrite poisson's equation as ∇⋅E = -ρ/ε? Which would say that the divergence of E isn't zero. Which would then imply that a potential can't exist for E because E is not conservative, right? So then, how would a potential exist?

Am I getting something mixed up in my logic? Any cleaning up you guys could do would be super appreciated.

 

[hutchphd]

You need to calm down and take a look at Maxwell's equations. Maybe after a good night's rest. Electrostatic E is conservative because the closed line integral (and therefore the curl) is zero. Find Maxwell!

 

[Delta2]

A vector field is conservative when its curl is zero everywhere. Its divergence can be anything we like and still be conservative as long as its curl remains zero.

 

[olgerm]

So I suddenly realized that I don't understand something about Poisson's equation, and more specifically, how it relates to the laplacian, which says ∇^2Φ = 0. I've read that poisson's equation, when looking in regions of no charge, reduce to laplacian, and yeah that makes sense, but why is Poisson's equation a thing at all in the first place?

What I mean by that is, can't I rewrite poisson's equation as ∇⋅E = -ρ/ε? Which would say that the divergence of E isn't zero.

If they are using coulomb gauge you can rewrite the equation ∇^2Φ=0 to ∇⋅E =0, because in coulomb gauge ∇^2Φ=-ρ/ε and ∇⋅E=-ρ/ε.

 

[vanhees71]

That's true only when you are working within electrostatics, i.e., all fields are time-independent and $\vec{j}=0$.

In the general case in Coulomb gauge you have
$$\vec{E}=-\vec{\nabla} \Phi + \frac{1}{c} \partial_t \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{A}, \quad \vec{\nabla} \cdot \vec{A}=0.$$
This solves the homogenous Maxwell equations identically. The inhomogeneous Maxwell equations now have to expressed with the potentials,
$$\vec{\nabla} \cdot \vec{E}=\rho \; \Rightarrow \; \Delta \Phi+\frac{1}{c} \partial_t \vec{\nabla} \cdot \vec{A} =\Delta \Phi=-\rho.$$
I.e., in the Coulomb gauge the $\Phi$ fulfills the Poisson equation as in electrostatics, but it's time dependent.
$$\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_{t} \vec{E}=\frac{1}{c} \vec{j}.$$
This translates into
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A})+\frac{1}{c} \partial_t \left (\frac{1}{c} \partial_t \vec{A}+\vec{\nabla} \Phi \right)$$
or using $\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla}(\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}$,
$$\Box \vec{A}=\frac{1}{c} (\vec{j}-\partial_t \vec{\nabla} \Phi).$$
The latter equation now is consistent with the Coulomb-gauge condition due to charge conservation (i.e., $\partial_t \rho+\vec{\nabla} \cdot \vec{j}=0$):
$$\vec{\nabla} \cdot (\vec{j}-\partial_t \vec{\nabla} \Phi) = -\partial_t \rho -\partial_t \Delta \rho=0.$$
One should make oneself clear that there's no action at a distance for the physical fields implied though $\Phi$ is implying such a thing, because the solution of Poisson's equation here reads as in electrostatics
$$\Phi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\rho(t,\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
Now $\vec{A}$ obeys an inhomogeneous wave equation, for which you have to use the retarded solution though. Now the source on the right-hand side includes the "nonlocal term" containing $\Phi$ thought. So both $\Phi$ and $\vec{A}$ include nonlocal terms.

It's a good excercise to check that what you get for the fields $\vec{E}$ and $\vec{B}$ which are observable ($\Phi$ and $\vec{A}$ as gauge-dependent quantities are never (!) directly observable), are the retarded fields as given by the Jefimenko equations, as it must be.

 

[olgerm]

@vanhees71 , was it answer to my post? It seems to me that in coulomb gauge
$\frac{\partial^2 \Phi}{\partial x^2}+\frac{\partial^2 \Phi}{\partial y^2}+\frac{\partial^2 \Phi}{\partial z^2}=-\frac{\rho}{\epsilon_0}$ and
$\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}=-\frac{\rho}{\epsilon_0}$ are valid for any EM field and any charges.

therefore in OP case where $\frac{\partial^2 \Phi}{\partial x^2}+\frac{\partial^2 \Phi}{\partial y^2}+\frac{\partial^2 \Phi}{\partial z^2}=0$
also $\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}=0$

 

[HomogenousCow]

Maxwell's equations in terms of 3 vectors are gauge invariant; they hold in arbitrary gauge.

 

[olgerm]

Maxwell's equations in terms of 3 vectors are gauge invariant; they hold in arbitrary gauge.

$\frac{\partial^2 \Phi}{\partial x^2}+\frac{\partial^2 \Phi}{\partial y^2}+\frac{\partial^2 \Phi}{\partial z^2}=-\frac{\rho}{\epsilon_0}$
is not gauge gauge invariant.

 

[olgerm]

@vanhees71 , maybe you meant to answer me in this thread to a post, that I deleted.

 

[vanhees71]

https://www.physicsforums.com/threa...mply-action-at-a-distance.977652/post-6241497