Do Maxwell's equations in integral form imply action at a distance?

[Delta2]

Lets take for example Gauss's law in integral form. Suppose at time $t$ we have charge $q(t)$ (at the center of the gaussian sphere) enclosed by a gaussian sphere that has radius $R>>c\Delta t$. At time $t+\Delta t$ the charge is $q(t+\Delta t)$ and if we apply gauss's law in integral form we find that the flux through that gaussian sphere of radius $R>>c\Delta t$ is $\oint E(t+\Delta t) \cdot dA=\frac{q(t+\Delta t)}{\epsilon_0}$ that is the flux and hence the electric field at a distance much greater than $c\Delta t$ have been "informed" of the change of the charge from $q(t)$ to $q(t+\Delta t)$.

I think similar examples can be given with Maxwell-Faraday's law and Maxwell-Ampere's law all in integral form.

I know that as long as the fields are continuously differentiable, the integral form of the equations is equivalent to the differential form, from which we can conclude the wave equation for the fields, hence no instantaneous action. So there must be a catch here. What's the catch ?

 

[Delta2]

The only catch I can think is that in order for the charge(located at the center) enclosed by this huge gaussian sphere to change by $\Delta q=q(t+\Delta t)-q(t)$ we must move (since charge cannot be created or destroyed - conservation of charge principle)charge $\Delta q$ from the outside of the gaussian sphere to its center ( or from the center to the outside) and this must be done with speed $v \geq \frac{R}{\Delta t}\geq c$ and this simply cannot happen if we accept the fact that we can't move charge faster than lightspeed.

 

[Dale]

Summary: Integral form of Maxwell's equations seem to imply instantaneous action at a distance.

What's the catch ?

Charge is locally conserved.

Think about that a bit and see if you can explain why local charge conservation resolves this concern.

Edit: never mind, I see you already did that!

 

[vanhees71]

Since the integral form (be careful, they are usually stated under often not explicitly mentioned assumptions on the volumes, surfaces, and curves along which the integrals are taken; often it's tacitly assumed they are at rest in the considered reference frame!) is equivalent to the differential form, they are just expressions of the same theory, and there's no instantaneous action at a distance whatsoever since classical electrodynamics is a relativistic field theory.

 

[olgerm]

Integral form of Maxwell's equations seem to imply instantaneous action at a distance.
Lets take for example Gauss's law in integral form. Suppose at time $t$ we have charge $q(t)$ (at the center of the gaussian sphere) enclosed by a gaussian sphere that has radius $R>>c\Delta t$. At time $t+\Delta t$ the charge is $q(t+\Delta t)$ and if we apply gauss's law in integral form we find that the flux through that gaussian sphere of radius $R>>c\Delta t$ is $\oint E(t+\Delta t) \cdot dA=\frac{q(t+\Delta t)}{\epsilon_0}$ that is the flux and hence the electric field at a distance much greater than $c\Delta t$ have been "informed" of the change of the charge from $q(t)$ to $q(t+\Delta t)$.So there must be a catch here. What's the catch ?

If charge is the not in centre of the sphere, the flux throught every point of the sphere might not homogeneous so the $\vec{E}$ field throught the sphere is not same and change of does not mean that $\vec{E}$ must change on every point on the sphere.
If charge is in centre of the sphere, then it takes more time than $\frac{R}{c}$ to move it out of sphere, BUT
you can take a new sphere so that charge is in centre of new sphere, with radius $R_2$ in time $t+_\Delta t$ and in time t charge is in centre of sphere1. Point $\vec{X_{reciver}}$ is on both sphere. By calculating $\vec{E}$ field with assumation, that $\vec{E}$vector projection on the surface of sphere is homogeneous, you get that $|\vec{E}|(t)=\frac{q}{4*\pi*R_1^2*\epsilon_0}$ and $|\vec{E}|(t)=\frac{q}{4*\pi*R_2^2*\epsilon_0}$. It seems like it was possible to send signals faster than light this way, but it is not true because assumation, that $\vec{E}$vector projection on the surface of sphere is homogeneous is not true, because moving the charge creates $\vec{j}$ that creates preferred direction.

 

[Delta2]

Come to think of it, with reverse logic we can conclude that Maxwell's equations imply that we can't move electric charge faster than the speed of the EM-waves which is $\frac{1}{\sqrt{\epsilon_0\mu_0}}=c$.
Because if we were able to move charge faster than the speed of the EM-waves, then this example shows that Maxwell's equations are not consistent, that is from gauss's law we can infer transfer of signal with bigger than the speed of $\frac{1}{\sqrt{\epsilon_0\mu_0}}$, while all Maxwell's equations together give us the wave equation with speed constant $\frac{1}{\sqrt{\epsilon_0\mu_0}}$