- #1

- 6,002

- 2,627

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 ?

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 ?

Last edited: