Recent content by Bob44

The Helmholtz theorem states that any vector field can be uniquely decomposed into longitudinal (curl free) and transverse (divergence free) parts $$\mathbf{E}=\mathbf{E}_\parallel+\mathbf{E}_\perp.\tag{1}$$ We take the Coulomb gauge $$\nabla \cdot \mathbf{A}_C=0\tag{2}$$ so that $$...

But the tangential magnetic field produced by the spark due to Stokes law is still instantaneous at all distances. This seems to contradict causality regardless of the radiation that later travels from the spark at the speed of light.

Let us take the Ampere-Maxwell law $$\nabla \times \mathbf{B} = \mu_0\,\mathbf{J}+\frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t}.\tag{1}$$ Assume we produce a spark that is so fast that the ##\partial \mathbf{E}/\partial t## term in eqn.##(1)## has not yet been produced by Faraday’s law...

Thanks for the reference! https://arxiv.org/abs/physics/0204034

But how could the retarded transverse field cancel the instantaneous longitudinal field as it takes a finite time for the transverse field to get to the observation point whereas the longitudinal field is there already?

Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by...

I set up an electrodynamics experiment with an axial symmetry which causes the vector potential to vanish. The Lorenz gauge $$\nabla \cdot \mathbf{A} = - \frac{1}{c^2} \frac{\partial \phi}{\partial t}$$ is usually employed in electrodynamic situations so that the scalar and vector potentials...

Imagine that two charged particles, with charge ##+q##, start at the origin and then move apart symmetrically on the ##+y## and ##-y## axes due to their electrostatic repulsion. The ##y##-component of the retarded Liénard-Wiechert vector potential at a point along the ##x##-axis due to the two...

I guess I should use ##\epsilon##, ##\mu## and ##c## appropriate to the material in the coaxial cable.

Is that also true if I derive Poisson’s equation from Gauss’s law and solve that? $$\nabla \cdot (-\nabla \phi-\frac{\partial \mathbf{A}}{\partial t})=\frac{\rho}{\epsilon_0}$$ If ##\partial_t \mathbf{A}=0## due to constant current then we just have Poisson’s equation which has instantaneous...

Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following...

This argument is another version of my previous post. Imagine that we have two long vertical wires connected either side of a charged sphere. We connect the two wires to the charged sphere simultaneously so that it is discharged by equal and opposite currents. Using the Lorenz gauge...

Interesting. I guess as ##q(t)## is not arbitrary your example does not break causality as we cannot use it to signal faster than light.

I’ve just used a spherical Gaussian surface with the same center as the charged sphere but with a larger radius so that it encloses the charged sphere. By Gauss’ law the total electric flux through the Gaussian surface is equal to the total charge inside divided by ##\epsilon_0##. By symmetry...

Hi, I'm a Physics graduate who is still interested in the big questions of physics! Bob