Bob44
- 4
- 0
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 ##\nabla\cdot\mathbf{A}+(1/c^2)\partial \phi/\partial t=0##, Maxwell's equations have the following retarded wave solutions in the scalar and vector potentials.
Starting from Gauss's law
$$\nabla \cdot \mathbf{E}=\frac{\rho}{\epsilon_0}.\tag{1}$$
We substitute in the potentials to give
$$\nabla \cdot (-\nabla \phi - \frac{\partial \mathbf{A}}{\partial t})=\frac{\rho}{\epsilon_0}.\tag{2}$$
Now for every contribution to ##\frac{\partial \mathbf{A}}{\partial t}## from an element of current density ##\mathbf{j}## at position ##(3)## on the upper wire there is an equal and opposite contribution from an element of current density ##\mathbf{j}## at position ##(4)## on the lower wire.
Therefore the scalar potential ##\phi## at point ##(1)## is simply due to the charge density ##\rho## on the sphere at position ##(2)##:
$$\nabla^2 \phi=-\frac{\rho}{\epsilon_0}.\tag{3}$$
Poisson's equation ##(3)## has an instantaneous general solution given by:
$$\phi(1,t)=\int \frac{\rho(2,t) dV_2}{4\pi\epsilon_0 r_{12}}.\tag{4}$$
This is different from the expected retarded solution for the scalar potential wave equation:
$$\phi(1,t)=\int \frac{\rho(2,t-r_{12}/c) dV_2}{4\pi\epsilon_0 r_{12}}.\tag{5}$$
What is the reason for this discrepancy?
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 ##\nabla\cdot\mathbf{A}+(1/c^2)\partial \phi/\partial t=0##, Maxwell's equations have the following retarded wave solutions in the scalar and vector potentials.
Starting from Gauss's law
$$\nabla \cdot \mathbf{E}=\frac{\rho}{\epsilon_0}.\tag{1}$$
We substitute in the potentials to give
$$\nabla \cdot (-\nabla \phi - \frac{\partial \mathbf{A}}{\partial t})=\frac{\rho}{\epsilon_0}.\tag{2}$$
Now for every contribution to ##\frac{\partial \mathbf{A}}{\partial t}## from an element of current density ##\mathbf{j}## at position ##(3)## on the upper wire there is an equal and opposite contribution from an element of current density ##\mathbf{j}## at position ##(4)## on the lower wire.
Therefore the scalar potential ##\phi## at point ##(1)## is simply due to the charge density ##\rho## on the sphere at position ##(2)##:
$$\nabla^2 \phi=-\frac{\rho}{\epsilon_0}.\tag{3}$$
Poisson's equation ##(3)## has an instantaneous general solution given by:
$$\phi(1,t)=\int \frac{\rho(2,t) dV_2}{4\pi\epsilon_0 r_{12}}.\tag{4}$$
This is different from the expected retarded solution for the scalar potential wave equation:
$$\phi(1,t)=\int \frac{\rho(2,t-r_{12}/c) dV_2}{4\pi\epsilon_0 r_{12}}.\tag{5}$$
What is the reason for this discrepancy?