I'm sorry, I don't understand how the integral can go from gauge independent to gauge dependent in the process. I think your gauge transformation is not correct. You are forgetting the transformation to the electric potential.
If we go back to the form
$$\frac{dW}{dt}=-\int_V...
The vector potential energy expression must be gauge invariant, because ##\vec{E}## is gauge invariant in
$$\frac{dW}{dt}=\int_V \vec{E}\cdot \vec{J}$$
from which the expression is derived. The gauge transformation does not only transform the vector potential, but also the electric potential...
There is a 1-to-1 correspondance between the magnitudes of the current density and the vector potential. This is enough and the transformation of variables is perfectly valid. One might even make a stronger statement that, given some boundary conditions, there is a 1-to-1 correspondance between...
But the work done to produce the system is exactly the energy of the system.
Which transformation of variables are you talking about? Replacing the electric field with the derivative of the vector potential? What does this have to do with the current density?
I think more or less static situation could be created with an appropriate RC circuit. But in any case, we can try to compute the result based on the electromagnetic theory, regardless of whether it is possible to construct an infinite wire in reality.
Here is an argument for the vector...
So are you saying that we should not even try to compute the inductance (or energy) of an infinite wire?
I disagree that an infinite wire is the limiting case of a large circuit. In the infinite circuit case the vector potential at the other wire due to the other becomes infinite (like you...
Right, but my suspicion is that the surface integral cannot be neglected for the infinite wire.
I don't understand. By using the vector potential of an infinite wire given in the first post, the energy (per unit length) is clearly finite.
I agree that a loop (or two parallel wires) in the limit of infinite radius (distance) has an infinite inductance per unit length. However, I think the case of a single infinitely long wire is different.
It is true that in deriving the vector potential form for the energy, one considers the...
See, for example:
Griffiths, Introduction to Electrodynamics (1999), section 7.2.4
Vanderlinde, Classical Electrodynamic Theory (2005), section 4.1.4
Jackson, Classical Electrodynamics (1999), section 5.16
The ##B^2##-term might be defined as the magnetic energy term, but it gives the correct...
I think the vector potential version of the energy is more fundamental. It can be derived with some rather general arguments. And then from that one finds the ##B^2##-energy by neglecting that one term.
I'm not convinced that the ##B^2##-energy term can be derived easily from Poynting's...
Let's begin with the total work (against the EMF) required to create the currents of a system:
$$W=\frac{1}{2} \int \vec{A}\cdot\vec{J}d^3r$$
If you plug in the vector potential of an infinite wire (given in the first post) with a current density
$$\vec{J}=\frac{I}{\pi R^2}\hat{e}_z$$
then the...
DaleSpam, thank you for your response.
I think I should restate my assumptions. Let's assume we have a finite size circular cross-section. The center of this cross-section moves along the elliptical path so that the normal of the circular disc always points in the direction of the ellipse...
This thread is continuation to:
https://www.physicsforums.com/threads/self-inductance-of-a-straight-current-carrying-wire.461622/
It's a really old thread, but I happened to come across the same problem a while back and since the old thread does not give the correct answer, I decided to create...