Vector potential and energy calculations in magnetostatics

[Mbert]

I have some trouble with the calculation of energy in magnetostatics, using the vector potential A. From the classic formula that uses B*H, I find the expression (in magnetostatics) in terms of A and J (current density):

$$\begin{align}W &=\frac{1}{2}\int_V{\vec{B}\cdot\vec{H}{\rm d}V}\\ &=\frac{1}{2}\int_V{\left(\vec{\nabla}\times\vec{A}\right)\cdot\vec{H}{\rm d}V}\\ &=\frac{1}{2}\int_V{\left(\vec{\nabla}\times\left(\vec{A}+\vec{\nabla}\psi\right)\right)\cdot\vec{H}{\rm d}V}\\ &=\frac{1}{2}\int_V{\left(\vec{A}+\vec{\nabla}\psi\right)\cdot\left(\vec{\nabla}\times\vec{H}\right){\rm d}V}\\ &=\frac{1}{2}\int_V{\left(\vec{A}+\vec{\nabla}\psi\right)\cdot\vec{J}{\rm d}V}\end{align}$$

Since the vector potential A is defined up to a gradient of some scalar field (the divergence of B is still 0), from the above equation, we can see that the energy will be depending on the gradient. However, there is only one energy that can be calculated for a given B and H. How can I find the value of the gradient to match the "real" energy?

thanks

 

[vanhees71]

Of course the total energy of the electromagnetic field must be gauge invariant, and indeed it is.

In magnetostatics, all fields are time-independent. From the continuity equation,

$$\partial_t \rho+\vec{\nabla} \cdot \vec{J}=\vec{\nabla} \cdot \vec{j}=0,$$

one concludes that the stationary current density must be source free and thus, using Green's integral theorem and the assumption that $\vec{j}$ vanishes at infinity, you get

$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; (\vec{\nabla} \psi) \cdot \vec{J} = -\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; \psi \vec{\nabla} \cdot \vec{J}=0.$$

 

[Mbert]

If I define in 2D a rectangular conductor domain in air (infinitely long in Z), where J is oriented outside the plane in Z, it is only necessary to evaluate the scalar product A*J inside the conductor domain, since elsewhere J=0. If the vector potential is also outside the plane in Z, but I choose the additive gradient to be along -Z (and also very large in module so that the resulting vector potential becomes negative), the energy becomes negative. How can this be possible if the total energy is gauge invariant?

 

[Bill_K]

Because in that situation J does not vanish at infinity, and you have a large contribution to the surface integral over the end caps at Z = ± ∞

 

[Mbert]

But we also have div J = 0 in this case, since we have only a Jz component that does not depend on Z. So the equation of the previous poster should still be true:
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; (\vec{\nabla} \psi) \cdot \vec{J} = -\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \; \psi \vec{\nabla} \cdot \vec{J}=0$$