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## Main Question or Discussion Point

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):

[tex]\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}[/tex]

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

[tex]\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}[/tex]

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