- #1

jostpuur

- 2,116

- 19

[tex]

L=-\frac{1}{2}\int d^3x\;(\partial_{\mu}A_{\nu}(x))(\partial^{\mu} A^{\nu}(x)) - \sum_{k=1}^N \Big(q_k A^0(x_k) - q_k v_k\cdot A(x_k) + m_k\sqrt{1-|v_k|^2}\Big)

[/tex]

(I'm just coping this from my notes. I'm sure the not gauge fixing Lagrangian with [itex]F^{\mu\nu}[/itex] could be used as well.)

then the energy of the system (Hamilton's function, but now as function of [itex]\partial_0 A^{\mu}[/itex] and [itex]v_k[/itex]) turns out to be

[tex]

H=-\frac{1}{2}\int d^3x\;\Big((\partial_0 A_{\mu}(x))(\partial^0 A^{\mu}(x)) + (\nabla A_{\mu}(x))\cdot(\nabla A^{\mu}(x))\Big) + \sum_{k=1}^N\Big(\frac{m_k}{\sqrt{1-|v_k|^2}} + q_k A^0(x_k)\Big)

[/tex]

Now I see a problem with the Coulomb's potential. Suppose we fix some configurations of two point charges, and ask what are the energies in these configurations. The energy, corresponding to the Coulomb's potential, comes from two different locations. It comes from the term [itex]q_k A^0(x_k)[/itex] in the Hamiltonian, and it comes also from the energy density of the field it self as explained in the posts #26 and #37 of this thread What exactly is spin? Doesn't this mean that the potential energy of the system is in fact twice the amount given by the Coulomb's potential?

Of course from the practical point of view this doesn't matter, because we cannot experimentally measure any correct scale of energy and charges, but that still seems disturbing from theoretical point of view.