# Coulomb potential twice problem

1. Feb 21, 2008

### jostpuur

If I describe a system by a Lagrange's function

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

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

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

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

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 $q_k A^0(x_k)$ 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.

2. Feb 23, 2008

### pam

There is a derivation of the E^2 form of energy density from the qA^0 form.
The two are equivalent forms, and not to be added.

3. Feb 23, 2008

### jostpuur

It would be very nice to think they are equivalent, but on the other hand all these physical theories are supposed to arise from some Lagrange's function. When I start with the L defined in the OP, this already implies what the energy H is going to be, and as you can see, the energy density E^2 and potential term qA^0 are added in this expression for energy. It would not be consistent at this point to make an ad hoc modifications to the energy.

4. Mar 1, 2008

### jostpuur

No, the potential energy is three times the Coulomb's potential! With two particles in locations x1 and x2, the A^0(x1) gives energy amount of the Coulomb's potential with the A^0-component given by particle 2, and A^0(x2) gives the Coulomb's potential for the second time with A^0-component given by particle 1. Integration over the energy density of the field gives the Coulomb's potential for the third time.

I think this is starting to make sense. Earlier, I was being confused about how the energy density of the field could result in a Coulomb's attraction, because the field configuration changes with a delay after particles' position change. As a result, also the energy density of the field changes with a delay. It would be strange if such energy could result in an interaction, because the particles experience forces due to their ability to move to the lower potentials in more local sense. The answer to this paradox seems to be, that the Coulomb's force does not rise from the energy density of the field, but from the A^0(x1) and A^0(x2) terms.

5. Jun 3, 2008

### jostpuur

It seems I was not right on this thing. The Hamiltonian I solved is not very useful, because the term

$$(\partial_0 A_{\mu}(x))(\partial^0 A^{\mu}(x)) + (\nabla A_{\mu}(x))\cdot(\nabla A^{\mu}(x))$$

doesn't have any obvious meaning.

I redid the calculation starting with the Lagrangian

$$L=-\frac{1}{4}\int d^3x\;F_{\mu\nu}(x)F^{\mu\nu}(x) \;-\; \sum_{k=1}^N \Big(q_k A^0(x_k) \;-\; q_k v_k\cdot A(x_k) \;+\; m_k\sqrt{1-|v_k|^2}\Big)$$

and the Hamiltonian turns out to be

$$H = \frac{1}{2}\int d^3x\;\Big(|E(x)|^2 \;+\; |B(x)|^2\Big) \;+\; \sum_{k=1}^N \frac{m_k}{\sqrt{1-|v_k|^2}}.$$

There is no interaction term. The terms $$A^0(x_k)$$ and $$v_k\cdot A(x_k)$$ appear in the intermediate steps, but when one simplifies the expression so that the energy density depends only on the electric and magnetic fields, and not on the potentials explicitly, the terms $$A^0(x_k)$$ and $$v_k\cdot A(x_k)$$ vanish.