Subsections


D.37 Forces by particle exchange derivations


D.37.1 Classical energy minimization

The energy minimization including a selecton is essentially the same as the one for only spoton and foton field. That one has been discussed in chapter A.22.1 and in detail in {A.2}. So only the key differences will be listed here.

The energy to minimize is now

\begin{displaymath}
\frac{\epsilon_1}{2}\int \left(\nabla\varphi\right)^2 {\,\...
...ec r}-{\skew0\vec r}_{\rm {e}})\Big){\,\rm d}^3{\skew0\vec r}
\end{displaymath}

So the only real difference in the variational analysis is

\begin{displaymath}
s_{\rm {p}}\delta^3_\varepsilon({\skew0\vec r}-{\skew0\vec...
...\delta^3_\varepsilon({\skew0\vec r}-{\skew0\vec r}_{\rm {p}})
\end{displaymath}

That means that the Poisson equation now becomes

\begin{displaymath}
- \nabla^2\varphi({\skew0\vec r}) =
\frac{s_{\rm {p}}}{\...
...\delta^3_\varepsilon({\skew0\vec r}-{\skew0\vec r}_{\rm {e}})
\end{displaymath}

Since the Poisson equation is linear, the solution is $\varphi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\varphi^{\rm {p}}+\varphi^{\rm {e}}$. Here $\varphi^{\rm {p}}$ is the foton field (A.107) produced by the spoton as before, and $\varphi^{\rm {e}}$ is a similar expression, but using the selecton sarge and distance from the selecton:

\begin{displaymath}
\varphi^{\rm {p}} = \frac{s_{\rm {p}}}{4 \pi \epsilon_1 \v...
...epsilon_1 \vert{\skew0\vec r}- {\skew0\vec r}_{\rm {e}}\vert}
\end{displaymath}

The energy lowering is now

\begin{displaymath}
-{\textstyle\frac{1}{2}} \int \Big(\varphi^{\rm {p}}({\ske...
...c r}-{\skew0\vec r}_{\rm {e}})\Big) {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

Multiplying out, you get, of course, the energy lowerings for the spoton and selecton in isolation. But you also get two additional interaction terms between these sarges. These two terms are equal; the selecton field $\varphi^{\rm {e}}$ evaluated at the position of the spoton times spoton sarge is the same as the spoton field $\varphi^{\rm {p}}$ at the selecton times selecton sarge. So it is seen that each term contributes half to the Koulomb energy as claimed in the text.

The foton field energy is still half of the particle-field interaction energies and of opposite sign. That is why the energy change is half of what you would expect from the interaction of the particles with each other’s field: the other half is offset by changes in field energy.


D.37.2 Quantum energy minimization

This derivation includes the selecton in the spoton-fotons system analyzed in {A.22.3}. Since the analysis is essentially unchanged, only the key differences will be highlighted.

If an selecton is added to the system, the system wave function becomes

\begin{displaymath}
\psi_{\varphi\rm {pe}} = C_0 \psi_{\rm {p}} \psi_{\rm {e}}...
...
\quad\quad \vert C_0\vert^2 + \vert C_1\vert^2 + \ldots = 1
\end{displaymath}

The demon can hold the selecton in its other hand. The Hamiltonian will now of course include a term for the selecton in isolation, as well as an interaction with the foton field. These are completely analogous to the corresponding spoton terms.

So the energy to be minimized for the ground state becomes

\begin{displaymath}
E = E_{\rm {p}} + E_{\rm {e}} +
\vert C_1\vert^2 \hbar \...
... {e}}\big\rangle
\Big\vert\vert C_1\vert\cos(\alpha+\beta)
\end{displaymath}

If this is minimized as in {A.22.3}, the energy is

\begin{displaymath}
E = E_{\rm {p}} + E_{\rm {e}} - \frac{1}{2\epsilon_1{\cal ...
...0\vec r}_{\rm {e}}}\big\vert\psi_{\rm {e}}\big\rangle \vert^2
\end{displaymath}

The square absolute value of a quantity can be found as the product of that quantity times its complex conjugate. That gives the same energy lowering as for the lone spoton, and a similar term for a lone selecton. However, there is an additional term

\begin{displaymath}
- \frac{s_{\rm {p}}s_{\rm {e}}}{2\epsilon_1{\cal V}k^2}
...
...ec r}_{\rm {p}}}\big\vert\psi_{\rm {p}}\big\rangle
\right)
\end{displaymath}

If you write out the inner product integrals over the selecton coordinates explicitly, this becomes

\begin{displaymath}
- \int s_{\rm {e}} \psi_{\rm {e}}^*({\skew0\vec r}_{\rm {e...
...{\skew0\vec r}_{\rm {e}}) {\,\rm d}^3{\skew0\vec r}_{\rm {e}}
\end{displaymath}

Summed over all ${\vec k}$, the second term inside the square brackets gives the same answer as the first; that is because opposite ${\vec k}$ values appear equally in the summation. Looking at the first term, the summation over ${\vec k}$ produces again the spoton potential $\varphi^{\rm {p}}_{\rm {cl}}$, but now evaluated at the position of the selecton. That then shows the additional energy lowering to be

\begin{displaymath}
- \int s_{\rm {e}} \psi_{\rm {e}}^*({\skew0\vec r}_{\rm {e...
...{\rm {e}}({\skew0\vec r}_{\rm {e}}) {\,\rm d}^3{\skew0\vec r}
\end{displaymath}

Except for the differences in notation, that is the same selecton-spoton interaction energy as found in {A.22.1}.


D.37.3 Rewriting the Lagrangian

The rules of engagement are as follows:

Consider first the square magnetic field:

\begin{displaymath}
{\cal B}^2 = \sum_i (A_{{\overline{\overline{\imath}}},{\o...
...} - A_{{\overline{\imath}},{\overline{\overline{\imath}}}})^2
\end{displaymath}

Expanding out the square, that is equivalent to

\begin{displaymath}
{\cal B}^2 = \sum_i ( A_{{\overline{\overline{\imath}}},{\...
...th}}} A_{{\overline{\imath}},{\overline{\overline{\imath}}}})
\end{displaymath}

The summation indices can now be cyclically redefined to give an equivalent sum over $i$ equal to

\begin{displaymath}
{\cal B}^2 = \sum_i (A_{i,{\overline{\overline{\imath}}}}^...
...ne{\overline{\imath}}}} A_{{\overline{\overline{\imath}}},i})
\end{displaymath}

The terms can be combined in sets of three as

\begin{displaymath}
{\cal B}^2 = A_{i,j}^2 - A_{i,j}A_{j,i}
\end{displaymath}

Here summation over $i$ and $j$ is now understood.

The square electric field is

\begin{displaymath}
{\cal E}^2=(-\varphi_i-A_{i,t})^2 = A_{i,t}^2 + 2 A_{i,t}\varphi_i + \varphi_i^2
\end{displaymath}

All together, that gives

\begin{eqnarray*}
{\cal E}^2 - c^2 {\cal B}^2 & = & \displaystyle
A_{i,t}^2 ...
... 2 A_{i,i}\varphi_t
+ c^2 A_{i,j}A_{j,i} - c^2 A_{i,i}A_{j,j}
\end{eqnarray*}

as can be verified by multiplying out and simplifying. The right hand side in the first line is the self-evident electromagnetic Lagrangian density, except for the factor $\epsilon_0$$\raisebox{.5pt}{$/$}$​2. The second line is the square of the Lorentz condition quantity. The final line can be written as a sum of pure derivatives:

\begin{displaymath}
2 (A_i \varphi_i)_t - 2 (A_i \varphi_t)_i
+ c^2 (A_i A_{j,i})_j - c^2 (A_i A_{j,j})_i
\end{displaymath}

Pure derivatives do not produce changes in the action, as the changes in the potentials disappear on the boundaries of integration.


D.37.4 Coulomb potential energy

The Coulomb potential energy between charged particles is typically derived in basic physics. But it can also easily be verified from the conventional electromagnetic energy (A.143). In the steady case, there is only the electric field, due to the Coulomb potential. The energy may then be written as

\begin{displaymath}
\frac{\epsilon_0}{2} \int{\cal E}_{\rm {C}}^2{\,\rm d}^3{\...
...kew0\vec r};t)\rho({\skew0\vec r};t){\,\rm d}^3{\skew0\vec r}
\end{displaymath}

where the first equality comes from the definition of the electric field, the second from integration by parts and the third one from the first Maxwell equation. Substitution of the Coulomb potential in terms of the charge distribution as given in {A.22.8},

\begin{displaymath}
\varphi_{\rm {C}}({\skew0\vec r};t) = \int_{{\rm all\ }{\u...
...ine{\skew0\vec r}}\vert}{\,\rm d}^3{\underline{\skew0\vec r}}
\end{displaymath}

now gives the Koulomb potential energy $V_{\rm {C}}$ for a continuous charge distribution:

\begin{displaymath}
V_{\rm C} = {\textstyle\frac{1}{2}} \int_{{\rm all\ }{\ske...
... {\,\rm d}^3{\skew0\vec r}{\rm d}^3{\underline{\skew0\vec r}}
\end{displaymath}

For point charges, the charge distribution is by definition

\begin{displaymath}
\rho({\skew0\vec r};t) = \sum_{i=1}^I q_i \delta^3({\skew0\vec r}-{\skew0\vec r}_i)
\end{displaymath}

Here $\delta^3$ is the three-di­men­sion­al delta function, ${\skew0\vec r}_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec r}_i(t)$ the position of point charge $i$, and $q_i$ its charge.

Recall that the delta function picks out the value at ${\skew0\vec r}_i$ from whatever it is integrated against. Using this twice on the Coulomb potential energy above,

\begin{displaymath}
V_{\rm C} = {\textstyle\frac{1}{2}} \sum_{i=1}^I \int_{{\r...
..._0\vert{\skew0\vec r}_i-{\skew0\vec r}_{{\underline i}}\vert}
\end{displaymath}

That is the Coulomb potential energy $V_{\rm {C}}$ for point charges.

Note again that physically all the energy is inside the electromagnetic field. There is no energy of interaction of the charged particles with the field. If equal charges move closer together, they increase the energy in the electromagnetic field. That requires work.