WikiMechanics

Quarks are indestructible and the overall quantity of any type of quark is conserved. So the quark coefficients of atoms and exchange particles are related like

Changes in the net number of Z-type quarks in $\mathbf{B}$ are

$\begin{align} \partial \Delta n^{\mathsf{Z}} \equiv \Delta n^{\mathsf{Z}}_{\it{f}} - \Delta n^{\mathsf{Z}}_{\it{i}} &= \left[ n^{\sf{\bar{z}}}_{\it{f}} - n^{\sf{z}}_{\it{f}} \right] - \left[ n^{\sf{\bar{z}}}_{\it{i}} - n^{\sf{z}}_{\it{i}} \right] \\ &= \left[ n^{\sf{\bar{z}}}_{\it{f}} - n^{\sf{\bar{z}}}_{\it{i}} \right] - \left[ n^{\sf{z}}_{\it{f}} - n^{\sf{z}}_{\it{i}} \right] \\ &= \partial n^{\sf{\bar{z}}} - \partial n^{\sf{z}} \\ &= n^{\sf{\bar{z}}}\left(\sf{X}\right) - n^{\sf{z}}\left(\sf{X}\right) \\ &= \Delta n^{\mathsf{Z}}\left(\sf{X}\right) \end{align}$

And this is true for any sort of quark, so we can just drop the Z to write

Changes to the magnetic radius of atom $\mathbf{B}$ are given by

$\begin{align} \partial \rho_{m} \equiv \left[ \rho_{m} \right]_{f} - \left[ \rho_{m} \right]_{i} &= \left[ \frac{ {\Delta}n^{\mathsf{A}}_{\it{f}} U^{\mathsf{a}} - {\Delta}n^{\mathsf{M}}_{\it{f}} U^{\mathsf{m}} }{k_{\sf{F}}} \right] - \left[ \frac{ {\Delta}n^{\mathsf{A}}_{\it{i}} U^{\mathsf{a}} - {\Delta}n^{\mathsf{M}}_{\it{i}} U^{\mathsf{m}} }{k_{\sf{F}}} \right] \\ &= \frac{ {\Delta}n^{\mathsf{A}}_{\it{f}} - {\Delta}n^{\mathsf{A}}_{\it{i}}}{k_{\sf{F}}} U^{\mathsf{a}} - \frac{ {\Delta}n^{\mathsf{M}}_{\it{f}} - {\Delta}n^{\mathsf{M}}_{\it{i}} }{k_{\sf{F}}} U^{\mathsf{m}} \\ &= \frac{ \partial \Delta n^{\mathsf{A}}}{k_{\sf{F}}} U^{\mathsf{a}} - \frac{ \partial \Delta n^{\mathsf{M}} }{k_{\sf{F}}} U^{\mathsf{m}} \\ &= \frac{ \Delta n^{\mathsf{A}} \left(\sf{X}\right) \, U^{\mathsf{a}} - \Delta n^{\mathsf{M}} \left(\sf{X}\right) \, U^{\mathsf{m}} }{k_{\sf{F}}} \\ &= \rho_{m} ^{\sf{X}} \end{align}$

And similar relationships hold for the other components as well so that

$\partial \overline{\rho}^{\mathbf{B}} \equiv \overline{\rho}^{\mathbf{B}}_{\it{f}} -\overline{\rho}^{\mathbf{B}}_{\it{i}} = \left( \partial \rho_{m}^{\mathbf{B}}, \partial \rho_{e}^{\mathbf{B}}, \partial \rho_{z}^{\mathbf{B}} \right) = \left( \rho_{m}^{\sf{X}}, \rho_{e}^{\sf{X}}, \rho_{z}^{\sf{X}} \right)$

But the radius vector of X is defined by $\ \overline{\rho}^{\sf{X}} \equiv \left( \rho_{ m}^{\sf{X}}, \rho_{e}^{\sf{X}}, \rho_{ z}^{\sf{X}} \right)$ thus

The norm of a radius vector is given by

$\rho = \sqrt{ \vphantom{\sum^{2}} \; k_{mm} \, \rho_{m}^{ 2} + k_{ee} \, \rho_{e}^{2} + k_{zz} \, \rho_{z}^{2} + 2 k_{em} \, \rho_{e} \rho_{m} + 2 k_{ez} \, \rho_{e} \rho_{z} + 2 k_{mz} \, \rho_{m} \rho_{z} \; }$

where $k$ stands for various dimensionless numbers in the quark metric. So

Now let $A = \rho^{2}$ and set an average radius by $2 \tilde{\rho} = \rho_{i} + \rho_{f}$ then

$\partial A = A_{f} - A_{i} = \rho^{2}_{f} - \rho^{2}_{i} = \left( \rho_{f} + \rho_{i} \right) \left( \rho_{f} - \rho_{i} \right) = 2 \tilde{\rho} \, \partial \rho$

The average value of $\rho$ can also be written as $2\tilde{\rho} = 2\rho_{i} + \partial \rho = 2\rho_{f} - \partial \rho$. Then if $\partial \rho$ is a lot smaller than both of $\rho_{i}$ and $\rho_{f}$ we obtain $\tilde{\rho} \simeq \rho_{i} \simeq \rho_{f}$ and so can drop the tilde and subscripts to write

$\partial A = 2 \rho \, \partial \rho$

That is, when an interaction causes changes in a particle's shape that are negligible compared to its radii, then we can use the usual result from differential calculus. Similarly Equation (1) implies that

$\begin{align} \rho \, \partial \rho &= k_{mm} \, \rho_{m} \partial \rho_{m} + k_{ee} \, \rho_{e} \partial \rho_{e} + k_{zz} \, \rho_{z} \partial \rho_{z} \\ &+ \ k_{em} \left( \rho_{e} \partial \rho_{m} + \rho_{m} \partial \rho_{e} \right) + k_{ez} \left( \rho_{e} \partial \rho_{z} + \rho_{z} \partial \rho_{e} \right) + k_{mz} \left( \rho_{m} \partial \rho_{z} + \rho_{z} \partial \rho_{m} \right) \end{align}$

Recall that $k_{zz} \equiv 1$ and rearrange terms to get

The work required to form a particle is $W \equiv \rho k_{\sf{F}}$ so the work done by the interaction is

$\begin{align} \partial W \equiv W_{\it{f}} - W_{\it{i}} = \partial \rho \, k_{\sf{F}} = \frac{ \rho \, \partial \rho \, }{\rho} k_{\sf{F}} \end{align}$

Using Equation (2) this can be written as

$\begin{align} \partial W &= \partial \rho_{m} \frac{ k_{mm} \rho_{m} + k_{em} \rho_{e} + k_{mz} \rho_{z}}{ \rho } k_{\sf{F}} \\ &+ \ \partial \rho_{e} \frac{ k_{em} \rho_{m} + k_{ee} \rho_{e} + k_{ez} \rho_{z} }{ \rho } k_{\sf{F}} \\ &+ \ \partial \rho_{z} \frac{ k_{mz} \rho_{m} + k_{ez} \rho_{e} + \rho_{z} }{ \rho } k_{\sf{F}} \end{align}$

Then in terms of $F_{m}$, $F_{e}$ and $F_{c}$ which note the forces that hold the atom together

$\begin{align} \partial W = F_{m} \partial \rho_{m} + F_{e} \partial \rho_{e} - F_{c} \partial \rho_{z} \end{align}$

Or more explictly, the work done on an atom is related to attributes of the exchange particle by