WikiMechanics

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• to assess a two-body mechanical system in an isotropic three-dimensional space requires that both bodies be at least as large as atoms
• but if we make extra assumptions about angular momenta, and limit what sort of interactions are allowed, then we can limit considerations to just a two-dimensional problem
• we can make two-dimensional models of electrons and protons by associating a magnetic field with the proton
• so the smallest two-body, two-dimensional problem we can consider in an isotropic space is an electron orbiting a proton, i.e. the Bohr model

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Bohr Balmer

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The positions and trajectories of some simple particles cannot be well known, or even well-defined. To have a properly defined position a particle must contain enough of the right sort of quarks to establish its spatial orientation. But some of the models that we have discussed cannot satisfy this requirement so their positions cannot be assigned without making further assumptions. For example we cannot state the position of a solitary photon. And this uncertain quality can be observed when Young's double slit experiment is performed at low light levels.

One common way of dealing with this issue to to assume that a sub-atomic particle has been absorbed by an atom that does have a well-defined position. Then both particles are supposedly in the same place. Another possibility is to conjecture additional fields to align a particle's orientation. Such presumptions are codified in various three-dimensional arrangements that assign quarks to sub-orbital events by convention. These designs are called sub-atomic particle models. For example, here are some ways of representing protons and electrons in a three-dimensional space.

Proton

$\Large{ k }$	$\large{ \delta _{\hat{m}} }$	$\large{ \delta _{\hat{e}} }$	$\large{ \delta _{\theta} }$	$\large{ \sf{P}_{\it{k}} }$
1	+1	0	+1	$\sf{d} \ \ \bar{\sf{t}} \sf{b}$
2	0	-1	+1	$\,$
3	-1	0	+1	$\sf{d} \ \ \bar{\sf{t}} \sf{b}$
4	0	+1	+1	$\,$
5	+1	0	-1	$\sf{d} \ \ \bar{\sf{t}} \sf{b}$
6	0	-1	-1	$\,$
7	-1	0	-1	$\sf{d} \ \ \bar{\sf{t}} \sf{b}$
8	0	+1	-1	$\,$

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