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Collisions and Explosions

Here is an archetypal vignette for Newtonian mechanics. Two compound atoms called \mathbf{A} and \mathbf{B} have an interaction with each other by swapping \mathsf{X} another particle which is called the exchange particle. The interaction is caused when \mathbf{A} emits \mathsf{X} at event \mathbf{A}_{i} which is called the initial event of the interaction. This is written as

\mathbf{A}_{\, i-1} \to 	\mathbf{A}_{i} + \mathsf{X}_{i}

Particle \mathsf{X} then has an effect on \mathbf{B} by being absorbed at event \mathbf{B}_{f} which is called the final event of the interaction. We express this by writing

\mathbf{B}_{f} + \mathsf{X}_{f} \to \mathbf{B}_{f+1}

For EthnoPhysics, the interaction is described using three repetitive chains of historically ordered events written as

\Psi \! \left( \bar{r}, t \right)^{\mathbf{A}} = \left( \mathbf{A}_{1}, \, \mathbf{A}_{2} \, \ldots \, \mathbf{A}_{i} \, \ldots \, \mathbf{A}_{f} \, \ldots \right)

\Psi^{\mathsf{X}} = \left( \mathsf{X}_{i} \, \ldots \, \mathsf{X}_{f} \right)

\Psi \! \left( \bar{r}, t \right)^{\mathbf{B}} = \left( \mathbf{B}_{1}, \, \mathbf{B}_{2} \, \ldots \, \mathbf{B}_{i} \, \ldots \, \mathbf{B}_{f} \, \ldots  \right)

Since \mathbf{A} and \mathbf{B} are composed from atoms, we assume that they can be described by space-time events with a position  \bar{r} and time of occurrence  t. We do not assume that \mathsf{X} is an atom, rather we often take it to be a photon or a graviton. So we cannot always describe \mathsf{X} using a trajectory. And the position of \mathsf{X} is well-defined only for the initial and final events where it is included as part of an atom. Overall, the interaction is characterized by the following quantities.

Momentum Change by Emission

The interaction is caused when \mathbf{A} emits \mathsf{X} at an event \mathbf{A}_{\, i} which is called the initial event of the interaction

\mathbf{A}_{\, i-1}  \to \mathbf{A}_{i} + \mathsf{X}_{i}

Momentum is conserved so a total over all momenta  \overline{p} are the same before and after the interaction

\overline{p}_{i-1}^{\mathbf{A}} = \overline{p}_{i}^{\mathbf{A}} + \overline{p}_{i}^{\mathsf{X}}

And the change in \mathbf{A}‘s momentum due to the emission of \mathsf{X} is given by

\Delta \overline{p}^{\mathbf{A}} = \overline{p}_{i}^{\mathbf{A}} - \overline{p}_{i-1}^{\mathbf{A}}

so that

\Delta \overline{p}^{\mathbf{A}} = - \, \overline{p}^{\mathsf{X}}

Momentum Change by Absorption

Particle \mathsf{X} has an effect on \mathbf{B} by being absorbed at event \mathbf{B}_{f} which is called the final event of the interaction.

\mathbf{B}_{f} + \mathsf{X}_{f} \to \mathbf{B}_{f+1}

Momentum is conserved so a total over all momenta  \overline{p} are the same before and after the interaction

\overline{p}_{f}^{\mathbf{B}} + \overline{p}^{\mathsf{X}}_{f} = \overline{p}_{f+1}^{\mathbf{B}}

And the change in \mathbf{B}‘s momentum due to the emission of \mathsf{X} is given by

\Delta \overline{p}^{\mathbf{B}} = \overline{p}_{f+1}^{\mathbf{B}} - \overline{p}_{f}^{\mathbf{B}}

so that

\Delta \overline{p}^{\mathbf{B}} = \overline{p}^{\mathsf{X}}

Elapsed Time between Events

The difference in the time of occurrence  t between initial and final events is

\Delta t = t_{f}^{\, \mathbf{B}} - t_{i}^{\, \mathbf{A}}

If the frame of reference is inertial and \mathsf{X} is isolated (apart from the interaction under consideration) then the period  \widehat{\tau} of \mathsf{X} does not vary between events. So the elapsed time is given by

\Delta t = \left( \rule{0px}{10px} f-i \right) \widehat{\tau}^{\mathsf{X}}

or in terms of the frequency  \nu

\Delta t = \dfrac{f-i}{\nu^{\mathsf{X}}}

Then using Planck’s postulate gives the elapsed time in terms of the mechanical energy  E as

\Delta t = \dfrac{h\left(f-i \right)}{E^{\mathsf{X}}}

Distance between Events

The distance between initial and final events is given by

\ell \equiv \left\| \,   \bar{r}^{\, \mathbf{B}}_{f} - \bar{r}^{\, \mathbf{A}}_{i} \right\|

If the frame of reference is inertial and \mathsf{X} is isolated (apart from the interaction under consideration) then the wavelength  \lambda does not vary so that

\ell = \left( \rule{0px}{10px} f-i \right) \lambda^{\mathsf{X}}

Then using de Broglie’s expression for the wavelength in terms of the momentum  p gives

\ell = \dfrac{h \left( f-i \right)}{p^{\,\mathsf{X}}}