Editing Special relativity (section)

=== Elastic collisions ===
Examination of the collision products generated by particle accelerators around the world provides scientists evidence of the structure of the subatomic world and the natural laws governing it. Analysis of the collision products, the sum of whose masses may vastly exceed the masses of the incident particles, requires special relativity.<ref name="Idema_2022">{{cite web |last1=Idema |first1=Timon |title=Mechanics and Relativity. Chapter 14: Relativistic Collisions |url=https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_Mechanics_and_Relativity_(Idema)/14%3A_Relativistic_Collisions |website=LibreTexts Physics |date=17 April 2019 |publisher=California State University Affordable Learning Solutions Program |access-date=2 January 2023}}</ref>

In Newtonian mechanics, analysis of collisions involves use of the [[conservation of mass|conservation laws for mass]], [[conservation of momentum|momentum]] and [[conservation of energy|energy]]. In relativistic mechanics, mass is not independently conserved, because it has been subsumed into the total relativistic energy. We illustrate the differences that arise between the Newtonian and relativistic treatments of particle collisions by examining the simple case of two perfectly elastic colliding particles of equal mass. (''Inelastic'' collisions are discussed in [[Spacetime#Conservation laws]]. Radioactive decay may be considered a sort of time-reversed inelastic collision.<ref name="Idema_2022"/>)

Elastic scattering of charged elementary particles deviates from ideality due to the production of [[Bremsstrahlung]] radiation.<ref name="Nakel_1994">{{cite journal |last1=Nakel |first1=Werner |title=The elementary process of bremsstrahlung |journal=Physics Reports |date=1994 |volume=243 |issue=6 |pages=317–353 |doi=10.1016/0370-1573(94)00068-9|bibcode=1994PhR...243..317N }}</ref><ref>{{cite book |last1=Halbert |first1=M.L. |editor1-last=Austin |editor1-first=S.M. |editor2-last=Crawley |editor2-first=G.M. |title=The Two-Body Force in Nuclei |date=1972 |publisher=Springer |location=Boston, MA. |chapter=Review of Experiments on Nucleon-Nucleon Bremsstrahlung}}</ref>

==== Newtonian analysis ====
[[File:Elastic collision of moving particle with equal mass stationary particle.svg|thumb|Figure 6–2. Newtonian analysis of the elastic collision of a moving particle with an equal mass stationary particle]]
Fig. 6-2 provides a demonstration of the result, familiar to billiard players, that if a stationary ball is struck elastically by another one of the same mass (assuming no sidespin, or "English"), then after collision, the diverging paths of the two balls  will subtend a right angle. (a) In the stationary frame, an incident sphere traveling at 2'''v''' strikes a stationary sphere. (b) In the center of momentum frame, the two spheres approach each other symmetrically at ±'''v'''. After elastic collision, the two spheres rebound from each other with equal and opposite velocities ±'''u'''. Energy conservation requires that {{abs|'''u'''}} = {{abs|'''v'''}}. (c) Reverting to the stationary frame, the rebound velocities are {{nowrap|'''v''' ± '''u'''}}. The dot product {{nowrap|1=('''v''' + '''u''') ⋅ ('''v''' − '''u''') = '''v'''<sup>2</sup> − '''u'''<sup>2</sup> = 0}}, indicating that the vectors are orthogonal.<ref name="Rindler0"/>{{rp|26–27}}

==== Relativistic analysis ====
[[File:Relativistic elastic collision of equal mass particles.svg|thumb|Figure 6–3. Relativistic elastic collision between a moving particle incident upon an equal mass stationary particle]]
Consider the elastic collision scenario in Fig.&nbsp;6-3 between a moving particle colliding with an equal mass stationary particle. Unlike the Newtonian case, the angle between the two particles after collision is less than 90°, is dependent on the angle of scattering, and becomes smaller and smaller as the velocity of the incident particle approaches the speed of light:

The relativistic momentum and total relativistic energy of a particle are given by
{{NumBlk2||<math>\quad\quad \vec{p} = \gamma m \vec{v} \quad \text{and} \quad E = \gamma m c^2 </math>  |6-4}}

Conservation of momentum dictates that the sum of the momenta of the incoming particle and the stationary particle (which initially has momentum = 0) equals the sum of the momenta of the emergent particles:
{{NumBlk2||<math>\quad\quad \gamma_1 m \vec{v_1} + 0 = \gamma_2 m \vec{v_2}  +  \gamma_3 m \vec{v_3} </math>  |6-5}}

Likewise, the sum of the total relativistic energies of the incoming particle and the stationary particle (which initially has total energy mc<sup>2</sup>) equals the sum of the total energies of the emergent particles:
{{NumBlk2||<math>\quad\quad \gamma_1 m c^2 + m c^2 = \gamma_2 m c^2 + \gamma_3 m c^2 </math>  |6-6}}

Breaking down ({{EquationNote|6-5}}) into its components, replacing <math>v</math> with the dimensionless {{tmath|1= \beta }}, and factoring out common terms from ({{EquationNote|6-5}}) and ({{EquationNote|6-6}}) yields the following:<ref name="Champion_1932" group="p"/>
{{NumBlk2||<math>\quad\quad \beta_1 \gamma_1 = \beta_2 \gamma_2 \cos{\theta} +  \beta_3 \gamma_3 \cos{\phi} </math>  |6-7}}
{{NumBlk2||<math>\quad\quad \beta_2 \gamma_2 \sin{\theta} = \beta_3 \gamma_3 \sin{\phi} </math>  |6-8}}
{{NumBlk2||<math>\quad\quad \gamma_1 + 1 = \gamma_2 + \gamma_3 </math>  |6-9}}

From these we obtain the following relationships:<ref name="Champion_1932" group="p">{{cite journal |last1=Champion |first1=Frank Clive |title=On some close collisions of fast β-particles with electrons, photographed by the expansion method. |journal=Proceedings of the Royal Society of London. Series A, Containing Papers of a Mathematical and Physical Character |year=1932 |volume=136 |issue=830 |pages=630–637 |publisher=The Royal Society Publishing |doi=10.1098/rspa.1932.0108 |bibcode=1932RSPSA.136..630C |s2cid=123018629 |doi-access=free }}</ref>
{{NumBlk2||<math>\quad\quad \beta_2 = \frac{\beta_1 \sin{\phi}}{ \{ \beta_1^2 \sin^2{\phi} +  \sin^2(\phi + \theta )/\gamma_1^2 \}^{1/2} }    </math>  |6-10}}
{{NumBlk2||<math>\quad\quad \beta_3 = \frac{\beta_1 \sin{\theta}}{ \{ \beta_1^2 \sin^2{\theta} +  \sin^2(\phi + \theta )/\gamma_1^2 \}^{1/2} }    </math>  |6-11}}
{{NumBlk2||<math>\quad\quad \cos{(\phi + \theta)} = \frac{ (\gamma_1 - 1) \sin{\theta} \cos{\theta} }{ \{ (\gamma_1 + 1)^2 \sin^2 \theta + 4 \cos^2 \theta  \}^{1/2} }    </math>  |6-12}}
For the symmetrical case in which <math> \phi = \theta</math> and {{tmath|1= \beta_2 = \beta_3 }}, ({{EquationNote|6-12}}) takes on the simpler form:<ref name="Champion_1932" group="p"/>
{{NumBlk2||<math>\quad\quad \cos{\theta} = \frac{\beta_1}{ \{ 2/\gamma_1 + 3 \beta_1^2 - 2 \}^{1/2} }    </math>  |6-13}}