awarnes Posted March 18, 2017 Posted March 18, 2017 I've been doing some research into Einstein's equation E2=(mc2)2 +(pc)2 but apart from in nucleur reactions, where you can use the simpler E=mc2 as momentum=0, I have been unable to find any applications. Thank you in advance
Tom O'Neil Posted March 18, 2017 Posted March 18, 2017 (edited) I've been doing some research into Einstein's equation E2=(mc2)2 +(pc)2 but apart from in nucleur reactions, where you can use the simpler E=mc2 as momentum=0, I have been unable to find any applications. Thank you in advance I believe momentum is shared between the four force as in angular momentum. Here is an online e=mc2 calculator. http://www.endmemo.com/physics/einsteinlaw.php http://www.scienceforums.net/topic/103840-if-gravity-is-a-separate-force-then-how-can-it-exist-without-the-other-3/ Edited March 18, 2017 by Tom O'Neil
awarnes Posted March 18, 2017 Author Posted March 18, 2017 I believe momentum is shared between the four force as in angular momentum. Here is an online e=mc2 calculator. http://www.endmemo.com/physics/einsteinlaw.php http://www.scienceforums.net/topic/103840-if-gravity-is-a-separate-force-then-how-can-it-exist-without-the-other-3/ Am i right in believing then that there are no real-world applications of the equation. That it is only used in further derivations of other equations?
zztop Posted March 18, 2017 Posted March 18, 2017 I've been doing some research into Einstein's equation E2=(mc2)2 +(pc)2 but apart from in nucleur reactions, where you can use the simpler E=mc2 as momentum=0, I have been unable to find any applications. Thank you in advance The equation is central to all experiments involving relativistic particle collisions, Compton effect, inverse Compton effect, etc.
Sensei Posted March 18, 2017 Posted March 18, 2017 (edited) I've been doing some research into Einstein's equation E2=(mc2)2 +(pc)2 This equation can be written as: [math]E=m_0 c^2 \gamma[/math] [math]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/math] when v=0 it simplifies to [math]\gamma= 1[/math]. giving: [math]E=m_0 c^2[/math] but apart from in nucleur reactions, where you can use the simpler E=mc2 as momentum=0, I have been unable to find any applications. Annihilation of matter-antimatter. Pair production. Decay of unstable particle (lepton,meson,baryon). Suppose so you have gamma photon with energy 1.5 MeV. It's slightly more than required for pair production (electron-positron) 1.022 MeV. But energy has to be conserved. So you have: [math]E_{src}=1.5 MeV[/math] [math]E_{dst}=m_e c^2 \gamma + m_e c^2 \gamma[/math] [math]E_{src}=E_{dst}[/math] [math]1.5 MeV=m_e c^2 \gamma + m_e c^2 \gamma[/math] 1.5 MeV - 1.022 MeV = (approximately) 0.5 MeV And it's double kinetic energy of electron-positron. So each will have K.E. = 0.25 MeV (calculate gamma for it as exercise) [math]KE=m_0 c^2 \gamma-m_0 c^2[/math] Edited March 18, 2017 by Sensei
timo Posted March 18, 2017 Posted March 18, 2017 (edited) As a side-note to the thread: The equation you are citing is the energy-momentum relation for a free particle. It is actually not called "Einstein's equation" (by physicists). The reason I am stressing this is because there is a different mathematical relation that is actually called "Einstein's equation", namely the relation between spacetime-curvature and spacetime-content. That doesn't invalidate your question, of course. For your actual question: I think zztop's answer perfectly fits what I'd have said. If you consider how important conservation of energy and conservation of momentum are it is only one step further to realize how important an equation relating energy to momentum can be. Edited March 18, 2017 by timo 1
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