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'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, 20178 yr by Tom O'Neil

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?

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.

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, 20178 yr by Sensei

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, 20178 yr by timo