Derivation of E=Mc2? For the common man

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  • Thread starter jonatron5
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  • #1
jonatron5
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Ok this is perhaps the single most famous equation known to man. And I have basically zero idea how it came to be. I have a slight background in classical physics (a couple college classes)

I know it states that energy is equal to mass multiplied by the speed of light squared.

I have tried to read online postings on it but have been unable to follow.

So is it possible to get a step by step derivation of this equation so that even a lay person could understand it?
 

Answers and Replies

  • #3
LevLandau
14
2
by the momentum of a particle we can mean the vector p=∂L/∂v, p=mv/√1-v^2/c^2, for small velocities this expression goes over into the classical p=mv. for v=c the momentum becomes infinite. after dp/dt=m/√1-v^2/c^2×dv/dt, if the velocity changes only in magnitude, that is, if the force is parallel to the velocity, dp/dt=m/(1-v^2/c^2)^1/2+dv/dt. the energy of the particle - E=p×v-L, then E=mc^2/√1-v^2/c^2, v=0 ⇒ E=mc^2
 
  • #4
SiennaTheGr8
489
189
Here's Einstein's original derivation in translation: https://www.fourmilab.ch/etexts/einstein/E_mc2/www/

To make sense of it, you have to either accept the relativistic Doppler equation he invokes or seek out a derivation of that, first. I like this one: https://www.lsw.uni-heidelberg.de/users/mcamenzi/DopplerAberration.pdf (though you'll need to be comfortable with the basics of special relativity, especially the Lorentz transformation).
 

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