Can you derive #E=mc^2# ?

1 Answer
Nam D.
Feb 7, 2018

Please refer to the explanation below.

Explanation:

This was taken from:

http://www.emc2-explained.info/Emc2/Deriving.htm#.WnrzqJNubwc

This proof uses calculus and some basic physics.

First, we are going to express kinetic energy in the integral form. That is

#KE=int_0^sF \ ds#

In Newton's Second Law of Motion, it states that force can be expressed like this,

#F=ma#

Know that, acceleration is the change of velocity over a certain amount of time, so the equation becomes

#F=m(dv)/dt#

#:.F=(d(mv))/dt#

Now, we can rewrite the integral in terms of

#KE=int_0^s(d(mv))/dt \ ds#

Since velocity is the displacement over time, we can write,

#v=(ds)/dt#

and the integral becomes

#=int_0^(mv)v \ d(mv)#

Now here comes the crucial observation, we have to know the relativistic energy equation, (I might derive it another time), but know that

#m=(m_0)/(sqrt(1-v^2/c^2))#

So, the integral now transforms into

#=int_0^(v)vd[(m_0)/(sqrt(1-v^2/c^2))]#

For here, we have to use integration by parts.

#intu \ dv=uv-intv \ du#

#:.=(m_0v^2)/(sqrt(1-v^2/c^2))-m_0int_0^(v)(v \ dv)/(sqrt(1-v^2/c^2))#

#=(m_0v^2)/(sqrt(1-v^2/c^2))+[m_0c^2*(m_0v)/(sqrt(1-v^2/c^2))]_0^v#

Since the limit of #v# is #c# (the speed of light), we can let #v=c#, and we get

#=(m_0c^2)/(sqrt(1-v^2/c^2))-m_0c^2#

Remember that #m=(m_0)/(sqrt(1-v^2/c^2))# from ages ago? We can plug that back in to get

#=mc^2-m_0c^2#

After thousands of derivations... we get #KE=mc^2-m_0c^2#.

Rearranging, #mc^2=KE+m_0c^2#

When #KE=0#, the stationary object still has that #m_0c^2# energy. When it has that #E_0# energy, it has rest mass, called #m_0#. (this concept is kinda complicated)

Substituting those values, we get

#E=KE+E_0#

when

#E_0=m_0c^2#

Finally, we find that

#E=mc^2# for rest, #E=(m_0c^2)/(sqrt(1-v^2/c^2))# for moving objects

Phew....

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