We know that,
Work done (W) is
directly proportional to the force applied (F) on an object to move to a displacement (s).
So, we get that,
W=F*s
But, we know that, energy (E) is equal to the work done (W).
Therefore,
E=F*s
Now,
If force (F) is applied, there is small change in displacement (ds) and energy (dE).
So, we get that,
dE=F*ds
We know that, energy (E) is integral of force (F) and displacement (s).
So, we get,
E=int F*ds ---(1)
Now, we know that, force (F) is the rate of change of momentum (p).
So,
F=d/dt(p)
F=d/dt(m*v)
therefore F=m*d/dt(v) ---(2)
Now,
Putting (2) in (1), we get,
E=int(m*d/dt(v)+v*d/dt(m))*ds
=intm*dv(d/dt(s))+v*dm(d/dt(s)) because{here, d/dt(s)=v}.
therefore E=intmv*dv+v^2dm ---(3).
Now, from relativity,we get relativistic mass (m) as,
m=m_0/sqrt(1-v^2/c^2)
It can be written as,
m=m_0(1-v^2/c^2)^(-1/2)
Now,
Differentiating the equation w.r.t velocity (v), we get,
=>d/(dv)(m)=m_0(-1/2)(1-v^2/c^2)^(-3/2)(-2v/(c^2))
=m_0v/c^2(1-v^2/c^2)^(-3/2)
=m_0v/c^2(1-v^2/c^2)^(-1/2)*(1-v^2/c^2)^(-1)
=v/(c^2(1-v^2/c^2))*m_0(1-v^2/c^2)^(-1/2)
=(vc^2)/(c^2(c^2-v^2))*m
{because m_0(1-v^2/c^2)=m}
So,d/(dv)m=(mv)/c^2-v^2
Now,
Cross-multiplying, we get,
=>dm(c^2-v^2)=mv*dv
=>c^2dm-v^2dm=mv*dv
=>c^2dm=mv*dv+v^2dm---(4)
Now,
Putting (4) in (3), we get that,
E=intc^2dm
Here,
We know (c) is constant
So,
E=c^2intdm ---(5)
Now, from constant rule,
=int dm
=m ---(6)
Now,
Putting (6) in (5), we get,
E=c^2int dm
E=c^2*m
therefore E=mc^2
___ Hence, Proved.
Phew...