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...#
