Question

For Any energy - mass relation :-
Prove that :-
The change in ∆E = MC^2 ​

Answer 1

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Answer 2

${\boxed{\sf\:{To\;Prove\;:E=mc^{2}}}}$

★Here we have :-

E = Energy

m = Mass

c = Speed of light (299,792,458 m/s)

★By work energy principle work is eqivalent to Energy :-

Work = Energy

★As we know that :-

${\boxed{\int F\times dx}}$

★We may also write it as :-

${\boxed{\sf\:{dW = F\times dx}}}$

dE = F × dx

★According to Newton second law of motion :-

$\tt{\rightarrow dE =\dfrac{dp}{dt}\times dx ....(1)$

Here p is linear momentum hence

${\boxed{\sf\:{p = mv}}}$

dp = mdv + vdm

★Substitute the value of dp in Equation (1)

$\tt{\rightarrow dE =(mdv+vdm)\dfrac{dx}{dt}$

dE = mvdv + v²dm ...... (2)

$\text{Einstein\;Equation\;for\;change\;in\;mass}$

$\tt{\rightarrow m=\dfrac{m_{0}}{\sqrt{1-(v^{2}/c^{2})}}$

★It can be also written as :-

$\tt{\rightarrow m=m_{0}1-(\dfrac{v^{2}}{c^{2}})^{-1/2}}$

$\tt{\rightarrow\dfrac{dm}{dv}=m_{0}(\dfrac{-1}{2})(1-\dfrac{v^{2}}{c^{2}})^{-3/2}\times(\dfrac{-2v}{c^{2})}}$

$\tt{\rightarrow\dfrac{dm}{dv}=m_{0}(1-\dfrac{v^{2}}{c^{2}})^{-1/2}\times(1-\dfrac{v^{2}}{c^{2})^{-1}\times\dfrac{v}{c^{2}}}$

$\tt{\rightarrow\dfrac{dm}{dv}=m\times(\dfrac{c^{2}-v^{2}}{c^{2}})^{-1}\times\dfrac{v^{2}}{c^{2}}}$

$\tt{\rightarrow\dfrac{dm}{dv}=m_{0}(\dfrac{-1}{2})(1-\dfrac{v^{2}}{c^{2}})^{-3/2}\times(\dfrac{-2v}{c^{2}}$

$\tt{\rightarrow\dfrac{dm}{dv}=m\times\dfrac{c^{2}}{c^{2}-v^{2}\times\dfrac{v}{c^{2}}}$

c²dm - v²dm = mvdv

c²dm = mvdv + v²dm ........ (3)

${\boxed{\sf\:{From\; Equation\;(1)}}}$

c²dm = dE

${\boxed{\int dE = c^{2}\int dm}}$

∆E = ∆m × c²

${\boxed{\sf\:{E=mc^{2}}}}$