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Is this derivation of E = mc^2 correct? ... I have some doubt at the red place ...

[itex]\large{ F = \frac{dp}{dt} = \frac{d}{dt}(mv) }[/itex]

[itex]\large{ F = v\frac{dm}{dt} + m\frac{dv}{dt} }[/itex]

Let this force cause a displacement dx

[itex]\large{ dW = F \cdot dx }[/itex]

Assuming body was initially at rest and this work is converted into kinetic energy and increase it by dK

[itex]\large{ dK = F\cdot dx }[/itex]

[itex]\large{ dK = v\frac{dm}{dt}dx + m\frac{dv}{dt}dx }[/itex]

[itex]\large{ dK = mvdv + v^2dm }[/itex] ---Equation 1

Now using eqn

[itex]\large{ m = \frac{m_o}{ \Large{ \sqrt{1-\frac{v^2}{c^2}} } } }[/itex]

Squaring both sides,

[itex]\large{ m^2= \frac{{m_o}^2}{1-\frac{v^2}{c^2}} }[/itex]

[itex]\large{ m^2c^2 - m^2v^2 = {m_o}^2c^2 }[/itex]

differentiating the expression

[itex]\large{ 2mc^2dm - 2mv^2dm - 2vm^2dv = 0 }[/itex]

[itex]\large{ c^2dm = mvdv - v^2dm }[/itex]

Using this in eqn 1

[itex]\large{ dK = c^2dm }[/itex]

Integrating

[itex]\large{ K = \int_0^K{dK} = \int_{m_o}^{m}{c^2dm} }[/itex]< --- HERE

[itex]\large{ K = c^2(m - m_o) }[/itex]

Total energy of body,

[itex]\large{ E = K + m_o c^2 }[/itex]

[itex]\large{ E = c^2(m - m_o) + m_o c^2 }[/itex]

[itex]\large{ E = mc^2 = \frac{m_o c^2}{ \Large{ \sqrt{1-\frac{v^2}{c^2}} } } }[/itex]

Also [itex]\large{ K = E - m_o c^2 = (m - m_o)c^2 }[/itex]

[itex]\large{ \Delta E = \Delta m c^2 }[/itex]

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# Homework Help: E=mc^2 Derivation

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