# E equal to mc squared derivation

Here is how we derive mass energy equivalence

Force is defined as rate of change of momentum i.e.,

$F = \frac{d}{dt}(m\nu)$ ….. (1)

According to the theory of relativity, both mass and velocity are variable. Therefore

$F = \frac{d}{dt}(m\nu) = m \frac{d\nu}{dt} + \nu \frac{dm}{dt}$ ….. (2)

Let the force F displace the body through a distance dx. Then, the increase in the kinetic energy $(dE_k)$ of the body is equal to the work done $(Fdx)$. Hence,

$dE_k = F dx = m \frac{d\nu}{dt}dx + \nu \frac{dm}{dt}dx$

or, $dE_k = m\nu d\nu + \nu^2 dm$ ….. (3)

According to the law of variation of mass of velocity

$m = \frac{m_0}{\sqrt{1-\frac{\nu^2}{c^2}}}$ ….. (4)

Squaring both sides, $m^2 = \frac{m_0^2}{1-\frac{\nu^2}{c^2}}$

or, $m^2c^2 = m_0^2c^2 + m^2\nu^2$

Differentiating,

$c^2 2m dm = m^2 2\nu d\nu + \nu^2 2m dm$

or, $c^2 dm = m\nu d\nu + \nu^2 dm$ ….. (5)

From equation (3) and (5)

$dE_k = c^2 dm$ ….. (6)

Thus, a change in K.E. $dE_k$ is directly proportional to a change in mass  dm. When a body is at rest, its velocity is zero, (K.E. = 0) and $m = m_0$. When its velocity is $\nu$, its mass becomes m. Therefore, integrating equation (6)

$E_k = \int_0^{E_k}dE_k = c^2 \int_{m_0}^{m} dm = c^2(m-m_0)$

Therefore, $E_k = mc^2 - m_0c^2$ ….. (7)

This is the relativistic formula for K.E. When the body is at rest, the internal energy stored in the body is $m_0c^2$. $m_0c^2$ is called the rest mass energy. The total energy (E) of the body is the sum of K.E. $(E_k)$ and rest mass energy $m_0c^2$

So, $E = E_k + m_0c^2 = (mc^2 - m_0c^2) + m_0c^2 = mc^2$

This is Einstein’s mass energy relation.

This relation states a universal equivalence between mass and energy. It means that mass may appear as energy and energy as mass.

The relationship $E = mc^2$ between energy and mass forms the basis of understanding nuclear reactions such as fission and fusion. These reactions take place in nuclear bombs and reactors. When a uranium nucleus is split up, the decrease in its total rest mass appears in the form of an equivalent amount of K.E. of its fragments.