Question

# Derive an expression $E = m{c^2}$ , where the symbols have their usual meaning.

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Hint: The above problem is based on the conversion of the mass into energy. The problem can be solved by using the concept of rest mass and work-energy theorem. The rest mass is defined as the mass of a stationary object. The work-energy theorem states that the work done by an object is the same as the change in the kinetic energy of the object.

Let us assume that a particle of mass m is moving with speed v. The rest mass of the particle is ${m_0}$ .
The rest mass of the object is given as:
${m_0} = m\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}}$
Here, c is the speed of the light.
$m_0^2 = {m^2}\left( {1 - \dfrac{{{v^2}}}{{{c^2}}}} \right)$
$m_0^2{c^2} = {m^2}{c^2} - {m^2}{v^2}$
${m^2}{c^2} - {m^2}{v^2} = m_0^2{c^2}$
Differentiate the above expression.
$\left( {2m} \right)\left( {dm} \right){c^2} - \left( {2m} \right)\left( {dm} \right){v^2} - 2v\left( {dv} \right){m^2} = \left( 0 \right){c^2}$
${c^2}dm = {v^2}dm + mvdv......\left( 1 \right)$
The momentum of the object is given as:
$P = mv$
The force acting on the object is the same as the change in the momentum of the object with respect to time.
$F = \dfrac{{dP}}{{dt}}$
Substitute mv for P in the above expression.
$F = \dfrac{{d\left( {mv} \right)}}{{dt}}$
$F = m\dfrac{{dv}}{{dt}} + v\dfrac{{dm}}{{dt}}$
If the object is displaced to distance ds by the force on the object, then the work done by the object is given as:
$dW = F \cdot ds$
Substitute the value of F in the above expression.
$dW = \left( {m\dfrac{{dv}}{{dt}} + v\dfrac{{dm}}{{dt}}} \right)ds$
$dW = \left( {mdv} \right)\dfrac{{ds}}{{dt}} + \left( {vdm} \right)\dfrac{{ds}}{{dt}}$
The change in the position is the same as the speed of the object, so substitute v for $\dfrac{{ds}}{{dt}}$ in the above expression.
$dW = \left( {mdv} \right)v + \left( {vdm} \right)v$
$dW = mvdv + {v^2}dm$
According to the work-energy theorem the work given by the above expression will be equal to the change in the kinetic energy.
$dK = dW$
Substitute the work done by the object in above expression.
$dK = mvdv + {v^2}dm......\left( 2 \right)$
Comparing the expression (1) and expression (2).
$dK = {c^2}dm......\left( 3 \right)$
Integrating the above expression for change in kinetic energy from 0 to K and rest mass ${m_0}$ to mass of object in motion m.
$\int\limits_0^K {dK} = \int\limits_{{m_0}}^m {{c^2}dm}$
$\left( K \right)_0^K = {c^2}\left( m \right)_{{m_0}}^m$
$K - 0 = {c^2}\left( {m - {m_0}} \right)$
$K = \left( {m - {m_0}} \right){c^2}$
The total energy of the object is the same as the kinetic energy of the object and rest mass energy of the object.
$E = K + {m_0}{c^2}.....\left( 4 \right)$
Substitute the value of kinetic energy in the expression (4).
$E = \left( {m - {m_0}} \right){c^2} + {m_0}{c^2}$
$E = m{c^2} - {m_0}{c^2} + {m_0}{c^2}$
$E = m{c^2}$
Therefore, the expression for mass-energy relation is proved.

Note
Always remember that the total energy of the object is equal to the sum of energy of the object in motion and energy of the object at rest. The rest mass of the object remains always constant and mass of the object varies only during the motion of the object.