Feeds:
Posts

## Derivation of E=mc2

There are quite a few ways to derive Einstein’s famous equation $E=mc^{2}$. I am going to show you what I consider to be the simplest way.  Feel free to comment if you think you know of an easier way.

We will start off with the relationship between energy, force and distance. We can write

$dE = F dx \text{ (1) }$

Where $dE$ is the change in energy, $F$ is the force and $dx$ is the distance through which the object moves under that force.  But, force can also be written as the rate of change of momentum,

$F = \frac{dp}{dt}$

Allowing us to re-write Equation (1) as

$dE = \frac{dp}{dt}dx \rightarrow dE = dp \frac{dx}{dt} = vdp \text{ (2) }$

Remember that momentum $p$ is defined as

$p =mv$

In classical physics, mass is constant. But this is not the case in Special Relativity, where mass is a function of velocity (so-called relativistic mass).

$m = \frac{ m_{0} }{ \sqrt{ ( 1 - v^{2}/c^{2} ) } } \text{ (3) }$

where $m_{0}$ is defined as the rest mass (the mass of an object as measured in a reference frame where it is stationary).

Assuming that both $m \text{ and } v$ can change, we can therefore write

$dp =mdv + vdm$

This allows us to write Equ. (2) as

$dE = vdp = v(mdv + vdm) = mvdv + v^{2}dm \text{ (4) }$

Differentiating Equ. (3) with respect to velocity we get

$\frac{dm}{dv} = \frac{d}{dv} \left( \frac{ m_{0} }{ \sqrt{ (1 - v^{2}/c^{2}) } } \right) = m_{0} \frac{d}{dv} (1 - v^{2}/c^{2})^{-1/2}$

Using the chain rule to differentiate this, we have

$\frac{dm}{dv} = m_{0} \cdot - \frac{1}{2} (1 - v^{2}/c^{2})^{-3/2} \cdot (-2v/c^{2}) = m_{0} (v/c^{2}) \cdot (1 - v^{2}/c^{2})^{-3/2} \text{ (5) }$

But, we can write

$(1 - v^{2}/c^{2})^{-3/2}$ as $(1-v^{2}/c^{2})^{-1/2} \cdot (1-v^{2}/c^{2})^{-1}$

This allows us to write Equ. (5) as

$\frac{dm}{dv} = m_{0} (v/c^{2}) \cdot (1 - v^{2}/c^{2})^{-1} \cdot (1 - v^{2}/c^{2})^{-1/2}$

From the definition of the relativistic mass in Equ. (3), we can rewrite this as

$\frac{dm}{dv} = \frac{ m v }{ c^{2} }(1-v^{2}/c^{2})^{-1}$

Which is

$\frac{dm}{dv} = \frac{ m v }{ c^{2} } \left( \frac{c^{2}}{c^{2}} - \frac{ v^{2}}{c^{2} } \right)^{-1} = \frac{ m v }{ c^{2} } \left( \frac{c^{2}-v^{2}}{c^{2}} \right)^{-1} = \frac{ m v }{ c^{2} } \left( \frac{c^{2}}{c^{2}-v^{2}} \right)$

$\frac{dm}{dv} = \frac{ m v }{ (c^{2}-v^{2}) } \text{ (6) }$

So we can write

$c^{2}dm - v^{2}dm = mvdv$

Substituting this expression for $mvdv$ into Equ. (4) we have

$dE = vdp = vd(mv) = mvdv + v^{2}dm = c^{2}dm - v^{2}dm + v^{2}dm$

So

$dE = c^{2} dm$

Integrating this we get

$\int_{E_{0}}^{E} dE = c^{2} \int_{m_{0}}^{m} dm$

So

$E - E_{0} = c^{2} ( m - m_{0} ) = mc^{2} - m_{0}c^{2}$

$E - E_{0} = mc^{2} - m_{0}c^{2}$

This tells us that an object has rest mass energy $E_{0} = m_{0}c^{2}$ and that its total energy is given by

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

where $m$ is the relativistic mass.