Zelaron Gaming Forum - Deriving e=mc^2

We begin with the definition of relativistic mass:

$\reverse m = \frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ .

We can now define relativistic momentum as:

$\reverse \vec{p} = \frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\vec{v}$ .

Since force is defined as

$\reverse \vec{F} = \frac{d\vec{p}}{dt}$

we can define force in the relativistic context analogously:

$\reverse \vec{F} =\frac{d\vec{p}}{dt} =\frac{d}{dt} \left[ \frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\vec{v} \right] = \frac{d}{dt} \left[m\vec{v} \right]$ .

In classical physics work is defined as

$\reverse W = \int\limits_{r} \! \vec{F} \! \cdot \! d\vec{r}$ .

We can reduce this to a problem in a single dimension for simplicity, and recall that the kinetic energy added to a body is equal to the work done on the body by an external force:

$\reverse W = \int_{0}^{x} \! Fdx = E_{k}$ .

From here we can say:

$\reverse \int_{0}^{x} \! Fdx = \int_{0}^{x} \! \frac{dp}{dt}dx = \int_{0}^{x} \! \frac{d}{dt} \left[ mv \right] dx = \int_{0}^{t} \! \frac{d}{dt} \left[ mv \right] vdt$ .

Applying the definition of relativistic mass, the last expression becomes:

[

$\reverse \int_{0}^{t} \! \frac{d}{dt} \left[ mv \right] vdt = \int_{0}^{v} \! vd \! \left[ \frac{m_{0}v}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \right] = m_{0} \int_{0}^{v} \! \left( \frac{v}{\sqrt{1-\frac{v^{2}}{c^{2}}}} + \frac{v^{3}c}{\left( 1-\frac{v^{2}}{c^{2}} \right)^{3/2} \right) \! dv$ .

Integrating through, we have

$\reverse E_{k} = m_{0}c^{2} \left( \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} - 1 \right) = mc^{2}-m_{0}c^{2}$

The term on the left,

$\reverse mc^{2}$

, is the total energy of the body after an external force is applied, the term on the right,

$\reverse m_{0}c^{2}$

is the rest energy, and the final expression is the kinetic energy of the body.

The kinetic energy can be expressed more clearly, in my opinion, as

$\reverse E_{k} = (m-m_{0})c^{2} = \Delta m c^{2}$ .

And there you have it. Mass is energy!