Einstein's Mass-Velocity Equation

Physical Law
The mass $m$ of a body is not constant.

It varies with the body's velocity, according to the equation:
 * $\displaystyle m = \frac {m_0}{\sqrt{1 - \frac {v^2}{c^2}}}$

where:
 * $v$ is the magnitude of the velocity of the body;
 * $c$ is the speed of light;
 * $m_0$ is the rest mass of the body.

The value $m$ is known as the relativistic mass of the body.

The factor $\displaystyle \frac 1 {\sqrt{1 - \frac {v^2}{c^2}}}$ is known as the Lorentz Factor.

Proof
Imagine a comet that flies towards a planet on which you are resting. The comet's velocity $u$ towards the planet be much smaller than the speed of light. Now imagine the impact caused by the comet striking the planet as a deformation of the planet. That impact can be seen as proportional to the momentum of the comet which is
 * $\displaystyle p = m * u $.

If someone else watches the crash from a space ship passing by with a remarkable velocity (for example $v = 0.7c$) he will find that the comet is much slower than it is from your resting perspective on the planet. This is due to the time dilatation, which is
 * $\displaystyle t' = t * \sqrt{1 - \frac {v^2}{c^2}}$

Because the time that the one in the space ship mesures is less, the comet will need more time to cover a certain distance. Thus, its velocity seems smaller from the perspective of the space ship (Note: The space ship's trajectory be perpendicular to the comet's trajectory towards the planet, so there is no length contraction parallel to the trajectory of the comet). The velocity measured from the perspective of the planet is:
 * $\displaystyle u = \frac {ds}{dt}$

where the velocity that is mesured from the space ship is:
 * $\displaystyle u' = \frac {ds}{dt'}$

and as we know from the time dilatation, the term for $u'$ is thus:
 * $\displaystyle u' = u * \sqrt{1 - \frac {v^2}{c^2}}$

The observer in the space ship will nevertheless find out, that the impact is equal to the one observed by the resting person. The comet's momentum from the perspective of the space ship is:
 * $\displaystyle p' = m' * u' $

And because the measured momentums from both observers are the same, you can write:
 * $\displaystyle p = p'$
 * $\displaystyle m * u = m' * u' $
 * $\displaystyle m' = m * \frac {u}{u'} $
 * $\displaystyle m' = m * \frac {u}{u * \sqrt{1 - \frac {v^2}{c^2}}} $
 * $\displaystyle m' = \frac {m}{\sqrt{1 - \frac {v^2}{c^2}}}$

Note
Einstein himself suggested in later years that it may be misleading to consider the concept of the relativistic mass. He suggested that it may be better to apply the Lorentz Factor to the momentum instead.