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The fundamental equation of Classical Newtonian dynamics is

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dp/dt = F

where p is momentum and F is force.
In classical mechanics momentum is m_{0}v, where m_{0} is rest mass and v is velocity.
Thus

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m_{0}(dv/dt) = F

If F is constant then

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v(t) − v_{0} = Ft

## Relativistic Corrections

As is well known Special Relativity requires mass to be adjusted for velocity; i.e.,

####
m = m_{0}/(1−β²)^{½}

where β is velocity relative to the speed of light c; i.e., β=v/c.

What is less well known is that relativistic momentum is not mv, where
m is mass adjusted for velocity. Instead relativistic momentum P is given by

####
P = mv/(1−β²) = m_{0}v/(1−β²)^{3/2}

This can be put into the nondimensional form

####
P/(m_{0}c) = β/(1−β²)^{3/2}

The dynamic equation is then

####
(dP/dt)/(m_{0}c) = d[ β/(1−β²)^{3/2}]/dt = F/(m_{0}c)

For convenience let F/(m_{0}c) be denoted by Φ. The solution is of course

####
β/(1−β²)^{3/2} = Φt

But rather than trying to solve this equation for β it is better to find a numerical solution.
The differential form of the above equation is

####
dβ/(1−β²)^{3/2} + (3/2)β/(1−β²)^{5/2}(2β)dβ = Φdt

or, equivalently

dβ[1/(1−β²)^{3/2} + 3β²/(1−β²)^{5/2}]= Φdt

This expression may be used to find the increment in β resulting from an increment in time.

As can be seen β asymptotically approaches the limiting value of 1.0.

## An Illustration

Suppose an electron is ejected at a velocity 99 percent of the velocity of light from a nucleus of mass 10,000 MeV.
The electron with a rest mass of 0.511 Mev will appear to have a mass of 5.11=0.511/√(1-0.99) MeV. Its momentum
will not be its relativistic mass times its velocity but 100=1/(1-0.99) times that value. Ignoring the neutrino, the recoil velocity
required to counterbalance the momentum of the ejected electron would be roughly 100 times what would be
expected if the proper formula for relativistic momentum is not taken into account.