Let m₀ be the rest mass for a particle. If its velocity is v then its relativistic mass m is

m = m₀/(1 − β²)^½

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

The total energy of the particle is its kinetic energy K(v) plus its potential energy V(x). The kinetic energy component of the Hamiltonian function is expressed as a function of its momentum. Relativistic kinetic energy is K(v) = (m − m₀)c²

So far everything is as expected. The problem is that relativistic momentum in the direction of travel is not simply mv. Instead it is

p = mv/(1 − β²) = m₀v/(1 − β²)^3/2
or, equivalently
p/c = m₀β/(1 − β²)^3/2

For the details on this matter see Relativistic Momentum

The solution for β proceeds as follows.

(p/c)²(1 − β²)³ = m₀²β²

This is an equation in β². Let

γ = (1 − β²)
and hence
β² = 1 − γ

The equation to be solved is then the cubic

(p/c)²γ³ = m₀²(1 − γ)

In standard form this is

γ³ + (m₀c/p)²γ − (m₀c/p)² = 0

Such an equation is called a depressed cubic equation. The quantity (m₀c/p) is intriguingly interesting. It is a dimensionless number in the nature of a ratio of momenta.

One real solution to

z³ + az + b = 0
is
z = (−b/2 + ((b/2)² + (a/3)³)^½)^⅓ + (−b/2 − ((b/2)² + (a/3)³)^½)^⅓

The equation for γ is such that b=−a. The above solution reduces to

z = (a/2 + ((a/2)² + (a/3)³)^½)^⅓ + (a/2 − ((a/2)² + (a/3)³)^½)^⅓

With a equal to (m₀c/p)²

γ = ((m₀c/p)² /2 + (((m₀c/p)² /2)² + ((m₀c/p)² /3)³)^½)^⅓ +
((m₀c/p)² /2 − (((m₀c/p)² /2)² + ((m₀c/p)² /3)³)^½)^⅓

The kinetic energy of the particle is

K(p) = m₀c²(1/γ(p)^½ − 1)

The relativistic Hamiltonian function H(p, z) for a particle in a potential field V(x) is then given by

H(p, z) = m₀c²(1/γ(p)^½ − 1) + V(x)