In a mechanical system, there ∃ an integral S, the *action*, which has an extremum for the actual classical motion. This means that the variation δS for the classical path is equal to 0.
We seek first the action for a free material particle (not suffering external forces). The action cannot depend on our choice of sor. In other words, it must be a Lorentz invariant. Why such requirement? Because if it would depend on our choice of reference system, for some systems the action would be an extremum while for others not, which would contradict the principle. In each sor, the action for which δS=0 must give the classical path.
Since the action is a Lorentz invariant, it must depend on a scalar. Or perhaps on combinations of tensors with components that are not dependent on the sor, like those of the δ tensor.
The integrand must be a differential of first order. Why?
The only scalar of this kind for a free particle is ds or b·ds with b being a Lorentz invariant constant which must characterise the particle. We can write
S = -b∫{ds;a;b},
where the integral is along a world line from event a to event b. The time for event a is t₁ and the time for event 2 is t₂.
Since we have explicitly forced S to reach a minimum and since we have done this though an explicit minus sign, b must be a positive constant.
We also know that ∫{ds;a;b} has a maximum value along a straight world line and that it diminishes as the path departs from being straight (it is the proper time read by the travelling clock). But the travelling clock will not give a negative reading, so the integral must always be positive.
The action can be represented as the time integral of the *Lagrangian* between t₁ and t₂,
S = ∫{Ldt;t₁;t₂}.
Combining the two expressions up to now, and knowing that ds = c·dt·√, we arrive to
S = -∫{b·c·√·dt;t₁;t₂}.
Here, √ is (1-v²/c²)¹⸍², where v is the velocity of the particle. So the Lagrangian is
L = -b·c·√.
We need to find what is b. In classical mechanics, what characterises a particle is its mass. Plus the mass is constant, positive and Lorentz invariant. We can go to the classical limit, for which √ expands as (binomial theorem) ≃ 1 + v²/(2c²). This means that
L = -bc√ ≃ -bc·(1 + v²/(2c²)) = -bc + b·v²/(2c).
Constant terms like -bc don't affect the equations of motions, so we can forget about it. And we know that the classical Lagrangian for a free particle is L=m·v²/2, so we obtain b=m·c. Finally, we can write
S = -mc∫{ds;a;b} = -mc²Δτ(a,b)
and the Lagrangian is
L = -mc²√ = -mc²·(1-v²/c²)¹⸍².