Oh, ok. So, I guess what you meant was that the principle of least action is equivalent to Newton's laws or D'Alambert's principle. This is only true for purely mechanical systems. Nevertheless, the principle of least action is valid for a far wider range of phenomena. For example, as was pointed out above, due to the finite speed of propagation of interactions in Nature, the concept of action at a distance, which is essential in the notion of a potential energy, loses its meaning. One needs to ascribe a field as a continuum with its own degrees of freedom and its own action. The interaction between the particle and the field is then described by another term in the action.
On another note, in the first part of Landau Lifgarbagez, they derive the Lagrangian for a free particle on principle of homogeneity of space and time and isotropy of space and invariance (up to a total differential as I pointed out in #5) of the action differential w.r.t. to infinitesimal Galilean transformations.
Let us follow this route in deriving the Lagrangian but w.r.t. to Lorentz invariance. Again, the Lagrangian has to be a function of the square of the velocity of the particle L = L(v^{2}).
The components of the velocity transform during an infinitesimal Lorentz transformation (with \beta = \epsilon \rightarrow 0):
<br />
v'_{x} = \frac{v_{x} - c \, \epsilon}{1 - \frac{v_{x} \, \epsilon}{c}} = (v_{x} - c \, \epsilon) \, \left(1 + \frac{v_{x} \, \epsilon}{c} + o(\epsilon)\right) = v_{x} - c \, \left(1 - \frac{v^{2}_{x}}{c^{2}} \right) \, \epsilon + o(\epsilon)<br />
<br />
v'_{y} = \frac{v_{y} \, \sqrt{1 - \epsilon^{2}}}{1 - \frac{v_{x} \, \epsilon}{c}} = (v_{y} + o(\epsilon)) \left(1 + \frac{v_{x} \, \epsilon}{c} + o(\epsilon)\right) = v_{y} + \frac{v_{x} \, v_{y}}{c} \, \epsilon + o(\epsilon)<br />
<br />
v'_{z} = v_{z} + \frac{v_{x} \, v_{z}}{c} \, \epsilon + o(\epsilon)<br />
The square of the speed then expands as:
<br />
v'^{2} = v'^{2}_{x} + v'^{2}_{y} + v'^{2}_{z} = v^{2}_{x} - 2 \, c \, v_{x} \, \left(1 - \frac{v^{2}_{x}}{c^{2}}\right) \, \epsilon + v^{2}_{y} + 2 \, \frac{v_{x} \, v^{2}_{y}}{c} \, \epsilon + v^{2}_{z} + 2 \, \frac{v_{x} \, v^{2}_{z}}{c} \, \epsilon + o(\epsilon) = v^{2} - 2 c \, v_{x} \, \left(1 - \frac{v^{2}}{c^{2}}\right) \, \epsilon + o(\epsilon)<br />
Thus, the Lagrange's function has an expansion:
<br />
L' \equiv L(v'^{2}) = L(v^{2}) - 2 c \, \epsilon \, v_{x} \, \frac{\partial L}{\partial v^{2}} \, \left(1 - \frac{v^{2}}{c^{2}}\right) + o(\epsilon)<br />
But, we are not done yet. We also have to transform the time differential up to linear powers in \epsilon:
<br />
dt' = \frac{dt - \epsilon \, \frac{dx}{c}}{\sqrt{1 - \epsilon^{2}}} = dt - \frac{dx}{c} \, \epsilon + o(\epsilon)<br />
Then, for the action differential, we get:
<br />
L' \, dt' = L \, dt - \epsilon \, \left[\frac{L}{c} \, dx + 2 c \, \left(1 - \frac{v^{2}}{c^{2}}\right) \, \frac{\partial L}{\partial v^{2}} \, v_{x} \, dt\right] + o(\epsilon)<br />
The term in the square bracket is a total differential of a function of coordinates and time (remember that v_{x} \, dt = dx along the trajectory of the particle) if and only if:
<br />
\frac{L}{c} + 2 c \, \left(1 - \frac{v^{2}}{c^{2}}\right) \, \frac{\partial L}{\partial v^{2}} = K<br />
where K is a constant independent on the speed of the particle. This can be rearranged as an inhomogeneous first order linear ordinary differential equation (we make the substitution u = v^{2}/c^{2} for the argument, which gives \frac{d L}{d u} = c^{2} \, \frac{\partial L}{\partial v^{2}}):
<br />
2 (1 - u) \, \frac{d L}{d u} + L = c \, K<br />
<br />
\frac{d L}{d u} + \frac{1}{2(1 - u)} \, L = \frac{c \, K}{2 (1 - u)}<br />
The integrating factor is:
<br />
\mu(u) = \exp{\left(\int{\frac{du}{2(1- u)}}\right)} = \exp{\left[-\frac{1}{2} \, \mathrm{Log}{(1 - u)}\right]} = (1 - u)^{-1/2}<br />
Then, the equation becomes:
<br />
\frac{d}{d u}\left((1 - u)^{-1/2} \, L\right) = c \, K \, (1 - u)^{-3/2}<br />
Integrating once:
<br />
(1 - u)^{-1/2} \, L = C_{1} - c \, K \, \frac{(1 - u)^{-1/2}}{(-1/2)}<br />
<br />
L = C_{1} \, (1 - u)^{1/2} + 2 \, c \, K<br />
The additive constants are unimportant in the definition of the Lagrangian. Therefore we may omit the second constant term.
<br />
L = C_{1} \, \sqrt{1 - \frac{v^{2}}{c^{2}}}<br />
Expanding for small speeds (v/c \ll 1), we get:
<br />
L = C_{1} \, \left[1 - \frac{v^{2}}{2 c^{2}} + o\left(\frac{v^{2}}{c^{2}}\right)\right]<br />
Again, we may ignore any constant terms that arise. Comparing this with the expression for the Lagrangian of a free particle (kinetic energy) in classical mechanics:
<br />
L_{\mathrm{cl}} = \frac{m v^{2}}{2}<br />
we conclude that:
<br />
C_{1} = -m \, c^{2}<br />
and we finally get the well known result.
