Also, an easier way to go about things might be the following:
We're supposed to extremize S = -m∫L(x,v,t)dt = -m∫(1/γ) dt. We can just plug this Lagrangian into the Euler-Lagrange equation. A trick to make this easier is to realize that if a Lagrangian L satisfies the Euler-Lagrange equation, then in general any function of the Lagrangian f(L) also satisfies the Euler-Lagrange equation as long as ∂/∂λ (∂f/∂L)=0, where λ is your parameter (I can't think of an example where this wouldn't be the case, but I included it just to be safe). The reason is simple:
If:
∂L/∂q = ∂/∂λ (∂L/∂(∂q/∂λ))
, then by chain rule we get that:
∂f/∂q = (∂f/∂L)(∂L/∂q) = (∂f/∂L)(∂/∂λ)(∂L/∂(∂q/∂λ)) = ∂/∂λ (∂f/∂(∂q/∂λ))So, instead of using L(x,v,t) = 1/γ, we can use f(L) = L2 = γ-2 = (1-v2):\frac{\partial }{\partial \vec{x}}[1-\vec{v}\cdot \vec{v}]=0
and
\frac{\partial }{\partial t} \frac{\partial }{\partial \vec{v}}[1-\vec{v}\cdot \vec{v}]=\frac{\partial }{\partial t}[-2\vec{v}]=-2\frac{\partial \vec{v}}{\partial t}
So we find that:
\frac{\partial \vec{v}}{\partial t}=0.
This is equivalent to what I did in my last post. They both say that objects will move inertially (not accelerate).
One more thing to point out is that dt/γ = dτ, where τ is proper time. This means that the action S = -m∫dτ = -mΔτ. So extremizing the action is equivalent to extremizing the proper time elapsed. Some call this the principle of extremal proper time, and it generalizes nicely to general relativity. Geodesics can be found in this way, by extremizing ∫gμν(dxμ/dλ)(dxν/dλ)dλ.