# Physical intuition for the Geodesic equation derivation via Action Principle

The most commom derivation I've seen of the geodesic equation of a massive particle is by the use of the Variational Principle. My problem is that I can't realize what the meaning of find a spacetime path (the geodesic) such that the proper time is extremized. (If the signature is $$(+,-,-,-)$$ it should be a maximum as some textbooks say.)

I understood that the action integral must be proportional to the line element $$ds$$ because we need that all the observers compute the same value of action to obtain the same equations of motion.

What I don't understand is the physical meaning of finding a maximum proper time instead a minimum, and what physical implications it leads to. How can I conclude that what I need to find a geodesic equation is maximize proper time of the massive particle? If possible, make an analogy with the Minkowski Space.

1. Let us for simplicity consider Minkowski space although the generalization to curved spacetime is straightforward. Lorentz invariance suggests that the Lagrangian one-form for a massive point particle should be $$\mathbb{L}~=~ f(\dot{x}^2)\mathrm{d}\lambda, \qquad \dot{x}^2~:=~\eta_{\mu\nu} \dot{x}^{\mu}\dot{x}^{\nu}~>~0, \qquad \dot{x}^{\mu}~:=~\frac{dx^{\mu}}{d\lambda}, \qquad x^0~\equiv~ct, \tag{1}$$ for some function $$f$$. Here $$\lambda$$ denotes a world-line (WL) parameter, and a dot denotes differentiation wrt. $$\lambda$$. We chose Minkowski signature $$(+,-,-,-)$$ so that timelike vectors have positive length.

2. WL reparametrization invariance implies that the function $$f~\propto ~\sqrt{\cdot}\tag{2}$$ is proportional to a square root.

3. In other words, the variational principle finds stationary paths for timelike arc-length (or equivalently, proper time multiplied with $$c$$): $$c\tau~=~\int_{\lambda_i}^{\lambda_f}\!\sqrt{\dot{x}^2}\mathrm{d}\lambda. \tag{3}$$ The corresponding Euler-Lagrange (EL) equations are the geodesic equations. In Minkowski space the geodesics are just straight lines.

4. Let us impose boundary conditions (BCs) $$x(\lambda_i)~=~x_i\qquad\text{and}\qquad x(\lambda_f)~=~x_f. \tag{4}$$ By changing coordinate system, we may assume that $${\bf x}_i={\bf x}_f.$$ Finally let us choose static gauge $$\lambda=t$$.

5. Then eq. (3) becomes $$c\tau~=~\int_{t_i}^{t_f}\!\sqrt{c^2-\dot{\bf x}^2}\mathrm{d}t. \tag{5}$$ Eq. (5) is clearly maximal for $$\dot{\bf x}={\bf 0}$$, i.e. a particle at rest, which is also what a free particle would do with the given BCs. This answers OP's question.