Why is the action integral of relativity particles $S = -mc\int ds$? [duplicate]

Question

In my classical mechanic course material, it states that

(In context of relativity) The path of a particle is called its "world line". Each world line can be noted mathematically using the parametric equation $x^\mu=x^\mu(\tau)$. Where $x^\mu$ is the position four-vector and $\tau$ is a Lorentz invariant. Symmetricity shows that the action integral of such particle can only be $$S = -mc\int ds = -mc \int d\tau \left(\frac{dx^\mu}{d\tau}\frac{dx_\mu}{d\tau} \right)^{1/2}$$

My questions are:

1. What is this symmetricity that the paragraph is talking about?
2. What is the Lagrangian here?
3. Why is the action integral given as $S = -mc\int ds$? As most action I have encountered are in the form of $\int L(q,\dot{q},t) \mathrm{d}t$

Answer 1

1. By symmetricity the course material is apparently pointing out that (up to a multiplicative constant) the only$^1$ (local) Lorentz-invariant & coordinate-independent quantity associated with a (possibly virtual) world-line $$\lambda~\mapsto~ x^{\mu}(\lambda) $$ between 2 spacetime events is the arc length $\Delta s=c\Delta \tau$.

2. Hence it is natural to chose the action functional $$S[x]~=~\int_{\lambda_i}^{\lambda_f} \!d\lambda~L(x,\dot{x})$$ to be the elapsed proper time $\Delta \tau$ (up to a multiplicative constant), cf. e.g. this related Phys.SE post. Here $\lambda$ is a world-line parameter.

3. The Lagrangian is $$L(x,\dot{x})~=~ -mc\sqrt{\dot{x}^2},\qquad \dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}, \qquad \dot{x}^{\mu}~:=~\frac{dx^{\mu}}{d\lambda},$$ with signature $(+,-,-,-)$,

--

$^1$ We are excluding possible higher-order invariants not suitable for a first-order action principle.

Answer 2

The canonical way can be checked on WIKI. Here I would like to show you a hand-wavy solution:

First, we anticipate our action to be relativistically invariant, so dt, the integral element of non-relativistic action, is invalid, the best as well as the simplest supplant is d\tau, the so-called proper time.

Then in the context of special relativity, we naively suppose particles are moving inertially, and the velocity can hence be settled as 0 after a linear coordinate transformation leaving the action invariant. Hamiltonian, as well as Lagrangian, is, as relativistic mechanics, mc^2.

Adding them up to an extra conventional minus sign is the solution.