Why is integral of relativistic action $-\alpha \int_{a}^{b} \, \mathrm ds$ minimised with respect to $\mathrm ds$?

Question

While reading some aspects concerning the conclusion that bodies follow geodesics of spacetime, I ran into relativistic action on p.24 in chapter 2 $\S8$ of the 2nd volume of Landau & Lifshitz: $$S = -\alpha \int_{a}^{b} \, \mathrm ds, \tag{$\ast$} \label{eq:1}$$ where $\alpha$ does represent a positive constant to be determined by comparison with the classical Lagrangian of a free particle (as a side note, the minus sign is essential for the action to have a minimum). My question is: why is integral $\eqref{eq:1}$ minimised with respect to $\mathrm ds$? One clever justification I have read (in Landau & Lifshitz) is that the action must be invariant with respect to Lorentz transformations. The problem is that I don't understand why it is precisely $\mathrm ds$ that has been chosen as scalar.

One possible explanation I tried to give (about whose correctness I am not at all sure) is that, choosing $\mathrm ds$ as scalar and expressing the latter as a function of $\mathrm dt$, and after finding the value of $\alpha$ by comparison with the classical Lagrangian, we obtain the relativistic Lagrangian. Once the latter is derived as a function of velocity, the particle's impulse is obtained. This value could be obtained by other means without involving the Lagrangian and, therefore, retracing the path backwards - and considering the known impulse of the particle - we obtain the action as per $\eqref{eq:1}$. Does what I say make sense or am I wrong and is there a deeper reason?

Answer 1

My question is: why...

We can not really answer "why" nature chooses to do this.

But, if you are looking for some intuitive justification, you can compare your worldline integral to an ordinary path length integral in real space. Minimal path length corresponds to a straight line. So, we might suspect that a "straight line" in space-time is the simplest trajectory and therefore we might be motivated to consider minimizing $$ \int ds\;. $$

Also, $\int ds$ has the virtue of being invariant to Lorentz transformations, analogous to how the path length of a line in real space is invariant under rotations.

One clever justification I have read (in Landau & Lifshitz) is that the action must be invariant with respect to Lorentz transformations.

This is not really a justification, but is more of something we want to be true in our theory, a priori. This is motivation to choose the form $\int ds$, not an explanation of why it is minimized by nature.

The problem is that I don't understand why it is precisely $\mathrm ds$ that has been chosen as scalar.

An invariant space-time length is chosen because we want our physics to not depend on the frame of reference.

One possible explanation I tried to give (about whose correctness I am not at all sure) is that, choosing $\mathrm ds$ as scalar and expressing the latter as a function of $\mathrm dt$, and after finding the value of $\alpha$ by comparison with the classical Lagrangian, we obtain the relativistic Lagrangian. Once the latter is derived as a function of velocity, the particle's impulse is obtained. This value could be obtained by other means without involving the Lagrangian and, therefore, retracing the path backwards - and considering the known impulse of the particle - we obtain the action as per $\eqref{eq:1}$. Does what I say make sense or am I wrong and is there a deeper reason?

It seems to me like your reasoning is backwards. We seek an invariant form for the action. Thus we choose $$ S=\alpha \int ds\;, $$ which is invariant. This is very simple and is also invariant, so we are pleased with ourselves for this choice.

But we do not know the value of $\alpha$, so we take the non-relativistic limit and use our knowledge that $$ S_{NR}=\int dt \frac{1}{2}m\dot x^2\;, $$ is well-known to correctly generate the equations of motion for a non-relativistic free particle. This lets us fix $\alpha=-mc^2$.

Thus, we assert that $$ S=-mc^2\int ds $$ is the correct relativistic action for a free particle. And we are pleased with ourselves for our elegance and simplicity.