Trying to understand relativistic action of a massive point particle

Question

I got badly lost in derivation of relativistic formulas for energy and momentum.

I stumbled upon relativistic action as follows (which should explain relativistic motion of a classical particle):

$$ S = \int Cds=C\int_{t_i}^{t_f}\sqrt{c^2-(x')^2}dt $$

Where $C$ is some constant (depends on what kind of physics we put in equation) and $s$ is relativistic interval. Later on Lagrangian $$ L(x')\equiv C\sqrt{c^2-(x')^2} $$ is used in derving relativistic energy and momentum.

I am familiar with Lagrangians and symmetry rules which connect energy and momentum to Lagrangian formalism. What I do not understand is this action - weren't action sum over time? Why all of a sudden it is sum over relativistic interval?

Answer 1

The action is commonly written in terms of $ds$ because it is a Lorentz scalar. $dt$ is not a Lorentz scalar, but $dt \sqrt{1 - v^2} = dt / \gamma = ds$ is, so you can write the action as an integral over time if you want: $$ S = - m \int ds = - m \int dt \sqrt{1 - v^2} $$ We can check this in the non-relativistic limit: $$ S = - m \int dt \sqrt{1 - v^2} \approx \int dt \left(-m + \frac{1}{2} m v^2 \right) $$ The constant $-m$ does not affect the equations of motion, so it can be removed, and we are left with $L = m v^2 / 2$, as expected.

The speed of light $c=1$ above.

Answer 2

The action is a time integral, just as you wrote. However, it's also a (line) integral of proper distance. This form is convenient when you make the jump to GR, because $ds$ has an obvious generalization whereas $\sqrt{c^2-x'^2}$ is less obvious.