# What is the intuitive concept of the action of a relativistic point particle? [duplicate]

The action of a relativistic point particle is its negative rest energy along its worldline, the parameter being its own proper time. $$S = - mc^2 \int d\tau$$ (see Wikipedia)

Action is energy multiplied with time. How can I imagine the "action" of an unaccelerated point particle which does not seem to deploy any kind of energy?

Further explanation: An oven may emit a certain number of Jsec in a given time, but not an unaccelerated point particle.

The invariant interval is the object to be extremized. The infinitesimal interval $$ds^2~=~c^2dt^2~-~dx^2~-~dy^2~-~dz^2$$ is also $$ds^2~=~\left[c^2~-~\left(\frac{dx}{dt}\right)^2~-~\left(\frac{dy}{dt}\right)^2~-~\left(\frac{dx}{dt}\right)^2\right]dt^2~=~(c^2~-~v^2)dt^2$$ We may then take the square root of this and multiply by the invariant mass $$mc^2ds~=~mc^2\sqrt{1~-~v^2/c^2}dt$$ Now assume that the velocity is small, and use binomaial theorem $$mc^2ds~\simeq~mc^2\left(1~-~\frac{v^2}{2c^2}\right)dt~=~mc^2dt~-~\frac{1}{2}mv^2dt.$$ We can see that the last term is the kinetic energy with the wrong sign. To correct for this we must then define the action with the negative sign.