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.


1 Answer 1


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.

  • $\begingroup$ This is not my question, I know that. I am referring to the units of action (energy multiplied with time) $\endgroup$
    – Moonraker
    Jul 13, 2016 at 21:02
  • $\begingroup$ @Moonraker Those units are the same as in nonrelativistic mechanics. First understand these concepts there before moving to relativistic mechanics. $\endgroup$
    – Ryan Unger
    Jul 13, 2016 at 22:23

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