Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

all Lorentz observers watching a particle move will compute the same value for the quantity

$$ds^2 = -(c \, dt)^2 + dx^2 + dy^2 + dz^2,$$ $$ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu},$$ and ''ds/c'' is then an infinitesimal proper time. For a point particle not subject to external forces (''i.e.,'' one undergoing inertial motion), the relativistic action is:

$$S = -m_oc \int ds.$$

Remark: An edit to the comment. we can write a Lagrangian for a relativistic particle, which will be valid even if the particle is traveling close to the speed of light. To preserve Lorentz invariance, the action should only depend upon quantities that are the same for all (Lorentz) observers.

share|cite|improve this question
up vote 4 down vote accepted

The action

$$S= - E_0 ~ \Delta \tau $$

of a relativistic massive particle is minus the rest energy $E_0=m_0c^2$ times the change $\Delta \tau=\tau_f-\tau_i$ in proper time.

share|cite|improve this answer
brilliant!, this is better – Neo Nov 25 '12 at 6:41

You do not explain what do you mean by "better answer" but an alternative form is obtained by using $ds = \gamma^{-1} cdt$ with $\gamma$ the time-dilation factor. Then $$S = -mc \int ds = \int - \frac{mc^2}{\gamma} dt$$ from where you can obtain the Lagrangian, recalling that $S = \int L dt$.

share|cite|improve this answer
isn't m relativistic mass? – Neo Nov 23 '12 at 21:02
$m$ is mass. Relativistic mass is given by $m_\mathrm{rel}=m\gamma$. – juanrga Nov 24 '12 at 0:07
beside that you take the gamma factor but it is inverse $\gamma^{-1} $ – Neo Nov 24 '12 at 6:40
You are right. Mistake corrected. Thank you! – juanrga Nov 24 '12 at 11:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.