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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.

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Could you state your question more clearly, please? –  Claudius Nov 23 '12 at 20:27
Still not getting what the question is - the title is a question, but the body gives the answer –  twistor59 Nov 23 '12 at 20:40
@twistor59 im looking for better answer –  Neo Nov 23 '12 at 20:49

2 Answers 2

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.

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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 $S = \int L dt$.

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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

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