# Is there a Lorentz invariant action for a free multi-particle system?

I want to write down a Lorentz-invariant action of free multi-particle systems.

I know that a Lorentz-invariant action for each particle might be expressed as $$S[\vec{r}]=\int dt L(\vec{r}(t),\dot{\vec{r}}(t),t)=-\int dt\, m\sqrt{c^2-\Big(\frac{d\vec{r}(t)}{dt}\Big)^2}$$ or $$S[r]=-\int d\sigma\, m\sqrt{\frac{dr^\mu(\sigma)}{d\sigma} \eta_{\mu\nu} \frac{dr^\nu(\sigma)}{d\sigma}}=-\int m c d\tau.$$ In the second line, I parametrized the four vector $$r^\mu$$ including time $$t$$ by a parameter $$\sigma$$. The last expression assumes $$\sigma$$ coincides with the proper time $$\tau$$.

I tried to extend this analysis to (noninteracting) multi-particle systems. If we define $$r_i^\mu(\sigma)=\begin{pmatrix} c t(\sigma)\\ x_i(\sigma)\\ y_i(\sigma)\\ z_i(\sigma)) \end{pmatrix}\quad...(*)$$ ($$i$$ is the label distinguishing particles), we may write $$S[r]=-\int d\sigma\, \sum_{i=1}^Nm_i\sqrt{\frac{dr_i^\mu(\sigma)}{d\sigma} \eta_{\mu\nu} \frac{dr_i^\nu(\sigma)}{d\sigma}}=-\sum_{i=1}^N\int m_i c d\tau_i.$$ To go to the last expression, $$\sigma$$ is chosen as the proper time for each particle. Indeed, this expression is in Landau's classical theory of fields textbook as Eq (27.2).

My problem is that $$t$$ in $$(*)$$ is common among all particles. Thus, the Lorentz transformation $$r_i'{}^\mu(\sigma)=\begin{pmatrix} c t'(\sigma)\\ x_i'(\sigma)\\ y_i'(\sigma)\\ z_i'(\sigma)) \end{pmatrix}=\Lambda^\mu_{\,\,\nu}r_i^\nu(\sigma)$$ for different $$i$$ gives mutually contradicting expressions for $$t'$$. This is why I couldn't verify the Lorentz invariance of the second last expression. Should we introduce different $$t'$$ for each particle? (i.e., $$t_i'$$).

I looked up Goldstein Sec 7.10 but it does not really give me an answer.

• Hint: Sum over particles. Jan 31 at 13:17
• Thanks. I edited my question so that my confusion is clearer. Jan 31 at 20:35
• I think I found a solution. As $t$ was just an integration valuable, I could have set $t\to t_i$ from the begging, and then it is natural to write $t_i'$ after Lorentz transformation. Thanks to different $t_i$ and $t_i'$ for each particle, there is no contradiction anymore. Thanks again for your help. Feb 1 at 0:26