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.

  • 4
    $\begingroup$ Hint: Sum over particles. $\endgroup$
    – Qmechanic
    Jan 31 at 13:17
  • $\begingroup$ Thanks. I edited my question so that my confusion is clearer. $\endgroup$
    – watahoo
    Jan 31 at 20:35
  • $\begingroup$ 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. $\endgroup$
    – watahoo
    Feb 1 at 0:26


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.