1
$\begingroup$

The relativistic action is

$$ S=- m \int_a^b d s. $$

With metric $ds^2=dx^2 - dt^2$, we get:

$$ \begin{align} S&=\pm m \int_a^b \sqrt{dx^2-dt^2}\\ &=\pm mc\int_a^b dt\sqrt{\left(\frac{dx}{dt}\right)^2-1}\\ &=\pm mc\int_a^b dt\sqrt{\dot{x}^2-1}. \end{align} $$

Developing $\sqrt{\dot{x}^2-1}$ as a series around $\dot{x}^2$, we get:

$$ \sqrt{\dot{x}^2-1}=i - \frac{i \dot{x}^2}{2} + O[\dot{x}]^4. $$


If we had instead taken the metric to be $ds^2=dt^2-dx^2$, we would have obtained the usual non-relativistic limit:

$$ \sqrt{1-\dot{x}^2}=1 - \frac{ \dot{x}^2}{2} + O[\dot{x}]^4. $$

So, why is $ds^2=dx^2 - dt^2$ "incorrect", but $ds^2=dt^2-dx^2$ "correct"? What does the imaginary term mean?

$\endgroup$
  • $\begingroup$ You can always multiply the action by a constant, so there is no difference. $\endgroup$ – knzhou Sep 26 '19 at 18:48
  • $\begingroup$ The trajectory of a massive point particle is always timelike $dt^2-dx^2>0$. $\endgroup$ – Qmechanic Sep 26 '19 at 18:54
  • $\begingroup$ @Qmechanic: The OP is using the -+++ metric. $\endgroup$ – user4552 Sep 26 '19 at 23:17
  • $\begingroup$ @knzhou: That should be an answer. $\endgroup$ – user4552 Sep 26 '19 at 23:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.