# Imaginary term solution in the limit $v\ll c$ of relativistic Lagrangian

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?

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