Variational Principle for Free Particle Motion (Relativistic)

Question

This is the same problem as someone asked before: Problem understanding something from the variational principle for free particle motion (James Hartle's book, chapter 5)

The question below:

Here is the problem. First, we have $$L = \biggl[\biggl(\frac{dt}{d\sigma}\biggr)^2 - \biggl(\frac{dx}{d\sigma}\biggr)^2 - \biggl(\frac{dy}{d\sigma}\biggr)^2 - \biggl(\frac{dz}{d\sigma}\biggr)^2\biggr]^\frac{1}{2} \tag{5.57}$$ then, $$-\frac{d}{d\sigma}\biggl(\frac{\partial L}{\partial(dx^\alpha/d\sigma)}\biggr) + \frac{\partial L}{\partial x^\alpha} = 0 .\tag{5.58}$$ And considering that $x^1 = x$ we conclude that $$\frac{d}{d\sigma}\biggl[\frac{1}{L}\frac{dx^1}{d\sigma}\biggr] = 0. \tag{5.60}$$

I can't understand how eq (5.60) came. I don't know how he got this equation. I tried solving this, but I can't get the equation. Even a hint on how to proceed will be appreciated.

Answer 1

I hope the notation of writing $L = \sqrt{\eta_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}$ is clear as this makes the following calculations very compact.

\begin{equation} \frac{\partial L}{\partial \dot{x}^{\alpha}} = \frac{1}{2}\left(\eta_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}\right)^{-1/2}\times 2\dot{x}^{\alpha} = \frac{\dot{x}^{\alpha}}{L} \end{equation} The above is just a consequence of chain rule. To be explicit the Minkowski metric was introduced $\eta_{\mu\nu} = diag(1,-1,-1,-1)$ and $\dot{x}^{\alpha}= \frac{d x^{\alpha}}{d \sigma}$. Also $\frac{\partial L}{\partial x^{\alpha}} = 0$. The answer they proved follows.