This is something that has been bothering me for a little while. The usual procedure that I've seen is to write the proper time as the line integral $$\tau=\int_\gamma d\tau$$ along some curve $\gamma$. That curve that minimizes this is a geodesic (assuming Levi-Civita connection here). The definition $d\tau^2=-g_{\mu\nu}dx^\mu dx^\nu$ leads to $$\frac{d\tau}{d\lambda}=\sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\mu}{d\lambda}}.$$ So, using the standard rule for doing line integrals, we have $$\int_\gamma d\tau=\int_{\lambda_i}^{\lambda_f}\sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\mu}{d\lambda}}d\lambda,$$ where $\gamma(\lambda_i),\gamma(\lambda_f)$ are the beginning and end points of the curve. It can be easily verified that this is parameterization invariant. It is standard to normalize $\gamma$ $$g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-1.$$ Thus it would seem that $$\int_\gamma d\tau=\int_{\lambda_i}^{\lambda_f}\sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\mu}{d\lambda}}d\lambda=\int_{\tau_i}^{\tau_f}d\tau=\tau_f-\tau_i.$$ Thus $\gamma$ does not even play a role it seems. Thus when we introduce a one-parameter family $\{\gamma_\varepsilon\}$ and take the variational derivative, we get 0 arbitrarily. What gives? How can we vary this integral without changing the bounds?
Answer
Comments to the question (v1):
Note first of all that there isn't a canonical/unique prescription to assign values to $\tau_i$ and $\tau_f$ for a given path $\gamma$.
In particular, it is not assumed that $\tau_i$ and $\tau_f$ are kept fixed during the variation.
In contrast, the parameter endpoints $\lambda_i$ and $\lambda_f$, and the path endpoints $\gamma_i$ and $\gamma_f$, are kept fixed during the variation.
Only the difference $\Delta\tau = \tau_f - \tau_i$ is important. In fact $$\tag{1}\Delta\tau~=~\int_{\lambda_i}^{\lambda_f} \!d\lambda ~\sqrt{-g_{\mu\nu}(\gamma(\lambda)) ~\dot{\gamma}^{\mu}(\lambda)~\dot{\gamma}^{\nu}(\lambda)}$$ is the proper time of the path and precisely the functional that we want to vary.
This functional (1) does depend on the path $\gamma$.
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