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?
1 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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$\begingroup$ I had suspected 1. and 2. But this raises another question. Can we apply the standard Euler-Lagrange methods to an integral where the bounds change? If not, then isn't Carroll's derivation (Spacetime and Geometry, p. 107.) of the geodesic equation false? $\endgroup$ Jan 14, 2015 at 20:09
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1$\begingroup$ @0celo7 : Carroll's derivation is OK in principle. Although: 1. He seems to forget to mention that the parameter endpoints $\lambda_i$ and $\lambda_f$ are kept fixed 2. He confusingly calls the parameter $\lambda$ for $\tau$. The parameter $\lambda$ only has an interpretation as proper time for the stationary path in the massive case, not for arbitrary paths. $\endgroup$– Qmechanic ♦Jan 14, 2015 at 20:31
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$\begingroup$ Eq. 3.48 is $\delta\tau=-\tfrac{1}{2}\int\delta f d\tau$. What are the bounds on that integral? Presumably, in Eq. 3.45, the bounds on the $\lambda$ integral are something like $\lambda_0$ and $\lambda_1$. Also, I've never quite understood how we can vary $f$ when it is constant. $\endgroup$ Jan 14, 2015 at 20:39
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1$\begingroup$ 1. Yes. 2. $f$ depends on the path. Moreover, $f$ is not constant as a function of $\lambda$ for arbitrary paths in general. (Btw. note that Carroll [from eq. (3.49) on] replaces the integrand $\sqrt{-f}$ with the square. The squaring trick is discussed in e.g. this Phys.SE post.) $\endgroup$– Qmechanic ♦Jan 14, 2015 at 20:47
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$\begingroup$ If $\lambda$ is affine, then isn't $f$ constant in general? How does Eq. 3.49 work? The integrand is just -1/2. I've always seen the energy functional used with an arbitrary parameterization and then proven that the parameterization must necessarily be affine (i.e. prove $\dot x^2=$const). Using proper time from the start doesn't seem to make any sense. $\endgroup$ Jan 14, 2015 at 21:00