2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

Comments to the question (v1):

  1. 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 $\gamma$.

  2. In particular, it is not assumed that $\tau_i$ and $\tau_f$ are kept fixed during the variation.

  3. 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.

  4. 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.

  5. This functional (1) does depend on the path $\gamma$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Ryan Unger Jan 14 '15 at 20:09
  • 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 '15 at 20:31
  • $\begingroup$ 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. $\endgroup$ – Ryan Unger Jan 14 '15 at 20:39
  • 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 '15 at 20:47
  • $\begingroup$ 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. $\endgroup$ – Ryan Unger Jan 14 '15 at 21:00

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.