5
$\begingroup$

The geodesic equation is derived from the Euler-Lagrange equation, which (as I understand it) is a necessary but not sufficient condition to ensure that the geodesic is a minimum.

The introductory GR books I have looked at do not bother with sufficiency.

How do we know for sure that the geodesic is a minimum and not a maximum or inflection?

$\endgroup$
1
  • 5
    $\begingroup$ Why do you believe that the geodesic is always a minimum and never a maximum or an inflection? $\endgroup$ Dec 26, 2011 at 16:52

2 Answers 2

12
$\begingroup$

In general, the geodesics are "stationary points" which means that they satisfy the appropriate Euler-Lagrange equations. This condition is more fundamental than being a minimum or maximum or a saddle point or some marginal situation involving inflections. Being a minimum etc. is a combination of the fundamental condition specified by the Euler-Lagrange equations; and some kind of an inequality for the matrix of second functional derivatives.

Geodesics in spaces with the positively definite Euclidean signature – not spacetimes! – will tend to be minima of the proper length.

In the Minkowski spacetime, almost all timelike geodesics are actually maxima rather than minima of the proper time measured along the geodesics. That's easy to see: any other timelike curve connecting the two end points is closer to being null (whose proper time is zero), so its proper time is shorter than the time along the straight path (the twin who stays home in the twin paradox will be older). This gets generalized to the curved Minkowski-signature spacetime, too: one can see that if he divides the long curved path to many pieces because each piece de facto follows the same rules as the paths in a flat space.

However, spacelike geodesics are saddle points. Perturbations of the geodesic in the transverse spacelike directions make the proper length longer while perturbations of the geodesic in the timelike transverse direction are shorter.

Null geodesics can't really be uniquely defined as maxima or minima at all: any null curve connecting the same two end points (and there are infinitely many) has the same length, namely zero, while both timelike and spacelike curves are "longer" although the unit of length differs by a factor of $i$. Relatively to some variations, null geodesics will probably be inflections. In that case, the appropriate Euler-Lagrange equations are the only simple way to describe what's special about a geodesic relatively to nearby paths.

The paragraphs above were about "minima vs maxima vs saddle points". There's one more question, whether the stationary points may be global minima or maxima. Saddle points can't be "global" so this possibility is only relevant for timelike geodesics. In many cases "close" to the straight lines in a flat spaces and comparable situations in curved space, the geodesic is a global maximum of the proper time along the curve. In general enough spacetimes, nothing like that can be said.

$\endgroup$
1
  • $\begingroup$ Thanks. I see I had a somewhat misconceived notion of what the geodesic is. $\endgroup$
    – ben
    Dec 27, 2011 at 16:13
0
$\begingroup$

A geodesic is the curved-space generalization of the notion of a straight line in Euclidean flat space (the one of a lifetime). We all know what a straight line is: it's the path of shortest distance between two points.

We know that massive particles move on timelike paths. So in name of simplicity let's consider timelike paths, where the proper time functional is defined by $$\tau=\int \left(-g_{\mu \nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}\right)^{1/2}d\lambda$$ where the integral is over the path parametrized by $x^{\mu}(\lambda)$. To search for shortest-distance paths, we could do the usual calculus of vaiations treatment to seek critical points of this functional. They will turn out to be de curves of maximum proper time, consistent with the discussion of the twin paradox (moving clocks run slow, i.e. the person who stays on Earth will be older than the person who arrives from space travel). If you do the computing, we see that we obtain precisely the geodesic equation, but with the specific choice of Christoffel connection, $$\frac{d^2x^{\rho}}{d\tau^2}+\frac{1}{2}g^{\rho\sigma}(\partial_\mu g_{\nu\sigma} + \partial_\nu g_{\sigma\mu} - \partial_\sigma g_{\mu\nu})\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=0$$ or $$\frac{d^2x^\rho}{d\tau^2}+\Gamma^\rho_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=0$$

Now, we define the interval $ds^2=-d\tau^2$, that is $$ds^2=g_{\mu\nu}dx^\mu dx^\nu$$ then, it is straightforward to note that geodesic minimizes the interval. If you consider a Euclidean flat space, the the geodesic reduces to the shortest path between two points, that is, the geodesic minimizes the distance between these points.

In summary, geodesics are critical points of some functional (either iterval or proper time in this case), which can maximizase (in the case of timelike paths) or minimizase (in the case of spacelike paths), all this in the context of general relativity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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