# Are all null curves of a Lorentzian metric extrema?

Suppose $g$ is a Lorentzian metric on $\mathbb{R}^4$, then consider the variational problem of finding extrema of

$$F(\gamma) = \int_a^b \| \dot{\gamma}(s) \|_g ds$$

The so-called "null curves" are defined as curves for which the integrand $\| \dot{\gamma}(s) \|_g$ is zero, i.e. they have zero "speed" measured by $g$.

Are such curves necessarily extrema of $F$?

NOTE: I changed the wording of the question: I replaced "null geodesic" with "null curve"

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You should read the definition of null geodesic before asking. –  Anton Petrunin Mar 8 '13 at 17:34
As far as I understand a "null geodesic" is a curve for which $\| \dot{\gamma}(s) \|_g = 0$. Is that not the case? –  user7807 Mar 8 '13 at 17:38
I guess that's technically only a null curve. I guess I should re-phrase my question as to whether any null curve is a null geodesic... –  user7807 Mar 8 '13 at 17:40
Clearly not every null curve is a null geodesic. –  José Figueroa-O'Farrill Mar 8 '13 at 17:41
OK, thanks! I don't know, I'm 100% far from an expert on this, but it came up in a discussion today –  user7807 Mar 8 '13 at 17:43
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Actually, your notation is causing some confusion. In one very real sense (probably not your intended one) the answer to your question is yes, not no which is probably the answer to the question that you intended to ask.

It depends on what you mean by $\|\dot\gamma(s)\|_g$. If you borrow this notation directly from the standard notation in Riemannian geometry, where $$\|\dot\gamma(s)\|_g = \sqrt{\langle \dot\gamma(s),\dot\gamma(s)\rangle_g}\ ,$$ what this often means in Lorentzian geometry is that you only consider curves $\gamma$ as admissable if this makes sense, i.e., only if $\langle \dot\gamma(s),\dot\gamma(s)\rangle_g\ge0$. Unfortunately, there is also a different convention in which $$\|\dot\gamma(s)\|_g = \sqrt{-\langle \dot\gamma(s),\dot\gamma(s)\rangle_g}\ ,$$ and one only considers curves $\gamma$ as admissable if this makes sense, i.e., only if $\langle \dot\gamma(s),\dot\gamma(s)\rangle_g\le0$. (It depends on the sign conventions one chooses, and, unfortunately, there seem to be as many sign conventions in use as there are choices of sign, each defended by its practitioners as the only sane convention, but that's another story.)

At any rate, if you adopt one of the above conventions and you only consider admissable curves (which is standard in the Calculus of Variations), then, yes, all null curves are extrema, almost by definition; it's not a question of geodesics, which is a different matter.

Sad to say, there is a third convention in pseudo-Riemannian geometry. Some books and papers define $$\|\dot\gamma(s)\|_g = \langle \dot\gamma(s),\dot\gamma(s)\rangle_g\ .$$ (They don't like the square that you would expect to see based on common mathematical convention because $\|\dot\gamma(s)\|_g^2$ suggests that the quantity is nonnegative, so they break the conventional distinction between 'norm' and 'inner product' instead. This often leads to dreadful confusion when one tries to read the papers and frequent mistakes involving the so-called Cauchy inequality, but that's another story, too.)

If that is your convention, then the answer to your question is no, because, in this case, all sufficiently smooth curves are admissable, and all the extrema are geodesics. However, in dimensions higher than $2$ (in particular, in dimension $4$), not all null curves are geodesics, as the above commenters have noted.

Oh, one more (pedantic, I'm afraid) comment: While it's true that all extrema are geodesics in this latter convention, I'm sure you know that not all geodesics are extrema (they are always critical points of the functional, but 'extrema' is a stronger condition). Moreover, in pseudo-Riemannian geometry, null geodesics (and, a fortiori, null curves) are never extrema, because you can always perturb the curve slightly to make the functional either positive or negative.

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