I'm reading Caroll's Lectures on GR 2 on pages 71-72, he states:

Let’s now explain the earlier remark that timelike geodesics are maxima of the proper time. The reason we know this is true is that, given any timelike curve (geodesic or not), we can approximate it to arbitrary accuracy by a null curve. To do this all we have to do is to consider “jagged” null curves which follow the timelike one:

As we increase the number of sharp corners, the null curve comes closer and closer to the timelike curve while still having zero path length. Timelike geodesics cannot therefore be curves of minimum proper time, since they are always infinitesimally close to curves of zero proper time; in fact they maximize the proper time.

My question is, if the geodesic is infinitesimally close to a null curve, shouldn't it also have zero path length? Why does this imply maximizing proper time?

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    $\begingroup$ I think the real problem at hand is actually the same as in the case of Euclidean geometry. See this link. One way to answer this has to do with the fact that making a curve "infinitely crinkled" leads to many mathematical pathologies. (For example, a circle is "infinitesimally close" to the infinitely crinkled curve depicted in the link, but we don't say the circle has pi=4.) $\endgroup$ Apr 10, 2020 at 19:26
  • $\begingroup$ Let $\nabla_{\dot\gamma}\dot\gamma=0$ be a geodesic $\gamma$ on $M$, where $M$ is connected and geodesically complete. At $T(M)_p$, given $\gamma(0)$ and $\dot\gamma(0)$, there exists a unique maximal geodesic at $T(M)_{p}$. Then one can use the exponential map to drag $\dot\gamma$ along the geodesic from $T(M)_p$ to any $T(M)_{p}$ as long its on the geodesic aad it's timelike. I don't see the need for a null geodesic to follow a geodesic or to show it's maximal. $\endgroup$ Apr 11, 2020 at 23:14

6 Answers 6


The fact that the curve doesn't have zero path length is identical to the following 'proof' that $\pi=4$.

PI paradox

A detailed explanation can be found in this link, but the main idea is that the black line doesn't become a tangent line in the limit. This means the perimeter of the circle and jagged line aren't equal in the limit.

To why this implies it is maximal: a regular function has the property that in its maximum $f(x+\delta x)\leq f(x)$. Here $x$ maximizes $f$ and $\delta x$ is a small (or infinitessimal) quantity. For the proper time this argument is less obvious because it depends on the entire path. You can define it as a functional: an object which takes a function as input and outputs a scalar. $$\Delta \tau[x^\mu]=\int d\lambda\sqrt{-\eta_{\mu\nu}\frac {d x^\mu}{d\lambda}(\lambda)\frac {d x^\nu}{d\lambda}(\lambda)}$$

Our argument can then be extended to

$$\cases{\Delta \tau[x^\mu+\delta x^\mu]<\Delta \tau[x^\mu] & $x^\mu$ is a maximum\\ \Delta \tau[x^\mu+\delta x^\mu]>\Delta \tau[x^\mu] & $x^\mu$ is a minimum}$$

Now $\delta x^\mu(\lambda)$ it not a constant anymore but a function. In our case it is the offset between our geodesic and the jagged approximation of the geodesic. Since the proper time is positive and $\Delta \tau[x^\mu+\delta x^\mu]=0$ we have that $x^\mu$ must be a maximum.

Note: forgive me if I made mistakes, has been a while since I did any GR.

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    $\begingroup$ What still is not clear to me is why the argument that all geodesics are $\textit{infinitesimally}$ close to a null curve is either true nor useful, since when you vary a path, the resultant proper time should be infinitesimally close to the original one. This is not the case since geodesics have finite proper time and null curves $0$ so I don't think the maximum-minimum argument is valid under these conditions. $\endgroup$
    – user728261
    Apr 12, 2020 at 12:04
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    $\begingroup$ @AccidentalTaylorExpansion really wonderful your counterexample! But I do not think that the maximisation property can be proved by invoking the existence of approximating piecewise smooth null geodesics... $\endgroup$ Aug 5, 2022 at 9:19

I think that all Caroll meant by this remark was that the stationary point has to be a maximum not a minimum of proper time. It can't be a minimum since there are nearby paths with lower (namely, zero) proper time. So given that the path is one with a stationary value of total proper time between the given events, then it must be a maximum.

  • $\begingroup$ How do you know that the nearby paths have lower proper time? I understand this can be seen from the reversed triangle inequality in Minkowski spacetime, but I was just wondering whether there's some more intuitive reasoning which I missed. $\endgroup$
    – Sanjana
    Sep 8, 2023 at 23:42
  • $\begingroup$ It suffices to note that there are nearby paths with zero proper time. Proper time is non-negative, so you can't get lower than that. Hence the only way to get a stationary value which is a minimum is if you are dealing with a null path. $\endgroup$ Sep 9, 2023 at 10:03
  • $\begingroup$ How do you know proper time is non-negative? e.g. In $(+,-,-,-)$ signature $ds^2=d\tau^2$ and for spacelike intervals $d\tau^2<0$ in which case proper time is not even real. Why are we excluding this possibility in the diagram itself? $\endgroup$
    – Sanjana
    Sep 9, 2023 at 17:08
  • $\begingroup$ The argument only concerns timelike geodesics. $\endgroup$ Sep 9, 2023 at 17:58
  • $\begingroup$ I was just asking that, in the construction why do we use null jagged curves only and not spacelike curves also. I understand that your point holds if we consider null and timelike curves as in the original question/Carroll's book. I was just wondering if this would hold if we also included a spacelike curve in addition. $\endgroup$
    – Sanjana
    Sep 9, 2023 at 20:36

If the geodesic is infinitesimally close to a null curve, shouldn't it also have zero path length?

This is a similar paradox to the "$\pi = 4$ paradox". Take a semi-circle of radius 1. We know it has length $\pi$. But we can approximate it by superposing the circle on a grid of pixels and tracing it out in a jagged way. As you make the grid finer, you can keep approximating the circle with arbitrarily nearby circles, each of which has a total length of 4. Thus, $\pi = 4$?

The important mathematical takeaway is that the limit of a function of a sequence is not always equal to the function of the limit of the sequence. Here, the sequence is the sequence of ever-closer arbitrarily nearby circles, which converges to the circle. The function takes an element of that sequence and spits out its length. So the former is 4, the latter is $\pi$.

Physically, the important takeaway is to remember to what order a quantity contributes to a sum. As the size of the elements you're cutting up your path into gets smaller, does the thing you're using to approximate the actual length of the small path segment really tend to the true path segment length? Turns out, "angle" information contributes to $O(h)$ to the actual length of the small path segment, so to $O(1)$ to the final integral. Whoops!

Why does this imply maximising proper time?

Well, what's the condition for the path to minimise the proper time? It means that no matter what small disturbance $ \epsilon(\tau) $ you pick around that extremal path, the path that results from adding that disturbance to the path produces a new path that has longer proper time. So the strategy, if you want to show that the path doesn't minimise the proper time, is to find a nearby path that has shorter proper time.

Here Caroll suggests one possible such path, built up from null segments. Is this a convincing argument? Well, I haven't read the text, but I do have some possible objections. First of all, showing it is not a minimum is not the same as showing it is a maximum. Could it not be some kind of "critical point" in the path-space? Maybe someone more familiar with the mathematical technicalities can show this isn't the case.

I would also object that the non-differentiabilities of the jagged path could conceivably contribute non-negligibly to the final length. I think you could fix that part of the argument by "smoothing out" those bends and show that the contribution is bounded by the size of those smoothed-out turns. However, there would be some technicalities to do with which limit is taken first: the cutting up of the path into many kinks, and the "pinching out" of the bends into kinks. I've also got in mind the twin paradox, in which this exact issue can come up, and you have to think carefully about why the contribution to the accelerating twin's proper time when they're turning can be neglected as the turnaround time goes to zero.


First, Carroll should have said "arbitrarily close", not "infinitesimally close" (whatever that means).

Second, the length of the geodesic cannot be a local minimum if there are curves arbitrarily close by for which the length is smaller (in this case, zero).

Third, once you've convinced yourself that the geodesic must be either a local maximum or a local minimum for the length of curves connecting two given points, and once you've ruled out the possibility of a minimum, the only thing left is a maximum. Presumably Carroll has at some earlier point in the discussion offered some argument for the "must be one or the other" part.


if the geodesic is infinitesimally close to a null curve, shouldn't it also have zero path length?

It depends on what you mean by "infinitesimally close". If you approximate a curve $\gamma$ with a piecewise smooth curve $\gamma'$ made of segments, whose endpoints (corners) of each segment stay on $\gamma$, then the length of $\gamma'$ tends to the length of $\gamma$ when the number of corners tends to infinity (it is true with some other technical hypotheses). That is not the case for the considered situation, evidently. When the endpoints of (some of) the "infinitesimal segments" do not belong to the curve, then there is no correlation between the two lengths (in the limit). See the explicit beautiful counterexample provided in @AccidentalTaylorExpansion's answer.

Why does this imply maximizing proper time?

It does not imply, in fact! I do not know the rest of the reasoning offered by the book, so I do not kwow how the text may conclude along that line. I guess that the argument of the approximating piecewise smooth null curves is only exploited to show that timelike geodesics cannot minimize the length because there is always a shorter curve with the same endpoints of the considered geodesic segment. (The fact that the shorter curve is not smooth is not a problem: Notice that a piecewise smooth null curve can be made everywhere smooth by slight changes around the corners without substantially increase its length.)

In general, timelike geodesics do not maximise the length. It is instead true in geodesicslly normal neighborhoods of one of the endpoints which include the whole line. The proof, in curved spacetime, is by no means obvious since it is based on a subtle result known as the "Gauss lemma".

The fact that timelike geodesics can be approximated by piecewise smooth null curves is valid also outside normal neighborhoods and therefore it cannot be the reason why timelike geodesics maximize the length.

A proof of the maximisation property can be found for instance in O'Neill's book on semi Riemannian geometry.

In Minkowski spacetime (which is a normal neighborhood of every point in its own right!), the considered property of timelike geodesics is easy to be established and there is a direct almost selfevident proof of mine in PSE. The extension to curved spacetime of that idea needs, as said, the Gauss lemma.


By variational method, a geodesic connecting two points in spacetime extremizes the proper time, which is either a maximum or a minimum. If a world line connecting the same two points can be found with smaller proper time, then this geodesic's proper time must be a maximum. Caroll's jagged line shows that there exists a world line as close to null like as necessary such that its proper time is small enough to be less than the geodesic's proper time. But it seems that this jagged line doesn't have to entwine around the geodesic to prove this argument.


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