# Maximum time difference between clocks in a gravity field

From Surely you must be joking, Mr Feynman.

You blast off in a rocket which has a clock on board, and there's a clock on the ground. The idea is that you have to be back when the clock on the ground says one hour has passed. Now you want it so that when you come back, your clock is as far ahead as possible. According to Einstein, if you go very high, your clock will go faster, because the higher something is in a gravitational field, the faster its clock goes. But if you try to go too high, since you've only got an hour, you have to go so fast to get there that the speed slows your clock down. So you can't go too high. The question is, exactly what program of speed and height should you make so that you get the maximum time on your clock?

This assistant of Einstein worked on it for quite a bit before he realized that the answer is the real motion of matter. If you shoot something up in a normal way, so that the time it takes the shell to go up and come down is an hour, that's the correct motion. It's the fundamental principle of Einstein's gravity--that is, what's called the "proper time" is at a maximum for the actual curve.

How is this result proven?

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– Raskolnikov Apr 9 '11 at 11:25

## 3 Answers

To elaborate on Marek's (correct) answer, since it seems that math is the issue that @Casebash is having:

Start with an integral representing the time elapsed on the moving observer's clock in terms of the stationary observer's coordinates (I'm supressing G and c below. Feel free to replace all $t$'s with $c\,t$'s, and all $M$'s with $\frac{GM}{c^{2}}$'s):

$$\begin{equation} \Delta \tau = \int d\tau \sqrt{{\dot t}^{2}(1-\frac{2M}{r})- \frac{{\dot r}^{2}}{1-\frac{2M}{r}}- r^{2} {\dot \theta}^{2} - r^{2}\sin^{2}\theta {\dot \phi}^{2}} \end{equation}$$

Where $(t,r,\theta,\phi)$ are all considered to be functions of $\tau$ denoting the moving observer's position at time $\tau$. Our problem amounts to looking for the path beginning at $(t_{0},R,\Theta,\Phi)$ and ending at $(t_{f},R,\Theta,\Phi)$ that maximizes this integral, subject to the constraint that the quantity under the square root will be equal to $1$ when the calculation is complete. To make our calculations easier, note that the square root is a monotonic function over all its domain, so we might as well maximize

$$\begin{equation} \frac{1}{2}\int d\tau \left({\dot t}^{2}(1-\frac{2M}{r})- \frac{{\dot r}^{2}}{1-\frac{2M}{r}}- r^{2} {\dot \theta}^{2} - r^{2}\sin^{2}\theta {\dot \phi}^{2}\right) \end{equation}$$

which is considerably simpler. It would take a whole page to carefully vary (meaning that we basically, but not quite, take the derivative of the function with respect to) this with respect to the four independent functions. So, I'm going to just give you a taste of the task by varying with respect to $\theta$. The path of maximum "moving clock" time (heretofore called 'proper time') will be the one for which these variations are zero. The other four functions follow just as easily. The variation of this integral with respect to $\theta$ gives us

$$\begin{align} \delta \Delta\tau |{}_{\theta}&=\int d\tau\left(-r^{2}{\dot \theta}\delta {\dot \theta}-r^{2}\sin\theta \cos \theta {\dot \phi}^{2}\delta \theta\right)\\ 0&=\int d \tau \left(-\frac{d}{d\tau}\left({\dot \theta} r^{2}\delta \theta\right) + \left({\ddot \theta}r^{2} + 2 r {\dot r}{\dot \theta}-r^{2}\sin\theta \cos \theta {\dot \phi}^{2}\right)\delta \theta\right)\\ &=\left(\dot \theta(t_{0})r(t_{0})^{2}\delta \theta (t_{0})-\dot \theta(t_{f})r(t_{f})^{2}\delta \theta (t_{f})\right)\\ &\quad + \int d\tau\; r^{2}\left({\ddot \theta} + \frac{1}{r}2{\dot r} {\dot \theta}- \sin \theta \cos \theta {\dot \phi}^{2}\right)\delta \theta \end{align}$$

The first term, that we obtained by integrating the total derivative, vanishes since the variation of $\theta$ is zero at the endpoints $t_{0}$ and $t_{f}$ (our space of states is paths beginning and ending at a fixed value of $\theta$. If the variation will be zero, then the remaining stuff under the integral must turn out to be zero if it is going to be a minimum for an arbitrary fixed-point variation (if the integrand isn't zero, just imagine making the variation a hundred kajillion only at the point where it is nonzero--clearly this isn't an extremum). Therefore, the $\theta$ variable of the maximum proper time path must satisfy ${\ddot \theta} + \frac{1}{r}2{\dot r} {\dot \theta}- \sin \theta \cos \theta {\dot \phi}^{2}=0$. It turns out that this equation is exactly the same equation you get for geodesic motion of a particle, which is the path followed by an observer being acted upon only by gravity. A simple calculation of the first integral for any other path (and the fact that it will come out higher) will then show you that this, in fact, is the maximum time path to travel.

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In the last line, how did the multiple of {\delta \theta} change from only affecting the {2 r \dot r \dot \theta} to affecting everything inside the brackets? – Casebash Apr 10 '11 at 1:41
Ah, okay, the issue is actually with the line before that. The other variables are missing a `{\delta \theta}`. BTW, how did you know to rewrite the equation to involve the derivative of {\dot \theta r^2 \delta \theta}? – Casebash Apr 10 '11 at 1:49
@Casebash: this is standard procedure in calculus of variations. The integral Jerry started with is known as Lagrangian and equations he obtains are called Euler-Lagrange equations. As to how did he know that: it's a standard trick. You want to remove every $\delta \dot \theta$ because you are not perturbing in these variables. That's why you need integration by parts to produce just $\delta \theta$. Then the whole integrand will be linear in $\delta \theta$ and you can "factor it out", obtaining local equations. – Marek Apr 10 '11 at 6:19

As Raskolnikov says, it's the principle of least action (more precisely extremal action but let's put that off till later). Consider the task of finding the shortest path between two points in Euclidean space. We call such a special path geodesic and it's obvious that here it is just plain old line segment between the two points. But to prove this statement is not so trivial (one needs calculus of variations at hand), so let's concentrate on just the intuitive picture.

If you try to deform the line a little to make the total path shorter, you'll find it's not possible. Reason is that infinitesimally, each such deformation will correspond to a triangle with longest side $c$ being the part of the line and $a$ and $b$ lines comprising the perturbed path. As per basic Euclidean geometry we have $c < a + b$.

In Minkowski space-time, the picture is very similar. Straight lines are again geodesics although now they extremize space-time intervals or equivalenty proper time along the path (at least if the particle is massive). Again, if you repeat the above argument you will obtain an inequality but with the sign reversed because time component has different signature than the space components. That means that time will be maximized along the geodesic.

The above can be generalized to curved spaces and space-times. E.g. on a sphere you'll find the shortest path between two points are parts of great circles. And in fact, our argument will work too because infinitesimally sphere looks like a plane (as I am sure you know from your day to day experience living on the surface of Earth) so our arguments holds. And similarly for space-times and proper times.

A note on the word extremal is in order. If you inspect the sketches of the proofs I made, you'll see that we always make only local (infinitesimal) perturbations around the extremal paths to see if there is something shorter/longer around. That means that we only find local extrema. Also, these paths need not exist or be unique between every pair of points (consider the sphere again and all the geodesics between north and south poles). This is good to keep in mind but I won't delve into general mathematical theory as it's messy and not needed usually.

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1. What do you mean by "proper time along the time"? 2. How do you extremize a space-time interval? For example, what units is it measured in? – Casebash Apr 9 '11 at 12:05
Also, what do you mean by the time component having "a different signature"? – Casebash Apr 9 '11 at 12:09

2
@Casebash: thanks, that was just a typo. There are probably few more lurking about. As for the signature, special (and general) relativity are based on the concept of metric which (in contrast with usual geometry) is not positive definite. The space-time interval is written as $ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2$. This is just a special relativity version of Pythagorean theorem. But I am not going to explain to you all of special (not to mention general) relativity, sorry. If you want to delve deeper into these matters ask a separate question. – Marek Apr 9 '11 at 12:14

There's a way to see the answer in the OP logically, no math. In SR, clocks that "feel" acceleration, like the ground clock does, run slower than floating clocks (ones in free fall) do. Therefore the clock that floats during the entire experiment will elapse the most time. Only a clock "shot up in a normal way", to use your words, floats during the entire experiment. Search for the Non-Mathematical Proof of Gravitational Time Dilation for an elaboration. The falling clock in that experiment is one half of a symmetrical experiment involving a clock "shot up in a normal way".

The clock on the ground is analogous to the traveling twin that thrusts its rocket engines in the twin paradox. The clock "shot up in the normal way" is analogous to the stay-at-home twin that floats "stationary" in space, waiting for the traveling twin to return. Any clock shot up, but not in the normal way, will be a bit less like the stay-at-home twin and a bit more like the traveling twin.

BTW, the reason that a floating clock elapses more time than a clock that "feels" acceleration, when they start and end at the same location, is also given by SR: The clocks move relative to each other, completing trips between when they start and end at the same location. The trip travel distance of the clock that "feels" acceleration is shortened by length contraction predicted by SR. (A ruler that freely falls downward and near the clock on the ground shortens as it falls, as measured in that clock's frame, because the ruler's speed is increasing as measured in that frame. The ruler measures a portion of the remaining distance the clock has to travel.) Therefore that clock travels less distance at the same average speed relative to the other clock, elapsing less time than the floating clock does.

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Oops, should be "the reason that a floating clock elapses more time than a clock that "feels" acceleration" – finbot Apr 12 '11 at 20:25
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