Geodesics: paths of longest proper time

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 8; Problem 8.1.

One of the central problems in general relativity is the determination of geodesic curves, or just geodesics for short. These are the curves that a free particle (that is, a particle upon which no force acts, where ‘force’ in this case excludes gravity, since the effects of gravity are felt entirely through the curvature of space-time) will follow in a curved space-time.

To analyze this problem, it is useful to start, as always, with flat space-time in special relativity. In this case, a free particle travels in a straight line with constant velocity. If we look at this particle in its rest frame and place it at spatial coordinates {x=y=z=0}, its world line (in 2-d) is the {t} axis. That is, the particle starts off at the origin and just moves up the {t} axis with a constant {x} coordinate of {x=0}.

Now suppose we consider another particle that is not free, that is, it is acted on by some forces and as a result its world line is not a straight line. Suppose, however, that both particles start off at event {A}, which is at the origin at {t=0}, and also both pass through the event {B} at {x=0}, {t=1}. The free particle’s world line between these two events is a straight line, as we’ve said, while the non-free particle’s world line will be some path that can wander around any way it likes (as long as its speed doesn’t exceed 1), provided that it comes back to meet the first particle at event {B}.

An invariant quantity for each particle is the proper time (that is, the time as measured by a clock that moves with the particle) that elapses between {A} and {B}. In the case of the free particle, the proper time is also the time coordinate in the measurement frame since the particle is at rest, so the proper time interval {\Delta\tau} is just the time interval {\Delta t=1}. For the other particle, its proper time will not be the same as {\Delta t}, since the particle has a non-zero speed for part or all of its world line.

Intuitively, you might think that the non-free particle’s proper time would be greater than that of the free particle, since obviously the path length of its world line is larger. However, remember that the invariant interval in relativity is calculated using the metric {\eta_{ij}}, whose 00 component is {-1}, so intuition leads you astray here. In the non-free particle’s frame, its velocity is zero, so {-ds^{2}=d\tau^{2}}. In the free particle’s frame (assuming motion only in the {x} direction):

\displaystyle   -ds^{2} \displaystyle  = \displaystyle  dt^{2}-dx^{2}\ \ \ \ \ (1)
\displaystyle  \displaystyle  = \displaystyle  dt^{2}\left(1-\beta^{2}\right) \ \ \ \ \ (2)

Thus the relation between the two time intervals is

\displaystyle  d\tau=\sqrt{1-\beta^{2}}dt \ \ \ \ \ (3)

That is, the proper time interval for the moving particle is less than the time interval measured in the free particle’s frame. In fact, if we integrate this between the times {t_{0}} and {t_{1}} (the time interval as measured by the free particle at rest), we get

\displaystyle  \Delta\tau=\int_{t_{0}}^{t_{1}}\sqrt{1-\beta^{2}}dt \ \ \ \ \ (4)

The square root is always between 0 and 1, so the maximum possible value of this integral is if {\beta=0} for the entire world line; that is, the particle is at rest. In other words, the free particle has the maximum proper time of all possible world lines that connect two events.

If the space-time metric had {\eta_{00}=+1} instead of {-1}, then the integral would be {\int_{t_{0}}^{t_{1}}\sqrt{1+\beta^{2}}dt}, and then the minimum proper time would occur when {\beta=0}, which is what our intuition (falsely) told us was true. Thus the non-intuitive result is purely the result of the metric having a negative time component.

Don’t confuse this result with the time dilation principle, since they are considering different cases. Time dilation is an effect that occurs when two different observers (one moving at a speed {v} relative to the other) measure the time interval between two events on the same world line. In the current case, we’re looking at the difference in proper time between two different world lines, as analyzed in the same inertial frame. This is why the limits on the integral in 4 are the same for all world lines; we’re using the same time coordinate {t} (the time in the free particle’s rest frame) in every case.

In flat space, then, the geodesic curve between two events is the one having the largest proper time for an object moving along that curve. The problem is therefore how to find a curve that maximizes the proper time.

Readers familiar with classical mechanics may recognize this problem as one that can be solved using the calculus of variations. We won’t go through the derivation here, but the idea is that if we can identify a function called the Lagrangian {L}, which depends on a set of generalized coordinates {q_{i}} and generalized velocities {\dot{q}_{i}} (where the dot indicates a time derivative) then we can define the action as

\displaystyle  S\equiv\int_{t_{A}}^{t_{B}}Ldt \ \ \ \ \ (5)

The idea is that we need to minimize the action to find the path actually travelled by the object in the given time interval, and we do that by varying the path it follows until we minimize {S}. It turns out that this leads to the set of differential equations

\displaystyle  \frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}_{i}}\right)-\frac{\partial L}{\partial q_{i}}=0 \ \ \ \ \ (6)

The solutions of these equations give the coordinates and velocities as functions of time; that is, they give the trajectory followed by the object.

In classical mechanics, the Lagrangian is the difference between the kinetic and potential energies: {L=T-V}. In relativity, as we’ve seen above, the Lagrangian is the proper time interval. (If you haven’t seen this before, you can take it as god-given at this point. The important thing to understand is that, to find the path actually followed by an object, we need to find something we can maximize or minimize (the action in classical mechanics, the proper time in relativity), and then write that quantity as an integral over some parameter that describes the path followed.)

We’ve already found the solution of the Lagrangian problem for a free particle in flat space: the particle travels between the two events in a straight line at a constant velocity. The fundamental assumption that is made is that we can generalize this method of solution to curved space-time.

First, we use the fact that the proper time interval {\Delta\tau} between two events {A} and {B} is given by integrating the invariant interval over the world line of the object. That is

\displaystyle  \Delta\tau=\int_{A}^{B}\sqrt{-ds^{2}} \ \ \ \ \ (7)

The minus sign occurs because {ds^{2}=dx^{2}+dy^{2}+dz^{2}-dt^{2}}, so in the rest frame of the object {ds^{2}=-d\tau^{2}}.

Now suppose we have a general world path (we’ll call it a ‘path’ rather than a ‘line’ since in general it need not be a straight line) that is described by a parameter {\sigma} that ranges from 0 at event {A} to 1 at event {B}. That is, all four coordinates on the world path are functions of {\sigma}: {x^{i}=x^{i}\left(\sigma\right)}. Then in a general curved space-time, we can say

\displaystyle   ds^{2} \displaystyle  = \displaystyle  g_{ij}dx^{i}dx^{j}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  g_{ij}\left(\frac{dx^{i}}{d\sigma}d\sigma\right)\left(\frac{dx^{j}}{d\sigma}d\sigma\right)\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  g_{ij}\frac{dx^{i}}{d\sigma}\frac{dx^{j}}{d\sigma}d\sigma^{2} \ \ \ \ \ (10)

Then for a given world path, the proper time that elapses along that path as we move from {A} to {B} is

\displaystyle  \Delta\tau=\int_{0}^{1}\sqrt{-g_{ij}\frac{dx^{i}}{d\sigma}\frac{dx^{j}}{d\sigma}}d\sigma \ \ \ \ \ (11)

Remember that both the metric and the coordinates are, in general, functions of {\sigma}, so they can all vary as we move along the world path. This equation is in precisely the right form for a Lagrangian treatment. The proper time interval {\Delta\tau} plays the role of the action {S} above, and the square root term becomes the Lagrangian {L}. The parameter {\sigma} plays the role of {t} in the classical equation.

Before we continue the analysis by plugging this form of the Lagrangian into 6 to get a differential equation involving the metric and coordinates, it’s worth looking at an example of a relativistic system that can be analyzed without the full mathematical toolbox.

Consider an experimental setup in which we fire a laser downwards from the top of a tower of height {z} on the Earth’s surface. One of the fundamentals of general relativity is the principle of equivalence, which states that a frame at rest in a gravitational field is equivalent to an accelerating frame in flat space. One way of visualizing this is to think of what it would feel like to be in an elevator in free space. As the elevator accelerates, you will feel a force pressing you into the floor in the same way you feel a force pressing you onto the ground if you stand still on the Earth’s surface.

Applying this principle to our experiment, we see that in the time the light takes to reach the ground, the lab frame will have ‘accelerated’ from rest to a speed {\beta=gz}, where {g} is the acceleration of gravity. (Remember we’re using {c=1}, so the time {t} taken to reach the ground is {z/c=z}.) That is, the observer on the ground is effectively moving towards the light source with speed {\beta} so the light will appear blue-shifted, with the relation between the emitted wavelength {\lambda_{e}} and observed wavelength {\lambda} being

\displaystyle  \frac{\lambda}{\lambda_{e}}=\sqrt{\frac{1-\beta}{1+\beta}} \ \ \ \ \ (12)

(For a blue-shift, we must have {\lambda<\lambda_{e}}.)

If {\beta\ll1}, we can approximate this by

\displaystyle   \sqrt{\frac{1-\beta}{1+\beta}} \displaystyle  \approx \displaystyle  \left(1-\frac{1}{2}\beta\right)\left(1+\left(-\frac{1}{2}\right)\left(\beta\right)\right)\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  1-\beta+\mathcal{O}\left(\beta^{2}\right)\ \ \ \ \ (14)
\displaystyle  \displaystyle  \approx \displaystyle  1-gz \ \ \ \ \ (15)

Therefore

\displaystyle   1-\frac{\lambda}{\lambda_{e}} \displaystyle  \approx \displaystyle  gz\ \ \ \ \ (16)
\displaystyle  \frac{\lambda_{e}-\lambda}{\lambda_{e}} \displaystyle  \approx \displaystyle  gz \ \ \ \ \ (17)

The ground observer measures the time interval {dt=\lambda/c=\lambda} between successive crests of the light wave. Since this is the same light wave as that emitted by the laser, the time interval between successive crests as measured at the laser must be {dT=\lambda_{e}=\frac{\lambda_{e}}{\lambda}\lambda=\frac{\lambda_{e}}{\lambda}dt\approx\frac{1}{1-gz}dt\approx\left(1+gz\right)dt}.

Now suppose that we fire a clock upwards from the ground, and that at a height of {z} it has a speed {v}. Due to time dilation, the proper time {d\tau} between two infinitesimally close events measured by this moving clock will be shorter than the interval {dT} measured by a clock at rest at height {z} according to {d\tau=\sqrt{1-v^{2}}dT}. (Remember that moving clocks run slow, so the elapsed time on the moving clock is less than on the laser’s clock.) Therefore, the proper time for the clock that travels from the ground up to the laser is related to the time measured by the clock on the ground by

\displaystyle   d\tau \displaystyle  = \displaystyle  \sqrt{1-v^{2}}dT\ \ \ \ \ (18)
\displaystyle  \displaystyle  \approx \displaystyle  \sqrt{1-v^{2}}\left(1+gz\right)dt\ \ \ \ \ (19)
\displaystyle  \displaystyle  \approx \displaystyle  \left(1-\frac{1}{2}v^{2}\right)\left(1+gz\right)dt\ \ \ \ \ (20)
\displaystyle  \displaystyle  \approx \displaystyle  \left(1-\frac{1}{2}v^{2}+gz\right)dt \ \ \ \ \ (21)

This gives us the Lagrangian as

\displaystyle   L \displaystyle  = \displaystyle  1-\frac{1}{2}v^{2}+gz\ \ \ \ \ (22)
\displaystyle  \displaystyle  = \displaystyle  1-\frac{1}{2}\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right)+gz \ \ \ \ \ (23)

We can now write out the Lagrangian differential equations from 6:

\displaystyle   \frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x} \displaystyle  = \displaystyle  -\ddot{x}=0\ \ \ \ \ (24)
\displaystyle  \frac{d}{dt}\frac{\partial L}{\partial\dot{y}}-\frac{\partial L}{\partial y} \displaystyle  = \displaystyle  -\ddot{y}=0\ \ \ \ \ (25)
\displaystyle  \frac{d}{dt}\frac{\partial L}{\partial\dot{z}}-\frac{\partial L}{\partial z} \displaystyle  = \displaystyle  -\ddot{z}+g=0 \ \ \ \ \ (26)

The solutions are

\displaystyle   x\left(t\right) \displaystyle  = \displaystyle  x_{1}t+x_{0}\ \ \ \ \ (27)
\displaystyle  y\left(t\right) \displaystyle  = \displaystyle  y_{1}t+y_{0}\ \ \ \ \ (28)
\displaystyle  z\left(t\right) \displaystyle  = \displaystyle  \frac{1}{2}gt^{2}+z_{1}t+z_{0} \ \ \ \ \ (29)

where {x_{0},x_{1}} and so on are constants of integration. The motion is therefore the standard parabolic trajectory, with constant speed in the {x} and {y} directions and the parabolic motion in the {z} direction. For a gravitational field pulling downwards, {g=-9.8\mbox{m sec}^{-2}} at the Earth’s surface, so the moving clock slows down as it rises.

17 thoughts on “Geodesics: paths of longest proper time

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  2. JERRY

    Glenn, I’m confused about the first statement in the third paragraph,,,.
    “An invariant quantity for each particle is the proper time…”
    The particle which has a force acting on it is not in an Inertial frame, so does it obey a Lorentz transformation? Is dS still an invariant?

    Reply
    1. growescience

      The argument is essentially the same as that used in defining the derivative in calculus. If we consider an infinitesimal segment of an object’s world path, the object is essentially moving in an inertial frame for that segment (the argument in calculus is that if we consider a short enough interval, any continuous function appears as a straight line segment over that interval). The object’s proper time {d\tau} over that segment is the time as measured by a clock attached to the object, so the interval measured over that segment is {ds^{2}=-d\tau^{2}}. If the object is moving at a speed {\beta} relative to another inertial frame at that point, then the elapsed time as measured in the inertial frame is {dt=d\tau/\sqrt{1-\beta^{2}}}. The total proper time over a finite interval is obtained by integrating over that interval, so we get {\Delta\tau=\int\sqrt{1-\beta^{2}}dt}. The interval {ds^{2}} is an invariant for each infinitesimal interval along the world path, since at each point, we’re comparing two inertial frames.

      I know the argument does sound a bit dodgy, but if you’ve managed to convince yourself that the definition of an integral in calculus as the limit of a sum of rectangular areas is valid, it’s really the same argument here.

      Reply
      1. José Victor

        To contribute to the understanding of this crucial question, I repeat what I found in Carrol, Chapter 3 Curvature (Pag.110), the following and perfect argument:
        “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 all we have to do is consider “jagged” null curves that follow the timelike one, as portrayed in Figure 3.3( in the book). As we increase the numer of sharp corners, the null curve comes closer and closer to the timelike curve while still having zero path length. Timelike geodesic cannot therefore be curves de minimum proper time, since they always infinitesimally close to curves of les proper time(zero, inf fact); actually they maximize the proper time. This is how you can rmember which twin paradox ages more – the one who stays home is basicially on a geodesic, and therefore experiences more proper time. Of course even this is being little cavalier; actually every time we say “maximize” or “minimize” we sould add the modifier “locally”. Often the cas is que bewtween two points on a manifold there is more than one goedesic. Por instance, on S^2 we can draw a great cicle through any points, and imagine travelling between them either the short or the long way around. One of these is obviously longer than the other, althoug both are stationary points of the length funcional.”
        JVictor

        Reply
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      1. José Victor

        Dear Rowe,

        In his deduction, it follows that the equation of motion in the z-direction is
        Z (t) = (1/2) gt ^ (2) + z_1 t + z_0

        The velocity in the z direction (t) is
        (1)

        Z ̇ (t) = gt + z_1 (2)

        Where we did z_0 = 0
        Applying the Hamilton principle to (1), one has to action between 0 and t:
        S = z_1 / 2 t (3)

        Where I considered t as proper time along the parabolic geodesic. So the minimum action corresponds to the maximum proper time? It is as I interpret, according to the calculations above.
        The question is: where am I making confusion? Grateful.

        Reply
  8. cyril

    “Don’t confuse this result with the time dilation principle, since they are considering different cases.”
    I would say that it is still linked to time dilation effect, even if we are considering two different paths as mentionned in your post, in the sense that:
    1) we still use the same formula to explain both phenomena
    2) the time dilation is real, even if not reciprocal (as in plain correct SR)

    From an inertial observer, processes taking place on ANY moving objects (inertial or not) will appear slowed down for this observer, due to the unique formula ds^2 = c^2dt^2 – dx^2.
    When looking at the same path from a ‘rest’ observer and another inertial ‘moving’ observer (prime), we assume spacetime distance ds^2 = c^2dt^2 – dx^2 equals ds’^2 = c^2dt’^2 (dx’=0) so c^2dt^2 – dx^2 = c^2dt’^2 so necessarily with dx^2>0 we get dt>dt’ (slow down of the moving clock dt’)

    In your example we use exactly the same formula but from the same inertial observer (no prime coordinate there) by posing dt(free particule)=dt(non free particule) so ds^2(nonfree)0) so dtau(nonfree)^2<dtau(free)^2 (maximization of the 'rest' proper time) as ds^2(free) = cdtau(free)^2 and ds^2(nonfree) = cdtau(nonfree)^2

    I would rather say that the specificicty of the situation in that case is that the time dilation is NOT symmetric for that the above formula can not be applied from the non inertial observer= non-free particle's point of view (cf Twin Paradox)

    Reply

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