# Curved space/ spacetime

1. Dec 18, 2005

### Ratzinger

The GR equ. tells us that a test particle will follow a geodesic line in spacetime, which is not a geodesic line in space. Usually space is flat, but that does not imply the geodesic line of a test particle in space is a straight line (right?). Since if I throw a ball the gravity of earth makes the ball go a parabolic line.

So why go balls curved trajectories in the presence of earth gravity? What does GR say about little balls in earth's gravity? Can we speak here about space or spacetime curvature that causes its parabolic trajectory?

I know my thinking went somewhere utterly wrong here. Could someone clearify?
Thanks

Last edited: Dec 18, 2005
2. Dec 18, 2005

### JesseM

In flat spacetime, a geodesic is always a straight worldline. In flat space but not flat spacetime (like a spatially flat expanding universe), this isn't true, although I'm not sure whether the particle's path through space would still always be straight in this case.
The earth definitely curves spacetime appreciably, if you tried to model it as flat you wouldn't get those curved trajectories. A geodesic path always maximizes the proper time, ie the time elapsed on a clock that takes that path, so one way of thinking about the path of a thrown object is that there's a tension between the clock "wanting" to spend as much time as possible at greater heights because it ticks slower when it's closer to the ground due to gravitational time dilation, but not "wanting" to move upwards too quickly because then velocity-based time dilation will slow it down. The parabolic path is the one that ideally balances these opposing tensions to maximize the number of ticks of the clock that travels that path (I saw this explanation in one of Feynman's books).

3. Dec 18, 2005

### Garth

First, the ball does not follow anything in 'space', if by space you mean a space-like hypersurface of simultaneity (defined in the observers frame of reference). Anything on that hyper-surface is fixed in time and does not move, you have taken a snap shot of it.
Secondly, you are seeing the ball follow a parabola in 4D space-time, the problem is your time dimension is not to scale.

Suppose you throw a ball so that it lands 100 metres away, after rising to 2.5 metres altitude before falling to the ground ~ 1.5 seconds later. To visualise the trajectory with the time dimension to scale, the time duration would be equal to the distance to the Moon (1.5 light seconds away).

So, visualise your parabola stretched out 100 metres in one direction, 2.5 metres in another and ~ 384,000,000 metres in the other, it is pretty straight! The Earth's gravitational field only deviates from flat space-time by a factor of around 7x10-10 ($\frac{GM}{rc^2}$).

I hope this has helped.

Garth

Last edited: Dec 18, 2005
4. Dec 18, 2005

### JesseM

But you can also do a "foliation" of spacetime into a series of spacelike hypersurfaces, and then see how an object's position in space changes at different instants. Of course there is no unique way to divide spacetime into a series of instants, that depends on your choice of coordinate system.

5. Dec 18, 2005

### robphy

6. Dec 18, 2005

### pervect

Staff Emeritus
Basically it is "curved time", the fact that the rate at which clocks ticks depends on the altitude of the clock, that causes objects to follow a curved trajectory.

Formally, one can do the analysis with "only" the calculus of variations.

We say that the path an object takes makes the proper time an extremum. Suppose we have a particle following a trajectory r(t) in (for example) the Schwarzschild metric. Then we can write (using geometric units, in which c=1 for convenience)

$$d\tau^2 = g_{00} dt^2 - g_{11} dr^2$$
$$\tau = \int \sqrt{ g_{00} - g_{11} \left( \frac{dr}{dt} \right)^2} dt$$

where g_00 and g_11 are functions of r. For weak fields and slow velocities, we can ignore the effects of g_11 entirely, and assume that it is unity. Thus we are taking into account only g_00, which represents the fact that clocks tick at different rates at different altitudes. We then have a basic problem in the calculus of variations

http://mathworld.wolfram.com/CalculusofVariations.html

i.e. we are extremizing the intergal

$$\int L(r,\dot{r},t) dt \hspace{.5 in} \dot{r} = \frac{dr}{dt}$$

with
$$L(r,\dot{r},t) = \sqrt{g_{00}(r) - \dot{r}^2}$$

which imples that the falling body obeys the Euler-Lagrange equations

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) = \frac{\partial L}{\partial r}$$

You can solve this in detail if you like (for the specific examle you'll need to look up how g_00 depends on r by looking up the Schwarzschild metric). T

he only points I want to make are very general ones, that the "force" term, $\partial L/\partial r$, results from the fact that g_00 is a function of r, and is in fact proportional to $\partial g_{00}/\partial r$, and that the momentum term is very close to the SR form for momentum, with some small corrections for the metric coefficient g_00.

You can also include the effects of g_11 if you're really ambitious, and show why they are small for low velocities/weak fields.

Last edited: Dec 18, 2005
7. Dec 18, 2005

### Ratzinger

Thanks, gentlemen. You are all great people.

And did I mention recently how much I like this site?

8. Dec 18, 2005

### pmb_phy

You don't have to have a curved spacetime to have a ball follow a parabolic trajectory in space. In fact a ball will follow a curved spatial trajectory in a uniform gravitational field and a uniform g-field has zero spacetime curvature.

By the way, particles follow geodesics of [bextremal[/b aging, not maximal.

Pete

9. Dec 18, 2005

### JesseM

Is this is basically the same as noting that in SR an object will follow a curved trajectory in an accelerated coordinate system, and then also noting that the equivalence principle of GR means you can explain the observations of any accelerated observer in terms of a g-field?
What are examples of spacetimes where a geodesic corresponds to minimal aging as opposed to maximal? Would they come up in any realistic physical situations?

10. Dec 19, 2005

### pmb_phy

11. Dec 19, 2005

### pervect

Staff Emeritus
I should probably note yet again, that "curvature", especially in scare quotes, does not necessarily mean a non-zero Riemann curvature tensor.

Speaking losely, it can apply to any situation where the metric coefficients vary with position.

For an example of this usage, see for instance MTW pg 187, chapter 7, section 3. The title of this chapter is "Gravitational redshift implies space-time is curved".

Reading this section, it is clear that the "curvature" being described in this chapter does necessarily mean a non-zero curvature tensor. MTW, and many other authors, use the term "curvature" to describe any situation where the metric coefficients vary with position.

In fact gravitational red-shift does occur in situations where the Riemann curvature tensor is zero, such as the Rindler metric of an accelerating observer (also sometimes called a uniform gravitational field).

One can either believe that several well-recognized texts are "wrong" and go to great efforts to "correct" them, or one can believe that "curved space-time" does not always mean "a non-zero curvature tensor". I have chosen the later course, pmb has chosen the previous course.

12. Dec 19, 2005

### JesseM

But we already know that particles that start at the same position with different initial speeds and directions will follow different geodesics--just think of two particles orbiting the sun whose orbits intersect in two distinct locations, if the orbits were arranged right they could meet at each crossing and yet (correct me if I'm wrong) they could have aged different amounts between the two events of their meeting, but I don't think this would contradict the idea that each particle's path between the two events was the one that maximized its aging, given each particle's speed and direction at the first crossing. So why does the photon example show that the slower-than-light particle's proper time was minimized? Would it be true in this case that if you had another particle with the same initial speed and direction as the first one, but which had a non-gravitational force acting on it so it deviated from the first particle's path before reuniting with it later, that the particle deviating from the geodesic path would have actually aged more rather than less as long as they were both inside the photon sphere? Would the twin paradox work in reverse inside the photon sphere, in other words?

13. Dec 20, 2005

### pmb_phy

The principle of extremal aging refers to two events. I.e. the proper time as measured on a clock which passes through both events.
The proper time along the geodesic of any photon is zero and no positive number has a value less than zero. I could tell you to look in the glossary of Taylor and Wheeler's Exploring Black Holes if you don't believe me but note that I'm the one who wrote that glossary.
I havne't calculated it bu that would seem so. Note that the principle of extremal aging refers to geodesics not arbitrary worldlines.

Pete

14. Dec 20, 2005

### George Jones

Staff Emeritus
I'm not going to give specific examples for spacetime, but I can give a specific example for space, and I can talk a little about what happens in general.

First, consider the analgous situation in 2-dimensional spaces (not spacetimes). For points p and q in flat R^2, a geodesic (i.e., a straight line) is a local minimum (in terms of length) for curves that run from p to q. For point p and q in curved 2-dimensional spaces (i.e., positive definite Reimannian manifolds), it is not always the case that a geodesic that joins p and q is a local minimum in terms of length.

As an example, consider S^2, the 2-dimensional surface of a 3-dimensional ball, and, for concreteness, take the surface to be the surface of the Earth. Take p to be the north pole and q to be Greenwich England.

The shortest route from the north pole to anywhere is along the appropriate line of longitude, and the shortest route from the north pole to Greewich is south along the 0 line of longitudeprime (prime meridian). This route is a geodesic and a local minimum for length. Call it the short geodesic.

There is, however, another geodesic route that starts at the north pole and ends at Greenwich. From the north pole, go south along the 180 line of longitude to the south pole pole and then north along the 0 line of longitude from the south pole to Greenwich. This route is also a geodesic, but clearly is not a local minimum for length. Call this the long geodesic. Taken together, the short and long geodesics comprise the unique great circle that runs through both the north pole and Greenwich.

How can the diffence between the short geodesic and the long geodesic be characterized? First consider the long geodesic. Any geodesic that starts the north pole, and that is infintesimally close to the long geodesic crosses the long geodesic before it gets to Greenwich, i.e., at the south pole. (Note: these "close" geodesic start at he north pole, but do not go through Greewich). The south pole is what is called a conjugate point.

Now consider the short geodesic. Any geodesic that starts at the north pole, and that is infinitesimally close to the short geodesic does *not* cross the short geodesics between the north pole and greenwich. The short geodesic does not have any conjugate points.

This gives the required characterization: a geodesic between any point p and q of a 2-dimensional space is a local minimum for length if and only if the geodesic has no conjugate points.

Something similar is true for spacetime: a timelike geodesic between events p and q is a local maximum for proper time if and only if there are no conjugate points (to p or q) between p and q. See Wald for a proof. This is very much related to the singularity theorems of Penrose and Hawking, as the focusing nature of gravity (assuming an appropriate energy condition) often causes conjugate points to occur.

Regards,
George

15. Dec 20, 2005

### RandallB

GR says almost nothing here.
You shouldn’t expect so much from GR & space-time they say very little about parabolic curves of balls, or Galileo’s cannonballs, on earth. GR affects compared to the predominantly Newtonian effects are just too small.

First - in the ideal case the paths are NOT parabolic – they are elliptical as they attempt to establish an elliptical orbit around the center of the earth. Of course the surface of the earth gets in the way but before it hits it is basically on an orbital path.

Second – We never get to see the “ideal” because AIR gets in the way. Due to the air resistance the horizontal speed is slowly but only partly reduced all the way to infinity.
Maximum speed in the vertical is also limited by terminal velocity. Since the horizontal speed is never 100% eliminated the curve only approaches a vertical line. Thus it’s parabolic.

Guys, you don’t have to create the most complicated answers, just a little insight to a proper concept is so much simpler.
I’m sure Occam would agree.

16. Dec 20, 2005

### JesseM

Yes, but doesn't it refer to taking two paths between the two events, starting with the same initial speed and direction from the first event, one path being a geodesic and one being a non-geodesic? I don't see why it would be relevant that, starting with different initial speed and directions from the first event, you could have two different paths which were both geodesics and which would reunite at a different event with different proper time elapsed. Again, this would seem to be possible for two different orbits around the sun that cross at two points--in this case you might have an observer A who ages less than an observer B between their two meetings, but that wouldn't prove that observer A's path did not maximize his aging given his initial speed and direction, would it?
But I don't see how this is relevant to the issue of extremal aging--if you could address my example of the two orbits that intersect twice it would help me out.
Again, I thought it referred to a variational principle where you show that any deviations from the geodesic path would give a non-extremal aging (similar to showing that variations from the path that an object takes in classical mechanics would give a non-extremal value for the action integral).

17. Dec 20, 2005

### JesseM

GR reduces to Newtonian mechanics in the limit of small spacetime curvature and small velocities, just like SR reduces to Newtonian mechanics in the limit of small relative velocities. So when dealing with a mass like the earth, the geodesic path GR predicts for a cannonball should be almost identical to the parabolic path predicted by Newtonian mechanics.

18. Dec 20, 2005

### pmb_phy

No.
That's the idea.

The variation from the path is infinitesimal and I believe it yields a maximum aging when the events are close together. The idea is that extremal aging tells you that amoung all the worldlines which pass through two events, those which are geodesics are those for which the proper time is extremal.

Your example of crossing orbits is another example of two geodesics which may have different proper times associated with them and each geodescic is one for which the proper time is an extremal. This means that if you vary either path just a tad then the proper time would be either slightly higher or slightly lower than the one on the geodesic. If you chose alpha as an index to paramaterize worldlines which are close to the geodesic then plotter the proper time vs the index alpha then you'd see that those values of alpha for which the curve has a zero derivative corresponds to a geodesic. The curve doesn't have to have a maximum there. It could have a minimum of a point of inflection.

Pete

19. Dec 20, 2005

### JesseM

But doesn't your description below that you take a geodesic path and then "vary it just a tad" match what I was saying above? Isn't the idea that if you vary it slightly from the true geodesic path, you will always get a smaller proper time? (assuming you're dealing with a spacetime where extremal = maximal) In contrast, if you take two different initial velocities then you can have two different valid geodesics, but the fact that the second geodesic may have more proper time between the event of the particles departing and the event of their reuniting than the first geodesic does not show that the first geodesic was not a maximal one, correct? So in order for the comparison of two paths to be at all relevant to showing that one path has maximal proper time, it seems critical that both paths start out parallel (same initial velocity) and then one path is a non-geodesic one.
Why? Do you agree that if you have two geodesic paths A and B that intersect at two events, then the fact that path A may have greater proper time in no way disproves the idea that path B was a maximal one? If so, then in what way is comparing two geodesic paths relevant to deciding whether one of them is maximal or not?
But even if you want the variation to be infinitesimal, by having the particles start out with infinitesimally different initial velocities, it seems you could have two infinitesimally different paths which were each valid geodesics, and yet the fact that path A had infinitesimally greater proper time would not show that path B was not a maximal one. Am I wrong about that? If not, then again, it seems the relevant comparison is between two paths which start out with the same initial velocity, but then one departs slightly from the geodesic path, and in this case you are guaranteed that the non-geodesic path will have a slightly smaller proper time (again, assuming the geodesic path was one that maximized the proper time rather than minimized it, I'm not objecting to the idea that there are spacetimes where a geodesic is minimal rather than maximal).
But not "extremal when compared to other geodesic paths between the same two events", just "extremal when compared to non-geodesic paths that start out parallel to the geodesic path", it seems to me.
But see my example of two infinitesimally close geodesics--since they are both geodesics, it can't be true that they are each extremal when compared with each other, which again is why I always understood it as critical that you only compare nearby paths which start out with exactly the same initial velocity, and then you will see that given that initial velocity, there is a unique extremal path that starts out tangent to that velocity. Comparing paths with different initial velocities would seem to be apples and oranges, it would be of no use in showing that a given path is "extremal".

20. Dec 20, 2005

### JesseM

It occurs to me that my thinking could be going wrong here, perhaps there are no spacetimes where given a geodesic between two events, it would be possible to find another arbitrarily close geodesic between the same events...this wouldn't be true in flat spacetime, for example. I was thinking about gravitational lensing and how geodesic paths starting at different angles behind the lens will be focused at a single point in front of the lens, but I don't know if objects travelling these paths will arrive at the same time as well, so that you really have a collection of arbitrarily-close geodesics between the same two points in spacetime. I was also thinking about great circles on a sphere, but perhaps geodesics in spacetime are fundamentally different from geodesics in space in this way. If indeed there are no other geodesics among the paths arbitrarily close to a given geodesic, then I suppose the geodesic could be the extremal one out of all of these paths, even if the other paths did not start with the same initial velocity. Do you know the answer to this question about whether a geodesic path between two events can have arbitrarily close geodesic "neighbors" between the same two events?